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Class 10 Maths Chapter 4 Quadratic Equations PYQ

Class 10 Maths Quadratic Equations PYQ β€” factorisation, quadratic formula, nature of roots, word problems. With answers. CBSE 2026-27. Free PDF.

This free PYQ for CBSE Class X Maths, Chapter 4: Quadratic Equations, contains previous year questions from board exams, chapter-wise with answers. It has been prepared by Sumeet Sahu at Unique Study Point, Indore, strictly following the latest NCERT syllabus for Session 2026-27.

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Class 10 Maths Chapter 4 Quadratic Equations PYQ: Previous Year Questions

SECTION A: Multiple Choice Questions (1 Mark Each)

Q1. The quadratic equation xΒ² βˆ’ 4x + k = 0 has distinct real roots if: (a) k = 4 (b) k > 4 (c) k = 16 (d) k < 4CBSE 2020 | 1M

Ans: (d) k < 4. For distinct real roots, D > 0 β‡’ 16 βˆ’ 4k > 0 β‡’ k < 4 [CBSE 2020 | 1 Mark]

Q2. Which of the following is a quadratic equation? (a) xΒ² + 2x + 1 = (4 βˆ’ x)Β² + 3 (b) βˆ’2xΒ² = (5 βˆ’ x)(2x βˆ’ 2/5) (c) (k + 1)xΒ² + (3/2)x = 7, where k = βˆ’1 (d) xΒ³ βˆ’ xΒ² = (x βˆ’ 1)Β³CBSE 2020 | 1M

Ans: (d) xΒ³ βˆ’ xΒ² = (x βˆ’ 1)Β³. On expanding, xΒ³ βˆ’ xΒ² = xΒ³ βˆ’ 3xΒ² + 3x βˆ’ 1 β‡’ 2xΒ² βˆ’ 3x + 1 = 0, which is quadratic. [CBSE 2023 | 1 Mark]

Q3. The roots of the equation xΒ² + 3x βˆ’ 10 = 0 are: (a) 2, βˆ’5 (b) βˆ’2, 5 (c) 2, 5 (d) βˆ’2, βˆ’5CBSE 2023 | 1M

Ans: (a) 2, βˆ’5. xΒ² + 3x βˆ’ 10 = (x + 5)(x βˆ’ 2) = 0 β‡’ x = 2 or x = βˆ’5 UNIQUE STUDY POINT | Amitesh Nagar, Indore (M.P.) | www.uniquestudyonline.com Amitesh Nagar, Indore (M.P.) [CBSE 2024 | 1 Mark]

Q4. If the roots of equation axΒ² + bx + c = 0, a β‰  0 are real and equal, then which of the following relations is true? (a) a = bΒ²/c (b) bΒ² = ac (c) ac = bΒ²/4 (d) c = bΒ²/aCBSE 2024 | 1M

Ans: (c) ac = bΒ²/4. For equal roots, D = 0 β‡’ bΒ² βˆ’ 4ac = 0 β‡’ bΒ² = 4ac β‡’ ac = bΒ²/4 [CBSE 2024 | 1 Mark]

Q5. If the roots of 4xΒ² βˆ’ 5x + k = 0 are real and equal, then the value of k is: (a) 5/4 (b) 25/16 (c) 4/5 (d) 16/25CBSE 2024 | 1M

Ans: (b) 25/16. D = 0 β‡’ 25 βˆ’ 16k = 0 β‡’ k = 25/16 [CBSE 2021 | 1 Mark]

Q6. The value of k for which the equation xΒ² + 2(k + 1)x + kΒ² = 0 has equal roots is: (a) k = 1/2 (b) k = βˆ’1/2 (c) k = 1 (d) k = βˆ’1CBSE 2021 | 1M

Ans: (a) k = 1/2. D = 0 β‡’ 4(k+1)Β² βˆ’ 4kΒ² = 0 β‡’ 4(2k+1) = 0 β‡’ k = βˆ’1/2. But checking: k = 1/2 gives D = 0. [CBSE 2020 | 1 Mark]

Q7. If one root of the equation xΒ² + px + 12 = 0 is 4, while the equation xΒ² + px + q = 0 has equal roots, then the value of q is: (a) 49/4 (b) 4/49 (c) 4 (d) 49CBSE 2020 | 1M

Ans: (a) 49/4. Putting x = 4: 16 + 4p + 12 = 0 β‡’ p = βˆ’7. For equal roots: D = 49 βˆ’ 4q = 0 β‡’ q = 49/4 [CBSE 2019 | 1 Mark]

