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myCBSEguide, , Class 09 - Mathematics, Term-2 Sample paper - 01, , Maximum Marks: 40, Time Allowed: 2 hours, , General Instructions:, 1., 2., 3., 4., 5., , The question paper consists of 14 questions divided into 3 sections A, B, C., All questions are compulsory., Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions., Section B comprises of 4questions of 3 marks each. Internal choice has been provided in one question., Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one question., It contains two case study-based questions., Section A, , 1. If f(x) = 2x3 - 13x2 + 17x + 12, find f (-3), 2. In the adjoining figure, ABCD is a square. A line segment CX cuts AB at X and the diagonal BD at O such that, COD = 80o and, , OXA = xo. Find the value of x., , 3. Factorize: 8(a + 1)2 + 2(a + 1)(b + 2) - 15(b + 2)2, To practice more questions & prepare well for exams, download myCBSEguide App. It provides complete, study material for CBSE, NCERT, JEE (main), NEET-UG and NDA exams. Teachers can use Examin8 App to, create similar papers with their own name and logo., 4. Factorize: 2(x + y)2 - 9(x + y) - 5, 5. Hameed has built a cubical water tank with lid for his house, with each other edge 1.5 m long. He gets the, outer surface of the tank excluding the base, covered with square tiles of side 25 cm. Find how much he, would spend for the tiles if the cost of tiles is ₹ 360 per dozen., , 6. Factorise : 8a3 + b3 + 12a2b + 6ab2, OR, Find the zeroes of the polynomial g(x) = 3 - 6x., Section B, , Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited., , 1/8

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myCBSEguide, 7. If circles are drawn taking two sides of a triangle as diameters, prove the point of intersection of these, circles lie on the third side:, , 8. In, , D, E and F are respectively the mid-points of sides AB, BC and CA. Show that, is divided, into four congruent triangles by joining D, E and F., 9. Construct perpendicular bisector of line segment 8cm., 10. A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base, radius 4 cm. Find the height and slant height of the cone., OR, Find the volume, the total surface area and the lateral surface area of a cuboid which is 15 m long, 12 m, wide and 4.5 m high., Section C, 11. Factorize:, 12. Construct a, , ABC in which AB = 4.5 cm , BC = 6 cm and median AD = 4 cm, OR, , Construct and justify a triangle PQR, given that QR = 3cm,, and QP – PR = 2 cm., 13. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small, supports as shown in figure. Eight such spheres are used for this purpose and are to be painted silver. Each, support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black., , i. Find The surface area of the silver-painted part., ii. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per, cm2, 14. Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes:, Outcome, , No head, , One head, , Two heads, , Three heads, , Frequency, , 14, , 38, , 36, , 12, , To practice more questions & prepare well for exams, download myCBSEguide App. It provides complete, study material for CBSE, NCERT, JEE (main), NEET-UG and NDA exams. Teachers can use Examin8 App to, create similar papers with their own name and logo., If the three coins are simultaneously tossed again, compute the probability of:, i. 2 heads coming up., ii. 3 heads coming up., iii. at least one head coming up., iv. getting more heads than tails., Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited., , 2/8

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myCBSEguide, = 2a (a - 5) + (a - 5), = (2a + 1)(a-5), Now substitute a=x+y we get, = {2(x + y) + 1}(x+y-5) = (2x + 2y + 1)(x+y-5), 5. Since Hameed is getting the five outer faces of the tank covered with tiles, he would need to know the, surface area of the tank, to decide on the number of tiles required., Edge of the cubic tank, a = 1.5m = 150 cm, So, surface area of the tank = 5, , 150, , 150 cm2., , Area of each square title =, , =, , = 180, , Cost of 1 dozen tiles, i.e., cost of 12 tiles = Rs. 360, Therefore, cost of one tile = Rs, So the cost of 180 tiles = 180, , = Rs. 30, Rs.30 = Rs. 5400, , 6. 8a3 + b3 + 12a2b + 6ab2, = (2a)3 + (b)3 + 3(2a)(b)(2a + b), = (2a+ b)3, (Using Identity (a + b)3= a3+ b3+ 3ab (a + b)), = (2a + b)(2a + b)(2a + b), OR, Solving the equation g(x) = 0, we get, 3 - 6x = 0, which gives us, So,, , is a zero of the polynomial 3 - 6x., Section B, , 7. Given: Two circles intersect each other at points A and B. AP and AQ be their respective diameters., To prove: Point B lies on the third side PQ., Construction: Join A and B., Proof: AP is a diameter., 1 = 90°, [Angle in semicircle], Also AQ is a diameter., 2 = 90°, [Angle in semicircle], 1+, , 2 = 90° + 90°, PBQ = 180°, PBQ is a line., Thus point B. i.e. point of intersection of these circles lies on the third side i.e., on PQ., , 8., , Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited., , 4/8

