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Why?, , To unlock the imagination, and come up with ideas, To remember facts and, gures easily, To make clearer and, better notes, To concentrate and save, time, To plan with ease and ace, exams, , which help make schoolwork fun and cut homework time in half !!, , Mind Maps for each chapter show the breakthrough system of planning and note-taking, , with a blank sheet of paper, coloured pens and, your creative imagination!, , How?, , AN INTERACTIVE MAGICAL TOOL, , When?, , MIND MAP, , Result, , Learning made simple, ‘a winning combination’, , What?, , presenting words and, concepts as pictures!!, , anytime, as frequently as you like, till it becomes a habit!, , LEARNING MADE SIMPLE, , OSWAAL BOOKS
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2, 3 are irrational, , that the prime factorisation of q is not of, the form of 2n5m where n, m are non-negative, integers. Then, x has a decimal expansion, which is non-terminating repeating, , p, 5. Let x = —, q be a rational number, such, , p, 4. Let x = —, q be a rational number such, that the prime factorisation of q is of the, form 2n 5 m where n, m are non-negative, integers. Then, x has a decimal expansion, which terminates., , 3. Let x be a rational number whose, decimal expansion terminates. Then x can be, p, expressed in the form, —, q where p & q are, coprime, the prime factorisation of q is of, the form 2n 5 m where n, m are nonnegative integers, , 2., , 1. Let p be a prime number., If p divides a2, then p divides a,, where a is a positive integer, , For any two positive integers,, a and b, HCF (a, b) × LCM (a, b) =a × b, For Example, f(x) = 3x2y, g(x) = 6xy2, HCF = 3xy, LCM = 6x2y2, , Euclid's, , Every composite number, can be expressed as a product of, primes, and this factorisation is unique,, apart from the order in which the, prime factors occur, , Composite Number, x = P1×P2× P3...×Pn,, where P1P2 ... Pn are, prime numbers, , Step 3: Continue the process, till the remainder is zero, , Step 2: If r = zero, d is the, HCF of c and d If r 0, apply Euclid' s, Division to d and r, , Step 1: Apply Euclid’s Division, Lemma, to c & d. c = dq + r, , Steps to obtain the HCF of two, positive integers, say c and d,, with c > d, , Division Algorithm, , Division Lemma, Given positive integers a, and b, there exist unique integers, q and r satisfying, a = bq +r ; 0 < r < b, , Fundamental Theorem of Arithmetic, , Real Numbers, , Prime Factorization Method, , Theorems, , CHAPTER : 1 Real Numbers, , 2 ], Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X
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Relationship-Zeroes and, , Division Algorithm, , Number of Zeroes 2, , Number of Zeroes 1, Number of Zeroes 0, , Case3- Graph, does not cut x-axis, , x, , Types, , x, , Highest power of, x in Polynomial, p (x), , y = x2–3x–4, , y, , y, , Parabola, , ax2+bx+c, a≠0, ax3+bx2+cx+d, a≠0, , 1, 2, 3, , Linear, Quadratic, Cubic, , ax+b, , Polynomial Degree General Form, , Degree of Polynomial, , Quadratic Polynomial, , Graphical Representation, , Polynomials, , Case2- Graph cuts, x-axis at exactly one point, , Graphically, , Zeroes of Polynomial, , Coefficient of Polynomials, , Case1- Graph cuts, x-axis at 2 points, , ax3 + bx2 + cx + d, Sum of zeroes,, b, α+ β+ γ = –, a, Sum