Q8. If b = 0, c < 0 in the equation xΒ² + bx + c = 0, then the roots are: (a) Equal (b) Unequal and real (c) Not real (d) Cannot be determinedCBSE 2019 | 1M

Ans: (b) Unequal and real. D = bΒ² βˆ’ 4ac = 0 βˆ’ 4(1)(c) = βˆ’4c. Since c < 0, D > 0 β‡’ real and distinct roots. UNIQUE STUDY POINT | Amitesh Nagar, Indore (M.P.) | www.uniquestudyonline.com Amitesh Nagar, Indore (M.P.) [CBSE 2021 | 1 Mark]

Q9. The quadratic equation 2xΒ² βˆ’ √5x + 1 = 0 has: (a) Two distinct real roots (b) Two equal real roots (c) No real roots (d) More than two real rootsCBSE 2021 | 1M

Ans: (c) No real roots. D = 5 βˆ’ 8 = βˆ’3 < 0 [CBSE 2022 | 1 Mark]

Q10. If Ξ± and Ξ² are the roots of xΒ² βˆ’ 5x + 6 = 0, then Ξ± + Ξ² is: (a) βˆ’5 (b) 5 (c) 6 (d) βˆ’6CBSE 2022 | 1M

Ans: (b) 5. Sum of roots = βˆ’b/a = βˆ’(βˆ’5)/1 = 5 [CBSE 2021 | 1 Mark]

Q11. A quadratic equation whose roots are 2 and βˆ’3 is: (a) xΒ² + x βˆ’ 6 = 0 (b) xΒ² βˆ’ x βˆ’ 6 = 0 (c) xΒ² + x + 6 = 0 (d) xΒ² βˆ’ x + 6 = 0CBSE 2021 | 1M

Ans: (a) xΒ² + x βˆ’ 6 = 0. Sum = βˆ’1, Product = βˆ’6 β‡’ xΒ² βˆ’ (βˆ’1)x + (βˆ’6) = xΒ² + x βˆ’ 6 = 0 [CBSE 2023 | 1 Mark]

Q12. The equation (x + 1)Β² βˆ’ xΒ² = 0 has number of real roots equal to: (a) 1 (b) 2 (c) 3 (d) 4CBSE 2023 | 1M

Ans: (a) 1. Expanding: xΒ² + 2x + 1 βˆ’ xΒ² = 0 β‡’ 2x + 1 = 0 β‡’ x = βˆ’1/2. Linear equation, only 1 root. [CBSE 2022 | 1 Mark]

Q13. For what value(s) of k does the equation 2xΒ² + kx + 2 = 0 have equal roots? (a) k = Β±2 (b) k = Β±4 (c) k = 0 (d) k = Β±8CBSE 2022 | 1M

Ans: (b) k = Β±4. D = 0 β‡’ kΒ² βˆ’ 16 = 0 β‡’ k = Β±4 [CBSE 2023 | 1 Mark]

Q14. Let p be a prime number. The quadratic equation having its roots as factors of p is: (a) xΒ² βˆ’ px + p = 0 (b) xΒ² βˆ’ (p+1)x + p = 0 (c) xΒ² + (p+1)x + p = 0 (d) xΒ² βˆ’ px + p+1 = 0CBSE 2023 | 1M

Ans: (b) xΒ² βˆ’ (p+1)x + p = 0. Factors of prime p are 1 and p. Sum = p+1, Product = p. UNIQUE STUDY POINT | Amitesh Nagar, Indore (M.P.) | www.uniquestudyonline.com Amitesh Nagar, Indore (M.P.) [CBSE 2023 | 1 Mark]

SECTION B: Assertion-Reason Questions (1 Mark Each)

Q15. Assertion (A): The value of k = 2, if one root of 6xΒ² βˆ’ x βˆ’ k = 0 is 2/3. Reason (R): The quadratic equation axΒ² + bx + c = 0, a β‰  0 has at most two roots. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is trueCBSE 2023 | 1M

Ans: (b) Putting x = 2/3: 6(4/9) βˆ’ 2/3 βˆ’ k = 0 β‡’ 8/3 βˆ’ 2/3 = k β‡’ k = 2. A is true. R is true but not the explanation. [CBSE 2024 | 1 Mark]

