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Contents, þ One Day Revision, , 1-10, , þ The Qualifiers, , 11-28, , þ CBSE Question Bank, , 29-41, , þ Latest CBSE Sample Paper, , 42-55, , Sample Paper 1, , 59-69, , Sample Paper 2, , 70-82, , Sample Paper 3, , 83-96, , Sample Paper 4, , 97-110, , Sample Paper 5, , 111-123, , Sample Paper 6, , 124-136, , Sample Paper 7, , 137-148, , Sample Paper 8, , 149-161, , Sample Paper 9, , 162-172, , Sample Paper 10, , 173-185, , Watch Free Learning Videos, Video Solutions of CBSE Sample Papers, Chapterwise Important MCQs, CBSE Case Based MCQs, CBSE Updates, Much more valuable content will be available regularly

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Syllabus, Mathematics (Standard) CBSE Class 10 (Term I ), No., , Unit Name, , Marks, , I, , Number Systems, , II, , Algebra, , III, , Coordinate Geometry, , IV, , Geometry, , V, , Trigonometry, , VI, , Mensuration, , VII, , Statistics, , Probability, , Total, Internal Assessment, Total, , Internal Assessment, , Marks, , Total Marks, , Periodic Tests, Multiple Assessments, , marks for, the term, , Portfolio, Student Enrichment Activities-practical work, , UNIT I NUMBER SYSTEMS, . Real Number, Fundamental Theorem of Arithmetic statements after reviewing work done, earlier and after illustrating and motivating, through examples. Decimal representation, of rational numbers in terms of, terminating non-terminating recurring, decimals., , UNIT II ALGEBRA, . Polynomials, Zeroes of a polynomial. Relationship, between zeroes and coefficients of, quadratic polynomials only., , . Pair of Linear Equations in Two Variables, Pair of linear equations in two variables and, graphical method of their solution,, consistency inconsistency. Algebraic conditions, for number of solutions. Solution of a pair of, linear equations in two variables algebraically by substitution and by elimination. Simple, situational problems. Simple problems on, equations reducible to linear equations., , UNIT III COORDINATE GEOMETRY, . Coordinate Geometry, LINES In two-dimensions Review: Concepts of, coordinate geometry, graphs of linear, equations. Distance formula. Section formula, internal division

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UNIT IV GEOMETRY, . Triangles, Definitions, examples, counter examples, of similar triangles., . Prove If a line is drawn parallel to one, side of a triangle to intersect the other, two sides in distinct points, the other two, sides are divided in the same ratio., ., , Motivate If a line divides two sides of a, triangle in the same ratio, the line is, parallel to the third side., , ., , Motivate If in two triangles, the, corresponding angles are equal, their, corresponding sides are proportional and, the triangles are similar., , ., , Motivate If the corresponding sides of, two triangles are proportional, their, corresponding angles are equal and the, two triangles are similar., , ., , Motivate If one angle of a triangle, is equal to one angle of another triangle, and the sides including these angles are, proportional, the two triangles are, similar., , ., , Motivate If a perpendicular is drawn, from the vertex of the right angle of a, right triangle to the hypotenuse, the, triangles on each side of the, perpendicular are similar to the whole, triangle and to each other., , ., , ., , Motivate The ratio of the areas of two, similar triangles is equal to the ratio of, the squares of their corresponding sides., Prove In a right triangle, the square on, the hypotenuse is equal to the sum of the, squares on the other two sides., , ., , Motivate In a triangle, if the square on, one side is equal to sum of the squares, on the other two sides, the angle, opposite to the first side is a right angle., , UNIT V TRIGONOMETRY, . Introduction to Trigonometry, Trigonometric ratios of an acute angle of a, right-angled triangle. Proof of their, existence well defined . Values of the, trigonometric ratios of, ,, and, ., Relationships between the ratios., Trigonometric Identities, Proof and applications of the identity sin A, + cos A = . Only simple identities to be, given, , UNIT VI MENSURATION, . Areas Related to Circles, Motivate the area of a circle; area of sectors, and segments of a circle. Problems based, on areas and perimeter circumference of, the above said plane figures. In calculating, area of segment of a circle, problems should, be restricted to central angle of, and, only. Plane figures involving triangles,, simple quadrilaterals and circle should be, taken., , UNIT VII STATISTICS, , PROBABILITY, , . Probability, Classical definition of probability. Simple, problems on finding the probability, of an event.

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MCQs Preparation Tips, Focus on Theory, , Learn to Identify Wrong Answers, , MCQs can be formed from any part or, line of the chapter. So, strong command, on theory will increase your chances to, solve objective questions correctly and, quickly., , The simplest trick is, observe the options, first and take out the least possible one, and repeat the process until you reach, the correct option., , Practice of Solving MCQs, Cracking an MCQ-based examination, requires you to be familiar with the, question format, so continuous practice, will make you more efficient in solving, MCQs., , Speed & Accuracy, In MCQ-based examination, you need, both speed and accuracy, if your, accuracy is good but speed is slow then, you might attempt less questions, resulting in low score., , Analyse your Performance, During the practice of MCQs, you can, identify your weak & strong, topics/chapter by analysing of incorrect, answers, in this way you will get an, awareness about your weaker topics., , Practice through Sample Papers, Solving more & more papers will make, you more efficient and smarter for, exams. Solve lots of Sample Papers, given in a good Sample Papers book., , Attempting MCQs in Exams, 1. Read the paper from beginning to, end & attempt those questions first in, which you are confident. Now move, on to those questions which requires, thinking and in last attempt those, questions for which you need more, attention., , 2. Read instructions of objective, questions carefully and find out what, is being asked, a bit carelessness can, lead you to incorrect answer., , 3. Tick/Write down the correct option, only while filling the OMR sheet., , Step by step solution is not required, in MCQ type questions, it is a waste, of time, you will not get extra marks, for this., , 4. Most of the time, you need not to, solve the MCQ completely to get the, correct option. You can start thinking, in reverse order and choose the best, fit option., , 5. As there is no negative marking for, incorrect answers, so don't leave any, question unanswered. Use your, guess if you have not exact idea, about the correct answer.

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CBSE Sample Paper Mathematics Standard Class X (Term I), , 1, , ONE DAY, , REVISION, Revise All the Concepts in a Day, Just Before the Examination..., , ●, , Real Number, , Prime, Coprime and Composite numbers, Prime numbers are those numbers, which have no, factors other than 1 and the number itself., e.g. 2, 3, 5, 7, 11, …, Coprime numbers are those numbers, which do not, have any common factor other than 1., e.g. 2 and 9 are coprime numbers., Composite numbers are those numbers, which have, atleast 1 factor other 1 and the number itself., e.g. 4, 6, 24, …, , Factor Tree, A chain of factors which is represented in the form of a, tree, is called factor tree., , Fundamental Theorem of Arithmetic, , \ Composite number = Product of prime numbers, Relation between Numbers and their HCF, and LCM, (i ) For any two positive integers a and b, the relation, between these numbers and their HCF and LCM is, Þ, or, , HCF (a, b) ´ LCM (a, b) = a ´ b, a´b, HCF (a, b) =, LCM (a, b), a´b, LCM (a, b) =, HCF (a, b), , or LCM (a, b, c ), a ´ b ´ c ´ HCF (a, b, c ), =, HCF (a, b) ´ HCF (b, c ) ´ HCF (c, a), , Real Numbers, A number, which is either rational or irrational, is called, a real number., , Rational Numbers, , p, , where p, q are, q, integers and q ¹ 0, is called a rational number., , A number that can be expressed as, , Irrational Numbers, , p, ,, q, where p, q are integers and q ¹ 0, is called an irrational, number., , A number that cannot be expressed in the form, , Useful Theorems, Theorem 1 Let p be a prime number and a be a, positive integer. If p divides a2, then p divides a., Theorem 2 2 is irrational number, then 2 2 is, irrational number., , Decimal Expansions of Rational Numbers, 1. Terminating Decimal Expansion The number, which terminates (i.e. ends completely) after a finite, , ONE DAY REVISION, , Fundamental theorem of arithmetic states that every, composite number can be written (factorised) as the, product of primes and this factorisation is unique, apart, from the order in which the prime factors occur. It is, also called unique factorisation theorem., , (ii ) For any three positive integers a, b and c, the, relation between these numbers and their HCF and, LCM is, HCF (a, b, c ), a ´ b ´ c ´ LCM (a, b, c ), =, LCM (a, b) ´ LCM (b, c ) ´ LCM (c , a)

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02, , CBSE Sample Paper Mathematics Standard Class X (Term I), , number of steps in the process of division, is said, to be terminating decimal expansion. e.g. 1.25,, 3.14, etc., 2. Non-terminating Decimal Expansion, The number which does not terminate in the, process of division, is said to be non-terminating, decimal expansion., There are following two types of non-terminating, decimal expansions, (i ) Non-terminating Repeating Expansion, The number, which does not terminate but, repeats the particular number again and again, in the process of division, is said to be, non-terminating repeating decimal or recurring, decimal expansion. The repeated digit is, denoted by bar ‘-’, 1, e.g., = 0.333 K = 0. 3, 3, (ii ) Non-terminating Non-repeating Decimal, Expansion The number, which neither terminates, nor repeats the particular number in the process, of division, is said to be a non-terminating, , ●, , e.g. 1. 030030003..., 3, etc., , Important Theorems on Decimal, Expansion of Rational Numbers, Theorem 3 Let x be a rational number whose decimal, expansion terminates. Then, x can be expressed in the, form p / q, where p, q are coprimes and the prime, factorisation of q is of the form 2 n5m , where n and m are, non-negative integers., Theorem 4 (Converse of Theorem 3) Let x = p / q, be a rational number, such that the prime factorisation, of q is of the form 2 n5m , where n and m are, non-negative integers. Then, x has a decimal, expansion, which terminates., Theorem 5 Let x = p / q be a rational number,, such that the prime factorisation of q is not of the form, 2 n5m , where n and m are non-negative integers. Then, x, has a decimal expansion, which is non-terminating, repeating (recurring)., , Polynomials, , A polynomial in one variable x, is an algebraic, expression of the form, p( x ) = an x n + an- 1x n- 1 + an- 2 x n- 2, + ... + a2 x 2 + a1x + a0, where n is a positive integer and constants, a0, a1, a2,..., an are known as coefficients of, polynomial., , Degree of a Polynomial, The highest power (exponent) of x in a polynomial f ( x ),, is called the degree of the polynomial f ( x )., , Types of Polynomials, (i) Linear Polynomial A polynomial of degree one, is, called linear polynomial., (ii) Quadratic Polynomial A polynomial of degree two,, is called quadratic polynomial., , ONE DAY REVISION, , non-repeating decimal expansion. These numbers are, called irrational numbers., , (iii) Cubic Polynomial A polynomial of degree three, is, called cubic polynomial., (iv) Biquadratic Polynomial A polynomial, of degree four, is called biquadratic polynomial., , Value of a Polynomial at Given Point, If p( x ) is a polynomial and a is a real value, then the, value obtained by putting x = a in p( x ) , is called the, value of p( x ) at x = a and it is denoted by p(a )., , Zeroes of a Polynomial, A real number k is said to be a zero of a polynomial, f ( x ), if f (k ) = 0., , Geometrical Meaning of the Zeroes of a Polynomial, The geometrical meaning of the zeroes of a polynomial, means that the curve intersect the X-axis, the, intersection point is said to be zeroes of the curve., , Relationship between Zeroes and, Coefficients of a Polynomial, The zeroes of a polynomial are related to its, coefficients., (i) For a Linear Polynomial The zero of the linear, polynomial ax + b is, b, Constant term, ., - =a, Coefficient of x, (ii) For a Quadratic Polynomial Let a and b be the, zeroes of quadratic polynomial p( x ) = ax 2 + bx + c ,, a ¹ 0, then, \Sum of zeroes, a + b, Coefficient of x, b, =–, =a, Coefficient of x 2, and product of zeroes, ab, Constant term, c, =, =, Coefficient of x 2 a, , Formation of Quadratic and, Cubic Polynomials, If a and b are the zeroes of a quadratic polynomial,, then quadratic polynomial will be k [x 2- (sum of, zeroes) x + product of zeroes], i.e. k [x 2 - (a + b )x + ab], where k is some constant.

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03, , CBSE Sample Paper Mathematics Standard Class X (Term I), , ●, , Pair of Linear Equations in Two Variables, , Two linear equations in the same two variables, say x, and y, are called pair of linear equations, (or system of pair equations) in two variables., The general form of pair of linear equations in two, variables x and y is, a1 x + b1 y + c 1 = 0, and, a2 x + b2 y + c 2 = 0,, where a1, b1, c 1 and a2, b2, c 2 are all real numbers, and a12 + b12 ¹ 0, a22 + b22 ¹ 0., , Solution of a Pair of Linear Equations, in Two Variables, Any pair of values of x and y which satisfies both the, equations, a1x + b1y + c 1 = 0 and a2 x + b2 y + c 2 = 0, is, called a solution of a given pair of linear equations., , Solution of a Pair of Linear, Equations by Graphical Method, Let us consider a pair of linear equations in two, variables, a1x + b1y + c 1 = 0 anda2 x + b2 y + c 2 = 0., To find the solution graphically, there are three cases, arise, Case I When the graph of system of linear equations, will represent two intersecting lines, then coordinates of, point of intersection say, (a, b) is the solution of the pair of linear equations. This, is called consistent pair of linear equations., Case II When the graph of system of linear equations, will represent two parallel lines, then there is no point of, intersection and consequently there is no pair of values, of x and y which satisfy both equations. Thus, given, system of equations have no solution. This is called, inconsistent pair of linear equations., Case III When the graph of system of linear equations, will represent coincident or overlapping lines, there are, infinitely many common points. Thus, the given system, of equations have infinitely many solutions., Such pair of linear equations is called dependent pair, of linear equations and it is always consistent., , Nature of Lines and Consistency, , Compare, Graphical, the ratios representation, a1 b1, ¹, a2 b2, , Intersecting, lines, , Algebraic, interpretation, , Consistency, , Exactly, one, solution, (unique), , System is, consistent, , Coincident, lines, , Infinitely, many, solutions, , System is, consistent, (dependent), , Parallel, lines, , No solution, , System is, inconsistent, , Algebraic Methods for Solving a Pair of Linear, Equations, There are three methods for solving a pair of linear, equations, , 1. Substitution Method, In this method, value of one variable can be found out, in terms of other variable from one of the given, equation and this value is substituted in other equation,, then we get an equation in one variable, which can be, solved easily., , 2. Elimination Method, In this method, one variable out of the two variables is, eliminated by making the coefficients of that variable, equal in both the equations., After eliminating that variable, the left equation is an, equation in another variable, which can be solved, easily., Value of one variable obtained in this way can be, substituted in any one of the two equations to find the, value of other variable., , Equations Reducible to a Pair of, Linear Equations, Sometimes, equations are not linear but they can be, reduced to a pair of linear equations by making some, suitable substitutions., 1, 1, (i) If the given equations involve and , then put, x, y, 1, 1, = p and = q to convert into linear form., x, y, 1, 1, and, ,, (ii) If the given equations involve, x±a, y±b, 1, 1, then put, = p and, = q to convert into, x±a, y±b, linear form., 1, 1, (iii) If the given equations involve, and, ,, x+ y, x-y, 1, 1, then put, = p and, = q to convert into, x+ y, x-y, linear form., , ONE DAY REVISION, , The nature of lines and consistency corresponding to, linear equations a1x + b1y + c 1 = 0 and, a2 x + b2 y + c 2 = 0, is shown in the table given below, , a1 b1, =, a2 b2, c, = 1, c2, a1 b1 c 1, = ¹, a2 b2 c 2

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04, , CBSE Sample Paper Mathematics Standard Class X (Term I), , ●, , Coordinate Geometry, , Cartesian System, , Section Formulae, , The system used to describe the position of a point in a, plane, is called cartesian system. In cartesian system,, there are two mutually perpendicular straight lines XX¢, and YY ¢, which intersect each other at origin point O., Y, , 1, X', , –3 –2, , P (x, y), , 90°, , –1 O, –1, , 1, , 2, , 3, , Internal Division of a Line Segment, Let A ( x1, y1) and B( x2, y2 ) are two points and P ( x, y ) is, a point on the line segment joining A and B such that, AP : BP = m1 : m2, then point P is said to divide line, segment AB internally in the ratio m1 : m2., , 3, 2, , In section formula, we find the coordinates of a point, which divides the given line segment internally (or, externally) in a given ratio., , m2, , X, , (x1 , y1 ), A, , –2, Y', , The horizontal line XOX¢ is called X-axis (or abscissa), and the vertical line YOY ¢ is called Y-axis (or ordinate)., , Distance between Two Points, in a Cartesian Plane, The distance between any two points P( x1, y1) and, Q( x2, y2 ) is given by, PQ = ( x2 - x1)2 + ( y2 - y1)2, PQ = ( x1 - x2 )2 + ( y1 - y2 )2, , Collinear Points, When three or more than three points lie on a same, line, then they are called collinear points., Suppose A, B and C are three points, then the, condition for collinearity of three points is, , ONE DAY REVISION, , AC + CB = AB, , or, , BA + AC = BC, , P, , æ m1x2 + m2 x1 m1y2 + m2 y1 ö, ,, ÷., ç, m1 + m2 ø, è m1 + m2, Generally, for finding internal division ratio, we consider, P divides AB in the ratio k : 1, then the coordinates of, the point P will be, æ kx2 + x1 ky2 + y1 ö, ,, ÷., ç, è k +1, k +1 ø, Coordinates of Mid-point of Line Segment, If the point P divides the line segment equally, i.e. 1 : 1, then the coordinates of P will be, æ x 1 + y1 y1 + y2 ö, ,, ç, ÷ . This is also called mid-point, è 2, 2 ø, formula., Note Trisection of the line segment means, a line is, divided into three equal line segment, , AB + BC = AC, or, , B, (x2 , y2 ), , The coordinates of point P are given by, , –3, , or, , m1, , A, , i.e., , P, , AP = PQ = QB., , Q, , B

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05, , CBSE Sample Paper Mathematics Standard Class X (Term I), , ●, , Triangles, , Similar Polygons, , (i) AAA Similarity Criterion, , Two polygons of the same number of sides are similar, if, (i) all the corresponding angles are equal and, (ii) all the corresponding sides are in the same ratio (or, proportion)., D, , H, , C, , and, , 6, A, , G, , 8, , B, , 9, E, , F, , 12, , If only one condition from (i) and (ii) is true for two, polygons, then they cannot be similar., , Similar Triangles, (i) their corresponding angles are equal and, (ii) their corresponding sides are proportional., Symbolically it can be represented by the, symbol ‘~’., P, , Q, , R, , e.g. In DABC and DPQR, if, ÐA = ÐP, ÐB = ÐQ , ÐC = ÐR, and, , (ii) SSS Similarity Criterion, If in two triangles, three sides of one triangle are, proportional (i.e., in the same ratio) to the three sides of, the other triangle, then their corresponding angles are, equal and hence the two triangles are similar., , If one angle of a triangle is equal to one angle of the, other triangle and the sides including these angles are, proportional, then the two triangles are similar., Theorem 1 If a perpendicular is drawn from the vertex, of the right angle of a right angled triangle to the, hypotenuse, then triangles on both sides of the, perpendicular are similar to the whole triangle and to, each other., , A, , C, , Note If two angles of one triangle are respectively, equal to two angles of another triangle, then the two, triangles are similar. AAA similarity criterion can be, consider as AA similarity criterion., , (iii) SAS Similarity Criterion, , Two triangles are said to be similar, if, , B, , In two triangles, if corresponding angles are equal,, then their corresponding sides are proportional and, hence the two triangles are similar., , AB BC AC, ., =, =, PQ QR PR, , Theorem 2 (Pythagoras Theorem) In a right angled, triangle, the square of the hypotenuse is equal to the, sum of the squares of the other two sides., Theorem 3 (Converse of Pythagoras Theorem) In a, right angled triangle, the square of the hypotenuse is, equal to the sum of the squares of the other two sides., AC 2 = AB 2 + BC 2, , Then, DABC is similar to DPQR., , A, , Conversely If DABC is similar to DPQR, then, , and, , ÐA = ÐP, ÐB = ÐQ, ÐC = ÐR, AB BC AC, =, =, PQ QR PR, , Basic Proportionality Theorem (BPT), , Theorem 2 (Converse of Basic Proportionality, Theorem) If a line divides any two sides of a triangle in, the same ratio, then the line must be parallel to the, third side., , Criteria for Similarity of Triangles, We have some criteria for congruency of two triangles, involving only three pairs of corresponding parts, (elements) of two triangles. Similarly, we have some, criteria for similarity of two triangles, which are given, below:, , C, , Area of Similar Triangles, Theorem 1 The ratio of the areas of two similar, triangles is equal to the square of the ratio of their, corresponding sides., P, , A, , B, , C Q, , R, 2, , ar (DABC ) ( AB)2 æ AC ö, æ BC ö, =, =ç, ÷ =ç, ÷, è QR ø, ar (DPQR ) (PQ ) è PR ø, , 2, , ONE DAY REVISION, , Theorem 1 (Thales Theorem) If a line is drawn parallel, to one side of a triangle to intersect the other two sides, in distinct points, then the other two sides are divided, in the same ratio., , B

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06, , CBSE Sample Paper Mathematics Standard Class X (Term I), , ●, , Introduction to Trigonometry, , Trigonometric Ratios, The ratios of the sides of a right angled triangle with, respect to its acute angles, are called trigonometric, ratios., Trigonometric ratios are also called T-ratio., Trigonometric ratios of ÐA in right angled D ABC are, defined below., , A Popular Technique to Remember, PBP, T-ratios i.e., HHB, Pandit (P ), , Badari (B ), , Prasad (P ), , Har (H ), , Har (H ), , Bholay (B ), , X, , Hy, po, , ten, u, , se, , (H, ), , Side opposite to Ð A, [i.e. Perpendicular (P)], , C, , A, B, Side adjacent, to ÐA [i.e. Base (B)], , (i) sin A =, , Side opposite to ÐA æ, Pö, çi.e. ÷, è, Hypotenuse, Hø, , BC, AC, AB, Side adjacent to ÐA æ, Bö, (ii) cos A =, çi. e. ÷ =, è, ø, AC, Hypotenuse, H, =, , Side opposite to ÐA æ, P ö BC, çi. e. ÷ =, è, Side adjacent to ÐA, B ø AB, , (iii) tan A =, , (iv) cosec A =, (v) sec A =, (vi) cot A =, , Hypotenuse, H ö AC, æ, çi. e. ÷ =, Side opposite to ÐA è, P ø BC, , Hypotenuse, H ö AC, æ, çi. e. ÷ =, Side adjacent to ÐA è, B ø AB, , AB, Side adjacent to ÐA æ, Bö, çi. e. ÷ =, Side opposite to ÐA è, BC, Pø, , Side adjacent to Ð A, [i.e. Base (B)], , ten, us, , e(, , H), , C, , Hy, po, , ONE DAY REVISION, , Similarly trigonometric ratios of ÐC are, AB, BC, (a) sin C =, (b) cos C =, AC, AC, AB, AC, (c) tan C =, (d) cosecC =, BC, AB, AC, BC, (e) sec C =, (f ) cot C =, BC, AB, , A, B, Side opposite to ÐC, [i.e. Perpendicular (P)], , H, , Z, , P, , B, , Y, , P, B, P, , cos q = , tan q =, H, H, B, H, H, B, Þ cosec q = , sec q = , cot q =, P, B, P, , Then, sin q =, , where, P is perpendicular, B is base and H is, hypotenuse., , Important Points, (i) In an isosceles right DABC, right angled, at B, the trigonometric ratios obtained by taking, either ÐA or ÐC, both give the same value., (ii) The value of each of the trigonometric ratios of an, angle does not depend on the size of the triangle. It, only depends on the angle., (iii) It is clear that the values of the trigonometric ratios, of an angle do not vary with the lengths of the sides, of the triangle, if the angle remains the same., (iv) As, the hypotenuse is the longest side in a right, angled triangle, the value of sin A or cos A is always, less than 1 (or in particular equal to 1) whereas the, value of sec A or cosec A is always greater than or, equal to 1., , Relation Between Trigonometric Ratios, (i) sin A =, , 1, 1, , cosec A =, cos ec A, sin A, , (ii) cos A =, , 1, 1, , sec A =, sec A, cos A, , (iii) tan A =, , 1, 1, , cot A =, cot A, tan A, , sin A, cos A, cos A, (v) cot A =, sin A, , (iv) tan A =

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07, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Important Points, , Values of Trigonometric Ratios for, Some Specific Angles, , (i) The value of sin q increase from 0 to 1 and cos q, decrease from 1 to 0, where 0 £ q £ 90°., , 0°, , 30°, , 45°, , 60°, , 90°, , sin q, , 0, , 1, 2, , 1, 2, , 3, 2, , 1, , cos q, , 1, , 3, 2, , 1, 2, , 1, 2, , 0, , (iv) In the case of cosec q, the values decrease from ¥, to 1, where 0 £ q £ 90°., , tan q, , 0, , 1, 3, , 1, , 3, , ¥, , (v) In the case of sec q, the values increase from 1 to, ¥, where 0 £ q £ 90° ., , cosec q, , ¥, , 2, , 2, , 2, 3, , 1, , (vi) Division by 0 is not allowed, since 1/0 is, indeterminate (not defined)., , sec q, , 1, , 2, 3, , 2, , 2, , ¥, , 1, , 1, 3, , Angles, , cot q, , ¥, , 3, , (ii) In the case of tan q, the values increase from 0 to ¥,, where 0 £ q £ 90°., (iii) In the case of cot q, the values decrease from ¥ to, 0, where 0 £ q £ 90°., , Trigonometric Identity, For any acute angle q, we have, (i) sin 2 q + cos 2 q = 1, , (ii) sec 2 q - tan 2 q = 1, , (iii) 1 + cot 2 q = cosec 2 q, , 0, , Note sin 2 q = (sin q)2 but sin q2 ¹ (sin q)2., The same is true for all other trigonometric ratios., , Here, ¥ = undefined, , Representation of a Trigonometric Ratio in Terms of Any Other Trigonometric Ratio, sin q, sin q, , cos q, tan q, , cot q, , sin q, , (1 - sin 2 q), sin q, 2, , (1 - sin q), (1 - sin 2 q), sin q, 1, , sec q, cosec q, , 2, , (1 - sin q), 1, sin q, , cos q, (1 - cos 2 q), cos q, (1 - cos 2 q), cos q, cos q, 2, , (1 - cos q), 1, cos q, 1, 2, , (1 - cos q), , tan q, , cot q, , tan q, , 1, 2, , (1 + tan q), , 2, , (1 + cot q), cot q, , 1, 2, , (1 + tan q), tan q, 1, tan q, , 2, , (1 + cot q), 1, cot q, , sec q, (sec 2 q - 1), sec q, 1, sec q, , cosec q, 1, cosec q, (cosec 2 q - 1), cosec q, 1, , (sec 2 q - 1), , (cosec 2 q - 1), , 1, cot q, , (1 + tan 2 q), , (1 + cot 2 q), cot q, , (1 + tan 2 q), tan q, , (1 + cot 2 q), , (sec 2 q - 1), , cosec 2 q - 1, cosec q, , sec q, , (cosec 2 q - 1), , sec q, (sec 2 q - 1), , cosec q, , ONE DAY REVISION

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08, , CBSE Sample Paper Mathematics Standard Class X (Term I), , ●, , Areas Related to Circles, , Circle, , Circular Ring, , A circle is the locus of a point which moves in a plane, such a way that its distance from a fixed point remains, the same. The fixed point is called the centre and the, given constant distance is known as radius of the, circle., , It is a plane figure bounded by the circumference of two, concentric circles of two different radii., , R, , O, , Ra, , di, , us, , r, , O, Diameter, , A, , B, , Area of ring = p(R 2 - r 2 ) sq units., , Sector of a Circle, Circumference (Perimeter) of a Circle, The distance covered by travelling once around a circle, is called the circumference or the length of boundary, of a circle., \ Circumference = p ´ Diameter = p ´ 2r units, where, r is the radius of circle., , The region enclosed by two radii and the, corresponding arc of a circle is called the sector of, circle., In the figure, unshaded region OACBO is called the, major sector and shaded region OAPBO is called the, minor sector of a circle., , Length of an Arc of a Sector, , Area of a Circle, The space occupied in a circular region is called area, of a circle., , The arc corresponding to a sector is called the arc of, the sector., , \ Area of a circle = pr 2 sq units, , C, Major, sector, O, , Semi-circle, A diameter divides the circle into two parts, each part is, called semi-circle., , r, A, , r, A, , B, P l, Minor sector, , B, , 2 pr, (i) Perimeter of semi-circle =, + 2r, 2, = (pr + 2 r ) units, 1, (ii) Area of semi-circle = (pr 2 ) sq units, 2, , Quadrant of a Circle, , Length of an arc of a sector, l =, , q, ´ 2 pr, 360°, , Area of Sector of a Circle, (i) Area of the sector =, , q, ´ pr 2, 360°, , (ii) Area of sector in terms of length of arc =, , If a circle is divided into four equal parts, then each part, of a circle is said to be quadrant of a circle., , ONE DAY REVISION, , q, , 1, lr, 2, , (iii) Area of the major sector = pr 2 - Area of minor, sector, (iv) Area of minor sector = pr 2 - Area of major sector, , A, , r, , B, , 2 pr, + 2r, 4, ö, æ pr, =ç, + 2 r ÷ units, ø, è2, æ pr 2 ö, (ii) Area of a quadrant = ç, ÷ sq units, è 4 ø, (i) Perimeter of a quadrant =, , Note, (i) If q = 180°, then sector becomes a semi-circular, 1, region and its area = pr 2., 2, (ii) If q = 90°, then sector becomes a quadrant of a, 1, circle and its area = pr 2., 4

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09, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Segment of a Circle, The region bounded by a chord and the corresponding, arc of the circle is called the segment of the circle., , (i) Area of a segment, , D, Major, segment, O, , = Area of corresponding sector, - Area of triangle formed by chord, and the radii of the circle., , q, A, , (ii) Area of major segment, B, , P, Minor segment, , ●, , The segment containing the minor arc is called the, minor segment and the segment containing the major, arc is called the major segment., , = pr 2 - Area of minor segment, (iii) Area of minor segment, = pr 2 - Area of major segment, , Probability, , Probability is the study of the chances (or likelihood), of events happening. By means of probability, the, chance (or likelihood) of events is measured by a, number lying from 0 to 1., , Experiment, An operation which produces some well defined, outcomes, is called an experiment., e.g. Tossing a coin, throwing a dice, etc., (i) Random experiment If an experiment is repeated, under identical conditions and they do not produce, the same outcomes every time, then it is said to be, random, (or probabilistic) experiment., (ii) Deterministic experiment If an experiment is, repeated under identical conditions and they, produce the same outcomes every time, then it is is, said to be deterministic experiment., An event for an experiment is the collection of some, outcomes of the experiment. We generally denote it by, capital letter E., e.g. Getting an even number in a single throw of a die, is an event. This event would consist of three, outcomes, namely 2, 4 and 6., , Elementary Event, , Occurrence of an Event, An event E associated to a random experiment is, said to be occur (or happen) in a trial, if the outcome, of trial is one of the outcomes that favours E., e.g. If a die is rolled and the outcome of a trial is 4, then, we say that event getting an even number has, happened (or occurred)., , If E is an event associated with a random experiment,, then probability of E, denoted by P(E), represents the, chance of occurrence of event E., e.g. If E denotes the event of getting an even number in, a single throw of a die, then P(E) represents the chance, of occurrence of event E, i.e. the chance of getting 2, 4, or 6., , Compound Event, A collection of two or more elementary events, associated with an experiment is called a compound, event. e.g. In the random experiment of tossing of two, coins simultaneously, if we define the event of getting, exactly one head, then it is a collection of elementary, events (or outcomes) HT and TH. So, it is a compound, event., , Equally Likely Outcomes, The outcomes of a random experiment are said to be, equally likely, when each outcome is as likely to occur, as the other, i.e. when we have no reason to believe, that one is more likely to occur than the other., e.g. When a die is thrown, all the six outcomes, i.e. 1,, 2, 3, 4, 5 and 6 are equally likely to appear. So, the, outcomes 1, 2, 3, 4, 5 and 6 are equally likely outcomes., , Favourable Outcomes, The outcomes which ensure the occurrence of an, event are called favourable outcomes to the event. e.g., The favourable outcomes to the event of getting an, even number when a die is thrown are 2, 4 and 6., , Complement of an Event/Negation of, an Event, Let E be an event associated with a random, experiment. Then, we can define the complement of, event E or negation of event E, denoted by E,, as an event which occurs if and only if E does, not occur., , ONE DAY REVISION, , An event having only one outcome of the random, experiment is called an elementary event. e.g. In, tossing of a coin, the possible outcomes are head (H ), and tail (T ). Getting H or T are known as elementary, events., , Probability of an Event, (or Probability of occurrence of an Event)

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10, , CBSE Sample Paper Mathematics Standard Class X (Term I), , e.g. Let E be the event of getting an even number in a, single throw of a die. Then, its complement can be define, as event E of getting an odd number, as E is consisting 2,, 4 and 6. Therefore, E would consist 1, 3 and 5., Note E and E are called complementary events., , Theoretical (Classical) Definition of Probability, Let us assume all the outcomes of an experiment are, equally likely and E is an event associated with the, experiment, then the theoretical probability, (or classical probability) of the event E is given by, Number of outcomes favourable to E, P(E ) =, Total number of outcomes, n (E ), =, n (S ), (i) Probability of an event can never be negative., (ii) The sum of the probabilities of complementary events, of an experiment is 1., i.e. If E and E are complementry events., Then, P( E ) + P( E ) = 1or P( E ) = 1 - P( E ), or, , P( E ) = 1 - P( E ), , ONE DAY REVISION, , where, P(E) represents the probability of, occurrence of an event E and P( E ) represents the, probability of non-occurrence of an, event E ., , Impossible Event, An event which is impossible to occur, is called an, impossible event and probability of impossible, event is always zero., e.g. In throwing a die, there are only six possible, outcomes 1, 2, 3, 4, 5 and 6. Let we are interested, in getting a number 7 on throwing a die. Since, no, face of the die is marked with 7. So, 7 cannot come, in any throw. Hence, getting 7 is an impossible, event., 0, Then, P ( getting a number 7 ) = = 0, 6, , Sure Event or Certain Event, An event which is sure to occur, is called a sure, event or certain event and probability of sure event, is always 1. e.g. Suppose we want to find the, probability of getting a number less than 7 in a, single throw of a die having numbers 1 to 6 on its, six faces., We are sure that, we shall always get a number, less than 7, whenever we throw a die. So, getting a, number less than 7 is a sure event., 6, Then, P (getting a number less than 7) = = 1, 6

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CBSE Sample Paper Mathematics Standard Class X (Term I), , 11, , THE, , QUALIFIERS, Chapterwise Set of MCQs to Check Preparation, Level of Each Chapter, , 1. Real Numbers, Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct., Select the correct option as your answer., , 1. If two positive integers a and b are written as a = x 4 y 2 and b = x 2 y 3 , where x and y are, prime numbers, then LCM ( a , b) is, (a) x 8 y 6, , (b) x 6 y 5, , (c) x 4 y 3, , (d) None of these, , 2. The HCF and LCM of two numbers are 23 and 828. When the first number is, completely divided by 2 the quotient is 23. The other number is, (a) 420, (b) 410, (c) 414, (d) 425, , 3. Two natural numbers whose difference is 66 and the LCM is 360, are, (a) 180 and 114, (c) 120 and 54, , (b) 90 and 24, (d) 130 and 64, , 4. The HCF and LCM of the smallest composite number and the smallest prime number, are respectively., (a) 2 and 2, (c) 4 and 4, , (b) 2 and 4, (d) 8 and 4, , 5. If HCF of two numbers is 4 and their product is 160, then their LCM is, (a) 40, (c) 80, , (b) 60, (d) 120, , (a) 16, (c) 20, , (b) 8, (d) 10, , 7. The sum of two numbers is 528 and their HCF is 33, then the number of pairs satisfying, the given condition is, (a) 5, (c) 4, , (b) 3, (d) 2, , THE QUALIFIER, , 6. If LCM = (32, a) = 64 and HCF (32, a) = 4, then a is equal to

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12, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 8. The ratio of LCM and HCF of second smallest prime number and second smallest, composite number is, (a) 2 : 5, (c) 1 : 2, , (b) 2 : 1, (d) 5 : 2, , 9. If the LCM of two prime number a and b (a > b) is 253, then the value of 8b - 3a is, (a) 16, (c) 19, , (b) 17, (d) 18, , 2 5, + , then x is, 36 6, (a) irrational, (c) whole number, , (b) rational, (d) integer, , 10. If x 2 = 1 +, , 11. Prime factors of the denominator of a rational number with the decimal expansion, 62.47 are, (a) 2 and 35, (c) 3 and 5, , (b) 2 and 5, (d) 4 and 5, , 12. The decimal expansion of the rational number, places of decimal?, (a) 1, (c) 4, , 53, 3, , 2 ´5, , , will terminate after how many, , (b) 3, (d) 2, , 1, should be multiplied, so that its decimal expansion, 17, terminator after one decimal place is, 17, 17, (a), (b), 100, 10, 100, 10, (d), (c), 17, 17, p, 14. The rational form of 0.325 is in the form of , then q - p is, q, , 13. The smallest number by which, , (a) 640, (c) 668, , (b) 650, (d) 670, , 15. What smallest number must be multiplied in the denominator, so that the decimal, number, , 14588, will be terminated?, 625, , THE QUALIFIER, , (a) 4, (c) 16, , (b) 18, (d) 20, , Answers, 1. (c), 6. (b), 11. (b), , 2. (c), 7. (c), 12. (b), , 3. (b), 8. (b), 13. (b), , 4. (b), 9. (c), 14. (c), , 5. (a), 10. (a), 15. (c), , For Detailed Solutions, Scan the code

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13, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 2. Polynomials, Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct., Select the correct option as your answer., , 1. If 2 is a zero of polynomial p(x) = 5x 2 + 3x - 4 k, then the value of k is, 15, 2, 11, (c), 2, , (b), , (a), , 13, 2, , (d) None of these, , 2. The graph of the polynomial p(x) is given as below,, Y, , X¢, , –4 –3 –2 –1 0 1, , 2, , 3, , X, , Y¢, , The zeroes of p( x) are, (a) - 3, - 2 , 0 and 0, (c) - 3, - 2 , 0 and 2, , (b) - 2 , 0, 2 and 3, (d) - 3, 0, 2 and 3, , 3. The number of zeroes of given polynomial graph are, Y, , X¢, , X, , Y¢, , (a) 4, , (b) 3, , (c) 2, , (d) 1, , 4. Which of the following is not the graph of a quadratic polynomial?, Y, , (a), , (b), , X, , X¢, , X, , Y¢, , Y¢, , Y, , Y, , X¢, , (d), , X, , X¢, , X, , Y¢, , Y¢, 2, , 5. If one zero of the polynomial x - 4 x + 1 is 2 - 3, then the other zero is, (a) 2 - 3, , (b) 2 + 3, , (c) 3 - 2, , (d) None of these, , THE QUALIFIER, , (c), , X¢, , Y

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14, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 6. The number of polynomials having zeroes as 3 and - 7 is, (a) 11, , (b) 2, , (c) 3, , (d) more than 2, , 2, , 7. If p(x) = ax + bx + c and a + b + c = 0, then one zero is, c, a, b, (c), a, , b, a, , (b) -, , (a), , (d) Cannot determined, , 1, 4, (b) 4x 2 + 13x - 3, (d) 4x 2 - 13x + 3, , 8. Find a quadratic polynomial, whose zeroes are 3 and ., (a) 2 x 2 + 13x - 3, (c) 4x 2 - 13x - 3, , 9. If 2 and 3 are zeroes of polynomial 3x 2 - 2kx + 2m, then the values of k and m are, respectively, 9, (a) 15 and, 2, , (b), , 15, and - 9, 2, , (c), , 15, and 9, 2, , (d) None of these, , 10. If one zero of the polynomial (a 2 + 4)x 2 + 9x + 4 a is the reciprocal of the other, then the, value of a is, (a) 2 and 3, , (b) - 2, , (d) - 2 and - 3, , (c) 2, , 11. If zeroes of the polynomial p(x) = - 8x 2 + (k + 5)x + 36 are negative to each other, then, the value of k is, (a) - 5, , (b) 5, , (c) 4, , (d) 3, , 12. If a and b are zeroes of the polynomial p(x) = x 2 - p(x + 1) + c such that (a + 1) (b + 1) = 0,, then the value of c is, (a) - 1, , (c) -, , (b) 1, , 1, 2, , (d), , 1, 2, , 13. Suppose a and b are zeroes of the quadratic polynomial p(x) = x 2 - (k + 5)x + 3(2k - 3), such that a + b =, (a), , 19, 3, , ab, . Then, the value of k is, 2, 19, 19, (b) (c), 4, 4, , (d) None of these, , 14. If one zero of the polynomial p(x) = 2x 2 - 5x - (2k + 1) is twice the other zero, then the, value of k is, 17, (a), 9, , (b) -, , 17, 9, , (c), , 9, 17, , (d) None of these, , THE QUALIFIER, , 15. If the square of difference of the zeroes of the quadratic polynomial p(x) = x 2 + px + 45, is equal to 144, then the value of p are, (a) ± 9, (b) ± 12, , (c) ± 15, , (d) ± 18, , Answers, 1. (b), 6. (d), 11. (a), , 2. (c), 7. (b), 12. (a), , 3. (b), 8. (d), 13. (c), , 4. (c), 9. (c), 14. (b), , 5. (b), 10. (c), 15. (d), , For Detailed Solutions, Scan the code

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15, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 3. Pair of Linear Equations in Two Variables, Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct., Select the correct option as your answer., , 1. The line x = h and y = k, where h ¹ 0 and k ¹ 0, are, (a) parallel, (c) overlapping, , (b) intersecting, (d) None of these, , 2. The pair of equations x = 0 and x = 4 has, (a) no solution, (c) two solutions, , (b) one solution, (d) infinitely many solutions, , 3. If the system of equations 2x + ky = 8 and x + y = 6 has no solution, then the value of k is, (a) 2, , (b) -, , (c) - 2, , (d), , 3, 2, , 3, 2, , 4. One equation of a pair of dependent linear equation is 3x + 5y = 4. Then, second, equation can be, (a) 15x + 25y = 15, (c) - 15x - 25y = 20, , (b) 15x + 25y = 20, (d) 15x - 25y = 20, , 5. If the lines given by 2x + ky = 1 and 3x - 5y = 7 has unique solution, then the value of k is, (a) all real values of k, , (b) all real values of k except k =, , 10, (c) all real value of k except k =, 3, , (d) None of these, , - 10, 3, , 6. If the system of equations (k + 2)x + 21y - 3k = 0 and 4 x + 7 y - 10 = 0 has infinitely many, solutions, then value of k is, (a) 11, (c) 12, , (b) 10, (d) - 12, , 7. The pair of linear equations 2x + 3y = 5 and 4 x + 6y = 10 is, (a) consistent, (c) dependent consistent, , (b) inconsistent, (d) None of these, , 8. The nature of the lines representing the linear equations 2x - y = 3 and 4 x - y = 5 is, (a) intersecting, (c) coincident, , (b) parallel, (d) None of these, , (a) 10 and 8, (c) 1 and 1, , 10. The solution of the system of equations, (a) a and b, (c) - a and - b, , (b) 8 and 13, (d) None of these, , x y, + = 2 and ax - by = a 2 - b 2 is, a b, (b) b and a, (d) - b and - a, , THE QUALIFIER, , 9. The solution of the pair of systems 7 x - 4 y = 3 and x + 2y = 3 is

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16, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 11. The area of figure formed by the lines x = - 3 and y = 2 and along with coordinate axes, is, (a) 5 sq units, (c) 3 sq units, , (b) 6 sq units, (d) None of these, , 12. The graph of linear equations 2x + y = 6 and 4 x - 2y = 4 is shown below. Find the area of, triangle formed by lines and X-axis., Y, 6, 5, 4, , 4x – 2y – 4 = 0, , 3, C, , 2, 1, X¢, , –5 –4 –3 –2 –1 O, , A D, 1 2, –1, –2, , B, 3 4, , 5, , X, , 2x + y = 6, , –3, –4, Y¢, , (a) 3 sq units, (c) 1 sq units, , (b) 2 sq units, (d) None of these, , 13. The solution of 7 x + y = 343 and 343 x - y = 7 is, 4, 5, and y =, 3, 3, -5, 4, and y =, (c) x =, 3, 3, , (a) x =, , (b) x =, , 5, 4, and y =, 3, 3, , (d) None of these, , 14. The solution of the pair of equations, (a) x = 4 and y = 9, (c) x = 2 and y = 3, , 2, 3, 4, 9, +, = 2 and, = - 1 is, x, y, x, y, (b) x = - 4 and y = - 9, (d) None of these, , 15. The sum of the digits of a two-digit number is 9. Also nine times this number is twice, , THE QUALIFIER, , the number obtained by reversing the order of the digits. The original number is, (a) 17, (b) 16, (c) 18, (d) 15, , Answers, 1. (b), 6. (b), 11. (b), , 2. (a), 7. (c), 12. (b), , 3. (a), 8. (a), 13. (b), , 4. (b), 9. (c), 14. (a), , 5. (b), 10. (a), 15. (c), , For Detailed Solutions, Scan the code

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17, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 4. Coordinate Geometry, Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct., Select the correct option as your answer., , 1. The distance between the points A(- 1, - 5) and B(- 6, 7) is, (a) 12 units, (c) 14 units, , (b) 13 units, (d) 15 units, , 2. If the distance between the points A(x , 2) and B(3, - 6) is 10 units, then the positive, value of x is, (a) 6, (c) - 3, , (b) 9, (d) 5, , 3. A triangle with vertices A(4 , 0), B(- 1, - 1) and C(3, 5) is/an, (a) equilateral triangle, (b) isosceles triangle, (c) right triangle, (d) right isosceles triangle, , 4. In the given figure, the area of DABC (in sq. units) is, Y, C, , 4, 3, 2, X¢, , 1, , A, –4 –3 –2 –1 O, , 1, –1, , D, 2, , 3, , 4, , 5B, , X, , –2, –3, Y¢, , (a) 18 sq units, (c) 14 sq units, , (b) 12 sq units, (d) 16 sq units, , 5. If three points A(0, 0), B(3, 3) and C(3, l) form an equilateral triangle, then l is equal to, (a) 2, , (b) - 3, , (c) - 4, , (d) ± 3, , 6. Find the points on Y-axis which is equidistant from two points A(- 3, 4) and B(3, 6) on, (b) P(0, 5), (d) None of these, , 7. If the points A(1, - 1), B(5, y) and C(9, 5) are collinear, then the value of y is, (a) 2, (c) - 2, , (b) 3, (d) 4, , THE QUALIFIER, , the same plane., (a) P(0, - 5), (c) P(0, ± 5)

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18, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 8. Find the equation of the perpendicular bisector of line segment joining points A(- 2, 3), and B( 4 , 7)., (a) 3x + 2 y = 13, (c) 2 x - 3y = 13, , (b) 3x - 2 y = 13, (d) 3x + 2 y = - 13, , 9. Find the ratio, in which the line segment joining points A(- 3, 10) and B(6, - 8) is divided, by C( - 1, 6)., (a) 3 : 4, (c) 7 : 2, , (b) 2 : 7, (d) None of these, , 10. The vertices of a parallelogram in order are A(2, 3), B(3, y), C(x , 6) and D(4 , 5). Then,, ( x , y) is equal to, (a) ( - 5, 4), (c) (5, 4), , (b) (4, 5), (d) ( 4, - 5), , 11. Find the coordinates of the centroid of a triangle, whose vertices are A(0, 6),B(8, 12) and, C( 8, 0)., 16, (a) æç , - 6ö÷, è3, ø, (c) (16, 2), , 16, (b) æç , 6ö÷, è3 ø, (d) None of these, , 12. If the centroid of triangle formed by the points P(a , b), Q(b , c) and R(c, a) is a origin., Then, the value of a + b is equal to, (a) 3c, (c) - c, , (b) c, c, (d), 3, , 13. The point P which divides the line segment joining the points A(2, - 5) and B(5, 2) in the, ratio 2 : 3, lies in the quadrant, (a) IV, (c) II, , (b) III, (d) I, , 14. If vertices of a DABC are A(5, 1), B(1, 5) and C(- 3, - 1). Then, find the length of median, AD., (a) 6 units, (c) 37 units, , (b) 35 units, (d) 41 units, , 15. Point P divides the line segment joining R(- 6, 10) and S(3, - 8) in the ratio l : 1. If point P, , THE QUALIFIER, , lies on the line 2x - y + 4 = 0, then the value of l is, (a) - 1, (b) 1, 1, 1, (d) (c), 2, 2, , Answers, 1. (b), 6. (b), 11. (b), , 2. (b), 7. (a), 12. (c), , 3. (d), 8. (a), 13. (a), , 4. (c), 9. (b), 14. (c), , 5. (d), 10. (c), 15. (b), , For Detailed Solutions, Scan the code

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19, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 5. Triangles, Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct., Select the correct option as your answer., , 1. In the figure, if MN ||QR, PM = 6 cm, QM = 8 cm and QR = 28 cm, then MN is equal to, P, 6 cm, M, , N, , 8 cm, Q, , R, , 28 cm, , (a) 20 cm, (c) 12 cm, , (b) 32 cm, (d) 16 cm, , 2. In the given figure, DE|| BC. If AD = x + 2, DB = x - 2, AE = x + 3 and EC = x - 4, then the, value of x, C, E, , A, , D, , 3, (a), 2, 3, (c), 5, , B, , (b), (d), , 3, 8, -2, 3, , 3. In DPQR is such that PQ = 4 cm, QR = 3 cm PR = 3.5 cm. If DPQR ~ DXYZ and YZ = 5 cm,, then perimeter of DXYZ is, P, , X, , cm, , 4c, , m, , 3.5, , Q, , R Y, , 3 cm, , (a) 17.5 cm, (c) 18.5 cm, , 5 cm, , Z, , (b) 16 cm, (d) 22.5 cm, , 4. In the given figure ÐM = ÐN and, , PM PN, , then DPQR is, =, MQ NR, P, , Q, , (a) equilateral triangle, (c) right triangle, , N, , R, , (b) isosceles triangle, (d) None of these, , THE QUALIFIER, , M

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20, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 5. In the given figure ÐP = ÐT, PQ = 6 cm, QR = 15 cm, PR = 12 cm and RT = 4 cm, then the, value of RS is, P, 12 cm, , 6 cm, , R, , S, , 15 cm, , Q, , 4c, , m, T, , (a) 6 cm, , (b) 5 cm, , (c) 8 cm, , (d) 9 cm, , 6. In the given figure ÐMNP = 90° and ON ^ MP. If ON = 12 cm and MO = 6 cm, then the, value of OP is, M, 6 cm, O, 12 cm, P, , N, , (a) 16 cm, , (b) 36 cm, , (c) 24 cm, , (d) 18 cm, , 7. In DABC,ÐA is obtuse angle, PB ^ PC and QC ^ QB, then the value of AB ´ AQ is, Q, , P, A, , C, , B, , (a) AP ´ AB, , (b)AC ´ AP, , (c) AB ´ AC, , (d) AQ ´ AP, , 8. If two coconut trees 15 m and 25 m high are 70 m apart, then the height of the point of, intersection of the line joining the top of each tree to the foot of the opposite tree is, A, , P, R, h, , B, , THE QUALIFIER, , (a) 9, , 19, m, 56, , (b) 9, , Q, , S, , 20, m, 56, , (c) 9, , 21, m, 56, , (d) 9, , 22, m, 56, , 9. Diagonal of a trapezium ABCD intersect each other at the point O, AB||CD and, AB : CD = 2 : 3, then the ratio of the areas of DAOB and DCOD is, D, , C, O, , A, , (a) 4 : 9, (c) 16 : 36, , B, , (b) 9 : 16, (d) None of these

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21, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 10. Two isosceles triangles have equal vertical angles and their area are in the ratio 49 : 64,, then the ratio of their corresponding heights., (a) 8 : 7, (b) 9 : 6, (c) 7 : 9, , (d) 7 : 8, , 11. In DABC, AD is bisector of ÐA, if BD = 6 cm, DC = 8 cm and AB = 6 cm, then AC is, , 6c, , m, , A, , B 6 cm D 8 cm C, , (a) 12 cm, , (b) 4.5 cm, , (c) 7 cm, , (d) 6 cm, , 12. The hypotenuse of a right triangle is 6 m more than the twice of the shortest side. If the, third side is 2 m less than the hypotenuse, then the sides of triangle is, (a) 10, 12 and 14, (b) 10, 24 and 26, (c) 12, 13 and 15, (d) None of these, , 13. In the given figure, PR = 12 cm and PL = 8 cm, then QM is, P, , 12cm, 8cm, , M, Q 6 cm, , L, , (a) 6 cm, , (b) 9 cm, , R, , (c) 4 cm, , (d) 5 cm, , 14. An equilateral triangle is inscribed in a circle of radius 8 cm, then the side of, equilateral triangle is, (a) 16 cm, (b) 4 3 cm, , (c) 14 cm, , (d) 12 cm, , 15. In the given figure, AD ^ BC, then AC 2 + 2BC × BD is, A, , B, , (a) AB2 + BC 2, (c) AD 2, , C, , D, , (b) AB2 + AD 2, (d) AB2 - BC 2, , 2. (d), 7. (b), 12. (b), , 3. (a), 8. (c), 13. (c), , 4. (b), 9. (a), 14. (b), , 5. (b), 10. (d), 15. (a), , For Detailed Solutions, Scan the code, , THE QUALIFIER, , Answers, 1. (c), 6. (c), 11. (d)

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22, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 6. Trigonometry, Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct., Select the correct option as your answer., 3, 1. If secq = , then the value of 2 cos 2 q + 2 cot 2 q - 9, 2, - 91, 91, 44, 90, (c), (d), (a), (b), 15, 15, 44, 15, 2. If 3 cot A = 4 cos A, then the relation between sec A and tan A is, (b) 3 sec 2 A - 4 tan 2 A = 0, (d) 3 sec A - 4 tan A = 0, , (a) 4 sec A = 3 tan A, (c) 4 sec 2 A - 3 tan 2 A = 0, , 3. In the given figure, ÐP = q and ÐR = f,, R, f, x +2, , x, , q, P, , Q, , then the value of x 3 + x 2 tan q is, (a) - x 2, , (b), , - x2, , (c), , 2, , 2, x2, , 4. The positive minimum value of cosec q is, (a) 0, , (b) 1, 1, (d), 2, , (c) 2, , 4, 5, , 5. If - x tan 45° sin 60° + cos 60°× cot 45° = , then the value of x is, 3, 10, 3, (c), 5, (a), , (b), (d), , - 3, 5, - 3, 10, , 6. The value of (sin 30°× cos 60° + cos 30° sin × 60° ) is, (a) 2, , THE QUALIFIER, , (c) 1, , 1, 2, 1, (d), 4, (b), , 7. If DABC is right angled at C, then the value of sin(A + B) is, (a) 1, (c) - 1, , (b) 0, 1, (d), 2, , (d), , x2, 2

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CBSE Sample Paper Mathematics Standard Class X (Term I), , 23, , 8. In an acute angled DABC, if cot(A + B - C) = 1 and cosec(B + C - A) = 2, then the angles, A, B and C is, 105° 135°, (a), ,, and 75°, 2, 2, 75° 75°, 135°, (c), ,, and, 2, 2, 2, , 75°, 135°, and, 2, 2, 105° 135°, 75°, (d), ,, and, 2, 2, 2, , (b) 75° ,, , é 1 - sin 2 q ù, ú, 2, êë 1 + cos q úû, , 4, 3, , 9. If cot q = , then the value of ê, 16, 41, 25, (c), 16, , (b), , (a), , 16, 35, , (d) None of these, , æ 1 + tan 2 q ö, ÷, ÷, 2, è cot q ø, 1, (b) 2, (d) 0, , 10. If cot q = 1, then the value of (1 + sin 2 q) - çç, 1, 2, (c) 1, (a), , 11. The value of, , 1, 1, +, - 2 is, 1 + cos q 1 - cos q, , (a) 2 sec 2 q, (c) 2 secq, , (b) 2 cot 2 q, (d) sec q × tan q, , 12. If m = cos q - sin q and n = cos q + sin q, then the value of sec 2 q is, ( m + n) 2, , (a), (c), , (b), , 2, 4, , 4, , (d) None of these, , ( m + n) 2, , 13. The value of 1 +, , ( m + n) 2, , tan 2 a, is, 1 + sec a, (b) cosec a, (d)cot a, , (a) seca, (c) cos a, , 1 ö, æ, 1 ö, 3, ÷ = x, then the value of ç tan q ÷ is, ø, è, tan q, tan 3 q ø, (a) x 2 ( x + 2 ), (b) x( x 2 - 2 ), 2, (c) x( x + 3), (d) x( x 2 - 3), , 14. If æç tan q è, , 15. The value of 2(sin 6 q + cos 6 q) - 3(sin 4 q + cos 4 q) + 1 is, (b) 1, (d) 2, , Answers, 1. (a), 6. (c), 11. (b), , 2. (d), 7. (a), 12. (c), , 3. (d), 8. (b), 13. (a), , 4. (b), 9. (a), 14. (c), , 5. (b), 10. (b), 15. (a), , For Detailed Solutions, Scan the code, , THE QUALIFIER, , (a) 0, (c) - 1

