Official CBSE Assignments for Class 10 Mathematics
Explore structured practice materials through the CBSE Class 10 Maths Real Numbers to Probability Revision Assignment with Solutions Set 07. Tailored for Class 10 learners, utilizing these Mathematics assignments ensures thorough preparation and strengthens foundational knowledge before final CBSE evaluations.
Solved Practice Assignments for Mathematics
View or download the dedicated CBSE Class 10 Maths Real Numbers to Probability Revision Assignment with Solutions Set 07 resource below. Engaging with these assignments under focused study conditions ensures continuous academic progress and mastery of the 2026-27 curriculum.
Real Numbers and Polynomials
1. Find the HCF of 1656 and 4025 by Euclid's division algorithm.
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Answer: 4025 = 1656 × 2 + 713; 1656 = 713 × 2 + 230; 713 = 230 × 3 + 23; 230 = 23 × 10 + 0. The remainder is now 0, so the HCF is 23.
Teacher's Note: Use the key term in your answer.
2. Prove that 4 − 3√2 is irrational.
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Answer: Suppose 4 − 3√2 = r is rational. Then √2 = (4 − r)/3, which is rational, since 4 and r are rational. But √2 is irrational, a contradiction. So 4 − 3√2 is irrational.
Teacher's Note: Give an example if the question allows.
3. The HCF of 65 and 117 is expressible as 65m − 117. Find m. Also find the LCM by prime factorisation.
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Answer: 65 = 5 × 13 and 117 = 3² × 13, so the HCF is 13. Then 65m − 117 = 13 gives 65m = 130, so m = 2. LCM = 3² × 5 × 13 = 585.
Teacher's Note: Write a short, clear answer.
4. For any positive integer n, prove that n³ − n is divisible by 6.
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Answer: n³ − n = (n − 1)n(n + 1), the product of three consecutive integers. Among them one is divisible by 3 and at least one is even. So the product is divisible by 2 × 3 = 6.
Teacher's Note: Remember this for your exam.
5. Find the quadratic polynomial whose zeroes are 3 and −4.
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Answer: Sum = −1, product = −12. The polynomial is x² + x − 12.
Teacher's Note: Learn this point and write it in your own words.
6. If α and β are zeroes of a quadratic polynomial with α + β = 24 and α − β = 8, find the polynomial and verify the relationship.
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Answer: Adding, 2α = 32, so α = 16 and β = 8. The polynomial is x² − 24x + 128. Check: sum of zeroes = 24 = −(−24)/1, product = 128 = 128/1.
Teacher's Note: Use the key term in your answer.
7. If f(x) = 3x⁴ − 9x³ + x² + 15x + k is divisible by 3x² − 5, find k and the other two zeroes.
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Answer: Dividing, the quotient is x² − 3x + 2 and the remainder is k + 10. So k = −10. The other zeroes come from x² − 3x + 2 = (x − 1)(x − 2): they are 1 and 2.
Teacher's Note: Give an example if the question allows.
Linear Equations, Quadratic Equations and AP
9. For which value of k is the pair 3x + y = 1 and (2k − 1)x + (k − 1)y = 2k + 1 inconsistent?
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Answer: For no solution: 3/(2k − 1) = 1/(k − 1) ≠ 1/(2k + 1). From the first equality, 3k − 3 = 2k − 1, so k = 2. Then 1/(2k + 1) = 1/5, which is not equal to 1, so k = 2.
Teacher's Note: Write a short, clear answer.
10. Solve 2x + 3y = 12 and x − y − 1 = 0 graphically.
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Answer: From the second, y = x − 1. Points: (1, 0), (2, 1), (3, 2) for x − y = 1, and (0, 4), (3, 2), (6, 0) for 2x + 3y = 12. The two lines meet at (3, 2), so x = 3 and y = 2. The line 2x + 3y = 12 meets the x-axis at (6, 0), and x − y = 1 meets it at (1, 0); the shaded region is the triangle with vertices (1, 0), (6, 0) and (3, 2).
Teacher's Note: Remember this for your exam.
11. Raghav scored 70 marks, getting 4 marks for each right answer and losing 1 mark for each wrong answer. With 5 marks for a right answer and a loss of 2 for a wrong one, he would have scored 80. How many questions were there? Which values would he violate by unfair means?
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Answer: Let the right answers be x and the wrong ones y. 4x − y = 70 and 5x − 2y = 80. From the first, y = 4x − 70. Then 5x − 8x + 140 = 80, so x = 20 and y = 10. There were 30 questions. By unfair means he would violate honesty and integrity.
Teacher's Note: Learn this point and write it in your own words.
12. Find k for which 9x² − 3kx + k = 0 has real roots.
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Answer: D = 9k² − 36k ≥ 0, so 9k(k − 4) ≥ 0. Hence k ≤ 0 or k ≥ 4.