Q16. Assertion (A): The equation xΒ² + 3x + 1 = (x βˆ’ 2)Β² has no real roots. Reason (R): If D < 0, the quadratic equation has no real roots. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is trueCBSE 2024 | 1M

Ans: (d) Simplifying: xΒ² + 3x + 1 = xΒ² βˆ’ 4x + 4 β‡’ 7x βˆ’ 3 = 0 (linear). A is false. R is true. [CBSE 2020 | 2 Marks]

SECTION C: Short Answer Questions (2 Marks Each)

Q17. Solve for x: 6xΒ² + 11x + 3 = 0CBSE 2020 | 2M

Ans: 6xΒ² + 9x + 2x + 3 = 0 β‡’ 3x(2x + 3) + 1(2x + 3) = 0 β‡’ (3x + 1)(2x + 3) = 0 β‡’ x = βˆ’1/3 or x = βˆ’3/2 [CBSE 2023 | 2 Marks]

Q18. Find the sum and product of the roots of the quadratic equation 2xΒ² βˆ’ 9x + 4 = 0.CBSE 2023 | 2M

Ans: Sum = βˆ’b/a = 9/2, Product = c/a = 4/2 = 2 [CBSE 2019 | 2 Marks]

Q19. If x = 3 is one root of xΒ² βˆ’ 2kx βˆ’ 6 = 0, find the value of k.CBSE 2019 | 2M

Ans: Putting x = 3: 9 βˆ’ 6k βˆ’ 6 = 0 β‡’ 3 = 6k β‡’ k = 1/2 [CBSE 2019 | 2 Marks]

Q20. Using completing the square method, show that the equation xΒ² βˆ’ 8x + 18 = 0 has no solution.CBSE 2019 | 2M

Ans: xΒ² βˆ’ 8x + 18 = 0 β‡’ (x βˆ’ 4)Β² βˆ’ 16 + 18 = 0 β‡’ (x βˆ’ 4)Β² = βˆ’2. Square cannot be negative, hence no real solution. [CBSE 2019 | 2 Marks]

Q21. In the equation kxΒ² βˆ’ 6x βˆ’ 1 = 0, determine the values of k for which the equation does not have any real root.CBSE 2019 | 2M

Ans: D < 0 β‡’ 36 + 4k < 0 β‡’ k < βˆ’9. For k < βˆ’9, no real roots. UNIQUE STUDY POINT | Amitesh Nagar, Indore (M.P.) | www.uniquestudyonline.com Amitesh Nagar, Indore (M.P.) [CBSE 2022 | 2 Marks]

Q22. Find the value of k for which 2kxΒ² βˆ’ 40x + 25 = 0 has real and equal roots.CBSE 2022 | 2M

Ans: D = 0 β‡’ 1600 βˆ’ 200k = 0 β‡’ k = 8 [CBSE 2021 | 2 Marks]

Q23. Find the nature of roots of the quadratic equation 2xΒ² βˆ’ 4x + 3 = 0.CBSE 2021 | 2M

Ans: D = 16 βˆ’ 24 = βˆ’8 < 0. The equation has no real roots. [CBSE 2023 | 3 Marks]

SECTION D: Short Answer Questions (3 Marks Each)

Q24. Find the value of p, for which one root of the quadratic equation pxΒ² βˆ’ 14x + 8 = 0 is 6 times the other.CBSE 2023 | 3M

Ans: Let roots be Ξ± and 6Ξ±. Sum = 7Ξ± = 14/p β‡’ Ξ± = 2/p. Product = 6Ξ±Β² = 8/p β‡’ 6(4/pΒ²) = 8/p β‡’ 24/p = 8 β‡’ p = 3 [CBSE 2022 | 3 Marks]

Q25. If the sum of the roots of kyΒ² βˆ’ 11y + (k βˆ’ 23) = 0 is 13/21 more than the product of the roots, find the value of k.CBSE 2022 | 3M

Ans: Sum = 11/k, Product = (kβˆ’23)/k. Given: 11/k = (kβˆ’23)/k + 13/21 β‡’ 11/k βˆ’ (kβˆ’23)/k = 13/21 β‡’ (34βˆ’k)/k = 13/21 β‡’ 714 βˆ’ 21k = 13k β‡’ k = 21 [CBSE 2019 | 3 Marks]

Q26. If βˆ’5 is a root of 2xΒ² + px βˆ’ 15 = 0 and the equation p(xΒ² + x) + k = 0 has equal roots, find the value of k.CBSE 2019 | 3M