of products of the, zeroes taken two at a time, c, αβ + βγ + γα = a, Product of zeroes, d, αβγ = –, a, , of Cubic Polynomial, , Cubic, α, β and γ are zeroes, , ax2 + bx + c, Then, Sum of zeroes,, α+ β = – ba, Product of zeroes, c, αβ = a, , zeroes of Quadratic, Polynomial, , Quadratic, α and β are, , If p(x) and g (x) are, two polynomials with, g(x) ≠ 0, then –, p(x)= g(x) × q(x) + r (x), where, r (x) = 0 or, degree of r (x) < degree, of g (x), , CHAPTER : 2 polynomials, , Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X, , [ 3
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By Substitution, , x, , y, , 5, , 3, , (–46)(3)–(–74)(2), 1, =, 2(5)–3(3), , =, , 3, , 2, , =, , i.e. x = 8 and y = 10, , –222+230, , x, , y, 1, = 10–9, –138+148, y, y, x, x, 1, 1, 1, = = ⇔ = and, =, 8 10, 8, 1, 1, 10, 1, , 3(–74)–5(–46), , –74, , 5, , Then,, , –46, , 3, , Solution: By cross-multiplication method, x, y, 1, , Solve: 2x+3y–46 = 0 –(i), 3x+5y–74 = 0 –(ii), , By Cross-Multiplication, , – (i), Solve: x+3y = 6, 2x+3y = 12 – (ii), Now, Adding equation (i) and (ii), 3x = 18 or x = 6, Again, from (i) ×2– (ii), 3y = 0 or, y = 0, Hence, x = 6, y = 0, , By Elimination, , Solution: From equation (ii), x = 3–2y, substitute value of x in eq. (i), 7(3–2y)–15y = 2, 19, –29y = –19 ⇔ y =, 29, Now, from x = 3 – 2y, 19, 49, =, x=3–2, 29, 29, , Solve: 7x–15y = 2 –(i), x+2y = 3 –(ii), , Algebraic Interpretation : No solution, – Inconsistent, , a1 b1 c1, Compare the Ratios = a = ≠, 2, b2 c2, , a1, 1 b1, 2 c1, –4, a 2 = 2 , b 2 = 4 , c 2 = –12, , Pair of Lines = x+2y–4 = 0, 2x+4y–12 = 0, , Parallel Lines, , Graphical Representation, , Graphical Representation, , in Two Variables, , Pair of Linear Equations, , Algebraic Methods, , Algebraic Interpretation = Infinitely, many solutions – Dependent, , a1, 2 b1, 3 c1, –3, a 2 = 4 , b 2 = 6 , c 2 = –18, a, b1 c1, Compare the Ratios = a = = c, 2, 2, b2, , Pair of Lines = 2x+3y–9 = 0, 4x+6y–18 = 0, , Intersecting Lines, , Coincident Lines, , Graphical Representation, , Algebraic Interpretation : Exactly, one solution – consistent (unique), , Graphical Representation, , Each solution (x, y), corresponds, to a point on the line representing, the equation and vice-versa, , a1, 1 b 1 –2 c 1, 0, a 2 = 3 , b 2 = 4 , c 2 = –20, a1 b1, Compare the Ratios = a ≠, 2 b2, , Pair of Lines = x–2y = 0, 3x+4y–20 = 0, , Solution Graphically, , General Form, , a1x+b1y+c1 = 0, a2x+b2y+c2 = 0, a1,b1,c1,a2,b2,c2, – Real numbers, , CHAPTER : 3 pair of linear equations in two variables, , 4 ], Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X
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By Completing the Square, , 2, , –, , 1, =0, 16, , (3x–2) = 0 or (2x+1) = 0, –1, 2, or x =, x=, 3, 2, 2 –1, Roots are ,, 3 2, , The roots of 6x2–x–2=0, , Solution: 6x2+3x–4x–2 = 0, 3x(2x+1)–2(2x+1) = 0, (3x–2)(2x+1) = 0, , Find roots of 6x2–x–2=0, , By Factorization, , Quadratic Equation, , Solution of a, , 5, 1, 5, 1, =, x–, =±, ⇔ x–, 4, 4, 4, 16, 1, 1, 5, 5, x= +, or x = –, 4, 4, 4, 4, 3, x=, or x = 1, 2, , 2, , Solution:, , 2x2–5x+3 = 0, 3, 5, x2– x+ = 0, 2, 2, 2, 2, 5, x – 5 – 5 + 3 = 0 ⇔ x–, 4, 4, 2, 4, , Solve: 2x2–5x+3 = 0, , For quadratic equation, ax2+bx+c = 0,, b2–4ac is Discriminant (D), , Nature of Roots, , General Form, , 3. D < 0, No real roots (imaginary), , 2. D = 0, Two equal real roots, , 2a, , –b ± b2–4ac, , Roots of ax2+bx+c = 0 are given by, , ax2+bx+c = 0, a, b, c – real numbers, a≠ 0, , Equation of degree 2,, in one variable, , Quadratic Formula, , 1. D > 0, Two distinct real root, , Equations, , Quadratic, , Meaning, , CHAPTER : 4 quadratic equations, , Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X, , [ 5