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24, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 7. Area Related to Circles, Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct., Select the correct option as your answer., , 1. If the area of circular field is 30800 m 2 , then the perimeter of the field is, (a) 720 m, (c) 360 m, , (b) 360 2 m, (d) None of these, , 2. The radii of two circles are 14 cm and 7 cm, respectively, then the radius of the circle, which has circumference equal to sum of the circumference of the two circles., (a) 21 cm, (b) 19 cm, (c) 18 cm, (d) 20 cm, , 3. The number of revolution made by circular wheel of radius 1.4 m in rolling a distance, 176 m is, (a) 40, (c) 30, , (b) 20, (d) 15, , 4. The ratio of the outer and inner circumference of a circular path is 24 : 23. If the path is, 5 m wide, then the diameter of the inner circle is, (a) 210 m, (b) 220 m, (c) 200 m, (d) 230 m, , 5. The circumference of circle is 110 cm. The side of a square inscribed in the circle is, 35, cm, 2, 35, (c), 2 cm, 2, , (a), , (b) 35 2 cm, (d) 70 cm, , 6. The short and long hands of a clock are 10 cm and 12 cm along, respectively. The sum, of the distances travelled by their tips in one days is, (a) 1282 cm, (b) 1789 cm, (c) 1936 cm, (d) 1848 cm, , 7. In a circle of radius 14 cm, an arc subtends an angle of 60° at the centre, then the length, of the arc and area of sector are, , THE QUALIFIER, , 14, , cm, , O, , 44, 305, cm and, cm 2, 3, 3, 88, 616, cm and, (c), cm 2, 3, 3, , (a), , 60°, , (b), , 22, 154, cm and, cm 2, 3, 3, , (d) None of these, , 8. The length of minute hand of a clock is 16 cm, then the area swept by the minute hand, in 1 min is, (a) 15.75 cm 2, (c) 10.27 cm 2, , (b) 17.65 cm 2, (d) 19.85 cm 2

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25, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 9. In the given figure, sectors of two concentric circle of radii 14 cm and 7 cm are shown,, then the area of the shaded region is, 7 cm, 30°, 14 c, , m, , (a) 35.5 cm 2, (c) 38.5 cm 2, , (b) 36.5 cm 2, (d) 42.5 cm 2, , 10. The minute hand of a clock is 12 cm long, then the area of the face of the clock, described by the minute hand in 30 min is, (a) 220 cm 2, (b) 226. 28 cm 2, (c) 246 cm 2, , (d) 315 cm 2, , 11. An chord 16 cm long is drawn in a circle whose radius is 16 cm. Then, the area of, segment is, , O, qq, A, , (a) 80 cm 2, , B, , M, 16 cm, , (b) 60 cm 2, , (c) 23.24 cm 2, , (d) 86.14 cm 2, , 12. In the adjoining figure OACBO respresents a quadrants of a circle of radius 3.5 cm with, centre O, then the area of shaded portion, A, C, 3 cm, , O, , B, , 3.5 cm, , (a) 4. 37 cm 2, (c) 10. 5 cm 2, , (b) 6. 25 cm 2, (d) None of these, , 13. In the given figure, ABCD is a square of side 8 cm with E , F, G and H as the mid-points, of sides AB , BC , CD and DA respectively. Then, area of the shaded portion is, E, B, , H, , F, , D, , G, , C, , (a) 32 cm 2, , (b) 64 cm 2, , (c) 98 cm 2, , (d) 84 cm 2, , THE QUALIFIER, , A

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26, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 14. In the given figure PQRS is a square of length 20 2 cm. If DPEQ is an isosceles triangle, inscribed in the semi-circle with diameter PQ, then the area of the shaded region is, P, , Q, , S, , R, , E, , (a) 328 cm 2, (c) 600 cm 2, , (b) 428 cm 2, (d) 628 cm 2, , 15. In the given figure, ABCD is a square of side 8 cm and A, B , C and D are centres of equal, circle touching externally in pairs, then the area of the shaded region., , 88, cm 2, 7, 78, (c), cm 2, 7, , D, , C, , A, , B, , 96, cm 2, 7, 92, (d), cm 2, 7, , (a), , (b), , Answers, 1. (d), 6. (c), 11. (c), , 2. (a), 7. (d), 12. (a), , 3. (b), 8. (c), 13. (a), , 4. (d), 9. (c), 14. (b), , 5. (c), 10. (b), 15. (b), , For Detailed Solutions, Scan the code, , 8. Probability, Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct., Select the correct option as your answer., 1. Two fair coins are tossed, then the probability of getting at the least one head is, 3, 4, 1, (c), 2, , THE QUALIFIER, , (a), , 1, 4, 3, (d), 8, (b), , 2. In a simultaneous tossing of three coins, the probability of getting at most one head is, 3, 4, 3, (c), 8, (a), , 1, 2, 2, (d), 3, (b)

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27, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 3. A card is drawn from a well shuffled deck of 52 cards, then the probability of getting, the jack of heart is, 1, (a), 13, 1, (c), 52, , (b), , 1, 26, , (d) None of these, , 4. Cards marked with numbers 5 to 75 are placed in a box and mixed thoroughly., One card is drawn from the box. Then, the probability that the number on the card is, even is, 35, 35, (a), (b), 71, 70, 36, 36, (c), (d), 71, 70, , 5. A die is thrown once, then the probability of getting a number lying between 3 and 6 is, 1, 3, 2, (c), 3, , 5, 6, 1, (d), 6, (b), , (a), , 6. A die is thrown once, then the probability of getting a number which is not a factor of, 30 is, 1, (a), 6, 2, (c), 3, , (b), , 1, 3, , (d) None of these, , 7. A letter of English alphabets is chosen at random. Then, the probability that it is a letter, of the word ‘ARIHANT’ is, 4, (a), 13, 7, (c), 26, , 9, 26, 5, (d), 26, (b), , 8. A child has a die whose six faces show the letters as given below, A, , B, , C, , D, , E, , Z, , O, , The die is thrown once, then the probability of getting Z is, 2, 1, (a), (b), 3, 3, 1, 1, (d), (c), 6, 2, divisible by 8 is, 6, (a), 49, 3, (c), 49, , 5, 49, 2, (d), 49, (b), , THE QUALIFIER, , 9. An integer is chosen at random between 1 and 100, then the probability that it is

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28, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 10. The probability of guessing the correct answer to certain question is, , p, , . If the probability, 10, 1, of not guessing the correct answer to same questions is , then the value of p is, 4, 15, 8, (a), (b), 2, 3, 15, (d) None of these, (c), 4, , 11. A number x is chosen at random from the numbers -3, - 2, - 1, 0, 1, 2, 3, then the, probability that x 2 £ 4 is, 5, (a), 7, 5, (c), 7, , 4, 7, 6, (d), 7, (b), , 12. Suppose, you drop a die at random on the square region shown in figure. What is the, probability that it will land inside the circle of diameter 1 m ?, 5m, 5m, , p, 24, p, (c), 40, , (b), , (a), , p, 100, , (d) None of these, , 13. 10 defective pens are accidentally mixed with 140 good ones. It is not possible to just, look at a pen and tell whether or not it is defective. One pen is taken out at random, from this lot, the probability that the pen taken out is a good one is, 7, 6, (a), (b), 15, 15, 14, 15, (d), (c), 15, 14, , 14. A box contains 80 discs, which are numbered from 1 to 80. If one disc is drawn at, random from the box, then the probability that it bears a number divisible by 5 is, 2, 8, (a), (b), 5, 45, 3, (d) None of these, (c), 5, , THE QUALIFIER, , 15. In a family of 3 children, then the probability of having atmost one boy is, 5, 8, 7, (c), 8, , 1, 8, 3, (d), 8, (b), , (a), , Answers, 1. (a), 6. (a), 11. (c), , 2. (b), 7. (c), 12. (b), , 3. (c), 8. (c), 13. (c), , 4. (a), 9. (a), 14. (d), , 5. (a), 10. (a), 15. (d), , For Detailed Solutions, Scan the code

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29, , CBSE Sample Paper Mathematics Standard Class X (Term I), , CBSE, QUESTION BANK, Case Study Based Questions, Real Numbers, , (iv) 7 ´ 11 ´ 13 ´ 15 + 15 is a, (a) Prime number, (b) Composite number, (c) Neither prime nor composite, (d) None of the above, , 1. To enhance the reading skills of grade X, students, the school nominates you and two, of your friends to set up a class library., There are two sections- Section A and, Section B of grade X. There are 32 students in, section A and 36 students in section B., , (v) If p and q are positive integers such that, p = ab 2 and q = a2b, where a and b are, prime numbers, then the LCM ( p , q) is, (a) ab, (c) a 3 b 2, , (b) a 2 b 2, (d) a 3 b 3, , Sol., , (i) What is the minimum number of books, you will acquire for the class library, so, that they can be distributed equally, among students of Section A or Section B?, (a) 144, (c) 288, , (b) 128, (d) 272, , (ii) If the product of two positive integers is, equal to the product of their HCF and, LCM is true, then the HCF (32, 36) is, (b) 4, , (c) 6, , (d) 8, , (iii) 36 can be expressed as a product of its, primes as, (a) 22 ´ 3 2, , (b) 21 ´ 3 3, , (c) 23 ´ 31, , (d) 20 ´ 3 0, , (iv) (b) 7 ´ 11 ´ 13 ´ 15 + 15 = 15 ´ (7 ´ 11 ´ 13 + 1), Since, the number is divisible by a number other than, itself and 1., Hence, it is a composite number., (v) (b) Given, p = ab 2 and q = a2 b, LCM ( p, q ) = Product of the greatest power of each, prime factor involved in the numbers, with highest, power = a2 ´ b 2, , CBSE QUESTION BANK, , (a) 2, , (i) (c) Given, number of students in Section A = 32, Number of students in Section B = 36, The minimum number of books acquire for the class, library = LCM of (32, 36), = 2 ´2 ´2 ´2 ´2 ´ 3 ´ 3, = 2 5 ´ 32, = 32 ´ 9 = 288, (ii) (b) Given, product of the two numbers, = LCM ´ HCF, \, 32 ´ 36 = LCM (32, 36) ´ HCF (32, 36), Þ, 32 ´ 36 = 288 ´ HCF (32, 36), 32 ´ 36, =4, Þ HCF (32, 36) =, 288, (iii) (a) The prime factors of 36 are, 36 = 2 ´ 2 ´ 3 ´ 3 = 2 2 ´ 32

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30, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 2. A seminar is being conducted by an, Educational Organisation, where the, participants will be educators of different, subjects. The number of participants in, Hindi, English and Mathematics are 60, 84, and 108 respectively., , (iii) (a) LCM of (60, 84, 108) = Product of the greatest, power of each prime factor involved in the numbers, with highest power, = 2 2 ´ 33 ´ 5 ´ 7 = 4 ´ 27 ´ 35 = 3780, (iv) (d) Now, HCF (60, 84, 108) ´ LCM (60, 84, 108), = 12 ´ 3780 = 45360, (v) (d) Number 108 can be expressed as a product of its, prime as 2 2 ´ 33 ., , 3. A, , Mathematics Exhibition is being, conducted in your School and one of your, friends is making a model of a factor tree. He, has some difficulty and ask for your help in, completing a quiz for the audience. Observe, the following factor tree and answer the, following :, , (i) In each room the same number of, participants are to be seated and all of, them being in the same subject, hence, maximum number participants that can, accommodated in each room are, (a) 14, , (b) 12, , (c) 16, , x, 5, y, , (d) 18, , (ii) What is the minimum number of rooms, required during the event?, (a) 11, , (b) 31, , (c) 41, , (b) 3680, , (d) 21, , (a) 15005, (c) 56920, , (c) 4780 (d) 4680, , (a) 23, , (b) 35360, (d) 45360, , (a) 2 ´ 3, , 2, , (c) 22 ´ 3 2, , 3, , (b) 2 ´ 3, , (b) 22, , (c) 11, , (d) 19, , (iii) What will be the value of z ?, (a) 22, , (i) (b) Given, number of students in each subject are, Hindi = 60, English = 84 and Mathematics = 108., The prime factors of each subject students are, , The maximum number of participants that can, accommodated in each room, = HCF (60, 84, 108), = Product of the smallest power of each common, prime factor involved in the numbers, = 2 2 ´ 3 = 12, (ii) (d) The minimum number of rooms required during, the event is, Total number of participants, =, Maximum participants in one room, 252, =, = 21rooms., 12, , (d) 19, , (v) The prime factorisation of 13915 is, (a) 5 ´ 113 ´ 13 2, , 60 = 2 ´ 2 ´ 3 ´ 5 = 2 2 ´ 3 ´ 5, 108 = 2 ´ 2 ´ 3 ´ 3 ´ 3 = 2 2 ´ 33, , (c) 17, , (a) Composite number, (b) Prime number, (c) Neither prime nor composite, (d) Even number, , (d) 22 ´ 3 3, , 84 = 2 ´ 2 ´ 3 ´ 7 = 2 2 ´ 3 ´ 7, , (b) 23, , (iv) According to Fundamental Theorem of, Arithmetic 13915 is a, , 3, , Sol., , CBSE QUESTION BANK, , (b) 13915, (d) 17429, , (ii) What will be the value of y ?, , (v) 108 can be expressed as a product of its, primes as, 3, , z, , (i) What will be the value of x ?, , (iv) The product of HCF and LCM of 60, 84, and 108 is, (a) 55360, (c) 45500, , 253, 11, , (iii) The LCM of 60, 84 and 108 is, (a) 3780, , 2783, , 2, , (c) 5 ´ 11 ´ 23, , (b) 5 ´ 113 ´ 23 2, (d) 5 ´ 112 ´ 13 2, , Sol., (i) (b) x = 5 ´ 2783 = 13915, (ii) (c) We have, 2783 = y ´ 253, 2783, y=, = 11, Þ, 253, (iii) (b) We have, 253 = 11 ´ z, 253, z=, = 23, Þ, 11, (iv) (a) Here, 13915 = 5 ´ 2783, Since, 13915 has factor other than 1 and the number, itself. It is a composite number., (v) (c) 13915 = 5 ´ 11 ´ 11 ´ 23 = 5 ´ 112 ´ 23

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31, , CBSE Sample Paper Mathematics Standard Class X (Term I), Sol., , Polynomials, 1. The below picture are few natural examples, of parabolic shape which is represented by a, quadratic polynomial. A parabolic arch is an, arch in the shape of a parabola., , (i) (c) In the standard form of quadratic polynomial, ax2 + bx + c; ‘a’ is a non-zero real number, and b, and c are any real number., (ii) (d) In a quadratic polynomial, if roots are equal, then, discriminant, D = 0., 1, (iii) (b) Given, a and are the zeroes of quadratic, a, polynomial 2 x2 - x + 8k., Now, product of zeroes,, 1, Constant term, a´ =, a Coefficient of x2, 8k, 2 1, Þk= =, 1=, Þ, 2, 8 4, 2, (iv) (c) Given equation is x + 1 = 0., x2 = - 1, y = x2 +1, , In structures, their curve represents an, efficient method of load, and so can be, found in bridges and in architecture in a, variety of forms., (i) In the standard form of quadratic, polynomial, ax 2 + bx + c , a, b and c are, (a) All are real numbers, (b) All are rational numbers., (c) a is a non-zero real number, b and c are, any real numbers., (d) All are integers., , (ii) If the roots of the quadratic polynomial, are equal, where the discriminant, D = b 2 - 4 ac, then, (a) D > 0, (c) D ³ 0, , (b) D < 0, (d) D = 0, , 1, are the zeroes of the quadratic, a, polynomial 2 x 2 - x + 8 k, then k is, , (iii) If a and, , (a) 4, , (b), , 1, 4, , (c), , -1, 4, , (0, 1), , Hence, it is neither touches nor intersects X-axis., (v) (c) Given, sum of roots = - p, 1, and product of roots = p, \ Required quadratic polynomial, = k [ x2 - (Sum of roots) x + Product of roots], é, 1ö, æ 1 öù, æ, = k ê x2 - (- p)x + ç - ÷ ú = k ç x2 + px - ÷, pø, è p øû, è, ë, , 2. An asana is a body posture, originally and, still a general term for a sitting meditation, pose, and later extended in hatha yoga and, modern yoga as exercise, to any type of pose, or position, adding reclining, standing,, inverted, twisting, and balancing poses., In the figure, one can observe that poses, can be related to representation of quadratic, polynomial., , (d) 2, , (iv) The graph of x 2 + 1 = 0, , (v) If the sum of the roots is -p and product, 1, of the roots is - , then the quadratic, p, polynomial is, æ, ö, x, (a) k çç - px 2 + + 1÷÷, p, è, ø, æ 2, 1ö, (c) k çç x + px - ÷÷, pø, è, , æ, ö, x, (b) k çç px 2 - - 1÷÷, p, è, ø, æ 2, 1ö, (d) k çç x - px + ÷÷, pø, è, , CHAKRASANA, , TRIKONASANA, , CBSE QUESTION BANK, , (a) Intersects X-axis at two distinct points., (b) Touches X-axis at a point., (c) Neither touches nor intersects X-axis., (d) Either touches or intersects X-axis.

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32, , CBSE Sample Paper Mathematics Standard Class X (Term I), (v) (b) Let p( x) = 4 3 x2 + 5 x - 2 3, , (i) The shape of the poses shown is, , = 4 3 x2 + (8 - 3)x - 2 3, [by splitting middle term], , (a) Spiral, (b) Ellipse, (c) Linear, (d) Parabola, , = 4 3 x2 + 8 x - 3 x - 2 3, , (ii) The, graph, of, parabola, downwards, if ………… ., (a) a ³ 0, (c) a < 0, , = 4 x( 3 x + 2 ) -, , opens, , = (4 x -, , (4 x -, , \, , (iii) In the graph, how many zeroes are there, for the polynomial?, , Þ, Þ, , (b) 1, , (c) 2, , 4x -, , 3 )( 3 x + 2 ) = 0, , 3 = 0 and 3 x + 2 = 0, 3, 2, and x = x=, 4, 3, , 3. Basketball and soccer are played with a, , 4, , 1, , (a) 0, , 3 )( 3 x + 2 ), , For finding the zeroes, put p( x) = 0, , (b) a = 0, (d) a > 0, , –2, , 3( 3 x + 2 ), , (d) 3, , (iv) The two zeroes in the below shown graph, are, , spherical ball. Even though an athlete, dribbles the ball in both sports, a basketball, player uses his hands and a soccer player, uses his feet. Usually, soccer is played, outdoors on a large field and basketball is, played indoor on a court made out of wood., The projectile (path traced) of soccer ball, and basketball are in the form of parabola, representing quadratic polynomial., , Y, , X¢, , –2 –1, , 1 2 3 4, , X, , Y¢, , (a) 2, 4, (c) - 8, 4, , (b) - 2, 4, (d) 2, - 8, , (v) The zeroes of the quadratic polynomial, 4 3 x 2 + 5 x - 2 3 are, 2, 3, ,, 3 4, 2, 3, (b) ,, 3 4, 2, 3, (c), ,4, 3, 2, 3, (d) ,4, 3, , CBSE QUESTION BANK, , (a), , Sol., , (i) The shape of the path traced shown is, , (i) (d) The shape of given poses are parabolic., (ii) (c) The graph of parabola opens downwards, if a < 0., (iii) (c) Number of zeroes is equal to number of times, intersects parabola on the X-axis., \ Number of zeroes = 2., (iv) (b) The curve intersect X-axis at points x = - 2, and x = 4., Hence, two zeroes in the given graph are - 2 and 4., , (a) Spiral, (b) Ellipse, (c) Linear, (d) Parabola, , (ii) The, graph, of, parabola, downwards, if ………… ., (a) a = 0, (c) a > 0, , (b) a < 0, (d) a ³ 0, , opens

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33, , CBSE Sample Paper Mathematics Standard Class X (Term I), , (iii) Observe the following graph and answer., , Pair of Linear Equations in, Two Variables, 1. A test consists of ‘True’ or ‘False’ questions., , 6, , One mark is awarded for every correct, 1, answer while mark is deducted for every, 4, wrong answer. A student knew answers to, some of the questions. Rest of the questions, he attempted by guessing. He answered 120, questions and got 90 marks., , 2, –4, , –3, , –2, , –1, , 1, , 2, , 3, , 4, , –2, , –6, , Type of, question, , In the above graph, how many zeroes are, there for the polynomial?, (a) 0, (c) 2, , (b) 1, (d) 3, , (iv) The three zeroes in the above shown, graph are, (a) 2 , 3 , - 1, (c) - 3 , - 1, 2, , (b) - 2, 3, 1, (d) -2, - 3 , - 1, , (v) What will be the expression of the, polynomial?, , True/False, , Marks given for Marks deducted, correct answer for wrong answer, 1, , 0.25, , (i) If answer to all questions he attempted by, guessing were wrong, then how many, questions did he answer correctly?, (ii) How many questions did he guess?, (iii) If answer to all questions he attempted by, guessing were wrong and answered 80, correctly, then how many marks he got?, , (c) x3 + 2x2 + 5 x - 6, , (iv) If answer to all questions he attempted by, guessing were wrong, then how many, questions answered correctly to score 95, marks?, , (d) x3 + 2x2 + 5 x + 6, , Sol. Let the number of questions whose answer is known, , (a) x3 + 2x2 - 5 x - 6, (b) x3 + 2x2 - 5 x + 6, , Sol., , = x3 - (Sum of zeroes) x2, + (Sum of product of two zeroes at a time) x, - (Product of three zeroes), = x3 - (-2 ) x2 + (-5)x - (6), = x3 + 2 x2 - 5 x - 6, , (i), (ii), (iii), , (iv), , CBSE QUESTION BANK, , (i) (d) The shape of the path traced shown in the given, figure is the form of parabola., (ii) (b) The graph of parabola opens downwards, if a < 0., (iii) (d) In the given graph, we see that curve intersect the, X-axis at three points. Hence, number of zeroes in the, given polynomial are 3., (iv) (c) The given curve intersect the X-axis at points, x = - 3, -1and 2., Hence, three zeroes in the given graph are -3, - 1, 2., (v) (a) Since, given polynomial has three zeroes., So, it will be a cubic polynomial., Now, sum of zeroes = - 3 - 1 + 2 = - 2, Sum of product of two zeroes at a time, = - 3 ´ (-1) + (-1) ´ 2 + 2 ´ (-3), = 3-2 - 6= - 5, and product of all zeroes = - 3 ´ - 1 ´ 2, =6, \ Required cubic polynomial, , to the student be x and questions attempted by, guessing be y., Then,, … (i), x + y = 120, 1, and, … (ii), x - y = 90 Þ 4 x - y = 360, 4, On adding Eqs. (i) and (ii), we get, 480, = 96, 5 x = 480 Þ x =, 5, Put x = 96 in Eq. (i), we get, 96 + y = 120 Þ y = 120 - 96 = 24, He answered 96 questions correctly., He guesses only 24 questions., In out of 120 questions attempted 80 answered are, correct and 40 guessing answered are wrong., 1, Then, he got the marks = 80 - of 40, 4, 1, = 80 - ´ 40 = 80 - 10 = 70, 4, According to the given condition,, 1, 1, x - of (120 - x) = 95 Þ x - ´ (120 - x) = 95, 4, 4, Þ, 4 x - 120 + x = 380, Þ, 5 x = 500, Þ, x = 100, Hence, he answered correctly 100 questions to score, 95 marks.

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34, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 2. Amit is planning to buy a house and the, layout is given below. The design and the, measurement has been made such that, areas of two bedrooms and kitchen together, is 95 sq m., , 5m, , x, , 2, , y, , Bedroom 1, , Bath, room, , Kitchen, , 2m, Living room, 5m, , Bedroom 2, 15 m, , But, it is also given, the cost of laying tiles in, kitchen at the rate of `50 per m 2 ., \Total cost of laying tiles in the kitchen = 35 ´ 50, = ` 1750, , 3. It is common that Governments revise travel, fares from time to time based on various, factors such as inflation ( a general increase, in prices and fall in the purchasing value of, money) on different types of vehicles like, auto, Rickshaws, taxis, Radio cab etc. The, auto charges in a city comprise of a fixed, charge together with the charge for the, distance covered. Study the following, situations, , Based on the above information, answer the, following questions:, (i) Form the pair of linear equations in two, variables from this situation., (ii) Find the length of the outer boundary of, the layout., (iii) Find the area of each bedroom and, kitchen in the layout., , Name of, the city, , Distance travelled, (km), , Amount, paid (in `), , (iv) Find the area of living room in the layout., , City A, , 10, , 75, , 15, , 110, , 8, , 91, , 14, , 145, , (v) Find the cost of laying tiles in kitchen at, the rate of ` 50 per sq m., , City B, , CBSE QUESTION BANK, , Sol., (i) From the given figure we see that area of two, bedrooms = 2(5 x) = 10x m 2, Area of kitchen = 5 ´ y = 5 y m2, According to the question,, Area of the two bedrooms and area of kitchen, = 95, \, 10 x + 5 y = 95, Þ, 2 x + y = 19 [divide both sides by 5] … (i), Also, length of the home = 15 cm, … (ii), \, x + 2 + y = 15 Þ x + y = 13, Hence, pair of linear equations is, 2 x + y = 19 and x + y = 13, (ii) The length of the outer boundary of the layout, = 2(l + b ) = 2(15 + 12 ), = 2(27 ) = 54 m, (iii) On solving Eqs. (i) and (ii), we get, x = 6 and y = 7, \Area of each bedroom = 5 ´ x = 5 ´ 6 = 30 m2, and area of kitchen = 5 ´ y = 5 ´ 7 = 35 m 2, (iv) Area of living room = 15 ´ (5 + 2 ) - Area of bedroom 2, = 15 ´ 7 - 5 ´ 6, = 105 - 30 = 75 m 2, (v) Since, area of kitchen = 5 ´ y, = 5 ´ 7 = 35 m2, , Situation 1 In city A, for a journey of 10 km,, the charge paid is ` 75 and for a journey of 15, km, the charge paid is ` 110., Situation 2 In a city B, for a journey of 8 km,, the charge paid is ` 91 and for a journey of 14, km, the charge paid is ` 145., Refer situation 1, (i) If the fixed charges of auto rickshaw be, `x and the running charges be ` y km/h,, the pair of linear equations representing, the situation is, (a) x + 10y = 110, x + 15 y = 75, (b) x + 10y = 75 , x + 15 y = 110, (c) 10x + y = 110, 15 x + y = 75, (d) 10x + y = 75, 15 x + y = 110, , (ii) A person travels a distance of 50 km., The amount he has to pay is, (a) ` 155, (b) ` 255, (c) ` 355, (d) ` 455

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35, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Refer situation 2, (iii) What will a person have to pay for, travelling a distance of 30km ?, (a) ` 185, (c) ` 275, , (b) ` 289, (d) ` 305, , (iv) The graph of lines representing the, conditions are: (situation 2), Y, 25, , (20, 25), , 20, 15, , (a), , 10, (30, 5), , 5 (0, 5), X¢, , 0, –5, , 5 10 15 20 25 30 35, , X, , –10, Y¢, Y, 25, 20, 15, , (b), , 10 (0, 10), 5, X¢, , (20, 10), (12.5, 0), , 0, –5, , 5 10 15 20 25 30 35, (5, –10), , –10, Y¢, , X, , 1. Vijay is trying to find the average height of a, tower near his house. He is using the, properties of similar triangles.The height of, Vijay’s house, if 20 m when Vijay’s house, casts a shadow 10m long on the ground., , 20, 15, , (11, 10) (19, 9), , 10, 5, X¢, , - 5 y = - 35 Þ y = 7, \, x + 10 ´ 7 = 75, Þ, x = 75 - 70 = 5, To travel a distance of 50 km, a person has to pay, amount = x + 50 y = 5 + 50 ´ 7, = 5 + 350 = ` 355, (iii) (b) As per the situation 2, the pair of linear is, … (i), x + 8 y = 91, and, … (ii), x + 14 y = 145, On solving Eqs. (i) and (ii), we get, x + 8 y = 91, x + 14 y = 145, - - 6 y = - 54, Þ, y=9, Put y = 9 in Eq. (i), we get, x + 8 ´ 9 = 91, Þ, x = 91 - 72 = 19, To travel a distance of 30 km, a person has to pay, amount = x + 30 ´ y, = 19 + 30 ´ 9 = 19 + 270 = ` 289, (iv) (c) In situation 2, the intersection point of two lines is, (19, 9), which is shown in figure (c)., , Similar Triangles, , (25, –10), , Y, 25, , (c), , (ii) (c) On solving the above equations, we get, x + 10 y = 75, x + 15 y = 110, - -, , (5, 10), , 0, –5, , 5 10 15 20 25 30 35, , X, , At the same time, the tower casts a shadow, 50 m long on the ground and the house of, Ajay casts 20 m shadow on the ground., , Y¢, Y, 25, 15, , (d), , 10, , (0, 10), , (15, 15), , (35, 10), Vijay’s, house, , 5, X¢, , 0, –5, , 5 10 15 20 25 30 35, (15, 5), , X, , Y¢, , Sol., (i) (b) As per the situation 1, the pair of linear equation, representing the situation is, x + 10 y = 75, and, x + 15 y = 110, , Tower, , Ajay’s, house, , (i) The height of the tower is, (a) 20 m, (c) 100 m, , (b) 50 m, (d) 200 m, , (ii) What will be the length of the shadow of, the tower when Vijay’s house casts a, shadow of 12 m?, (a) 75 m, (c) 45 m, , (b) 50 m, (d) 60 m, , CBSE QUESTION BANK, , 20

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36, , CBSE Sample Paper Mathematics Standard Class X (Term I), , (iii) What is the height of Ajay’s house?, (a) 30 m, (c) 50 m, , 2. Rohan wants to measure the distance of a, , (b) 40 m, (d) 20 m, , pond during the visit to his native. He marks, points A and B on the opposite edges of a, pond as shown in the figure below. To find, the distance between the points, he makes a, right-angled triangle using rope connecting, Bwith another point C are a distance of 12 m,, connecting C to point D at a distance of 40 m, from point C and the connecting D to the, point A which at distance of 30 m from D, such the Ð ADC = 90°., , (iv) When the tower casts a shadow of 40 m,, same time what will be the length of the, shadow of Ajay’s house?, (a) 16 m, (c) 20 m, , (b) 32 m, (d) 8 m, , (v) When the tower casts a shadow of 40 m,, same time what will be the length of the, shadow of Vijay’s house?, (a) 15 m, (c) 16 m, , (b) 32 m, (d) 8 m, , 12 m, , B, , A, , C, , Sol., 30 m, , (i) (c) Let CD = h m be the height of the tower. Let, BE = 20 m be the height of Vijay’s house and GF be, the height of Ajay’s house., , 40 m, D, , D, , E, , G, , A 10 m B, 50 m, , CBSE QUESTION BANK, , (iii), , (iv), , (v), , (a) Similarity of triangles, (b) Thales Theorem, (c) Pythagoras Theorem, (d) Area of similar triangles, , h, , 20 m, , (ii), , (i) Which property of geometry will be used, to find the distance AC ?, , C, , F 20 m H, 50 m, , DACD ~ ABE, AC CD, \, =, AB EB, h, 50, =, Þ h = 100 m, Þ, 10 20, (d) Given AB = 12 m, let AC = h, In similar DABE and DACD,, AB BE 12, 20, =, Þ, =, AC CD, h 100, 12 ´ 100, h=, Þ, = 12 ´ 5 = 60 m, 20, (b) Let height of Ajay’s house be GF = h1, Since, DHFG ~ DHCD, HF FG, =, \, HC CD, h, 20, = 1, Þ, 50 100, 20 ´ 100, h1 =, = 40 m, Þ, 50, (a) Given, HC = 40 cm, Let length of the shadow of Ajay’s hour be HF = l m, Since, DHFG ~ DHCD, HF FG, \, =, HC CD, 40 ´ 40, l, 40, = 16 m, =, Þl=, Þ, 100, 40 100, (d) Given, AC = 40 cm, Let length of the shadow of Vijay’s house be AB = l m, Since,, DABE ~ ACD, AB EB, \, =, AC CD, 20 ´ 40, l, 20, =, Þ h=, = 8m, Þ, 100, 40 100, , (ii) What is the distance AC ?, (a) 50 m, (c) 100 m, , (b) 12 m, (d) 70 m, , (iii) Which is the following does not form a, Pythagoras triplet?, (a) (7,24,25), (c) (5,12,13), , (b) (15,8,17), (d) (21,20,28), , (iv) Find the length AB?, (a) 12 m, (c) 50 m, , (b) 38 m, (d) 100 m, , (v) Find the length of the rope used., (a) 120 m, (c) 82 m, , (b) 70 m, (d) 22 m, , Sol., (i) (c) To find the distance AC in the given figure, we, use Pythagoras theorem., (ii) (a) In right DADC, use Pythagoras theorem,, AC = ( AD)2 + (CD)2 = (30)2 + (40)2, = 900 + 1600 = 2500 = 50 m, (iii) (d) (a) Now, 242 + 7 2 = 576 + 49 = 625 = (25)2 ,, which forms a Pythagoras triplet, (b) (15)2 + (8)2 = 225 + 64 = 289 = (17 )2 ,, which forms a Pythagoras triplet, (c) (12 )2 + (5)2 = 144 + 25 = 169 = (13)2 ,, which form a Pythagoras triplet., (d) (20)2 + (21)2 = 400 + 441 = 881¹ (28)2 ,, which does not form a Pythagoras triplet., (iv) (b) Since, AC = 50 m, \, AB = AC - BC = 50 - 12 = 38 m, (v) (c) The length of the rope used = BC + CD + DA, = 12 + 40 + 30 = 82 m

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37, , CBSE Sample Paper Mathematics Standard Class X (Term I), , (v) The length of AB in the given figure, , 3. A scale drawing of an object is the same, shape at the object but a different size. The, scale of a drawing is a comparison of the, length used on a drawing to the length it, represents. The scale is written as a ratio., The ratio of two corresponding sides in, similar figures is called the scale factor., Scale factor, Length in range, =, Corresponding length in object, If one shape can become another using, revising, then the shapes are similar. Hence,, two shapes are similar when one can, become the other after a resize, flip, slide or, turn. In the photograph below showing, the side view of a train engine. Scale factor is, 1 : 200., , This means that a length of 1 cm on the, photograph above corresponds to a length, of 200 cm or 2 m, of the actual engine. The, scale can also be written as the ratio of two, lengths., (i) If the length of the model is 11cm, then, the overall length of the engine in the, photograph above, including the, couplings(mechanism used to connect) is, (a) 22 cm, (c) 220 m, , (b) 220 cm, (d) 22 m, , (ii) What will affect the similarity of any two, polygons?, (a) They are flipped horizontally, (b) They are dilated by a scale factor, (c) They are translated down, (d) They are not the mirror image of one, another., , (a) 0.7 m, (c) 0.07 cm, , (b) 0.7 cm, (d) 0.07 m, , (iv) If two similar triangles have a scale factor, 5 : 3 which statement regarding the two, triangles is true?, (a) The ratio of their perimeters is 15 : 1, (b) Their altitudes have a ratio 25 :15, (c) Their medians have a ratio 10 : 4, (d) Their angle bisectors have a ratio 11 : 5, , x cm, B, , C, , 3 cm, , 4 cm, D, , E, , 6 cm, , (a) 8 cm, (c) 4 cm, , (b) 6 cm, (d) 10 cm, , Sol. Given, scale factors = 1 : 200, It means that length of 1 cm on the photograph, above corresponds to a length of 200 cm (or 2 m) of, the actual engine., (i) (d) Since, length of the model is 11 cm., Therefore, the overall length of the engine, = 11 ´ 200, = 2200 cm, = 22 m, (ii) (d) The similarity of any two polygons will affect that, they are not the mirror image of one another., (iii) (a) The actual width of the door = 0.35 ´ 200 cm, = 70 cm, = 07, . m, (iv) (b) If two similar triangles have a scale factor 5 : 3,, then their altitudes have a ratio 25 : 15., (v) (c) In the given BC || DE., \, Þ, Þ, Þ, Þ, Þ, , DABC ~ ADE,, AB BC, =, AD DE, x, 3, =, x+ 4 6, x, 1, =, x+ 4 2, 2x = x + 4, x = 4 cm, , Coordinate Geometry, 1. In order to conduct Sports Day activities in, your School, lines have been drawn with, chalk powder at a distance of 1 m each, in a, rectangular shaped ground ABCD, 100, flowerpots have been placed at a distance of, 1 m from each other along AD, as shown in, given figure below. Niharika runs 1/4 th the, distance AD on the 2nd line and posts a, 1, green flag. Preet runs th distance AD on, 5, the eighth line and posts a red flag., , CBSE QUESTION BANK, , (iii) What is the actual width of the door, if the, width of the door in photograph is 0.35, cm?, , A

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38, , CBSE Sample Paper Mathematics Standard Class X (Term I), , D, , C, , (iii) (c) Distance between these flags = GR, = (8 - 2 )2 + (25 - 20)2, = (6)2 + (5)2, = 36 + 25 = 61 m, (iv) (a) The point at which Rashmi should post her blue, flag is the mid-point of the line joining these points., Let this point be A( x, y)., 2+ 8, 25 + 20, , y=, x=, 2, 2, 10, 45, x=, = 5, y =, = 22.5, 2, 2, Hence, A( x, y) = (5, 22.5), (v) (a) Let the point at which Joy post his flag be B( x, y)., , G, R, , 1:3, , 2, 1, , G(2, 25), B, , A, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , (i) Find the position of green flag, (a) (2, 25), (c) (25, 2), , (b) (2, 0.25), (d) (0, - 25), , B, , R(8, 20), , 1´ 8 + 3 ´2, 1 ´ 20 + 3 ´ 25, , y=, 1+ 3, 1+ 3, 14, 95, x=, = 3.5, y =, = 23.75 » 24, 4, 4, Hence, B( x, y) = (3.5, 24), , Then, x =, , (ii) Find the position of red flag, (a) (8, 0), (c) (8, 20), , (b) (20, 8), (d) (8, 0.2), , (iii) What is the distance between both the, flags?, (a) 41 m, (c) 61 m, , (b) 11 m, (d) 51 m, , (iv) If Rashmi has to post a blue flag exactly, halfway between the line segment joining, the two flags, where should she post her, flag?, (a) (5, 22.5), (c) (2, 8.5), , (b) (10, 22), (d) (2.5, 20), , (v) If Joy has to post a flag at one-fourth, distance from green flag ,in the line, segment joining the green and red flags,, then where should he post his flag?, (a) (3.5, 24), (c) (2.25, 8.5), , (b) (0.5, 12.5), (d) (25, 20), , CBSE QUESTION BANK, , Sol., (i) (a) It can be observed that Niharika posted, 1, the green flag at th distance of AD, 4, 1, i.e. ´ 100 = 25 m from the starting point, 4, of 2nd line., Therefore, the coordinates of this point G is, (2, 25)., 1, (ii) (c) Preet posted red flag at th distance of AD,, 5, 1, i.e. ´ 100 = 20 m from the starting point of 8th line., 5, Therefore the coordinates of this point R is, (8, 20)., , Areas Related to Circles, 1. Pookalam is the flower bed or flower, pattern designed during Onam in Kerala., It is similar as Rangoli in North India, and Kolam in Tamil Nadu. During the, festival of Onam , your school is planning to, conduct a Pookalam competition. Your, friend who is a partner in competition,, suggests two designs given below. Observe, these carefully., A, , B, , A, , B, , D, , C, , C, , Design I This design is made with a circle of, radius 32 cm leaving equilateral triangle, ABC in the middle as shown in the given, figure., Design II This Pookalam is made with, 9 circular design each of radius 7 cm., Refer Design I, (i) The side of equilateral triangle is, (a) 12 3 cm, (c) 48 cm, , (b) 32 3 cm, (d) 64 cm, , (ii) The altitude of the equilateral triangle is, (a) 8 cm, (c) 48 cm, , (b) 12 cm, (d) 52 cm

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39, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Refer Design II, (iii) The area of square is, (a) 1264 cm 2, (c) 1830 cm 2, , (b) 1764 cm 2, (d) 1944 cm 2, , (iv) Area of each circular design is, (a) 124 cm 2, (c) 144 cm 2, , (b) 132 cm 2, (d) 154 cm 2, , (v) Area of the remaining portion of the, square ABCD is, (a) 378 cm 2, (c) 340 cm 2, , (b) 260 cm 2, (d) 278 cm 2, , Sol., , Directions (i-ii), Given, radius of the circle = 32 cm, Let the side of the equilateral DABC be a cm., Let h be the height of the triangle., , Now, area of one circle = pr 2 = p (7 )2, 22, =, ´ (7 )2 = 154 cm 2, 7, \Area of nine circles = 9 ´ 154 = 1386 cm 2, Area of square ABCD = (Side)2 = (42 )2 = 1764 cm2, Hence, area of the remaining portion of the handkerchief, = Area of square - Area of nine circles, = 1764 - 1386 = 378 cm 2, (iii) (b) 1764 cm 2, (iv) (d) 154 cm 2, (v) (a) 378 cm 2, , 2. A brooch is a small piece of jewellery which, has a pin at the back so it can be fastened on a, dress, blouse or coat. Designs of some, brooch are shown below. Observe them, carefully., , A, , a, , a, , O, h, , A, B, , a/2, , D, , a/2, , C, , We know that in an equilateral triangle, centroid and, circumcentre coincide., 2, \, AO = h cm, 3, [Q centroid divides the median in the ratio 2 : 1], which is equal to the radius of circle., 2, …(i), \, h = 32 Þ h = 48 cm, 3, Now, we draw a perpendicular from vertex A to, side BC which bisects BC at D., In right angled DADB,, AB2 = BD2 + AD2, , [by Pythagoras theorem], , 2, , Þ, Þ, , Þ, , a2, 3a2, æaö, a2 = ç ÷ + h2 Þ h2 = a2 =, 4, 4, è2 ø, 2, a, 3, [from Eq. (i)], (48)2 =, 4, a2 = 3072, a = 3072, = 32 3 cm, (i) (b) 32 3 cm, (ii) (c) 48 cm, , [taking positive square root], , Directions (iii-v), Given, radius of each circle, r = 7 cm, [Q diameter = 2 ´ radius], \Diameter of circle, d = 14 cm, In the given figure, horizontal three circles touch each other., \Length of a side of square = 3 ´ Diameter of one circle, = 3 ´ 14 = 42 cm, , Design A Brooch A is made with silver wire, in the form of a circle with diameter 28 mm., The wire used for making 4 diameters which, divide the circle into 8 equal parts., Design B Brooch b is made two colours_, Gold and silver. Outer part is made with, Gold. The circumference of silver part is, 44 mm and the gold part is 3mm wide, everywhere., Refer to Design A, (i) The total length of silver wire required is, (a) 180 mm, (c) 250 mm, , (b) 200 mm, (d) 280 mm, , (ii) The area of each sector of the brooch is, (a) 44 mm 2, (c) 77 mm 2, , (b) 52 mm 2, (d) 68 mm 2, , Refer to Design B, (iii) The circumference of outer part (golden) is, (a) 48.49 mm, (c) 72.50 mm, , (b) 82.2 mm, (d) 62.86 mm, , CBSE QUESTION BANK, , Þ, , B

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40, , CBSE Sample Paper Mathematics Standard Class X (Term I), , (iv) The difference of areas of golden and, silver parts is, (a) 18p mm 2, (c) 51p mm 2, , (b) 44p mm 2, (d) 64p mm 2, , (v) A boy is playing with brooch B. He makes, revolution with it along its edge. How, many complete revolutions must it take, to cover 80 p mm ?, (a) 2, (c) 4, , Probability, 1. On a weekend Rani was playing cards with, her family. The deck has 52 cards.If her, brother drew one card., , (b) 3, (d) 5, , Sol., (i) (b) Given, diameter of circle, d = 28 mm, [Qd = 2 r], \Circumference of circle = pd, 22, =, ´ 28 = 88 mm, 7, Now, length of 4 diameters = 4 ´ 28 = 112 mm, Total length of the silver wire = pd + 4 d, = 88 + 112 = 200 mm, (ii) (c) Here, we see that total circle is divided into, 8 parts., 1, \ Area of each sector = ´ Area of circle, 8, 1, = ´ pr 2, 8, 1 22, = ´, ´ 14 ´ 14, 8 7, = 77 mm 2, , (i) Find the probability of getting a king of, red colour., (a), , (b), , 1, 13, , (c), , 1, 52, , (d), , 1, 4, , (ii) Find the probability of getting a face card., (a), , 1, 26, , (b), , 1, 13, , (c), , 2, 13, , (d), , 3, 13, , (iii) Find the probability of getting a jack of, hearts., (a), , 1, 26, , (b), , 1, 52, , (c), , 3, 52, , (d), , 3, 26, , (iv) Find the probability of getting a red face, card., (a), , Directions (iii-iv), , 1, 26, , 3, 13, , (b), , 1, 13, , (c), , 1, 52, , (d), , 1, 4, , (v) Find the probability of getting a spade., (a), r, 3m, , CBSE QUESTION BANK, , R, , (b), , 1, 13, , (c), , 1, 52, , (d), , 1, 4, , Sol. Total number of cards in one deck of cards is 52., Silver, , m, , 1, 26, , Gold, , We have, circumference of silver part = 44 mm, \, 2 pr = 44, 44, r=, = 7 mm, Þ, 22, 2´, 7, \, R = r + 3= 7 + 3, = 10 mm, (iii) (d) Circumference of golden part = 2pR, 22, =2 ´, ´ 10, 7, = 62.86 mm, (iv) (c) Difference of areas = pR 2 - pr 2 = p(R 2 - r 2 ), = (102 - 7 2 ) p = 51p mm 2, (v) (c) Required number of revolutions, Distance covered, =, Circumference, 80p, 80p, =, =, =4, 2 pR 2 p ´ 10, , \ Total number of outcomes = 52, (i) (a) Let E1 = Event of getting a king of red colour, \ Number of outcomes favourable to E1 = 2, [Q there are four kings in a deck of playing, cards out of which two are, red and two are black], Hence, probability of getting a king of red colour,, 2, 1, =, P(E1 ) =, 52 26, (ii) (d) Let E2 = Event of getting a face card, \ Number of outcomes favourable to E2 = 12, [Q in a deck of cards, there are, 12 face cards, namely 4 kings,, 4 jacks, 4 queens], Hence, probability of getting a face card,, 12, 3, =, P(E2 ) =, 52 13, (iii) (b) Let E3 = Event of getting a jack of heart, \ Number of outcomes favourable to E3 = 1, [Q there are four jack cards in a deck,, namely 1 of heart, 1 of club,, 1 of spade and 1 of diamond], Hence, probability of getting a jack of heart,

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41, , CBSE Sample Paper Mathematics Standard Class X (Term I), 1, 52, (iv) (*) Let E4 = Event of getting a red face card., \Number of outcomes favourable to E4 = 6, [Q in a deck of cards, there are 12 face cards, out of which 6 are red cards], Hence, probability of getting a red face card,, 6, 3, =, P(E4 ) =, 52 26, (v) (d) Let E5 = Event of getting a spade, \ Number of outcomes favourable to E5 = 13, [Q in a deck of cards, there are 13 spades,, 13 clubs, 13 hearts and 13 diamonds], Hence, probability of getting a spade,, 13 1, =, P(E5 ) =, 52 4, , the top face of the dice is less than or, equal to 12 ?, , P(E3 ) =, , 2. Rahul and Ravi planned to play Business, (board game) in which they were supposed, to use two dice., , (a) 1, , (b), , 5, 36, , (c), , 1, 18, , (d) 0, , (iv) Rahul got next chance. What is the, probability that he got the sum of the two, numbers appearing on the top face of the, dice is equal to 7 ?, (a), , 5, 9, , (b), , 5, 36, , (c), , 1, 6, , (d) 0, , (v) Now it was Ravi’s turn. He rolled the, dice. What is the probability that he got, the sum of the two numbers appearing on, the top face of the dice is greater than 8 ?, (a) 1, , (b), , 5, 36, , (c), , 1, 18, , (d), , 5, 18, , Sol. Total number of out comes = 6 ´ 6 = 36, (i) (b) Let E1 = Event of getting sum 8, \Number of favourable outcomes to E1 = 5, i.e. (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), 5, P(E1 ) =, \, 36, (ii) (d) Let E2 = Event of getting sum 13., Since, we can’t get sum more than 12., \, , P(E2 ) =, , 36, =0, 0, , (iii) (a) Let E3 = Event of getting sum less than or equal, to 12., \Number of favourable outcomes = 36, , (i) Ravi got first chance to roll the dice. What, is the probability that he got the sum of, the two numbers appearing on the top, face of the dice is 8?, (a), , 1, 26, , (b), , 5, 36, , (c), , 1, 18, , (d) 0, , (ii) Rahul got next chance. What is the, probability that he got the sum of the two, numbers appearing on the top face of the, dice is 13?, 1, (c), 18, , (d) 0, , (iii) Now it was Ravi’s turn. He rolled the, dice. What is the probability that he got, the sum of the two numbers appearing on, , \ P(E3 ) =, , 36, =1, 36, , (iv) (c) Let E4 = Event of getting sum 7., \Number of favourable outcomes to E2 = 6, i.e. (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) (6, 1), 6, 1, \, =, P(E4 ) =, 36 6, (v) (d) Let E5 = Event of getting sum greater than 8 i.e., getting sum equal to 8, 9, 10, 11, 12, \Favourable outcomes = 10 i.e. (3, 6), (4, 5), (5, 4),, (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)., 10, 5, =, P(E5 ) =, \, 36 18, , CBSE QUESTION BANK, , (a) 1, , 5, (b), 36, , As sum of all the out comes is less than or equal to 12.

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42, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Latest CBSE, , SAMPLE PAPER, Latest Sample Question Paper for Class XII (Term I), Issued by CBSE on 2 Sept 2021, , Mathematics Class 10 (Term I), Instructions, 1. The question paper contains three parts A, B and C., 2. Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., 3. Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., 4. Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions., 5. There is no negative marking., Maximum Marks : 40, , Time : 90 Minutes, , Section A, Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 1. The ratio of LCM and HCF of the least composite and the least prime numbers is, (a) 1 : 2, , (b) 2 : 1, , (c) 1 : 1, , (d) 1 : 3, , 2. The value of k for which the lines 5x + 7 y = 3 and 15x + 21y = k coincide is, (a) 9, , (b) 5, , (c) 7, , (d) 18, , Latest CBSE SAMPLE PAPER, , 3. A girl walks 200m towards East and then 150m towards North. The distance of the girl, from the starting point is, (a) 350 m, (b) 250 m, , (c) 300 m, , (d) 225 m, , 4. The lengths of the diagonals of a rhombus are 24cm and 32cm, then the length of the, altitude of the rhombus is, (a) 12 cm, (b) 12.8 cm, , (c) 19 cm, , (d) 19.2 cm, , 5. Two fair coins are tossed. What is the probability of getting at the most one head?, (a), , 3, 4, , (b), , 1, 4, , (c), , 1, 2, , (d) 3/8, , 6. DABC ~ DPQR. If AM and PN are altitudes of DABC and DPQR respectively and AB 2 :, PQ 2 = 4 : 9, then AM : PN =, (a) 16 : 81, (b) 4 : 9, , (c) 3 : 2, , (d) 2 : 3, , (c) 45°, , (d) 30°, , 7. If 2 sin 2 b - cos 2 b = 2, then b is, (a) 0°, , (b) 90°

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43, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 8. Prime factors of the denominator of a rational number with the decimal expansion, 44.123 are, (a) 2, 3, , (b) 2, 3, 5, , (c) 2, 5, , (d) 3, 5, , 9. The lines x = a and y = b are, (a) intersecting, (c) overlapping, , (b) parallel, (d) None of these, , 10. The distance of point A(- 5, 6) from the origin is, (a) 11 units, (c) 11 units, , (b) 61 units, (d) 61 units, , 11. If a 2 = 23 / 25, then a is, (a) rational, (c) whole number, , (b) irrational, (d) integer, , 12. If LCM (x , 18) = 36 and HCF (x , 18) = 2, then x is, (a) 2, , (b) 3, , (c) 4, , (d) 5, , 13. In DABC right angled at B, if, tan A = 3, then cos A cos C - sin A sin C is equal to, (b) 0, (c) 1, , (a) - 1, , (d) 3 / 2, , 14. If the angles of DABC are in ratio 1 : 1 : 2, respectively (the largest angle being angle C),, then the value of, (a) 0, , sec A, tan A, is, cosec B cot B, (b) 1/2, , (c) 1, , (d) 3 / 2, , 15. The number of revolutions made by a circular wheel of radius 0.7 m in rolling a, distance of 176 m is, (a) 22, (b) 24, , (c) 75, , (d) 40, , 16. DABC is such that AB = 3 cm, BC = 2 cm, CA = 2.5 cm. If DABC ~ DDEF and EF = 4 cm, then, perimeter of DDEF is, (a) 7.5 cm, (b) 15 cm, , (c) 22.5 cm, , (d) 30 cm, , A, , D, , E, , B, , (a) 7 cm, , 18. If 4 tan b = 3, then, (a) 0, , C, , (b) 6 cm, , 4 sin b - 3 cos b, , (c) 4 cm, , (d) 3 cm, , (c) 2/3, , (d), , =, , 4 sin b + 3 cos b, (b) 1/3, , 3, 4, , Latest CBSE SAMPLE PAPER, , 17. In the figure, if DE|| BC, AD = 3 cm, BD = 4 cm and BC = 14 cm, then DE equals

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44, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 19. One equation of a pair of dependent linear equations is - 5x + 7 y = 2., The second equation can be, (a) 10x + 14y + 4 = 0, (c) - 10x + 14y + 4 = 0, , (b) - 10x - 14y + 4 = 0, (d) 10x - 14y = - 4, , 20. A letter of English alphabets is chosen at random. What is the probability that it is a, letter of the word ‘MATHEMATICS’?, (a) 4/13, (b) 9/26, , (c) 5/13, , (d) 11/26, , Section B, Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 21. If sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of, such numbers are, (a) 2, , (b) 3, , (c) 4, , (d) 5, , 22. Given below is the graph representing two linear equations by lines AB and CD, respectively. What is the area of the triangle formed by these two lines and the, line x = 0?, Y, A 4, 3, C, , 2, 1, X¢, , –4 –3, , –2 –1 0, , 1, , 2, , 3, , B, 4, , X, , –1, D –2, –3, –4, Y¢, , Latest CBSE SAMPLE PAPER, , (a) 3 sq units, , (b) 4 sq units, , (c) 6 sq units, , (d) 8 sq units, , 23. If tan a + cot a = 2, then tan 20 a + cot 20 a is equal to, (a) 0, , (b) 2, , (c) 20, , (d) 2 20, , 24. If 217 x + 131y = 913, 131x + 217 y = 827, then x + y is, (a) 5, , (b) 6, , (c) 7, , (d) 8, , 25. The LCM of two prime numbers p and q ( p > q) is 221. Find the value of 3p - q., (a) 4, , (b) 28, , (c) 38, , (d) 48, , 26. A card is drawn from a well shuffled deck of cards. What is the probability that the, card drawn is neither a king nor a queen?, (a) 11/13, (b) 12/13, , (c) 11/26, , (d) 11/52, , 27. Two fair dice are rolled simultaneously. The probability that 5 will come up at least, once is, (a) 5/36, , (b) 11/36, , (c) 12/36, , (d) 23/36

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45, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 28. If 1 + sin 2 a = 3 sin a cos a, then values of cot a are, (a) - 1, 1, , (b) 0, 1, , (c) 1, 2, , (d) - 1, - 1, , 29. The vertices of a parallelogram in order are A(1, 2), B (4, y), C(x , 6) and D(3, 5). Then,, ( x , y) is, (a) (6, 3), , (b) (3, 6), , (c) (5, 6), , (d) (1, 4), , 30. In the given figure, ÐACB = ÐCDA, AC = 8 cm and AD = 3 cm, then BD is, C, , 8 cm, , A 3 cm D, , (a) 22/3 cm, , B, , (b) 26/3 cm, , (c) 55/3 cm, , (d) 64/3 cm, , 31. The equation of the perpendicular bisectors of line segment joining points A(4 , 5) and, B( - 2, 3) is, (a) 2 x - y + 7 = 0, (c) 3x - y - 7 = 0, , (b) 3x + 2 y - 7 = 0, (d) 3x + y - 7 = 0, , 32. In the given figure, D is the mid-point of BC, then the value of, , cot y, cot x, , is, , A, x, y, , C, , (a) 2, , D, , (b) 1/2, , B, , (c) 1/3, , (d) 1/4, , 33. The smallest number by which 1/13 should be multiplied, so that its decimal expansion, (c) 10/13, , (d) 100/13, , 34. Sides AB and BE of a right triangle, right angled at B are of lengths 16 cm and 8 cm, respectively. The length of the side of largest square FDGB that can be inscribed in the, DABE is, A, , 16 cm, D, , F, , B, , (a) 32/3 cm, , (b) 16/3 cm, , G, 8 cm, , E, , (c) 8/3 cm, , (d) 4/3 cm, , Latest CBSE SAMPLE PAPER, , terminates after two decimal places is, (a) 13/100, (b) 13/10

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46, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 35. Point P divides the line segment joining R(- 1, 3) and S(9, 8) in ratio k :1. If P lies on the, line x - y + 2 = 0, then value of k is, (b) 1/2, , (a) 2/3, , (c) 1/3, , (d) 1/4, , 36. In the figure given below, ABCD is a square of side 14 cm with E , F , G and H as the, mid-points of sides AB, BC, CD and DA respectively. The area of the shaded portion is, A, , E, , H, , B, , F, , D, , G, , (a) 44 cm 2, (c) 98 cm 2, , C, , (b) 49 cm 2, (d) 49p / 2 cm 2, , 37. Given below is the picture of the Olympic rings made by taking five congruent circles, of radius 1cm each, intersecting in such a way that the chord formed by joining the, point of intersection of two circles is also of length 1cm. Total area of all the dotted, regions assuming the thickness of the rings to be negligible is, , B, , A, , Latest CBSE SAMPLE PAPER, , (a) 4( p / 12 - 3 / 4) cm 2, (b) ( p / 6 - 3 / 4) cm 2, (c) 4( p / 6 - 3 / 4) cm 2, (d) 8 ( p / 6 - 3 / 4) cm 2, , 1, 2, (a) p = r = 2, (b) p = r = - 2, (c) p = 2 and r = - 2, (d) p = - 2 and r = 2, , 38. If 2 and are the zeroes of px 2 + 5x + r, then, , 39. The circumference of a circle is 100 cm. The side of a square inscribed in the circle is, (a) 50 2 cm, (b) 100 / p cm, (c) 50 2 / p cm, (d) 100 2 / p cm, , 40. The number of solutions of 3 x + y = 243 and 243 x - y = 3 is, (a) 0, (c) 2, , (b) 1, (d) infinite