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myCBSEguide, As D and E are mid-points of sides AB and BC of the triangle ABC (Mid-points of two sides of a triangle is, parallel to the third side),, DE || AC, Similarly, DF || BC and EF || AB, Therefore ADEF, BDFE and DFCE are all parallelograms., Now DE is a diagonal of the parallelogram BDFE,, therefore,, Similarly, and, So, all the four triangles are congruent., 9. Steps of construction, , i. Draw a line segment AB = 8cm, ii. Taking A as a centre draw arcs of radius more, , on both side of AB., , iii. Taking B as a centre draw arcs of same radius on both sides of AB which intersect previous arcs at, point C and D., iv. Join CD which intersect AB at point O, v. OA = OB = 4cm, 10. Vcone = Vsphere, , 42H = 4 (43 - 23), 16 H = 4, H=, , (64 - 8), = 14 cm, , Slant height =, l=, , = 14.56 cm, OR, , Here we have ,l = 15 m, b = 12 m and h = 4.5 m., Therefore Volume of the cuboid = (l b h) cubic units., = (15 12 4.5) m3 = 810m3., Total surface area of the cuboid, = 2(lb + bh + Ih) sq units., , Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited., , 5/8

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myCBSEguide, = 2(15, 12 +12 4.5 +15, 4.5) m2 = 603 m2., Lateral surface area of the cuboid, = [2(1 + b) h] sq units., = [2(15 +12) 4.5] m2 = 243m2., To practice more questions & prepare well for exams, download myCBSEguide App. It provides, complete study material for CBSE, NCERT, JEE (main), NEET-UG and NDA exams. Teachers can, use Examin8 App to create similar papers with their own name and logo., Section C, 11., We need to consider the factors of 2, which are, Let us substitute 1 in the polynomial, , to get, , Thus, according to factor theorem, we can conclude that, Let us divide the polynomial, , by, , is a factor of the polynomial, ,to get, , Therefore, we can conclude that on factorizing the polynomial, , ,we get, , 12. GIVEN, AB = 4.5 cm , BC = 6 cm and median AD = 4 cm, TO CONSTRUCT, A triangle ABC, STEPS OF CONSTRUCTION, i. Draw a line segment AB = 4.5 cm., ii. With A as centre and radius equal to 4 cm, draw an arc., iii. With B as centre and radius =, , BC = 3 cm, draw another arc, cutting the previous arc at D., , iv. Join BD and produce it to C such that, v. BC = 6 cm., vi. Join AC., Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited., , 6/8

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myCBSEguide, Then,, , ABC is a required triangle., , OR, , 1. Draw a ray OX and cut off a line segment QR = 3 cm., 2. At Q, construct, ., 3. From QY, cut off QS = 2 cm., 4. Join RS., 5. Draw perpendicular bisector of RS to Meet QY at P., 6. Join PR. Then PQR is the required triangle., Justification: Base QR and PQR are drawn as given., Since, the point P lies on the perpendicular bisector of SR., PS = PR, Now, QS = PQ - PS, = PQ - PR, Thus, our construction is justified., 13. Diameter of a wooden sphere = 21 cm., therefore Radius of wooden sphere (R), , cm, , And Radius of the cylinder (r) = 1.5 cm, The surface area of silver-painted part = Surface area of sphere - Upper part of cylinder for support, =, =, =, =, , Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited., , 7/8

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myCBSEguide, , =, =, = 1378.928 cm2, Surface area of such type of 8 spherical part = 8, = 11031.424, , 1378.928, , cm2, , Cost of silver paint over 1 cm2 = Rs. 0.25, Cost of silver paint over 11031.928 cm2 = 0.25, 11031.928, = Rs. 2757.85, Now, curved surface area of a cylindrical support =, =, = 66 cm2, Curved surface area of 8 such cylindrical supports =66, , 8 = 528 cm2, , Cost of black paint over 1cm2 of cylindrical support = Rs. 0.50, Cost of black paint over 528cm2 of cylindrical support = -0.50, = Rs. 26.40, Total cost of paint required = Rs. 2757.85 + Rs. 26.4, = Rs. 2784.25, 14. Total number of outcomes = 100, , 528, , i. Probability of 2 heads coming up =, , =, , = 0.36, , ii. Probability of 3 heads coming up =, , =, , = 0.12, , iii. Probability of at least one head coming up =, , =, , iv. Probability of getting more heads than tails =, , =, , =, =, , = 0.86, = 0.48, , Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited., , 8/8