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n(a+l), n(1+n), =, 2, 2, , If a, b, c, are in AP,, a+c, b=, 2, b is arithmetic mean, , i.e., n–1 =, , 87, = 29, 3, n = 30, , 2-digit numbers divisible by, are 12, 15, 18, ... 99, a = 12, d = 3, an = 99, an = a + (n–1)d, 99 = 12 + (n–1)3, , How many 2-digit numbers, are divisible by 3?, , sn =, , Sum of first n positive integers, Let sn = 1 + 2 + 3 + ... n, a = 1, last term l = n, , When first term and common, differnce are given :, n, Sn = (2a+(n–1) d ), 2, a – first term, d – common difference, n – total terms, , Arithmetic mean, , Examples, , When first & last terms are given :, n, n, (a+l), Sn = (a+an) or Sn =, 2, 2, a – first term, n – total terms, an – nth term, l – last term, , Sum (s), , Progressions, , Arithmetic, , General form, , a, a+d, a+2d, a+3d, ...a+(n –1) d, , From beginning, an = a+(n–1)d, Here, a – first term, d – common difference, an – nth term, , Nth term, , From the end, an = l – (n–1) d, Here, l – last term, d – common difference, an – nth term, , •Fixed number in arithmetic, progression which provides the, to and fro terms by adding/subtracting, from the present number., •Can be positive or negative., , Common Difference, , Definition, , List of numbers in which each term is, obtained by adding a fixed number to the preceding, term except the first term. Fixed number is, called common difference., , CHAPTER : 5 arithmetic progressions, , 6 ], Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X
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2, , Example, , 2, , 2, , x1 + x2 , y1 + y2, , 1, , 1+2, , 1(–7) + 2(2), , ,, , Q=, , i.e., (–4, 2), , 2+1, , 2(–7) + 1(2), , ,, , Coordinate of Q, , i.e., (–1, 0), , =, , Coordinate of P, , 2+1, , 2(4) + 1(–2), , 1+2, , 1(4) + 2(–2), , Find point of Trisection of, line segment AB, A(2, –2) and B(–7, 4), , R, , 3, , Area of Triangle, , ,, , Internally (+), Externally (–), , m1y2 ± m2y1, m1 ± m2, , m1x2 ± m2x1, m1 ± m2, , PQ = (x – x )2 + (y2 – y1)2, 2, 1, , √, , Meaning, , Y’, , Centroid, , (–,–), , II, Quadrant, , (+,+), , X’, , G=, , G, , C, (x3, y3), , Y, , abscissa, , X, , ordinate, , x1+x2+x3 y1+y2+y3, ,, 3, 3, , B, (x2, y2), , A (x1, y1), , Example, , Y’, Horizontal = x–axis (Abscissa), Vertical = y–axis (Ordinate), , (+,–), , = 34, , Since, AB = BC = CD = DA and AC = BD., All four sides and diagonals are equal Hence,, ABCD is a square, , = 68, , = 68, , = 34, , BC = (4 + 1)2 + (2 + 1)2 = 34, CD = (–1 + 4)2 + (–1 –4)2 = 34, , AB = (1 – 4)2 + (7 – 2)2, , √, √, √, DA =√(1 + 4)2 + (7 – 4)2, AC =√(1 + 1)2 + (7 + 1)2, BD =√(4 + 4)2 + (2 – 4)2, , Are the following points vertices of a, square : (1, 7), (4, 2), (–1, –1), (–4, 4)?, A (1, 7); B = (4, 2); C = (–1, –1); D = (–4, 4), , Coordinate axis, , Distance formula, , Lines (In TwoDimensions), , Section formula, , Mid-point Line Segment, , 1, Area = |[x1(y2 – y3) + x2(y3 – y1), 2 + x (y – y )]|, , (–,+), , Y, , I, Study of, Quadrant, algebraic X’, X, equations, III, IV, on graphs Quadrant Quadrant, , CHAPTER : 6 lines (in two dimensions), , Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X, , [ 7