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47, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Section C, Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted., 41-45 are based on Case Study-1., , CASE STUDY 1, The figure given alongside shows the path of a diver, when she takes a jump from the diving, board. Clearly, it is a parabola., , Annie was standing on a diving board, 48 ft above the water level. She took a dive into the pool., Her height (in ft) above the water level at any time ‘t’ in seconds is given by the polynomial h( t ), such that, h( t ) = - 16t 2 + 8t + k, , 41. What is the value of k?, (a) 0, (c) 48, , (b) - 48, (d) 48/( - 16), , 42. At what time will she touch the water in the pool?, (a) 30 s, , (b) 2 s, , (c) 1.5 s, , (d) 0.5 s, , 43. Rita’s height (in ft) above the water level is given by another polynomial p(t) with, zeroes - 1 and 2. Then, p( t) is given by, (b) t 2 + 2 t - 1, (d) - 24t 2 + 24t + 48, , 44. A polynomial q(t) with sum of zeroes as 1 and the product as - 6 is modelling Anu’s, height (in ft) above the water at any time t ( in s). Then, q( t) is given by, (a) t 2 + t + 6, (c) - 8t 2 + 8t + 48, , (b) t 2 + t - 6, (d) 8t 2 - 8t + 48, , 45. The zeroes of the polynomial r(t) = - 12t 2 + (k - 3)t + 48 are negative of each other., Then, k is, (a) 3, , (b) 0, , (c) - 1.5, , (d) - 3, , Latest CBSE SAMPLE PAPER, , (a) t 2 + t - 2, (c) 24t 2 - 24t + 48

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48, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 46-50 are based on Case Study-2., , CASE STUDY 2, A hockey field is the playing surface for the game of hockey. Historically, the game was played, on natural turf (grass) but nowadays it is predominantly played on an artificial turf., It is rectangular in shape-100 yards by 60 yards. Goals consist of two upright posts placed, equidistant from the centre of the backline, joined at the top by a horizontal crossbar. The inner, edges of the posts must be 3.66 m (4 yards) apart, and the lower edge of the crossbar must be 2.14, m (7 ft) above the ground., Each team plays with 11 players on the field during the game including the goalie. Positions you, might play include:, Forward As shown by players A, B, C and D., Midfielders As shown by players E, F and G., Fullbacks As shown by players H, I and J., Goalie As shown by player K., Using the picture of a hockey field below, answer the questions that follow, 8, 7, A, , 6, 5, H, , F, , 4, 60 yards, , B, , 3, 2, K, , I, , E, , 1, , O, , 0, –6 –5 –4 –3 –2 –1, J, , 1, –1, , 2, , 3, , 4 5, , 6, , 7, , 8, , 9 10, , 11 12 13 1415, , C, , –2, –3 G, , D, , –4, –5, 100 yards, , –6, , 46. The coordinate of the centroid of DEHJ are, (a) ( - 2 / 3, 1), , (b) (1, - 2 / 3), , (c) (2 / 3, 1), , (d) ( - 2 / 3, - 1), , Latest CBSE SAMPLE PAPER, , 47. If a player P needs to be at equal distances from A and G, such that A, P and G are in, straight line, then position of P will be given by, (a) ( -3 / 2 , 2 ), (b) (2 , - 3 / 2 ), (c) (2 , 3 / 2 ), , (d) ( -2 , - 3), , 48. The point on X-axis equidistant from I and E is, (a) (1 / 2 , 0), , (b) (0, - 1 / 2 ), , (c) ( -1 / 2 , 0), , (d) (0, 1 / 2 ), , 49. What are the coordinates of the position of a player Q such that his distance from K is, twice his distance from E and K , Q and E are collinear?, (a) (1, 0), (b) (0, 1), (c) ( -2 , 1), , (d) ( -1, 0), , 50. The point on Y-axis equidistant from B and C is, (a) ( -1, 0), , (b) (0, - 1), , (c) (1, 0), , (d) (0, 1), , ANSWERS, 1. (b), 11. (b), 21. (c), , 2. (a), 12. (c), 22. (c), , 3. (b), 13. (b), 23. (b), , 4. (d), 14. (a), 24. (a), , 5. (a), 15. (d), 25. (c), , 6. (d), 16. (b), 26. (a), , 7. (b), 17. (b), 27. (b), , 8. (c), 18. (a), 28. (c), , 9. (a), 19. (d), 29. (a), , 10. (d), 20. (a), 30. (c), , 31. (d), 41. (c), , 32. (b), 42. (b), , 33. (a), 43. (d), , 34. (b), 44. (c), , 35. (a), 45. (a), , 36. (c), 46. (a), , 37. (d), 47. (c), , 38. (b), 48. (a), , 39. (c), 49. (b), , 40. (b), 50. (d)

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49, , CBSE Sample Paper Mathematics Standard Class X (Term I), , SOLUTIONS, 1. (b) We have, least composite number is 4 and, the least prime number is 2., \LCM (4, 2) = 4, , We have, AC = 32 cm and BD = 24 cm., Since, diagonals of a rhombus bisects each, other at right angle., , and HCF (4, 2) = 2, So, LCM (4, 2) : HCF (4, 2) = 4 :2 = 2 : 1, 2. (a) We know that, if a1 x + b1 y + c1 = 0 and, a2 x + b2 y + c2 = 0 coincide, then, a1 b1 c1, =, =, a2 b2 c2, So, if 5 x + 7 y - 3 = 0 and 15 x + 21y - k = 0, coincide, then, 5, 7 -3, =, =, 15 21 -k, 1 1 3, Þ, = =, 3 3 k, 1 3, \, =, 3 k, Þ, k =9, 3. (b) From the figure, we have, Distance of the girl from the starting point, = OB, N, B, 150 m, O, , W, , Let ABCD be a rhombus., , 200 m A, , E, , S, , OB2 = OA2 + AB2, = (200 )2 + (150 )2, , AB 2 = OA 2 + OB 2, [by Pythagoras theorem], 2, , 2, , AB = 16 + 12 2, [Q OA =, , 1, 1, AC and OB = BD], 2, 2, , = 256 + 144 = 400, \, , AB = 20 cm, 1, AC ´ BD, 2, 1, = ´ 32 ´ 24, 2, , Now, area of a rhombus =, , = 384 cm 2, Also, area of a rhombus = Side ´ Altitude, \, Þ, , Side ´ Altitude = 384, 384, Altitude =, = 19.2 cm, 20, , 5. (a) When two fair coins are tossed, then, possible outcomes are HH , HT ,TH ,TT. The, favourable outcomes of getting at the most, one head are HT ,TH ,TT., \Required probability, Favourable number of outcomes, =, Total number of outcomes, 3, =, 4, AB2 4, 6. (d) Given,, =, PQ2 9, AB 2, Þ, =, PQ 3, We know that, if two triangles are similar, then, the ratio of corresponding altitudes is equal to, the ratio of corresponding sides., AM AB 2, =, =, \, PN PQ 3, , = 40000 + 22500, = 62500, OB = 62500 = 250, \, So, required distance is 250 m., , 7. (b) We have,, , 4. (d), D, , C, , c, 32O, , A, , 2 sin 2 b - cos 2 b = 2, Þ 2 sin 2 b - (1 - sin 2 b) = 2, [Q cos 2 q + sin 2 q = 1], , m, 24, , cm, , B, , Þ 2 sin 2 b - 1 + sin 2 b = 2, Þ 3 sin 2 b = 3 Þ sin 2 b = 1, Þ, sin 2 b = sin 2 90 °, Þ, b = 90 °, , Latest CBSE SAMPLE PAPER, , Now, In DOAB,, By Pythagoras theorem,, , \

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50, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 44123, 1000, Now, prime factors of denominator, = (10 )3, = (2 ´ 5 )3, \ Prime factors are 2, 5., 9. (a) Line x = a is a line parallel to Y-axis and, y = b is a line parallel to X-axis. So, both lines, are intersecting line., 10. (d) Distance of the point A( - 5 , 6 ) from the, origin O(0, 0) is given as, 8. (c) We have, 44.123 =, , OA = ( - 5 -0 )2 + (6 - 0 )2, [from distance formula], 2, , = ( - 5 ) + (6 )2, , 14. (a) We have,, Ratio of angles of DABC are 1 : 1 : 2, \, , ÐA = x , ÐB = x, , and ÐC = 2 x, We know that, sum of all angles of a triangle is, 180°., \, , ÐA + ÐB + ÐC = 180 °, , Þ, , x + x + 2 x = 180 °, , Þ, , 4 x = 180 °, , Þ, , x = 45 °, , So, ÐA = 45 °, ÐB = 45 ° and ÐC = 90 °, sec A, tan A, Now,, cosec B cot B, =, , = 25 + 36, = 61 units, , =, , 11. (b) We have,, 23, a2 =, 25, 23, a=, Þ, 5, , =0, 15. (d) We have,, 23, will, 5, , Radius of circle, r = 0.7m, Distance travelled in one revolution, = Circumference of the circle = 2 pr, 22, =2 ´, ´ 0.7, 7, = 4.4 m, , also be a irrational number. So, a is an, irrational number., 12. (c) We know that,, LCM ( a, b) ´ HCF ( a, b) = a ´ b, \LCM ( x , 18 ) ´ HCF ( x , 18 ) = x ´ 18, 36 ´ 2 = x ´ 18, , Þ, x=4, 13. (b) We have,, , Latest CBSE SAMPLE PAPER, , tan A = 3, Þ, , tan A = tan 60 °, , Þ, , A = 60 °, , [Q tan60 ° = 3 ], , Again, in DABC, ÐA + ÐB + ÐC = 180 °, Þ, , 60 ° + 90 ° + ÐC = 180 °, , Þ, , ÐC = 30 °, , Now, cos A cos C - sin A sin C, = cos 60 ° cos 30 ° - sin 60 ° sin 30 °, 1, 3, 3 1, = ´, ´, 2, 2, 2, 2, 3, 3, =, 4, 4, =0, , 2 1, 2 1, , = 1 -1, , Since, 23 is a irrational number, so, , Þ, , sec 45 °, tan 45 °, cosec 45 ° cot 45 °, , Now, Number of revolutions, Total distance travelled, =, Distance travelled in one revolution, 176, =, = 40, 4.4, 16. (b) We know that, if two triangles are similar, then the ratio of their perimeter is equal to, the ratio of their corresponding sides., Now, DABC ~ DDEF, Perimeter of DABC BC, \, =, Perimeter of DDEF EF, AB + BC + CA, BC, Þ, =, Perimeter of DDEF EF, Þ, , Perimeter of DDEF, ( AB + BC + CA) ´ EF, =, BC, (3 + 2 + 2 .5 ) ´ 4, =, 2, = 7.5 ´ 2 = 15 cm

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51, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 17. (b) In DABC,, DE || BC, \, Ð1 = Ð3 and Ð2 = Ð4, [corresponding angles], A, , D, , 1, , 3, , 2, , E, 4, , B, , C, , Now, in DADE and DABC, Ð1 = Ð3, Ð2 = Ð4, By AA similarity criterion, DADE ~ DABC, AD DE, \, =, AB BC, DE, 3, =, [Q AB = AD + BD], Þ, 3 + 4 14, 3 ´ 14, DE =, Þ, Þ DE = 6 cm, 7, 18. (a) We have,, 4 tanb = 3, 3, tanb =, Þ, 4, 4 sin b - 3 cos b, Now,, 4 sin b + 3 cos b, sin b 3 cos b, 4, cos b cos b, =, 4 sin b 3 cos b, +, cos b, cos b, [Q dividing numerator and, denominator by cosb], 4 tan b - 3, éQ tan q = sin q ù, =, êë, 4 tan b + 3, cos q úû, , =0, 19. (d) Second equation can be obtained by, multiplying or dividing the first equation by, a non-zero number., Now,, - 5x + 7y = 2, Þ, - 5x + 7y - 2 = 0, [multiply by -2], Þ, - 2(- 5x + 7y - 2 ) = 0, Þ, 10 x - 14 y + 4 = 0, Þ, 10 x - 14 y = - 4, , [divide both sides by 81], So, two coprime numbers whose sum is 15 are, (1, 14), (2, 13), (4, 11) and (7, 8)., So, there are 4 possible number of pairs., 22. (c) Required area = Area of DACD, 1, = ´ Base ´ Altitude, 2, 1, = ´6´2, 2, [Base = AD = 4 - ( - 2 ) = 6 and, height = Distance of point C, from Y-axis = 2], = 6 sq units, 23. (b) We have,, tan a + cot a = 2, 1, tan a +, =2, Þ, tan a, Þ, tan 2 a + 1 = 2 tan a, Þ, tan 2 a - 2 tan a + 1 = 0, Þ, (tan a - 1)2 = 0, Þ, tan a = 1, 1, 1, Again, cot a =, = =1, tan a 1, Now, tan 20 a + cot 20 a = (1)20 + (1)20, =1 + 1=2, 24. (a) We have,, 217 x + 131y = 913, , … (i), , 131x + 217 y = 827, , … (ii), , On adding Eqs. (i) and (ii), we get, 348 x + 348 y = 1740, Þ, Þ, , 348 ( x + y ) = 1740, x+y=, , 1740, =5, 348, , Latest CBSE SAMPLE PAPER, , 3, -3, 4, =, 3, 4´ +3, 4, 3 -3, =, 3+3, 4´, , 20. (a) Number of possible outcomes, = Number of letters of English alphabets, = 26, Favourable outcomes are letters of the word, ‘MATHEMATICS’, i.e., M, A, T, H, E, I, C, S, \Number of favourable outcomes = 8, 8, 4, So, required probability =, =, 26 13, 21. (c) We have, HCF of two numbers is 81., So, numbers are 81x and 81y, where x and y, are coprime., Now, according to the question, 81x + 81y = 1215, Þ, x + y = 15

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52, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 25. (c) We know that,, , 30. (c), , LCM of two prime numbers, , C, , = Product of the numbers., \, , p ´ q = 221, , Þ, , p ´ q = 13 ´ 17, , 8 cm, , Þ, , p = 17, q = 13, , \, , 3 p - q = 3 ´ 17 - 13, , [Q p > q], A 3 cm D, , = 51 - 13 = 38, 26. (a) We know that, there are 4 kings and, 4 queens in a well shuffled deck of 52 cards., 52 - 8 44 11, \ Required probability =, =, =, 52, 52 13, 27. (b) When two fair dice are rolled, simultaneously, then total number of possible, outcomes will be 36., Now, favourable outcome are those in which, 5 will come up atleast once i.e. (5, 1), (5, 2),, (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5),, (4, 5), (6, 5)., \Total number of favourable outcomes = 11, 11, \Required probability =, 36, 28. (c) We have,, 1 + sin 2 a = 3 sin a cos a, Þ sin 2 a + cos 2 a + sin 2 a = 3 sin a cos a, , B, , In DACD and DABC, ÐA = ÐA, , (common), (given), , ÐADC = ÐACB, So, by AA similarity criterion,, DACD ~ DABC, AC AD, =, AB AC, 64, 8, 3, = Þ AB =, 3, AB 8, 64, AD + BD =, 3, 64, BD =, - AD, 3, 55, 64, cm, =, -3 =, 3, 3, , \, Þ, Þ, Þ, , 31. (d) Let l be the perpendicular bisector of, line segment joining the points A( 4 , 5 ), and B( - 2 , 3 ). Let P( x , y ) be any point of l., , [Q sin 2 q + cos 2 q = 1], P (x, y), , 2, , Þ 2 sin a - 3 sin a cos a + cos 2 a = 0, 2, , Þ 2 sin a - 2 sin a cos a - sin a cos a, + cos 2 a = 0, A (4, 5), , Þ 2 sin a( sin a - cos a), , B (–2, 3), , - cos a(sin a - cos a) = 0, Þ (sin a - cos a )(2 sin a - cos a ) = 0, , Latest CBSE SAMPLE PAPER, , Þ, Þ, , \, , sin a = cos a or 2 sin a = cos a, 1 = cot a or 2 = cot a, éQ cos q = cot q ù, úû, êë sin q, cot a = 1 or 2, , 29. (a) We know that, diagonals of a, parallelogram bisects each other. So,, Mid-point of AC = Mid-point of BD, æ1+ x 2 + 6ö æ 4 + 3 y + 5ö, Þ ç, ,, ,, ÷ =ç, ÷, 2 ø è 2, 2 ø, è 2, y+5, 1+ x 7, = and 4 =, Þ, 2, 2, 2, , l, , So, P will be equidistant from A and B., \, PA = PB, Þ ( x - 4 )2 + ( y - 5 )2 = ( x + 2 )2 + ( y - 3 )2, Þ ( x - 4 )2 + ( y - 5 )2 = ( x + 2 )2 + ( y - 3 )2, Þ x 2 - 8 x + 16 + y 2 -10 y + 25, = x2 + 4x + 4 + y2 - 6y + 9, Þ, , - 12 x - 4 y + 28 = 0, , Þ, , 3x + y - 7 = 0, , which is required equation., 32. (b), A, , x, y, , [comparing x and y coordinate], Þ, , x = 6 and y = 3, , \ ( x , y ) is equal to (6, 3)., , C, , D, , B

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53, , CBSE Sample Paper Mathematics Standard Class X (Term I), , In DABC, cot y =, , Base, AC, =, Perpendicular BC, , Base, AC, =, Perpendicular CD, æ AC ö, cot y çè BC ÷ø, =, cot x æ AC ö, ÷, ç, è CD ø, CD CD, =, =, BC 2 CD, , and in DADC, cot x =, , \, , [Q D is the mid-point of BC, so CD = BD], 1, =, 2, 1, 13, 1, 33. (a) (a), ´, =, = 0.01, which is, 13 100 100, terminate after two decimal places., 1 13 1, (b), ´, =, = 0.1, which is terminate after, 13 10 10, one decimal place., 1 10 10, (c), ´, =, = 0.05917..., which is not, 13 13 169, terminate., 1 100 100, (d), ´, =, = 0.5917..., which is not, 13 13 169, terminate., 34. (b) Let the side of the square be x cm., Then, AF = AB - FB = (16 - x ) cm, and GE = BE - BG = (8 - x ) cm, A, , 16 - x, x, =, x, 8-x, , Þ, Þ, , (16 - x ) (8 - x ) = x 2, , Þ 128 - 8 x - 16 x + x 2 = x 2, 24 x = 128, 128, x=, Þ, 24, 16, cm, Þ, x=, 3, 35. (a) Since, P( x1 , y1 ) divides the line segment, Þ, , joining R( - 1, 3 ) and S(9 , 8 ) in the ratio k :1., k:1, R (–1, 3), , P (x1, y1), , S (9, 8), , æ 9k - 1 8k + 3 ö, ÷, Then, P ( x1 , y1 ) = çç, ,, ÷, è k +1 k +1 ø, [from section formula], Also, P lies on the line x - y + 2 = 0,, So, P satisfies it, 9k - 1 8k + 3, + 2 =0, \, k +1, k +1, Þ, , 9k - 1 - 8k - 3 + 2 k + 2 = 0, , 3k - 2 = 0, 2, k=, Þ, 3, 36. (c) Let the side of square be a., Þ, , A, , E, , B, , H, , F, , 16 cm, D, D, , G, , C, , Area of shaded region, = Area of semi-circle, , G, B, , 8 cm, , E, , + (Area of half square, - Area of two quadrants), , Now, in DAFD and DDGE, ÐF =ÐG = 90 ° [BGDFis a square], ÐADF = ÐDEG, [corresponding angles, as FD||BE], So, by AA similarity criterion,, , \, , DAFD ~ DDGE, AF FD, =, DG GE, , 2, , =, , 1 a, 1 æaö, 1, pç ÷ + a2 - 2 ´ pæç ö÷, 4 è2 ø, 2 è2 ø, 2, , 2, , [Q Radius of semi-circle], 1 2 1 2 1 2, = pa + a - pa, 8, 2, 8, 1 2 1, 2, [Q) a = 14cm], = a = (14, 2, 2, = 98 cm 2, , Latest CBSE SAMPLE PAPER, , F

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54, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 37. (d) We have, OA = OB = AB = 1 cm, , D, , C, r, , B, , O, , A, , A, , a, , O, , r, , B, , a, , From figure, we have,, AB2 + BC 2 = AC 2, , \ DOAB is an equilateral triangle of side 1 cm., \Required area = 8 ´ Area of one segment, ù, é 60 °, 3, =8ê, ´ p(1)2 ´ (1)2 ú, 4, û, ë360, [Q q = 60 °, r = 1 cm, side of triangle = 1 cm], ép, 3ù, =8ê cm 2, 4 úû, ë6, 1, 38. (b) We have, 2 and are the zeroes of the, 2, polynomial px 2 + 5 x + r., Coefficient of x, \ Sum of zeroes = Coefficient of x 2, =, Þ, , 2+, , -( 5 ), p, , 1 -5, =, 2, p, , Þ, , 5 -5, =, 2, p, , Þ, , p = -2, , Latest CBSE SAMPLE PAPER, , =, 2´, , Constant term, Coefficient of x 2, r, p, , 1 r, =, 2 p, r, p, , Þ, , 1=, , Þ, , r= p, = -2, , 39. (c) Let the radius of the circle be r cm and, side of the square be a cm., Then,, , Þ, , 2 pr = 100, 100, cm, r=, 2p, , a2 + a2 = (2 r)2, , Þ, , 2 a 2 = 4 r2, , Þ, , a 2 = 2 r2, , Þ, , a= 2r, , Þ, , a= 2 ´, , 100 50 2, cm, =, 2p, p, , 40. (b) We have,, 3 x + y = 243, Þ, , 3x + y = 35, , Þ, , x+ y=5, , …(i), , Again, 243 x - y = 3, Þ, , 3 5( x - y ) = 3, , Þ, , 5( x - y ) = 1, 1, x-y=, 5, , Þ, , and product of zeroes =, , Þ, , Þ, , …(ii), , On adding Eqs. (i) and (ii), we get, 1, 2x = 5 +, 5, 26, Þ, 2x =, 5, 13, Þ, x=, 5, 13, in Eq. (i), we get, On putting, x =, 5, 13 12, y=5-x=5=, 5, 5, 13, \, x=, 5, 12, and, y=, 5, So, number of solution is one., 41. (c) Initially, when t =0, h( t ) = 48 ft, \, , h(0 ) = 48, , Þ, , - 16(0 )2 + 8(0 ) + k = 48, , Þ, , k = 48

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55, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 42. (b) When Annie touches the pool, her height, h( t ) = 0 ft. Let the time be t s., \, , h( t ) = 0, , Þ, , - 16 t 2 + 8 t + k = 0, , Þ, , - 16 t 2 + 8 t + 48 = 0, , [Q k = 48], , 2, , Þ, , 2 t -t - 6 = 0, , Þ, , 2 t2 - 4t + 3t - 6 = 0, , Þ, , 2 t( t - 2 ) + 3 ( t - 2 ) = 0, , Þ, , ( t - 2 ) (2 t + 3 ) = 0, , Þ, , t =2 , -, , Þ, , t =2 s, , [divide by - 8], , 3, 2, , [Q t can’t be negative], 43. (d) We have, - 1 and 2 are the zeroes of p( t )., \, , p( t ) = l (t 2 - (sum of zeroes) t, + product of zeroes), 2, , = l ( t - ( - 1 + 2 ) t + ( - 1) ´ 2 ), = l (t2 - t - 2 ), , E (2, 1), F (1, 5), G (1, - 3), H (- 2, 4),, I (- 1, 1), J ( - 2 , - 2 ), K (- 4, 1), 46. (a) Centroid of DEHJ, æ2 -2 -2 1 + 4 -2 ö, =ç, ,, ÷, 3, 3, ø, è, 2 ö, æ, = ç - , 1÷, è 3 ø, 47. (c) If P needs to be at equal distance from, A(3, 6) and G(1, - 3) such that A, P and G are, collinear, then P will be the mid-point of AG., æ3 + 1 6 -3 ö æ 3 ö, \ Coordinate of P = ç, ,, ÷ = ç2 , ÷, 2 ø è 2ø, è 2, 48. (a) Let the point of X-axis which is, equidistant from I ( -1,1) and E(2 ,1) be, P ( x ,0 ). Then,, PI = PE, 2, , Þ ( x + 1) + (0 - 1)2 = ( x - 2 )2 + (0 - 1)2, Þ, , ( x + 1)2 + 1 = ( x - 2 )2 + 1, , Now, when t =0, height = 48, , Þ, , ( x + 1)2 = ( x - 2 )2, , p(0 ) = 48, , Þ, , x2 + 2 x + 1 = x2 - 4x + 4, , l(0 - 0 - 2 ) = 48, , Þ, , \, Þ, , 2, , l =- 24, , Þ, , p( t ) = - 24 ( t 2 - t - 2 ), , \, , = - 24 t 2 + 24 t + 48, 44. (c) We have,, q( t ) = b( t 2 - (Sum of zeroes) t, + Product of zeroes), = b( t 2 - t - 6 ), Now, when t = 0, height = 48, \, , q(0 ) = 48, , 6x = 3, 1, Þ, x=, 2, 1, So, required point is æç , 0 ö÷ ., è2 ø, , 49. (b) Let the coordinates of the position of the, player Q be ( x , y ). Since distance of Q from, K( - 4 , 1) is twice the distance from E(2 , 1) and, K , Q and E are collinear. So Q divides the line, segment KE in 2 : 1., 2:1, , Þ b(0 2 - 0 - 6 ) = 48, q( t ) = - 8 t 2 + 8 t + 48, , 45. (a) We have, r( t ) = - 12 t 2 + ( k - 3 )t + 48, Since, zeroes of the polynomial r( t ) are, negative of each of other., \ Sum of zeroes = 0, (Coefficient of t ), Þ, =0, Coefficient of t 2, -( k - 3 ), =0, Þ, -12, Þ, Þ, Solutions (46-50), , Q (x, y), , E (2, 1), , \Coordinates of Q, æ 2 ´ 2 + 1 ´ (- 4) 2 ´ 1 + 1 ´ 1 ö, ÷, = çç, ,, ÷, 2 +1, 2 +1, è, ø, = (0, 1), 50. (d) Let the point on Y-axis which is, equidistant from B (4, 3) and C (4, - 1) be, T (0 , y ). Then, TB = TC, 2, , k -3 =0, , Þ, , ( 4 -0 ) + (3 - y )2 = ( 4 -0 )2 + ( - 1 - y )2, , k =3, , Þ, , 16 + 9 - 6 y + y 2 = 16 + 1 + 2 y + y 2, , Þ, , - 8y = - 8, y =1, , From the given figure, we have coordinates, , Þ, , of A (3, 6), B (4, 3), C (4, - 1), D (3, - 4),, , \ Required point is (0, 1)., , Latest CBSE SAMPLE PAPER, , \, , K (–4, 1), , b = -8, , Þ

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59, , CBSE Sample Paper Mathematics Standard Class X (Term I), , SAMPLE PAPER 1, MATHEMATICS (Standard), A Highly Simulated Practice Questions Paper, for CBSE Class X (Term I) Examination, , Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions., There is no negative marking., Maximum Marks : 40, Time allowed : 90 minutes, , Roll No., , Section A, Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 1. The product of a non-zero rational and an irrational number is, (a) always rational, (c) always irrational, , 2., , (b) rational or irrational, (d) zero, , 2 times the distance between (0, 5) and ( -5, 0) is ………, (a) 10, , (b) 8, , (c) 17, , (d) 14, , 3. The circumference of a circle of diameter 21 cm is, (a) 66 cm, , (b) 69 cm, , (c) 63 cm, , (d) 68 cm, , 4. If a, b are the zeroes of the polynomial f (x) = x 2 - p(x + 1) - c such that (a + 1)(b + 1) = 0,, then c is equal to, (a) 1, , (b) 0, , (c) -1, , (d) 2, , (a) 2 x + 7y = 11, , (b) 4x - 2 y = 5, , (c) x - 3y = 5, , (d) 3x - 4y = 8, , 6. 119 2 - 111 2 is, (a) prime number, (c) an odd prime number, , (b) composite number, (d) an odd composite number, , 7. The probability that a non-leap year has 53 Sunday’s, is, (a), , 2, 7, , (b), , 5, 7, , (c), , 6, 7, , (d), , 1, 7, , SAMPLE PAPER 1, , 5. Which of the equation has solution as x = 2, y = 1?

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60, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 8. If 3 is one zero of the polynomial f (x) = 9x 2 - 3(a - 1)x + 5, then the value of a is, (a), , 81, 5, , (b), , 95, 9, , (c), , 40, 9, , (d) None of these, , 9. In which quadrant does the point (-3, 5) lie?, (a) I, , (b) II, , (c) III, , (d) IV, , 10. The sum of two numbers is 137 and their difference is 43. The situation can be, algebrically represented as, (a) x - y = 137, x + y = 180, (c) x + y = 137,x - y = 43, , (b) 2( x + y) = 137, 2( x - y) = 43, (d) x + y = 43, x - y = 137, , 11. If the sum of the circumference and the radius of a circle is 51 cm, then the radius of, the circle (in cm) is, (a) 154, , (b) 44, , (c) 14, , (d) 7, , 12. If DABC ~ DDEF such that AB = 91, . cm and DE = 6.5 cm. If the perimeter of DDEF is 25, cm, then the perimeter of DABC is, (a) 36 cm, (b) 30 cm, 3, , (c) 34 cm, , (d) 35 cm, 3, , n, , 2, , 13. If a = 2 ´ 3, b = 2 ´ 3 ´ 5, c = 3 ´ 5 and LCM (a , b , c) = 2 ´ 3 ´ 5, then n =, (a) 1, , (b) 2, , (c) 3, , (d) 4, , 14. A paper is in the form of a rectangle ABCD in which AB = 18 cm and BC = 14 cm. A semi, circular portion with BC as diameter is cut off. Find the area of the remaining paper., (b) 165 cm 2, (d) None of these, , (a) 175 cm 2, (c) 145 cm 2, , 15. If the zeroes of the quadratic polynomial x 2 + (m + 1)x + n are 4 and 5, then, (a) m = -20, n = -80, (c) m = -10, n = 20, , (b) m = -20, n = 80, (d) m = 20, n = -10, , a, è3, ø, then the value of ‘a’ is, (a) 12, (b) - 6, , (c) - 12, , 16. If Qæç , - 4 ö÷ is the mid-point of the segment joining the points P(6, - 5) and R(2, - 3),, (d) - 4, , 17. If am ¹ bl then the pair of equations ax + by = c and lx + my = n, (a) has a unique solution, (c) has infinitely many solutions, , (b) has no solution, (d) may or may not have a solution, , SAMPLE PAPER 1, , 18. One card is drawn at random from a well-shuffled deck of 52 cards. What is the, probability of getting a face card?, 1, 3, (b), (a), 26, 26, , (c), , 3, 13, , (d), , 4, 13, , 19. The coordinates of the point which is reflection of point (-3, 5) in X-axis are, (a) (3, 5), , (b) (3, -5), , (c) (-3, - 5), , 20. 2.13113111311113…… is, (a) an integer, (c) an irrational number, , (b) a rational number, (d) None of these, , (d) (-3, 5)

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61, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Section B, Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 21. The diameter of a wheel is 1.26 m. How far will it travel in 400 revolutions?, (a) 2670 m, , (b) 2880 m, , (c) 1980 m, , (d) 1584 m, , 2, , 22. If p - q + r = 0 then a zero of the polynomial px + qx + r is, (a) 0, (c) -1, , (b) 1, (d) None of these, , 23. For the pair of linear equations 47 x + 31y = 18 and 31x + 47 y = 60 the value of x + y is, (a) 1, , (b) 0, , (c) -4, , (d) 7, , 24. If LCM = 350, product of two numbers is 25 ´ 70, then their HCF = 5., (a) 12, , (b) 15, , (c) 5, , (d) 10, , 25. The distance of the point (2, 11) from the X-axis is, (a) 11 units, (c) 13 units, , (b) 2 units, (d) 12 units, , 26. A bag contains 3 red and 7 black balls. A ball is taken out of the bag at random. What is, the probability of getting a black ball?, 3, (a), 10, 1, (c), 10, , (b), , 7, 10, , (d) None of these, , 27. The area of a quadrant of a circle whose circumference is 44 cm is, (a) 24 cm 2, , (b) 28 cm 2, , (c) 35.5 cm 2, , (d) 38.5 cm 2, , 28. If one zero of the polynomial p(x) = (k 2 + 9)x 2 + 9x + 6k is the reciprocal of the other, zero, then k is, (a) - 2, , (b) 3, , (c) 2, , (d) - 3, , 29. Two vertices of a triangle are (- 3, 5) and (7 , - 4). If its centroid is (2, - 1), then the third, vertex is, (a) (2, 4), , (b) ( - 2 , 4), , (c) (2 , - 4), , (d) ( - 2 , - 4), , 30. A number x is chosen at random from the numbers - 3, - 2, - 1, 0, 1, 2, 3 the probability, (b), , 2, 7, , (c), , 3, 7, , (d), , 1, 7, , 31. There are deer and peacock in the zoo. By counting heads, they are 52, the number of, their legs is 176. Number of peacock are, (a) 12, (b) 16, , (c) 20, , (d) 36, , 32. Which of the following rational numbers have terminating decimal?, 16, 50, (a) i and ii, (i), , 5, 18, (b) ii and iii, , (ii), , 2, 21, (c) i and iii, , (iii), , 7, 250, (d) i and iv, , (iv), , SAMPLE PAPER 1, , that| x |< 3 is, 5, (a), 7

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62, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 33. The minute hand of a clock is 6 cm long. Then, the area of the face of the clock, described by the minute hand in 35 minutes is, (a) 265 cm 2, (b) 266 cm 2, (c) 264 cm 2, , (d) None of these, , 34. If the sum of the zeroes of the quadratic polynomial kx 2 + 2x + 3k is equal to their, product, then k is equal to, -2, 1, (b), (a), 3, 3, , (c), , -1, , (d), , 3, , 2, 3, , 35. Find the value of a so that the point (3, a) lies on the line represented by 2x - 3y = 5., (a), , 1, 3, , (b), , 1, 5, , (c), , 1, 4, , (d), , 1, 6, , 36. The graphical representation of x - 2y + 4 = 0 and 3x + 4 y + 2 = 0 will be, (a) intersecting, (c) coincident, , (b) parallel, (d) None of these, , 37. In DABC, ÐB = 90° and BD ^ AC. If AC = 9 cm and AD = 3 cm, then BD is equal to, (a) 2 2 cm, , (b) 3 2 cm, , (c) 2 3 cm, , (d) 3 3 cm, , 38. If the radius of a circle is increased by 25% then its circumference will increases by, (a) 25%, , (b) 50%, 2, , (c) 75%, , (d) 100%, , 2, , 39. The value of 4 sin 60° + 3 tan 30° - 8 sin 45° cos 45° is, (a) 0, , (b) 1, , (c) 2, (d) 5, 1, 40. The smallest rational number by which should be multiplied so that its decimal, 3, expansion terminates after one place of decimal, is, 3, 1, 3, (b), (c) 3, (d), (a), 10, 10, 100, , Section C, Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted., 41-45 are based on Case Study-1., , Case Study 1, Two hotels are at the ground level on either side of a mountain. On moving a certain distance, towards the top of the mountain two huts are situated as shown in the figure. The ratio between, the distance from hotel 2 to hut Band that of hut B to mountain top is 3 : 7., , SAMPLE PAPER 1, , Mountain top, , 10 miles, Hut-B, Hut-A, , Ground level, , Hotel-1, , Hotel-2

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63, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 41. What is the ratio of the perimeters of the triangle formed by both hotels and mountain, top to the triangle formed by both huts and mountain top?, (a) 5 : 2, (b) 10 : 7, (c) 7 : 3, , (d) 3 : 10, , 42. The distance betweeen the hotel 1 and hut A is, (a) 2.5 miles, , (b) 29 miles, , (c) 4.29 miles, , (d) 1.5 miles, , 43. If the horizontal distance between the hut A and hut B is 8 miles, then the distance, between the two hotels is, (a) 2.4 miles, (b) 11.43 miles, , (c) 9 miles, , (d) 7 miles, , 44. If the distance from mountain top to hut A is 5 miles more than that of distance from, hotel 2 to mountain top, then what is the distance between hut B and mountain top, (a) 3.5 miles, (b) 6 miles, (c) 5.5 miles, (d) 4 miles, , 45. Which property of geometry will be used to find the distance between hut B and, mountain top?, (a) Congruent of triangles, (c) pythagoras theorem, , (b) Thales theorem, (d) None of these, , 46-50 are based on Case Study-2., , Case Study 2, Children were playing a game by making some right angled triangles on the plane sheet of, paper. They took a right angled triangle with two of its sides AC = 25 cm, BC = 20 cm and, ÐABC = 90°. With the help of right angled triangle, solve the following questions., , 25, , cm, , A, , C, , 90°, , B, , 20 cm, , 46. Using the above data, the value of sin A is, (a), , 12, 13, , (b), , 3, 5, , (c), , 5, 13, , (d), , 4, 5, , (c), , 3, 5, , (d), , 13, 5, , (c), , 4, 5, , (d), , 3, 5, , (c), , 4, 5, , (d), , 12, 13, , (d), , 9, 13, , 47. Using the above data, the value of sin C is, (a), , 12, 13, , (b), , 13, 12, , 48. Using the above data, the value of tan C is, 12, 5, , (b), , 3, 4, , 49. Using the above data, the value of cos A is, (a), , 3, 5, , (b), , 3, 4, , 50. Using the above data, the value of, (a), , 9, 4, , (b) 4, , tan A + sin A, , is, tan A - sin A, 12, (c), 9, , SAMPLE PAPER 1, , (a)

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OMR SHEET, , SP 1, , Roll No., Sub Code., , Student Name, , Instructions, Use black or blue ball point pens and avoid Gel & Fountain pens for filling the OMR sheet., Darken the bubbles completely. Don’t put a tick mark or a cross mark, half-filled or over-filled bubbles will not be read, by the software., Correct, , ✔, , ✗, , Incorrect, , Incorrect, , Incorrect, , Do not write anything on the OMR Sheet., Multiple markings are invalid., , 1, , 18, , 35, , 2, , 19, , 36, , 3, , 20, , 37, , 4, , 21, , 38, , 5, , 22, , 39, , 6, , 23, , 40, , 7, , 24, , 41, , 8, , 25, , 42, , 9, , 26, , 43, , 10, , 27, , 44, , 11, , 28, , 45, , 12, , 29, , 46, , 13, , 30, , 47, , 14, , 31, , 48, , 15, , 32, , 49, , 16, , 33, , 50, , 17, , 34, , Check Your Performance, Total Questions:, Total Correct Questions:, , If Your Score is, , Score Percentage =, , Total Correct Questions, Total Questions, , × 100, , Less than 60%, > Average (Revise the concepts again), Greater than 60% but less than 75% > Good (Do more practice), Above 75%, > Excellent (Keep it on)

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65, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Answers, 1. (c), 11. (d), 21. (d), , 2. (a), 12. (d), 22. (c), , 3. (a), 13. (b), 23. (a), , 4. (a), 14. (a), 24. (c), , 5. (a), 15. (c), 25. (a), , 6. (b), 16. (a), 26. (b), , 7. (d), 17. (a), 27. (d), , 8. (b), 18. (c), 28. (b), , 9. (b), 19. (c), 29. (c), , 10. (c), 20. (c), 30. (a), , 31. (b), 41. (b), , 32. (d), 42. (c), , 33. (d), 43. (b), , 34. (b), 44. (a), , 35. (a), 45. (b), , 36. (a), 46. (d), , 37. (b), 47. (c), , 38. (a), 48. (b), , 39. (a), 49. (a), , 40. (a), 50. (b), , SOLUTIONS, 1. Product of a non-zero rational and an irrational, number is always irrational., 3, 3 2, e.g. ´ 2 =, (irrational), 4, 4, 2. Given points are (0, 5) and (- 5, 0)., Let the distance between the points be d., By distance formula,, d = ( -5 - 0 )2 + (0 - 5 )2, d = 25 + 25 = 50 = 5 2 units, 2 d = 5 2 ´ 2 = 10 units, \, 21, 3. Radius of the circle, r = cm, 2, Now, circumference of circle = 2 pr, 22 21, =2 ´, ´, = 66 cm, 7 2, 4. Given, a, b are the zeroes of the polynomial, f ( x ) = x 2 - p( x + 1) - c, = x 2 - px - p - c = x 2 - px - ( p + c), -p, Coefficient of x, a +b===p, \, 2, 1, Coefficient of x, -( p + c), Constant term, and ab =, =, = - ( p + c), 2, 1, Coefficient of x, Now, it is given that ( a + 1)(b + 1) = 0, Þ, ab + a + b + 1 = 0, Þ, - ( p + c) + p + 1 = 0, Þ, -c + 1 = 0, \, c =1, 5. Put x = 2 , y = 1 in given equations, we get, , 6. We know that,, a2 - b2 = ( a + b)( a - b), 2, \ 119 - 1112 = (119 + 111)(119 - 111), = 230 ´ 8 = 2 4 ´ 5 ´ 23, , 7. A non-leap year has 365 days., In 365 days, there are 52 weeks and 1 day., In 52 weeks, the number of Sundays will be 52., 1 remaining day can be Sunday, Monday,, Tuesday, Wednesday, Thursday, Friday,, Saturday., We can have any one of these days out of, 7 days., Hence, out of these 7 outcomes, the favourable, outcome is 1, Therefore, probability of getting 53 Sundays in, 1, a non-leap year =, 7, 8. Given, 3 is one zero of the polynomial, f ( x ) = 9 x 2 - 3 ( a - 1)x + 5, \, f (3 ) = 0, Þ, 0 = 9(3 )2 - 3( a - 1)3 + 5, Þ, 0 = 81 - 9 a + 9 + 5, Þ, 9 a = 95, 95, a=, \, 9, 9. Given point is ( -3 , 5 ), Y, II, (–, +), , I, (+, +), , III, (–, –), , IV, (+, –), , X¢, , X, , Y¢, , The point has negative x-coordinate and, positive y-coordinate., Hence, it lies in II quadrant., 10. Let the two numbers be x and y, where x > y, Given, sum of numbers = 137, \, x + y = 137, And difference of numbers = 43, \, x - y = 43, , SAMPLE PAPER 1, , (a) LHS = 2 (2 ) + 7(1) = 11 = RHS, (b) LHS = 4(2 ) - 2 (1) = 6 ¹ RHS, (c) LHS = 2 - 3(1) = - 1 ¹ RHS, (d) LHS = 3(2 ) - 4(1) = 2 ¹ RHS, Only 2 x + 7 y = 11 satisfies the given values., Hence, 2 x + 7 y = 11 is the required equation., , Here, the expression has more than two factor., Hence, it is a composite number.

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66, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 11. Let r be the radius of the circle. Then,, circumference = 2 pr cm, According to the question,, 2 pr + r = 51, 44, æ, Þ, + 1 ö÷ = 51, rç, 7, ø, è, æ 44 + 7 ö, rç, Þ, ÷ = 51, è 7 ø, 7 ´ 51, Þ, r=, = 7 cm, 51, 12. Given, DABC ~ DDEF, AB = 9.1, DE = 6.5 cm, and perimeter of DDEF = 25 cm, Perimeter of DABC AB, \, =, Perimeter of DDEF DE, Perimeter of DABC 9.1, =, 25, 6.5, 91 ´ 25, = 35 cm, Þ Perimeter of DABC =, 65, 13. Given, a = 2 3 ´ 3,, b =2 ´3´ 5, and, c = 3n ´ 5, \ LCM ( a, b, c) = 2 3 ´ 3 n ´ 5, On comparing with the given LCM, we get, n =2, 14. Area of rectangle ABCD = AB ´ BC, = 18 ´ 14 = 252 cm 2, C, , D, , 14 cm, , A, , 18 cm, , Radius of semi-circle ( r) =, , B, , 1, 1, BC = ´ 14 = 7cm, 2, 2, , 1 2, pr, 2, 1 22, = ´, ´ 7 2 = 77 cm 2, 2, 7, Area of remaining portion = Area of rectangle, - Area of semi-circle, = 252 - 77 = 175 cm 2, , SAMPLE PAPER 1, , Area of semi-circle =, , 15. Given 4 and 5 are the zeroes of the quadratic, polynomial x 2 + ( m + 1)x + n., Coefficient of x, \Sum of zeroes = Coefficient of x 2, - ( m + 1), =, 1, , 4 + 5 = - m-1, m = - 10, Coefficient term n, and product of zeroes =, =, Coefficient of x 2 1, , Þ, Þ, , 4´ 5 =n, n = 20, æ x1 + x2 y1 + y2 ö, 16. Mid-point = ç, ,, ÷, 2 ø, è 2, æ6 + 2 - 5 -3 ö, ,, =ç, ÷, 2 ø, è 2, a, Þ æç , - 4 ö÷ = ( 4 , - 4 ), ø, è3, a, \, =4, 3, Þ, a = 12, Þ, Þ, , 17. Given equations can be written as, ax + by - c = 0, and lx + my - n = 0, As,, am ¹ bl, a b, ¹, l m, Condition for the pair of equations has unique, solution,, a b, ¹, l m, Hence, ax + by = c and lx + my = n has a unique, solution., 18. In 52 cards, the king, queen and Jack are three, face cards in each suit., So, total face cards = 12, 12, 3, \Probability of getting face card =, =, 52 13, 19. When the point is reflected in X-axis, its, x-coordinate remains same and y-coordinate, changes by negative sign., Hence, the point will be ( -3 , - 5 )., 20. The given decimal is non-terminating and, non-repeating decimal., Hence, it must be an irrational number., 1.26, 21. Radius of the wheel, r =, = 0.63 m, 2, [Q diameter = 2 (radius)], Distance travelled in one revolution is equal to, the perimeter of the wheel., 22, ´ 0.63 = 3.96 m, \Distance = 2 pr = 2 ´, 7, \Distance travelled in 400 revolutions, = 400 ´ 3.96 = 1584 m

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67, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 22. Let p( x ) = px 2 + qx + r be the polynomial, , 28. Given, p( x ) = ( k 2 + 9 )x 2 + 9 x + 6 k, , \ p( -1) = p( -1)2 + q ( -1) + r, p( -1) = p - q + r, According to the question,, p - q + r = 0, so clearly -1 is the zero of the, polynomial., 23. Given, 47 x + 31y = 18, , Let one zero be a then other zero is, , Þ, …(ii), , 24. We know that, LCM × HCF = Product of two numbers, Given LCM = 350,, Product of two numbers = 25 ´ 70, 350 ´ HCF = 25 ´ 70, \, 25 ´ 70, Þ, HCF =, =5, 350, 25. Plot the point (2, 11) in coordinate axes., Y, 2 units, , k 2 + 9 = 6k, k 2 - 6k + 9 = 0, ( k - 3 )2 = 0 Þ k = 3, , Þ, Þ, Þ, , 29. Let the coordinates of the third vertex be ( x , y ),, then centroid of triangle is, æ x -3 + 7 y + 5 - 4 ö, ,, ç, ÷ = (2 , - 1), 3, 3, è, ø, x, +, 4, y, +, 1, æ, ö, ,, Þ, ç, ÷ = (2 , - 1), 3 ø, è 3, x+4, y+1, = 2,, = -1, Þ, 3, 3, Þ, x + 4 = 6, y + 1 = - 3, Þ, x =2, y = - 4, \Coordinates of third vertex is (2 , - 4 )., 30. Total numbers = 7, , (2, 11), 11 units, , O, , Constant term, Coefficient of x 2, 1, 6k, a× = 2, a k +9, , Product of zeroes =, , …(i), , and 31x + 47 y = 60, On adding Eqs. (i) and (ii), we get, 78 x + 78 y = 78, On dividing both sides by 78, we get, x + y =1, , 1, ., a, , X, , From the above figure, the required distance is, 11 units., 26. Total number of balls in bag = 3 + 7 = 10, Total number of black balls in bag = 7, Probability of getting a black ball, Total number of black balls in bag, =, Total number of balls in bag, 7, =, 10, , Number x such that | x | < 3 are - 2 , - 1, 0 , 1, 2 ., \Total numbers of x such that | x | < 3 = 5, 5, \Probability =, 7, 31. There are deer and peacocks in a zoo. By, counting heads they are 52., The number of their legs is 176., Let there be x deer and y peacock., Then, 4 x + 2 y = 176, … (i), and, … (ii), x + y = 52, Multiply by 2 in Eq. (ii), we get, …(iii), 2 x + 2 y = 104, Subtract Eq. (iii) from Eq. (i), we get, 4 x + 2 y = 176, , 27. Given, circumference of circle = 25 cm, Þ, \, , 2 x + 2 y = 104, 2 x = 72, x = 36, y = 52 - 36 = 16, 4, , 32. (i), , 16, 2, =, 50 2 ´ 5 2, , (iii), , 2, 2, =, 21 3 ´ 7, , (ii), , [from Eq. (ii)], , 5, 5, =, 18 2 ´ 3 2, , (iv), , 7, 7, = 3, 250 5 ´ 2, , Only i and iv have denominator in the form of, 2 m ´ 5 n, hence i and iv have terminating, decimal., , SAMPLE PAPER 1, , We know that circumference of circle = 2 pr, 22, 44, Þ, ´ r Þ 44 =, ´r, 44 = 2 ´, 7, 7, \, r = 7 cm, Area of a quadrant of a circle, 1, = ´ p´ r´ r, 4, 1 22, = ´, ´7´7, 4 7, 11 ´ 7, =, = 38.5 cm 2, 2, , -

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68, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 33. Let r be the radius of the clock., Angle described by the minute hand in, 60 minutes = 360 °, Angle described by the minute hand in, °, 360, 35 minutes = æç, ´ 35 ö÷ = 210 °, ø, è 60, Required area = Area of sector with central, angle of 210°, q, = æç, ´ pr2 ö÷ cm 2, ø, è 360 °, 210 °, ´ p´ 6 ´ 6, =, 360 °, 210 ° 22, ´, ´6´6, =, 360 ° 7, = 66 cm 2, 34. Given polynomial = kx 2 + 2 x + 3 k, If ax 2 + bx + c is a polynomial then its sum of, -b, roots is given by, and product of zeroes is, a, c, given by ., a, -2, \ Sum of zeroes of given polynomial =, k, 3k, Product of zeroes =, =3, k, It is given that, sum of zeroes, = product of zeroes, -2, Þ, =3, k, -2, k=, Þ, 3, 35. Given point (3, a) lies on the line 2 x - 3 y = 5, , SAMPLE PAPER 1, , Q (3, a) satisfies the given equation., \, 2 ´3 -3´ a = 5, Þ, 6 - 3a = 5, Þ, 1 = 3a, 1, Þ, a=, 3, 36. On comparing the given equation with, standard equation, we get, a1 = 1, b1 = - 2 , c1 = 4, and, a2 = 3, b2 = 4, c2 = 2, For unique solution,, a1 b1, ¹, a2 b2, 1 -2, ¹, 3, 4, So, graphical representation of given lines are, intersecting., , Here, , 37. Given, AC = 9 cm, AD = 3 cm, CD = AC - AD = 9 - 3 = 6 cm, B, , 3 cm, A, , D, , C, 9 cm, , In DABC and DADB, ÐBAC = ÐBAD, ÐABC = ÐADB [each 90° angle], \, DABC ~ DADB, [by AA similarity criterion] … (i), In DABC and DBDC,, ÐABC = ÐBDC [each 90° angle], ÐACB = ÐBCD, \, DABC ~ DBDC, [by AA similarity criterion] … (ii), From Eqs. (i) and (ii) DADB ~ DBDC, BD AD, \, =, CD BD, BD2 = AD × CD = 3 ´ 6 = 18, BD = 3 2 cm, 38. Let original radius be R cm. Then original, circumference = (2 pR ) cm, New radius = 125 % of R cm, 125, 5R, cm, = æç, ´ R ö÷ cm =, 4, ø, è 100, 5R ö, New circumference = æç 2 p ´, ÷ cm, 4 ø, è, 5 pR, cm, =, 2, Increase in circumference, 5 pR, pR, cm, = æç, - 2 pR ö÷ =, ø 2, è 2, pR, 1, Percentage increase = æç, ´, ´ 100 ö÷%, è 2 2 pR, ø, = 25%, 1, 3, 39. Since, sin60 ° =, , tan30 ° =, 3, 2, 1, and cos 45 ° = sin 45 ° =, 2, \ 4 sin 2 60 ° + 3 tan 2 30 ° - 8 sin 45 ° cos 45 °, 2, , 2, æ 3ö, 1 ö, 1, 1, ÷÷ + 3 ´ æç, = 4 ´ çç, ´, ÷ -8 ´, 2, 2, è 3ø, è 2 ø, 4´3 3 8, =, + - =3 + 1 - 4 =0, 4, 3 2

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69, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 40. Now,, , 1 3, 1, ´, =, = 0.1, 3 10 10, , 45. Thales theorem will be used to find the, distance between hut B and mountain top., , 1, should be multiplied by 3/10, so that its, 3, decimal expansion terminates after one place, of decimal., , \, , Solutions (46-50), 46. In DABC , ÐB = 90 °, A, , Solutions (41-45), 25, , cm, , 41. Let DABC be the triangle formed by both hotels, and mountain top. DCDE is the triangle formed, by both huts and mountain top., Clearly DE || AB and So, DABC ~ DDEC, [By AAA similarity criterion], , 90°, , C, , sin A =, , C, , 20 cm, , BC Perpendicular side, =, Hypotenuse, AC, =, , 7, , E, 3, B, , D, A, , ( AB)2 + ( BC )2 = ( AC )2, AB = ( AC )2 - ( BC )2, = (25 )2 - (20 )2, = 625 - 400, = 225 = 15, AB 12 3, \ sin C =, =, =, AC 25 4, 48., , é AD = CD ´ EBù, CE ûú, ëê, , = 4.29 miles, , 44. Given, DC = 5 + BC, Clearly, BC = 10 - 5 = 5 miles, 7, 7, Now, CE = ´ BC = ´ 5 = 3.5 miles, 10, 10, , AB, BC, 15 3, Þ tan C =, =, 20 4, tan C =, , 49. cos A =, , AB 15 3, =, =, AC 25 5, , BC BC, tan A + sin A AB + AC, 50., =, tan A - sin A BC - BC, AB AC, 20 20, +, = 15 25, 20 20, 15 25, 25 + 15, =, 25 - 15, =, , 40, =4, 10, , SAMPLE PAPER 1, , 43. Since, DABC ~ DDEC, BC AB, =, EC DE, [Q Corresponding sides of similar, triangles are proportional], 10 AB, =, Þ, 7, 8, 80, Þ, = 11.43 miles, AB =, 7, , 20 4, =, 25 5, , 47. Use Pythagoras theorem, , Perimeter of DABC BC CE + BE, =, =, EC, Perimeter of DDEF EC, 7 + 3 10, =, =, 7, 7, 42. Since, DE||AB, therefore,, CD CE, =, AD EB, 10 7, =, Þ, AD 3, 10 ´ 3, Þ, AD =, 7, , B

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70, , CBSE Sample Paper Mathematics Standard Class X (Term I), , SAMPLE PAPER 2, MATHEMATICS (Standard), A Highly Simulated Practice Questions Paper, for CBSE Class X (Term I) Examination, , Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions., There is no negative marking., Maximum Marks : 40, Time allowed : 90 minutes, , Roll No., , Section A, Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 1. The smallest number by which 27 should be multiplied so as to get a rational, number is, (a) 27, , (b) 3 3, , (c) 3, , (d) 3, , 2. A quadratic polynomial can have at most ______ zeroes., (a) 0, (c) 2, , (b) 1, (d) infinite, , 3. The point of intersection of the coordinate axes is, (a) X-axis, (c) origin, , (b) Y-axis, (d) (1, 2), , 4. In a DABC, it is given that AB = 3 cm, AC = 2 cm and AD is the bisector of ÐA., , m, , A, , B, , (a) 3 : 4, , (b) 9 : 16, , m, , Ö3, c, , 2c, , SAMPLE PAPER 2, , Then, BD : DC =, , D, , C, , (c) 4 : 3, , (d) 3 : 2

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71, , CBSE Sample Paper Mathematics Standard Class X (Term I), , æ x2, , 5. If x = a sec q cos f , y = b sec q sin f and z = c tan q, then çç, æ, z2 ö, (a) çç 1 + 2 ÷÷, c ø, è, , æ, z2, (b) çç 1 - 2, c, è, , 2, èa, æ z2, ö, (c) çç 2 - 1÷÷, èc, ø, , ö, ÷÷, ø, , +, , y2 ö, ÷ is equal to, b 2 ÷ø, z2, (d) 2, c, , 6. A card is selected from a deck of 52 cards. The probability of its being a black face card, is, (a), , 3, 26, , (b), , 3, 13, , (c), , 2, 13, , (d), , 1, 2, , 7. If A = 2n + 13 , B = n + 7, where n is a natural number then HCF of A and B is, (a) 2, , (b) 1, , (c) 3, , (d) 4, , 1, 3, , 8. The sum and product of zeroes of a quadratic polynomial are respectively and - 2., Then the corresponding quadratic polynomial is, (a) 4x 2 + x - 4, (b) x 2 - 4x - 4, (c) 4x 2 - 4x - 1, (d) 3x 2 - x - 6, , 9. In the given figure P(5, - 3) and Q(3, y) are the points of trisection of the line segment, joining A( 4 , 7) and B(1, - 5). Then y equals, A(4, 7), , (a) 2, , P(5, –3), , Q(3, y), , (b) 4, , B(1, – 5), , (c) - 4, , (d) -, , 5, 2, , 10. In the given figure, PQ|| BC, find AQ., B, , C, , 4 cm, , P, , 6 cm, , Q, , 6 cm, A, , (a) 3.5 cm, (c) 9 cm, , (b) 4.5 cm, (d) 9.5 cm, , 11. If x cosq = 1 and tan q = y, then x 2 - y 2 is, (b) - 1, , (c) 3, , (d) 1, , 12. A girl calculates the probability of her winning the first prize in a lottery is 0.08. If 6000, tickets are sold, then the total number of tickets she bought is, (a) 40, (b) 240, (c) 480, , (d) 750, , 13. From the following rational number, which decimal expansion is terminating, is, (a), , 2, 15, , (b), , 11, 160, , (c), , 17, 60, , (d), , 6, 35, , SAMPLE PAPER 2, , (a) 2

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72, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 14. If zeroes a and b of a polynomial x 2 - 7 x + k are such that a - b = 1, then the value of k is, (a) 21, , (b) 12, , (c) 9, , (d) 8, , 15. If the point P(x , y) is a equidistant from L(5, 1) and M(-1, 5), then the relation between x, and y is, (a) 3x = 2 y, , (b) x = y, , 16. In DABC it is given that,, , (c) 2 x = 3y, , (d) 3x = 6y, , AB BD, if ÐB = 60° and ÐC = 60°, then ÐBAD is equal to, =, AC DC, A, , B, , (a) 30°, , C, , D, , (b) 40°, , (c) 45°, , (d) 50°, , 17. If sin A = cos A, 0°< A < 90°, then A is equal to, (a) 30°, , (b) 45°, , (c) 60°, , (d) 90°, , 18. The probability that it will rain tomorrow is 0.3. What is the probability that it will not, rain tomorrow?, (a) 0.3, (b) 0.2, 13, 19., is a, 1250, (a) terminating decimal fraction, (c) upto 2 decimal fraction, sin 60°+ cot 45° - cosec 30°, 20. The value of, is, sec 60° - cos 30° + tan 45°, 4 3 -9, 4 3 +9, (a), (b), 33, 33, , (c) 0.7, , (d) 0.07, , (b) non-terminating decimal fraction, (d) None of these, , (c), , 9 3 -4, 33, , (d), , 9 3 +4, 33, , Section B, Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , SAMPLE PAPER 2, , 21. The product of the HCF and LCM of the smallest prime number and the smallest, composite number is, (a) 2, (b) 4, , (c) 6, , (d) 8, , 22. If a and b are zeroes of the quadratic polynomial 2x 2 + kx + 4 and if a 2 + b 2 = 8, then, k is equal to, (a) ± 4 3, , (b) ± 3 3, , (c) ± 2 3, , 23. If 2x + 3y = 7 and (a + b)x + (2a - b)y = 21 has infinite solutions, then, (a) a = 1, b = 5, (c) a = - 1, b = 5, , (b) a = 5, b = 1, (d) None of these, , (d) ± 3

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73, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 24. The coordinates of one of the points of trisection of the line segment joining the points, P(7 , - 2) and Q(1, - 5) are, 13 1, (a) æç - , - ö÷, (b) (3, 5), 3ø, è 3, , 13, (c) æç - , - 3 ö÷, ø, è 3, , (d) ( - 5, - 3), , 25. A piece of wire 20 cm is bent into the form of an arc of a circle subtending an angle of, 60° at its centre, then the radius of the circle will be (in cm), 30, 40, 50, (b), (c), (a), p, p, p, , (d), , 60, p, , 26. In an isosceles triangle PQR, if PR = QR and PQ 2 = 2PR 2 , then ÐR is, (a) acute angle, (c) right angle, , (b) obtuse angle, (d) None of these, , xcosec 2 30° sec 2 45°, , = tan 2 60° - tan 2 30°, then x is equal to, 8 cos 2 45° sin 2 60°, (a) 1, (b) -1, (c) 2, , 27 If, , (d) 0, , 28. Two dice are thrown simultaneously. Then the number of possible outcomes for getting, the sum from 3 to 10 is, (a) 32, (b) 30, , (c) 34, , (d) 38, , 29. The sum of powers of prime factors of 196 is, (a) 1, , (b) 2, , (c) 4, , (d) 6, , 30. If a and b are the zeroes of the polynomial p(x) = 4 x 2 + 3x + 7, then the value of, (a), , 47, 28, , (b), , - 47, , - 28, , (c), , 28, , 47, , (d), , a b, + is, b a, , 28, 47, , 31. The distance between the points P(2, - 3) and Q(10, y) is 10 then the value of y is, (a) 3, - 9, , (b) 2, 7, , (c) 1, 3, , (d) 3, 9, , 32. In the given figure, if ABCD is a rhombus, then the value of x is, D, , C, x, , 4, , x, , O, , –1, , 5, B, , A, , (a) 3, , (b) 4, , (c) 5, , (d) 6, , (a) 1, , (b), , 3, 4, , (c), , 1, 2, , (d), , 1, 4, , 34. A letter is chosen at random from the English alphabets Find the probability that the, letter chosen succeeds V., 2, 5, (a), (b), 13, 26, , (c), , 1, 26, , (d), , 1, 2, , SAMPLE PAPER 2, , 33. If sin q - cos q = 0, then sin 4 q + cos 4 q is equal to

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74, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 35. Which of the following rational numbers have terminating decimal?, 16, 25, (a) (i) and (ii), (c) (i) and (iii), (i), , (ii), , 5, 18, , 2, 21, (b) (ii) and (iii), (d) (i) and (iv), , (iii), , (iv), , 7, 250, , 36. A quadratic polynomial whose one zero is 5 and product of the zeroes is 0, is, (a) x 2 - 5, , (b) x 2 - 5x, , (c) 5x 2 + 1, , a, è3 ø, then the value of a is, (a) - 4, (b) - 12, , (d) x 2 + 5x, , 37. If Pæç , 4 ö÷ is the mid-point of the line segment joining the points Q(- 6, 5) and R(- 2, 3),, (c) 12, , (d) - 6, , 38. The sides of a triangle are 30, 70 and 80 units. If an altitude is droped upon the side of, length 80 units, the larger segment cut off on this side is, (a) 62 units, (b) 63 units, (c) 64 units, , (d) 65 units, , 39. If x sin q = 1 and cot q = y, then which of the following is correct?, (a) x 2 + y 2 = 1, (c) y 2 - x 2 = 1, , (b) x 2 - y 2 = 1, (d) None of these, , 40. There are five cards in which the numbers are written as nine, ten, jack, queen and king, of hearts. These cards are well shuffled with their face downwards, one card is then, picked up at random. The probability that the drawn card is a king, is, 1, 2, 3, 4, (a), (b), (c), (d), 5, 5, 5, 5, , Section C, Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted., 41-45 are based on Case Study-1., , Case Study 1, , SAMPLE PAPER 2, , A earing is a small piece of jewellery which has a hook/pin at the back side so that it can be, fastened on ears. Designs of some earing are shown below. Observe them carefully., , A, , B, , C, , Design A Earing A is made with platinum wire in the form of a circle with diameter 28 mm., The wire used for making 4 diameters which divide the circle into 8 equal parts., Desion B Earing B is made two colours platinum and silver. Outer parts is made with platinum, The circumference of silver part is 88 mm and the platinum part is 7 mm wide everywhere., Observe the above designs and answer the following questions.

