CBSE Class 10 Mathematics Basic Sample Paper 3 2027 Solutions

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SAMPLE PAPER 03 Sample Paper 03 Class - 10th Exam - 2026 - 27 Mathematics - Basic Time : 3 Hours Max. Marks : 80 General Instructions : Read the following instructions carefully and follow them: 1. This question paper contains 38 questions. 2. This Question Paper is divided into 5 Sections A, B,…
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page 34 Cbse Class 10th Mathematics - Basic SP we have k 2 3 + x = and y 2 4 6 = + 2 10 = 5 = Substituting above value in x y 10 + − 0 = we have k 2 3 5 10 + + − 0 = k 2 3 + 5 = k 3 + 10 = & k 10 3 = − 7 = Thus (d) is correct option. 4. In the given factor tree what is the composite number x ? (a)…
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SAMPLE PAPER 03 8. ? tan tan 4 2 θ θ + = (a) sec sec 2 2 4 q q - (b) sec sec 2 2 4 q q - (c) sec sec 2 4 q q - (d) sec sec 4 2 q q - AAns :  tan tan 4 2 θ θ + ( ) tan tan 1 2 2 θ θ = + tan sec 2 2 # θ θ = ( ) sec sec 1 2 2 θ θ = − sec sec 4 2 θ θ = − Hence Proved Thus (d) is correct option. 9.…
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page 36 Cbse Class 10th Mathematics - Basic SP b b 2 1 6 3 2 1 = = and c c 2 1 18 9 2 1 = − − = Thus c c 2 1 b b c c 2 1 2 1 = = Hence, line are coincident. Thus (b) is correct option. 13. The zeroes of the polynomial p x x x 4 12 9 2 = − + ^ h will be (a) 2 3 and 2 3 (b) 3 2 and 3 1 (c) 3 2 and 3…
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SAMPLE PAPER 03 u v 21 47 + 110 = ...(1) and u v 47 21 + 162 = ...(2) Adding equation (1) and (2) we get u v 68 68 + 272 = u v + 4 = ...(3) Subtracting equation (1) from (2) we get u v 26 26 - 52 = u v - 2 = ...(4) Adding equation (3) and (4), we get u 2 6 = & u 3 = Substituting u 3 = in equation…
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page 38 Cbse Class 10th Mathematics - Basic SP Section - B Section B consists of 5 questions of 2 marks each. 21. If the point , P x y ^ h is equidistant from the points , Q a b b a + − ^ h and , R a b a b − + ^ h, then prove that bx ay = . AAns :  We have PQ PR = x a b y b a 2 2 − + + − − ^ ^ h h…
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SAMPLE PAPER 03 ( , )× LCM a b 18 306 1314 # = LCM ( , ) a b 18 306 1314 # = LCM ( , ) 306 1314 , 22 338 = 24. In a certain AP 32th term is twice the 12th term. Prove that 70th term is twice the 31st term. AAns :  Let the first term of an AP be a, common difference be d and nth term be an. Now we…
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page 40 Cbse Class 10th Mathematics - Basic SP We have tan 45º x h 50 = − x h 50 = − ...(1) and tan60º x h = 3 x h = x h 3 = ...(2) From (1) and (2) we have h 50 - h 3 = h 3 50 3 - h = h h 3 - 50 3 = h 3 1 - ^ h 50 3 = h 3 1 50 3 = − 2 50 3 3 = + ^ h ( ) 25 3 3 = + 75 25 3 = + . 118 25 = m Thus . h…
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SAMPLE PAPER 03 29. A solid is in the shape of a hemisphere surmounted by a cone. If the radius of hemisphere and base radius of cone is 7 cm and height of cone is 3.5 cm, find the volume of the solid. Take 7 22 π = ^ h AAns :  As per given information in question we have drawn the figure below,…
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page 42 Cbse Class 10th Mathematics - Basic SP Section - D Section D consists of 4 questions of 5 marks each. 32. If α and β are the zeroes of polynomial p x x x 3 2 1 2 = + + ^ h , find the polynomial whose zeroes are 1 1 α α + − and 1 1 β β + − . AAns :  We have p x x x 3 2 1 2 = + + ^ h Since α…
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SAMPLE PAPER 03 2 2 1 7 22 40 40 # # # # = 7 35200 = m2 Area of racing track = (area of outer rectangle + area of outer semicircles) – (area of inner rectangle + area of inner semicircles) 4240 7 35200 7 3180 19800 = + − + b l 1060 7 15400 7 7420 15400 = + = + 7 22820 3260 = = m2 Hence, area of…
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page 44 Cbse Class 10th Mathematics - Basic SP O If , tan sec 15 4 23 2 2 θ θ + = then find the value of sec cosec sin 2 2 θ θ θ + − ^ h . AAns :  We have tan sec 15 4 2 2 θ θ + 23 = tan tan 15 4 1 2 2 θ θ + + ^ h 23 = tan tan 15 4 4 2 2 θ θ + + 23 = tan 19 2θ 19 = tanθ tan 1 45º = = Thus θ 45º =…
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SAMPLE PAPER 03 (i) What is the distance of point A from origin ? (ii) What is the distance between A and B ? (iii) What is the distance between B and C ? (iv) A point D lies on the line segment between points A and B such that : : AD DB 4 3 = . What are the coordinates of point D ? AAns :  (i) It…
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page 46 Cbse Class 10th Mathematics - Basic SP (i) We prepare the following frequency distribution table. SO2 (in ppm) Class mark xi ^ h Frequency fi ^ h cf . . u x 0 04 0 10 i i = − f ui i 0.00- 0.04 0.02 4 4 2 - 8 - 0.04- 0.08 0.06 9 13 1 - 9 - 0.08- 0.12 0.10 9 22 0 0 0.12- 0.16 0.14 2 24 1 2…
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