CBSE Class 10 Mathematics Basic Sample Paper 24 2027 Solutions

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page 324 Cbse Class 10th Mathematics - Basic SP Sample Paper 24 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…
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SAMPLE PAPER 24 3. The quadratic equation x x 2 3 2 0 2 4 9 − + = has (a) Two distinct real roots (b) Two equal real roots (c) No real roots (d) More than 2 real roots AAns :  We have x x 2 3 2 4 9 2 − + 0 = Here a 2 = , b 3 2 =− , c 4 9 = Discriminant D b ac 4 2 = − 3 2 4 2 4 9 2 # # = − − ^ h 18…
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page 326 Cbse Class 10th Mathematics - Basic SP Using Pythagoras theorem we have AB2 ( ) ( ) AD BD 2 2 = + ( ) a 2 2 ( ) ( ) AD a 2 2 = + a a 4 2 2 - ( ) AD 2 = ( ) AD 2 a 3 2 = AD a 3 = Hence, the length of attitude is a 3 . Thus (a) is correct option. 8. A pair of dice is thrown once. What is the…
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SAMPLE PAPER 24 Thus (c) is correct option. 12. In the formula x a f f d i i i = + / / , for finding the mean of grouped data di’s are deviation from a of (a) Lower limits of the classes (b) Upper limits of the classes (c) Mid-points of the classes (d) Frequencies of the class marks AAns : …
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page 328 Cbse Class 10th Mathematics - Basic SP x x 3 2 2 3 - - ^ ^h h ( ) p x 0 = Solving we get , x 2 3 3 2 = Hence, zeroes are 3 2 and 2 3 . Thus (d) is correct option. 18. Select the quadratic polynomial p x ^ h with 3 and 5 2 - as sum and product of its zeroes, respectively. (a) x x 3 2 5 2 -…
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SAMPLE PAPER 24 AAns :  [Board Term-2 OD 2015] We redraw the given figure by joining O to P as shown below. PRO + PRQ 2 1 + = 2 120 60 c c = = Here OPR ∆ is right angle triangle, thus POR + PRO 90c + = − 90 60 30 c c c = − = Now OR PR sin30 2 1 c = = OR PR 2 = PR PR = + Since PR QR = , OR PR QR =…
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page 330 Cbse Class 10th Mathematics - Basic SP 3 5 + , q p = q 0 ! 5 q p 3 = − 5 q p q 3 = − Here 5 is irrational and q p q 3 - is rational. But rational number cannot be equal to an irrational number. Hence 3 5 + is an irrational number. O Show that any positive even integer can be written in the…
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SAMPLE PAPER 24 Since OA is radius and PA is tangent at A, . OA AP = Now in right angle triangle OAP ∆ , OP is equal to diameter of circle, thus OP OA 2 = OP OA 2 1 = sinθ 2 1 = 30 & c θ = Since PO bisect the angle APB + , Hence, APB + º 2 30 60 # c = = Now, in APB T , AP AB = PAB + PBA + = º 2 180…
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page 332 Cbse Class 10th Mathematics - Basic SP Total possible outcomes, ( ) n S 6 = (i) Probability of getting letter A, ( ) n E1 2 = . ( ) P E1 ( ) ( ) n S n E1 = 6 2 = 3 1 = (ii) Probability of getting letter C , ( ) n E2 3 = ( ) P E2 ( ) ( ) n S n E2 = 6 1 = O Possible outcomes are {HHH, HHT,…
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SAMPLE PAPER 24 Clearly, the two lines intersect at point , D 2 3 ^ h. Hence, x 2 = and y 3 = is the solution of the given pair of equations. The line CD intersects the x -axis at the point , E 4 0 ^ h and the line AB intersects the x -axis at the points , F 1 0 - ^ h. Hence, the co-ordinates of…
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page 334 Cbse Class 10th Mathematics - Basic SP ABD T ~ PQM T By CPST we have B + , Q + = In ABC T and , PQR T PQ AB QR BC = By SAS similarity we have B + , Q + = Thus ABC T ~ . PQR T Hence Proved. O We have redrawn the given figure as below. We have AB AD + BC CD = + AD BC CD AB = + − AD h d x = +…
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SAMPLE PAPER 24 BC 5 7 2 2 2 2 = − + − + ^ ^ h h 2 0 2 2 = − + ^ h 2 = CA 7 6 2 4 2 2 = − + − − ^ ^ h h 1 6 37 2 2 = + = AB BC 37 = = Since two sides of a triangle are equal in length, triangle is an isosceles triangle. Section - E Section E consists of 3 case study based questions of 4 marks each.…
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page 336 Cbse Class 10th Mathematics - Basic SP Now tan60c a h = 3 a h = a h 3 = Now tan30c a h 50 = + 3 1 a h 50 = + a 50 + h 3 = Substituting a h 3 = we have h 3 50 + h 3 = h 50 3 + h 3 = 50 3 h 2 = h 25 3 = . 25 1 732 # = . 43 3 = m a 2 h 3 = 3 25 3 25 = = m (i) Height of pyramid is 43.3 m. (ii)…
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SAMPLE PAPER 24 Median, Md l f F h N 2 = + − d n 525 x 500 20 50 36 100 # = + − − b l 25 ( ) x 14 5 # = − 25 x 70 5 = − x 5 70 25 9 = − = Now y 24 9 15 = − = (i) Thus x 9 = (ii) y 15 = (iii) Class 500-600 has the maximum frequency 20, therefore this is model class. *********** 
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