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75, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Refer to Design A, , 41. The total length of platinum wire required is, (a) 180 mm, , (b) 200 mm, , (c) 250 mm, , (d) 280 mm, , (c) 77 mm 2, , (d) 68 mm 2, , (c) 72.50 mm, , (d) 132 mm, , 42. The area of each sector of earing is, (a) 44 mm 2, Refer to Design B, , (b) 52 mm 2, , 43. The circumference of outer part platinum is, (a) 48.49 mm, , (b) 82.20 mm, , 44. The difference of areas of platinum and silver parts is, (a) 245 p mm 2, , (b) 44 p mm 2, , (c) 147 p mm 2, , (d) 64 p mm 2, , 45. A boy is playing with brooch B. He makes revolution with it along its edge. How many, complete revolutions must it take to cover 168 p mm?, (a) 2, (b) 3, (c) 4, , (d) 5, , 46-50 are based on Case Study-2., , Case Study 2, Palak went to a mall with her mother and enjoy rides on the giant wheel and play hoopla (a game, in which you throw a ring on the items kept in stall and if the ring covers any object completely, you get it). The number of times she played hoopla is half the number of times she rides the giant, wheel. If each ride costs ` 3 and a game of hoopla costs ` 4 and she spent ` 20 in the fair., , Based on the given information, give the answer of the following questions, , 46. The representation of given statement algebraically is, (a) x - 2 y = 0 and 3x + 4y = 20, (c) x - 2 y = 0 and 4x + 3y = 20, , (b) x + 2 y = 0 and 3x - 4y = 20, (d) None of these, , 47. Graphically, if the pair of equations intersect at one point, then the pair of equations is, (a) consistent, (c) Consistent or inconsistent, , (b) Inconsistent, (d) None of these, , (a) ( -4, - 2 ), , (b) (4, 3), , (c) (2, 4), , (d) (4, 2), , 49. Intersection points of the line x - 2y = 0 on X and Y-axes are, (a) (2, 0), (0, 1), , (b) (1, 0), (0, 2), , (c) (0, 0), , (d) None of these, , 50. Intersection points of the line 3x + 4 y = 20 on X and Y-axes are, 20, (a) æç , 0 ö÷, (0, 5), è3 ø, , (b) (2, 0), (0, 1), , 20, c) (5, 0), æç 0, ö÷, è 3 ø, , (d) None of these, , SAMPLE PAPER 2, , 48. The intersection point of two lines is

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OMR SHEET, , SP 2, , Roll No., Sub Code., , Student Name, , Instructions, Use black or blue ball point pens and avoid Gel & Fountain pens for filling the OMR sheet., Darken the bubbles completely. Don’t put a tick mark or a cross mark, half-filled or over-filled bubbles will not be read, by the software., Correct, , ✔, , ✗, , Incorrect, , Incorrect, , Incorrect, , Do not write anything on the OMR Sheet., Multiple markings are invalid., , 1, , 18, , 35, , 2, , 19, , 36, , 3, , 20, , 37, , 4, , 21, , 38, , 5, , 22, , 39, , 6, , 23, , 40, , 7, , 24, , 41, , 8, , 25, , 42, , 9, , 26, , 43, , 10, , 27, , 44, , 11, , 28, , 45, , 12, , 29, , 46, , 13, , 30, , 47, , 14, , 31, , 48, , 15, , 32, , 49, , 16, , 33, , 50, , 17, , 34, , Check Your Performance, Total Questions:, Total Correct Questions:, , If Your Score is, , Score Percentage =, , Total Correct Questions, Total Questions, , × 100, , Less than 60%, > Average (Revise the concepts again), Greater than 60% but less than 75% > Good (Do more practice), Above 75%, > Excellent (Keep it on)

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77, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Answers, 1. (c), 11. (d), 21. (d), , 2. (c), 12. (c), 22. (a), , 3. (c), 13. (b), 23. (b), , 4. (d), 14. (b), 24. (c), , 5. (a), 15. (a), 25. (d), , 6. (a), 16. (a), 26. (c), , 7. (b), 17. (b), 27. (a), , 8. (d), 18. (c), 28. (a), , 9. (c), 19. (a), 29. (c), , 10. (c), 20. (a), 30. (b), , 31. (a), 41. (b), , 32. (c), 42. (c), , 33. (c), 43. (d), , 34. (a), 44. (a), , 35. (d), 45. (c), , 36. (b), 46. (a), , 37. (b), 47. (a), , 38. (d), 48. (d), , 39. (b), 49. (c), , 40. (a), 50. (a), , SOLUTIONS, 1. As 27 = 3 ´ 3 ´ 3 = 3 3, So, if we multiply it by 3 it will become, 3 3 ´ 3 =3´3 =9, i.e a rational number., 2. The degree of a quadratic polynomial is 2., There are 2 zeroes of quadratic polynomial., The maximum number of zeroes of a, polynomial is equal to its degree, so quadratic, polynomial can have at most 2 zeroes., 3. Since, X-axis and Y-axis intersect at (0, 0). So,, point of intersection of the coordinate axes is, origin., 4. We know that the bisector of an angle of a, triangle divides the opposite side in the ratio of, the sides containing the angle., \, BD : DC = AB : AC, = 3 :2, , 6. We have, Total number of outcomes are 52., There are 6 black face cards in a deck of, 52 cards., So, favourable number of outcomes are 6., 6, 3, =, \Required probability =, 52 26, 7. We have, A = 2 n + 13 , B = n + 7, Here, we put n = 1, 2 , 3 ,K . Thus, we get, A = 2 ´ 1 + 13 = 15, B = 1 + 7 = 8, A = 2 ´ 2 + 13 = 17, B = 2 + 7 = 9, A = 2 ´ 3 + 13 = 19, B = 3 + 7 = 10, Here, we find that A and B are coprime., Hence, HCF ( A, B) = 1, 1, 8. Sum of zeroes =, 3, Product of zeroes = - 2, \The required polynomial is, k [ x 2 - ( sum of zeroes ) x, + (product of zeroes], 1, ö, æ, 2, = kçx - x -2 ÷, 3, ø, è, , 5. We, have, x, a, y, sec q sin f =, b, z, tan q =, c, , …(i), , sec q cos f =, , and, , …(ii), …(iii), , 9. From the figure PQ = QB, , Squaring and adding Eqs. (i) and (ii), we get, x2 y2, +, = sec2 q cos 2 f + sec2 q sin 2 f, a 2 b2, = sec2 q(cos 2 f + sin 2 f) = sec2 q, [Q cos 2 A + sin 2 A = 1], [Q sec2 A - tan 2 A = 1], , \, , æ x2 y2 ö æ, z2 ö, ç 2 + 2 ÷ = ç1 + 2 ÷, ça, b ÷ø çè, c ÷ø, è, , [from Eq. (iii)], , …(i), , The distance between P and Q, Q and B are, PQ = ( 5 - 3 )2 + ( -3 - y )2, and, , QB = (3 - 1)2 + ( y + 5 )2, , [using distance formula], On putting the values in Eq. (i), we get, ( 5 - 3 )2 + ( - 3 - y )2 = (3 - 1)2 + ( y + 5 )2, Þ, , (2 )2 + (3 + y )2 = (2 )2 + ( y + 5 )2, , Þ, , 4 + 9 + 6 y + y 2 = 4 + y 2 + 25 + 10 y, , Þ, , 6 y + 13 = 29 + 10 y, , Þ, Þ, , 4 y = - 16, y=-4, , SAMPLE PAPER 2, , = 1 + tan 2 q, æ, z2 ö, = çç 1 + 2 ÷÷, c ø, è, , Put k = 3, Required polynomial is 3 x 2 - x - 6

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78, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 10. In DABC, we have, PQ || BC, AQ AP, [by Thales Theorem], Þ, =, QC PB, AQ 6, Þ, =, 6, 4, 6´6, AQ =, = 9 cm, Þ, 4, 11. We have,, x cos q = 1, 1, cos q =, x, , Þ, Þ, , x=, , 1, = sec q and y = tan q, cos q, , Now, x 2 - y 2 = sec2 q - tan 2 q = 1, [Q sec2 q = 1 + tan 2 q ], 12. Given, total number of tickets sold = 6000, Let she bought x tickets., Then, probability of her winning the first prize, x, =, 6000, x, Þ, 0.08 =, 6000, Þ, x = 0.08 ´ 6000 = 480, Hence, she bought 480 tickets., 2, 2, 13. (a), , which is not in the form of, =, 15 3 ´ 5, 2 m ´ 5 n, so it is non terminating., 11, 11, (b), = 5, 160 2 ´ 5, The denominator of is in the form of 2 m ´ 5 n., Hence, its decimal expansion is terminating., 14. Given polynomial is x 2 - 7 x + k and a - b = 1, -b, a, - (- 7), a +b=, Þ, 1, Þ, a + b =7, Product of zeroes,, c, a ×b =, a, Þ, a ×b = k, , SAMPLE PAPER 2, , Sum of zeroes, a + b =, , Now,, Þ, Þ, \, , ( a + b)2 = ( a - b)2 + 4 × a × b, (7 )2 = 1 + 4 k, 49 - 1 = 4 k, 48, k=, = 12, 4, , 15. The distance between P and Lis, PL = ( x - 5 )2 + ( y - 1)2, = x 2 + 25 - 10 x + y 2 + 1 - 2 y, The distance between P and M is, PM = ( x + 1)2 + ( y - 5 )2, = x 2 + 1 + 2 x + y 2 + 25 - 10 y, [given], As, PM = PL Þ PM2 = PL2, 2, \ x + 26 - 10 x - 2 y + y 2, = x 2 + y 2 + 2 x - 10 y + 26, Þ, -12 x = - 8 y, [divide by 4], Þ, 3x = 2 y, AB BD, 16. Given, in DABC ,, =, AC DC, ÐB = 60 ° and ÐC = 60 °, We know that sum of angles of a triangle is, 180°., In DABC, ÐA + ÐB + ÐC = 180 °, Þ, ÐA + 60 ° + 60 ° = 180 °, Þ, ÐA = 180 ° - 120 ° = 60 °, AB BD, Now,, [Given], =, AC DC, Therefore, AD bisects BC., [By angle bisector theorem], 1, Then, ÐBAC = ÐA = 30 °., 2, Hence, the value of ÐBAD is 30°., 17. Given, sin A = cos A, 0 ° < A < 90 °, sin A, Þ, =1, cos A, Þ, tan A = 1, Þ, tan A = tan 45 °, \, A = 45 °, 18. Let A be the event of rain tomorrow., Then,, P( A) = 0.3, We know that,, P( A) + P( A ) = 1, Then, probability that it will not rain tomorrow, = 1 - 0.3 = 0.7, 19. Since, the factors of the denominator 1250 is of, the form 2 1 ´ 5 4 ., 13, is a terminating decimal., \, 1250, 13, 13, 13 ´ 2 3, Now,, = 1, = 4, 4, 1250 2 ´ 5, 2 ´ 54, 104, 104, = 0.0104, = 4 =, 10000, 10

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79, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 20. We have,, , sin 60 °+ cot 45 ° - cosec 30 °, sec60 ° - cos 30 ° + tan 45 °, 3, + 1 -2, 2, =, 3, 2+1, 2, 3 + 2 -4, 3 -2, 2, =, =, 4- 3 +2 6- 3, 2, 3 -2 6 + 3, =, ´, 6- 3 6+ 3, , [Q Condition for infinite solution;, Þ, , 2, 3, 1, =, =, a + b 2a - b 3, , \, , 2, 3, 3, 1, and, =, =, a + b 2a - b, 2a - b 3, , Þ, Þ, , =, , 4 3 -9, 33, , and A and B are the points of trisection., , P(–7, –2), , 22. We have,, a and b are the zeroes of 2 x 2 + kx + 4, then, -k, Coefficient of x, a +b==, 2, Coefficient of x 2, and, , ab =, , Constant term, 4, = =2, Coefficient of x 2 2, , 2:1, , A, , B, , Q(1, –5), , or, or, , æ 1 - 14 -5 - 4 ö, æ 13 - 9 ö, ç, ÷, ç 1 + 2 , 1 + 2 ÷ or A çè - 3 , 3 ÷ø, è, ø, 13, æ - , -3 ö, ÷, ç, ø, è 3, , 25. We have,, l = 20 cm, q = 60 °, Now, we know that, q, l=, ´ 2 pr, 360 °, 60 °, ´ 2 ´ p´ r, 20 =, Þ, 360 °, 20 ´ 6 60, cm, r=, =, Þ, 2p, p, 26. Given, PR = QR and PQ2 = 2 PR 2, Þ, , ( PQ)2 = PR 2 + PR 2, , Þ, , ( PQ)2 = ( QR )2 + ( PR )2, P, , 23. Given, 2 x + 3 y - 7 = 0 and, ( a + b)x + (2 a - b)y - 21 = 0 has infinite, solutions, then, -7, 2, 3, =, =, a + b 2 a - b - 21, , Q, , R, , By converse of pythagoras theorem, DPQR is, right angle triangle at angle R., , SAMPLE PAPER 2, , Now, it is given that, a 2 + b2 = 8, 2, Þ, ( a + b) - 2 ab = 8, 2, æ-kö, Þ, ç, ÷ -2 ´ 2 =8, è 2 ø, k2, Þ, = 12, 4, Þ, k 2 = 48, k =± 4 3, Þ, , 1:2, , The coordinates of point A are, æ 1 ´ 1 + 2 ´ ( -7 ) 1 ´ ( -5 ) + 2 ( -2 ) ö, ç, ÷, ,, ç, ÷, 1+2, 1+2, è, ø, , 21. Smallest prime number = 2, Smallest composite number = 4, Since, Product of HCF and LCM, = Product of numbers, \HCF ´ LCM = 2 ´ 4 = 8, , a = 5, b = 1, , 24. Let P(7 , - 2 ) and Q(1, - 5 ) be the given points, , [Q ( a + b)( a - b) = a2 - b2 ], 6 3 + 3 - 12 - 2 3, 36 - 3, , 4 a - 2 b = 3 a + 3 b and 2 a - b = 9, a = 5 b and 2 a - b = 9, , \, , [rationalise the denominator], ( 3 - 2 )(6 + 3 ), =, 6 2 - ( 3 )2, , =, , a1 b1 c1 ù, =, =, a2 b2 c2 úû

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80, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 27. We have,, xcosec2 30 ° sec2 45 °, = tan 2 60 ° - tan 2 30 °, 8 cos 2 45 ° sin 2 60 °, 2, x(2 )2 ( 2 )2, 1 ö, = ( 3 )2 - æç, Þ, ÷, 2, è 3ø, 1 öæ 3 ö, ÷÷, 8æç, ÷çç, è 2 øè 2 ø, 1 8x 8, 8x, Þ, =3 - Þ, =, 3, 3, 3 3, 3 8, Þ, x = ´ =1, 8 3, 28. Number of possible outcomes to get the sum as, 3 = 2 {i.e. (2 , 1), (1, 2 )}, Number of possible outcomes to get the sum as, 4 = 3 {i.e. (2 , 2 ), (1, 3 ), (3 , 1)}, Number of possible outcomes to get the sum as, 5 = 4 {i.e. (3 , 2 ), (2 , 3 ), ( 4 , 1), (1, 4 )}, Number of possible outcomes to get the sum as, 6 = 5 {i.e. ( 5 , 1), (1, 5 ), (3 , 3 ), ( 4 , 2 ), (2 , 4 )}, Number of possible outcomes to get the sum as, 7 = 6 {i.e. ( 4 , 3 ), (3 , 4 ), (6 , 1), (1, 6 ), ( 5 , 2 ), (2 , 5 )}, Number of possible outcomes to get the sum as, 8 = 5 {i.e. ( 4 , 4 ), (6 , 2 ), (2 , 6 ), ( 5 , 3 ), (3 , 5 )}, Number of possible outcomes to get the sum as, 9 = 4 {i.e. (6 , 3 ), (3 , 6 ), ( 5 , 4 ), ( 4 , 5 )}, Number of possible outcomes to get the sum as, 10 = 3 {i.e. ( 5 , 6 ), (6 , 4 ), ( 4 , 6 )}, \ The total posible outcomes, = 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 = 32, 29. We have,, , SAMPLE PAPER 2, , 2, 2, 7, 7, , 196, 98, 49, 7, 1, , \ 196 = 2 2 ´ 7 2, So, sum of powers of prime factors = 2 + 2 = 4, 30. We have, a and b are the zeroes of the, polynomial p( x ) = 4 x 2 + 3 x + 7, then, -3, 7, and ab =, a +b=, 4, 4, a b a 2 + b2, Now,, + =, b a, ab, ( a + b)2 - 2 ab, =, ab, 2, , æ -3 ö, æ7ö, ç, ÷ - 2ç ÷, 4 ø, è4ø, =è, 7, 4, , æ 9 - 14 ö, ÷, ç, = è 16 4 ø, æ7ö, ç ÷, è4ø, æ 9 - 56 ö, ÷, ç, 16 ø, =è, æ7ö, ç ÷, è4ø, - 47, =, 28, 31. Given, PQ = 10, [squaring both side], PQ2 = (10 )2, = 100, From the distance formula, , Þ, , PQ = (10 - 2 )2 + ( y + 3 )2, On squaring both sides, we get, (10 - 2 )2 + ( y + 3 )2 = 100 [Q PQ2 = 100 ], Þ, , ( y + 3 )2 = 100 - 64 = 36, , Þ, , y 2 + 9 + 6 y = 36, , Þ, , y 2 + 6 y - 27 = 0, , Þ, , y 2 + 9 y - 3 y - 27 = 0, , Þ, Þ, , y (y + 9) - 3 (y + 9) = 0, (y - 3) (y + 9) = 0, , Þ, , y = 3, y = - 9, , 32. Since ABCD is a rhombus, then, OA OB, =, OC OD, [Q Diagonals of rhombus bisect each other], 4, 5, =, Þ, x -1 x, Þ, Þ, , 4x = 5x - 5, x=5, , 33. We have,, sin q - cos q = 0, Þ, sin q = cos q, sin q, Þ, =1, cos q, Þ, , tan q = 1, , éQ tan q = sin q ù, êë, cos q úû, [Q tan 45 ° = 1], , Þ, tan q = tan 45 °, Þ, q = 45 °, 4, 4, Now, sin q + cos q = sin 4 45 ° + cos 4 45 °, 4, 4, 1 ö, æ 1 ö, = æç, ÷, ÷ +ç, è 2ø, è 2ø, 1 1 1, = + =, 4 4 2

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81, , CBSE Sample Paper Mathematics Standard Class X (Term I), , A, , 34. Total number of outcomes = 26, Letter succeeding V are W , X , Y , Z., Then, total number of favourable outcomes = 4, \Required probability, Number of favourable outcomes, =, Total number of possible outcomes, 4, 2, =, Þ Required probability =, 26 13, 35. (i), , 5, 5, =, 18 2 ´ 3 2, , (iii), , 2, 2, =, 21 3 ´ 7, , B, , 70 units, , x units, (80 – x) D, units, 80 units, , C, , Now, in DABD, h 2 = 30 2 - (80 - x )2, [use pythagoras theorem], , 7, 7, (iv), =, 250 2 ´ 5 3, Hence (i) and (iv) having terminating decimal, because denominators are the form of 2 m ´ 5 n ., 36. Let a and b are the zeroes of quadratic, polynomial., According to the question,, a =5, a ×b = 0, Þ, b =0, Now, required quadratic polynomial, = x 2 - ( a + b)x + a × b, = x2 - (5 + 0) x + 0, = x2 - 5x, a, 37. We have, Pæç , 4 ö÷ is the mid-point of Q( - 6 , 5 ), è3 ø, and R( - 2 , 3 ), then, æ a , 4ö =æ -6 -2 , 5 + 3 ö, ÷, ÷ ç, ç, 2 ø, è3 ø è 2, [using section formula], Þ, , h units, , Let us assume AB = 30 units, AC = 70 units and, BC = 80 units., Let h is the height of altitude which cuts BC, into two parts., Let us asume the first part be x so that the, second part will be (80 - x )., , 16, 16, =, 25 ( 5 )2 ´ 2 0, , (ii), , 30 units, , æ a , 4 ö = (- 4, 4), ÷, ç, è3 ø, , 38. Given, a triangle with sides 30, 70 and, 80 units., Altitude is droped upon the side of length, 80 units., , h 2 = 70 2 - x 2, 2, , Þ, 30 - (80 - x )2 = 70 2 - x 2, Þ 900 - (6400 + x 2 - 160 x ) = 4900 - x 2, Þ, 160 x = 10400, Þ, x = 65 units, 39. We have,, , x sin q = 1, , 1, sin q, Þ, x = cosec q, and, y = cot q, Now, we know that, cosec2 q - cot 2 q = 1, Þ, x2 - y2 = 1, x=, , Þ, , 40. Total number of cards = 5, Total number of king = 1, \ The required probability, Number of favourable outcomes, =, Total number of outcomes, 1, =, 5, Solutions (41-45), 41. Let r be the radius and d be the diameter., Here, total length of platinum wire used, = 2 pr + 4 d, 22, éQ r = 28 = 14 ù, =2 ´, ´ 14 + 4 ´ 28, úû, êë, 2, 7, = 88 + 112, = 200 mm, , SAMPLE PAPER 2, , On comparing the x-coordinate, we get, a, \, =-4, 3, Þ, a = - 12, , Also, in DACD,

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82, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 28, = 14 mm, 2, Since, circle is divided into 8 sectors., 360 °, Therefore, angle of each is, = 45 °., 8, 45 ° 22, Area of each sector =, ´, ´ 14 ´ 14, 360 ° 7, 1, = ´ 22 ´ 2 ´ 14, 8, , 42. Here, the radius of the sector =, , = 77 mm 2, 43. Let r be the radius of the silver part., Given, circumference of silver part = 88 mm, \, 2 pr = 88, 88 ´ 7, = 14 mm, r=, 2 ´ 22, Radius of outer part,, R = 7 + 14 = 21 mm, Circumference of outer part, = 2 pR, 22, =2 ´, ´ 21, 7, = 132 mm, 44. Let R be the radius of the platinum part and, r be the radius of the silver part., Then, R = 21 and r = 14, \Required difference of area of platinum and, silver parts = pR 2 - pr2, = p (21)2 - p (14 )2, = p ( 441 - 196 ), = 245 p mm 2, 45. Circumference of the brooch, B = 2 p´ 21 = 42 p mm, Number of revolutions = 168 p ¸ 42 p = 4, Solutions (46-50), , SAMPLE PAPER 2, , 46. Let x be the number of rides on the giant wheel, and y be the numebr of hoopla played by, Palak., , x, and 3 x + 4 y = 20, 2, Rewrite the above equations to represent, algebraically, …(i), x -2 y = 0, Then, y =, , 3 x + 4 y = 20, , …(ii), , 47. If the pair of equations intersect at only one, point. Then the pair of equations has a unique, solution and hence consistent., 48. From Eq. (i), x = 2 y, Put x = 2 y in Eq. (ii), we get, 3(2 y ) + 4 y = 20, Þ, , 10 y = 20, , Þ, , y =2, , Þ, x =2 ´2 = 4, Hence, the intersection point is (4, 2)., 49. Put y = 0 in the given equation., x -0 = 0 Þ x = 0, \ The point is (0, 0)., Put x = 0 in the given equation, x - 2 y = 0,, 0 -2y =0, Þ, y =0, \ The point is (0, 0)., Hence, we conclude that given line intersect X, and Y axes at only (0, 0)., 50. Put y = 0 in the given equation, we get, 3 x + 0 = 20, 20, x=, 3, 20, \The point is æç , 0 ö÷., è 3 ø, Put x = 0 in the given equation, we get, 0 + 4 y = 20, Þ, y=5, \ The point is (0, 5).

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83, , CBSE Sample Paper Mathematics Standard Class X (Term I), , SAMPLE PAPER 3, MATHEMATICS (Standard), A Highly Simulated Practice Questions Paper, for CBSE Class X (Term I) Examination, , Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions., There is no negative marking., Maximum Marks : 40, Time allowed : 90 minutes, , Roll No., , Section A, Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., 1. Given figure shows the graph of the polynomial f ( x) = ax 2 + bx + c, Y, , f(x) = ax2 + bx +c P, X¢, O, , X, – b, –D, 2a 4a, , Y¢, , (a) a < 0, b < 0 and c > 0, (c) a < 0, b < 0 and c < 0, , (b) a > 0, b < 0 and c > 0, (d) a > 0, b > 0 and c > 0, , 2. The sum of the numerator and denominator of a fraction, , x, is 8. If the denominator is, y, , 1, , 0 < a, b< 90°, then the value of cot(a + b) is, 3, 1, (b) 0, (c), (d) 1, 3, , 3. If tan a = 3 and tan b =, (a) 3, , SAMPLE PAPER 3, , 1, increased by 1, the fraction becomes . The situation can be represented algebraically as, 2, x +1 1, x, 1, (b) x + y = 8 and + 1 =, (a) x + y = 8 and, =, y, y, 2, 2, x, x, 1, x, 1, (c) = 8 and, (d) x + y = 8 and, =, =, y, y +1 2, y +1 2

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84, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 4. The distance of the point (- 3, 8) from the X-axis is, (a) 3 units, , (b) - 3 units, , (c) 8 units, , 5. The decimal expansion of the rational number, places of decimal?, (a) 1, , (b) 2, , 53, 2453, , (d) 5 units, , will terminate after how many, , (c) 3, , (d) 4, , 6. The perimeter of a square circumscribing a circle of radius a cm is, A, , F, , D, , G, , B, , O, , H, , C, , E, , (a) 8a cm, (c) 2 a cm, , (b) 4a cm, (d) 16a cm, , 7. A box contains cards numbered 6 to 50. A card is drawn at random from the box, the, probability that the drawn can has a number, which is a perfect square, is, 1, 2, (a), (b), 9, 3, 1, 7, (d), (c), 6, 18, , 8. A boy walks 8 m due East and 6 m due South. How far is he from the starting point?, (a) 31 m, , (b) 26 m, , (c) 62 m, , (d) 10 m, , 9. The graph of x 2 + 1 = 0, (a) Intersecting X-axis at two distinct points, (b) Touches X-axis at a point, (c) Neither touches nor intersect X-axis, (d) Either touches or intersect X-axis, , 10. If the pair of linear equations 3x + y = 3 and 6x + ky = 8 does not have a solution, then, , SAMPLE PAPER 3, , the value of k is, (a) 2, , (b) -3, , (c) 0, , (d) 1, , 11. In DPQR, if PS is the internal bisector of ÐP meeting QR at S and PQ = 13 cm,, QS = ( 3 + x) cm, SR = ( x - 3) cm and PR = 7 cm, then find the value of x., (a) 9 cm, (b) 10 cm, (c) 13 cm, (d) 12 cm, , 12. Two concentric circles form a ring. The inner and outer circumference of the ring are, 2, 3, m and 75 m respectively. Find the width of the ring., 7, 7, (a) 1 m, (b) 2 m, (c) 3 m, 50, , (d) 4 m

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85, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 13. In the given figure, line BD||CE. If AB = 1.5 cm, BC = 6 cm and AD = 2 cm. Find DE., E, , D, , m, 2c, , A, , 1.5 cm B, , C, , 6 cm, , (a) 6 cm, (c) 4 cm, , (b) 8 cm, (d) None of these, , 14. A child has a die whose six faces show the number as given below., 1, , 2, , 3, , 2, , 4, , 6, , The die is thrown once the probability of getting 5 is, 1, 2, (b), (c) 0, (a), 6, 5, , (d) 1, , 15. Which of the following is not a polynomial?, (a) 2 x -3 - 5 + 3x -1, , (b) x 3 + 2 x - 9, (2 x + 10), (d), ´ ( x 2 - 25), x+5, , (c) ( x - 2 ) 2 + 3x, , 16. The age of a daughter is one third the age of her mother. If the present age of mother is, x yr, then the age (in yr) of the daughter after 15 yr will be, x + 15, x, (a) + 15, (b), (c) x + 5, 3, 3, , (d), , x, -5, 3, , 17. Check the relation between the following triangles, P, , A, 83°, , 58°, , C, 39°, , R, , Q, , (a) similar by SAS, (c) similar by SSS, , 58°, B, , (b) similar by AAA, (d) similar by ASS, , (a) 2 19, , (b) 2 20, , (c) 2, , (d) 2 39, , 19. Evaluate 8 3 cosec 2 30° sin 60° cos 60° cos 2 45° sin 45° tan 30° cosec 3 45°., (a) 8, , (b) 4 3, , (c) 8 3, , (d) 16 3, , 20. The diameter of the wheel of a bus is 1.4 m. The wheel makes 10 revolutions is 5 s., The speed of the vehicle (in km/h) is, (a) 30, (b) 31, , (c) 31.68, , (d) 35, , SAMPLE PAPER 3, , 18. If sin x + cosec x = 2, then sin 19 x + cosec 20 x is equal to

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86, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Section B, Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., a b, 21. If a and b are the zeroes of a polynomial f (x) = px 2 + qx + r. Then, + + 2 is equal to, b a, q2, q2, q2, (b) 2, (c) 2, (d) None of these, (a), rp, r p, p r, , 22. If HCF (306, 657) = 9, what will be the LCM (306, 657)?, (a) 12338, , (b) 22338, , (c) 23388, , (d) 22388, , 23. The area of the triangle formed by the lines 2x + 3y = 12, x - y - 1 = 0 and X-axis, (as shown in figure), is, 2x + 3y – 12 = 0 Y, B(0, 4), D, X¢, , O, , (3, 2), , C(1, 0) (6, 0)A, , X, , E(0, –1), x–y–1=0, , (a) 7 sq units, , Y¢, , (b) 5 sq units, , (c) 6.5 sq units, , (d) 6 sq units, , 24. In the following figure, LM||BC and LN||CD, then which of the following relation, is true?, B, , M, L, , A, , C, , N, D, , (a), , AM AN, =, AB AD, , (b), , ML AL, =, BC AC, , (c) Both (a) and (b), , (d) None of these, , SAMPLE PAPER 3, , 25. From the given figure, the value of 25 (sin 2 q + 2 cos 2 q - tan q) is, P, , 10, , R, , cm, , 8 cm, , q, Q, , 6 cm, , (a), , 2, 3, , (b) -, , 2, 3, , (c), , 3, 2, , (d) -, , 3, 2

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87, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 26. If the radius of a circle is diminished by 10%, then its area is diminished by ………, (a) 29%, , (b) 19%, , (c) 15%, , (d) 9%, , 27. If DBAC is triangle with ÐA = 90°. From A, a perpendicular AD is drawn on BC. Which, one of the following is correct?, (a) Only DABC ~ DDAC, (c) Only DABC ~ DDBA, , (b) Only DDAC ~ DDBA, (d) All of these, , 28. A card is drawn from a pack of cards numbered 2 to 53. The probability that the, number of the card is a prime number less than 20 is, 2, 4, 5, (b), (c), (a), 13, 13, 13, , (d), , 8, 13, , 29. For what value of k, does - 4 is a zero of the polynomial x 2 - x - (2k + 2) ?, (a) 7, , (b) 8, , (c) 9, (d) 10, x, 7, 30. The probability of passing a certain test is . If the probability of not passing is , then, 24, 8, x is equal to, (a) 2, (b) 3, (c) 4, (d) 6, , 31. To place a pole vertical on the ground a guy attach a wire of length 26 m to it at a point, 10 m away from its foot. Then, the length of pole will be, (a) 10 m, (b) 28 m, (c) 20 m, , (d) 24 m, , 32. A two-digit number, where ten’s digit is greater than ones digit is obtained by either, multiplying sum of the digits by 8 and adding 1 or by multiplying the difference of, digits by 13 and adding 2. The number is, (a) 14, (b) 51, (c) 41, (d) 49, , 33. Two players Sania and Deepika play a tennis match. If the probability of Sania winning, the match is 0.68, then the probability of Deepika winning the match is, (a) 0.32, (b) 0.38, (c) 0.42, (d) 0.48, , 34. In an equilateral triangle DABC, G is the centroid. Each side of the triangle is 6 cm. The, length of AG is, (a) 2 2 cm, , (b) 3 2 cm, , (c) 2 3 cm, , 1, 1 öæ 1, 1 ö, +, ÷ç, ÷ is, è cos q cot q ø è cos q cot q ø, (b) -1, (c) 1, , (d) 3 3 cm, , 35. The value of æç, (a) 0, , (d) 2, , 36. The area of a circular field is 13.86 hectares, then the cost of fencing at the rate of, , (a) 2, , (b) 6, , (c) ` 5600, , (d) ` 5500, , (c) 4, , (d) 5, , 38. If the point C (k , 4) divides the join of points A(2, 6) and B(5, 1) in the ratio 1 : 3, then the, value of k is, (a) 11, , (b), , 29, 4, , (c), , 11, 4, , (d), , 9, 4, , SAMPLE PAPER 3, , ` 4.40 per metre is, (a) ` 5800, (b) ` 5808, tan 30° + cot 60°, 37. The value of, is, tan 30° (sin 30° + cos 60°)

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88, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 39. In the figure ABDCA represents a quadrant of a circle of radius 7 cm with centre A., Find the area of the shaded portion., C, , D, , 2 cm, , E, , A, , B, , 7 cm, , (a) 14 cm 2, (c) 38.5 cm 2, , (b) 24.5 cm 2, (d) 31.5 cm 2, , 1, 2, , 5, 4, , 40. If is a zeroes of the polynomial x 2 + kx - , then the sum of the zeroes is, (a) 2, , (b) - 2, , (c), , 1, 2, , (d) -, , 1, 2, , Section C, Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted., 41-45 are based on Case Study-1., , Case Study 1, A vegetable seller has 420 potatoes and 130 tomatoes. He wants to stack them in such a way that, each stack has same number and they take up the least area of the tray., , Based on the above information of a vegetable shop, answer the following questions, , 41. The total number of vegetables are, (a) 420, , (b) 130, , (c) 550, , (d) 290, , SAMPLE PAPER 3, , 42. The product of exponents of the prime factors of total number of vegetables is, (a) 2, , (b) 3, , (c) 5, , (d) 6, , 43. What is the number of vegetables that can be placed in each stack for this purpose?, (a) 45, , (b) 40, , (c) 10, , (d) 35, , 44. The sum of exponents of the prime factors of the number of vegetables that can be, placed in each stack for this purpose is, (a) 5, (b) 2, , (c) 4, , (d) 6, , 45. What is the total number of rows in which they can be placed?, (a) 15, , (b) 25, , (c) 35, , (d) 55

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89, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 46-50 are based on Case Study-2., , Case Study 2, Class X students of a school in Gomtinagar have been alloted a rectangular plot of a land for, gardening activity. Sapling of roses are planted on the boundary at a distance of 1m from each, other. There is a triangular grassy fountain in the plot as shown in the figure. The students are to, sow seeds of flowering plants on the remaining area of the plot., B, , C, P, , 6, 5, 4, 3, 2, 1, , R, , Q, A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15, , D, , Considering A as origin answer the following questions, , 46. What are the coordinates of P?, (a) (4, 6), (c) (4, 5), , (b) (6, 4), (d) (5, 4), , 47. What are the coordinates of R?, (a) (6, 5), (c) (6, 0), , (b) (5, 6), (d) (7, 4), , 48. The distance between points Q and R is, (a) 13 units, (c) 13 units, , (b) 2 3 units, (d) 10 3 units, , 49. DPQR is a/an, (a) right angled triangle, (c) scalene triangle, , (b) equilateral triangle, (d) None of these, , 50. The centroid of DPQR is, 13 13, (a) æç , ö÷, è2 2 ø, (c) (13, 13), , 13 13, (b) æç , ö÷, è3 3 ø, (d) None of these, , SAMPLE PAPER 3

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OMR SHEET, , SP 3, , Roll No., Sub Code., , Student Name, , Instructions, Use black or blue ball point pens and avoid Gel & Fountain pens for filling the OMR sheet., Darken the bubbles completely. Don’t put a tick mark or a cross mark, half-filled or over-filled bubbles will not be read, by the software., Correct, , ✔, , ✗, , Incorrect, , Incorrect, , Incorrect, , Do not write anything on the OMR Sheet., Multiple markings are invalid., , 1, , 18, , 35, , 2, , 19, , 36, , 3, , 20, , 37, , 4, , 21, , 38, , 5, , 22, , 39, , 6, , 23, , 40, , 7, , 24, , 41, , 8, , 25, , 42, , 9, , 26, , 43, , 10, , 27, , 44, , 11, , 28, , 45, , 12, , 29, , 46, , 13, , 30, , 47, , 14, , 31, , 48, , 15, , 32, , 49, , 16, , 33, , 50, , 17, , 34, , Check Your Performance, Total Questions:, Total Correct Questions:, , If Your Score is, , Score Percentage =, , Total Correct Questions, Total Questions, , × 100, , Less than 60%, > Average (Revise the concepts again), Greater than 60% but less than 75% > Good (Do more practice), Above 75%, > Excellent (Keep it on)

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91, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Answers, 1. (b), 11. (b), , 2. (d), 12. (d), , 3. (b), 13. (b), , 4. (c), 14. (c), , 5. (d), 15. (a), , 6. (a), 16. (a), , 7. (a), 17. (b), , 8. (d), 18. (c), , 9. (c), 19. (c), , 10. (a), 20. (c), , 21. (a), 31. (d), 41. (c), , 22. (b), 32. (c), 42. (a), , 23. (b), 33. (a), 43. (c), , 24. (c), 34. (c), 44. (b), , 25. (a), 35. (c), 45. (d), , 26. (b), 36. (b), 46. (a), , 27. (d), 37. (a), 47. (a), , 28. (a), 38. (c), 48. (c), , 29. (c), 39. (d), 49. (c), , 30. (b), 40. (b), 50. (b), , SOLUTIONS, 1. From the figure, the graph of polynomial p( x ), is a parabola open upwards. Therefore, a > 0,, y = ax 2 + bx + c, cuts Y-axis at P., On putting x = 0 in y = ax 2 + bx + c, we get, y=c, Hence, the coordinates of P are (0 , y ) or (0 , c). It is, clear that P lies on positive Y-axis, therefore, c >0., Also, x-coordinate of vertex should be positive, b, i.e. >0, 2a, Since, a > 0 therefore b should be negative, i.e. b < 0., 2. We have numerator and denominator of the, fraction be x and y respectively., Sum of numerator and denominator = 8, i.e., x + y =8, Denominator is increased by 1, i.e New, denominator = y + 1, x, 1, \ New fraction is, =, y +1 2, 3. We have, tan a = 3, Þ, a = 60 °, 1, Again,, tanb =, 3, , [Q tan60 ° = 3 ] …(i), , 7. Given numbers are 6, 7, 8, 9, ……, 50., Total number of possible outcomes, = 50 - 6 + 1 = 45, Favourable outcomes = Perfect square, numbers between 6 to 50 i.e. 9 , 16, 25, 36, 49, = 32 , 42 , 52 , 62 , 72, \Total number of favourable outcomes = 5, So, the required probability, Number of Favourable Outcomes, =, Total Number of Possible Outcomes, 5 1, =, =, 45 9, 8. Let A be the starting point., A, , 8m, , Thus, the rational number terminates after, 4 decimal places., , N, W, , b = 30 °, , The distance of any point ( x , y ) from X-axis is, equal to absolute value of y-coordinate., \ Distance ( - 3 , 8 ) from X-axis = 8 units., 53 ´ 5, 53, 5. 4 3 = 4, 2 5, 2 ´ 54, 265, 265, = 0.0265, =, =, (2 ´ 5 )4 10 4, , B, , E, , 6m, S, C, , In DABC,, , ÐB = 90 °, AC 2 = AB2 + BC 2, [by pythagoras theorem], , Þ, , AC 2 = 8 2 + 6 2, = 64 + 36 = 100 = 10 2, AC = 10 m, , Þ, 9. We have,, , x2 + 1 = 0, , x2 = - 1, It means, x cannot be real., , SAMPLE PAPER 3, , éQ tan30 ° = 1 ù …(ii), 3 úû, ëê, Adding Eqs. (i) and (ii), we get, a + b = 60 ° + 30 ° = 90 °, \ cot( a + b) = cot 90 °, =0, [Q cot 90 ° = 0 ], 4. The given point is ( - 3 , 8 )., \, , 6. Let ABCD be the square that inscribes a circle, with centre O, touching it at the points E, F , G, and H., Given, the radius of the circle = a cm, Thus, OE = OF = OG = OH = a cm, Hence, OG + OE = AD = BC, and OF + OH = AB = DC, So, each side of the square = 2 a cm, Hence, perimeter of the square = 4 (2 a) = 8 a cm

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92, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Hence, the graph is neither touches nor, intersect X-axis., 10. Given, 3 x + y = 3 and 6 x + ky = 8, , Þ, , On comparison the above equation with, standard equation, we get, a1 = 3, b1 = 1 and c1 = - 3, and, a2 = 6, b2 = k and c2 = - 8, Since, given equations have no solution., a1 b1 c1, =, ¹, \, a2 b2 c2, 3 1, Þ, = Þ k =2, 6 k, 11. Since, PS is the internal bisector of ÐP and it, meets QR at S., , 13, , cm, , P, , Q, , \, Þ, Þ, Þ, Þ, Þ, , 7, , cm, , (3 + x) cm S (x – 3) cm, , R, , PQ QS, =, PR RS, 13 3 + x, =, 7, x -3, 13( x - 3 ) = 7 (3 + x ), 13 x - 39 = 21 + 7 x, 6 x = 60, x = 10 cm, , 12. Let the inner and outer radii be r and R., 2 352, Then, 2 pr = 50 =, 7, 7, 352, 7, 1, æ, r =ç, ´, ´ ö÷ = 8 m, Þ, 22 2 ø, è 7, 3 528, and, 2 pR = 75 =, 7, 7, 528, 7, 1, Þ, R = æç, ´, ´ ö÷ = 12 m, 22 2 ø, è 7, \Width of the ring = ( R - r) = (12 - 8 ) m = 4 m, , SAMPLE PAPER 3, , \, , 13. In DADB and DAEC,, ÐADB = ÐAEC, [corresponding angles as BD || CE], ÐABD = ÐACE, [corresponding angles as BD || CE], [common], ÐA = ÐA, \, DADB ~ DAEC, [by AAA similarity criterion], AD AB, [by CPCT], =, Þ, AE AC, , Þ, Þ, Þ, , 2, 1.5, =, 2 + DE 1.5 + 6, 2, 1.5, =, 2 + DE 7.5, 2, 1, =, 2 + DE 5, 2 + DE = 10, DE = 8 cm, , 14. On the given dice, 5 is not printed. So, its, impossible event. Probability of impossible, event is 0., 15. A polynomial is an algebraic expression, containing two or more terms. Also, it requires, the exponents of the variables in the each term, should be positive integer., (a) 2 x - 3 - 5 + 3 x - 1 , it is not polynomial,, because it has negative power in second term., (b) x 3 + 2 x - 9, it is a polynomial., (c) ( x - 2 )2 + 3 x = x 2 + 4 - 4 x + 3 x, = x 2 - x + 4, it is a polynomial., 2 x + 10, 2 (x + 5), (d), ´ ( x 2 - 25 ), ´ ( x 2 - 25 ) =, x+5, (x + 5), = 2 x 2 - 50, it is a polynomial., 16. It is given, present age of mother = x yr, Since age of daughter is one third the age of, mother., x, \ Present age of daughter = yr, 3, x, After 15 yr age of the daughter = æç + 15 ö÷ yr, ø, è3, 17. In DABC,, ÐA + ÐB + ÐC = 180 °, [sum of all angles of a triangle is 180°], 83 ° + 58 ° + ÐC = 180 °, Þ, ÐC = 180 ° - (83 ° + 58 ° ), Þ, ÐC = 39 °, Now, In DPQR,, ÐP + ÐQ + ÐR = 180 °, Þ, 58 ° + ÐQ + 39 ° = 180 °, Þ, ÐQ = 180 ° - ( 58 ° + 39 ° ), Þ, ÐQ = 83 °, Now, In DABC and DQPR,, ÐA = ÐQ = 83 °, ÐB = ÐP = 58 °, ÐC = ÐR = 39 °, \, DABC ~ DQPR, [by AAA similarity criterion], 18. We have, sin x + cosec x = 2

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93, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Þ, Þ, Þ, Þ, Þ, \, , 1, =2, sin x, sin 2 x + 1 = 2 sin x, (sin x - 1)2 = 0, sin x = 1, cosec x = 1, sin19 x + cosec20 x = (1)19 + (1)20, =1 + 1 =2, sin x +, , 19. We have, 8 3 cosec2 30 ° sin 60 ° cos 60 °, cos 2 45 ° sin 45 ° tan 30 ° cosec3 45 °, æ 3 ö æ 1 ö æ 1 ö2 æ 1 ö æ 1 ö, 3, ÷÷ ç ÷ ç, = 8 3 (2 )2 çç, ÷ ç, ÷ç, ÷( 2), 2, 2, 2, 2, 3, ø, è, è, ø, è, ø, è, ø, ø, è, =8 3 ´ 4´, , 3 1 1, 1, 1, ´ ´ ´, ´, ´ 2 2 =8 3, 2, 2 2, 2, 3, , 20. The distance covered in one revolution is equal, to the circumference of the wheel., Circumference of the wheel = p ´ d = p´ 1.4 m, 22, =, ´ 1.4 = 4.4 m, 7, \ Distance covered in 10 revolutions, = 10 ´ 4.4 m = 44 m, Distance traveled 44, m/s, Speed =, \, =, Total time, 5, 44 60 ´ 60, . km/h, =, ´, = 3168, 5, 1000, 21. Given, quadratic polynomial is px 2 + qx + r., Since, a and b are the zeroes of quadratic, polynomial., -q, r, and a ×b =, \, a +b=, p, p, Now,, , =, 24. In DABC,, , ML|| BC, \ By basic proportionality theorem, we have, AM AL, …(i), =, AB AC, Again, in DADC,, NL|| DC, \ By basic proportionality theorem, we have, AN AL, …(ii), =, AD AC, From Eqs. (i) and (ii), we get, AM AN, =, AB AD, Again, in DAMLand DABC, [corresponding angles], ÐAML = ÐABC, [corresponding angles], ÐALM = ÐACB, \ By AA similar criterion,, DAML ~ DABC, \ By CPCT, we have, AM ML, …(iii), =, AB BC, From Eqs. (i) and (iii), we have, ML AL, =, BC AC, 25. In DPQR, ÐQ = 90 °, PQ = 8 cm, QR = 6 cm, and PR = 10 cm, \, , a 2 + b2 + 2ab (a + b)2, a b, =, + +2=, ab, ab, b a, , 23. From the figure, the required triangle is DACD., Here, base of triangle = 6 - 1 = 5, Height of triangle = 2, We know that,, 1, Area of DACD = ´ base × height, 2, , Perpendicular PQ 8 4, =, =, =, Hypotenuse, PR 10 5, , Base, QR 6 3, =, =, =, Hypotenuse PR 10 5, Perpendicular PQ 8 4, tan q =, =, = =, Base, QR 6 3, , and, , \ 25(sin 2 q + 2 cos 2 q - tan q ), 2, é 4 2, 3, 4ù, = 25 êæç ö÷ + 2 ´ æç ö÷ - ú, 3û, è5ø, ëè 5 ø, 9 4ù, 16, = 25 é + 2 ´, êë25, 25 3 úû, é 48 + 54 - 100 ù, = 25, úû, êë, 25 ´ 3, = 25, , é102 - 100 ù 2, =, êë 25 ´ 3 úû 3, , 26. Let radius of the circle be x., Given that the radius is diminished by 10%., x 9x, So, new radius = x - =, 10 10, , SAMPLE PAPER 3, , 22. We know that,, LCM ´ HCF = Product of two numbers, \ LCM (306, 657) ´ HCF (306, 657) = 306 ´ 657, LCM ´ 9 = 306 ´ 657, Þ, LCM = 22338, Þ, , sin q =, , cos q =, , 2, , æ qö, ç- ÷, 2, 2, ç p÷, ø =q ´ p=q, =è, 2, r pr, p, æ rö, ç ÷, ç p÷, è ø, , 1, ´ 5 ´ 2 = 5 sq units, 2

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94, , CBSE Sample Paper Mathematics Standard Class X (Term I), , …(i), …(ii), , C, 4, D, , 1, A, , 2, , 3, , B, , SAMPLE PAPER 3, , From Eqs. (i) and (ii), we have, Ð1 + Ð4 = Ð3 + Ð4, Þ, Ð1 = Ð3, Similarly,, Ð2 = Ð4, Now, in DABC and DDAC., ÐA = ÐD = 90 °, [common], ÐC = ÐC, By AA similarity criterion, DABC ~ DDAC, Again, in DDAC and DDBA, [proved above], Ð1 = Ð3, [proved above], Ð4 = Ð2, By AA similarity criterion,, DDAC ~ DDBA, Hence, DABC ~ DDAC ~ DDBA., 28. Given a pack of cards contain card numbered, 2 to 53., Prime number less than 20 are, 2, 3, 5, 7, 11, 13, 17, 19., Number of prime numbers = 8, Total number of cards = 53 - 2 + 1 = 52, \ Required probability, Number of favourable outcomes, =, Total number of outcomes, 8, 2, \ Required probability =, =, 52 13, 29. Let p( x ) = x 2 - x - (2 k + 2 ), , 30. Let E be the event passing the test P( E) =, Also, P (not passing the test) P( E ) =, , x, 24, , 7, 8, , Now, P( E) + P( E ) = 1, x 7, + =1, 24 8, 7, x, Þ, =1 8, 24, x 8 -7, Þ, =, 24, 8, x 1, =, Þ, 24 8, 24, Þ, Þ x =3, x=, 8, 31. Let height of vertical pole = AB, Length of wire = AC = 26 m, Distance from the base of the pole to the, another end of the wire = BC = 10 m, A, , m, , 27. From figure, we have, Ð1 + Ð4 = 90 °, and, Ð3 + Ð4 = 90 °, , Since, - 4 is zero of p( x )., \, p( - 4 ) = 0, Þ, ( - 4 )2 - ( - 4 ) - (2 k + 2 ) = 0, Þ, 16 + 4 - 2 k - 2 = 0, Þ, 2 k = 18, Þ, k =9, , 26, , Now, old area of the circle = px 2, 2, 9x, 81x 2, and new area of the circle = pæç ö÷ = p, 100, è 10 ø, 81, \Required change in area = px 2 - p, x2, 100, 19, =, px 2, 100, \Percentage at which area diminished, æ 19 px 2 ö, ÷, ç, ç 100 ÷, ø ´ 100 = 19%, è, =, px 2, , C, , 10 m, , B, , Since, the pole will be perpendicular (vertical), to ground., \, ÐABC = 90 °, Þ DABC is a right-angled triangle., Using pythagoras theorem,, Þ, Þ, Þ, , AC 2 = AB2 + BC 2, 26 2 = AB2 + 10 2, AB2 = 676 - 100 = 576, AB = 24 m, , 32. Let the number be 10y + x., Where x and y are unit’s and ten’s digit., According to the situation, 10 y + x = 8( x + y ) + 1, …(i), Þ, 7x - 2 y + 1 = 0, and, 10 y + x = 13( y - x ) + 2, …(ii), Þ, 14 x - 3 y - 2 = 0, Multiply Eq. (i) by 2, we get

