Class 10 Maths Chapter 4 Quadratic Equations Practice Paper 2

PRACTICE PAPER 02 - CHAPTER 04 QUADRATIC EQUATIONS (2025-26) SUBJECT: MATHEMATICS MAX. MARKS: 40 CLASS: X DURATION: 1½ hrs

General Instructions:

1. All questions are compulsory.

2. This question paper contains 20 questions divided into five Sections A, B, C, D and E.

3. Section A: 10 MCQs of 1 mark each. Section B: 4 questions of 2 marks each. Section C: 3 questions of 3 marks each. Section D: 1 question of 5 marks. Section E: 2 Case Studies of 4 marks each.

4. There is no overall choice.

5. Use of Calculators is not permitted. SECTION – A (Questions 1 to 10 carry 1 mark each)

1. The discriminant of the quadratic equation 2x² - 7x + 3 = 0 is:
(a) 25
(b) 49
(c) 73
(d) 37

2. For what value of k does the equation x² + 4x + k = 0 have real and equal roots?
(a) 2
(b) 4
(c) 8
(d) 16

3. The roots of the equation x² - 6x + 9 = 0 are:
(a) 3, 3
(b) 3, -3
(c) 6, 9
(d) -3, -3

4. The equation x² + 2x + 3 = 0 has:
(a) Two distinct real roots
(b) Equal real roots
(c) No real roots
(d) One real root

5. Using the quadratic formula, the roots of x² + 3x - 4 = 0 are:
(a) 1, -4
(b) -1, 4
(c) 2, -2
(d) 3, -4

6. The equation x² - 4x + 13 = 0 can be written in the form (x - 2)² + k = 0. What is the value of k?
(a) 5
(b) 9
(c) 13
(d) 17

7. If the quadratic equation px² - 2√5px + 15 = 0 has two equal roots, then the value of p is:
(a) ±3
(b) 3
(c) -3
(d) 0, 3

8. The nature of roots of the equation 2x² - 3x + 5 = 0 is:
(a) Real and distinct
(b) Real and equal
(c) Not real
(d) Cannot be determined

9. Assertion
(a) : The equation x² + x + 1 = 0 has no real roots. Reason (R): For a quadratic equation to have no real roots, its discriminant must be negative.
(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 true

10. Assertion
(a) : The quadratic formula is x = [-b ± √(b² - 4ac)] / 2a. Reason (R): This formula can be used to solve any quadratic equation.
(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 true SECTION – B (Questions 11 to 14 carry 2 marks each)

11. Find the discriminant of the equation 3x² - 5x + 2 = 0 and determine the nature of its roots.

12. For what value of k will the equation kx² + 6x + 1 = 0 have equal roots?

13. Solve the equation x² + 5x + 6 = 0 using the quadratic formula.

14. Express x² + 8x + 10 = 0 in the form (x + a)² = b and hence solve it. SECTION – C (Questions 15 to 17 carry 3 marks each)

15. Solve the quadratic equation x² - 4x - 5 = 0 by completing the square method.

16. Find the values of k for which the quadratic equation (k + 1)x² - 6(k + 1)x + 3(k + 9) = 0 has equal roots.

17. A quadratic equation has roots that are reciprocals of the roots of 2x² - 5x + 3 = 0. Form the new quadratic equation. SECTION – D (Question 18 carries 5 marks)

18. A rectangular park is 50 m long and 40 m wide. A path of uniform width is constructed around the outside of the park. If the area of the path is 1056 m², find the width of the path. SECTION – E (Questions 19 to 20 carry 4 marks each)

19. A shopkeeper buys a number of books for ₹1800. If he had bought 15 more books for the same amount, each book would have cost him ₹20 less.
(a) Taking the original number of books as x, form a quadratic equation. (2 marks)
(b) Find the original number of books bought. (1 mark)
(c) What was the original price per book? (1 mark)

20. 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.
(a) Let the speed of the stream be x km/h. Form a quadratic equation in x. (2 marks)
(b) Find the speed of the stream. (1 mark)
(c) Find the time taken to go upstream. (1 mark) DETAILED ANSWER KEY

SECTION A - ANSWERS

1. Answer:
(a) 25

Solution: D = b² - 4ac = (-7)² - 4(2)(3) = 49 - 24 = 25

2. Answer:
(b) 4

Solution: For equal roots: D = 0

16 - 4(1)(k) = 0 → 16 = 4k → k = 4

3. Answer:
(a) 3, 3

Solution: x² - 6x + 9 = (x - 3)² = 0 → x = 3 (repeated root)