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Summary, , ∠A = ∠D, ∠B = ∠E,, ∠C = ∠F, , ∆ABC ∆DEF, , ∠C = ∠F, , AB, BC, CA, If, =, =, DE, EF, FD, then, ∠A = ∠D, ∠B = ∠E,, , Triangles, , M, , C, , Q, , N, , P, , R, , Here ∆ABC ~ ∆PQR, AB 2 BC 2 CA 2, ar(ABC), =, =, =, PQ, ar(PQR), QR, RP, , B, , A, , The ratio of the areas of two similar, triangles is equal to the square of the ratio, of their corresponding sides, , Similarity, , theorem Pythagoras, , Right angled triangle, , Area of Similar Triangles, , Theorems, , then, AB= BC = AC, DF, DE EF, ∆ABC ∆DEF, , If, , 5. If one angle of a, triangle is equal to one, angle of the other triangle, and the sides including these, angles are proportional,, AB AC, & ∠A =∠D, =, then the two triangles are If, DE DF, similar.(SAS criterion), then, ∆ABC ~ ∆DEF, , 4. If in two triangles, sides, of one triangle are proportional, to (i.e., in the same ratio of ) the, sides of the other triangle, then, their corresponding angles are, equal and hence the two triangles, are similiar.(SSS criterion), , 3. If in two triangles,, corresponding angles are equal,, then their corresponding sides are in, the same ratio (or proportion) and, hence the two triangles are, similar.(AAA criterion), , AD AE, 2. If a line divides any, two sides of a triangle in the If DB= EC, same ratio, then the line is, parallel to the third side. then, DE || BC, , 1. If a line is drawn parallel to If, DE || BC, one side of a triangle to intersect the, other two sides in distinct points,, the other two sides are divided, AD AE, in the same ratio., =, then, DB EC, , In ∆ABC, let DE BC. Then,, (i) AD, AE, =, DB, EC, AC, (ii) AB, =, DB, EC, (iii) AD, AE, =, AB, EC, , CHAPTER : 7 triangles, , A, , C, , B, , BC 2 = AB 2+AC 2, , In right ∆ABC,, , 2. In a right triangle, the square, of the hypotenuse is equal to the, sum of the squares of the other, two sides., , B, , A, , C, , If AC 2 = AB 2+BC 2, then, B = 90°, , 3. In a triangle, if square of one, side is equal to the sum of the squares, of other two sides, then the angle opposite, the first side is a right angle., , D, , In right ∆ABC, BD⊥AC,, then, ∆ADB ~ ∆ABC, ∆BDC ~ ∆ABC, C, ∆ADB ~ ∆BDC, , B, , A, , C, , Q, , ∆ABC ~ ∆PQR, , P, , R, , (i) Corresponding angles are equal, (ii) Corresponding sides are in the same ratio, , A, , B, , 1. If a perpendicular is drawn from the vertex, of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are, similar to the whole triangle and to each other., , 8 ], Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X
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The lengths of, tangents drawn, from an external P, point to a circle, are equal, , The tangent at any, point of a circle is, perpendicular to, the radius through, the point of contact, , 3. There are exactly two, tangents to a circle through, a point lying outside, the circle., , 2. There is one and only, one tangent to a circle, passing through a point, lying on the circle., , 1. There is no tangent, to a circle passing, through a point lying, inside the circle., , Y, , P, , O, , P, , P, , R, PQ = PR, , Q, O, , OPQ = 90°, , T2, , T1, , P, , Q, , X, , Circles, , Only