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95, , CBSE Sample Paper Mathematics Standard Class X (Term I), , …(iii), 14 x - 4 y + 2 = 0, Now, subtract Eq. (iii) from Eq. (ii),, (14 x - 3 y - 2 ) - (14 x - 4 y + 2 ) = 0, Þ 14 x - 3 y - 2 - 14 x + 4 y - 2 = 0, Þ, y - 4 =0Þ y = 4, Put y = 4 in Eq. (i), 7x - 2 ´ 4 + 1 = 0 Þ 7x = 7 Þ x = 1, Hence, the number = 10 y + x = 10 ´ 4 + 1 = 41, 33. Let E be the event Sania win the match., So, probability of Sania winning the match, = P( E) = 0.68, Also,, P( E) + P( E ) = 1, \Required probability of Deepika winning, the match = P( E ), = 1 - P( E), = 1 - 0.68 = 0.32, 34. In an equilateral triangle,, AG : GD = 2 : 1, Þ, AG = 2 x and GD = x, A, , 6 cm, , 6 cm, G, , B, , Now,, , D, 6 cm, , C, , 3, a, 2, 3, =, ´ 6 = 3 3 cm, 2, AG + GD = 3 3, 2x + x =3 3, x = 3 cm, AG = 2 x = 2 3 cm, GD = x = 3 cm, AD =, , Again,, \, Þ, \, and, , éQ cot A = cos A ù, sin A ûú, ëê, æ 1 + sin q ö æ 1 - sin q ö, =ç, ÷ç, ÷, è cos q ø è cos q ø, =, , 1 - sin 2 q, cos 2 q, , cos 2 q, =1, cos 2 q, , [Q sin 2 A + cos 2 A = 1], , 36. Fencing is done on the circumference of the, circular fields., Let R be the radius of the circular field., Given, area of circular field, = 13.86 hec, = 13.86 ´ 10000 m 2, Þ, ( pR 2 ) = 138600, 7 ö, Þ, R 2 = æç 138600 ´, ÷, 22 ø, è, Þ, R 2 = 44100 m 2, Þ, R = 210 m, Circumference of the circular field, 22, = 2 pR = æç 2 ´, ´ 210 ö÷, 7, ø, è, = 1320 m, \ Cost of fencing = ` (1320 ´ 4.40 ) = ` 5808, 37. Given,, , tan 30 ° + cot 60 °, tan 30 ° (sin 30 ° + cos 60 ° ), 1, 1, +, 3, 3, =, 1 é1 1 ù, +, 3 êë2 2 úû, 2, 3, =, 1, ´1, 3, 2´ 3, =2, =, 3´1, , 38. Given, join of points A(2 , 6 ) and B( 5 , 1) divides, by point C in 1 : 3., AC : BC = 1 : 3, 3 ´ 2 + 1´ 5, Thus,, k=, 1+3, 6 + 5 11, =, =, 4, 4, 39. As area of quadrant is equal to the one fourth, of the total area of circle. Let r be the radius of, the circle., 1, \ Area of quadrant ABDCA = pr2, 4, 1 22, = ´, ´ (7 )2, 4 7, [Q r = 7 cm ], 77, = cm 2, 2, = 38.5 cm 2, In DABE,, , SAMPLE PAPER 3, , 35. Given, the numerical value, æ 1 + 1 öæ 1 - 1 ö, ÷, ÷ç, ç, è cos q cot q ø è cos q cot q ø, 1, sin q ö æ 1, sin q ö, = æç, +, ÷, ÷ç, è cos q cos q ø è cos q cos q ø, , =

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96, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Base = 7 cm, Height = 2 cm, 1, Area of DBAE = ´ base × height, 2, 1, = ´ AB´ AE, 2, 1, = ´ 7 ´ 2 = 7 cm 2, 2, Hence, area of the shaded portion, = Area of the quadrant ABDCA, - Area of DBAE, = (38.5 - 7 ) cm 2, = 31.5 cm 2, 1, is a zeroes of the polynomial, 2, 5, x 2 + kx - , then, 4, 2, æ 1 ö + 1 k - 5 =0, ç ÷, 2, 4, è2 ø, 1 1, 5, Þ, + k - =0, 4 2, 4, -4 1, + k =0, Þ, 4, 2, 1, -1 = - k, Þ, 2, Þ, k =2, -k, Now, sum of zeroes =, = - k = -2, 1, , 40. Since,, , Solutions (41-45), 41. The total number of vegetables, = 420 + 130 = 550, 42. Since, total number of vegetables is 550, Prime factorisation of 550 = 2 ´ 5 ´ 5 ´ 11, = 2 1 ´ 5 2 ´ 111, The exponents are 1, 2 and 1., \Product of exponents = 1 ´ 2 ´ 1 = 2, , SAMPLE PAPER 3, , 43. Prime factor of 420 = 2 2 ´ 3 ´ 5 ´ 7, and, 130 = 2 ´ 5 ´ 13, \ HCF (420, 130) = 2 ´ 5 = 10, So, the number of vegetables that can be, placed in each stack for this purpose is 10., , 44. The number of vegetables that can be placed in, ech stack = 10, Prime factor of 10 = 2 1 ´ 51, The exponents are 1 and 1., \ Sum of exponents = 1 + 1 = 2, 45. Total number of vegetables = 550, Number of vegetables that can be placed in, each stack = 10, 550, = 55, \ Number of rows of vegetables =, 10, Solutions (46-50), 46. The coordinates of P is (4, 6) as the point is, 4 units on X-axis and 6 unit on Y-axis., 47. The coordinates of R is (6, 5) as the point is, 6 units on X-axis and 5 units on Y-axis., 48. The coordinates of Q are (3, 2) and R are (6, 5)., \ The distance between points Q and R is, QR = (6 - 3 )2 + ( 5 - 3 )2, = (3 )2 + (2 )2, = 9 + 4 = 13 units, 49. Since, coordinates of vertices of a DPQR are, P( 4 , 6 ), Q(3 , 2 ) and R(6 , 5 )., Now, PQ = (3 - 4 )2 + (2 - 6 )2, = ( -1)2 + ( - 4 )2, = 1 + 16 = 17 units, PR = (6 - 4 )2 + ( 5 - 6 )2, = (2 )2 + ( -1)2, = 4 + 1 = 5 units, and QR = 13, Here, PQ ¹ QR ¹ PR, Hence, DPQR is a scalene triangle., æ4 +3 +6 6 +2 + 5ö, 50. The centroid of DPQR is ç, ,, ÷, 3, 3, è, ø, 13 13 ö, æ, i.e. ç , ÷., è 3 3 ø

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97, , CBSE Sample Paper Mathematics Standard Class X (Term I), , SAMPLE PAPER 4, MATHEMATICS (Standard), A Highly Simulated Practice Questions Paper, for CBSE Class X (Term I) Examination, , Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions., There is no negative marking., Maximum Marks : 40, Time allowed : 90 minutes, , Roll No., , Section A, Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 1. What will be the least possible number of the planks, if three pieces of timber 42 m,, 49 m and 63 m long have to be divided into planks of the same length?, (a) 5, (b) 6, (c) 7, (d) None of these, , 2. The value of k such that the polynomial x 2 - (k + 4)x - 2k + 3 has sum of its zeroes equal, to doubled of their product is, 2, 1, (a), (b) 5, 5, , (c), , 2, 5, , (d), , 4, 5, , 3. The lengths of the diagonals of a rhombus are 12 cm and 8 cm. The length of each side, of the rhombus is, (a) 12 cm, , (b) 2 13 cm, , (c) 14 cm, , (d) 17 cm, , a red card, is, 1, (a), 2, , (b), , 1, 13, , (c), , 1, 52, , (d), , 12, 13, , 5. The coordinates of the point which is reflection of point (- 3, 5) in X-axis is, (a) (3, 5), (c) ( - 3, - 5), , (b) (3, - 5), (d) ( - 3, 5), , SAMPLE PAPER 4, , 4. A card is accidently dropped from a pack of 52 playing cards. The probability that it is

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98, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 6. If 4 tan A = 3, then find the value of the following expression, 4 sin 2 A - 2 cos 2 A, 4 sin 2 A + 3 cos 2 A, (a), , 1, 21, , (c) -, , (b), , 1, 21, , 23, 41, , (d) Cannot be determined, , 7. Three runners running around a circular track, can complete one revolution in, 2, 3 and 4 h respectively. They will meet again at the starting point after, (a) 8 h, (b) 6 h, (c) 12 h, (d) 18 h, , 8. The diameter of a circle, whose area is equal to the areas of two circle of radii 12 cm, and 5 cm, is, (a) 17 cm, , (b) 13 cm, , (c) 26 cm, , (d) 34 cm, , 3, 2, , 2, 3, 13, 2, (b) x +, x+2, 6, , 9. The quadratic polynomial whose zeroes are and is, 13, x+1, 6, 13, (c) x 2 +, x+1, 6, , (a) x 2 -, , (d) None of these, , 10. If A = 30°, then the value of 3 cos A - 4 cos 3 A is, (a) 0, , (c) -1, , (b) 1, , (d) 2, , 11. In DABC, DE|| BC, then the value of x is, A, 2x, D, , x+2, E, , 2x – 3, , x–3, , B, , (a), , 4, 7, , C, , (b), , 5, 7, , (c), , 6, 7, , (d), , 8, 7, , 12. In a single throw of a die, the probability of getting a multiple of 2 is, (a), , 1, 2, , (b), , 1, 3, , (c), , 1, 6, , (d), , 2, 3, , SAMPLE PAPER 4, , 13. If the point P(k , 0) divides the line segment joining the points A(2, - 2) and B(- 7 , 4), in the ratio 1 : 2, then the value of k is, (a) 1, (b) 2, 17, 14. The decimal expansion of, is, 125, (a) 1.36, (c) 13.6, , (c) - 2, , (b) 0.136, (d) None of these, , 15. Sum of zeroes of f (x) = x 2 - 16 is, (a) 16, (c) 4, , (b) 0, (d) None of these, , (d) - 1

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99, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 16. The distance between the points (a sin q + b cos q, 0) and (0, a cos q - b sin q) is, (a) a 2 + b 2, , (b) a 2 - b 2, , (c) a 2 + b 2, , (d) a 2 - b 2, , 17. If x tan 30° cos 30° = sin 60° cot 60°, then x is equal to, (a) 1, , (b) 3, , (c), , 1, 2, , (d), , 1, 2, , 18. In the given figure, DABC ~ DAQP. If AB = 6 cm, BC = 8 cm and PQ = 4 cm, then AQ is, equal to, P, Q, A, B, C, , (a) 2 cm, , (b) 2.5 cm, , (c) 3 cm, , (d) 3.5 cm, , 19. A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball is, selected at random. What is the probability that the drawn ball is white or blue?, 1, 5, 7, (a), (b) 2, (c), (d), 2, 6, 12, , 20. If P is a point on Y-axis, whose ordinate is 3 and Q is a point (-5, 2), then the distance, PQ is, (a) 26 units, , (b) 24 units, , (c) 5 units, , (d) 65 units, , Section B, Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., 13, 21. The decimal expansion of is, 8, (a) 1.625, (b) 1.25, (c) 1.675, (d) 1.0625, , 22. If HCF (306, 657) = 9, then LCM (306, 657) is, (a) 22338, (c) 33228, , (b) 23238, (d) 32328, , 23. If x = r sin q cos f, y = r sin q sin f and z = r cosq, then, (b) x 2 + y 2 - z 2 = r 2, (d) z 2 + y 2 - x 2 = r 2, , 24. Two numbers ‘a’ and ‘b’ are selected successively without replacement in that order, a, is an integer, is, b, 17, (c), 90, , from the integers 1 to 10. The probability that, (a), , 17, 45, , (b), , 1, 5, , (d), , 8, 45, , 25. The centre of a circle is (2a , a - 7). If the circle passes through the point (1, - 9) and has, diameter 10 2 units, then the value of a is, (a) 9, (b) - 3, , (c) 3, , (d) ± 3, , SAMPLE PAPER 4, , (a) x 2 + y 2 + z 2 = r 2, (c) x 2 - y 2 + z 2 = r 2

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100, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 26. In the given figure, PB||CF and DP|| EF then, , AD, is equal to, DE, , C, , A, , 2cm, , 8cm, B, , P, , F, , D, E, , (a), , 3, 4, , (b), , 1, 3, , (c), , 1, 4, , (d), , 2, 3, , 27. The LCM of 3x 2 y, 4 xz 2 and 5y 2 z (where x is a positive integer) is, (a) 30xy 2 z 2, , (b) 15xyz 2, , (c) 30x 3 y 2 z, , (d) 60x 2 y 2 z 2, , 28. The probability of getting a number between 1 and 100 which is divisible by 1 and, itself only is, 29, 1, 25, 23, (b), (c), (d), (a), 98, 2, 98, 98, 1, 29. If a and are the zeroes of the quadratic polynomial x 2 - x + 8, then a is, 2, -1, 1, (c), (a) 4, (b), (d) 16, 4, 4, , 30. If, , sin 2 q - 3 sin q + 2, cos 2 q, , (a) 0°, , = 1, then q will be, (b) 30°, , (c) 45°, , (d) 60°, , AB BC CA, , then, =, =, YZ XZ XY, (b) DZXY ~ DABC, (d) DABC ~ DYZX, , 31. If in two triangles DABC and DXYZ such that, (a) DXYZ ~ DABC, (c) DCBA ~ DXYZ, , 32. If tan q =, (a), , 9, 41, , 40, , then the value of cosec q is, 9, 41, (b), 40, , (c), , 40, 9, , (d), , 9, 40, , 33. The ratio in which the segment joining the points (1, -3) and (4, 5) is divided by X-axis, is, (a) 3 : 5, , (b) 5 : 3, , (c) 4 : 7, , (d) 7 : 4, , 1, 4, , SAMPLE PAPER 4, , 34. The sum and product of the zeroes of a quadratic polynomial are - and, respectively. Then, the quadratic polynomial is, (a) 4x 2 + x + 1, (b) 3x 2 - x + 2, (c) 4x 2 - x - 1, , 1, 4, , (d) 3x 2 + x - 2, , 35. Which of the following cannot be the probability of an event?, (a), , 2, 5, , (b) -, , 1, 2, , (c) 0.1, , (d) 0.8, , 36. If a vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same, time a tower casts a shadow of 28 m long, then the height of the tower is, (a) 30 m, (b) 35 m, (c) 40 m, (d) 42 m

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101, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 37. In the given figure, two line segments AC and BD intersect each other at the point P, such that PA = 6 cm, PB = 3 cm,PC = 2.5 cm, PD = 5 cm, ÐAPB = 50° and ÐCDP = 30°., Then, ÐPBA is equal to, A, , 6c, , D, , m, 50°, , 3c, B, , (a) 30°, , (b) 60°, , m, , 5 cm 30°, , P, , 2.5, , cm, C, , (c) 80°, , (d) 100°, , 38. 5 ( 11 - 3) is an, (a) rational number, (c) undefined, , (b) irrational number, (d) None of these, , 39. 5 yr hence, the age of a man shall be 3 times the age of his son while 5 yr earlier the age, of the man was 7 times the age of his son. The present age of the man is, (a) 45 yr, (b) 50 yr, (c) 43 yr, (d) 40 yr, , 40. The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(1, - 3), and R( - 3, 0), then the coordinates of P are, 1 1, (a) æç , ö÷, (b) (14, 7), (c) (18, 9), è 5 10 ø, , (d) (10, 5), , Section C, Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted., 41-45 are based on Case Study-1., , Case Study 1, Major Dhyanchand National Stadium is a very popular multi-purpose sports stadium at, Mumbai. It has a capacity to seat 60000 people. The stadium is conducting the annual sports, competition soon. The curator of the stadium is asked to figure out the dimensions for carving, out some areas allotted for a hockey court and a shooting range, as shown in the given figure., , SAMPLE PAPER 4

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102, , CBSE Sample Paper Mathematics Standard Class X (Term I), , A, O, , Shooting, Range, , C, , Hockey, Court, , B, E, , D, , The shapes of the hockey court and the shooting range are a square and a triangle respectively., Both of the courts have a common edge that touches the centre of the stadium. The construction, of the shooting range is such that the angle at centre is 90°., The radius of the stadium is 180 m. (take p = 3.14), On the basis of above information, answer the following questions., , 41. What is the area allotted to shooting range?, (a) 12600 m 2, , (b) 16200 m 2, , (c) 18660 m 2, , (d) 16880 m 2, , 42. What is the area alloted to the hockey court?, (a) 16200 m 2, , (b) 22000 m 2, , (c) 20000 m 2, , (d) 16880 m 2, , 43. If the team of the curators managing the stadium likes to allot space for some more, sports, how much area is available to them?, (c) 60040 m 2, (a) 76980 m 2, (b) 95806 m 2, , (d) 69336 m 2, , 44. If the boundaries of the hockey court and shooting range are to be fenced, then what is, the required length (in m) of the fence?, (a) 400(2 + 5 2 ), (b) 180(2 + 3 2 ), , (c) 180(2 + 5 2 ), , (d) 300(2 + 3 2 ), , 45. If the cost of fencing is ` 6 per metre, what is the total cost (in `) of fencing?, (a) 1800(2 + 3 2 ), , (b) 1080(2 + 5 2 ), , (c) 1080(2 + 3 2 ), , (d) 2400(2 + 5 2 ), , 46-50 are based on Case Study-2., , Case Study 2, , SAMPLE PAPER 4, , A tour company provides taxi to their customers for travelling and the taxi charges in a city, consist of a fixed charge together with the charge for the distance covered. For a distance of 10, km, the charge paid is ` 165 and for a journey of 18 km, the charge paid is ` 277.

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103, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 46. The fixed charges and charges per km respectively are, (a) ` 25 and ` 10, (c) ` 25 and ` 14, , (b) ` 10 and ` 25, (d) ` 14 and ` 25, , 47. The taxi fare for a distance of 25 km would be, (a) ` 350, (c) ` 280, , (b) ` 375, (d) ` 310, , 9, 2, , 48. The pair of linear equations -3x + 4 y = 5 and x - 6y +, , 15, = 0 has, 2, , (a) unique solution, (b) iInfinitely many solutions, (c) no solution, (d) Cannot be determined, , 49. The value of k for which the pair of linear equations 2x + 3y = 7 and (k - 1)x + (k + 2)y = 3k, have infinitely many solutions is, (a) k = - 1, (c) k = - 7, , (b) k = 0, (d) k = 7, , 50. From the graph given below, the area of the triangle formed by the two lines and the, Y-axis is, Y, 7, 6, P (0, 5), , 5, 4, , 2y, x–, , Q (4, 2), 3, 2, , 3x, , A (0, 0) 1, X¢, , –1, , B (2, 1), , 0, –1, , 1, , 2, , 3, , 4, , 5, , +, , 4y, , =0, , =, 20, , 6, , X, , Y¢, , (a) 10 sq units, (c) 14 sq units, , (b) 5 sq units, (d) 20 sq units, , SAMPLE PAPER 4

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OMR SHEET, , SP 4, , Roll No., Sub Code., , Student Name, , Instructions, Use black or blue ball point pens and avoid Gel & Fountain pens for filling the OMR sheet., Darken the bubbles completely. Don’t put a tick mark or a cross mark, half-filled or over-filled bubbles will not be read, by the software., Correct, , ✔, , ✗, , Incorrect, , Incorrect, , Incorrect, , Do not write anything on the OMR Sheet., Multiple markings are invalid., , 1, , 18, , 35, , 2, , 19, , 36, , 3, , 20, , 37, , 4, , 21, , 38, , 5, , 22, , 39, , 6, , 23, , 40, , 7, , 24, , 41, , 8, , 25, , 42, , 9, , 26, , 43, , 10, , 27, , 44, , 11, , 28, , 45, , 12, , 29, , 46, , 13, , 30, , 47, , 14, , 31, , 48, , 15, , 32, , 49, , 16, , 33, , 50, , 17, , 34, , Check Your Performance, Total Questions:, Total Correct Questions:, , If Your Score is, , Score Percentage =, , Total Correct Questions, Total Questions, , × 100, , Less than 60%, > Average (Revise the concepts again), Greater than 60% but less than 75% > Good (Do more practice), Above 75%, > Excellent (Keep it on)

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105, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Answers, 1. (d), 11. (c), 21. (a), , 2. (c), 12. (a), 22. (a), , 3. (b), 13. (d), 23. (a), , 4. (a), 14. (b), 24. (c), , 5. (c), 15. (b), 25. (d), , 6. (a), 16. (c), 26. (b), , 7. (c), 17. (a), 27. (d), , 8. (c), 18. (c), 28. (c), , 9. (a), 19. (c), 29. (d), , 10. (a), 20. (a), 30. (b), , 31. (d), 41. (b), , 32. (b), 42. (a), , 33. (a), 43. (d), , 34. (a), 44. (b), , 35. (b), 45. (c), , 36. (d), 46. (c), , 37. (d), 47. (b), , 38. (b), 48. (b), , 39. (d), 49. (d), , 40. (a), 50. (a), , SOLUTIONS, 1. Length of the plank will be HCF (42, 49, 63), The prime factors of, 42 = 2 ´ 3 ´ 7, 49 = 7 ´ 7, and, 63 = 3 ´ 3 ´ 7, \ HCF (42, 49, 63) = 7, \ Required number of plank will be, 42 49 63, =, +, +, 7, 7, 7, = 6 + 7 + 9 = 22, 2. Let a and b are the zeroes of the polynomial, x 2 - ( k + 4 )x - 2 k + 3, \ Sum of zeroes = a + b = -, , Coefficient of x, 2, , Coefficient of x, [ -( k + 4 )], == k+4, 1, Constant term, and product of zeroes = ab = Coefficient of x 2, -2 k + 3, =, = -2 k + 3, 1, Now, it is given that, a + b = 2 ab, a + b = 2 ab, Þ, k + 4 = 2 ( -2 k + 3 ), Þ, k + 4 = -4 k + 6, Þ, k + 4k = 6 - 4, Þ, 5k = 2, 2, Þ, k=, 5, , C, , D, O, , m, , A, , m, , 4c, , 6c, , B, , = (6 )2 + ( 4 )2 = 36 + 16 = 52, AB = 52 = 2 13, \, \ The length of each side of rhombus is, 2 13 cm., 4. Total number of possible outcomes = 52, Number of red cards = 26, So, number of favourable outcomes = 26, 26 1, =, \ Required probability =, 52 2, 5. The reflection of any point P( x , y ) in X-axis is, given by ( x , - y ), \Reflection of (-3 , 5) in X-axis will be ( -3 , - 5 )., 6. Given, 4 tan A = 3, 3, tan A =, 4, 2, , Now,, , 2, , 4 sin A - 2 cos A, , 4 sin 2 A, , 2, = cos 2A, 2, 2, 4 sin A + 3 cos A 4 sin A, , -2, , +3, cos 2 A, [divide cos 2 A by numerator and denominator], =, , 4 tan 2 A - 2, 4 tan 2 A + 3, 2, , 3, 4 ´ æç ö÷ - 2, è4ø, , 9, -2, 9 -8, 1, 4, =, =, =, =, 2, 9, 9 + 12 21, 3, 4 ´ æç ö÷ + 3 4 + 3, è4ø, 7. The prime factors of 2, 3 and 4 are given as, 2 = 1´2, 3 = 1´3, 4 = 2 ´2, \ LCM (2, 3, 4) = 2 ´ 2 ´ 3 = 12, So, they will meet at the starting point after, 12 h., 8. Let the radius of circle be R and radii of given, circles are r1 = 12 cm and r2 = 5 cm, respectively., , SAMPLE PAPER 4, , 3. Let ABCD be a rhombus whose diagonals AC, and BD bisect each other at O at right-angled, 1, 1, Then, OA = AC = ´ 12 = 6 cm, 2, 2, 1, 1, and OB = BD = ´ 8 = 4 cm, 2, 2, Also, ÐAOB = 90°, , In DAOB, AB2 = OA2 + OB2, [by pythagoras theorem]

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106, , CBSE Sample Paper Mathematics Standard Class X (Term I), , pR 2 = pr12 + pr22, , Then,, , 2, , R =, , r12, , 13. A( 2, –2), , + r22, 2, , = (12 ) + ( 5 )2, 2, , 2, , R = (12 ) + ( 5 ) = 144 + 25, = 169 = 13 cm, Diameter of a circle = 2 R = 2 ´ 13, = 26 cm, 9. Given, zeroes of quadratic polynomial are, 3, 2, and . Then,, 2, 3, 3 2 9 + 4 13, Sum of zeroes (a + b ) = + =, =, 2 3, 6, 6, 3 2, and product of zeroes (ab ) = ´ = 1, 2 3, \ Required polynomial is x 2 - (a + b) x + ab, 13, i.e. x 2 x+1, 6, Ð A = 30°, , 10. Given,, \, , 3 cos A = 3 ´ cos 30° =, , 3´ 3 3 3, =, 2, 2, , and 4 cos 3 A = 4 ´ (cos 30° )3, 3, , æ 3ö, 3 3 3 3, =4´ç ÷ =4´, =, 2, 8, è 2 ø, \, , 3 cos A - 4 cos 3 A =, , 3 3 3 3, =0, 2, 2, , 11. Given, DE || BC, AD AE, =, BD CE, x+2, 2x, =, (2 x - 3 ) x - 3, , [by BPT theorem], , 2, , Þ 2 x - 6x = 2 x + 4x - 3x - 6, Þ, , 6 = 6x + x, 6, x=, 7, , \, , SAMPLE PAPER 4, , Hence, the value of x is, , 6, 7, , 12. Total number of possible outcomes = 6, Now, multiple of 2 are 2, 4, 6, \Favourable number of outcomes is 3., Now, Probability, Number of favourable outcomes, Total number of outcomes, 3 1, \ Required probability = =, 6 2, =, , æ 1 ´ ( -7 ) + 2 ´ 2 1 ´ 4 + 2 ´ ( -2 ) ö, P( k , 0 ) = ç, ,, ÷, 1+2, 1+2, ø, è, 7, 4, 4, 4, +, æ, ö, P( k , 0 ) = ç, ,, ÷, è 3, 3 ø, \, , P( k , 0 ) = ( - 1, 0 ), k = -1, , 17, 17 2 3, 136, = 3´ 3=, = 0.136, 125 5, 2, (10 )3, 17, is 0.136., Hence, decimal expansion of, 125, , 14. Now,, , 15. We have,, , f ( x ) = x 2 - 16, , = x 2 + 0 × x - 16, -Coefficient of x -(0 ), \Sum of zeroes =, =, =0, 1, Coefficient of x 2, 16. Given points are ( a sin q + b cos q , 0 ) and, (0 , a cos q - b sin q ), \By distance formula, d = ( x2 - x1 )2 + ( y2 - y1 )2, = (0 - a sin q - b cos q) 2 + ( a cos q - b sin q - 0 )2, a2 sin 2 q + b2 cos 2 q + 2 ab sin q cos q, + a2 cos 2 q + b2 sin 2 q - 2 ab sin q cos q, , = a2(sin 2 q + cos 2 q ) + b2(sin 2 q + cos 2 q), = a 2 + b2, , Þ 2 x( x - 3 ) = (2 x - 3 )( x + 2 ), , B(– 7, 4), 2, , By section formula, æ m x + m2x1 m1y2 + m2y1 ö, P( x , y ) = ç 1 2, ,, ÷, m1 + m2 ø, è m1 + m2, , =, , Then,, , 2, , P(k , 0), 1, , [Q sin 2 x + cos 2 x = 1], , 17. We have, x tan 30° cos 30° = sin 60° cot 60°, 1, 3, 3, 1, ´, =, ´, 3, 2, 2, 3, éQ tan 30° = 1 = cot 60° ù, ú, ê, 3, ú, ê, ê and sin 60° = 3 = cos 30° ú, û, ë, 2, 1 1, Þ, x ´ = Þx =1, 2 2, 18. Given, DABC ~ DAPQ , BC = 8 cm, PQ = 4 cm, and BA = 6 cm, Now,, DABC ~ DAQP, BC, AB, \, =, PQ AQ, x´, , [When two triangles are similar, then the, corresponding sides are in proportion]

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107, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Þ, Þ, , 8, 6, =, 4 AQ, AQ =, , z2 = r2 cos 2 q, Adding Eqs. (i), (ii) and (iii), we get, , 6´4, = 3 cm, 8, , 19. Total number of balls = 24, Now,, x + 2 x + 3 x = 24, Þ, 6 x = 24, Þ, x=4, \ Number of red balls = x = 4, Number of white balls = 2 x = 8, Number of blue balls = 3 x = 12, Now, Total number of possible outcomes = 24, Let E be the event that the ball drawn is white, or blue, \ Number of outcomes favourable to E, = 8 + 12 = 20, Number of favourable outcomes, P( E) =, Total number of outcomes, 20 5, =, =, 24 6, 20. Since, P lies on Y-axis and have ordinate as 3, \ The point P is (0, 3) and Q is ( -5 , 2 )., Now, The distance of PQ is, PQ = ( -5 - 0 )2 + (2 - 3 )2, = 25 + ( -1)2, 13 13, =, 8 23, 13 ´ 5 3, 1625 1625, = 3, =, =, = 1625, ., 3, 2 ´5, (10 )3 1000, 13, Hence, decimal expansion of, is 1625, ., 8, , 21. As,, , O, (2a, a – 7), , = 22338, x = r sin q cos f , y = rsin q sin f, … (i), … (ii), , A(1, – 9), , \, 2 ( OA) = 10 2 [Q Diameter = 2 2 given], Þ, OA = 5 2, OA2 = ( 5 2 )2 = 50, Þ, Þ (2 a - 1)2 + ( a - 7 + 9 )2 = 50, [using distance formula], (2 a - 1)2 + ( a + 2 )2 = 50, 4 a2 + 1 - 4 a + a2 + 4 + 4 a = 50, Þ, 5 a2 = 50 - 5, Þ, 5 a2 = 45 Þ a2 = 9, Þ, a= ±3, 26. In DACF, BP || CF, So, by Basic proportionality theorem, AB AP, =, BC PF, , SAMPLE PAPER 4, , 22. We know that,, LCM ( a, b) ´ HCF ( a, b) = a ´ b, \LCM (306, 657) ´ HCF (306, 657) = 306 ´ 657, 306 ´ 657, Þ LCM (306, 657) =, HCF (306 , 657 ), 306 ´ 657, =, 9, [Q HCF(306 , 657 ) = 9 ], , z = r cosq, 2, x = r2 sin 2 q cos 2 f, y 2 = r2 sin 2 q sin 2 f, , 24. Given, two numbers are selected without, replacement from integers 1 to 10., a, Now, favourable cases for which will be an, b, integer are, 2, 3, 4, 4, 5, = 2 ; = 3; = 4; = 2 ; = 5, 1, 1, 1, 2, 1, 6, 6, 8, 6, 7, = 6 ; = 3 ; = 2 ; = 7 ; = 8,, 3, 2, 1, 1, 1, 8, 9, 8, 9, 10, = 4 ; = 2 ; = 9 ; = 3;, = 10,, 4, 3, 2, 1, 1, 10, 10, = 5 and, =2, 5, 2, \Favourable number of cases = 17, 1 1, 1, 2 2, 2, Total cases are é , , ¼ ù , é , , ¼ ù, …, ê, ú, êë 2 3, 10 û ë 1 3, 10 úû, é 10 , 10 , 10 , ¼ 10 ù, 9 ûú, ëê 1 2 3, , 25. Let the circle be as given below with centre, O(2 a, a - 7 ) and A(1, - 9 ) be any point at the, circle., , PQ = 26 units, , and, \, , x 2 + y 2 + z2 = r2 sin 2 q cos 2 f + r2 sin 2 q sin 2 f, + r2 cos 2 q, 2, 2, 2, 2, 2, 2, 2, x + y + z = r sin q (cos f + sin f ) + r2 cos 2 q, x 2 + y 2 + z2 = r2 sin 2 q + r2 cos 2 q, [Q sin 2 x + cos 2 x = 1], x 2 + y 2 + z 2 = r2, , Total number of cases = 9 ´ 10 = 90, 17, \ Required probability =, 90, , = ( -5 )2 + ( -1)2, , 23. Given,, , … (iii)

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108, , CBSE Sample Paper Mathematics Standard Class X (Term I), , AP, 2, [Q BC = AC - AB], =, PF, 8 -2, AP 2, Þ, =, PF 6, AP 1, =, Þ, PF 3, Again, in DAEF, DP || EF, So, by Basic proportionality theorem, AD AP 1, =, =, DE PF 3, Þ, , 2, , 2, , 27. Factors of 3 x y = 3 ´ x ´ y,, and, , 4 xz2 = 2 2 ´ x ´ z2, 5 y 2z = 5 ´ y 2 ´ z, , LCM (3 x 2y , 4 xz2 and 5 y 2z), = 3 ´ 4 ´ 5 ´ x 2 ´ y 2 ´ z2, = 60 x 2y 2z2, 28. Number divisible by 1 and itself is a prime., Number of prime number between 1 and, 100 = 25, i.e., (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97), 25, \ Required probability =, 98, 29. We have,, 1, a and are the zeroes of the polynomial, 2, x 2 - x + 8, then, Constant term, Product of zeroes =, Coefficient of x 2, 1 8, \, a´ =, 2 1, a, =8, Þ, 2, Þ, a = 16, , SAMPLE PAPER 4, , 30. We have,, , sin 2 q - 3 sin q + 2, , 31. We know that if, two triangles are said to be, similar if their corresponding angles are equal, and the corresponding sides are proportion., X, , Y, , Z B, , C, , AB BC CA, =, =, YZ XZ XY, Then,, DABC ~ DYZX, 40 AB, 32. Given, tanq =, =, 9, BC, i.e., , A, , 40 k, , q, B, , 9k, , C, , Let BC = 9 k and AB = 40 k, By Pythagoras theorem,, AC =, , AB2 + BC 2, , = ( 40 k )2 + (9 k )2, = 1600 k 2 + 81k 2, = 1681k 2 = 41k, AC 41 K 41, So, cosec q =, =, =, AB 40 K 40, 33. Let P( x , 0 ) be any point on X-axis, which, divides the line segment joining points A(1, - 3 ), and B( 4 , 5 ) in the ratio k : 1, k:1, , =1, , cos 2 q, 2, Þ sin q - 3 sin q + 2 = cos 2 q, Þ sin 2 q + sin 2 q - 3 sin q + 2 = cos 2 q + sin 2 q, [Adding sin 2 q both side], 2, Þ 2 sin q - 3 sin q + 2 = 1 [Q sin 2 x + cos 2 x = 1], Þ 2 sin 2 q - 3 sin q + 1 = 0, Þ, 2 sin 2 q - 2 sin q - sin q + 1 = 0, Þ 2 sin q (sin q - 1) - 1(sin q - 1) = 0, Þ, (2 sin q - 1) (sin q - 1) = 0, Þ, 2 sinq - 1 = 0 or sinq - 1 = 0, 1, sinq = or sinq = 1, Þ, 2, Þ, q = 30° or q = 90°, \, q = 30°, [Qq = 90° does not satisfy the given equation], , A, , A(1, – 3), , \, , P(x, 0), , B(4, 5), , æ 4k + 1 5k - 3 ö, (x, 0) = ç, ,, ÷, è k +1 k +1 ø, , On equating y-coordinate both sides, we get, 5k - 3, 0=, k +1, 3, 5k - 3 = 0 Þ k =, 5, \ The required ratio is 3 : 5., -1, 34. Given, sum of the zeroes = a + b =, 4, 1, Product of the zeroes = ab =, 4, \ Quadratic polynomial

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109, , CBSE Sample Paper Mathematics Standard Class X (Term I), , = k [ x 2 - (a + b) x + ab ], -1, 1ù, é, = k ê x 2 - æç ö÷ x + ú, ø, è, 4, 4û, ë, = 4x2 + x + 1, , 38. We know that, product of a non-zero rational, and an irrational number is always irrational, number., [Q k = 4 ], , 35. As, we know that probability of any event is, lies from 0 to 1., So, probability can not be negative., Hence, option (b) is correct., 36. Let AB be the height of pole and BC be the, length of its shadow and DE be the height of, tower and EF be the length of its shadow such, that AB = 6 m, BC = 4 m and EF = 28 m., Let length of DE be x m., A, , D, , 6m, , B, , 4m, , C, , E, , 28 m, , F, , In DABC and DDEF,, ÐB = ÐE = 90°, ÐA = ÐD, \, Þ, , DABC ~ DDEF, AB BC, =, DE EF, 6, 4, =, x 28, , [Sun’s elevation], [by AA similarity], , Þ, , x = 6 ´ 7 = 42 m, 3, 6, AP 6, BP, 37. Given,, = and, =, =, PD 5, PC 2 .5 5, AP BP, Þ, =, PD PC, Now, in DAPB and DDPC, AP BP, =, DP CP, , So, 5( 11 - 3 ) is an irrational number., 39. Let the present age of man = x yr, and present age of son = y yr, After 5 yr, age of man = ( x + 5 ) yr, Age of son = ( y + 5 ) yr, According to the question,, x + 5 = 3( y + 5 ), x + 5 = 3 y + 15, x - 3 y = 10, x = 10 + 3 y, and 5 yrs earlier, Age of man = x - 5, and Age of son = y - 5, According to the question,, x - 5 = 7( y - 5 ), x - 7 y = - 30, Subtracting Eq. (i) from Eq. (ii), we get, - 4 y = - 40, y = 10 and x = 10 + 3 y = 10 + 3 ´ 10 = 40, Hence, present age of man = 40 yr., , equidistant from Q(1, - 3 ) and R( - 3 , 0 ). Then,, PQ = PR, (1 - x )2 + ( - 3 - y )2 = ( - 3 - x )2 + (0 - y )2, [by distance formula], On squaring both side, we get, 1 + x2 - 2 x + 9 + y2 + 6y, = 9 + x2 + 6x + y2, Þ 10 - 2 x + 6 y = 9 + 6 x, According to the question,, x = 2y, Þ 10 - 2 (2 y ) + 6 y = 9 + 6(2 y ), , then DAPB ~ DDPC, , Þ, , [by CPCT], Þ ÐPAB = ÐPDC = 30°, In DAPB ,, ÐAPB + ÐPBA + ÐBAP = 180°, Þ, 50° + ÐPBA + 30° = 180°, ÐPBA = 180° - 80° = 100°, , Þ, Þ, Þ, , 10 - 4 y + 6 y = 9 + 12 y, 10 - 9 = 12 y - 2 y, 10 y = 1, 1, y=, 10, , 1, 1, =, 10 5, 1 1, Hence, coordinates of P are æç , ö÷., è 5 10 ø, , Þ, , x=2´, , SAMPLE PAPER 4, , Þ ÐAPB = 50° and ÐCPD = 50°, , … (ii), , 40. Let the coordinates of P be ( x , y ). Since, P is, , and ÐAPB = ÐDPC = 50°,, [vertically opposite angles], [SAS similarity criterion], , … (i)

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110, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Solutions (41-45), , Solutions (46-50), , 41. The shape of the shooting range is right angle, triangle, O, 180 m, , 90°, 180 m, , A, , B, , 1, \ Area of triangle = ´ Base ´ Height, 2, In DOAB, Base = 180 m and Height = 180 m, 1, \ Area of triangle = ´ 180 ´ 180, 2, = 90 ´ 180 = 16200 m 2, 42. In the given figure, Let OC = CD = x, Using pythagoras theorem,, x 2 + x 2 = 180 2, 2 x 2 = 32400, x 2 = 16200, x = 90 2, , Þ, , As we know, area of square = ( side)2, Area of square, OCDE = x 2 = (90 2 )2, = 8100 ´ 2 = 16200 m 2, 43. The available area is the difference of the total, area of the circular stadium and the two courts, i.e., \Required area = pr2 - (Area of hockey court, + area of shooting range), = pr2 - (16200 + 16200 ), = 3.14 ´ 180 ´ 180 - 32400, = 101736 - 32400 = 69336 m 2, 44. From the figure, In DOAB, AB = OB2 + OA2, [by Pythagoras theorem], = (180 ) + (180 )2, , SAMPLE PAPER 4, , 2, , = 180 2 m, The required boundaries be the perimeter of, DOAB and ~ OCDE., Perimeter of (DOAB + ~OCDE), = 180 + 180 + 180 2 + ( 4 ´ 90 2 ), = 360 + 540 2 = 180(2 + 3 2 ) m, 45. Total cost = Total length of fencing, ´ Rate per metre, = 180(2 + 3 2 ) ´ 6, = ` 1080(2 + 3 2 ), , 46. Let the fixed charge be ` x and the charge per, km be ` y., Now, according to the situation, …(i), x + 10 y = 165, and, …(ii), x + 18 y = 277, From Eq. (i), we get, …(iii), x = 165 - 10 y, Now, put this value of x in Eq. (ii), we get, 165 - 10 y + 18 y = 277, 8 y = 277 - 165 = 112 Þ y = 14, Now, put y = 14 in Eq. (iii), we get, x = 165 - 10 y = 165 - 10(14 ), = 165 - 140 = 25, Fixed charge = ` 25, Charge per km = ` 14, 47. According to the condition,, Fair = ` ( x + 25 y ), Now, put x = 25 and y = 14, \ Fair = 25 + 25 ´ 14 = 25 + 350 = ` 375, 48. On comparing the given equation with, standard equation, a1 = - 3, b1 = 4 and c1 = - 5, 9, 15, a2 = , b2 = - 6 and c2 =, 2, 2, a1 -3, 2 b1, 4, 2, Now,, =, =- ,, =, =9, a2, 3 b2 -6, 3, 2, -5 -2, c, and 1 =, =, c2 15, 3, 2, a1 b1 c1, i.e, =, =, a2 b2 c2, Hence, the given pair of linear equations has, infinitely many solutions., 49. On comparing,, a1 = 2 , b1 = 3 and c1 = - 7, a2 = k - 1, b2 = k + 2 and c2 = -3 k, -7, a, 2 b1, 3, c, 7, ,, and 1 =, Now, 1 =, =, =, a2 k - 1 b2 k + 2, c2 -3 k 3 k, For infinitely many solutions,, a1 b1 c1, =, =, a2 b2 c2, 2, 7, Þ, =, Þ 6k = 7k - 7 Þ k = 7, k - 1 3k, 50. In DAPQ, Base = 5 and Height = 4, We know that,, 1, Area of triangle = ´ Base ´ height, 2, 1, \ Area of triangle = ´ 5 ´ 4 = 10 sq units., 2

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111, , CBSE Sample Paper Mathematics Standard Class X (Term I), , SAMPLE PAPER 5, MATHEMATICS (Standard), A Highly Simulated Practice Questions Paper, for CBSE Class X (Term I) Examination, , Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions., There is no negative marking., Maximum Marks : 40, Time allowed : 90 minutes, , Roll No., , Section A, Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 1. The ratio between HCF and LCM of 15, 20 and 5 is, (a) 1 : 9, , (b) 1 : 11, , (c) 1 : 12, , (d) 3 : 4, , 2. The graph of y = p(x) are given below, for polynomial p(x). Then, the number of, zeroes is, Y, , X¢, , X, , y = p(x), Y¢, , (a) 0, , (b) 2, , (c) 1, , (d) 3, , 3. If, , tan q - 1, = A sec 2 q + B tan q, then A + B is equal to, tan q - 1, , (a) 1, , (b) -1, , (c) 2, , (d) 3, , 4. The condition for which the equations ax + b = 0 and cx + d = 0 are consistent, is, (a) ad = bc, (c) ab - cd = 0, , (b) ad + bc = 0, (d) ab + cd = 0, , SAMPLE PAPER 5, , 3

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112, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 5. The room has its area as 120 m 2 and perimeter as 44 m, then the length and breadth of, the room are, (a) 11 m and 2 m, (c) 12 m and 10 m, , (b) 10 m and 2 m, (d) 12 m and 1 m, , 6. HCF of two number is 23 and their LCM is 1449. If one of the number is 161, then the, other number is, (a) 207, (c) 1449, , (b) 307, (d) None of these, , 7. Onkar and Neha play a badminton game. If the probability of Onkar winning the, match is 0.75, then what is the probability of Neha winning the match?, (a) 0.15, (b) 0.25, (c) 0.35, (d) 0.75, , 8. If 3 chairs and 1 table cost ` 1500 and 6 chairs and 1 table costs ` 2400, then the linear, equations to represent this situation is, (a) x + 3y = 1500 and 6x - y = 2400, (c) 3x + y = 1500 and 6x + y = 2400, , (b) 3x - y = 1500 and 6x - y = 2400, (d) x - y = 1500 and x + y = 2400, , 9. If p(x) is a polynomial of atleast one degree and p(k) = 0, then k is known as, (a) value of p( x), (c) constant term of p( x), , (b) zero of p( x), (d) None of these, , 10. A bag contains 6 green balls and n yellow balls. If probability of drawing a yellow ball, is five times that of a green ball, then n is equal to, (b) 20, (c) 30, 1, 11. If sin q = , then the value of sin q(sin q - cosecq) is, 2, 3, -3, 3, (b), (c), (a), 4, 4, 2, (a) 10, , (d) 40, , (d), , - 3, 2, , 12. Which of the following is irrational number?, (a) 0. 133, , (b) 5.329685 ......, , (c) 3.5428, , (d) 9.265, , 13. It is proposed to build a single circular park, whose area is equal to the sum of areas of, two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park, would be, (a) 10 m, (b) 15 m, (c) 20 m, (d) 24 m, , 14. If the point P(k , 0) divides the line segment joining the points A(2, - 2) and B(- 7 , 4) in, , SAMPLE PAPER 5, , the ratio 1 : 2, then the value of k is, (a) 1, (b) 2, , (c) - 2, , (d) - 1, , 15. If in two triangles DDEF and DPQR, ÐD = ÐQ and ÐR = ÐE, then which of the following, is not true?, EF DF, (a), =, PR PQ, , (b), , DE EF, =, PQ RP, , (c), , DE DF, =, QR PQ, , (d), , EF DE, =, RP QR, , 16. The area of two similar triangles are respectively 36 cm 2 and 64 cm 2 . The ratio of their, corresponding sides is, (a) 4 : 3, (b) 3 : 4, , (c) 3 : 5, , (d) 2 : 5

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113, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 17. The ratio of the areas of a circle and an equilateral triangle whose diameter and a side, are respectively equal, is, (b) p : 3, (a) p : 2, , (c) 3 : p, , (d) 2 : p, , 18. If sin q = cosecq = 1, then the value of sin q + cosecq is, (a) 2 5, , (b) 2, , (c) 3, , (d) 9, , 19. Ramesh buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at, random from a tank containing 5 male fish and 9 female fish. Then, the probability that, the fish taken out is a male fish, is, 5, 5, 6, 7, (b), (c), (d), (a), 13, 14, 13, 13, , 20. The perimeter (in cm) of a square circumscribing a circle of radius ‘a’ cm, is, (a) 8a, , (b) 4a, , (c) 2 a, , (d) 16a, , Section B, Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 21. The value of k, for which the given pair of linear equation 2x + 3y = 5 and 4 x + ky = 10, has infinite solution, is, (a) 1, (b) 3, , (c) 6, , (d) 0, , 22. If probability of success is 0.9%, then probability of failure is, (a) 0.91, , (b) 0.091, , (c) 0.92, , (d) 0.991, , 23. The number of zeroes lying between -4 and 4 of the polynomial f (x) whose graph is, given below is, Y, f(x), X¢, , –4, , –2, , O, , 2, , 4, , X, , Y¢, , (a) 2, , (b) 3, (c) 4, (, a, sin, q, b, cos, q, ), a, is equal to, 24. If tan q = , then, ( a sin q + b cos q), b, (a 2 + b 2 ), (a 2 - b 2 ), a2, (b), (c), (a) 2, (a - b 2 ), (a 2 + b 2 ), (a 2 + b 2 ), , (d) 1, , (d), , b2, (a 2 + b 2 ), , positive integers., (a) True, (c) Can’t say, , (b) False, (d) None of these, , 26. If the sum and product of zeroes of a quadratic polynomial is 3 and -2 respectively,, then the quadratic polynomial is, (a) x 2 - 3x + 2, (c) x 2 - 3x - 2, , (b) x 2 + 3x - 2, (d) x 2 - 5x - 4, , SAMPLE PAPER 5, , 25. Denominator of the decimal expression 3.434 is of the form 2 m ´ 5 n , where m and n are

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114, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 27. The system of equations 2x + 3y + 5 = 0 and 4 x + ky + 7 = 0 is inconsistent when k is equal, to, (a) 6, , (b) 7, , (c) 5, , (d) 3, , 28. In a game, the entry fee is `10. The game consists of tossing of 3 coins. If one or two, heads show, Amita win the game and gets entry fee. The probability that she gets the, entry fee is, 3, 3, 7, 5, (b), (c), (d), (a), 4, 8, 8, 8, 3, 1, 29. If 3x = secq and = tan q, then the value of x 2 - 2 is, x, x, 1, 1, (a) 9, (b), (c) 8, (d), 9, 8, 23, will terminate after …………, 30. The decimal expansion of the rational number 2, 2 ´5, decimal place(s)., (a) 3, (b) 2, (c) 4, (d) doesn’t terminate, , 31. If one of the zeroes of the quadratic polynomial x 2 - 3x + k is 2, then the value of k is, (a) 10, , (b) 2, , (c) - 7, , (d) - 2, , 32. The pair of equations x = a and y = b graphically represents lines, which are, (a) parallel, (c) coincident, , (b) intersecting at ( b, a), (d) intersecting at ( a, b), , 33. In the given figure, three sectors of a circle of radius 7 cm, making angles of 60°, 80°,, 40° at the centre are shown. The area (in cm 2 ) of the shaded region is, 80º, , 60º, , 40º, , (a) 77, , (b) 154, , (c) 44, , (d) 22, , 34. The distance between the points (m, - n) and (- m, n) is, (a) ( m2 + n 2 ) units, 2, , (b) (m + n) units, , 2, , (d) ( 2 m2 + n 2 ) units, , SAMPLE PAPER 5, , (c) (2 m + n ) units, , 35. In the given figure, AOB is a diameter of a circle with centre O. The value of tan A × tan B, will be, , C, 3 cm, 2 cm, A, , (a) 1, , (b) 2, , O, , B, , (c) 3, , (d) 3

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115, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 36. If a school has two sections A and B in class Xth. Section A has 32 students and, section B has 36 students. Then minimum number of books must be in library that, can be distributed equally in both section, is, (a) 288, (b) 278, (c) 268, (d) 258, , 37. In DABC and DDEF, ÐB = ÐE, ÐF = ÐC and AB = 3DE. Then, the two triangles are, (a) congruent but not similar, (b) similar but not congruent, (c) neither congruent nor similar, (d) congruent as well as similar, , 38. An arc of a circle is of length 5p cm and the sector bounds an areas of 20pcm 2 . Then,, the radius of the circle is, (a) 4 cm, (c) 12 cm, , (b) 8 cm, (d) 16 cm, , 39. The zeroes of the quadratic polynomial x 2 + kx + k, where k > 0, (a) are both positive, (c) are always equal, , (b) are both negative, (d) are always unequal, , 40. The sum of exponents of prime factors of 196 is, (a) 3, , (b) 4, , (c) 5, , (d) 2, , Section C, Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted., 41-45 are based on Case Study-1., , Case Study 1, Bhavna found the picture of Indian flag on the moon hosted by the chandrayan, which was lunar, exploration mission by the Indian space research organisation. She wondered, if there is any, relation between the height of the flag pole and the shadow of the flag., , from the far end of the shadow is, (a) 18.6 m, (c) 17.8 m, , (b) 20.4 m, (d) 15 m, , SAMPLE PAPER 5, , 41. A flag pole 18 m high casts a shadow 9.6 m long. The distance of the top of the pole

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116, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 42. Which concept is used to find the distance of the top of the pole from the far end of the, shadow?, (a) Pythagoras theorem, (c) Converse of thales theorem, , (b) Thales theorem, (d) Converse of pythagoras theorem, , 43. Now, if the flag pole 4 m high casts a shadow 3 m long, then the distance of the top of, the pole from the far end of the shadow is, (a) 5 m, (b) 8 m, (c) 6 m, (d) None of these, , 44. If the flag pole 15m high and the distance of the top of the pole from the far end of the, shadow is 25 m, then the length of shadow is, (a) 18 m, (b) 14 m, (c) 20 m, , (d) 12 m, , 45. If the distance of the top of the pole from the far end of the shadow is 13 m and the, length of a shadow is 5 m, then the height of the flag pole is, (a) 13 m, (b) 12 m, (c) 15 m, , (d) 17 m, , 46-50 are based on Case Study-2., , Case Study 2, Shivam went to the hospital near to his home for COVID-19 test along with his family members., The seats in the waiting area were as per the norms of distancing during this pandemic, (as shown in the given figure). His family members took their seats surrounded by the circular, area, 8, 7, 6, , Palak, , 5, , Shivam, , 4, 3, , Akshay, , 2, 1, O, , Malika, Enter, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 46. Considering O as the origin, what are the coordinates of O?, (a) (0,1), , (b) (1, 0), , (c) (0, 0), , (d) (-1, - 1), , 47. What is the distance between Palak and Shivam?, (a) 10 units, , (b) 2 5 units, , (c) 10 units, , (d) 8 units, , SAMPLE PAPER 5, , 48. What are the coordinates of seat of Akshay?, (a) (2, 3), , (b) (3, 2), , (c) (0, 3), , (d) (2, 0), , 49. What will be the coordinates of a point exactly between Akshay (A) and Malika (M), where a person can be seated?, (a) (3.5, 2.5), (b) (2.5, 5), , (c) (10, 5), , (d) (1.5, 0.5), , 50. Determine the shape of the figure while on joining the points where Shivam(S) family, members are seated., (a) rectangle, (c) parallelogram, , (b) square, (d) rhombus

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OMR SHEET, , SP 5, , Roll No., Sub Code., , Student Name, , Instructions, Use black or blue ball point pens and avoid Gel & Fountain pens for filling the OMR sheet., Darken the bubbles completely. Don’t put a tick mark or a cross mark, half-filled or over-filled bubbles will not be read, by the software., Correct, , ✔, , ✗, , Incorrect, , Incorrect, , Incorrect, , Do not write anything on the OMR Sheet., Multiple markings are invalid., , 1, , 18, , 35, , 2, , 19, , 36, , 3, , 20, , 37, , 4, , 21, , 38, , 5, , 22, , 39, , 6, , 23, , 40, , 7, , 24, , 41, , 8, , 25, , 42, , 9, , 26, , 43, , 10, , 27, , 44, , 11, , 28, , 45, , 12, , 29, , 46, , 13, , 30, , 47, , 14, , 31, , 48, , 15, , 32, , 49, , 16, , 33, , 50, , 17, , 34, , Check Your Performance, Total Questions:, Total Correct Questions:, , If Your Score is, , Score Percentage =, , Total Correct Questions, Total Questions, , × 100, , Less than 60%, > Average (Revise the concepts again), Greater than 60% but less than 75% > Good (Do more practice), Above 75%, > Excellent (Keep it on)

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118, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Answers, 1. (c), 11. (b), , 2. (c), 12. (b), , 3. (c), 13. (a), , 4. (a), 14. (d), , 5. (c), 15. (b), , 6. (a), 16. (b), , 7. (b), 17. (b), , 8. (c), 18. (b), , 9. (b), 19. (b), , 10. (c), 20. (a), , 21. (c), 31. (b), 41. (b), , 22. (d), 32. (d), 42. (a), , 23. (c), 33. (a), 43. (a), , 24. (b), 34. (c), 44. (c), , 25. (a), 35. (a), 45. (b), , 26. (c), 36. (a), 46. (c), , 27. (a), 37. (b), 47. (c), , 28. (a), 38. (b), 48. (a), , 29. (b), 39. (b), 49. (a), , 30. (b), 40. (b), 50. (a), , SOLUTIONS, 1. The prime factors of 15, 20 and 5 are given as, 15 = 3 ´ 5, 20 = 2 2 ´ 5, and, 5=5, \ HCF of (15, 20, 5) = 5, LCM of (15, 20, 5) = 2 2 ´ 3 ´ 5 = 60, HCF 5, 1, or 1 : 12, \ Required ratio =, =, =, LCM 60 12, 2. The graph of y = p( x ) cuts on X-axis at only one, point., So, the number of zeroes is only one., 3., , tan 3 q - 1 (tan q - 1)(tan 2 q + 12 + tan q ), =, tan q - 1, (tan q - 1), [Q a3 - b3 = ( a - b) ( a2 + b2 + ab)], 2, , = sec q + tan q [Q1 + tan 2 x = sec2 x ], Þ, sec2 q + tan q = A sec2 q + Btan q [given], On comparing, we get, A = 1 and B = 1, \ A + B=1 + 1 =2, , SAMPLE PAPER 5, , 4. Given equations are ax + b = 0 and cx + d = 0, On comparision with ax + by + c = 0, we have, a1 = a, b1 = 0 and c1 = b, a2 = c, b2 = 0 and c2 = d, For the system to be consistent,, a1 c1, =, a2 c2, a b, =, c d, or, ad = bc, 5. Given, area = 120 m 2 and perimeter = 44 m, Let the length be l and the breadth be b., …(i), \ Area, lb = 120 m2, and perimeter 2 ( l + b) = 44 m, Þ, l + b = 22, …(ii), Þ, l = 22 - b, On putting the value of Eq. (ii) in Eq. (i), Þ, (22 - b) b = 120, Þ, b2 - 22 b + 120 = 0, , Þ b2 - 12 b - 10 b + 120 = 0, Þ b( b - 12 ) - 10( b - 12 ) = 0, Þ, ( b - 12 )( b - 10 ) = 0, Þ, b = 10 or 12, \, l = 22 - 10 or l = 22 - 12, l = 12 m or l = 10 m, 6. Let the required number be x., As product of two numbers, = (HCF ´ LCM) of two number, \ x ´ 161 = 23 ´ 1449, 23 ´ 1449, = 207, x=, 161, 7. Probability of winning the game by Onkar, = 0.75, Total probability = 1, Here, P( A) + P (not A) = 1, So, probability of winning the game by Neha, = 1 - 0.75 = 0.25, 8. We have, the cost of 1 chair = ` x, and the cost of 1 table = ` y, Then, according to the given situation, 3 x + y = 1500, and, 6 x + y = 2400, 9. Let p( x ) = ax + b, Put x = k, then, p( k ) = ak + b = 0, [Q p( k ) = 0 ( given)], \ k is zero of p( x )., So, k is known as zero of p( x )., 10. Given, a bag contains 6 green balls and, n yellow balls., Total number of balls = 6 + n, So, the possible number of outcomes = 6 + n, 6, Now, P (drawing a green ball) =, 6+n, n, and P (drawing a yellow ball) =, 6+n, According to the question,, æ 6 ö, n, ÷ Þ n = 30, = 5çç, ÷, 6+n, è6 + nø