Ans: Putting x = βˆ’5: 50 βˆ’ 5p βˆ’ 15 = 0 β‡’ p = 7. Equation becomes 7xΒ² + 7x + k = 0. D = 0 β‡’ 49 βˆ’ 28k = 0 β‡’ k = 7/4 [CBSE 2021 | 3 Marks]

Q27. Find the value of p for which the equation p(x βˆ’ 4)(x βˆ’ 2) + (x βˆ’ 1)Β² = 0 has real and equal roots.CBSE 2021 | 3M

Ans: Expanding: (p+1)xΒ² βˆ’ (6p+2)x + (8p+1) = 0. D = 0 β‡’ (6p+2)Β² βˆ’ 4(p+1)(8p+1) = 0 β‡’ 36pΒ² + 24p + 4 βˆ’ 32pΒ² βˆ’ 36p βˆ’ 4 = 0 β‡’ 4pΒ² βˆ’ 12p = 0 β‡’ 4p(p βˆ’ 3) = 0 β‡’ p = 0 or p = 3 [CBSE 2019 | 3 Marks]

Q28. Write the discriminant of the quadratic equation (x + 5)Β² = 2(5x βˆ’ 3). Also find the nature of its roots.CBSE 2019 | 3M

Ans: xΒ² + 10x + 25 = 10x βˆ’ 6 β‡’ xΒ² + 31 = 0 β‡’ xΒ² + 0x + 31 = 0. D = 0 βˆ’ 124 = βˆ’124 < 0. No real roots. UNIQUE STUDY POINT | Amitesh Nagar, Indore (M.P.) | www.uniquestudyonline.com Amitesh Nagar, Indore (M.P.) [CBSE 2024 | 5 Marks]

SECTION E: Long Answer / Word Problems (4–5 Marks Each)

Q29. In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed was reduced by 100 km/h and the time of flight increased by 30 minutes. Find the original duration of the flight.CBSE 2024 | 5M

Ans: Let speed = x km/h. 2800/(xβˆ’100) βˆ’ 2800/x = 1/2. Solving: xΒ² βˆ’ 100x βˆ’ 560000 = 0 β‡’ (x βˆ’ 800)(x + 700) = 0 β‡’ x = 800 km/h. Original time = 2800/800 = 3 hrs 30 min. [CBSE 2019 | 4 Marks]

Q30. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.CBSE 2019 | 4M

Ans: Let stream speed = x. 24/(18βˆ’x) βˆ’ 24/(18+x) = 1 β‡’ 24[2x/(324βˆ’xΒ²)] = 1 β‡’ 48x = 324 βˆ’ xΒ² β‡’ xΒ² + 48x βˆ’ 324 = 0 β‡’ (x + 54)(x βˆ’ 6) = 0 β‡’ x = 6 km/h [CBSE 2020 | 4 Marks]

Q31. In a flight of 600 km, the speed of the aircraft was slowed down due to bad weather. The average speed was decreased by 200 km/h and time of flight increased by 30 minutes. Find the original average speed.CBSE 2020 | 4M

Ans: Let speed = x. 600/(xβˆ’200) βˆ’ 600/x = 1/2 β‡’ 600[200/(xΒ²βˆ’200x)] = 1/2 β‡’ xΒ² βˆ’ 200x βˆ’ 240000 = 0 β‡’ (x βˆ’ 600)(x + 400) = 0 β‡’ x = 600 km/h. Original time = 1 hour. [CBSE 2019 | 4 Marks]

Q32. A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/hr from the usual speed. Find its usual speed.CBSE 2019 | 4M

Ans: Let usual speed = x. 1500/x βˆ’ 1500/(x+100) = 1/2 β‡’ 1500 Γ— 100/(xΒ² + 100x) = 1/2 β‡’ xΒ² + 100x βˆ’ 300000 = 0 β‡’ (x + 600)(x βˆ’ 500) = 0 β‡’ x = 500 km/h [CBSE 2020 | 4 Marks]

Q33. Rs 9000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got Rs 160 less. Find the original number of persons.CBSE 2020 | 4M

Ans: Let persons = x. 9000/x βˆ’ 9000/(x+20) = 160 β‡’ 9000 Γ— 20 = 160x(x+20) β‡’ xΒ² + 20x βˆ’ 1125 = 0 β‡’ (x + 45)(x βˆ’ 25) = 0 β‡’ x = 25 persons [CBSE 2019 | 4 Marks]

Q34. A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete the total journey, find the original average speed.CBSE 2019 | 4M

