CBSE Class 10 Mathematics Basic Sample Paper 5 2027 Solutions

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SAMPLE PAPER 05 Sample Paper 05 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 62 Cbse Class 10th Mathematics - Basic SP 3. If radii of two concentric circles are cm 4 and cm 5 , then the length of each of one circle which is tangent to the other circle, is (a) cm 3 (b) cm 6 (c) cm 9 (d) cm 1 AAns :  Let C be the centre of two concentric circles C1 and C2, whose radii…
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SAMPLE PAPER 05 7. The sum and product of zeroes of a quadratic polynomial are 6 and 9 respectively. The quadratic polynomial will be (a) x x 9 6 2 + − (b) x x 6 9 2 + + (c) x x 6 9 2 − + (d) x x 6 9 2 + − AAns :  Sum of zeroes, α β + 6 = Product of zeroes αβ 9 = Now ( ) p x ( ) x x 2 α β αβ = − +…
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page 64 Cbse Class 10th Mathematics - Basic SP 12. For what value of k , the pair of linear equations , kx y 4 3 − = x y 6 12 9 − = has an infinite number of solutions ? (a) k 2 = (b) k 2 ! (c) k 3 ! (d) k 4 = AAns :  [ We have kx y 4 3 - - 0 = and x y 6 12 9 - - 0 = where, , a k b 1 1 = ,c 4 3 1…
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SAMPLE PAPER 05 2 1 BD 6 = & BD m 12 = Thus length of wire is 12 m. Thus (a) is correct option. 16. Which term of an AP, 21, 42, 63, 84, ... is 210 ? (a) 9th (b) 10th (c) 11th (d) 12th AAns :  (b) 10th Let nth term of given AP be 210, First term, a 21 = Common difference, d 42 21 = − 21 = and an…
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page 66 Cbse Class 10th Mathematics - Basic SP (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but…
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SAMPLE PAPER 05 Class Frequency . . c f 55-60 6 19 60-65 6 25 65-70 3 28 70-75 2 30 N 30 = We have N 30 = ; 15 N 2 = Cumulative frequency just greater than N 2 is 19 and the corresponding class is 55-60. Thus median class is 55-60. Now l 55 = , f 6 = , F 13 = , h 5 = Median, Md l f F h N 2 # = + −…
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page 68 Cbse Class 10th Mathematics - Basic SP Class Class Mark Frequency f x i i 11-13 12 3 36 13-15 14 6 84 15-17 16 9 144 17-19 18 13 234 19-21 20 f f 20 21-23 22 5 110 23-25 24 4 96 Total f 40 + f 704 20 + We have f i / f 40 = + f x i i / f 704 20 = + Mean, M f f x = i i i / / 18 f f 40 704 20…
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SAMPLE PAPER 05 Let A + 1θ = , B + 2θ = and C + 3θ = . Now, area which can be grazed by the horses is the sum of the areas of three sectors with central angles 1θ , 2θ and 3θ each with radius r 7 = m. r r r 360 360 360 2 1 2 2 2 3 c c c π θ π θ π θ + + ( ) r 360 2 1 2 3 c π θ θ θ = + + ...(1) From…
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page 70 Cbse Class 10th Mathematics - Basic SP Now AC 5 2 6 1 2 2 = − + + ^ ^ h h 3 7 2 2 = + 9 49 = + 58 = BD 5 2 1 6 2 2 = − + − − ^ ^ h h 3 7 2 2 = + 9 49 = + 58 = Since AC BD 58 = = the diagonals of rectangle ABCD are equal. Mid-point of AC , , 2 2 5 2 1 6 = + − + b l , 2 7 2 5 = b l Mid-point…
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SAMPLE PAPER 05 x y 2 3 − + 0 = x y 3 5 1 − + 0 = AAns :  We have x y 2 3 − + 0 = ...(1) x y 3 5 1 − + 0 = ...(2) Multiplying equation (1) by 5,and subtracting (2) from it we have x 7 14 =− x 7 14 2 = − =− Substituting the value of x in equation (1) we get x y 2 3 − + 0 = y 2 2 3 − − + ^ h 0 = y 4…
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page 72 Cbse Class 10th Mathematics - Basic SP O Amit, standing on a horizontal plane, find a bird flying at a distance of 200 m from him at an elevation of 30c. Deepak standing on the roof of a 50 m high building, find the angle of elevation of the same bird to be 45c. Amit and Deepak are on…
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SAMPLE PAPER 05 Now AMB + CMD + = Also, ABM + CDM + = 90 = c So, by AA similarity criterion, AMB T CDM T + (ii) As ABM T CDM T + we obtain, CD AB DM BM = . AB 1 8 .1 5 5 = AB . . 1 8 6 1 5 5 # = = m Thus, the height of the pole is 5 metres. (iii) From the geometry of diagram we have MCD + i = tan…
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page 74 Cbse Class 10th Mathematics - Basic SP (iii) ( ) h t t t 16 240 544 =− + + 2 ( ) t t 16 15 544 =− − + 2 ( . . ) t t 16 15 7 5 7 5 544 2 2 =− − + − + 2 ( . ) . t 16 7 5 16 7 5 544 2 # =− − + + 2 ( . ) t 16 7 5 900 544 =− − + + 2 ( . ) t 16 7 5 1444 =− − + 2 From above equation it is clear…
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