Teacher's Note: Use the key term in your answer.
13. A motor boat with speed 24 km/h in still water takes 1 hour more to go 32 km upstream than to return downstream. Find the speed of the stream.
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Answer: Let the speed of the stream be x. 32/(24 − x) − 32/(24 + x) = 1, so 64x = 576 − x², i.e. x² + 64x − 576 = 0, which gives (x + 72)(x − 8) = 0. The speed is 8 km/h.
Teacher's Note: Give an example if the question allows.
14. The sum of the squares of two consecutive multiples of 7 is 637. Find them.
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Answer: Let them be 7n and 7(n + 1). 49n² + 49(n + 1)² = 637 gives n² + (n + 1)² = 13, so n² + n − 6 = 0 and n = 2 (or −3). The multiples are 14 and 21 (or −21 and −14).
Teacher's Note: Write a short, clear answer.
15. If x − 2, 4x − 1 and 5x + 2 are in AP, find x.
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Answer: 2(4x − 1) = (x − 2) + (5x + 2), so 8x − 2 = 6x and x = 1/3.
Teacher's Note: Remember this for your exam.
16. The angles of a triangle are in AP and the greatest is twice the least. Find the angles.
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Answer: Let them be a − d, a and a + d. The sum is 3a = 180, so a = 60. Then 60 + d = 2(60 − d), so d = 20. The angles are 40°, 60° and 80°.
Teacher's Note: Learn this point and write it in your own words.
17. Which term of the AP 121, 117, 113, … is its first negative term?
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Answer: aₙ = 121 − 4(n − 1) < 0 gives n − 1 > 30.25, so n = 32. The 32nd term is 121 − 124 = −3.
Teacher's Note: Use the key term in your answer.
18. The 17th term of an AP is 5 more than twice its 8th term. If the 11th term is 43, find the nth term.
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Answer: a + 16d = 5 + 2(a + 7d), so 2d = a + 5. Also a + 10d = 43. Substituting a = 2d − 5: 12d = 48, so d = 4 and a = 3. So aₙ = 3 + (n − 1)4 = 4n − 1.
Teacher's Note: Give an example if the question allows.
19. State and prove the Basic Proportionality Theorem.
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Answer: Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides, it divides them in the same ratio.
Proof: In △ABC, let DE ∥ BC with D on AB and E on AC. Join BE and CD, and draw DM ⊥ AC and EN ⊥ AB. Then ar(△ADE)/ar(△BDE) = (½·AD·EN)/(½·DB·EN) = AD/DB. Also ar(△ADE)/ar(△CED) = (½·AE·DM)/(½·EC·DM) = AE/EC. △BDE and △CED are on the same base DE and between the same parallels DE and BC, so they have equal area. Hence AD/DB = AE/EC.
Teacher's Note: Write a short, clear answer.
Triangles and Coordinate Geometry
20. State and prove the Pythagoras Theorem.
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Answer: Statement: In a right triangle, the square on the hypotenuse equals the sum of the squares on the other two sides.
Proof: In right △ABC, right-angled at B, draw BD ⊥ AC. Then △ADB ~ △ABC (common angle A, right angles), so AD/AB = AB/AC, i.e. AB² = AD·AC. Similarly △BDC ~ △ABC, so CD/BC = BC/AC, i.e. BC² = CD·AC. Adding: AB² + BC² = AC(AD + CD) = AC².
Teacher's Note: Remember this for your exam.
21. The perpendicular from A on BC of △ABC meets BC at D such that DB = 3CD. Prove that 2AB² = 2AC² + BC².
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Answer: BC = DB + CD = 4CD. From the right triangles, AB² = AD² + DB² and AC² = AD² + CD². So AB² − AC² = DB² − CD² = 9CD² − CD² = 8CD² = 8(BC/4)² = BC²/2. Hence 2AB² = 2AC² + BC².
Teacher's Note: Learn this point and write it in your own words.
22. Prove that (3, 0), (6, 4) and (−1, 3) are vertices of a right-angled isosceles triangle.
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Answer: AB² = 3² + 4² = 25, BC² = 7² + 1² = 50, CA² = 4² + 3² = 25. So AB = CA, and AB² + CA² = 50 = BC². The triangle is right-angled at A and isosceles.
Teacher's Note: Use the key term in your answer.
23. The x-coordinate of P is twice its y-coordinate. P is equidistant from Q(2, −5) and R(−3, 6). Find P.
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Answer: Let P = (2y, y). (2y − 2)² + (y + 5)² = (2y + 3)² + (y − 6)². So 5y² + 2y + 29 = 5y² + 45, hence y = 8 and P = (16, 8).
Teacher's Note: Give an example if the question allows.