one common, point between circle A, and PQ line., , Tangent and tangent point, , Theorems, , Facts, , Q, , P, O, , r, , CHAPTER : 8 circles, , Two common, points between, line PQ and, circle., B, , Q, , Q, , P, , r, O, , A P, , No common, point between, line PQ and, circle., , Non-intersecting line, , Secant, , Definition, , O, , r, , P, , The locus of a point, p, equidistant from a fixed, r, point. Fixed Point is a radius, centre & separation of, O, centre, points in the radius, of circle., , Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X, , [ 9
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3. Draw line parallel to B4 C, from B3 intersecting BC at C'., Draw line || to AC from C', intersecting AB at A', A'BC' is the required triangle, , 2. Locate 4 points (greater of 3, and 4 in 3 ) on BX at equal, 4, distance from each other, (BB1 = B1 B2 = B2 B3 = B3 B4), Join B4 C, , 1. Draw any, ray BX making, an acute angle, with BC, , Construct a triangle similar to a given, ∆ABC with, sides 3 of the corresponding sides of ∆ABC, 4, , Step 3. Join PQ, and PR, required, tangents to, the circle, , Step 2. M as centre and, radius = MO draw a, circle, intersecting, given circle at, Q and R, , Step 1. Join PO, and bisect it at, mid-point M, , Given: Circle with centre O and point P outside it., , triangle, , 3. Locate A1, A2, A3, (m = 3) on AX and, B1, B2 (n = 2) on BY, Join A3 B2, intersecting, AB at C, AC : CB = 3 : 2, , 2. Draw ray BY || AX, , Method 1, , Step 3. Through, A3 (m = 3), draw, line parallel to BA5, cutting AB at C, , Step 2. Locate 5 points, A1, A2, A3, A4, A5 at equal, distances, (A1 = A2 = A3 = A4 = A5)., Join BA5, , AC : CB = 3 : 2, , Given: Line segment, ratio (3 : 2), , Step 1. Draw any ray AX,, Making acute angle, with line segment AB, , Given: Line segment, ratio (3 : 2), , Method 2, , of line in ratio’s, , Line Segment Division, , Definition, , To draw, geometrical shapes using, compasses, ruler etc, , 1. Draw any ray AX making, an acute angle with line, segment AB, , Constructions, , Triangle similar to given, , Tangent to circle, , CHAPTER : 9 constructions, , 10 ], Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X
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cos (90° – A) = sin A, , Note: How to learn the relation, P, “Some people have” sin =, H, tan (90° – A) = cot A, B, cot (90° – A) = tan A “Curly Brown Hair” cos =, H, sec (90° – A) = cosec A, cosec (90° – A) = sec A“through proper Brushing”tan = P, B, , sin (90° – A) = cos A, , Values, , Introduction, to Trigonometry, and Trigonometric, Identities, , Complementary Angles, , Example, , √1 – sin2A, , Trigonometric Identities, , cos2 A + sin2 A =1, 1 + tan2 A = sec2 A;0 ≤ A ≤ 90°, cot2 A + 1 = cosec2 A;0 ≤ A ≤ 90°, , cos2A = 1 – sin2 A i.e. cos A =, sin A, sin A, tan A =, =, cos A √1 – sin2A, , Solution : Since, cos2 A + sin2 A = 1, , Express tan A, cos A in terms of sin A, , Not (∞), defined, 1, , cosec A, sec A, , Not (∞), cot A, defined, , 0, , tan A, , 1, , 0, , sin A, cos A, , 0°, , ∠A, , 1, 2, , 3, , 2, 1, √3, , 3, , 3, , 2, , 2, , 3, , 2, 1, 2, , 1, 2, , 1, 2, , 1, , 2, , 2, , 3, , 1, , 2, , 3, , 2, , 3, , 60°, , 1, , Side, opposite, to ∠A, , C, , 0, , Not (∞), defined, , 1, , Not (∞), defined, , 0, , 1, , 90°, , B, Side opposite to ∠C, , p, Hy, , se, nu, ote, , 45°, , A, , =, , BC, AC, AB, Cosine of ∠A =, AC, BC, Tangent of ∠A=, AB, AC, Cosecant of ∠A =, BC, AC, Secant of ∠A =, AB, AB, Cotangent of ∠A =, BC, Sine of ∠A, , 30°, , Trigonometry Ratio, , Trigonometry, , Study of relationships between, the sides & angles of a right triangle, , CHAPTER : 10 introduction to trigonometry and trigonometric identities, , Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X, , [ 11