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119, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 11. We have, sin q =, , 1, 2, , \, , é, 1 ù, êQ sin x =, ú, cosec x û, ë, 1 1, Now, sin q (sin q - cosec q ) = æç - 2 ö÷, 2è2, ø, 1 - 4 -3, 1, = -1 =, =, 4, 4, 4, cosec q = 2, , 12. The number which is non-terminating and, non-repeating are irrational number., Here, only 5.329685 ...... is the required answer., , DF ED FE, =, =, QP RQ PR, , Hence, except option (b), all options are true., 16. Let DABC and DDEF are two similar triangles, such that ar ( DABC ) = 36 cm 2 and, ar ( DDEF ) = 64 cm 2, We know that ratio of area of two similar, triangles is equal to square of ratio of their, corresponding sides., ar ( DABC ) ( AB)2, =, Q, ar ( DDEF ) ( DE)2, , 13. Let the radius of new park be r m., Given, diameters of other circular parks are, d1 = 16 m and d2 = 12 m, d 16, =8 m, Þ, r1 = 1 =, 2, 2, d, 12, and, r2 = 2 =, =6 m, 2, 2, According to the question,, , Þ, , pr2 = pr12 + pr22, pr2 = p(8 )2 + p(6 )2, pr2 = 64 p + 36 p, pr2 = 100 p, r = 10 m, , \, Þ, Þ, Þ, , 14. When a point P( x , y ) divides (internally) the, line segment joining the points A( x1 , y1 ) and, B( x2 , y2 ) in the ratio m : n, then by section, formula, mx2 + nx1, my2 + ny1, and y =, x=, m+ n, m+ n, 1, , (k, 0), P, , Þ, , pr2, 3 2, a, 4, 2, d, pæç ö÷, = è2 ø, 3 2, a, 4, a2, p´, 4, =, 3 2, a, 4, p, =, 3, =, , B(–7, 4), , æ 1 ´ (- 7) + 2 ´ 2 1 ´ 4 + 2 ´ (- 2 ) ö, ÷, P( k , 0 ) = çç, ,, ÷, 1+2, 1+2, è, ø, æ -7 + 4 4 - 4 ö, =ç, ,, ÷ = ( - 1, 0 ), 3 ø, è 3, \, k = -1, , \ Required ratio = p : 3, , Q, , 18. We have,, , D, , F, , \DDEF - DQRP, , \, Þ, , E, , P, , [Q d = a], , R, , [by AA similarity criterion], , sin q = cosec q = 1, sin q + cosec q = 1 + 1, sin q + cosec q = 2, , 19. There are 5 + 9 = 14 fish in the tank., \ Total number of possible outcomes = 14, Out of 14 fish, there are 5 male fish., So, total number of favourable outcomes = 5, , SAMPLE PAPER 5, , 15. It is given that in DDEF and DPQR,, ÐD = ÐQ andÐR = ÐE, , 2, , 17. We are given that diameter of circle and side of, an equilateral triangle are equal., Let d and a be the diameter and side of circle, and equilateral triangle respectively., \ d=a, We know that,, Area of circle = p r2, 3 2, Area of an equilateral triangle =, a, 4, Area of circle, Ratio =, Area of equilateral triangle, , 2, , A(2, –2), , 36 æ AB ö, =ç, ÷, 64 è DE ø, AB 6 3, = =, DE 8 4

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120, , CBSE Sample Paper Mathematics Standard Class X (Term I), , sin q, cos q, -b, cos, cos q, q, =, sin q, cos q, a, +b, cos q, cos q, , \ Required probability, Number of favourable outcomes, =, Total outcomes, 5, =, 14, Hence, the probability that the fish taken out, as a male fish is 5/14., 20. Radius of circle = a cm, So, diameter of circle = 2 a, \Side of square diameter of circle = 2 a, Perimeter of square = 4 ´ side, = 4 ´ 2 a = 8 a cm, P, , A, , S, , a, O, a, Q, , B, , R, , 21. We know that,, If a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0, has infinite solution, then, a1 b1 c1, =, =, a2 b2 c2, , SAMPLE PAPER 5, , So, 2 x + 3 y - 5 = 0 and 4 x + ky - 10 = 0, has infinite solution, then, 2 3 -5, = =, 4 k - 10, 3 1, =, Þ, k 2, Þ, k =6, 0.9, 22. Given, probability of success = 0.9% =, 100, Since, probability of failure, + probability of success = 1, Probability of failure, = 1 - probability of success, 9, 0.9, =1 =11000, 100, 1000 - 9 991, =, =, = 0.991, 1000, 1000, 23. The graph of y = f ( x ) cuts on X-axis at 4 points., So, that number of zeroes lying between, -4 and 4 is 4., a, 24. We have, tan q =, b, ( a sin q - b cos q ), Now,, ( a sin q + b cos q ), , a, , [dividing numerator and, denominator by cos q], =, , ( a tan q - b), ( a tan q + b), , a 2 - b2, a, -b, a 2 - b2, b, =, = 2 b 2 = 2, a, a + b2, a´ + b a + b, b, b, 3434, 3434, 1717, 25. Here, 3.434 =, = 3, = 2, 3, 1000 2 ´ 5, 2 ´ 53, m, Denominator is of the form 2 ´ 5 n, m = 2 and n = 3 (positive integers), a´, , 26. Given, sum of zeroes = 3, and product of zeroes = -2, Quadratic polynomial, = x 2 - (sum of zeroes)x + (product of zeroes), = x 2 - 3 x + ( -2 ) = x 2 - 3 x - 2, 27. For the given system to be inconsistent, a1 b1 c1, =, ¹, a2 b2 c2, On comparision the given equation with the, standard equation, we get, a1 = 2 , b1 = 3 and c1 = 5, a2 = 4, b2 = k and c2 = 7, 2 3 5, = ¹, \, 4 k 7, Þ, k =6, 28. In case of tossing three coins., Total number of possible outcomes = 8, Favourable outcomes are, { HHT , HTH , THH , HTT , THT , TTH } i.e 6., 6 3, \ Required probability = =, 8 4, 29. Given,, , 3x = sec q, , …(i), 9 x 2 = sec2 q, 3, and, = tan q, x, 9, …(ii), = tan 2 q, x2, Subtracting Eq. (ii) from Eq. (i), we get, 9, 9 x 2 - 2 = sec2 q - tan 2 q, x, 1, Þ, [Q sec2 x - tan 2 x = 1], 9æç x 2 - 2 ö÷ = 1, x, ø, è, 1 1, 2, x - 2 =, Þ, 9, x

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121, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 30. Since, the denominator is of the form 2 n ´ 5 m., , 34. Distance between points ( m, - n) and ( - m, n), , So, it is terminating., 23, 23, Now, 2, =, = 115, ., 1, 20, 2 ´5, 23, So, 2, will terminate after 2 places of, 2 ´ 51, , = ( x2 - x1 )2 + ( y2 - y1 )2, [by Distance formula], = ( - m - m)2 + ( n + n)2, = ( -2 m)2 + (2 n)2, , decimal., , = 4 m2 + 4 n2, , 31. Let f ( x ) = x 2 - 3 x + k, , = 2 m2 + n2 units, , Since, 2 is a zeroes of f ( x )., \, f (2 ) = 0, Þ, 2 2 - 3(2 ) + k = 0, Þ, 4 -6 + k =0, Þ, k =2, , 35. In DABC, ÐC = 90 °, BC 2, =, AC 3, AC 3, =, tan B =, BC 2, 2 3, \tan A × tan B = ´ = 1, 3 2, tan A =, , 32. We know that, x = a is the equation of a, straight line parallel to the Y-axis at a distance, of ‘a’ units from it., Similarly, y = b is the equation of a straight line, parallel to the X-axis at a distance of ‘b’ units, from it., So, the pair of equations x = a and y = b, graphically represents lines which are, intersecting at ( a, b) as shown below, Y, (0, b), , (a , b ), y=b, , X¢, , 36. Required number of books will be LCM of 32, and 36. Now,, 32 = 2 5 and 36 = 2 2 ´ 3 2, \ LCM (32, 36) = 2 5 ´ 3 2 = 288, So, minimum number of books is 288., 37. In DABC and DDEF, we have, [given], ÐB = ÐE, [given], ÐF = ÐC, Þ, DABC ~ DDEF, [by AA similarity criterion], D, , X, (a, 0), , A, , x=a, , 0, , Y¢, , Hence, the two lines are intersecting at ( a, b)., 33. Radius of circle, r = 7 cm, Now, Area of shaded region, , B, , C, , E, , F, , Since, AB and DE are corresponding sides., But, [given], AB = 3 DE, We know that, two triangles are congruent,, if they have the same shape and size., But here, AB = 3 DE i.e., two triangles are, not of same size., \DABC is not be congruent to DDEF., Hence, the two triangles are similar but not, congruent., 38. Let r be the radius of the circle and q be the, angle formed by arc of the sector., q, Then, length of arc =, ´ 2 pr = 5 p, 360 °, q, and, area of sector =, ´ pr2 = 20 p, 360 °, , … (i), … (ii), , SAMPLE PAPER 5, , = Area of three sectors, q, q, q, = 1 pr2 + 2 pr2 + 3 pr2, 360 °, 360 °, 360 °, éQ Area of sector = q pr2 ù, úû, êë, 360 °, 2, pr, ( q1 + q 2 + q 3 ), =, 360 °, 22, 1, =, ´, ´ 7 ´ 7(60 ° + 80 ° + 40 ° ), 7 360 °, 1, = 11 ´, ´ 7 ´ 180 °, 180 °, = 77 cm 2, , [angle in a semi-circle]

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122, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Dividing Eq. (ii) by Eq. (i), we get, , Þ, , q, pr2, 20 p, 360 °, =, q, ´ 2 pr 5 p, 360 °, r, =4, 2, , 43. Let AB be a flag pole of the height 4m and BC, be its shadow is 3 m., A, , 4m, , r = 8 cm, , Þ, , C, , 39. Let a and b be the zeroes of x 2 + kx + k., Then, a + b = - k and ab = k, Since, k > 0, a + b < 0 and ab > 0, This is possible only when a and b are both, negative., 40. Given number is 196., 2, , Prime factors of 196 = 2 ´ 7, , 2, , So, the exponents are 2 and 2., \ Required sum = 2 + 2 = 4, Solution (41-45), , In right triangle DABC,, ÐB = 90 °, 2, \, BC + AB2 = AC 2, [from Pythagoras theorem], 3 2 + 4 2 = AC 2, 9 + 16 = AC 2, 25 = AC 2, AC = 25 = 5 m, , Þ, Þ, Þ, Þ, 44. In DPQR,, , ÐQ = 90 °, P, , m, , 41. Let AB be a flag pole of height 18 m and BC be, its shadow of 9.6 m long, As pole is vertical, so ÐABC = 90 °, , B, , The required distance of the top of the pole, A from the far end C of the shadow is AC., In right angled DABC, we have, AC 2 = AB2 + BC 2, [by Pythagoras theorem], 2, , Let PQ be a flag pole of the height 15 m and, the QR be a shadow of the flag pole., \, PQ2 + QR 2 = PR 2, [from Pythagoras theorem], 15 2 + QR 2 = 25 2, Þ, QR 2 = 625 - 225 = 400, QR = 20 m, 45. In DABC, use pythagoras theorem,, A, , 2, , Þ, , AC = (18 ) + (9.6 ), , Þ, , AC 2 = 324 + 92 .16, , Þ, , AC 2 = 416.16, , Þ, , SAMPLE PAPER 5, , 2, , Q, , m, , 9.6 m, , R, , 13, , C, , 15 m, , 25, , A, , 18 m, , B, , 3m, , AC = 416.16, , = 20.4, Hence, the required distance is 20.4 m., 42. Pythagoras theorem is used to find the, distance of the top of the pole from the far end, of the shadow., , B, , 5m, , C, , AC = AB2 - BC 2, AC = (13 )2 - ( 5 )2, = 169 - 25 = 144 = 12 m

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123, , CBSE Sample Paper Mathematics Standard Class X (Term I), , PS = (3 - 6 )2 + (6 - 5 )2, , Solution (46-50), 46. The coordinates of the origin is O(0, 0)., 47. The coordinates of Palak and Shivam are, P(3, 6) and S(6, 5) respectively., The distance between, PS = (3 - 6 )2 + (6 - 5 )2, , = 9 + 1 = 10 units, AM = (2 - 5 )2 + (3 - 2 )2, = 9 + 1 = 10 units, P(3, 6), , S(6, 5), , A(2, 3), , M(5, 2), , = 9+1, = 10 units, 48. The coordinates of Akshay are (2, 3)., 49. The coordinates of Akshay (A) and Malika (M), are (2, 3) and (5, 2) respectively., By using mid-point formula, coordinates of, point exactly between A and M is, æ2 + 5 3 + 2 ö æ7 5ö, ,, ç, ÷=ç , ÷, 2 ø è2 2 ø, è 2, or, , (3.5, 2.5), , AS = (6 - 2 )2 + ( 5 - 3 )2, = 4 2 + 2 2 = 16 + 4, = 20 = 2 5 units, and PM = ( 5 - 3 )2 + (2 - 6 )2, , 50. The coordinates of the families member are, A(2, 3), P(3, 6), M(5, 2) and S(6, 5)., The distance between these points are, 2, , 2, , AP = (2 - 3 ) + (3 - 6 ), , = 1 + 9 = 10 units, MS = ( 5 - 6 )2 + (2 - 5 )2, = 1 + 9 = 10 units, , = (2 )2 + ( - 4 )2, = 4 + 16, = 20 = 2 5 units, Here, AP = PS = MS = AM = 10 units, and diagonals AS = PM = 2 5 units, Hence, shape of the figured formed by joining, points is a square., , SAMPLE PAPER 5

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124, , CBSE Sample Paper Mathematics Standard Class X (Term I), , SAMPLE PAPER 6, MATHEMATICS (Standard), A Highly Simulated Practice Questions Paper, for CBSE Class X (Term I) Examination, , Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions., There is no negative marking., Maximum Marks : 40, Time allowed : 90 minutes, , Roll No., , Section A, Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 1. The square root of all positive integers are, (a) rational number, (c) may be rational or irrational number, , (b) irrational number, (d) None of these, , 2. If - 2 is one of the zeroes of the polynomial x 2 - x + k, then the value of k is, (a) - 6, , (b) - 8, , (c) 8, , (d) - 7, , 3. If the pair of lines are coincident, then we say that pair of lines is consistent and it has a, (a) unique solution, (c) infinite solutions, , (b) no solution, (d) None of these, , 4. In the given figure, DPMN ~ DPQR and PM = 3 cm, PQ = 4 cm, PN = 6 cm and, , SAMPLE PAPER 6, , PR = 8 cm, then relation between MN and QR is, P, , M, , Q, , (a) parallel, (c) intersect, , N, , R, , (b) perpendicular, (d) None of these

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125, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 5. If the points A(4 , 3) and B(x , 5) are on the circle with centre O(2, 3) , then the value of x is, (a) 5, , (b) 6, , é sin 2 q, , cos 2 q ù, is, 2, 2 ú, 1, +, q, 1, +, q, cot, tan, ë, û, (a) sin 2 q - cos 2 q, (b) sin 2 q, , 6. The value of ê, , (c) 2, , (d) 4, , (c) - 1, , (d) cos 2 q, , -, , 7. The HCF and LCM of two numbers are 33 and 264 respectively. When the first number, is completely divided by 2 the quotient is 33. The other number is, (a) 66, (b) 130, (c) 132, (d) 196, , 8. The parabola representing a quadratic polynomial f (x) = ax 2 + bx + c open downwards, when, (a) a > 0, , (b) a < 0, , (c) a = 0, , (d) a > 1, , 9. The values of x and y which satisfy the equations 2x + y + 1 = 0 and 2x - 3y + 8 = 0 are, (a) 1 and 2, , (b), , 11, 7, and, 8, 4, , (c) -, , 11, 7, and, 8, 4, , (d) 2 and 3, , 10. If in two right triangles, one of the acute angles of one triangle is equal to an acute, angle of the other, then two triangles are similar, then criteria of similarity is, (a) AA similarity, (b) SAS similarity, (c) ASA similarity, (d) None of these, , 11. If A = (a 2 , 2a) and B = æç, , 1, , èa, , (a) 2, , 2, 3, (a) 30°, , 12. If sin 4q =, , 2, , (b), , 3´4, 12, , ,-, , 1, 1, 2ö, is equal to, +, ÷ and S = (1, 0), then, SA SB, aø, , 1, 2, , 1, 3, , (c) 1, , (d), , (c) 15°, , (d) 60°, , , then the value of q is, (b) 45°, 2, , 13. If one zero of the quadratic polynomial 2x - 6kx + 6x - 7 is negative of the other, then k, is equal to, (a) - 1, , (b) 1, , (c) 0, , (d) -, , 1, 2, , 14. In school, there are two sections, section A and section B of class X. There are, 32 students in section A and 36 students in section B. Find the minimum number of, books required for their class library so that they can be distributed equally among, students of section A or section B., (b) 388, , (c) 208, , (d) None of these, , 15. If the pair of linear equations 2x + 3y = 11 and(m + n)x + (2m - n) y - 33 = 0 has infinitely, many solutions, then the values of m and n are, (a) 5 and 1, , (b) 1 and 2, , (c) - 1 and 5, , (d) 1 and - 5, , 16. One equation of a pair of dependent linear equations is - 5x + 7 y = 2. The second, equation can be, (a) 10x + 14y + 4 = 0, (c) - 10x + 14y + 4 = 0, , (b) - 10x - 14y + 4 = 0, (d) 10x - 14y = - 4, , SAMPLE PAPER 6, , (a) 288

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126, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 17. A girl walks 500 m towards East and then 1200 m towards North, then the travelling, distance from the starting point is, (a) 1100 m, (b) 1200 m, 4, 18. The value of sin q = is, 3, (a) possible, (c) in special case, if is possible, , (c) 1300 m, , (d) 1400 m, , (b) not possible, (d) None of these, , 19. The distance of the point (- 12, 5) from the origin is, (a) 12 units, , (b) 5 units, , (c) 13 units, , (d) 169 units, , 20. DABC is an isosceles triangle in which ÐC = 90°. If BC = 2 cm, then the value of AB is, (a) 4 2 cm, , (b) 2 2 cm, , (c) 4 cm, , (d) 2 cm, , Section B, Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 21. On Delhi road, three consecutive traffic lights change after 36, 42 and 72 seconds. If the, lights are first switched on at 9.00 am, at what time will they change simultaneously?, (a) 9:08:04, (b) 9:08:24, (c) 9:08:44, (d) None of these, , 22. If the zeroes of the quadratic polynomial ax 2 + x + a are equal, then a 2 is, (a), , 1, 2, , (b) 4, , (c) 2, , (d), , 1, 4, , m, n, , 23. Given that sin q = , then cot q - tan q is, (a), , n, m2 - n 2, , (b), , m2 - n 2, , (c), , m, , n, m2 - n 2, , (d), , n 2 - 2 m2, m n 2 - m2, , 24. If in two triangles DDEF and DPQR, ÐD = ÐQ and ÐR = ÐE, then which of the following, is not true?, EF DF, (a), =, PR PQ, DE DF, (c), =, QR PQ, , DE EF, =, QR QP, EF DE, (d), =, RP QR, , (b), , SAMPLE PAPER 6, , 25. If the distance between the points (x , - 1) and (3, 2) is 5, then the value of x is, (a) - 7 or - 1, , (b) - 7 or 1, , (c) 7 or 1, , (d) 7 or - 1, , p, , 26. The rational form of 0. 08 is in the form of , then ( p 2 + q 2 ) is, q, , (a) 1850, , (b) 1008, , (c) 2041, , (d) 3056, , 27. If both zeroes of the quadratic polynomial x 2 - 2kx + 2 are equal in magnitude but, opposite in sign, then value of k is, 1, 1, (a), (b) 2, 2, , (c) 0, , (d) - 1

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127, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 28. A number consists of two-digits. The sum of the digits is 12 and the unit’s digit, when, divided by the ten’s digit gives the result as 3. The number is, (a) 84, (b) 48, (c) 93, , (d) 39, , 29. Given DABC ~ DDEF. If AB = 4 cm, BC = 3.5 cm, CA = 2.5 cm and DF = 7.5 cm., Then, perimeter of DDEF is, (b) 30 cm, , (a) 20 cm, , (c) 35 cm, , (d) 47 cm, , 30. If 5x - 3y + 2k = 0 is a median of the triangle whose vertices are at points A(-1, 3), B(0, 4), and C( -5, 2). The value of k is, (a) 19, (b) 9.5, , (c) 20.5, , (d) 21.5, , 31. If the HCF of 408 and 1032 is expressible in the form 1032 ´ 2 + 408 ´ p, then the value of, p is, (a) 5, , (b) - 5, , (c) 4, , (d) - 4, , 32. If DABC ~ DDEF and DABC is not similar to DDEF then which of the following is not, true?, (a) BC × DF = AC × EF, (c) BC × DE = AB× EE, , (b) AB× DF = AE × DE, (d) BC × DE = AB× FE, , 33. If x = 3 and y = 1 is the solution of the line 2x + y - q 2 - 3 = 0. The value of q is, (a) ± 2, , (b) 2, , (c) - 2, , (d) - 1, , 34. The line segment joining the points (3, - 1) and (- 6, 5) is trisected. The coordinates of, point of trisection are, (a) (3, 3), (b) ( - 3, 3), , (c) (3, - 3), , (d) ( - 3, - 3), , 35. If the point (x , y) is equidistant from the points (2, 1) and (1, - 2), then, (a) x + 3y = 0, , 36. If, , (b) 3x + y = 0, , (c) x + 2 y = 0, , (d) 3x + 2 y = 0, , 1, 1, +, = k cosec 2q, then the value of k is, 1 + cos q 1 - cos q, , (a) 1, , (b) - 1, , (c) 2, , (d) 1/2, , 37. There is a circular path around a sports field. Priya takes 18 min to drive one round of, the field. Harish takes 12 min. Suppose they both start at the same point and at the same, time and go in the same direction. After how many minutes will they meet?, (a) 36 min, (b) 18 min, (c) 6 min, (d) They will not meet, , 38. If D and E are respectively the points on the sides AB and AC of a DABC such that, , 39. If the sum of the zeroes of the polynomial g(x) = ( p 2 - 23) x 2 - 2x - 12 is 1, then p takes, the value(s), (a) 23, (b) - 23, (c) 2, 5, 40. If sin q = , then the value of 2 sec q × tan q is, 13, 65, 45, 65, (a), (b), (c), 30, 25, 75, , (d) ± 5, , (d), , 65, 72, , SAMPLE PAPER 6, , AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then, length of DE (in cm) is, (a) 2.5, (b) 3, (c) 5, (d) 6

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128, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Section C, Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted., 41-45 are based on Case Study-1., , Case Study 1, Shivani bought a pendulum clock for her living room. The clock contains a small pendulum of, length 15 cm. The minute hand and hour hand of the clock are 18 cm and 12 cm long, respectively., , Based on the above information, answer the following questions., , 41. The area swept by the minute hand in 10 min are, (a) 22.24 cm 2, (c) 44 cm 2, , (b) 169.71 cm 2, (d) 44.42 cm 2, , 42. If the pendulum covers a distance of 22 cm in the complete oscillation, then the angle, described by pendulum at the centre is, (a) 40°, (c) 45°, , (b) 42°, (d) 48°, , 43. The angles described by hour hand in 10 min are, (a) 5°, (c) 15°, , (b) 10°, (d) 20°, , SAMPLE PAPER 6, , 44. The area swept by the hour hand in 1 h are, (a) 7.68 cm 2, (c) 8.86 cm 2, , (b) 8.2 cm 2, (d) 37.71 cm 2, , 45. The area swept by the hour hand between 11 am and 5 pm are, (a) 452.52 cm 2, (c) 70 cm 2, , (b) 62cm 2, (d) 72 cm 2

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129, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 46-50 are based on Case Study-2., , Case Study 2, On a weekend Rakhi was playing cards with her family. The deck has 52 cards and her brother, drew one card., , 46. Find the probability of getting a queen of red colour., (a), , 1, 26, , (b), , 1, 13, , (c), , 1, 52, , (d), , 1, 4, , (c), , 2, 13, , (d), , 3, 13, , 3, 52, , (d), , 3, 26, , (c), , 1, 52, , (d), , 1, 4, , (c), , 1, 52, , (d), , 1, 4, , 47. Find the probability of getting an ace., (a), , 1, 26, , (b), , 1, 13, , 48. Find the probability of getting a jack of diamond., (a), , 1, 26, , (b), , 1, 52, , (c), , 49. Find the probability of getting a red face card., (a), , 3, 26, , (b), , 1, 13, , 50. Find the probability of getting a club., (a), , 1, 26, , (b), , 1, 13, , SAMPLE PAPER 6

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OMR SHEET, , SP 6, , Roll No., Sub Code., , Student Name, , Instructions, Use black or blue ball point pens and avoid Gel & Fountain pens for filling the OMR sheet., Darken the bubbles completely. Don’t put a tick mark or a cross mark, half-filled or over-filled bubbles will not be read, by the software., Correct, , ✔, , ✗, , Incorrect, , Incorrect, , Incorrect, , Do not write anything on the OMR Sheet., Multiple markings are invalid., , 1, , 18, , 35, , 2, , 19, , 36, , 3, , 20, , 37, , 4, , 21, , 38, , 5, , 22, , 39, , 6, , 23, , 40, , 7, , 24, , 41, , 8, , 25, , 42, , 9, , 26, , 43, , 10, , 27, , 44, , 11, , 28, , 45, , 12, , 29, , 46, , 13, , 30, , 47, , 14, , 31, , 48, , 15, , 32, , 49, , 16, , 33, , 50, , 17, , 34, , Check Your Performance, Total Questions:, Total Correct Questions:, , If Your Score is, , Score Percentage =, , Total Correct Questions, Total Questions, , × 100, , Less than 60%, > Average (Revise the concepts again), Greater than 60% but less than 75% > Good (Do more practice), Above 75%, > Excellent (Keep it on)

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131, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Answers, 1. (c), 11. (c), 21. (b), , 2. (a), 12. (c), 22. (d), , 3. (c), 13. (b), 23. (d), , 4. (a), 14. (a), 24. (b), , 5. (c), 15. (a), 25. (d), , 6. (a), 16. (d), 26. (c), , 7. (c), 17. (c), 27. (c), , 8. (b), 18. (b), 28. (d), , 9. (c), 19. (c), 29. (b), , 10. (a), 20. (b), 30. (b), , 31. (b), 41. (b), , 32. (c), 42. (b), , 33. (a), 43. (a), , 34. (b), 44. (d), , 35. (a), 45. (a), , 36. (c), 46. (a), , 37. (a), 47. (b), , 38. (c), 48. (b), , 39. (d), 49. (a), , 40. (d), 50. (d), , SOLUTIONS, = sin 2 q ´ sin 2 q - cos 2 q ´ cos 2 q, , 1. Clearly, 9 = ± 3, which is rational number., , = sin 4 q - cos 4 q, , Hence, square root of all positive integers may, be rational or irrational number., , = (sin 2 q - cos 2 q ) (sin 2 q + cos 2 q ), , 2, , 2. Given, polynomial x - x + k, , [Q a2 - b2 = ( a - b) ( a + b)], , 2, , Let p( x ) = x - x + k, and - 2 is one of the zeroes of the polynomial., \ p( - 2 ) = 0, Þ, 0 = ( - 2 )2 - ( - 2 ) + k, Þ, 0=4+2 + k, Þ, k = -6, 3. If the pair of lines are coincident, then it has, infinite number of solutions and hence,, consistent., 4. Given, DPMN ~ DPQR, PM PN, 3 6, 3 3, Therefore,, =, Þ = Þ =, PQ PR, 4 8, 4 4, , 5. Since, A and B lie on the circle having centre O., Then, the distance between points A and B, from the centre are same as they are radius of, the circle., B (x, 5), , 7. Given, HCF = 33 and LCM = 264, Let the other number be x., First number = 2 ´ 33 = 66, Q Product of numbers = HCF ´ LCM, \, x ´ 66 = 33 ´ 264, 33 ´ 264, Þ, x=, 66, = 132, open downwards and a > 0, then parabola open, upwards., 9. The given equation can be written as, 2x + y + 1 =0, 2x + y = -1, , Þ, and, Þ, , 2 x - 3y = - 8, , i.e., OA = OB, From the distance formula,, , 4y = 7, 7, y=, 4, , On putting y =, , 2, , (2 ) + 0 = ( x - 2 ) + 4, , On squaring both sides, we get, 4 = ( x - 2 )2 + 4, 2, Þ, (x - 2 ) = 0 Þ x - 2 = 0 Þ x = 2, 6. Given,, , Þ, , 2, , sin q, cos q, 1 + cot 2 q 1 + tan 2 q, , Þ, 2, , =, , Þ, , 2, , sin 2 q, cos 2 q éQ1 + cot A = cosec Aù, ú, ê, 2, cosec q sec2 q êë 1 + tan 2 A = sec2 A úû, , \, , 7, in Eq. (ii), we get, 4, , 7, 2 x - 3æç ö÷ = - 8, è4ø, 21, -8, 2x =, 4, 21 - 32, 11, 2x =, =4, 4, 11, x=8, 11, 7, and y =, x=8, 4, , SAMPLE PAPER 6, , ( 4 - 2 )2 + (3 - 3 )2 = ( x - 2 )2 + ( 5 - 3 )2, , 2, , …(ii), , Subtract Eq. (ii) from Eq. (i), we get, , Þ, , Þ, , …(i), , 2 x - 3y + 8 = 0, , O (2, 3), , 2, , [Q sin 2 A + cos 2 A = 1 ], , 8. If a < 0 in f ( x ) = ax 2 + bx + c, then parabola, , By converse of Basic proportionality theorem,, we can say that MN || QR., , (4, 3) A, , = sin 2 q - cos 2 q

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132, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 10. Let two right angled triangle be DLMO and, DRST., T, , L, , \, , M, , R, , S, , ÐL = ÐR = 90 ° and ÐM = ÐS [acute angle], [AA similarity], DLMO ~ DRST, , 11. The distance between points S and A; S and B, are, SA = ( a2 - 1)2 + (2 a)2, = a4 + 1 - 2 a2 + 4 a2, , Hence, the minimum number of books, required to distributed equally among, students of section A and section B are 288., 15. The pair of linear equation has infinitely many, solutions, if, a1 b1 c1, =, =, a2 b2 c2, Here, on comparison the given equation with, standard equations, we get, , = a4 + 2 a2 + 1, = ( a2 + 1)2, , a1 = 2 , b1 = 3 and c1 = - 11, , = ( a2 + 1) units, 2, , and, , 1, -2, - 0 ö÷, SB = æç 2 - 1 ö÷ + æç, ø, è a, ø, èa, 1, 2, 4, = 4 +1- 2 + 2, a, a, a, , 2, , 1, 1, 2, + 1 + 2 = æç 2 + 1 ö÷, 4, a, a, ø, èa, 2, æ1+ a ö, 1, = 2 + 1 = çç 2 ÷÷ units, a, è a ø, 1, 1, 1, 1, \, +, = 2, +, SA SB a + 1 1 + a2, a2, , \, , a2 = m + n, b2 = 2 m - n and c2 = - 33, - 11, 2, 3, =, =, m + n 2 m - n - 33, , Þ, , 2, 3, 1, =, =, m+ n 2m- n 3, 2, 1, 3, 1, = and, =, m+ n 3, 2 m- n 3, , 2, , =, , =, , 1 + a2, a2, 1, =, =1, +, 2, 1 + a2, a +1 1+ a, 2, , 3, 2, sin 4 q =, 3, 3, 3, Then,, sin 4 q =, 2, , 12. Given,, , SAMPLE PAPER 6, , Þ, Þ, , 4 q = 60 °, q=, , Þ, , m + n = 6 and 2 m - n = 9, , …(i), , On adding Eqs. (i), (ii) and (iii), we get, 3 m = 15, Þ, m= 5, Putting m = 5 in Eq. (i), we get, 5 + n =6 Þ n =1, 16. In a pair of dependent linear equations, second, equation is equivalent to the k times the first, equation. Thus, the second equation is, k( - 5 x + 7 y - 2 ) = 0, On putting k = - 2, we get, , é, 3ù, êQ sin60 ° = 2 ú, û, ë, , 60 °, = 15 °, 4, , 13. Given, one zero of the quadratic polynomials is, negative of the other., Let one zero be a. Then, other zero be - a., b, Sum of zeroes = a, - (- 6k + 6) 6k - 6, =, \ a + (- a) =, 2, 2, 0 = 6k - 6, 6k = 6 Þ k = 1, , 10 x - 14 y + 4 = 0, or, , 10 x - 14 y = - 4, , Hence, 10 x - 14 y = - 4 is the second equation., 17. Let a girl starts from point O and walks 500 m, towards East, then 1200 m towards North., N, W, , N, E, , S, , 1200 m, , O, , 14. It is given that, 32 students in section A and, 36 students in section B., The minimum number of books required is the, LCM of 32 and 36., As,, 32 = 2 5, 36 = 2 2 ´ 3 2, \ LCM (32, 36) = 2 5 ´ 3 2 = 288, , O, , 500 m, , E

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133, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 22. Given, Quadratic polynomial = ax 2 + x + a, , By Pythagoras theorem,, 2, , 2, , ON = OE + NE, , 2, , ON 2 = ( 500 )2 + (1200 )2, ON 2 = 250000 + 1440000, = 1690000, or, ON = 1690000 = 1300 m, Hence, the girl is 1300 m far from the starting, point., Perpendicular 4, 18. As we know, sin q =, =, Hypotenuse, 3, Here, sin q is greater than 1, which is not, possible., 4, Hence, sin q = is not possible., 3, 19. The coordinates of origin is O(0, 0)., \The distance of the point ( -12 , 5 ) from, O(0, 0) = (0 + 12 )2 + (0 - 5 )2, , Let the zeroes be a and a., 1, Sum of zeroes = a, 1, a+ a=a, 1, 2a = a, a, Product of the zeroes = = 1, a, a× a = 1, , …(i), , a2 = 1, Squaring of Eq. (i), then we get, 1, 4 a2 = 2, a, 1, a2 = 2, 4a, 1, a2 =, 4´1, , = (12 )2 + ( - 5 )2, , …(ii), , [Q a2 = 1 from Eq. (ii)], a2 =, , = 144 + 25 = 169 = 13 units, 20. Given, DABC is an isosceles triangle, A, , 1, 4, , m, n, In right triangle DABC, ÐC = q and ÐB = 90 °, , 23. Given, sin q =, , A, , C, , and, \, In DABC,, , 2 cm, , B, , ÐC = 90 °, AC = BC = 2 cm, , 90º, , Let, , Þ, Þ, , 21. The factors are, , Here, LCM of 36, 42 and 72 = 504 s, i.e. 8 min 24 s, The lights are first switched on at 9 am (given), \The required time will be after 8 min 24 s or, 9:08:24., , AB = mk and AC = nk. Then,, , [by pythagoras theorem], Þ, , BC = k n2 - m2, , 2, 2, BC k n - m, =, AB, mk, AB, mk, and tan q =, =, BC k n2 - m2, , cot q =, , \ cot q - tan q =, =, , n2 - m2, , n2 - m2 - m2, m n2 - m2, , =, , m, , -, , m, , n2 - 2 m2, m n2 - m2, , 2, , n - m2, , SAMPLE PAPER 6, , 36 = 2 2 ´ 3 2, 42 = 2 ´ 3 ´ 7, 72 = 2 3 ´ 3 2, \ LCM (36, 42, 72) = 2 3 ´ 3 2 ´ 7 = 504, , C, , BC = AC 2 - AB2 = ( nk )2 - ( mk )2, , AB2 = AC 2 + BC 2, [by Pythagoras theorem], AB2 = (2 )2 + (2 )2, AB2 = 4 + 4 = 8, AB = 8 = 2 2 cm, , q, , B

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134, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 24., , D, , Þ, , P, , - (- 2 k ), 1, 0 =2k, k =0, , a + (- a) =, , Þ, Þ, , 28. Let two-digits number be xy in expanded form, E, , F, , Q, , In DDEF and DQRP,, ÐD = ÐQ, ÐE = ÐR, From AA-criterion,, DDEF ~ DQRP, Þ, ÐF = ÐP, ED FE DF, \, =, =, RQ PR QP, , R, , [given], , [by CPCT], , Hence, option (b) is not true., 25. Let P ( x , - 1) and Q(3 , 2 ) by the given point,, then PQ = 5., ( x - 3 )2 + ( - 1 - 2 )2 = 5, ( x - 3 )2 + 9 = 25, 2, x - 6 x + 9 + 9 = 25, x2 - 6x - 7 = 0, ( x - 7 ) ( x + 1) = 0, x = 7 or x = - 1, , Þ, Þ, Þ, Þ, Þ, , of xy = 10 x + y, Given, sum of digits = 12, i.e., x + y = 12, and unit’s digit divided by ten’s digit,, y, =3, x, Thus,, y = 3x, , SAMPLE PAPER 6, , p = 4 and q = 45, \, , p2 + q2 = ( 4 )2 + ( 45 )2, = 16 + 2025, p + q = 2041, 2, , 2, , 27. Given, zeroes of the quadratic polynomial are, equal in magnitude but opposite in sign., Let the zeroes of quadratic polynomial be a,, then the other zero be - a, Coefficient of x, Sum of zeroes = Coefficient of x 2, , …(ii), , On putting y = 3 x in Eq. (i), x + 3 x = 12, Þ, 4 x = 12, Þ, x =3, and, y =9, So, the required number is 3(10 ) + 9 = 39., 29. Given, DABC ~ DDEF, A, , 4 cm, , 26. Let x = 0.08, then, …(i), x = 0.08888 ...., On multiplying Eq. (i) by 10, we get, …(ii), \, 10 x = 0.8888, On subtracting Eq. (i) from Eq. (ii), we get, 10 x - x = 0.888 ...... - 0.0888 ......, 9 x = 0.8000, 8000, x=, 9 ´ 10000, 8, 4, =, =, 90 45, p, On comparing with , we get, q, , …(i), , B, , D, , 2.5 cm, , 3.5 cm, , C, , 7.5 cm, , E, , F, , Perimeter of DABC AC, =, Perimeter of DDEF DE, AB + BC + AC, AC, Þ, =, Perimeter of DDEF DF, 4 + 3.5 + 2 .5, 2 .5, Þ, =, Perimeter of DDEF 7.5, 10, 1, Þ, =, Perimeter of DDEF 3, Þ Perimeter of DDEF = 3 ´ 10 = 30, Hence, perimeter of DDEF is 30 cm., , \, , 30. The coordinates of the centroid G of DABC are, -1 + 0 - 5 3 + 4 + 2 ö, ,, Gæç, ÷, 3, 3, ø, è, -6 9 ö, æ, i.e., Gç, , ÷ or G( -2 , 3 ), è 3 3ø, Since, G lies on the median line, 5x - 3y + 2 k = 0, \ 5( -2 ) - 3(3 ) + 2 k = 0, Þ, - 10 - 9 + 2 k = 0, Þ, 2 k = 19, 19, k=, = 9.5, Þ, 2

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135, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 35. Let the points be P( x , y ), A(2 , 1) and B(1, -2 )., Given, P is equidistant from A and B i.e. the, distance between A and P, B and P are equal., , 31. Prime factors of given numbers are, 408 = 2 3 ´ 3 ´ 17, and, 1032 = 2 3 ´ 3 ´ 23, HCF (408, 1032) = 2 3 ´ 31 = 24, But HCF express in the form, 1032 ´ 2 + 408 ´ p = 24, Þ, 2064 + 408 ´ p = 24, Þ, 408 p = - 2040, 2040, p==-5, Þ, 408, 32., , The distance AP = ( x - 2 )2 + ( y - 1)2, and, , \, AP = BP, Þ, AP2 = BP2, Þ ( x - 2 )2 + ( y - 1)2 = ( x - 1)2 + ( y + 2 )2, Þ x2 - 4x + 4 + y2 - 2 y + 1, = x2 - 2 x + 1 + y2 + 4y + 4, Þ, 2 x + 6y = 0 Þ x + 3y = 0, , D, , A, , 36., , B, , C, , E, , F, , II, , III, , 33. As, x = 3 and y = 1 is the solution of, 2 x + y - q2 - 3 = 0, When x = 3 and y = 1,, 2 ´ 3 + 1 - q2 - 3 = 0, , 34., , 2, = 2 cosec2 q, sin 2 q, Þ 2 cosec2 q = k cosec2 q, Þ, k =2, =, , Taking I and II ratio, we get, AB BC, =, ED EF, Þ, AB× EF = ED × BC, Taking II and III ratio, we get, BC AC, =, Û BC × DF = AC × EF, EF DF, Taking I and III ratio, we get, AB AE, =, Û AB× DF = AE × ED, ED DF, Hence, (a), (b) and (d) is correct but option (c), is not true., , Þ, \, , 1 - cos q + 1 + cos q, 1, 1, +, =, 1 + cos q 1 - cos q (1 - cos q )(1 + cos q), 2, =, 1 - cos 2 q, [Q sin 2 A + cos 2 A = 1], , Given, DABC ~ DDEF, AB BC AC, \, =, =, ED EF DF, I, , 38. In the given figure, DE || BC, then by basic, proportionality theorem, we have, A, 2 cm, , (3, –1), , D, , B, , Q, , Since, the line segment AB is trisected., \, PB : BQ = 2 : 1, \Coordinates of B are, æ 2 ( - 6 ) + 1(3 ) 2 ( 5 ) + 1( - 1) ö, ÷, = çç, ,, ÷, 2 +1, 2 +1, è, ø, æ - 12 + 3 10 - 1 ö, =ç, ,, ÷, 3, 3 ø, è, 9 9, = æç - , ö÷ = ( -3 , 3 ), è 3 3ø, , Þ, Þ, Þ, , 7.5 cm, , C, , AD AE DE, =, =, DB EC BC, AD DE, =, DB CB, 2 DE, =, 3 7.5, 2 ´ 7.5, DE =, 3, = 2 ´ 2 .5 = 5 cm, , SAMPLE PAPER 6, , B, , E, , 3 cm, , (–6, 5), A, , [given], , 37. Given, Priya takes 18 min whereas Harish, takes 12 min to drive one round of the field, Let us find the LCM of 18 and 12., We have, factors of 18 and 12 are, 18 = 2 ´ 3 ´ 3 = 2 ´ 3 2, and, 12 = 2 ´ 2 ´ 3 = 2 2 ´ 3, LCM (18, 12) = 2 2 ´ 3 2 = 36, Hence, after 36 min Priya will be at the, starting point after completing 2 rounds of the, field and Harish also be at same point after, 3 rounds., , 4 - q2 = 0 Þ 4 = q2, q=±2, , P, , BP = ( x - 1)2 + ( y + 2 )2

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136, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 39. Let a and b be the zeroes of the polynomial,, , 44. Angle made by hour hand in 1 h =, , g( x ) = ( p2 - 23 )x 2 - 2 x - 12, ( -2 ), 2, Then, ( a + b) = - 2, =, p - 23 p2 - 23, , Also, r = 12 cm, \ Area swept by hour hand in 1 h, = Area of sector having central angle 30°, 30 ° ö, = pr2 ´ æç, ÷, è 360 ° ø, 1 264, 22, = 37.71 cm 2, =, ´ 12 ´ 12 ´, =, 12, 7, 7, , Also given, sum of zeroes = a + b = 1, 2, Þ, = 1 Þ p2 - 23 = 2 Þ p2 = 25, 2, p - 23, \, , p=± 5, , 5, 40. Given, sin q =, 13, Let be DABC be right angle,, triangle, we get, ÐB = 90 ° and ÐC = q, \ AB = 5 k and AC = 13 k, 2, , 45. Number of hours from 11 am to 5 pm = 6, Area swept by hour hand in 1 hour = 37.71 cm 2, \ Area swept by hour hand in 6 h, = 37.71 ´ 12 = 452 .52 cm 2, , A, , q, B, , C, , 2, , BC = AC - AB, , = (13 k )2 - ( 5 k )2 = 169 k 2 - 25 k 2, = k 144 = 12 k, AB 5, AC 13, and sec q =, So, tan q =, =, =, BC 12, BC 12, 13, 5 65, \2 sec q × tan q = 2 ´, ´, =, 12 12 72, 41. Angles made by minutes hand in 60 min = 360 °, \ Angle made by minute hand in 10 min, 360 °, ´ 10 = 60 °, =, 60 °, Length of minute hand = 18 cm, [given], \ Area swept by minute hand in 10 min, = Area of sector having central angles 60°, 60 ° ö 22, 1, = pr2 æç, ´ 18 ´ 18 ´, ÷=, 6, è 360 ° ø 7, 1188, = 169.71 cm 2, =, 7, , SAMPLE PAPER 6, , 360 °, = 30 °, 12, , 42. Let r be the length of the pendulum., 1, Given, r = 15 cm and l = (22 ) = 11 cm, 2, q ö, æ, We know that, l = 2 pr ç, ÷, è 360 ° ø, 11 ´ 360 °, Þ, q=, 22, 2´, ´ 15, 7, 90 ° ´ 7, =, = 6 ° ´ 7 = 42 °, 15, 43. Angle made by hour hand in 12 h = 360 °, 1, \ Angle made by hour hand in 10 min or h, 6, 360 ° 1 ö, = æç, ´ ÷ = 5°, 6ø, è 12, , 46. Total number of cards in one deck of cards is 52., Total number of outcomes = 52, Let E1 = Event of getting a queen of red colour, \Number of outcomes favourable of E1 = 2, [Q there are four queens in a deck of, playing cards out of which two, are red and two are black], Hence, probability of getting a queen of red, 2, 1, colour, P( E1 ) =, =, 52 26, 47. Let E2 = Event of getting an ace., \Number of outcome favourable to E2 = 4, [Q in a deck of cards, there are 4 ace cards], Hence, probability of getting an ace,, 4, 1, =, P( E2 ) =, 52 13, 48. Let E3 = Event of getting a jack of diamond, \ Number of outcomes favourable to E3 = 1, [Q there are four jack cards in a deck, namely 1, of heart, 1 of club, 1 of spade and 1 of diamond], Hence, probability of getting a jack of, 1, diamond =, 52, 49. Let E4 = Event of getting a red face card., \ Number of outcomes favourable to E4 = 6, [Q in a deck of cards, there are 12 face cards out, of which 6 are red cards], Hence, probability of getting a red face card,, 6, 3, =, P( E4 ) =, 52 26, 50. Let E5 = Event of getting a club., \ Number of outcomes favourable to E5 = 13, [Q in a deck of cards, there are 13 spades,, 13 clubs, 13 hearts and 13 diamonds], Hence, probability of getting a club,, 13 1, P( E5 ) =, =, 52 4

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137, , CBSE Sample Paper Mathematics Standard Class X (Term I), , SAMPLE PAPER 7, MATHEMATICS (Standard), A Highly Simulated Practice Questions Paper, for CBSE Class X (Term I) Examination, , Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions., There is no negative marking., Maximum Marks : 40, Time allowed : 90 minutes, , Roll No., , Section A, Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 1. The value of k, for which the pair of linear equations kx + y = k 2 and x + ky = 1 have, infinitely many solution, is, (a) ± 1, (b) 1, , (c) -1, , (d) 2, , 2. If p is the probability of happening of an event and q is the probability of, non-happening of an event, then p + q is equal to, (b) 1, (c) - 1, , (a) 0, , (d) 2, , 3. In the following figure, O is the centre of the circle. The area of the sector OAPB is, , 5, 12, , part of the area of the circle. Find the value of x., O, , A, , B, P, , (a) 130°, , (b) 60°, , (c) 45°, , (d) 150°, , (b) 2 sin A, , (c) 2 cos A, , (d) sec A, , cos A, 4. æç, + sin A ö÷ is, è cot A, (a) cot A, , ø, , 5. If DABC is similar to DPRQ, then which of the following is true?, (a) AB = PQ, , (b) AC = PQ, , (c) AB = QR, , (d) BC = PR, , SAMPLE PAPER 7, , x

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138, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 6. The distance of the point (3, 5) from the X-axis is, (a) 3 units, , (b) 5 units, , (c) 8 units, , (d) 4 units, , 7. The external and internal diameters of a circular path are 12 m and 8 m respectively., The area of the circular path is, (a) 9p m 2, (b) 16p m 2, , (c) 20p m 2, , (d) 36p m 2, , 8. Which of the following is not an irrational number?, (a) 7 5, (c) ( 7 - 3) - 7, , (b) 2 + 2 2, (d) 3 + 2, , 9. An integer is chosen between 0 to 50. What is the probability that it is divisible by 4?, (a), , 12, 49, , (b), , 13, 49, , (c), , 1, 7, , (d), , 4, 49, , 10. If x × tan 45°× cot 60° = sin 30°× cosec 60 °, then the value of x is, (a) 1, , (b), , 1, 4, , (c), , 1, 2, , (d) 3, , 11. The ratio in which the X-axis divides the line segment joining A(3, 6) and B(12, - 3) is, (a) 2 : 1, , (b) 1 : 2, , (c) - 2 : 1, , (d) 1 : - 2, , 12. The values of x and y in x - y + 1 = 0 and 3x + 2y - 12 = 0 are, (a) 2, - 3, , (b) - 2, 3, , (c) - 2 , - 3, , (d) 2, 3, , 13. The diameter of a wheel is 1.26 m. How long will it travel in 500 revolutions?, (a) 1492 m, , (b) 2530 m, , (c) 1980 m, , (d) 2880 m, , 14. If cosec 2q(1 + cos q) (1 - cos q) = l, then the value of l is, (b) cos 2 q, , (a) 0, , (c) 1, , (d) - 1, , 15. What is the probability of getting a king in a well shuffled pack of 52 cards?, (a), , 1, 13, , (b), , 1, 52, , (c), , 1, 26, , (d) 0, , 16. If the distance between A(k , 3) and B (2, 3) is 5, then the value of k is, (a) 5, , (b) 6, , (c) 7, , æ 4 sin q - cos q ö, ÷ is equal to, ÷, è 4 sin q + cos q ø, 1, 1, (b), (c), 3, 2, , (d) 8, , 17. If 4 tan q = 3, then çç, , SAMPLE PAPER 7, , (a), , 2, 3, , (d), , 3, 4, , 18. In a right-angled triangle, the square of the hypotenuse is equal to the, (a) Sum of other two sides, (b) Sum of squares of other two sides containing right angle, (c) Square of the perpendicular, (d) Square of the base, , 19. For which value(s) of p, will the lines represented by the following pair of linear, equations be parallel, 3x - y - 5 = 0, 6x - 2y - p = 0

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139, , CBSE Sample Paper Mathematics Standard Class X (Term I), , (a) all real values except 10, 5, (c), 2, , (b) 10, 1, (d), 2, , 20. If two angles of a triangle are equal to the corresponding two angles of another, triangle, then in such case two triangles can be called similar., (a) True, (b) False, (c) Can’t say, (d) None of these, , Section B, Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 21. If A(5, 3), B (11, - 5) and P (12, y) are the vertices of a right triangle, right angled at P,, then y is equal to, (a) - 2 or 4, , (b) - 2 or - 4, , (c) - 4 or 2, , (d) 2 or 4, , 22. In DPQR right angled at Q, QR = 3 cm and PR - PQ = 1 cm., The value of sin 2 R + cosec R is, 4, 4, (b), (a), 5, 9, , (c), , 189, 100, , (d), , 198, 100, , 23. One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The, probability that the selected ticket has a number, which is a multiple of 5, is, 1, 3, 4, 1, (a), (b), (c), (d), 5, 5, 5, 3, , 24. In a right-angled DABC, ÐC = 35° and in another right-angled DPQR, ÐR = 35°., Then relation between the two triangles is, (a) Congruent, (b) Equal, (c) Similar, , (d) No relation, , 25. The value of k for which the system of equations x + y - 4 = 0 and 2x + ky = 3, has no, solution is, (a) 2, , (b) 8, , (c) 3/4, , (d) - 2, , 26. Suppose, O is the centre of a circle of radius 5 cm. The chord AB subtends an angle of, 60° at the centre. Area of the shaded portion is equal to (approximately), , O, 60°, A, , B, , 27. The perimeter of the triangle formed by the points (0, 0), (2, 0) and (0, 2) is, (a) (1 - 2 2 ) units, , (b) (2 2 + 1) units, , (c) (4 + 2 ) units, , (d) (4 + 2 2 ) units, , 28. After how many places, the decimal expansion of the rational number, , 43, 2 ´ 53, , terminate, (a) 1, , (b) 2, , (c) 3, , (d) 4, , 4, , will, , SAMPLE PAPER 7, , (b) 62.78 cm 2, (d) 67.75 cm 2, , (a) 50 cm 2, (c) 49.88 cm 2

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140, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 29. Three unbiased coins are tossed together, then the probability of getting, exactly 1 head is, 1, (a), 8, , (b), , 2, 8, , (c), , 3, 8, , (d), , 1, 2, , 30. Equations 3x + 4 y + 5 = 0 and 6x + 8y + 9 = 0 represents a pair of ............ lines., (a) intersecting, (c) parallel, , (b) coincident, (d) None of these, , 31. In the given figure, if D is mid-point of BC, then value of, , cot y, cot x, , is, , A, x, y, , C, , (a) 2, , (b), , B, , D, , 1, 4, , (c), , 1, 3, , (d), , 1, 2, , 32. Name the criteria of similarity by which following triangles are similar, R, c, , 4.5, , 3, , 6, , 4, , A, , B, , 5, , (a) SSS, , P, , Q, , 7.5, , (b) SAS, , (c) AAA, , (d) ASA, , 33. The ratio in which the line segment joining the points (6,- 8) and (- 3, 10) is divided by, ( - 1, 6) is, (a) 2 : 3, , (b) 2 : 5, , (c) 7 : 2, , (d) 2 : 7, , 34. A letter is chosen at random from the letter of the word ‘ASSASSINATION’, then the, probability that the letter chosen is a vowel is in the form of, (a) 5, , (b) 6, , 6, , then x is equal to, 2x + 1, , (c) 7, , (d) 8, , SAMPLE PAPER 7, , 35. The area of the circle that can be inscribed in the square of side 6 cm is, (a) 18 p cm 2, , (b) 12 p cm 2, , (c) 9 p cm 2, , (d) 14 p cm 2, , 36. Name the criteria of similarity by which following triangles are similar., R, c, 6, A, , (a) Not similar, , 9.5, , 6, , 4, , 53°, , B, , (b) ASS, , P, , 53°, , Q, , (c) SSS, , (d) SAS

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141, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 37. The father’s age is six times his son’s age. Four years hence, the age of the father will be, four times his son’s age. The present ages, in years, of the son and the father are,, respectively., (a) 4 and 24, (b) 5 and 30, (c) 6 and 36, (d) 3 and 24, , 38. What is the supplementary angle of the central angle of a semicircle?, (a) 0°, , (b) 90°, , (c) 180°, , (d) 360°, , 39. P(5, - 3) and Q(3, y) are the points of trisection of the line segment joining A(7 , - 2) and, B(1, - 5), then y equals, (b) 4, , (a) 2, , (c) - 4, , (d) - 5 / 2, , 40. A race track is in the form of a ring whose inner circumference is 352 m and outer, circumference is 396 m. Then, width of the track is, (a) 7 m, (b) 9 m, (c) 11 m, , (d) 17 m, , Section C, Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted., 41-45 are based on Case Study-1., , Case Study 1, The Republic day parade is celebrated on 26th January every year in India. The day is celebrated, in the form of parades and other military shows in the national capital New Delhi as well as in, all headquarters of army., Parade I An Army contingent of 616 members is to march behind an army band of 32 members, in parade. The two groups are to march in the same number of columns., Parade II An Army contingent of 1000 members is to march behind an army band of 56 members, in parade. The two groups are to march in the same number of columns., , Refer to Parade I, (a) 2 1 ´ 141 ´ 22 1, (c) 2 3 ´ 71 ´ 111, , (b) 2 2 ´ 111 ´ 141, (d) 2 4 ´ 7 2 ´ 111, , 42. The HCF of 32 and 616 is, (a) 8, Refer to Parade II, , (b) 16, , (c) 18, , (d) 12, , (c) 8000, , (d) 9000, , 43. The LCM of 56 and 1000 is, (a) 6000, , (b) 7000, , SAMPLE PAPER 7, , 41. Number 616 can be expressed as a product of its prime factors as

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142, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 44. Number 1000 can be expressed as a product of its prime factors as, (a) 2 3 ´ 5 3, , (b) 2 2 ´ 5 4, , (c) 2 4 ´ 5 2, , (d) 2 3 ´ 5 4, , 45. The maximum number of columns in which army can march is, (a) 6, , (b) 10, , (c) 12, , (d) 8, , 46-50 are based on Case Study-2., , Case Study 2, Basketball and football are played with a spherical ball. Even though an athlete dribbles the ball, in both sports, a basketball player uses his hands and a football player uses his feet. Usually,, football is played outdoors on a large field and basketball is played indoor on a court made out, of wood. The projectile (path traced) of football and basketball are in the form of parabola, representing quadratic polynomial., , 46. The shape of the path traced shown in, (a) Spiral, , (b) Ellipse, , (c) Linear, , (d) Parabola, , 47. The graph of parabola opens downwards, if ……… ., (a) a = 0, , (b) a < 0, , (c) a > 0, , (d) a ³ 0, , 48. Observe the following graph and answer., Y, 6, , X¢, , 2, –4, , –3, , –2, , 2, , –1, , 3, , 4, , X, , –2, –6, , SAMPLE PAPER 7, , Y¢, , In the above graph, how many zeroes are there for the polynomial?, (a) 0, (b) 1, (c) 4, (d) 3, , 49. The four zeroes in the above shown graph are, (a) 2, 3, - 1, 4, (c) - 3, - 1, 2, 4, , (b) - 2, 3, 1, 4, (d) - 2 , - 3, - 1, 4, , 50. Which will be the expression of the quadratic polynomial?, (a) x 4 + 2 x 2 - 5x - 6, (c) 2 x 2 + 5x - 6, , (b) x 3 + 2 x 2 - 5x + 6, (d) 5x + 6

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OMR SHEET, , SP 7, , Roll No., Sub Code., , Student Name, , Instructions, Use black or blue ball point pens and avoid Gel & Fountain pens for filling the OMR sheet., Darken the bubbles completely. Don’t put a tick mark or a cross mark, half-filled or over-filled bubbles will not be read, by the software., Correct, , ✔, , ✗, , Incorrect, , Incorrect, , Incorrect, , Do not write anything on the OMR Sheet., Multiple markings are invalid., , 1, , 18, , 35, , 2, , 19, , 36, , 3, , 20, , 37, , 4, , 21, , 38, , 5, , 22, , 39, , 6, , 23, , 40, , 7, , 24, , 41, , 8, , 25, , 42, , 9, , 26, , 43, , 10, , 27, , 44, , 11, , 28, , 45, , 12, , 29, , 46, , 13, , 30, , 47, , 14, , 31, , 48, , 15, , 32, , 49, , 16, , 33, , 50, , 17, , 34, , Check Your Performance, Total Questions:, Total Correct Questions:, , If Your Score is, , Score Percentage =, , Total Correct Questions, Total Questions, , × 100, , Less than 60%, > Average (Revise the concepts again), Greater than 60% but less than 75% > Good (Do more practice), Above 75%, > Excellent (Keep it on)

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144, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Answers, 1. (b), 11. (a), , 2. (b), 12. (d), , 3. (d), 13. (c), , 4. (b), 14. (c), , 5. (b), 15. (a), , 6. (b), 16. (c), , 7. (c), 17. (c), , 8. (c), 18. (b), , 9. (a), 19. (a), , 10. (a), 20. (a), , 21. (c), 31. (d), 41. (c), , 22. (c), 32. (a), 42. (a), , 23. (a), 33. (c), 43. (b), , 24. (c), 34. (b), 44. (a), , 25. (a), 35. (c), 45. (d), , 26. (d), 36. (a), 46. (d), , 27. (d), 37. (c), 47. (b), , 28. (d), 38. (a), 48. (c), , 29. (c), 39. (c), 49. (c), , 30. (c), 40. (a), 50. (c), , SOLUTIONS, 1. The system of given equations will have, infinitely many solutions, when, a1 b1 c1, =, =, a2 b2 c2, On comparision the given equation with, standard equation, we get, a1 = k, b1 = 1, c1 = - k 2, a2 = 1, b2 = k, c2 = - 1, k 1 -k 2, = =, -1, 1 k, k 1, Consider,, = Þ k2 = 1, 1 k, Þ, k =±1, 1 k2, and, =, 1, k, Þ, k3 = 1 Þ k = 1, Hence, the common solution is k = 1., 2. Given,, The probability of happening of an event is, P( E) = p, and probability of non-happening of an event, is P( E ) = q, We know that,, P( E) + P( E ) = 1, Þ, p + q =1, 3. Let r be the radius of the circle., Area of circle = pr, , 2, , x, ´ pr2, 360 °, Given, area of sector OAPB, 5, =, ´ Area of circle, 12, On putting the values, we get, x, 5, ´ pr2 =, ´ pr2, 360 °, 12, x, 5, Þ, =, 360 ° 12, 5 ´ 360 °, x=, = 150 °, Þ, 12, , cos A, + sin A, cot A, cos A, = æç, ´ sin A ö÷ + sin A, è cos A, ø, = sin A + sin A = 2 sin A, , 4. We have,, , é\cot A = cos A ù, êë, sin A úû, , 5. Since, DABC ~ DPRQ, Therefore by CPCT,, AB = PR, BC = RQ, and, AC = PQ, 6. The point (3, 5) is shown in the figure, Y, 3 unit, , (3, 5), 5 unit, X, , From the figure, the distance is 5 units., 7. Let d1 and d2 be external and internal diameters., Given, d1 = 12 m and d2 = 8 m, Area of the path can be given by the difference, of the area of internal and external circles., , d1, d2, , SAMPLE PAPER 7, , Area of sector OAPB =, , 2, é d 2, d ù, \Area of the circular path = p êæç 1 ö÷ - æç 2 ö÷ ú, ëè 2 ø è 2 ø û, 2, 2, é 12, 8 ù, = pêæç ö÷ - æç ö÷ ú = p[(6 )2 - ( 4 )2 ], ëè 2 ø è 2 ø û, , = p{36 - 16} = 20 p m 2, 8. ( 7 - 3 ) - 7 = 7 - 3 - 7 = - 3, Since, - 3 is a rational number., So, ( 7 - 3 ) - 7 is a rational number.