24. Find the ratio in which the line 2x + y = 4 divides the join of A(2, −2) and B(3, 7). Also find the point of intersection.
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Answer: Let the ratio be k : 1. The point is ((3k + 2)/(k + 1), (7k − 2)/(k + 1)). On the line: 2(3k + 2) + (7k − 2) = 4(k + 1), so 13k + 2 = 4k + 4 and k = 2/9. The ratio is 2 : 9, and the point is (24/11, −4/11).
Teacher's Note: Write a short, clear answer.
25. Find the point on the x-axis equidistant from (2, −5) and (−2, 9).
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Answer: Let the point be (x, 0). (x − 2)² + 25 = (x + 2)² + 81, so −8x = 56 and x = −7. The point is (−7, 0).
Teacher's Note: Remember this for your exam.
26. If sin 3A = cos(A − 26°), where 3A is acute, find A.
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Answer: cos(90° − 3A) = cos(A − 26°), so 90° − 3A = A − 26°, 4A = 116° and A = 29°.
Teacher's Note: Learn this point and write it in your own words.
27. Evaluate sin 25° cos 65° + cos 25° sin 65°.
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Answer: This is sin(25° + 65°) = sin 90° = 1.
Teacher's Note: Use the key term in your answer.
Trigonometry
28. If x = a sin A + b cos A and y = a cos A − b sin A, prove x² + y² = a² + b².
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Answer: x² + y² = a² sin²A + b² cos²A + 2ab sinA cosA + a² cos²A + b² sin²A − 2ab sinA cosA = a²(sin²A + cos²A) + b²(cos²A + sin²A) = a² + b².
Teacher's Note: Give an example if the question allows.
29. What happens to the value of cos A when A increases from 0° to 90°?
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Answer: The value of cos A decreases from 1 to 0.
Teacher's Note: Write a short, clear answer.
30. If the height and the shadow of a man are equal, find the angle of elevation of the Sun.
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Answer: tan θ = height/shadow = 1, so θ = 45°.
Teacher's Note: Remember this for your exam.
31. An observer 1.5 m tall is 20.5 m from a tower 22 m high. Find the angle of elevation of the top of the tower from his eye.
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Answer: The height of the tower above the eye is 22 − 1.5 = 20.5 m, and the distance is 20.5 m. So tan θ = 20.5/20.5 = 1 and θ = 45°.
Teacher's Note: Learn this point and write it in your own words.
32. A man on a tower sees a car moving towards him at uniform speed. The angle of depression changes from 30° to 45° in 12 minutes. How soon will the car reach the tower?
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Answer: Let the height be h. The distances are h√3 (at 30°) and h (at 45°). The car covers h(√3 − 1) in 12 minutes, so it covers h in 12/(√3 − 1) = 6(√3 + 1) ≈ 16.4 minutes, i.e. about 16 minutes 24 seconds.
Teacher's Note: Use the key term in your answer.
33. Prove that the tangents drawn from an external point to a circle are equal in length.
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Answer: Let PA and PB be tangents from P to a circle with centre O. OA ⊥ PA and OB ⊥ PB, OA = OB (radii) and OP is common. So △OAP ≅ △OBP by RHS, and PA = PB.
Teacher's Note: Give an example if the question allows.
34. From a point Q, 13 cm from the centre of a circle, the tangent PQ is 12 cm. Find the radius.
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Answer: The radius is perpendicular to the tangent, so r² = 13² − 12² = 25 and r = 5 cm.
Teacher's Note: Write a short, clear answer.
Circles, Areas and Volumes
35. △ABC circumscribes a circle of radius 4 cm; BD = 8 cm and DC = 6 cm, where D is the point of contact on BC. Find AB and AC.
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Answer: Let the other tangent lengths from A be x. Then AB = x + 8, AC = x + 6, BC = 14, and s = x + 14. Area = r·s = 4(x + 14), and by Heron's formula area = √(s·x·8·6). So 16(x + 14)² = 48x(x + 14), hence x + 14 = 3x and x = 7. AB = 15 cm and AC = 13 cm.
Teacher's Note: Remember this for your exam.
39. If the area and circumference of a circle are numerically equal, find the diameter.
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Answer: πr² = 2πr, so r = 2 and the diameter is 4 units.
Teacher's Note: Learn this point and write it in your own words.
40. A rectangle ABCD has AB = 20 cm and BC = 14 cm. A semicircle with BC as diameter is cut off. Find the area of the remaining part.
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Answer: Rectangle = 20 × 14 = 280 cm². Semicircle = ½ × (22/7) × 7² = 77 cm². Remaining area = 280 − 77 = 203 cm².
Teacher's Note: Use the key term in your answer.
41. Find the area of the minor segment of a circle of radius 14 cm when the central angle is 60°. Also find the area of the major segment.