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Distance, , AB, 15, , i.e., AB = 15 √3m, , √3 =, , In ABC ∠B = 90°, ∠ C = 60°, AB, Here, tan 60° =, BC, , Determine height of object AB, , Object Height, , Examples, , From figure, AB = AD + DB, In right ∆APD ∠A = 30°, ∠D = 90°, PD, i.e., AD = 3 √3 m, tan 30° =, AD, In right ∆BPD ∠B = 45°, ∠D = 90°, PD, i.e., BD = 3, tan 45° =, BD, ∴ AB = (3√3 + 3)m = 3 (√3 +1)m, , Determine width AB, , Measuring Angles, , Heights and, Distances, , A, , gle, An, , n, sio, res, ep, , B, , C, , D, (ii), , h, , lag, , C, , DF, , A, , h ...(ii), tan 60º = —, x+200, , 30°, 60°, x B 200 m C, Find x and h, h ...(i), —, tan 60º =200, , h, , Distance between two objects, D, , B, x, Find flag length, h ...(i), tan = —, x, h+DC, tan = —, x ...(ii), , A, , Ratios (To determine), , Angle of Elevation is equal, to Angle of Depression, , D, of, , 90°, , (I) BD is a tree, AC = DC, if CD is broken, , height / length of an object, , Application–Trigonometric, , CHAPTER : 11 heights and distances, , 12 ], Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X
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Q, , 14 cm, , (), , Area of shaded region = Area of ABCD–, Area of 4 circles, = (196 – 154) = 42 cm2, , =, , 154, × 4cm2, 4, = 154 cm2, , Area of square ABCD = 14×14 cm2, = 196 cm2, 14, Diameter of each circle,D =, = 7 cm, 2, 7, For each circle, radius (r) = cm, 2, 2, Area of 1 circle = r, 2, 22, 7, 2, Area of 4 circles = 4 × ×, 2 cm, 7, , Find Area of, shaded region, , Example, , P, , =, , Q, , 1, Area of Sector = ×L×r, 2, , × r2 – area of ∆OAB, 360°, r2 1 2, =, –, r sin , 360° 2, , , , Area = Area of the corresponding sector–, Area of the corresponding triangle, , Formula, , Circles, , Areas Related to, , Area - Combination of figures, , T, Area of T = Area of P + Area of Q, , P, , Meaning, , L=, , 360°, , , , 360°, , × circumference, × 2 r, , Length of arc, , , × r2, , ×area of circle, , L=, , 360°, , , , 360°, , Area, , Portion of, the circular region, Segment, enclosed between, a chord and the, corresponding arc, Minor Segment, , Meaning, , A=, , A=, , , , Portion of the, circular region enclosed, by two radius and the, corresponding arc, , • Area = r2, , • Circumference = × diameter = 2 r, , Formula, , Sector, , Meaning, , Segment, , Circle, , CHAPTER : 12 areas related to circles, , Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X, , [ 13
