CBSE Class 10 Mathematics Basic Sample Paper 11 2027 Solutions

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SAMPLE PAPER 11 Sample Paper 11 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 146 Cbse Class 10th Mathematics - Basic SP 4. A sphere is melted and half of the melted liquid is used to form 11 identical cubes, whereas the remaining half is used to form 7 identical smaller spheres. The ratio of the side of the cube to the radius of the new small sphere is (a) 3 4 / 1 3 b…
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SAMPLE PAPER 11 Thus (a) is correct option. 9. If the point ( , ) P k 0 divides the line segment joining the points ( , ) A 2 2 - and ( , ) B 7 4 - in the ratio 1 : 2, then the value of k is (a) 1 (b) 2 (c) 2 - (d) 1 - AAns :  As per question statement figure is shown below. k ( ) ( ) 1 2 1 7 2 2…
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page 148 Cbse Class 10th Mathematics - Basic SP 13. The cumulative frequency table is useful in determining (a) Mean (b) Median (c) Mode (d) All of these AAns :  Cumulative frequency is defined as a running total of frequencies. It is helpful in finding the mean, median and mode. Thus (d) is…
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SAMPLE PAPER 11 and αβ 6 = Now ( ) p x ( ) x x 2 α β αβ = − + + ( ) x x 5 6 2 = −− + x x 5 6 2 = + + Thus (a) is correct option. 19. Assertion : Common difference of the AP 5 - , 1 - , 3, 7, .......... is 4. Reason : Common difference of the AP , , ,.......... a a d a d 2 + + is given by d a a 2 1…
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page 150 Cbse Class 10th Mathematics - Basic SP P (number of each die is even), ( ) P E1 ( ) ( ) n S n E1 = 36 9 = 4 1 = (ii) Sum of numbers is 5 in (1, 4) (2, 3) (3, 2) (4, 1) ( ) n E2 4 = P (sum of numbers appearing on two dice is 5) P E2 ^ h n S n E2 = ^ ^ h h 36 4 = 9 1 = 23. In the given…
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SAMPLE PAPER 11 Since XB AX YC AY =Y Hence XY is not parallel to . BC 25. In given figure, AB is the diameter of a circle with centre O and AT is a tangent. If AOQ 58c + = , find ATQ + . AAns :  We have AOQ + 58c = Since angle ABQ + and AOQ + are the angle on the circumference of the circle by the…
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page 152 Cbse Class 10th Mathematics - Basic SP 28. Find HCF and LCM of 16 and 36 by prime factorization and check your answer. AAns :  Finding prime factor of given number we have, 16 2 2 2 2 24 # # # = = 36 2 2 3 3 2 3 2 2 # # # # = = HCF(16, 36) 2 2 # = 4 = LCM (16, 36) 2 3 4 2 # = 16 9 # = 144…
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SAMPLE PAPER 11 30. A well of diameter 4 m is dug 14 m deep. The earth taken out is spread evenly all around the well to form a 40 m high embankment. Find the width of the embankment. AAns :  [Board Term-2 2012] Depth of well, d 14 = m, Radius, r 12 = m. Volume of earth taken out, r h 2 π ( ) 7 22…
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page 154 Cbse Class 10th Mathematics - Basic SP Section - D Section D consists of 4 questions of 5 marks each. 32. Find the ratio in which the y-axis divides the line segment joining the points ,1 4 - - ^ h and ,5 6 - ^ h. Also find the coordinates of the point of intersection. AAns :  Let points…
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SAMPLE PAPER 11 AR AQ = BQ BP = Now AR AD DR = − 23 5 18 = − = cm AQ AR 18 = = cm QB AB AQ = − 29 18 11 = − = cm PB QB 11 = = Now OQB + OPB 90c + = = because radius is always perpendicular to tangent. Thus OP OQ PB BQ = = = So, POQB is a square. Hence, r OP PB 11 = = = cm 35. The table below show…
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page 156 Cbse Class 10th Mathematics - Basic SP 700-800 9 65+ x+y 800-900 7 72+ x+y 900-1000 4 76+ x+y N 100 = Here median is 525, which lies between class 500- 600. Thus median class is 500-600. Now, x y 76 + + 100 = x y + 100 76 24 = − = ...(1) Median, Md l f F h N 2 = + − d n 525 x 500 20 2 100…
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SAMPLE PAPER 11 Rajesh has been given the task of designing a boiler for NTPC. Boiler consist of a cylindrical part in middle and two hemispherical part at its both end. The cross section of boiler is given below. Length of cylindrical part is the 3 times of radius of hemispherical part. (i) Which…
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page 158 Cbse Class 10th Mathematics - Basic SP (i) Let x be the number of horses hired and y be the number of elephant hired, then we have x y 25 + = and x y 50 100 1950 + = & x y 2 39 + = (ii) Solving equations x y 25 + = and x y 2 39 + = we get x 11 = and y 14 = . Number of horses hired = 11…
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