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145, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 9. Total number of possible outcomes = 49, [Q Numbers 0 and 50 are not, included in the outcomes], Favourable outcomes are {4, 18, 12 ..., 48}, i.e. 12., Required probability, Number of favourable outcomes, =, Total number of outcomes, 12, =, 49, 10. We have,, x × tan 45 ° × cot 60 ° = sin 30 ° cosec 60°, 1, 1 2, 1, Þ, Þx=, x ´ 1., = ´, ´ 3, 3 2, 3, 3, \, , x =1, , 11. Let P( x , 0 ) be the point of intersection of X-axis, with the line segment joining A(3 , 6 ) and, B(12 , - 3 ) which divides the line segment AB in, the ratio l : 1., Using section formula,, æ 12 l + 3 - 3 l + 6 ö, ÷, ( x , 0 ) = çç, ,, l + 1 ÷ø, è l+1, Now, equating the y component on both sides,, we get, - 3l + 6, 2, =0 Þ l =, l+1, 1, So, X-axis divides AB in the ratio 2 : 1., 12. The given equation can be written as, x - y = -1, and, , 3 x + 2 y = 12, , 14. Given,, cosec2 q(1 + cos q )(1 - cos q ) = l, cosec2 q (1 - cos 2 q ) = l, , Þ, , [Q a2 - b2 = ( a - b)( a + b)], 2, , cosec q ´ sin 2 q = l, [Q1 - cos 2 q = sin 2 q], , Þ, , cosec2 q ´, , Þ, , l =1, , Þ, 15. Total outcomes = 52, , Favourable outcomes of getting a king = 4, 4, 1, Probability =, =, 52 13, 16. The distance between A and B is, AB = ( k - 2 )2 + (3 - 3 )2, [By distance formule], \, , 2, , [given], , (k - 2 ) = 5, 2, , Squaring both sides, we have ( k - 2 ) = 25, k -2 = ± 5, Þ, k =2 ± 5, Þ, k = 7 or - 3, 17. Given,, , 4 tan q = 3, 3 BC, Þ, tan q = =, 4 AB, Let BC = 3 k and AB = 4 k, In right DABC,, AC = AB2 + BC 2, , …(iii), , = ( 4 k )2 + (3 k )2, = 16 k 2 + 9 k 2, = 25 k 2 = 5 k, C, , 3k, q, , 13. Diameter of the wheel = 1.26 m, Radius of the wheel,, 1.26, r=, = 0.63 m, 2, Distance travelled in one revolution, = Perimeter of the wheel, = 2 pr, 22, =2 ´, ´ 0.63 = 3.96 m, 7, , é, 1, 1 ù, = l êQ sin q =, ú, 2, cosec q û, sin q, ë, , A, , \, , Now,, , 4k, , B, , BC 3, AB 4, = and cos q =, =, AC 5, AC 5, 3 4, 4 sin q - cos q 4 ´ 5 - 5, =, 4 sin q + cos q 4 ´ 3 + 4, 5 5, 12 - 4 8 1, =, =, =, 12 + 4 16 2, , sin q =, , SAMPLE PAPER 7, , On multiplying Eq. (i) by 2, we get, 2x -2y = -2, Adding Eqs. (ii) and (iii), we get, 5 x = 10, \, x =2, Putting x = 2 in Eq. (i), we get, 2 -y = - 1, Þ, - y = - 3 or y = 3, \, x = 2 and y = 3, , …(i), …(ii), , Thus, distance travelled in 500 revolutions, = 500 ´ 3.96 m = 1980 m

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146, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 18. By definition of Pythagores theorem,, In a right-angled triangle, the square of the, hypotenuse is equal to the sum of square of, other two sides containing right angle., 19. The given pair of linear equations will be, parallel, when, a1 b1 c1, =, ¹, a2 b2 c2, Here, on comparison the given equations with, standard equation, we get, a1 = 3, b1 = - 1, c1 = - 5, a2 = 6, b2 = - 2 , c2 = - p, 3 -1 -5, 1 5, Þ ¹, ¹, =, \, 6 -2 -p, 2 p, p ¹ 10, , Þ, , 20. If two angles of a triangle are equal to the, corresponding two angles of another triangle, then in such case two triangles are similar by, AA similarity criteria., 21. Given, DAPB is a right triangle, right angled, at P., …(i), \, AB2 = AP2 + BP2, The distance between vertices of a triangle are, AB = (11 - 5 )2 + ( - 5 - 3 )2, , Þ, ( PR + PQ)(1) = 9, [Q PR - PQ = 1], Þ, PR + PQ = 9, On solving PR + PQ = 9 and PR - PQ = 1,, we get, PR = 5 cm and PQ = 4 cm, PQ 4, \ sin R =, =, PR 5, So, sin 2 R + cosec R, 2, 4, 1, 16 5 64 + 125 189, = æç ö÷ +, =, + =, =, 4 / 5 25 4, 25 ´ 4, 100, è5ø, 23. Total number of outcomes = 40, Multiples of 5 from 1 to 40 are (5, 10, 15, 20, 25,, 30, 35, 40), So, number of favourable outcomes = 8, \ Required probability, Number of favourable outcomes, =, Total noumber of outcomes, 8, 1, =, =, 40 5, 24. In DABC and DPQR,, ÐC = ÐR = 35 °, ÐB = ÐQ = 90 °, \ DABC ~ DPQR [by AA similarity criterian], C, , AP = (12 - 5 )2 + ( y - 3 )2, , 35°, , BP = (12 - 11)2 + ( y + 5 )2, \From Eq. (i),, (11 - 5 )2 + ( - 5 - 3 )2 = (12 - 5 )2 + ( y - 3 )2, + ( 12 - 11)2 + ( y + 5 )2, 2, , 2, , 2, , 2, , 6 + (- 8) = 7 + y - 6y + 9, + 1 + y 2 + 10 y + 25, 2, Þ 2 y + 4 y + 84 = 100, Þ, y2 + 2 y - 8 = 0, Þ ( y + 4 ) ( y - 2 ) = 0 Þ y = - 4 or 2, Þ, , 22. Since, DPQR is right angled triangle., , SAMPLE PAPER 7, , P, , 4 cm, , Q, , 5 cm, , 3 cm, , R, , From Pythagoras theorem,, PR 2 = PQ2 + QR 2, Þ, QR 2 = PR 2 - PQ2, [Q QR = 3], Þ, 3 2 = PR 2 - PQ2, Þ, PR 2 - PQ2 = 9, Þ ( PR + PQ) ( PR - PQ) = 9, {Q a2 - b2 = ( a + b)( a - b)}, , R, 35°, , 90°, A, , 25. Given,, , 90°, B, , P, , Q, , x + y - 4 =0, , and, 2 x + ky - 3 = 0, Here, on comparison the above equation with, standard equation, we get, a1 = 1, b1 = 1, c1 = - 4, a2 = 2 , b2 = k , c2 = - 3, For no solution,, a1 b1 c1, =, ¹, a2 b2 c2, 1 1 -4, Þ, = ¹, 2 k -3, 1 1, 1 4, Þ, = and ¹, 2 k, k 3, 3, Þ, k = 2 and k ¹ Þ k = 2, 4, 26. In DOAB, ÐAOB = 60 °, and, , OA = OB, , \ DAOB is an equilateral triangle., , [given]

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147, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 3, 25 3, ´ 52 =, cm 2, 4, 4, 3, (side) 2 ], [Q Area of equilateral D =, 4, Area of shaded portion = Area of circle, - Area of triangle, 25 3, = pr2 4, So, Area of triangle =, , æ 22, 25 3 ö, ÷ cm 2, = çç, ´ 5´ 5 4 ÷ø, è 7, = 78.57 - 10.82 = 67.75 cm 2, 27. Let the points A (0, 0), B(2, 0) and C(0, 2) be, the vertices of the required triangle., Then, the distance between points is, AB = (2 - 0 )2 + (0 - 0 )2 = 4 + 0 = 2, BC = (0 - 2 )2 + (2 - 0 )2 = 4 + 4 = 8 = 2 2, CA = (0 - 0 )2 + (0 - 2 )2 = 0 + 4 = 2, \ Required perimeter = 2 + 2 2 + 2, = ( 4 + 2 2 ) units, 43, 28. The given rational number is 4, 2 ´ 53, As the denominator is of the form 2 n ´ 5 m, where m and n are integers., So, it has terminating decimal expansion, 43 ´ 5, 43, 215, Now, 4, = 4, =, 3, 4, 2 ´5, 2 ´5, (2 ´ 5 )4, 215, 215, = 4 =, = 0.0215, 10000, 10, 29. Total possible outcomes are, { HHH , HHT , HTH ,THH ,TTH , THT , HTT ,TTT}, i.e. 8, Let E be the event of getting exactly 1 head., \ Outcomes favourable to E are, {TTH ,THT , HTT } i.e. 3, Favourable outcomes, 3, \ P( E) =, =, Total possible outcomes 8, 30. The condition for parallel lines is, , On comparing the given equations with, standard equation, we get, a1 = 3 , b1 = 4 , c1 = 5, Here,, Þ, , a2 = 6 , b2 = 8 , c2 = 9, a1 3 1 b1 4 1, c, 5, = = , = = and 1 =, a2 6 2 b2 8 2, c2 9, a1 b1 c1, =, ¹, a2 b2 c2, , Hence, the equation represents parallel lines., , 32. In DABC and DPQR,, 3, 5, 4 2, =, = =, Q, 4.5 7.5 6 3, AC AB BC, \, =, =, PR PQ RQ, So, by SSS similarity criteria, DABC ~ DPQR., 33. According to the situation,, , P, (6,–8), , k:1, R, (–1,6), , Q, (–3,10), , By section formula,, æ -3 k + 6 10 k - 8 ö, ç, ÷, ç k + 1 , k + 1 ÷ = ( -1,6 ), è, ø, - 3k + 6, 10 k - 8, = - 1 and, =6, Þ, k +1, k +1, On further solving, we get, - 3k + 6 = - k - 1, and, 10 k - 8 = 6 k + 6, 7, 7, k = and k =, Þ, 2, 2, \ Required ratio = 7 : 2, 34. There are 13 letters in the word, ‘ASSASSINATION’ out of which one letter can, be chosen in 13 ways., Hence, total number of outcomes = 13, There are 6 vowels in the given word., 6, Hence, required probability =, 13, 6, 6, But given that,, =, 2 x + 1 13, Þ, Þ, Þ, , 2 x + 1 =13, 2 x = 12, x =6, , SAMPLE PAPER 7, , a1 b1 c1, =, ¹, a2 b2 c2, , 31. In this figure, if D is mid-point of BC, then, DC = DB, AC, …(i), In DACD,, cot x =, CD, AC, …(ii), and In DACB, cot y =, CB, Since, D is the mid-point of BC, Þ, BC = 2 CD, Dividing Eq. (ii) by Eq. (i), we get, AC, cot y CB CD CD 1, =, =, =, =, cot x AC CB 2 CD 2, CD, cot y 1, \, =, cot x 2

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148, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 35. Given, side of square = 6 cm, , 6 cm, , \Diameter of circle ( d ) = side of square, = 6 cm, d 6, \Radius of circle ( r) = = = 3 cm, 2 2, \Area of circle = pr2 = p(3 )2 = 9 p cm 2, , Solutions (41-45), , 36. In DABC and DPQR,, , and, , 40. Let the radius of inner track be r., 352, \, 2 pr = 352 Þ r =, 2p, Let the radius of outer track be R., 396, \, 2 pR = 396 Þ R =, 2p, Width of the track, 396 352 44, = R-r=, =, 2p, 2p 2p, 44 ´ 7, =7 m, =, 2 ´ 22, , ÐA = ÐP = 53º, AC BC, =, PR RQ, , 41. Prime factors of 616 = 2 3 ´ 7 ´ 11, 2 616, 2 308, 2 154, 7 77, 11 11, 1, , éQ 6 = 4 = 2 ù, êë 4 6 3 úû, , Here two sides are proportional and one angle, in equal but this angle is not between the sides., So, triangles are not similar., 37. Let the present age of father be x years and, present age of son be y years., According to the problem,, …(i), x = 6y, After 4 yr,, x + 4 = 4( y + 4 ), x + 4 = 4 y + 16, …(ii), Þ, x - 4 y = 12, Put x = 6 y in Eq. (ii), we get, 6 y - 4 y = 12, Þ, 2 y = 12, Þ, y =6, and, x = 6 y = 36, \Present age of son = 6 yr, Present age of father = 36 yr, , SAMPLE PAPER 7, , 360 °, 38. Central angle of a semicircle =, = 180 °, 2, Now, the supplementary angle will be, = 180 ° - 180 ° = 0 °, , 42. Factors of 32 = 2 5, Factors of 616 = 2 3 ´ 7 ´ 11, \HCF (32, 616) = 2 3 = 8, 43. Factors of 56 = 2 3 ´ 7, Factors of 1000 = 2 3 ´ 5 3, \LCM( 56 ,1000 ) = 2 3 ´ 5 3 ´ 7 = 7000, 44. Prime factor of 1000 = 2 3 ´ 5 3, 45. HCF of 56 and 1000 is the required number of, column in which army can march, Factors of 56 = 2 3 ´ 7, Factors of 1000 = 2 3 ´ 5 3, \ HCF (56, 1000) = 2 3 = 8, Solutions (46-50), 46. The shape of the path traced shown in the, given figure is the form of parabola., 47. The graph of parabola opens downwards,, if a < 0., 48. In the given graph, we see that curve intersect, the X-axis at four points. Hence, number of, zeroes in the given polynomial are 4., , 39. Here, AQ : BQ = 2 : 1., A, (7, –2), , P, (5, –3), , Q, (3, y), , B, (1, –5), , Then, by section formule,, 2 ´ ( - 5 ) + 1 ´ ( - 2 ) - 10 - 2 -12, =-4, y=, =, =, 2 +1, 3, 3, , 49. The given curve intersect the X-axis at points, x = - 3 , - 1, 2 and 4., Hence, four zeroes in the given graph are, - 3 , - 1, 2, 4., 50. Option (a) is biquadratic polynomial., Option (b) is cubic polynomial., Option (c) is quadratic polynomial., Option (d) is linear polynomial.

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149, , CBSE Sample Paper Mathematics Standard Class X (Term I), , SAMPLE PAPER 8, MATHEMATICS (Standard), A Highly Simulated Practice Questions Paper, for CBSE Class X (Term I) Examination, , Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions., There is no negative marking., Maximum Marks : 40, Time allowed : 90 minutes, , Roll No., , Section A, Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., 1242, is a decimal, 1., 343, (a) non-terminating repeating, (b) non-terminating non repeating, (c) terminating, (d) None of the above, , 2. A quadratic polynomial having 5 and - 3 as zeroes is, (a) x 2 - 2 x - 15, (c) x 2 + 2 x + 15, , (b) x 2 - 2 x + 15, (d) x 2 + 2 x - 15, , 3. The value of x in the given figure is, P, , Q, , (a) 5 cm, , (b) 4 cm, , 5 cm, , 35° 35° 8 cm, , S, , x cm R, , (c) 2 cm, , (d) 3 cm, , 4. If the circumference of two circles are in the ratio of 3 : 4, then the ratio of their, areas is, (a) 9 : 16, , (b) 16 : 9, , (c) 9 : 17, , (d) 7 : 17, , SAMPLE PAPER 8, , 10 cm

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150, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 5. The coordinates of the point which divides the line segment joining the points (4 , - 3), and (8, 5) in the ratio 1 : 3 internally are, (a) (4, 3), (b) (7, 3), , 6. The value of æç, , 11, , è cot q, , (a) 11, , 2, , -, , (c) (3, 5), , (d) ( 5, - 1), , 11 ö, ÷ is, cos 2 q ø, , (b) 0, , (c), , 1, 11, , (d) - 11, , 7. HCF of (2 3 ´ 3 2 ´ 5), (2 2 ´ 3 3 ´ 5 2 ) and (2 4 ´ 3 ´ 5 3 ´ 7) is, (a) 30, , (b) 48, , (c) 60, , (d) 105, , 8. The difference between the circumference and the radius of a circle is 37 cm. The area, of the circle is, (a) 149 cm 2, , (b) 154 cm 2, , (c) 121 cm 2, , (d) 169 cm 2, , 9. If a and b are the zeroes of f (x) = 2x 2 + 8x - 8, then, (a) a + b = ab, , (b) a + b > ab, , (c) a + b < ab, , (d) a + b + ab = 0, , 10. It is given that DABC ~ DDFE, ÐA = 50 °, ÐC = 30°, AB = 10 cm, AC = 15 cm and DF = 8 cm., Then, which of the following is true?, (a) DE = 12 cm and ÐF = 50°, (c) EF = 12 cm and ÐD = 100°, , (b) DE = 12 cm and ÐF = 100°, (d)EF = 12 cm and ÐD = 30°, , 3, 4, , 11. If tan q = , then cos 2 q - sin 2 q is equal to, (a), , 7, 25, , (b) 1, , (c), , -7, 25, , (d), , 4, 25, , 12. C is the mid-point of PQ, if P is (4 , x),C is (y, - 1) and Q is (- 2, 4), then x and y, respectively are, (a) - 6 and 1, , (b) - 6 and 2, , (c) 6 and - 1, , (d) 6 and - 2, , 13. The radius of a wheel is 0.25 m. The number of approximate revolutions it will make to, travel a distance of 11 km, is ____ ., (a) 5000, (b) 7000, , (c) 6000, , (d) 1000, , 14. Which of the following is the decimal expansion of an irrational number?, (a) 4.761, , (b) 0. 32, , (c) 5.010010002…, , SAMPLE PAPER 8, , 15. In two triangles DABC and DDEF, ÐA = ÐE and ÐB = ÐF. Then,, (a), , DE, DF, , (b), , ED, EF, , (c), , EF, ED, , (d) 6.030303…, , AB, is equal to, AC, EF, (d), DF, , 16. The sum and product of zeroes of a quadratic polynomial are 0 and 3 respectively., The quadratic polynomial is, (a) x 2 - 3, (c) x 2 - 3, sin x - cos x, 17. If 16 cot x = 12, then, equal, sin x + cos x, 1, 3, (b), (a), 7, 7, , (b) x 2 + 3, (d) None of these, , (c), , 2, 7, , (d) 0

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151, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 18. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is, (a) 22 : 7, , (b) 14 : 11, , (c) 7 : 22, , (d) 11 : 14, , 19. The centre of a circle is (2a , a - 7). The value of a, if the circle passes through the point, (11, - 9) and has diameter 10 2 units is, (a) 3 or 6, (b) 5 or 3, , (c) 7 or 4, , (d) None of these, , 20. ABCD is a trapezium such that BC || AD and AB = 4 cm. If the diagonals AC and BD, AO OB 1, =, = , then CD is equal to, OC OD 2, (b) 8 cm, (c) 9 cm, , intersect at O such that, (a) 7 cm, , (d) 6 cm, , Section B, Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 21. In the given figure, AD = 3 cm, BD = 4 cm and CB = 12 cm, then sin q equals, A, , 90°, q, C, , 3, (a), 4, , 5, (b), 13, , D, , 90°, B, , (c), , 4, 3, , (d), , 12, 5, , 22. A race track is in the form of a circular ring whose outer and inner circumferences are, 396 m and 352 m respectively. The width of the track is, (a) 63 m, (b) 56 m, (c) 7 m, , (d) 3.5 m, , 2, , 23. If a and b are the zeroes of the polynomial x - 5x + c and a - b = 3, then c is equal to, (a) 0, , (b) 1, , (c) 4, , (d) 5, , 24. The least number which when divided by 18, 24, 30 and 42 will leave in each case the, same remainder 1, would be, (a) 2520, (b) 2519, , (c) 2521, , (d) None of these, , 25. The coordinates of the point P which divides the joining the points A(2, 5) and B(3, - 5), in the ratio 2 : 3, are, (a) (1, 0), , 12, (b) æç , 1ö÷, è5 ø, , (c) (3, 0), , (d) (0, 1), , (a) 2, , (b) 3, , (c) 1, , (d) 4, , 27. A man goes 15 m due West and then 8 m due North. How far is he from the starting, point?, (a) 12 m, , (b) 17 m, , (c) 18 m, , (d) 24 m, , 28. If in DABC, DE|| BC, then, (a), , AD AE, =, AB AC, , (b), , AB AE, =, AD AC, , (c), , AB AC, =, AD AD, , (d) None of these, , SAMPLE PAPER 8, , 26. If sin A + cosec A = 2, then sin 2 A + cosec 2 A is equal to

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152, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 29. If a pair of linear equations is consistent, then the lines will be, (a) always coincident, (b) parallel, (c) always intersecting, (d) intersecting or coincident, , 30. If a and b are the zeroes of the polynomial f (x) = x 2 - p(x + 2) - q, then (a + 1) (b + 1) is, equal to, (a) q - 1, , (b) 1 - q - p, , (c) q, , (d) 1 + q, , 31. If P(- 2, 5) and Q(3, 2) are two points. The coordinates of the point R on PQ such that, PR = 3QR are, 4, (a) Ræç , 3ö÷, è3 ø, 1, (c) Ræç , 7ö÷, è3 ø, , 7 11, (b) Ræç , ö÷, è4 4 ø, (d) None of these, , 32. The LCM and HCF of two non-zero positive numbers are equal, then the numbers, must be, (a) prime, , (b) coprime, , (c) composite, , (d) equal, , 33. The area of the shaded region in the given figure, if AC = 24 cm, BC = 10 cm and O is the, centre of the circle, is [take p = 314, . ], A, O, , B, 2, , C, , (b) 145.33 cm 2, (d) None of these, , (a) 128.56 cm, (c) 248.16 cm 2, , 34. If in an equilateral triangle, the length of the median is 3 cm, then the length of the, side of equilateral triangle is, (b) 2 cm, , (a) 1 cm, , (c) 3 cm, , (d) 4 cm, , 35. If DABC is right angled at C, then the value of sin(A + B) is, (a) 0, , (b) 1, , (c), , 1, 2, , (d), , 3, 2, , SAMPLE PAPER 8, , 36. If one of the zeroes of a quadratic polynomial of the form x 2 + ax + b is the negative of, the other, then it, (a) has no linear term and the constant term is negative, (b) has no linear term and the constant term is positive, (c) can have a linear term but the constant term is negative, (d) can have a linear term but the constant term is positive, , 37. The area of the largest circle that can be drawn inside the given rectangle of length ‘a’, cm and breadth ‘b’ cm ( a > b) is, 1, 1, (a) p b 2 cm 2, (b) p b 2 cm 2, 2, 3, , (c), , 1, p b 2 cm 2, 4, , (d) p b 2 cm 2

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153, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 38. If X = 28 + (1 ´ 2 ´ 3 ´ 4 ´¼ ´ 16 ´ 28) and Y = 17 + (1 ´ 2 ´ 3 ´¼ ´ 17), then which of the, following is true?, (a) X is a prime number, (b) Y is a prime number, (c) X - Y is a prime number, (d) X - Y is a composite number, , 39. The value(s) of y for which the distance between the points P(y, 3) and Q(9, 10) is, (10 - y) units, is, (a) 10, (c) - 15, , (b) 20, (d) None of these, , 40. If cos 30°× tan 30° = sin a and a < 90°, then the value of tan 2a is, (a), , 1, 3, , (b) 3, , (c) 1, , (d) 0, , Section C, Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted., 41-45 are based on Case Study-1., , Case Study 1, Vikas is working with TCS, and he is sincer and dedicated to his work. He pay all his taxes on, time and invest the some amount of his salary in funds for his future., , 41. Consider the amount invested at 12% be p and at 10% be q. Then, formulate the, required linear equation for first condition., (a) 10 p + 12 q = 13400, (b) 8 p + 15q = 12000, (c) 12 p + 10q = 13000, (d) 5 p + 6q = 6000, , 42. Now, formulate the linear equation for the second condition?, (a) 12 p + 10q = 13000, (c) 5 p + 6q = 12000, , (b) 10 p + 12 q = 13400, (d) 8 p + 15q = 6000, , SAMPLE PAPER 8, , He invested some amount at the rate of 12% simple interest and some other amount at the rate of, 10% simple interest. He received yearly interest of ` 130. But, if he interchange the amounts, invested, he would have received ` 4 more as interest., Now, answer the following questions

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154, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 43. What is value of the variable p?, (a) ` 700, , (b) ` 500, , (c) ` 300, , (d) ` 150, , (c) ` 300, , (d) ` 150, , 44. What is the value of the variable q?, (a) ` 700, , (b) ` 500, , 45. If the rate of interest changes to 15% for first amount and 12% for second amounts,, how much amount of money is earned?, (a) ` 170, (b) ` 150, , (c) ` 148, , (d) ` 159, , 46-50 are based on Case Study-2., , Case Study 2, Sanjeev sees a game in a fair. He is keenly interested to play it. He asks the rules of game from, the owner. Owner says that you will move this spinner first, if it stops on a prime number, then, you are allowed to pick a marble from a bag which contain 14 white and 10 black marbles. Prizes, are given when a white marble is picked randomly., 2, , 4, , 8, , 7, , 13, , 5, 10, , 11, , Based on the above information, answer the following questions:, , 46. What is the probability that Sanjeev will not be allowed to pick a marble from the bag?, (a), , 3, 8, , (b), , 5, 8, , (c), , 1, 2, , (d) 1, , 47. What is the probability that Sanjeev will be allowed to pick a marble from a bag?, (a), , 3, 8, , (b), , 1, 2, , (c), , 5, 8, , (d), , 3, 7, , 48. The probability that Sanjeev will pick a black marble from the bag is, (a), , 7, 12, , (b), , 5, 12, , (c), , 1, 2, , (d), , 2, 3, , 5, 12, , (d), , 7, 12, , (d), , 7, 12, , 49. The probability that Sanjeev will get prize is, , SAMPLE PAPER 8, , (a), , 3, 5, , (b), , 3, 4, , (c), , 50. The probability that Sanjeev will pick a red marble from the bag is, (a) 1, , (b) 0, , (c), , 5, 12

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OMR SHEET, , SP 8, , Roll No., Sub Code., , Student Name, , Instructions, Use black or blue ball point pens and avoid Gel & Fountain pens for filling the OMR sheet., Darken the bubbles completely. Don’t put a tick mark or a cross mark, half-filled or over-filled bubbles will not be read, by the software., Correct, , ✔, , ✗, , Incorrect, , Incorrect, , Incorrect, , Do not write anything on the OMR Sheet., Multiple markings are invalid., , 1, , 18, , 35, , 2, , 19, , 36, , 3, , 20, , 37, , 4, , 21, , 38, , 5, , 22, , 39, , 6, , 23, , 40, , 7, , 24, , 41, , 8, , 25, , 42, , 9, , 26, , 43, , 10, , 27, , 44, , 11, , 28, , 45, , 12, , 29, , 46, , 13, , 30, , 47, , 14, , 31, , 48, , 15, , 32, , 49, , 16, , 33, , 50, , 17, , 34, , Check Your Performance, Total Questions:, Total Correct Questions:, , If Your Score is, , Score Percentage =, , Total Correct Questions, Total Questions, , × 100, , Less than 60%, > Average (Revise the concepts again), Greater than 60% but less than 75% > Good (Do more practice), Above 75%, > Excellent (Keep it on)

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156, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Answers, 1. (b), 11. (a), , 2. (a), 12. (a), , 3. (b), 13. (b), , 4. (a), 14. (c), , 5. (d), 15. (c), , 6. (d), 16. (b), , 7. (c), 17. (a), , 8. (b), 18. (b), , 9. (a), 19. (b), , 10. (b), 20. (b), , 21. (b), 31. (b), 41. (c), , 22. (c), 32. (d), 42. (b), , 23. (c), 33. (b), 43. (b), , 24. (c), 34. (b), 44. (a), , 25. (b), 35. (b), 45. (d), , 26. (a), 36. (a), 46. (a), , 27. (b), 37. (c), 47. (c), , 28. (a), 38. (d), 48. (b), , 29. (d), 39. (c), 49. (d), , 30. (b), 40. (b), 50. (b), , SOLUTIONS, 1., , æ 20 - 4 ö, =ç ,, ÷ = ( 5 , - 1), è4 4 ø, , 1242 1242, = 3, 343, 7, Here, Denominator has one factor namely 7,, which is other than 2 and 5., So, given rational number is a non-terminating, non-repeating decimal., , 6. We have,, 11, 11, 1, 1 ù, = 11é, cot 2 q cos 2 q, ëê cot 2 q cos 2 q ûú, = 11[tan 2 q - sec 2 q], , 2. Let a = 5 and b = - 3, , éQ sec2 q - tan 2 q = 1 ù, ú, ê, 2, 2, êë Þ tan q - sec q = - 1úû, , 2, , \Required polynomial is x - (a + b )x + ab, = x 2 - ( 5 - 3 )x + 5( - 3 ) = x 2 - 2 x - 15, 3. In DPSQ and DPSR,, ÐQPS = ÐSPR = 35°, ÐPSQ = ÐPSR, From AA similarity, DPSQ ~ DPSR, PQ SQ, =, \, PR SR, 10 5, Þ, =, 8, x, Þ, x = 4 cm, , 7. Given, (2 ´ 3 ´ 5 ), (2 2 ´ 3 3 ´ 5 2 ) and, [Q PS ^ QR ], , 4. Let r1 and r2 are radii of two circle. Then, according to the given condition,, 2 pr1 3, =, 2 pr2 4, r1 3, =, Þ, r2 4, , SAMPLE PAPER 8, , Now,, , 2, , 2, æ r1 ö, æ3 ö = 9, =, =, ÷, ç, ç, ÷, è4ø, 16, pr22 è r2 ø, , pr12, , = 11 ´ ( - 1) = - 11, 3, , 5. Given,, ( x1 , y1 ) = ( 4 , - 3 ) and ( x2 , y2 ) = (8 , 5 ), Let ( x , y ) be the coordinates of the point which, divides the line joining the points ( x1 , y1 ) and, ( x2 , y2 ) in ratio m : n = 1 : 3 internally., æ mx + nx1 my2 + ny1 ö, So, ( x , y ) = ç 2, ,, ÷, m+ n ø, è m+ n, æ 1(8 ) + 3( 4 ) 1( 5 ) + 3( - 3 ) ö, =ç, ,, ÷, 1+3, ø, è 1+3, , 2, , (2 4 ´ 3 ´ 5 3 ´ 7 ), \HCF of above expressions = 2 2 ´ 3 ´ 5, = 4 ´ 3 ´ 5 = 60, 8. Let r be the radius of the circle., Given, circumference of circle, - radius of the circle = 37., Þ, , 2 pr - r = 37, , Þ, , r(2 p - 1) = 37, r=, , Þ, , 37, 2p - 1, , 37 ´ 7, 37, =, = 7 cm, 22, 37, 2 æç ö÷ - 1, è7 ø, 22, \ Area of circle = pr2 =, ´ (7 )2 = 154 cm 2, 7, =, , 9. Given,, a and b are the zeroes of f ( x ) = 2 x 2 + 8 x - 8, Coefficienct of x, - (8 ), \a + b = =, =-4, 2, 2, Coefficient of x, -8, Constant term, and a × b =, =, =-4, 2, 2, Coefficient of x, So, a + b = a × b, 10. Given, DABC ~ DDFE

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157, , CBSE Sample Paper Mathematics Standard Class X (Term I), , D, , A, , m, 8c, , cm, , 50°, , 10, , 15 cm, 30°, , B, , =2 ´, , C, , 30°, , F, , E, , Then, ÐA = ÐD = 50°, ÐC = ÐE = 30°, \, ÐB = ÐF = 180° - (30° + 50° ) = 100°, AB AC, Also,, =, DF DE, 10 15, Þ, =, 8, DE, 15 ´ 8, Þ, = 12 cm, DE =, 10, Hence, DE = 12 cm and ÐF = 100°, AB 3, 11. We have, tanq =, =, AC 4, Let AB = 3 k and BC = 4 k, where k is positive, constant., \, AC 2 = BC 2 + AB2, Þ, AC 2 = 9 k 2 + 16 k 2 = 25 k 2, Þ, AC = 5 k, , A, , q, , 4k, , 3k, B, , AB 4 k 4, =, =, AC 5 k 5, BC 3 k 3, and, sinq =, =, =, AC 5 k 5, 2, 2, 4, 3, 16 9, 7, \ cos 2 q - sin 2 q = æç ö÷ - æç ö÷ =, =, è5ø, è5ø, 25 25 25, , \, , Q Number of revolutions, Distance covered by wheel, =, Circumference of wheel, 11 ´ 1000 ~, =, - 7000 (approximate), 1.57, 14. An irrational number has non repeating non, terminating decimal expansion., Here, (a) is terminating decimal expansion and, (b) and (d) are non-terminating but repeating, decimal expansion., \(c) is non terminating non-repeating decimal, expansion., So, (c) is the decimal expansion of irrational, number., 15. In DABC and DEFD,, [given], ÐA = ÐE and ÐB = ÐF, \, DABC ~ DEFD, [by AA similarity criterion], AB AC, AB EF, \, =, Þ, =, EF ED, AC ED, 16. Given, Sum of zeroes = 0, and product of zeroes = 3, We know that,, f ( x ) = x 2 - (a + b )x + a × b = x 2 - 0 × x + 3, So, quadratic polynomial is ( x 2 + 3 )., , C, , 5k, , cosq =, , 17. Let DABC be the right triangle such that, ÐB = 90° and ÐC = x, A, , 4k, , 12. Given, C is the mid-point of PQ i.e. P( 4 , x ) and, Q( - 2 , 4 )., , x, , B, , 3k, , C, , Given, cot x =, , AC =, , \, , AB2 + BC 2, [use Pythagoras theorem], , = 16 k 2 + 9 k 2, , 1:1, Q, (–2, 4), , 13. The distance covered in one revolution is equal, to the circumference of the wheel., Circumference of wheel = 2 pr, , = 25 k 2 = 5 k, AB 4 k 4, =, =, AC 5 k 5, BC 3 k 3, =, =, cos x =, AC 5 k 5, , So, sin x =, , …(i), …(ii), , SAMPLE PAPER 8, , On equating the coordinate, we get, x+4, y = 1 and -1 =, 2, \ x = - 6 and y = 1, C, (y, –1), , 5k, , 12 3, =, 16 4, BC 3, \ cot x =, =, AB 4, Let BC = 3 k and AB = 4 k, where k is positive, constant., , 4 -2 x + 4ö, Therefore ( y , - 1) = æç, ,, ÷, è 2, 2 ø, x + 4ö, ( y , 1) = æç1,, Þ, ÷, è, 2 ø, , P, (4, x), , 22, ´ 0.25 = 1.57, 7

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158, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 4, sin x - cos x 5 Now,, =, sin x + cos x 4 +, 5, , 3, 5 = 4-3 = 1, 3 4+3 7, 5, [Using Eqs. (i) and (ii)], sin x - cos x 1, is ., Hence, the value of, sin x + cos x 7, , 18. Let r be the radius of a circle and a be the side, of a square., Given, Perimeter of a circle = Perimeter of a, square, pr, …(i), \, 2 pr = 4 a Þ a =, 2, Area of circle = pr2, Area of square = a2, p r2, Area of a circle, pr2, Now,, =, =, Area of a sqaure ( a)2 æ pr ö 2, ç ÷, è2 ø, [From Eq. (i)], =, , pr, , 2, , p 2 r2 / 4, 4, 4, 28 14, = =, =, =, p 22 / 7 22 11, , \The required ratio will be 14 : 11., 19. By using the condition,, Distance between the centre (2 a, a - 7 ) and the, point P(11, - 9 ) = Radius of circle., C (2a, a–7), Radius, P (11, –9), , \ Radius of circle = (11 - 2 a)2 + ( - 9 - a + 7 )2, …(i), [Q Distance between two points ( x1 , y1 ), , SAMPLE PAPER 8, , and ( x2 , y2 ) = ( x2 - x1 )2 + ( y2 - y1 )2 ], Given that, length of diameter = 10 2, length of diameter, \Length of radius =, 2, 10 2, =, =5 2, 2, Put this value in Eq. (i), we get, 5 2 = (11 - 2 a)2 + ( - 2 - a)2, Squaring on both sides, we get, 50 = (11 - 2 a)2 + (2 + a)2, Þ, 50 = 121 + 4 a2 - 44 a + 4 + a2 + 4 a, Þ, 5 a2 - 40 a + 75 = 0, , [divide by 5], a2 - 8 a + 15 = 0, a - 5 a - 3 a + 15 = 0, [by factorization method], Þ, a (a - 5) - 3 (a - 5) = 0, Þ, (a - 5) (a - 3) = 0, Þ, a = 3, 5, Hence, the required values of a are 5 and 3., Þ, Þ, , 2, , 20. In DAOB and DCOD, we have, ÐAOB = ÐCOD, [vertically opposite angles], AO OB 1, [given], =, =, OC OD 2, A, , D, O, , B, , C, , \DAOB ~ DCOD [by SAS similarity criterion], AB AO, \, =, CD OC, 4, 1, Þ, = Þ CD = 8 cm, CD 2, 21. We have, AD = 3 cm, BD = 4 cm and CB = 12 cm, In DABD,, , AB2 = BD2 + AD2, [from Pythagoras theorem], 2, 2, AB = 4 + 3 2, AB = 25 = 5 m, Now, In DABC, AC 2 =, , AB2 + BC 2, , = ( 5 )2 + (12 )2, = 25 + 144, = 169 = 13 cm, [from Pythagoras theorem], In DABC, ÐB = 90º, AB 5, sinq =, =, BC 13, 22. Let R and r be the outer and inner radii of the, track respectively., Given, outer and inner circumference be 396 m, and 352 m., R, , r, , …(i), \, 2 pR = 396, and, …(ii), 2 pr = 352, On subtracting Eq. (ii) from Eq. (i), we get

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159, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 2 pR - 2 pr = 396 - 352, Þ, 2 p( R - r) = 44, \Width of the track = outer radius, - inner radius, 44, 44, =7m, = R- r=, =, 2 p æ2 ´ 22 ö, ÷, ç, è, 7 ø, 23. Given, a - b = 3, , … (i), , C, W, , S, B, , 25. Let the coordinates of P be ( x , y )., Using section formula,, æ mx + nx1 my2 + my1 ö, (x, y) = ç 2, ,, ÷, m+ n ø, è m+ n, , AC = ( AB)2 + ( BC )2, , 26. Given, sin A + cosec A = 2, Squaring both sides, we get, sin 2 A + cosec2 A + 2 sin A × cosec A = 4, 1, sin 2 A + cosec2 A + 2 × sin A ×, =4, Þ, sin A, Þ, sin 2 A + cosec2 A = 4 - 2 = 2, 27. Let A be starting point of man. A man goes 15, m West and 8 m North., , [Apply pythagoras theorem], AC = (15 )2 + (8 )2 = 225 + 64, = 289 = 17 m, 28. Again, in DABC, DE || BC, AD AE, [from thales theorem], \, =, DB EC, DB, EC, +1=, +1, Þ, AD, AE, AD + DB AE + EC, Þ, =, AD, AE, A, , D, , E, , B, , Þ, Þ, , C, , AB AC, =, AD AE, AD AE, =, AB AC, , 29. If a pair of linear equations is consistent, then it, has unique solution and infinitely solutions., For unique solution, a1 b1, ¹, a2 b2, \The lines will be intersecting., For infinitely solutions,, a1 b1 c1, =, =, a2 b2 c2, \The lines will be coincident., 30. Given, a and b are the zeroes of the, polynomial, f ( x ) = x 2 - p( x + 2 ) - q, = x 2 - px - (2 p + q), Coefficient of x, ( - p), \ a +b = ==p, 2, 1, Coefficient of x, and a × b = - (2 p + q), , SAMPLE PAPER 8, , æ 2 (3 ) + 3(2 ) 2 ( - 5 ) + 3( 5 ) ö, =ç, ,, ÷, 2 +3, ø, è 2 +3, 12, 5, = æç , ö÷, è 5 5ø, 12, = æç , 1ö÷, è 5 ø, , A, , 15 m, , The distance between starting point A and end, point C is, , \ Given polynomial is x - 5 x + c and, a , b are the zeroes of the polynomial., Coefficient of x, = - ( - 5 ) …(ii), \ a +b = Coefficient of x 2, , 24. The factors of the given numbers are, 18 = 2 ´ 3 2, 24 = 2 3 ´ 3, 30 = 2 ´ 3 ´ 5, 42 = 2 ´ 3 ´ 7, LCM (18, 24, 30, 42) = 2 3 ´ 3 2 ´ 5 ´ 7, = 2520, As the remainder is 1., Hence, the required number is 2520 + 1 = 2521, , E, , 8m, , 2, , Adding Eqs. (i) and (ii), we get, 2a = 8, Þ, a=4, Þ, b = 5-4 =1, constant term, And, a ×b =, coefficient of x 2, c, Þ, 4 ´1 = Þ c = 4, 1, , N

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160, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Now, (a + 1) (b + 1) = ab + a + b + 1, = - (2 p + q) + p + 1, = - 2 p - q + p + 1 = 1q, = - p- q+1, =1- p- q, , = 265.33 - 120, = 145.33 cm 2, 34. Let a be the side of equilateral triangle., Median is also the altitude of an equilateral, triangle., A, , 31. Given, coordinate of P( - 2 , 5 ) and coordinate, of Q(3 , 2 )., Since PR = 3 QR, 3, P, (–2, 5), , 1, R, , Altitude, , Q, (3, 2), , B, , Therefore point R divides PQ in the ratio 3 : 1., Then, coordinates of R are, é æ 1 ´ (- 2 ) + 3 ´ 3 ö æ 1 ´ 5 + 3 ´ 2 ö ù, R = êç, ÷, ç, ÷ú, 3+1, 1+3, ø è, øû, ëè, [by section formula], éæ - 2 + 9 ö æ 5 + 6 öù, R = êç, ÷ú, ÷, ç, ëè 4 ø è 4 øû, æ 7 11 ö, =ç, ÷, è 4, 4 ø, 32. Given, LCM and HCF are equal., Let two non-zero positive number are p and q., Then, HCF ( p, q) = LCM ( p, q), , [given], , Let HCF ( p, q) = k, Þ, , p = ka, , and, , q = kb, , Where a and b are natural numbers., Q HCF ´ LCM = product of numbers, k ´ k = ka ´ kb, Þ, a´ b=1, \a = b = 1 as they are natural number., Hence, p = q or the number must be equal., 33. Given, AC = 24 cm, BC = 10 cm, , SAMPLE PAPER 8, , a, , a, , We know that angle in a semi-circle = 90°, ( AB)2 = ( AC )2 + ( BC )2, = 24 2 + 10 2, = 576 + 100 = 676 cm, So, AB = 26 cm,, AO = 13 cm = radius of the circle, Area of shaded region = Area of semicircle, - Area of DABC, pr2 1, =, - ´ BC ´ AC, 2, 2, p ´ (13 )2 1, =, - ´ 10 ´ 24, 2, 2, 3.14 ´ 169, =, - 120, 2, , a/2, , D, , a/2, , C, , In DADC,, 2, , a, (Altitude)2 + æç ö÷ = a2, è2 ø, a2, ( 3 )2 +, = a2, 4, 12 + a2, = a2, Þ, 4, Þ, 12 + a2 = 4 a2, Þ, 3 a2 = 12, Þ, a2 = 4, Þ, a= ±2, [Q a = - 2 is not possible], \ Side of triangle = 2 cm, 35. Given,, A, , 90°, C, , B, , In DABC is right angled at C, ÐC = 90°, We know that, ÐA + ÐB + ÐC = 180°, [sum of interior angles of triangle], ÐA + ÐB = 180° - 90°, ÐA + ÐB = 90°, sin( ÐA + ÐB) = sin 90° = 1, 36. If one of the zeroes of the equation is negative, of the other, then the zeroes will be m and- m., Hence, sum of zeroes = m - m = - a Þ a = 0, Thus, there will be no linear term involved in, the equation., Also, the constant term = b = product of zeroes, = - m2, will be negative.

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161, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 37. The largest circle that can be drawn inside a, rectangle is possible when rectangle becomes a, square., \Diameter of the circle = Breadth of rectangle, =b, \Radius of the circle = b / 2, p ´ b2, Hence, area of the circle = pr2 =, cm 2, 4, 38. Given,, X = 28 + (1 ´ 2 ´ 3 ´ 4 ´ ¼ ´ 16 ´ 28 ), Y = 17 + (1 ´ 2 ´ 3 ´ ¼ ´ 17 ), X can be rewritten as, X = 28 + [28(1 ´ 2 ´ 3 ´ ¼ ´ 16 )], = 28 [1 + (1 ´ 2 ´ 3 ´ ¼ ´ 16 )], Similarly, Y = 17 + [17(1 ´ 2 ´ ¼ ´ 16 )], = 17[1 + (1 ´ 2 ´ 3 ´ ¼ ´ 16 )], Hence, both X and Y are composite number, and X - Y is also a composite number., 39. Given : P( y , 3 ) and Q(9 , 10 ), d = 10 units, We know that, 2, , 2, , 10 - y = (9 - y ) + (10 - 3 ), , 10 - y = 81 + y 2 - 18 y + 49, 10 - y = 130 + y 2 - 18 y, Squaring both side, (10 - y )2 = 130 + y 2 - 18 y, 100 + y 2 - 20 y = 130 + y 2 - 18 y, - 20 y + 18 y = 130 - 100, - 2 y = 30, 30, y== - 15, 2, 40. Given, cos 30°× tan 30° = sin a, , Þ, \, , 12 p + 10 q = 13000, , 42. Vikas received ` 4 extra, if he interchange the, investment amount., \According to the situation,, 10, 12, p+, q = 130 + 4, 100, 100, Þ, , 10 p + 12 q = 13400, , 43. Given, from above situation, …(i), 12 p + 10 q = 13000, …(ii), 10 p + 12 q = 13400, Adding Eqs. (i) and (ii), we get, 22 p + 22 q = 26400, …(iii), Þ, p + q = 1200, Now, subtracting both Eqs. (i) and (ii), we get, 2 p - 2 q = - 400, …(iv), Þ, p - q = - 200, Now, on adding Eqs. (iii) and (iv), we get, 2 p = 1000, Þ, p = ` 500, 44. Put the value of ‘p’ in Eq. (iii), , D = ( x2 - x1 )2 + ( y2 - y1 )2, , Þ, , Þ, , Solutions (41-45), 41. Vikas received ` 130 as profit., \According to the situation,, 12, 10, p+, q = 130, 100, 100, , 45. Now, the rate of interest is 15% and 12%., 15, 12, \, p+, q = earned money, 100, 100, 12, 15, Þ earned money =, ´ 500 +, ´ 700, 100, 100, = 75 + 84, = ` 159, Solutions (46-50), 46. Total numbers on the spinner = 8, Non-prime numbers are 4, 8, 10 i.e. 3, 3, So, P(non-prime number) =, 8, 3 8-3 5, 47. Required probability = 1 - =, =, 8, 8, 8, 48. Total number of marbles = 10 + 14 = 24, Number of black marbles = 10, 10, 5, =, \ P(getting a black marble) =, 24 12, 49. Number of white marbles = 14, \ P(getting a prize) =, , 14, 7, =, 24 12, , 50. There is no red marble in the bag., \ P (getting a red marble) = 0, , SAMPLE PAPER 8, , 3, 1, ´, = sina, 2, 3, 1, sina = Û a = 30°, 2, tan 2 a = tan(2 ´ 30° ), = tan60° = 3, , p = 1200 - 500 = ` 700

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162, , CBSE Sample Paper Mathematics Standard Class X (Term I), , SAMPLE PAPER 9, MATHEMATICS (Standard), A Highly Simulated Practice Questions Paper, for CBSE Class X (Term I) Examination, , Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions., There is no negative marking., Maximum Marks : 40, Time allowed : 90 minutes, , Roll No., , Section A, Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 1. The value of HCF (8, 9, 25) ´ LCM (8, 9, 25) is, (a) 50, , (b) 1800, , (c) 1810, , (d) 1500, , 2. If (0, 0), (3, 3) and (3, p) are vertices of an equilateral triangle, then value of p is, (b) 3, , (a) - 3, , (c) 3, , (d) - 3, , 3. If DABC ~ DDEF, such that DE = 2AB and BC = 8 cm, then the length of EF is, (a) 8 cm, , (b) 16 cm, , (c) 18 cm, , (d) 26 cm, , 4. If in a lottery, there are 5 prizes and 20 blanks, then the probability of getting a prize is, (a), , 2, 5, , (b), , 4, 5, , (c), , 1, 5, , (d) 1, , SAMPLE PAPER 9, , 5. The sum and the product of the zeroes of polynomial 6x 2 - 5 respectively are, (a) 0 and, , -6, 5, , (b) 0 and, , b, x, when x = - 5, then a + b equals, (a) - 1, (b) 0, , 6, 5, , (c) 0 and, , 5, 6, , (d) 0 and, , -5, 6, , 6. For the equation y = a + , where a and b are real numbers, if y = 1 when x = - 1 and y = 5, (c) 11, , (d) 10, , 7. If C(- 2, 1) is the mid-point of the line segment joining A(- 6, p) and B(2, p + 6), then the, value of p is, (a) 2, , (b) - 2, , (c) 0, , (d) 4

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163, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 8. Name the criteria of similarity by which following triangles are similar., R, C, 4.5, 3, A, , (a) SSS, , 54°, , B P, , 5, , 54°, 7.5, , (b) SAS, , Q, , (c) AAA, , (d) ASA, , 9. If LCM of (a , b) = 53 and HCF of (a , b) = 12, then product of a and b is, (a) 636, , (b) 666, , (c) 696, , (d) 646, , (c) 0, , (d) 1, , 10. If P(E) is 0.75, what is P(not E)?, (a) 0.35, , (b) 0.25, , 11. If ax + by + c = 0, where a , b and c are real numbers, then for which condition the, equation is said be a linear equation in two variables x and y., (a) a ¹ b, (b) a 2 = b 2, (c) a 2 + b 2 = 0, , (d) a 2 + b 2 ¹ 0, , 12. The number of polynomials having zeroes as - 2 and 5 is, (a) 1, , (b) 2, , (c) 3, , (d) more than 3, , 13. The perpendicular distance of the point P(4 , 2) from the Y-axis is, (a) 4 units, , (b) 6 units, , (c) 2 units, , (d) 8 units, , (c) 21 and 25, , (d) 10 and 40, , 14. The values of x and y in the given figure are, 4, y, , 3, x, 7, , (a) 10 and 14, , (b) 21 and 84, , 15. A card is drawn from a deck of 52 cards. The event E is that card which is not an ace of, heart. The number of outcomes favourable to E is, (a) 4, (b) 13, (c) 48, , (d) 51, , 16. Graphically ax + by + c = 0 represents a line. Every solution of the equation is a point, (a) on the line representing, (c) on the X-axis, , (b) not on the line representing it, (d) on the Y-axis, , 17. For any two similar triangles which of the following statements are valid, (b) their sides are equal, (d) None of these, , 1, are the zeroes of the polynomial ax 2 + bx + c, then value of c is, a, (a) 0, (b) a, (c) - a, (d) 1, 13, have decimal expansion, 19. The number, 3125, (a) terminating, (b) non-terminating repeating, (c) non-terminating non-repeating, (d) non-terminating, , 18. If a and, , SAMPLE PAPER 9, , (a) their sides are proportional, (c) their sides are parallel

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164, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 20. In the figure, M is the mid-point of line LN. Then, x + y is equal to, P(5, 2), , (7, 3)L, , (a) 7.5, , M(x, y), , (b) 3.5, , N(1, 4), , (c) 4.5, , (d) 5.5, , Section B, Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 21. If DABC and DDEF are similar triangles such that ÐA = 57 ° and ÐE = 73°, then ÐC is, equal to, (a) 50°, , (b) 60°, , (c) 70°, , (d) 80°, , 22. A quadratic polynomial, whose zeroes are - 2 and 4, is, (a) x 2 - 2 x + 8, , (b) x 2 + 2 x + 8, , (c) x 2 - 2 x - 8, , (d) 2 x 2 + 2 x - 24, , 23. The sum of the digits of a two-digit number is 9. If 27 is added to it, the digits of the, number get reversed. The number is, (a) 27, (b) 72, , (c) 45, , (d) 36, , 24. A school has five houses A, B , C , D and E. The class has 23 students, 4 from house A, 8, from house B, 5 from house C, 2 from house D and rest from house E. A single student, is selected at random for the class monitor. The probability that the selected student is, not from A, B and C is, 4, 6, 8, 17, (b), (c), (d), (a), 23, 23, 23, 23, , 25. The points (- 5, 0), (5, 0) and (0, 4) are the vertices of, (a) right triangle, , (b) equilateral triangle (c) isosceles triangle, , (d) scalene triangle, , 26. In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84, , SAMPLE PAPER 9, , and 108 respectively. Find the minimum number of rooms required, if in each room the, same number of participants are to be seated and all of them being in the same subject., (a) 210, (b) 21, (c) 12, (d) 3780, BD, is equal to, 27. DABC is a right triangle, right-angled at A and AD ^ BC. Then,, DC, AB ö 2, AB, AB ö 2, AB, (a) æç, (c) æç, (b), (d), ÷, ÷, AC, AD, è AC ø, è AD ø, , 28. ……… is a solution of pair of equations 3x - 2y = 4 and 2x + y = 5., (a) x = 2 and y = 1, , (b) x = 2 and y = 2, , (c) x = 3 and y = 1, , (d) x = 3 and y = 2, , 2, , 29. If 2 and 3 are zeroes of polynomial 3x - 2kx + 2m, then the value of k and m are,, respectively, 9, (a) and 15, 2, , (b), , 15, and 9, 2, , (c) 9 and, , 15, 2, , (d) 15 and 9, , 30. If four vertices of a parallelogram taken in order are (- 3, - 1), (a , b) , (3, 3) and (4, 3),, then a : b is equal to, (a) 1 : 4, , (b) 4 : 1, , (c) 1 : 2, , (d) 2 : 1

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165, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 31. In a simultaneously throw of a pair of dice. The probability of getting a double is, (a), , 5, 12, , (b), , 1, 3, , (c), , 1, 6, , (d), , 1, 2, , 32. Two alarm clocks ring their alarms at regular intervals of 50 s and 48 s. If they first, beep together at 12 noon, at what time will they beep again for the first time?, (a) 12 : 20 pm, (b) 12 : 12 pm, (c) 12 : 11 pm, (d) None of these, , 33. (cos 4 A - sin 4 A) is equal to, (a) 1 - 2 cos 2 A, , (b) 2 sin 2 A - 1, , (c) sin 2 A - cos 2 A, , (d) 2 cos 2 A - 1, , 34. It is given that, DABC ~ DEDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12, cm, then the sum of the remaining sides of the triangles is, (a) 23.05 cm, (b) 16.8 cm, (c) 6.25 cm, , (d) 24 cm, , 35. The point of intersection of the line 3x - 2y = 6 and the X-axis is, (a) (2, 0), , (b) (0, - 3), , (c) ( - 2 , 0), , (d) (0, 3), , 36. A fair dice is rolled. Probability of getting a number x such that 1 £ x £ 6 is, (a) 0, , (b) > 1, , (c) between 0 and 1, , (d) 1, , 37. If the points A(- 2, 5) and B(4 , 3) are equidistant from the Y-axis, then the coordinates of, that point are, (a) (0, - 1), , (b) (0, 1), , (c) (1, 0), , (d) None of these, , 38. Which of the following has a non-terminating repeating decimal expansion?, (a), , 3, 8, , (b), , 13, 125, , (c), , 7, 80, , (d), , 29, 343, , 39. If one of the zeroes of the quadratic polynomial (k - 1)x 2 + kx + 1 is - 3, then the value of, k is, (a), , 4, 3, , (b), , -4, , (c), , 3, , 2, 3, , (d), , -2, 3, , 40. In an equilateral triangle DABC , if AD ^ BC, then, (a) 2 AB2 = 3 AD 2, , (b) 4 AB2 = 3 AD 2, , (c) 3 AB2 = 4 AD 2, , (d) 3 AB2 = 2 AD 2, , Section C, Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted., 41-45 are based on Case Study-1., , 12, , cm, , A, , C, , 6Ö3 cm, , 6 cm, B, , SAMPLE PAPER 9, , Case Study 1, While eating sandwich, Chetna jokingly remarked that she can find out the value of any, trigonometric ratio, if just one ratio is known to her, as the sandwich is a right triangle.

