CBSE Class 10 Mathematics Basic Sample Paper 21 2027 Solutions

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SAMPLE PAPER 21 Sample Paper 21 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 284 Cbse Class 10th Mathematics - Basic SP Applying Pythagoras theorem in OPA T , we have OA2 OP AP 2 2 = + 100 AP 64 2 = + AP 2 100 64 = − 36 = & AP 6 = cm AB AP 2 = 2 6 # = 12 = cm Thus (b) is correct option. 3. An observer, 1.5 m tall is 20.5 away from a tower m 22 high, then the angle of…
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SAMPLE PAPER 21 Volume of spherical shell R r 3 4 3 4 3 3 π π = − ( ) R r 3 4 3 3 π = − Thus (d) is correct option. 7. While computing mean of grouped data, we assume that the frequencies are (a) evenly distributed over all the classes (b) centred at the class marks of the classes (c) centred at…
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page 286 Cbse Class 10th Mathematics - Basic SP ( ) ( ) BDE ABC area area T T ( ) ( ) BD BC 4 3 2 4 3 2 = ( ) BC BC 2 1 2 2 = ^ h BC BC 4 2 2 = 1 4 = : 4 1 = Thus (b) is correct option. 13. Consider the data: Class 65- 85 85- 105 105- 125 125- 145 145- 165 165- 185 185- 205 Frequency 4 5 13 20 14 7…
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SAMPLE PAPER 21 AAns :  [Board 2020 OD Basic] We have ( ) p x kx x k 2 3 2 = + + Comparing it by ax bx c 2 + + , we get a k = , b 2 = and c k 3 = . Sum of zeroes, α β + a b =− k 2 =− Product of zeroes, αβ a c = k k 3 = 3 = According to question, we have α β + αβ = k 2 - 3 = & k 3 2 =− Thus (d) is…
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page 288 Cbse Class 10th Mathematics - Basic SP Section - B Section B consists of 5 questions of 2 marks each. 21. A bag contains 3 red, 4 green and 5 white candles, one candle is drawn at random from the bag, find the probability that candle is not red. AAns :  Total number of possible outcomes…
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SAMPLE PAPER 21 Let O be the centre of the circle. As per given information we have drawn the figure below. We know that, the radius and tangent are perpendicular at their point of contact. Now, in isosceles triangle POQ we have POQ OPQ OQP + + + + + 180c = Equal sides subtend equal angles in…
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page 290 Cbse Class 10th Mathematics - Basic SP Since, TP TQ = , due to opposite angles of equal sides we have TQP + 90c θ = − From angle sum property of a triangle we can write, TPQ TQP PTQ + + + + + 180c = PTQ 90 90 c c + θ θ − + − + 180c = PTQ + 180 180 2 c c θ = − + PTQ + 2θ = Hence, PTQ + OPQ…
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SAMPLE PAPER 21 31. An integer is chosen between 70 and 100. Find the probability that it is (i) A prime number (ii) Divisible by 7 O Find the probability that 5 Sundays occur in the month of November of a randomly selected year. AAns :  [Board 2020 SQP Standard] There are 29 integer from 70 to…
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page 292 Cbse Class 10th Mathematics - Basic SP and y x 2 - 1 = L3 : y x 1 2 = + x 0 1 2 y x 1 2 = + 1 3 5 Plotting the above points and drawing lines joining them, we get the graphical representation: O Let the present age of Aftab be x years and the age of daughter be y years. 7 years ago…
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SAMPLE PAPER 21 AAns :  [Board 2020 OD Standard] We have redrawn the given figure as shown below. In ADE T and GBD T , we have DAE + BGD + = [each 90c] Due to corresponding angles we have ADE + GDB + = Thus by AA similarity criterion, ADE T GBD T + Now, in ADE T and FEC T , EAD + CFE + = [each…
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page 294 Cbse Class 10th Mathematics - Basic SP Thus AB2 AC BC 2 2 = + l l 2 2 = + l2 2 = Now 2 2 2 ^ h l2 2 = Thus l 2 = and r 2 = m Slant height of conical portion, l 2 = m Total surface area of toy rh r rl 2 2 π π π + + r 7 2 2 π = + + 6 @ m2 2 9 2 π = + 6 @ m2 2 9 2 π = + 6 @ m2 Section - E…
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SAMPLE PAPER 21 37. Climbers have to go . h h 63 4 2 3 − = m to the top. John and Priya went for a small picnic. After having their lunch Priya insisted to travel in a motor boat. The speed of the motor boat was 20 km/hr. Priya being a Mathematics student wanted to know the speed of the current. So…
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page 296 Cbse Class 10th Mathematics - Basic SP For h1 , tan60c h 150 1 = 3 h 150 1 = h1 150 3 # = . 150 1 732 # = . 259 8 = m For h2 , tan45c h 150 2 = 1 h 150 2 = h2 150 = m For h3 , tan30c h 150 3 = 3 1 h 150 3 = h3 3 150 50 3 = = . 50 1 732 # = . 86 6 = m h h 2 3 - . . 150 86 6 63 4 = − = m (i)…
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