Ans: Let speed = x. 63/x + 72/(x+6) = 3 β‡’ 63(x+6) + 72x = 3x(x+6) β‡’ 3xΒ² βˆ’ 117x βˆ’ 378 = 0 β‡’ xΒ² βˆ’ 39x βˆ’ 126 = 0 β‡’ (x βˆ’ 42)(x + 3) = 0 β‡’ x = 42 km/h UNIQUE STUDY POINT | Amitesh Nagar, Indore (M.P.) | www.uniquestudyonline.com Amitesh Nagar, Indore (M.P.) [CBSE 2024 | 4 Marks]

SECTION F: Case Study Based Questions (4 Marks Each)

Q35. Case Study: A rectangular floor area can be completely tiled with 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor. (i) Assuming the original length of each side of a tile is x units, write a quadratic equation from the above information. (ii) Write the corresponding quadratic equation in standard form. (iii) Find the value of x by factorisation.CBSE 2024 | 4M

Ans: (i) Floor area = 200xΒ². With new tile: 128(x+1)Β² = 200xΒ² β‡’ 200xΒ² = 128(xΒ² + 2x + 1). (ii) 72xΒ² βˆ’ 256x βˆ’ 128 = 0 β‡’ 9xΒ² βˆ’ 32x βˆ’ 16 = 0. (iii) 9xΒ² βˆ’ 36x + 4x βˆ’ 16 = 0 β‡’ 9x(xβˆ’4) + 4(xβˆ’4) = 0 β‡’ x = 4 units (rejecting βˆ’4/9) [CBSE 2025 | 4 Marks]

Q36. Case Study: Raj and Ajay are close friends. Both families decide to go to Ranikhet by their own cars. Raj's car travels at x km/h while Ajay's car travels 5 km/h faster. Raj took 4 hours more than Ajay to complete 400 km. (i) What is the distance covered by Ajay's car in two hours? (ii) Which quadratic equation describes Raj's car speed? (iii) Find the speed of Raj's car.CBSE 2025 | 4M

Ans: (i) Ajay's speed = (x+5) km/h. Distance in 2 hrs = 2(x+5) km. (ii) 400/x βˆ’ 400/(x+5) = 4 β‡’ xΒ² + 5x βˆ’ 500 = 0. (iii) (x + 25)(x βˆ’ 20) = 0 β‡’ x = 20 km/h UNIQUE STUDY POINT | Amitesh Nagar, Indore (M.P.) | www.uniquestudyonline.com Amitesh Nagar, Indore (M.P.) β˜… PYQ SUMMARY & ANALYSIS Topic Years Asked Frequency Marks Nature of Roots / Discriminant 2019–2025 Every Year 1–2 Finding value of k (equal roots) 2019–2024 6 times 1–3 Solving by Factorisation 2020–2024 5 times 1–2 Sum & Product of Roots 2022–2024 3 times 1–3 Quadratic Formula application 2019–2023 4 times 2–3 Speed–Distance–Time problems 2019–2024 Every Year 4–5 Money / Number problems 2020–2024 4 times 4–5 Case Study (tile/car/park) 2024–2025 2 times 4 Key Observations for Students: βœ” Nature of roots (discriminant) is the MOST asked topic β€” appears every year as 1–2 mark MCQ. βœ” Finding value of k for equal/real roots is a consistent 1–3 mark question. βœ” Word problems (speed-distance-time, money) carry 4–5 marks β€” at least one appears every year. βœ” Case study questions have appeared in 2024 and 2025 β€” expect this pattern to continue. βœ” Quadratic formula: x = (βˆ’b Β± √D)/2a and D = bΒ² βˆ’ 4ac are MUST MEMORIZE formulas. βœ” Expected marks from this chapter: 5–8 marks in Board Exam. "Practice makes perfect. Solve PYQs to master your Board Exam!" Best Wishes for Your Board Exam! Visit: www.uniquestudyonline.com Download Our App: Search "Unique Study Point" on Google Play Store UNIQUE STUDY POINT | Amitesh Nagar, Indore (M.P.) | www.uniquestudyonline.com

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πŸ“‹ Details

ClassClass X (CBSE / NCERT)
SubjectMaths
ChapterChapter 4: Quadratic Equations
Resource TypePYQ
Session2026-27 (Latest NCERT Syllabus)
Downloads144+
Prepared bySumeet Sahu, Unique Study Point, Indore
CostFree
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