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166, , CBSE Sample Paper Mathematics Standard Class X (Term I), , On the basis of above information answer the following question., , 41. The value of ÐC is equal to, (a) 30°, (c) 45°, , (b) 60°, (d) None of these, , 42. The value of ÐA is equal to, (a) 30°, (c) 45°, , (b) 60°, (d) None of these, , 43. The value of tan C is, 1, 3, (d) None of these, (b), , (a) 3, (c) 1, , 44. The value of tan C ´ tan A is, 1, 2, (d) None of these, , (a) 2, , (b), , (c) 1, , 45. If 3 tan A = 4, then the value of, (a) 4, , (b), , 3 sin A + 2 cos A, 3 sin A - 2 cos A, , 11, 15, , is, , (c), , 7, 15, , (d) 3, , 46-50 are based on Case Study-2., , Case Study 2, Mr. Jay purchase a design which he has to decorate. In the design there are three semi-circles A, B, and C having diameter 3 cm each, another larger semi-circle having diameter 9 cm and a circle D, of diameter 4.5 cm, D, E, A, , C, B, , On the basis of above information, answer the following questions., , 46. Total area of three semi-circles A, B and C is, , SAMPLE PAPER 9, , (a) 10.6 cm 2, , (b) 3.53 cm 2, , (c) 7.06 cm 2, , (d) 21 cm 2, , (b) 63.6 cm 2, , (c) 15.91 cm 2, , (d) 3182, . cm 2, , (c) 63.64 cm 2, , (d) 81 cm 2, , (c) 15.91 cm 2, , (d) 22.98 cm 2, , (c) 12.375 cm 2, , (d) 24.75 cm 2, , 47. Area of circle D is, (a) 8 cm 2, , 48. Area of largest semi-circle is, (a) 15.91 cm 2, , (b) 3182, . cm 2, , 49. Find the area of unshaded region., (a) 12.38 cm 2, , (b) 10.6 cm 2, , 50. Find the area of shaded region., (a) 3182, . cm 2, , (b) 22.98 cm 2

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OMR SHEET, , SP 9, , Roll No., Sub Code., , Student Name, , Instructions, Use black or blue ball point pens and avoid Gel & Fountain pens for filling the OMR sheet., Darken the bubbles completely. Don’t put a tick mark or a cross mark, half-filled or over-filled bubbles will not be read, by the software., Correct, , ✔, , ✗, , Incorrect, , Incorrect, , Incorrect, , Do not write anything on the OMR Sheet., Multiple markings are invalid., , 1, , 18, , 35, , 2, , 19, , 36, , 3, , 20, , 37, , 4, , 21, , 38, , 5, , 22, , 39, , 6, , 23, , 40, , 7, , 24, , 41, , 8, , 25, , 42, , 9, , 26, , 43, , 10, , 27, , 44, , 11, , 28, , 45, , 12, , 29, , 46, , 13, , 30, , 47, , 14, , 31, , 48, , 15, , 32, , 49, , 16, , 33, , 50, , 17, , 34, , Check Your Performance, Total Questions:, Total Correct Questions:, , If Your Score is, , Score Percentage =, , Total Correct Questions, Total Questions, , × 100, , Less than 60%, > Average (Revise the concepts again), Greater than 60% but less than 75% > Good (Do more practice), Above 75%, > Excellent (Keep it on)

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168, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Answers, 1. (b), 11. (d), , 2. (a), 12. (d), , 3. (b), 13. (a), , 4. (c), 14. (b), , 5. (d), 15. (d), , 6. (c), 16. (a), , 7. (b), 17. (a), , 8. (b), 18. (b), , 9. (a), 19. (a), , 10. (b), 20. (a), , 21. (a), 31. (c), 41. (a), , 22. (c), 32. (a), 42. (b), , 23. (d), 33. (d), 43. (b), , 24. (b), 34. (a), 44. (c), , 25. (c), 35. (a), 45. (d), , 26. (b), 36. (d), 46. (a), , 27. (a), 37. (b), 47. (c), , 28. (a), 38. (d), 48. (c), , 29. (b), 39. (a), 49. (d), , 30. (b), 40. (c), 50. (c), , SOLUTIONS, 1. Prime factorization of 8 , 9 and 25 are, 8 =2 ´ 2 ´ 2, 9 =3 ´ 3, and, 25 = 5 ´ 5, \HCF (8, 9, 25) = 1, and LCM (8, 9, 25) = 2 3 ´ 3 2 ´ 5 2 = 1800, \ HCF (8, 9, 25) ´ LCM (8, 9, 25) = 1 ´ 1800, = 1800, , b, x, Put x = - 1 and y = 1 in above equation., b, Þ a - b =1, 1=a+, -1, , 6. Given, y = a +, , Put x = - 5 and y = 5 in the given equation., b, …(ii), Þ 25 = 5 a - b, 5=a+, -5, Subtract Eq. (i) from Eq. (ii), , 2. Let the points be A(0 , 0 ), B(3 , 3 ) and C(3 , p),, which are the points of equilateral triangle., \, AB = AC, Þ (3 - 0 )2 + ( 3 - 0 )2 = (3 - 0 )2 + ( p - 0 )2, Þ, Þ, Þ, , [by distance formula], 9 + 3 = 9 + p2, p2 = 3, p=± 3, , As, p ¹ 3, otherwise two points will be same., \, p = - 3., 3. Since,, \, Þ, Þ, , DABC ~ DDEF, AB BC, =, DE EF, AB, 8, =, 2 AB EF, EF = 2 ´ 8 = 16 cm, , 4. Total number of outcomes = 5 + 20 = 25, , SAMPLE PAPER 9, , Number of favourable outcomes = 5, \Required probability, Number of favourable outcomes, =, Total number of outcomes, 5 1, =, =, 25 5, 2, , 5. Given polynomial is 6 x - 5, Coefficient of x, Sum of zeroes = Coefficient of x 2, 0, = - =0, 6, -5, Constant term, Product of zeroes =, =, 2, 6, Coefficient of x, , …(i), , Þ, \, \, , 4 a = 24, a =6, b =6 -1 = 5, a + b = 6 + 5 = 11, , 7. Given, C( -2 , 1) is the mid-point of line segment, joining A( -6 , p) and B(2 , p + 6 )., æ -6 + 2 p + p + 6 ö, ,, \ç, ÷ = ( -2 , 1), 2, ø, è 2, [by section formula], 2 p+6, =1 Þ 2 p+6 =2, Þ, 2, Þ, 2 p = - 4 Þ p = -2, 8. In DABC and DPQR,, AC AB, =, PR PQ, and, , éQ 3 = 5 = 2 ù, êë 4.5 7.5 3 úû, , ÐA = ÐP = 54 °, , \ DABC ~ DPQR (by SAS similarity criterion), 9. We know that, Product of two numbers a and b, = Product of their LCM and HCF, = 53 ´ 12 = 636, 10. Given, P( E) = 0.75, Since, P( E) + P (not E) = 1, Þ, , P (not E) = 1 - P( E) = 1 - 0.75, = 0.25, , 11. Equation ax + by + c = 0 is said to be a linear, equation in two variable, if a and b are, simultaneously not equal to zero i.e. a2 + b2 ¹ 0, As, a2 + b2 = 0 only, if a = 0 and b = 0

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169, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 12. Given, zeroes of a polynomial are - 2 and 5., \ Required polynomial, = k [ x 2 - ( sum of zeroes )x, + product of zeroes], = k [ x 2 - ( - 2 + 5 )x + ( - 2 ´ 5 )], = k [ x 2 - 3 x - 10 ], where k Î R., Hence, infinite number of polynomials exist., , 20. From the figure, mid-point of LN is given by, æ7 + 1 3 + 4ö, (x, y) = ç, ,, ÷ = (4, 3.5), 2 ø, è 2, Now, value of x + y = 4 + 3.5 = 7.5, 21. Given, DABC ~ DDEF,, ÐA = 57 ° and ÐE = 73 °, D, , A, , 13. Firstly, plot the point P( 4 , 2 ), , 57°, , Y, 2, , 4 units, , P(4, 2), , X¢, , 73°, , 2 units, , 1, , B, X, , O, , 1 2 3 4 5 6 7, Y¢, , Hence, the distance from P to the X-axis is, 4 units., 14. Given,, , y, x, , 4, 3, , 7, , Here, x = 3 ´ 7 = 21 and y = 4 ´ x = 4 ´ 21 = 84, 15. In a deck of 52 cards, there are 13 cards of heart, and 1 of them is ace of heart., So, number of favourable outcomes to E, = 52 - 1 = 51, 16. The solution set of the line ax + by + c = 0 is, collinear with the other points on the given line., Hence, the solution is on the given line which, represents it., 17. Two triangles are said to be similar, if, (i) their corresponding angles are equal., (ii) their corresponding sides are proportional., , 19. The prime factorisation of the denominator, must be in the form of 2 n ´ 5 m, then only the, rational number have terminating decimal, expansion., Here, 3125 = 5 5, 13, has terminating decimal expansion., \, 3125, , E, , F, , According to the question,, DABC ~ DDEF, Then,, ÐA = ÐD = 57 °, ÐB = ÐE = 73 °, ÐC = ÐF, We know that, sum of all the angles of a, triangle is equal to 180°., \, ÐA + ÐB + ÐC = 180 °, Þ, 57 ° + 73 ° + ÐC = 180 °, Þ, ÐC = 180 ° - 130 ° = 50 °, 22. Let zeroes of given polynomial be, a = - 2 and b = 4., Then, a + b = - 2 + 4 = 2 and ab = - 2 ´ 4 = - 8, Now, quadratic polynomial is, x 2 - ( a + b) x + ab = x 2 - 2 x - 8, 23. Let the digit at unit’s place be y and the digit at, ten’s place be x., So, the number will be 10x + y., According to first condition,, …(i), x + y =9, According to second condition,, 10 y + x = 10 x + y + 27, Þ, 9 y = 9 x + 27, [divide by 9] …(ii), Þ, y=x+3, Put the value of y in Eq. (i), we get, x + x + 3 =9, Þ, 2x =6Þ x =3, \, y =3 + 3 =6, So, the number is 10 ´ 3 + 6 i.e. 36., 24. Total number of students = 23, Number of students in house A = 4, Number of students in house B = 8, Number of students in house C = 5, Total number of students in house A, B and, C = 4 + 8 + 5 = 17, \Remaining students = 23 - 17 = 6, , SAMPLE PAPER 9, , Coefficient of x, Coefficient of x 2, 1 -b, Þ, a+ =, a, a, Constant term, Product of zeroes =, Coefficient of x 2, 1 c, Þ, a´ =, a a, c, 1= Þ c=a, Þ, a, , 18. Sum of zeroes = -, , C

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170, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Thus, number of outcomes favourable to the, given event = 6, \Required probability, Number of favourable outcomes 6, =, =, 23, Total number of outcomes, 25. Let the points be A( - 5 , 0 ), B( 5 , 0 ) and C(0 , 4 ), The distance between the points A and B, B and, C, C and A are, AB = ( 5 + 5 )2 + (0 - 0 )2, = 10 2 + 0 = 100 = 10 units, BC = (0 - 5 )2 + ( 4 - 0 )2, = ( - 5 )2 + 4 2 = 25 + 16, = 41 units, and, , CA = ( - 5 - 0 )2 + (0 - 4 )2, = ( -5 )2 + ( -4 )2, , Here,, , = 25 + 16 = 41 units, BC = CA, , \DABC is an isosceles triangle., 26. Prime factorization of 60, 84 and 108 are, 60 = 2 2 ´ 3 ´ 5, 84 = 2 2 ´ 3 ´ 7, 108 = 2 2 ´ 3 3, \HCF (60, 84, 108) = 2 2 ´ 3 = 12, Therefore, in each room 12 participants can be, seated., \Number of rooms required, Total number of participants, =, 12, 60 + 84 + 108 252, =, =, = 21, 12, 12, 27. In a right triangle DABC, right-angled at A and, AD ^ BC., In DCAB and DCDA,, C, , SAMPLE PAPER 9, , D, , A, , \, Þ, Þ, , B, , (each 90º), ÐCAB = ÐCDA, (common), ÐACB = ÐDCA, DCAB ~ DCDA, (by AA similarity criterion), BC AC, =, AC DC, …(i), AC 2 = BC × DC, , Similarly, DCAB ~ DADB, (by AA similarity criterion), AB BC, Þ, =, BD AB, …(ii), Þ, AB2 = BC × BD, Dividing Eq. (ii) by Eq. (i), we get, AB2, BC × BD BD, =, =, 2, BC × DC DC, AC, 28. Given, 3 x - 2 y = 4, and, 2x + y = 5, Multiply Eq (ii) by 2, we get, 4 x + 2 y = 10, Adding Eqs (i) and (iii), we get, 7 x = 14 Þ x = 2, From Eq. (ii), 4 + y = 5 Þ y = 1, \, x = 2 and y = 1, , …(i), …(ii), …(iii), , 29. For a quadratic polynomial ax 2 + bx + c, the sum, -b, c, of the zeroes =, and product of the zeroes =, a, a, Here, given polynomial is 3 x 2 - 2 kx + 2 m and, zeroes are 2 and 3, - (- 2 k ), Thus, 2 + 3 =, 3, 2k, 15, 2m, and 2 ´ 3 =, Þk=, 5=, Þ, 3, 2, 3, Þ, 6 ´ 3 = 2 mÞ m = 9, 15, Thus, k =, and m = 9, 2, 30. Let points be A( -3 , - 1), B( a, b), C(3 , 3 ) and D( 4 , 3 )., So, coordinates of the mid-point of AC =, coordinates of the mid-point of BD, [Q in parallelogram, diagonals, bisect each other], æ -3 + 3 -1 + 3 ö æ a + 4 b + 3 ö, ,, ,, Þç, ÷=ç, ÷, 2 ø, 2 ø è 2, è 2, æa+ 4 b+3ö, (0, 1) = ç, ,, Þ, ÷, 2 ø, è 2, a+4, b+3, = 0 and, =1, Þ, 2, 2, Þ, a = - 4 and b = - 1, a -4 4, Now,, = Þ a: b = 4 :1, =, b -1 1, 31. When two dice are tossed. Total possible, outcomes = 36, i.e., n( S) = 36, and total favourable outcomes (doublet), = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}, i.e. n( E) = 6, 6 1, \Required probability =, =, 36 6

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171, , CBSE Sample Paper Mathematics Standard Class X (Term I), , So, the probability of getting a number that is, more than or equal to 1 and less than or equal, to 6 is, Outcomes of Event E, P( E) =, Total number of outcomes, 6, = =1, 6, , 32. The prime factorisation of 50 and 48 are, 50 = 2 ´ 5 2, 48 = 2 4 ´ 3, \LCM (50, 48) = 2 4 ´ 3 ´ 5 2 = 1200, Here, LCM of 50 s and 48 s = 1200 s, i.e. 20 min, The first beep is at 12 noon (given), \Again, the beep is at 20 min later, or 12 : 20 pm, , 37. Let point on Y-axis are P(0 , y )., 2, , Þ (0 + 2 ) + ( y - 5 )2 = (0 - 4 )2 + ( y - 3 )2, , 33. Consider, cos 4 A - sin 4 A, , Þ, , = (cos 2 A)2 - (sin 2 A)2, = (cos 2 A - sin 2 A)(cos 2 A + sin 2 A), 2, , 2, , [Q cos 2 q + sin 2 q = 1], , = (cos A - sin A), 2, , 2, , [Q sin 2 q = 1 - cos 2 q ], , = cos A - (1 - cos A), , = cos 2 A - 1 + cos 2 A = 2 cos 2 A - 1, 34. Given, DABC ~ DEDF, E, , 12, , cm, , A, 5 cm, , 7 cm, , B, , CD, , 15 cm, , F, , Since, DABC ~ DEDF, We know that, two triangles are said to be, similar, if their corresponding sides are in, proportion., 5, 7, BC, \, =, =, 12 EF 15, I, , II, , 4 y = 29 - 25, 4, Þ, y = =1, 4, Hence, required coordinates onY-axis are (0, 1)., , 38. A fraction can be expressed in the form, terminating decimal, if denominator can be, expressed in the form of 2 n ´ 5 m, where n and m, are integers., Here,, 8 =23, 125 = 5 3, 80 = 2 4 ´ 5, 343 = 7 3, 29, So,, has non-terminating repeating decimal, 343, expansion., 39. Given, (- 3) is the zeroes of the polynomial, , III, , = EF + BC = 16.8 + 6.25 = 23.05 cm, …(i), , Equation of X-axis is y = 0, Put y = 0 in Eq. (i),, 3 x - 2 (0 ) = 6 Þ x = 2, Hence, required point of intersection is (2, 0)., 36. On rolling a dice at once the numbers that are, more than or equal to 1 and less than or equal, to 6 are 1, 2, 3, 4, 5, 6., , ( k - 1)x 2 + kx + 1, So, (- 3) must satisfy the equation, ( k - 1)x 2 + kx + 1 = 0, Þ ( k - 1) ( - 3 )2 + k( - 3 ) + 1 = 0, Þ, 9( k - 1) - 3 k + 1 = 0, Þ, 9k - 9 - 3k + 1 = 0, Þ, 6k = 8, 4, k=, Þ, 3, 40. Given, DABC is an equilateral, triangle and AD ^ BC, , A, , \ AB = BC = AC, ÐABC = ÐBAC = ÐACB = 60 °, and ÐADB = ÐADC = 90 °, B, C, D, In DABD,, ÐBAD + ÐABD + ÐADB = 180 °, [angle sum property of triangle], ÐBAD = 180 ° - (90 ° + 60 ° ) = 30 °, Similarly for DACD,, ÐCAD + ÐACD + ÐADC = 180 °, [angle sum property of triangle], ÐCAD = 180 ° - 90 ° - 60 ° = 30 °, , SAMPLE PAPER 9, , On taking I and II ratios, we get, 7 ´ 12 84, 5, 7, =, = 16.8 cm, Þ EF =, =, 5, 5, 12 EF, On taking I and III ratios, we get, BC, 5, =, 12 15, 5 ´ 15 25, Þ, BC =, =, = 6.25 cm, 12, 4, Now, sum of the remaining sides of triangle, 35. Given, 3 x - 2 y = 6, , 4 + y 2 + 25 - 10 y = 16 + y 2 + 9 - 6 y, , Þ, , 2, , [Q a - b = ( a - b) ( a + b)], 2, , PA2 = PB2, , Then,

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172, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Now, In DABD and DACD,, AB = AC, [given], ÐBAD = ÐCAD = 30 °, and, (common side), AD = AD, Thus, DABD and DACD are congruent by SAS, congruency criterion., [by CPCT], \, BD = DC, 1, 1, Þ, BC = AB, 2, 2, Now, DABD is a right angled triangle., , SAMPLE PAPER 9, , Therefore, AB2 = BD2 + AD2, [by Pythagoras theorem], 2, 1, Þ, AB2 = æç ABö÷ + AD2, ø, è2, AB2, 2, + AD2, Þ, AB =, 4, AB2, Þ, = AD2, AB2 4, Þ, 4 AB2 - AB2 = 4 AD2, Þ, 3 AB2 = 4 AD2, Solutions (41-45), In DABC, ÐB = 90 °, cm, AB = 6 cm, 12, BC = 6 3 cm, C, AC = 12 cm, 6Ö3 cm, AB 6, 41., =, sin C =, AC 12, 1, Þ, sin C =, 2, Þ, sin C = sin 30 °, \, Ð C = 30 °, BC, 42., sin A =, AC, 6 3, sin A =, Þ, 12, 3, Þ, sin A =, 2, Þ, sin A = sin 60 °, \, Ð A = 60 °, 6, 1, AB, 43. tan C =, =, =, 3, BC 6 3, AB BC, 44. tan C ´ tan A =, ´, =1, BC AB, 3 sin A + 2 cos A, 45. We have, 3 tan A = 4 and, 3 sin A - 2 cos A, , A, , 3 tan A + 2 4 + 2, =, 3 tan A - 2 4 - 2, 6, = =3, 2, =, , [Q3 tan A = 4 ], , Solutions (46-50), 46. In the given figure,, 3, cm, 2, 3, Total area of three semi-circles = pr2, 2, 1 22 3 3, éQ r = 3 ù, =3 ´ ´, ´ ´, êë, 2, 7 2 2, 2 úû, 594, = 10.6 cm 2, =, 56, , The, radius of each semi-circle =, , 47. In the given figure,, 4.5, = 2 .25 cm, 2, 22, \Area of circle, D = pr2 =, ´ 2 .25 ´ 2 .25, 7, 111.375, = 15.91 cm 2, =, 7, The radius of circle, D =, , 48. In the given figure,, 6 cm, , B, , Divide by cos A in numerator and denominator, 3 sin A 2 cos A, +, cos A, = cos A, 3 sin A 2 cos A, cos A, cos A, , The radius of largest semi-circle =, , 9, = 4.5 cm, 2, , \Area of largest semi-circle, 22 9 9 1782, = 63.64 cm 2, = pr2 =, ´ ´ =, 7 2 2, 28, 49. Area of unshaded region, = 2 ´ area of semi-circle A + area of circle D, 1 22 3 3 22, ´ 2 .25 ´ 2 .25, =2 ´ ´, ´ ´ +, 2, 7 2 2, 7, 99 111375, ., = 7.07 + 15.91 = 22 .98 cm 2, =, +, 14, 7, 50. The area of shaded region, = area of region E + area of region B, - area of region A - area of region C, - area of region D, 1, 1, 2, 2, = p( 4.5 ) + p(1.5 ), 2, 2, 2, 4.5 ö, 1, 1, - p(1.5 )2 - p(1.5 )2 - p æç, ÷, 2, 2, è 2 ø, 1 p, 1, 1, 2, 2, = p( 4.5 ) - p(1.5 ) - ´ ´ ( 4.5 )2, 2 2, 2, 2, ì, 1, ( 4.5 )2 ü, = ´ pí ( 4.5 )2 - (1.5 )2 ý, 2 þ, 2, î, 1, = ´ p [20.25 - 2.25 - 10.125], 2, 1 22, = ´, ´ 7.875 = 12.375 cm 2, 2 7

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173, , CBSE Sample Paper Mathematics Standard Class X (Term I), , SAMPLE PAPER 10, MATHEMATICS (Standard), A Highly Simulated Practice Questions Paper, for CBSE Class X (Term I) Examination, , Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions., There is no negative marking., Maximum Marks : 40, Time allowed : 90 minutes, , Roll No., , Section A, Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., , 1. If a and b be the zeroes of the polynomial p(x) = x 2 - 5x + 2, find the value of, 1 1, + - 3ab, a b, 3, (a) 2, , (b) -, , 5, 2, , (c) -, , 7, 2, , (d) -, , 9, 2, , 2. If a man goes 24 m towards East and then 10 m towards North. How far he is starting, from?, (a) 26 m, , (b) 17 m, , (c) 18 m, , (d) None of these, , 3. Given that HCF (96, 404) is 4, then the LCM (96, 404) is, (a) 9187, , (b) 9230, , (c) 9696, , (d) 10387, , 4. In what ratio does the point P(3, 4) divided the line segment joining the points A(1, 2), (b) 2 : 3, , 5. The maximum value of, (a) 0, , (c) 3 : 4, , (d) 1 : 1, , (c) 1, , (d), , 1, is, cosec q, , (b) - 1, , 3, 2, , 6. A wire is in the shape of a circle of radius 21 cm. It is bent in the form of a square., , SAMPLE PAPER 10, , and B( 6, 7)?, (a) 1 : 2

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174, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 22 ö, The side of the square is æç use p =, ÷, 7 ø, è, (a) 22 cm, (b) 33 cm, , (c) 44 cm, , (d) 66 cm, , 7. The probability of getting a bad egg from a lot of 400 eggs is 0.035. The number of bad, eggs in the lot is, (a) 7, , (b) 14, , (c) 21, , (d) 28, , 8. The zeroes of the polynomial f (x) = x 2 - 2 2x - 16 are, (a) 2 and - 2, (c) - 4 2 and 2 2, , (b) 4 2 and - 2 2, (d) 4 2 and 2 2, , 9. In a DABC it is given that AB = 6 cm, AC = 8 cm and AD is the bisector of A. Then,, BD : DC is, , B, , (a) 3 : 4, , cm, , 6c, , 8, , m, , A, , D, , C, , (b) 9 : 16, 2, , (c) 4 : 3, , (d) 3 : 2, , 2, , 10. If x = 3 sec q - 1 and y = tan q - 2, then x - 3y is equal to, (a) 3, , (b) 4, , (c) 8, , (d) 5, , 11. If the point (x , y) is equidistant from the points (2, 1) and (1, - 2), then, (a) x + 3y = 0, (c) x + 2 y = 0, , (b) 3x + y = 0, (d) 3x + 2 y = 0, , 12. If the radius of a circle is, (a) 72 cm 2, (c) 36 cm 2, , 7, cm, then the area of the circle is, p, (b) 49 cm 2, (d) 56 cm 2, , 13. A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at, random. The probability that the drawn ball is red, is, 5, 3, 7, (a), (b), (c), 15, 15, 15, , (d), , 1, 15, , 14. A quadratic polynomial whose one zero is 6 and sum of the zeroes is 0, is, (a) x 2 - 6x + 2, , (b) x 2 - 36, , (c) x 2 - 6, , (d) x 2 - 3, , SAMPLE PAPER 10, , 15. In the figure, values of x in terms of a , b and c are, L, , a, , P, x, 35°, , 35°, M, , (a), , ab, a+b, , b, , c, , N, , (b), , ac, b+ c, , K

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175, , CBSE Sample Paper Mathematics Standard Class X (Term I), , (c), , bc, a+b, , (d), , ac, a+b, , 16. In the figure, the area of the shaded portion is (use p = 314, . ), A, , 8 cm, , O, , 6 cm, B, , (a) 15.25 cm 2, , (b) 12.75 cm 2, , C, , (c) 18.05 cm 2, , (d) 20. 60 cm 2, , 17. sec A is equal to, (a), , 1, cot A, , (b), , 1, cosec A, , 1, , (c), , 1 + cot 2 A, , 1 + cot 2 A, , (d), , cot A, , p, . If the, 12, 3, probability of not guessing the correct answer to same question is , then value of p is, 4, (a) 3, (b) 4, (c) 2, (d) 1, , 18. The probability of guessing the correct answer to certain question is, , 19. The value of k, for which the system of equations x + (k + 1)y = 5 and (k + 1)x + 9y = 8k - 1, has infinitely many solutions, is, (b) 3, , (a) 2, , (c) 4, , (d) 5, , (c) cosec 2 A - 1, , (d) 1 - sin 2 A, , 2, , 20., , 1 + cot A, 1 + tan 2 A, (a) tan 2 A, , is equal to, (b) sec 2 A, , Section B, Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted., 1, 1, 21. If a and b are zeroes of the polynomial f (x) = ax 2 + bx + c, then 2 + 2 is equal to, a, b, 2, 2, 2, b - 2 ac, b - 2 ac, b + 2 ac, b 2 + 2 ac, (b), (c), (d), (a), a2, c2, a2, c2, , 22. If a circular grass lawn of 35 m in radius has a path 7 m wide running around it on the, (c) 1694 m 2, , (d) 3368 m 2, , 23. A game consists of tossing a coin 3 times and noting the outcomes each time. If getting, the same result in all the tosses is a successes, then probability of losing the game is, 3, 1, 3, (b), (c), (d) 1, (a), 4, 4, 8, , 24. Which of the following has a terminating decimal expansion?, (a), , 23, 200, , (b), , 17, 9, , (c), , 8, 75, , (d), , 3, 35, , SAMPLE PAPER 10, , outside, then the area of the path is, (a) 1450 m 2, (b) 1576 m 2

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176, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 25. The perimeter of the triangle formed by the points (0, 0), (2, 0) and (0, 2) is, (a) (1 - 2 2 ) units, (c) (4 + 2 ) units, , (b) (2 2 + 1) units, (d) (4 + 2 2 ) units, , 26. In a right triangle DABC right-angled at B, if P and Q are points on the sides AB and AC, respectively, then, (a) AQ 2 + CP 2 = 2( AC 2 + PQ 2 ), , (b) 2( AQ 2 + CP 2 ) = AC 2 + PQ 2, 1, (d) AQ1 + CP = ( AC + PQ), 2, , (c) AQ 2 + CP 2 = AC 2 + PQ 2, , 27. In the following figure, OABC is a square of side 7 cm. OAC is a quadrant of a circle, with O as centre. The area of the shaded region is, O, , C, , 7 cm, , A, , (a) 10.5 cm, , 2, , (b) 38.5 cm, , B, 2, , (c) 49 cm 2, , (d) 11.5 cm 2, , 28. If both zeroes of the quadratic polynomial x 2 - 2kx + 2 are equal in magnitude but, opposite in sign, then value of k is, (a) 0, , (b) 1, , (c), , 1, 2, , (d) -1, , 29. DABC is an equilateral triangle with each side of length 4 p. If AD ^ BC, then value of, AD is, (b) 2 3 p, , (a) 3, , 30. The value of, , sin q - 2 sin 3 q, 2 cos 3 q - cos q, , (a) cosec q, , (c) 2 p, , (d) 4 p, , , where q is acute angle, is, , (b) cot q, , (c) tan q, , (d) sin q cos q, , 31. In a circle of diameter 42 cm, if an arc subtends an angle of 60° at the cente where, 22, , then what will be the length of arc (in cm), 7, (a) 22, (b) 21, (c) 24, p=, , (d) 27, , SAMPLE PAPER 10, , 32. Seven face cards are removed from a deck of cards and the cards are well shuffled., Then, the probability of drawing a face card is, 1, 1, 2, (a), (b), (c), 8, 9, 7, , (d), , 3, 4, , 33. If x = p secq and y = q tan q, then, (a) x 2 - y 2 = p2 q 2, (c) x 2 q 2 - y 2 p2 =, , (b) x 2 q 2 - y 2 p2 = pq, 1, pq, , (d) x 2 q 2 - y 2 p2 = p2 q 2, , 34. The point which divides the line segment joining the points (7 , - 6) and (3, 4) in ratio, 1 : 2 internally lies in the

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177, , CBSE Sample Paper Mathematics Standard Class X (Term I), , (a) I quadrant, , (b) II quadrant, , (c) III quadrant, , (d) IV quadrant, , 35. If the distance between A(k , 3) and B(2, 3) is 5 units, then the value of k is, (a) 5, , (b) 6, , (c) 7, , 36. The zeroes of the polynomial f (x) = x 2 + x (a) -, , 1, 3, and, 2, 2, , (b), , 1, 3, and 2, 2, , (d) 8, , 3, are, 4, 3, (c) - and 1, 2, , (d) 1 and, , 3, 2, , 37. The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2, - 5), and R( - 3, 6). Then, the coordinates of P are, (a) (16, 8), (b) (14, 7), (c) (18, 9), , (d) (10, 5), , 38. If DABC ~ DDEF such that DE = 6 cm, EF = 4 cm, DF = 5 cm and BC = 8 cm, then, perimeter of DABC is, (a) 18 cm, , (b) 20 cm, , (c) 12 cm, , (d) 30 cm, , 39. The centroid of the triangle whose vertices are (3, - 7), (- 8, 6) and (5, 10) is, (a) (0, 3), , (b) (0, 9), , (c) (1, 3), , (d) (3, 5), , 40. If tan q + sin q = m and tan q - sin q = n, then m 2 - n2 is equal to, m, n, (d) None of these, , (a) mn, , (b), , (c) 4 mn, , Section C, Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted., 41-45 are based on Case Study-1., , Case Study 1, Prabhat is planning to buy a house and the layout is given below. The design and the, measurement has been made such that areas of two bedrooms and garden together is 95 sq m., , 5m, , p, , 2, , q, , Bedroom 1, , Bath, Room, , Garden, , 5m, , Bedroom 2, , Living Room, 15 m, , Based on the above information, answer the following questions:, , SAMPLE PAPER 10, , 2m

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178, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 41. Which is the correct equation in two variables from this situation., (a) p + q = 13, (c) p + q = 17, , (b) p + q = 15, (d) p + q = 19, , 42. Find the length of the outer boundary of the layout., (a) 86 m, , (b) 45 m, , (c) 34 m, , (d) 54 m, , 43. Find the area of each bedroom and garden in the layout., (a) Bedroom 20 m 2 and garden 25 m 2, (b) Bedroom 50 m 2 and garden 55 m 2, (c) Bedroom 30 m 2 and garden 35 m 2, (d) Bedroom 40 m 2 and garden 45 m 2, , 44. Find the area of living room in the layout., (a) 85 m 2, , (b) 65 m 2, , (c) 45 m 2, , (d) 75 m 2, , 45. Find the area of the plot., 3, 41 m 2, 2, (d) None of these, , (a) 3 41 m 2, , (b), , (c) 41 m 2, , 46-50 are based on Case Study-2., , Case Study 2, Indian Water Department were carrying out periodic inspection of water tanks at every water, pump to check carbonate number. They found some carbonate number at both the water tanks at, Shyam’s water pump and asked him to empty them immediately and to cover them with a lid., The water tanks have capacities of 420 L and 130 L. Shyam had no other option to empty the two, filled water tanks with the help of a bucket., , 420 Liters, 130 Liters, , SAMPLE PAPER 10, , 46. The maximum capacity of the bucket he should use so that no water remains in the, tanks is, (a) 10 L, , (b) 13 L, , (c) 130 L, , (d) 420 L, , 47. When the HCF (420, 130) is expressed as a linear combination of 420 and 130, i.e. HCF (420, 130) = 420x + 130y, then values of x and y satisfying the above relation are, (a) 3 and 1, (b) - 4 and 13, (c) 4 and - 13, (d) 2 and 3, , 48. The HCF of the smallest composite number and the smallest even number is, (a) 1, , (b) 2, , (c) 4, , (d) 8

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OMR SHEET, , SP 10, , Roll No., Sub Code., , Student Name, , Instructions, Use black or blue ball point pens and avoid Gel & Fountain pens for filling the OMR sheet., Darken the bubbles completely. Don’t put a tick mark or a cross mark, half-filled or over-filled bubbles will not be read, by the software., Correct, , ✔, , ✗, , Incorrect, , Incorrect, , Incorrect, , Do not write anything on the OMR Sheet., Multiple markings are invalid., , 1, , 18, , 35, , 2, , 19, , 36, , 3, , 20, , 37, , 4, , 21, , 38, , 5, , 22, , 39, , 6, , 23, , 40, , 7, , 24, , 41, , 8, , 25, , 42, , 9, , 26, , 43, , 10, , 27, , 44, , 11, , 28, , 45, , 12, , 29, , 46, , 13, , 30, , 47, , 14, , 31, , 48, , 15, , 32, , 49, , 16, , 33, , 50, , 17, , 34, , Check Your Performance, Total Questions:, Total Correct Questions:, , If Your Score is, , Score Percentage =, , Total Correct Questions, Total Questions, , × 100, , Less than 60%, > Average (Revise the concepts again), Greater than 60% but less than 75% > Good (Do more practice), Above 75%, > Excellent (Keep it on)

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180, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 49. The greatest number which divides, 285 and 1249 leaving remainders 9 and, 7, respectively is, (a) 138, (b) 276, (c) 1242, (d) 2484, , 50. The HCF of the smallest and largest, two-digit numbers is, (a) 0, (b) 1, (c) 2, (d) 990, , Answers, 1. (c), 11. (a), , 2. (a), 12. (b), , 3. (c), 13. (c), , 4. (b), 14. (b), , 21. (b), 31. (a), , 22. (c), 32. (b), , 23. (a), 33. (d), , 24. (a), 34. (d), , 41. (a), , 42. (d), , 43. (c), , 44. (d), , 5. (c), 15. (b), 25. (d), , 6. (b), 16. (a), 26. (c), , 7. (b), 17. (d), 27. (a), , 8. (b), 18. (a), 28. (a), , 9. (a), 19. (a), 29. (b), , 10. (c), 20. (c), 30. (c), , 35. (c), 45. (b), , 36. (b), 46. (a), , 37. (a), 47. (b), , 38. (d), 48. (b), , 39. (a), 49. (a), , 40. (c), 50. (b), , SOLUTIONS, 1. Given, a and b are the zeroes of the polynomial, 2, , p( x ) = x - 5 x + 2, Coefficient of x, \Sum of zeroes = Coefficient of x 2, (- 5), a +b==5, 1, Constant term, and product of zeroes =, Coefficient of x 2, 2, ab = = 2, 1, a +b, 1 1, Now, + - 3 ab =, - 3 ab, a b, ab, 7, 5, = - 3(2 ) = 2, 2, , 4 ´ LCM = 96 ´ 404, LCM = 24 ´ 404 = 9696, , 4. Let P(3 , 4 ) divides the line segment joining, A(1, 2 ) and B(6 , 7 ) in the ratio k : 1., 6k + 1, \x-coordinate of P is, =3, k +1, Þ, Þ, , [by section formula], 6k + 1 = 3k + 3 Þ 3k = 2, 2, k=, 3, , 5. We know that,, sin q =, , 1, cosec q, , sin q is maximum when q = 90 °, i.e., sin90 ° = 1, , 2. In DABC,, B, , Þ, Þ, , N, , Therefore, maximum value of, 10 m, , SAMPLE PAPER 10, , A, , W, , 24 m, , C, , E, S, , ÐA = 90 °, CB2 = CA2 + AB2, [by Pythagoras theorem], Þ, \, , CB2 = 24 2 + 10 2, = 576 + 100 = 676, SB = 26 m, , 3. Given, HCF (96, 404) is 4, We know that,, HCF ´ LCM = Product of two numbers, , 1, is 1., cosec q, , 6. As, length of wire is same so, both figures have, same perimeter., \Circumference of circle = Perimeter of the, square., Let r be the radius of the circle and a is the side, of the square., So,, 2 pr = 4 a, 22, Þ, ´ 21, [Q r = 21 cm], 4a = 2 ´, 7, Þ, 4 a = 132, Þ, a = 33 cm, 7. Here, total number of eggs = 400, Probability of getting a bad eggs = 0.035

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181, , CBSE Sample Paper Mathematics Standard Class X (Term I), , Number of bad eggs, = 0.035, Total number of eggs, , Þ, , Number of bad eggs, = 0.035, 400, Number of bad eggs = 400 ´ 0.035, 35, = 400 ´, = 14, 1000, , Þ, , 8. The zeroes of f ( x ) = x 2 - 2 2 x - 16 are, i.e., , f (x) = 0, x 2 - 2 2 x - 16 = 0, , Þ, , x 2 - 4 2 x + 2 2 x - 16 = 0, , Þ, , x( x - 4 2 ) + 2 2 ( x - 4 2 ) = 0, , Þ, , (x + 2 2 ) (x - 4 2 ) = 0, , Þ, , x = 4 2 or x = - 2 2, , 9. Given, AD is the angle bisector of ÐA., By angle bisector theorem, bisector of an angle, divides the opposite side in the same ratio as of, two remaining sides., BD AB, \, =, DC AC, BD 6, =, Þ, DC 8, BD 3, Þ, =, DC 4, 10. We have, x = 3 sec2 q - 1 and y = tan 2 q - 2, Consider, x - 3 y = (3 sec2 q - 1) - 3(tan 2 q - 2 ), = 3 sec2 q - 1 - 3 tan 2 q + 6, = 3(sec2 q - tan 2 q ) + 5, [Q sec2 A - tan 2 A = 1], =3 ´ 1 + 5, =3 + 5 =8, 11. Let the points be P( x , y ), A(2 , 1) and B(1, - 2 )., Since, P is equidistant from A and B., \ AP = BP Þ AP2 = BP2, Þ ( x - 2 )2 + ( y - 1)2 = ( x - 1)2 + ( y + 2 )2, [by distance formula], Þ x2 - 4x + 4 + y2 - 2 y + 1, , 13. Given, Number of black balls = 5, Number of red balls = 7, Number of white balls = 3, Total balls = 5 + 7 + 3 = 15, , P(red ball) =, , 7, 15, , 14. Given, Sum of the zeroes = 0, Let a and b be the zeroes of the quadratic, polynomial. Then,, a + b =0, Þ, 6 + b =0, Þ, b = -6, \Required polynomial, = x 2 - ( a + b)x + a × b, = x 2 - 0 × x + 6( - 6 ), = x 2 - 36, 15. In DKNP and DKML,, [given], ÐKNP = ÐKML = 35 °, [common], ÐK = ÐK, \, DKNP ~ DKML, [by AA similarity criterion], PN KN, Þ, =, LM KM, [Q corresponding sides of similar, triangles are proportional], x, c, x, c, Þ, =, Þ =, a KN + NM, a c+ b, ac, x=, Þ, b+c, 16. DABC is right-angled triangle, \In DABC,, AC 2 = AB2 + BC 2, [using Pythagoras theorem], Þ, AC 2 = (8 )2 + (6 )2 = 100, Þ, AC = 10 cm, Now, Area of shaded portion, = Area of half circle - Area of DABC, 1, 1, = pr2 - ´ BC ´ AB, 2, 2, 1, p é2 1, 6 ´=8 AC = 5 cm ù, = ( 5 )Q-r = ´OA, úû, ê, 2, 2 ë 2, 2, = 39.25 - 24 = 15.25 cm, 1, éQ sec q = 1 ù, 17. We know that, sec A =, cos A êë, cos q úû, Divide numerator and denominator by sin A, 1, 2, sin, A = cosec A = 1 + cot A, Þ, cos A, cot A, cot A, sin A, [Q cosec2 q - cot 2 q = 1], 18. Given, the probability of guessing the correct, p, answer to certain question is, and the, 12, probability of not guessing the correct answer, , SAMPLE PAPER 10, , = x2 - 2 x + 1 + y2 + 4y + 4, Þ, 2 x + 6y = 0, [divide by 3], Þ, x + 3y = 0, 7, 12. Radius of the circle, r =, cm, p, 2, 7 ö, 2, \ Area of the circle = pr2 = pæç, ÷ = 49 cm, è pø, , \

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182, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 3, to same question is ., 4, We know that,, P( E) + P( E ) = 1, p 3, Þ, + =1, 12 4, p, 3 1, =1 - =, Þ, 12, 4 4, 12, p=, =3, Þ, 4, , 22. Let r = radius of inner circle and R = radius of, outer circle., R, , 7m, r, , Area of the path = Area of outer circle - area of, inner circle., , 19. For infinite many solutions,, a1 b1 c1, =, =, a2 b2 c2, k +1, -5, 1, =, =, Þ, k +1, 9, - (8 k - 1), 1, 5, So,, =, k + 1 8k - 1, Þ, , 8k - 1 = 5k + 5, 3k = 6 Þ k =, , Þ, 20. We have,, , \Area of path = pR 2 - pr2 = p( R 2 - r2 ), 22, = [( 42 )2 - (35 )2 ], 7, [Q R = 42 and r = 35], 22, = ( 42 - 35 ) ( 42 + 35 ), 7, [Q a2 - b2 = ( a - b) ( a + b)], 22, =, ´ 7 ´ 77, 7, = 1694 m 2, , 6, =2, 3, , 1 + cot 2 A, 1 + tan 2 A, 1 + cot 2 A, éQ tan q = 1 ù, =, êë, 1, cot q úû, 1+, 2, cot A, cot 2 A (1 + cot 2 A), =, cot 2 A + 1, = cot 2 A, = cosec2 A - 1, [Q1 + cot 2 q = cosec2 q ], , 21. We have, a and b are zeroes of the polynomial, f ( x ) = ax 2 + bx + c, b, \ Sum of zeroes ( a + b) = a, c, and product of zeroes ( ab) =, a, Now,, , …(i), …(ii), , 1, 1 b2 + a 2 ( a + b)2 - 2 ab, + 2 =, =, 2, a, b, a 2b2, ( ab)2, , SAMPLE PAPER 10, , [Q( a + b)2 = a2 + b2 + 2 ab], æç, =è, , 35 m, , 2, , bö, c, ÷ -2, aø, a, 2, æ cö, ç ÷, è aø, [using Eqs. (i) and (ii)], , b2 2 c b2 - 2 ac, 2, b2 - 2 ac, a2, =a 2 a =, =, c, c2, c2, 2, 2, a, a, , 23. Let H and T represent Head and Tails, respectively., The possible outcomes when a coin is tossed, 3 times, {( HHH , HHT , HTH , THH , HTT ,, THT , TTH , TTT)}, Total number of possible outcomes = 8, Possible outcomes of winning = 2, So, possible outcomes of losing = 8 - 2 = 6, 6 3, \Required probability = =, 8 4, 23, 23, 24., =, 200 2 3 ´ 5 2, 17 17, =, 9 32, 8, 8, =, 75 3 ´ 5 2, 3, 3, =, 35 5 ´ 7, As we know that, if the denominator is in the, form of 2 m ´ 5 n, where m and n are integers,, then the number has terminating decimal, expansion., Hence, in the given option, only option (a) has, denominator in the form of 2 m ´ 5 n., 25. Let A(0 , 0 ), B(2 , 0 ) and C(0 , 2 ) be the vertices, of DABC., AB = (2 - 0 )2 + (0 - 0 )2, \

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183, , CBSE Sample Paper Mathematics Standard Class X (Term I), , = 4 + 0 = 2 units, [by distance formula], , \a + ( - a ) =, , é, Coefficient of x ù, ú, êQ Sum of zeroes = Coefficient of x 2 û, ë, , BC = (0 - 2 )2 + (2 - 0 )2, = 4+4= 8, and, , k =0, , Þ, , = 2 2 units, 2, , - (- 2 k ), Þ 0 =2k, 1, , 29. Given an equilateral triangle DABC in which,, , 2, , CA = (0 - 0 ) + (0 - 2 ), , A, , = 0 + 4 = 2 units, \Required perimeter = AB + BC + CA, = (2 + 2 2 + 2 ) units, , 4p, , 4p, , = ( 4 + 2 2 ) units, 26. Given, right triangle DABC right-angled at B., Join PQ, PC and AQ, A, , P, C, , Q, , B, , 2, , D, , C, , AB = BC = CA = 4 p and AD ^ BC., Since, in equilateral triangle, perpendicular, bisects the side., 1, 1, \ BD = DC or BD = BC = ( 4 p) = 2 p, 2, 2, In DADB, AB2 = AD2 + BD2, [by Pythagoras theorem], ( 4 p)2 = AD2 + (2 p)2, , In DABQ, by pythagoras theorem,, 2, , B, , 16 p2 = AD2 + 4 p2, , Þ, , 2, , AQ = QB + AB, Similarly, In DPCB,, , …(i), , PC 2 = BC 2 + PB2, and In DACB,, , … (ii), , AC 2 = AB2 + BC 2, and In DPQB,, , … (iii), , PQ2 = PB2 + QB2, Adding Eqs. (i) and (ii), we get, , … (iv), , Þ 16 p2 - 4 p2 = AD2 Þ 12 p2 = AD2, AD = 2 3 p, , \, 30. We have,, =, , AQ2 + PC 2 = QB2 + AB2 + BC 2 + PB2, AQ2 + PC 2 = AC 2 + PQ2, [from Eqs. (iii) and (iv)], , sin q(1 - 2 sin 2 q), cos q(2 cos 2 q - 1), , = tan q ×, , [1 - 2 (1 - cos 2 q )], (2 cos 2 q - 1), , [Q sin 2 A = cos 2 A - 1], (2 cos 2 q - 1), = tan q ×, = tan q, (2 cos 2 q - 1), 31. Given, diameter = 42 cm, 42, r=, = 21 cm, Þ, 2, and, q = 60 °, , O, 60°, , = 10.5 cm 2, 28. As, the polynomial is x 2 - 2 kx + 2 and its, zeroes are equal but opposite in sign., Let the zeroes of the polynomials are a and - a., , \Length of arc =, , q, ´ 2 pr, 360 °, 22, 60 °, ´2´, ´ 21 = 22 cm, =, 7, 360 °, , 32. Total number of possible outcomes, , SAMPLE PAPER 10, , 27. Given, OABC is a square with side 7 cm and, OAC is a quadrant of a circle with centre O., Now, Area of shaded portion = Area of square, - Area of quadrant, 1 2, 2, = ( Side) - pr, 4, 1, 22, = (7 )2 - ´, ´7´7, 4 7, 77, = ( 49 - 38.5 ), = 49 2, , sin q - 2 sin 3 q, 2 cos 3 q - cos q

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184, , CBSE Sample Paper Mathematics Standard Class X (Term I), , = 52 - 7 = 45, Remaining number of face cards, = 12 - 7 = 5, So, favourable number of outcomes = 5, , 36. The zeroes of the polynomial f ( x ) = x 2 + x are given by f ( x ) = 0, , 5 1, =, 45 9, , 4x2 + 4x - 3 = 0, , Þ, , 4x2 + 6x - 2 x - 3 = 0, [splitting middle term], , Þ, , (2 x - 1) (2 x + 3 ) = 0, , Þ, , 2, , … (i), , 2, , æyö, y = q tan q Þ çç ÷÷ = tan 2 q, è qø, x2 y2, sec2 q - tan 2 q = 2 - 2, p, q, , … (ii), , [subtract Eq. (ii) from Eq.(i)], 2, , 2, , Þ 2 x - 1 = 0 and 2 x + 3 = 0, 1, 3, Þ, x = and x = 2, 2, 37. Let the coordinates of P be ( x , y )., Then, x = 2 y, (given) … (i), Since, P is equidistant from Q(2 , - 5 ), and R( - 3 , 6 ), \, PQ = PR, , Since, sec q - tan q = 1, \, , x2 y2, =1, p2 q2, , Þ, , x 2 q2 - y 2 p2 = p2 q2, , (2 - x )2 + ( - 5 - y )2 = ( - 3 - x )2 + (6 - y )2, [by distance formula], Squaring both sides, we get, , 34. Given points are (7 , - 6 ) and (3, 4)., Let, x1 = 7, x2 = 3, y1 = - 6, y2 = 4, m = 1 and n = 2, By section formula, the coordinate of the point, which divides the line segment joining the, points (7 , - 6 ) and (3, 4) in the ratio 1 : 2, internally are, æ 1 ´ 3 + 2 ´ 7 1 ´ 4 + 2 ´ (- 6) ö, ç, ÷, ,, ç, ÷, 1+2, 1+2, è, ø, æ 3 + 14 4 - 12 ö, =ç, ,, ÷, 3 ø, è 3, 17 8, = æç , - ö÷, 3ø, è 3, Since, x-coordinate is positive and y-coordinate, is negative., So, the point lies in the IV quadrant., , SAMPLE PAPER 10, , 35. From distance formula,, D = ( x2 - x1 )2 + ( y2 - y1 )2, Here,, So,, , AB = 5 units, , [given], 2, , Þ, Þ, , 2, , AB = ( k - 2 ) + (3 - 3 ), = ( k - 2 )2 + 0, , Þ, , 3, =0, 4, , Þ 2 x(2 x + 3 ) - 1(2 x + 3 ) = 0, , æxö, 33. Given, x = p sec q Þ çç ÷÷ = sec2 q, è pø, , \, , x2 + x -, , Þ, , \ Required probability, Number of favourable outcomes, =, Total number of possible outcomes, =, , 3, 4, , ( 5 )2 = ( k - 2 )2, = (k - 2 ) = ± 5, k =2 ± 5, k = 7, - 3, , (2 - 2 y )2 + ( - 5 - y )2 = ( - 3 - 2 y )2 + (6 - y )2, [using (i)], 4 + 4 y 2 - 8 y + 25 + y 2 + 10 y, = 9 + 4 y 2 + 12 y + 36 + y 2 - 12 y, Þ 5 y 2 + 2 y + 29 = 5 y 2 + 45, Þ, 2 y = 16 Þ y = 8, Hence, the coordinates of P are (16, 8)., 38. Given, DABC ~ DDEF, and DE = 6 cm, EF = 4 cm, DF = 5 cm, and BC = 8 cm, D, A, 6 cm, , B, , 8 cm, , C, , E, , 5 cm, , 4 cm, , F, , AB BC AC, =, =, DE EF DF, [Q from Basic Proportionality Theorem], AB 8 AC, Þ, = =, 6, 4, 5, \, AB = 2 ´ 6 = 12 cm, and, AC = 5 ´ 2 = 10 cm, Now, Perimeter of DABC = AB + BC + CA, = 12 + 8 + 10 = 30 cm

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185, , CBSE Sample Paper Mathematics Standard Class X (Term I), , 39. If we have three vertices of triangle ( x1 , y1 ),, ( x2 , y2 ) and ( x3 , y3 ), then, Coordinate of centroid, æ x + x2 + x3 y1 + y2 + y3 ö, ,, =ç 1, ÷, 3, 3, è, ø, Here, x1 = 3, x2 = - 8 and x3 = 5, and y1 = - 7, y2 = 6 and y3 = 10, Hence, coordinate of centroid, æ 3 + ( - 8 ) + 5 - 7 + 6 + 10 ö, ,, =ç, ÷, 3, 3, ø, è, 9ö, æ, = ç 0, ÷, è 3ø, = (0 , 3 ), 40. Given, tan q + sin q = m and tan q - sin q = n, Now, m2 - n2 = (tan q + sin q )2 - (tan q - sin q )2, = 4 tan q × sin q, [Q ( a + b)2 - ( a - b)2 = 4 ab], , = 95, [given] …(i), \, 10 p + 5 q = 95, and, …(ii), p + q = 13, Multiply Eq. (ii) by 5 and then subtract Eq. (ii), from Eq. (i), 5 p = 30, Þ, p =6, \, q = 13 - 6 = 7, \Area of each bedroom = 5 ´ 6 = 30 m 2, Area of garden = 5 ´ 7 = 35 m 2, 44. Area of living room + Area of each bedroom, = 15 ´ 7 = 105 m 2, Since, area of each bedroom = 30 m 2, \Area of living room = 105 - 30, = 75 m 2, 45. Draw a perpendicular line from centre O to, the chord AB. Then,, , = 4 tan 2 q × sin 2 q, , D, , = 4 sin 2 q (sec2 q - 1), é, sin 2 A ù, 2, êQ tan A = cos 2 A ú, ú, ê, êand sec2 A = 1 ú, êë, cos 2 A úû, 1, = 4 sin 2 qæç, - 1 ö÷, ø, è cos 2 q, =4, , sin 2 q, - sin 2 q, cos 2 q, , = 4 tan 2 q - sin 2 q, = 4(tan q - sin q )(tan q + sin q ), = 4 mn, Solutions (41-45), 41. From the given figure, p + 2 + q = 15, Þ, p + q = 15 - 2 = 13, , 43. Area of two bedrooms = 2 ( 5 ´ p) = 10 p m 2, Area of garden = 5 q m 2, Given, Area of two bedrooms, + Area of two garden, , A, , AE = EB =, , B, , E, , 15, m, 2, , Also, BC = 12 m, 1, 12, OE = BC =, =6 m, \, 2, 2, In right DOEB,, OB = OE2 + EB2, [by pythagoras theorem], 15, = (6 )2 + æç ö÷, è2 ø, 225, = 36 +, 4, 144 + 225, =, 4, 1, 369, =, 2, 3, 41 m 2, =, 2, , 2, , Solutions (46-50), 46. Here, we have to find the HCF of 420 and 130., Prime factor of 420 = 7 ´ 3 ´ 2 2 ´ 5, Prime factor of 130 = 13 ´ 2 ´ 5, \HCF (420, 130) = 2 ´ 5 = 10, Hence, the maximum capacity of bucket is 10 L., 47. Given, HCF (420, 130) = 420 x + 130 y, , SAMPLE PAPER 10, , 42. From the figure, Length = 15 m, and Breadth = 2 + 5 + 5 = 12 m, Total length of outer boundary, = Perimeter of rectangle, = 2 (Lenght + Breadth), = 2 (15 + 12 ) = 2 ´ 27, = 54 m, , C, O

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186, , CBSE Sample Paper Mathematics Standard Class X (Term I), , i.e., 10 = 420 x + 130 y, (a) When x = 3 and y = 1;, 420 ´ 3 + 130 ´ 1 = 1260 + 130, = 1390 ¹ 10, so it does not satisfy, (b) When x = - 4 and y = 13;, 420 ´ ( - 4 ) + 130 ´ (13 ) = - 1680 + 1690, = 10, Hence, x = - 4 and y = 13, 48. Since, smallest composite number = 4, and smallest even number = 2, HCF (4, 2) = 2, \, 49. Subtract the remainders from the given, numbers and, then we find the HCF., \, 285 - 9 = 276, 1249 - 7 = 1242, Prime factor of 276 = 2 2 ´ 3 ´ 23, Prime factor of 1242 = 2 ´ 3 3 ´ 23, \HCF (276, 1242) = 2 ´ 3 ´ 23, = 138, 50. Since, smallest two-digit number = 10, , SAMPLE PAPER 10, , and largest two-digit number = 99, Prime factor of 10 = 2 ´ 5, Prime factor of 99 = 3 ´ 3 ´ 11, \HCF (10, 99) = 1