CBSE Class 10 Mathematics Basic Sample Paper 4 2027 Solutions

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SAMPLE PAPER 04 Sample Paper 04 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 48 Cbse Class 10th Mathematics - Basic SP (a) 30 cm (b) 20 cm (c) 10 cm (d) None of these AAns :  Let radii of the circles with centres P , Q and R are ,p q and r , respectively. Then, PQ p q = − 10 = ...(1) PR p r = − 8 = ...(2) and QR q r = + 12 = ...(3) Adding equation (2) and (3), we get,…
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SAMPLE PAPER 04 Thus complete factor tree is as given below. Thus (b) is correct option. 6. If , am bl = then what do you say about the solution of the pair of linear equations ax by c + = and lx my n + = ? (a) No solution (b) Unique solution (c) Infinitely solution (d) Can’t say anything AAns : …
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page 50 Cbse Class 10th Mathematics - Basic SP 10. ? tan tan 1 1 2 2 θ θ + − = (a) 1 (b) cos sin 2 2 q q - (c) sin2θ (d) cos2θ AAns :  tan tan 1 1 2 2 θ θ + − sec tan 1 2 2 θ θ = − sec sec tan 1 2 2 2 θ θ θ = − cos cos sin cos 2 2 2 2 # θ θ θ θ = − cos sin 2 2 θ θ = − Hence Proved Thus (b) is…
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SAMPLE PAPER 04 CA2 1 3 3 0 2 2 = − − + − ^ ^ h h 16 9 25 = + = AB2 CA2 = or, AB CA = Hence triangle is isosceles. Also, 25 25 + 50 = or, AB CA 2 2 + BC2 = Since Pythagoras theorem is verified, therefore triangle is a right angled triangle. Thus (d) is correct option. 16. The length of a string…
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page 52 Cbse Class 10th Mathematics - Basic SP We have redrawn the given figure as shown below. We have CD BE = x y + 7 = ...(1) Also, perimeter of ABCDE is 27 cm, thus AB BC CD DE AE + + + + 27 = ( ) ( ) ( ) x y x y x y 5 5 + − + + + − + 27 = x y 3 - 17 = ...(2) Adding equation (1) and (2) we have…
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SAMPLE PAPER 04 k 2 3 + x = and y 2 4 6 = + 2 10 = 5 = Substituting above value in x y 10 + − 0 = we have k 2 3 5 10 + + − 0 = k 2 3 + 5 = k 3 + 10 = & k 10 3 = − 7 = 23. Two coins are tossed together. Find the probability of getting both heads or both tails. AAns :  [Board Term-2 2012]…
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page 54 Cbse Class 10th Mathematics - Basic SP ED . cm 4 8 = Section - C Section C consists of 6 questions of 3 marks each. 26. Prove that : cot cosec cot cosec 1 1 θ θ θ θ − + + − sin cot 1 θ θ = + AAns :  LHS cot cosec cot cosec 1 1 θ θ θ θ = − + + − 1 1 sin cos sin sin cos sin 1 1 = − + + − θ θ…
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SAMPLE PAPER 04 O Find the area of shaded region shown in the given figure where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre. AAns :  Area of sector OAPB 360 90 10 25 2 π π = = ^ h cm2 Area of AOB ∆ 2 1 10 10 50 # # = = cm2 Area…
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page 56 Cbse Class 10th Mathematics - Basic SP There are 6 6 36 # = possible outcome. Thus sample space for two die is ( ) n S 36 = O (i) 5 will come up at least once Favourable case are ( , ),( , ),( , ),( , ),( , ), 5 1 5 2 5 3 5 4 5 5 ( , )( , ),( , ),( , ),( , ) 5 6 1 5 2 5 3 5 4 5 and (6,…
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SAMPLE PAPER 04 a b 3 3 2 + − ^ h a b 7 = + ^ h a b 9 3 6 + − a b 7 7 = + a b 2 4 - 6 = a b 2 - 3 = ...(2) Subtracting equation (1) from (2), b 7 - 7 =− b 1 = Substituting the value of b in equation (1), a 5 = Hence, , a b 5 1 = = . 33. Sides AB and AC and median AD of a triangle ABC are…
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page 58 Cbse Class 10th Mathematics - Basic SP Hence BD BO 2 2 4 8 # # = = = cm 34. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pipe, given that 1 cm3 of iron has…
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SAMPLE PAPER 04 Total seats are 300 more than previous seats so total number of seats are x 300 2 + . Thus ( ) x x 2 10 - x 300 2 = + x x 2 20 2 - x 300 2 = + x x 20 300 2 - - 0 = (ii) We have x x 20 300 2 - - 0 = x x x 30 10 300 2 − + − 0 = ( ) ( ) x x x 30 10 30 − + − 0 = ( )( ) x x 30 10 − + 0 =…
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page 60 Cbse Class 10th Mathematics - Basic SP (ii) Class 35-40 has the maximum frequency 78, therefore this is model class. Now l 35 = , f1 33 = , f 11 2 = , f 21 0 = , h 5 = Mode, Mo l f f f f f h 2 1 0 2 1 0 = + − − − d n 35 66 21 11 33 21 5 # = + − − − 35 33 12 5 # = + 35 11 20 = + . 35 1 82 =…
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