CBSE Class 10 Mathematics Basic Sample Paper 8 2027 Solutions

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SAMPLE PAPER 08 Sample Paper 08 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 104 Cbse Class 10th Mathematics - Basic SP AAns :  In OAL T , OA2 OL AL 2 2 = + a2 OL b 2 2 = + OL a b 2 2 = − Length of chord, AL 2 a b 2 2 2 = − Thus (d) is correct option. 4. One equation of a pair of dependent linear equations x y 5 7 2 − + = The second equation can be (a) x y 10 14 4 0 +…
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SAMPLE PAPER 08 an envelope is picked up out, then the probability that is contains no cash prize is (a) 0.65 (b) 0.69 (c) 0.54 (d) 0.57 AAns :  Total number of envelopes in the box 1000 = Number of envelopes containing cash prize 10 100 200 = + + 310 = Number of envelopes containing no cash 1000…
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page 106 Cbse Class 10th Mathematics - Basic SP We have sin cos θ θ + cos 2 θ = Dividing both sides by cosθ , we get cos sin cos cos θ θ θ θ + cos cos 2 θ θ = tan 1 θ + 2 = tanθ 2 1 = − Thus (a) is correct option. 10. Which of the following equations has the sum of its roots as 3 ? (a) x x 2 3 6 0…
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SAMPLE PAPER 08 Now 12 5 EF 7 = BC 15 = Taking first and second ratios, we get 12 5 EF 7 = EF 5 7 12 & # = . 16 8 = cm Taking first and third ratios, we get 12 5 BC 15 = BC 12 5 15 & # = .6 25 = cm Now, sum of the remaining sides of triangle, EF BC + . . 16 8 6 25 = + . 23 05 = cm Thus (a) is…
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page 108 Cbse Class 10th Mathematics - Basic SP AB is tangent at P and CP is radius at P . Tangent at any point of circle is perpendicular to the radius through the point of contact. Thus CP AB = Now, in right triangle PAC By Pythagoras theorem we have AP 2 AC PC 2 2 = − 5 4 2 2 = − 25 16 = − 9 =…
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SAMPLE PAPER 08 Find the sum of first 15 multiples of 8. AAns :  We have an S S n n 1 = − − an 1 - S S n n 1 2 = − − − Subtracting above equation we have a a n n 1 - - S S S S n n n n 1 1 2 = − − + − − − d . S S S 2 n n n 1 2 = − + − − O Let the first term be a 8 = , common difference be d 8 = ,…
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page 110 Cbse Class 10th Mathematics - Basic SP AAns :  ] We have RPQ + 50c = Since OPQ QPR + + + is right angle triangle, OPQ + 90 50 40 c c c = − = Since, OP OQ = because of radii of circle, we have OPQ + OQR 40c + = = In POQ ∆ we have POQ + ( ) OPQ OQP 180c + + = − + 180 40 40 c c c = − + ^ h…
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SAMPLE PAPER 08 29. Find the area of minor segment of a circle of radius 14 cm, when its centre angle is 60c. Also find the area of corresponding major segment. Use 7 22 π = . O In the given figure, find the area of the shaded region, enclosed between two concentric circles of radii 7 cm and 14 cm…
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page 112 Cbse Class 10th Mathematics - Basic SP (i) of getting a doublet Doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) which are 6 doublets. Number of favourable outcomes, ( ) n E1 6 = P (doublet), ( ) P E1 ( ) ( ) n S n E1 = 36 6 6 1 = = (ii) of getting a sum 10, of the numbers on…
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SAMPLE PAPER 08 Again, x y - 1 = intersects y - axis at ,0 1 - ^ h and x y 2 + 8 = y - axis at ,0 8 ^ h. O We have x y 2 3 + 7 = p q x p q y 1 2 2 + + + + + ^ ^ h h p q 4 1 = + + ^ h Comparing given equation to ab by c 0 + + = we have , , a b c 2 3 7 1 1 1 = = =− , , a p q b p q c p q 1 2 2 4 1 2 2…
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page 114 Cbse Class 10th Mathematics - Basic SP By AA similarity we have AEB T ~ CBA T BC CE AC BC = BC2 CE AC # = ...(2) Adding equation (1) and (2) we have AB BC 2 2 + AE AC CE AC # # = + ( ) AC AE CE = + AC AC # = Thus AB BC 2 2 + AC2 = Hence proved (2) As per given condition we have drawn the…
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SAMPLE PAPER 08 We have , , , , , P Q R 2 1 3 4 2 3 - - ^ ^ ^ h h h, , S 3 2 - - ^ h PQ 1 5 2 2 = + 26 = QR 5 1 2 2 = + 26 = RS 1 5 2 2 = + 26 = PS 5 1 2 2 = + 26 = Since all the four sides are equal, PQRS is a rhombus. Now PR 1 5 26 2 2 = + = 4 4 32 2 2 = + = PQ QR 2 2 + 2 26 52 32 2 # = = =Y ^ h…
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page 116 Cbse Class 10th Mathematics - Basic SP A water tower is located 60 meter from a building (see the figure). From a window in the building, an observer notes that the angle of elevation to the top of the tower is 60c and that the angle of depression to the bottom of the tower is 30c. (i) How…
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SAMPLE PAPER 08 (i) Since numbers of students are 100, p 15 18 21 29 + + + + 100 = p 83 + 100 = p 100 83 17 = − = (ii) Class 60-80 has the maximum frequency 29, therefore this is model class. Lower limit of this class is 60. Here, l 60 = , f 29 1 = , f 21 0 = , f 17 2 = and h 20 = Mode, Mo l h f f…
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