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h2 + (r1 – r2)2, , r2, , r1, , l, , Volume, , r2, , h, , r1, , l, , Volume, , ( (, , 1 cm, r= —, 30, Thickness = Diameter of the cross-section, 1, = — cm = 0.07 cm, 15, , 1, r2 = —, 900, , Quantity of 3-D space, enclosed by a hollow/closed solid., , Actual capacity = Apparent capacity, – Volume of hemisphere, = 98.125 – 32.71, = 65.42 cm3, , 2 3, – πr ,if r =2.5cm, Volume of hemisphere = 3, 3, 3, = 2, ×, 3.14, ×, (2.5), cm, =, 32.71 cm3, –, 3, , 5 cm, Find – Actual capacity of Cylindrical glass, Solution :– Apparent capacity, of the glass = r2h, = 3.14 × 2.5 × 2.5 × 5 cm3, = 98.125 cm3, , 5 cm, , Given – Inner diameter of the, Cylindrical glass = 5 cm, Height = 5 cm, , Combination of Solids, , Surface Area, , Sum of all of the, surface areas of the, faces of solid., , A copper rod – Diameter 1cm, length 8cm, converted into a wire of length 18m, Find the thickness of the wire., 12, Solution : Volume of the rod = 2 × 8cm3, = 2 cm3, Let, r is the radius of cross-section of the, wire, volume = ×r2 × 1800 cm3, ∴ ×r2 × 1800 = 2 , , Volumes, , Surface Areas and, , Conversion of Solids, , r2, , r1, , Frustum of, Cone, , 1, V = — h(r21 + r22 + r1r2), 3, , TSA = l (rl + r2)+ (r2l + r22 ), , Total Surface Area, , Curved Surface Area, , where l =, , CSA = πl (r1 + r2), , CHAPTER : 13 surface areas and volumes, , 14 ], Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X
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Mode=l +, , f1–f0, , 2f1–f0–f2, , ( (, , 3 Median = Mode + 2 Mean, , Representation of, cumulative frequencies with, respect to given class, intervals, , Meaning, , A Cumulative Frequency Graph, , ×h, , Relationship, , Emperical, , Ogive, , f, , ( (, n, – –cf, , Median = l + 2, , ×h, , Grouped Data, , Statistics, , Mean, , CHAPTER : 14 statistics, , n, , n, , n, , ∑fiui, Step Deviation –x =a + ——, i=1, ×h, n, ∑fi, i=1, x–a, Here, u = ——, h, , ∑fidi, –, i=1, Assumed Mean (Short cut) x =a + ——, n, ∑fi, i=1, Here, d = (x–a), , i=1, , [, , 2, , Upper class Lower class, +, limit, limit, , [, , Frequency obtained by adding, the frequencies of all the classes, preceding the giving class, , ∑fixi, i=1, Direct Method (Long cut) x– = ——, n, ∑fi, , Class Mark, , Cumulative Frequency, , Definition, , A collection,, analysis, interpretation, of quantitative data, , Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X, , [ 15
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Dice, , 4, 1, P(E) =— =—, 52 13, , When event having, probability to occur as 1, , Number of possible outcomes = 52, , Event having, only one outcome of the, experiment, , Elementary Event, , For event E,, –, complement event, P(E ) = 1 – P(E), , Complementary Event, , Definitions, , Value, , Number of outcomes, favorable to E, Number of all possible, outcomes of the, experiment, , Sum of probabilities of all elementary, events is 1., For events A, B, C; P(A) + P(B) + P(C) = 1, , P(E) =, , What actually happens, in an experiment, , Number of trials in, which the event happened, P(E) =, Total number of trials, , What we expect to, happen in an experiment, , 0 ≤ P(E) ≤ 1, , Experimental Probability, , Theoretical Probability, , Probability, , Sure or Certain Event, , Examples, , outcomes = 4, , Solution : Number of favorable, , What is the probability of getting, an ace from a pack of 52 cards?, , Card, , 1, Required Prob. P(E) = —, 2, , Solution : Total outcomes = 2, Favorable outcomes = 1, , When a coin is tossed, what, would be the probability of, appearing head ?, , Coin, , Number of favorable outcomes = 1, 1, P(E) = —, 36, , outcome = 62 =36, , Solution : Number of possible, , Two dice are rolled, what is probability, of getting 12 as a sum?, , CHAPTER : 15 probability, , 16 ], Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – X