4. Answer:
(c) No real roots

Solution: D = 4 - 4(1)(3) = 4 - 12 = -8 < 0 (No real roots)

5. Answer:
(a) 1, -4

Solution: x = [-3 ± √(9 + 16)]/2 = [-3 ± 5]/2

x = 1 or x = -4

6. Answer:
(b) 9

Solution: x² - 4x + 13 = (x - 2)² - 4 + 13 = (x - 2)² + 9

So k = 9

7. Answer:
(b) 3

Solution: For equal roots: D = 0

(-2√5p)² - 4(p)(15) = 0 20p² - 60p = 0 → 20p(p - 3) = 0 p = 3 (p ≠ 0 for quadratic)

8. Answer:
(c) Not real

Solution: D = 9 - 4(2)(5) = 9 - 40 = -31 < 0

9. Answer:
(a) Both A and R are true and R is the correct explanation of A

Solution: D = 1 - 4(1)(1) = -3 < 0 ✓ (No real roots)

R correctly explains why A is true

10. Answer:
(a) Both A and R are true and R is the correct explanation of A

Solution: The quadratic formula is correct and can solve any quadratic equation

SECTION B - ANSWERS

11. Solution: D = b² - 4ac = 25 - 4(3)(2) = 25 - 24 = 1 Since D > 0, the equation has two distinct real roots. Answer: Discriminant = 1, Nature: Two distinct real roots

12. Solution: For equal roots: D = 0 36 - 4(k)(1) = 0 36 = 4k Answer: k = 9

13. Solution: x = [-5 ± √(25 - 24)]/2 = [-5 ± 1]/2 x = -2 or x = -3 Answer: x = -2, -3

14. Solution: x² + 8x + 10 = 0 x² + 8x = -10 x² + 8x + 16 = -10 + 16 (x + 4)² = 6 x + 4 = ±√6 Answer: x = -4 ± √6

SECTION C - ANSWERS

15. Solution: x² - 4x - 5 = 0 x² - 4x = 5 x² - 4x + 4 = 5 + 4 (x - 2)² = 9 x - 2 = ±3 Answer: x = 5 or x = -1

16. Solution: For equal roots: D = 0 36(k + 1)² - 4(k + 1) × 3(k + 9) = 0 36(k + 1)² = 12(k + 1)(k + 9) 3(k + 1) = k + 9 3k + 3 = k + 9 2k = 6 Answer: k = 3

17. Solution: 2x² - 5x + 3 = 0 has roots α, β New equation has roots 1/α, 1/β Sum = 1/α + 1/β = (α + β)/αβ = (5/2)/(3/2) = 5/3 Product = (1/α)(1/β) = 1/αβ = 1/(3/2) = 2/3 New equation: x² - (5/3)x + 2/3 = 0 Answer: 3x² - 5x + 2 = 0

SECTION D - ANSWER

18. Solution: Let width of path = x m New dimensions: (50 + 2x) × (40 + 2x) Area of path = New area - Original area (50 + 2x)(40 + 2x) - 50 × 40 = 1056 2000 + 100x + 80x + 4x² - 2000 = 1056 4x² + 180x = 1056 x² + 45x - 264 = 0 (x + 57)(x - 4.64) ≈ 0 or using formula: x = [-45 ± √(2025 + 1056)]/2 = [-45 ± √3081]/2 x ≈ 4.64 m (taking positive value) Answer: Width of path ≈ 4.64 m or 33/7 m

SECTION E - ANSWERS

19. Solution:
(a) Original price per book = 1800/x New price per book = 1800/(x + 15) 1800/x - 1800/(x + 15) = 20 1800(x + 15) - 1800x = 20x(x + 15) 27000 = 20x² + 300x Equation: x² + 15x - 1350 = 0
(b) (x + 45)(x - 30) = 0 x = 30 (taking positive value) Number of books = 30
(c) Original price = 1800/30 = ₹60 Price per book = ₹60

20. Solution:
(a) Upstream speed = (18 - x) km/h Downstream speed = (18 + x) km/h Time upstream - Time downstream = 1 24/(18 - x) - 24/(18 + x) = 1 24(18 + x) - 24(18 - x) = (18 - x)(18 + x) 48x = 324 - x² Equation: x² + 48x - 324 = 0
(b) Using formula: x = [-48 ± √(2304 + 1296)]/2 x = [-48 ± 60]/2 = 6 km/h (taking positive) Speed of stream = 6 km/h
(c) Time upstream = 24/12 = 2 hours Time = 2 hours