CBSE Class 10 Mathematics Basic Sample Paper 10 2027 Solutions

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SAMPLE PAPER 10 Sample Paper 10 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 132 Cbse Class 10th Mathematics - Basic SP Thus OP 5 = cm Thus (b) is correct option. 3. The length of a string between a kite and a point on the ground is 85 m. If the string makes an angle θ with level ground such that tan 8 15 θ = , then the height of kite is (a) 75 m (b) 78.05 m (c) 226 m…
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SAMPLE PAPER 10 AAns :  Let f x ^ h x kx k 2 = + + , k 0 ! Comparing the given polynomial with , ax bx c 2 + + we get a 1 = , b k = and c k = . Again, let if , α β be the zeroes of given polynomial then, α β + k =− αβ k = Case 1: If k is negative, then αβ is negative. It means α and β are of…
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page 134 Cbse Class 10th Mathematics - Basic SP 10. In the given figure, if , , A º B º 90 90 + + = = . OB 4 5 = cm OA 6 = cm and AP 4 = cm then find . QB (a) 3 cm (b) 6 cm (c) 4.5 cm (d) 3.5 cm AAns :  In PAO T and QBO T we have A + B º 90 + = = Vertically opposite angle, POA + QOB + = Thus PAO T…
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SAMPLE PAPER 10 Number of multiples of 4 between 1 to 15 are 4, 8, 12 i.e. 3 favourable outcome. ( ) n E 3 = Required Probability, ( ) P E ( ) ( ) n S n E = 15 3 5 1 = = Thus (d) is correct option. 14. Each root of x bx c 0 2 − + = is decreased by 2. The resulting equation is x x 2 1 0 2 − + = ,…
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page 136 Cbse Class 10th Mathematics - Basic SP Edge of cube, a cm cm 4 20 5 = = Surface area a 6 2 cm 6 52 2 # = cm 150 2 = Thus (b) is correct option. 19. Assertion : The value of sec cot 10 80 2 2 - c c is 1. Reason : The value of sin30 2 1 = c . (a) Both assertion (A) and reason (R) are true…
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SAMPLE PAPER 10 AB CD cp 2 1 2 1 # # = = Also, Area of ABC T BC AC ab 2 1 2 1 # # = = Thus cp 2 1 ab 2 1 = cp ab = Proved 22. In the given figure, from a point P , two tangents PT and PS are drawn to a circle with centre O such that SPT 120c + = , Prove that OP PS 2 = . AAns :  We have SPT + 120c…
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page 138 Cbse Class 10th Mathematics - Basic SP (ii) No. of multiples of 7 from 11 to 123 is 16 i.e 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112 and 119. No. of all favourable outcomes. ( ) n E2 16 = P (number drawn card is multiple of 7) P E2 ^ h n S n E2 = ^ ^ h h 113 16 = O A…
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SAMPLE PAPER 10 A + D + = (Corresponding angles) 2 1 + 2 2 + = Also B + E + = (Corresponding angles) DQ AP DE AB = Hence Proved (2) Since ABC T ~ DEF T A + D + = and C + F + = 2 3 + 2 4 + = 3 + 4 + = By AA similarity we have CAP T ~ FDQ T O In the given figure, || DE AC and || . DF AE Prove that .…
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page 140 Cbse Class 10th Mathematics - Basic SP zeroes are 3α and 3β . AAns :  We have f x ^ h x x 2 3 1 2 = − + If α and β are the zeroes of x x 2 3 1 2 − + , then Sum of zeroes α β + a b = − 2 3 = Product of zeroes αβ a c = 2 1 = New quadratic polynomial whose zeroes are 3α and 3β is, ( ) p x ×…
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SAMPLE PAPER 10 x x 4 4 1 2 + + 0 = x 2 1 2 + ^ h 0 = or, x 2 1 =− Hence, roots are ,1 2 1 - - . 33. In figure O is the centre of a circle of radius 5 cm. T is a point such that OT 13 = cm and OT intersects circle at E . If AB is a tangent to the circle at E , find the length of AB, where TP and TQ…
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page 142 Cbse Class 10th Mathematics - Basic SP Mean, M f f x i i i = / / , 300 3 198 = . 10 66 = Class 11-13 has the maximum frequency 95, therefore this is model class. Now l 11 = , f 95 1 = , f 41 0 = , f 36 2 = , h 2 = Mode, Mo l f f f f f h 2 1 0 2 1 0 = + − − − d n 11 190 41 36 95 41 2 # = +…
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SAMPLE PAPER 10 (i) Consider the prices mentioned by Ramesh. If the price of one notebook is Rs. 25 and the price of one pen is Rs. 2.50 then, The cost of 4 notebooks would be : 4 25 100 # = Rs And the cost for 3 pens would be : . . 3 2 5 7 5 # = Rs Aditya should have paid . . 100 7 5 107 5 + = Rs.…
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page 144 Cbse Class 10th Mathematics - Basic SP ( ) r 1 6 2 2 π = + + ( ) ( ) 7 2 3 2 π = + ( ) m 9 7 2 2 π = + (iii) Volume of cylindrical part, r l 2 π = r r 3 2 $ π = r 3 3 π = Volume of conical part, r h 3 1 2 π = r r 3 1 2 π = r 3 1 3 π = Ratio of volume of cylindrical part to conical part r r…
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