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Answer: Sector = (60/360) × (22/7) × 14² = 308/3 ≈ 102.67 cm². Triangle = (√3/4) × 14² = 49√3 ≈ 84.87 cm². Minor segment ≈ 102.67 − 84.87 = 17.80 cm². Circle = 616 cm², so the major segment ≈ 616 − 17.80 = 598.2 cm².
Teacher's Note: Give an example if the question allows.
42. Find the difference of the areas of a sector of angle 120° and its major sector of a circle of radius 21 cm.
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Answer: The major sector has angle 240°. Difference = (240 − 120)/360 × (22/7) × 21² = (1/3) × 1386 = 462 cm².
Teacher's Note: Write a short, clear answer.
43. A cylindrical tub of diameter 12 cm and height 15 cm full of ice-cream is divided among 10 children in equal cones with hemispherical tops. The height of the cone is twice the base diameter. Find the diameter of the cone.
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Answer: Tub volume = π × 6² × 15 = 540π. For a cone of radius r (height 4r) with a hemisphere: volume = (1/3)πr²(4r) + (2/3)πr³ = 2πr³. For 10 cones: 20πr³ = 540π, so r³ = 27, r = 3 cm. The diameter is 6 cm.
Teacher's Note: Remember this for your exam.
44. Find the mass of a silver cone with base diameter 14 cm and height 51 cm. The density of silver is 10 g/cm³.
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Answer: Volume = (1/3) × (22/7) × 7² × 51 = 2618 cm³. Mass = 2618 × 10 = 26,180 g = 26.18 kg.
Teacher's Note: Learn this point and write it in your own words.
45. Gayatri placed 4 cubes, each of edge 20 cm, one above another. Find the surface area of the resulting cuboid.
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Answer: The cuboid is 20 cm × 20 cm × 80 cm. Surface area = 2(20×20 + 20×80 + 20×80) = 2(400 + 1600 + 1600) = 7200 cm².
Teacher's Note: Use the key term in your answer.
47. Find the mode if the mean and median are 10.5 and 9.6.
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Answer: Mode = 3 × median − 2 × mean = 28.8 − 21 = 7.8.
Teacher's Note: Give an example if the question allows.
49. A box has 20 cards numbered 1 to 20. Find the probability that the number drawn is divisible by 2 or 3.
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Answer: Divisible by 2: 10 numbers; by 3: 6 numbers; by both (6): 3 numbers. So 10 + 6 − 3 = 13 favourable outcomes. P = 13/20.
Teacher's Note: Write a short, clear answer.
50. A number x is chosen from 1, 2, 3, 4 and y from 1, 4, 9, 16. Find the probability that xy < 16.
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Answer: There are 16 pairs. xy < 16 for: x = 1 with y = 1, 4, 9 (3); x = 2 with y = 1, 4 (2); x = 3 with y = 1, 4 (2); x = 4 with y = 1 (1). That makes 8 pairs, so P = 8/16 = 1/2.
Teacher's Note: Remember this for your exam.
Statistics and Probability
MCQs
45. Answers with working
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Answer: Q2. Zeroes of (x + 5)(x − 7): (ii) −5, 7.
Q3. Sum of ages 45, and 3(F − S) = 45, so F − S = 15 and F = 30: (i) 30 years.
Q4. DE = (2/3)BC, so the area of △ADE = (2/3)² × 81 = 36 cm²: (ii).
Q6. Mode = 3 median − 2 mean: 45 = 3M − 54, so the median is 33: (iv).
Q8. 21 + (n − 1)(−3) = −81, so n = 35: (ii).
Q9. XY = √(13² − 5²) = 12 cm: (ii).
Q12. Point (x, 0): (x − 7)² + 36 = (x + 3)² + 16, so x = 3. The point is (3, 0): (ii).
Q13. r² = 24² + 7² = 625, so r = 25 cm: (iv).
Q14. Circumference = 22 km / 10,000 = 2,200,000 cm / 10,000 = 220 cm: (ii).
Teacher's Note: Learn this point and write it in your own words.
Free study material for Mathematics
All topics Printable Assignments & Solutions for Class 10 Mathematics
Download Assignment: All topics (Class 10 Mathematics)
Access structured practice assignments for All topics designed in alignment with the latest CBSE curriculum for Class 10 Mathematics. These printable sets cover objective and descriptive problem types to support thorough revision.
Key Advantages of Solving All topics Assignments
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Effective Strategy for Class 10 Mathematics Assignments
- Initial Reading: Begin by reading the NCERT book for Class 10 Mathematics to build a baseline understanding.
- Independent Testing: Attempt assignment questions unassisted, then verify work using provided answer keys.
- Supplementary Aids: Leverage revision notes and worksheets whenever you encounter difficult topics.
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