CBSE Class 10 Mathematics Quadratic Equations MCQs Set 28

Mathematics Objective Questions and Answers: Chapter 04 Quadratic Equations

Access targeted multiple-choice questions for Chapter 04 Quadratic Equations designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Download Chapter 04 Quadratic Equations MCQs with Answers

Navigate directly to the 50 objective questions for Chapter 04 Quadratic Equations using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: If α and β are roots of \( x^2 - 2x + 3 = 0 \), the equation whose roots are \( \frac{\alpha - 1}{\alpha + 1} \) and \( \frac{\beta - 1}{\beta + 1} \) is:
(a) \( 3x^2 - 2x + 1 = 0 \)
(b) \( 3x^2 + 2x + 1 = 0 \)
(c) \( x^2 - 2x + 3 = 0 \)
(d) \( 3x^2 - 2x - 1 = 0 \)
Show Answer & Explanation

Answer: (a) \( 3x^2 - 2x + 1 = 0 \)

Explanation:
1. Let y = (x − 1)/(x + 1); then x = (1 + y)/(1 − y).
2. Substituting in x² − 2x + 3 = 0: (1 + y)² − 2(1 + y)(1 − y) + 3(1 − y)² = 0.
3. Expanding: 6y² − 4y + 2 = 0, i.e. 3y² − 2y + 1 = 0.

Teacher's Note:
1. Find the sum and product of the new roots.
2. Use x² − (sum)x + product = 0.

Question 2: The roots of \( 4x^2 - 2(a^2 + b^2)x + a^2b^2 = 0 \) are:
(a) \( a^2,\ b^2 \)
(b) \( \frac{a}{2},\ \frac{b}{2} \)
(c) \( 2a^2,\ 2b^2 \)
(d) \( \frac{a^2}{2},\ \frac{b^2}{2} \)
Show Answer & Explanation

Answer: (d) \( \frac{a^2}{2},\ \frac{b^2}{2} \)

Explanation:
1. Factorise as (2x − a²)(2x − b²) = 0: the product is 4x² − 2(a² + b²)x + a²b². So x = a²/2 or b²/2.

Teacher's Note:
1. Try (2x − a²)(2x − b²).
2. Expand to check the middle term.

Question 3: The roots of \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \) are:
(a) \( a^2 + b^2,\ a^2 - b^2 \)
(b) \( \frac{a^2 + b^2}{4},\ \frac{a^2 - b^2}{4} \)
(c) \( \frac{a^2 + b^2}{2},\ \frac{a^2 - b^2}{2} \)
(d) \( \frac{a + b}{2},\ \frac{a - b}{2} \)
Show Answer & Explanation

Answer: (c) \( \frac{a^2 + b^2}{2},\ \frac{a^2 - b^2}{2} \)

Explanation:
1. D = 16a⁴ − 16(a⁴ − b⁴) = 16b⁴. So x = (4a² ± 4b²)/8 = (a² ± b²)/2.

Teacher's Note:
1. Compute the discriminant first.
2. √(16b⁴) = 4b².

Question 4: The solutions of \( \left(\frac{2x}{x-5}\right)^2 + \frac{10x}{x-5} - 24 = 0 \) (x ≠ 5) are:
(a) x = −4 and x = −15
(b) x = 4 and x = 15
(c) x = 4 and x = −15
(d) x = 3 and x = 12
Show Answer & Explanation

Answer: (b) x = 4 and x = 15

Explanation:
1. u² + 5u − 24 = (u − 3)(u + 8) = 0.
2. 2x/(x − 5) = 3 gives x = 15.
3. 2x/(x − 5) = −8 gives 10x = 40, x = 4.

Teacher's Note:
1. Notice 10x/(x − 5) = 5u.
2. Solve the quadratic in u, then return to x.

Question 5: The solutions of \( \frac{1}{a+b+x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \) (a, b, x ≠ 0, x ≠ −(a+b)) are:
(a) \( x = -a,\ x = -b \)
(b) \( x = a,\ x = b \)
(c) \( x = a + b,\ x = -(a + b) \)
(d) \( x = -a,\ x = b \)
Show Answer & Explanation

Answer: (a) \( x = -a,\ x = -b \)

Explanation:
1. 1/(a+b+x) − 1/x = −(a+b)/(x(a+b+x)).
2. Right side: (a+b)/(ab), so x(a+b+x) = −ab.
3. x² + (a + b)x + ab = (x + a)(x + b) = 0.

Teacher's Note:
1. Subtract 1/x from both sides.
2. Divide out the common factor a + b.

Question 6: If 2 is a root of \( 3x^2 - kx - 2 = 0 \), the value of k and the other root are:
(a) \( k = -5,\ \text{other root } \frac{1}{3} \)
(b) \( k = 5,\ \text{other root } \frac{1}{3} \)
(c) \( k = 3,\ \text{other root } -1 \)
(d) \( k = 5,\ \text{other root } -\frac{1}{3} \)
Show Answer & Explanation

Answer: (d) \( k = 5,\ \text{other root } -\frac{1}{3} \)

Explanation:
1. 12 − 2k − 2 = 0 gives k = 5.
2. Product of roots = −2/3.
3. Other root = (−2/3) ÷ 2 = −1/3.

Teacher's Note:
1. Substitute x = 2.
2. Use the product of roots for the other root.

Question 7: For which values of k does \( 9x^2 + 8kx + 16 = 0 \) have equal roots?
(a) \( k = 3\ \text{only} \)
(b) \( k = \pm 4 \)
(c) \( k = \pm 3 \)
(d) \( k = \pm 9 \)
Show Answer & Explanation

Answer: (c) \( k = \pm 3 \)

Explanation:
1. D = 64k² − 4(9)(16) = 64k² − 576 = 0, so k² = 9 and k = ±3.

Teacher's Note:
1. Set the discriminant to zero.
2. Include both signs.

Question 8: The roots of \( 9x^2 - 9(a+b)x + 2a^2 + 5ab + 2b^2 = 0 \) are:
(a) \( \frac{a + 2b}{3},\ \frac{a - 2b}{3} \)
(b) \( \frac{2a + b}{3},\ \frac{a + 2b}{3} \)
(c) \( \frac{2a - b}{3},\ \frac{2b - a}{3} \)
(d) \( \frac{2a + b}{9},\ \frac{a + 2b}{9} \)
Show Answer & Explanation

Answer: (b) \( \frac{2a + b}{3},\ \frac{a + 2b}{3} \)

Explanation:
1. D = 81(a² + 2ab + b²) − 72a² − 180ab − 72b² = 9a² − 18ab + 9b² = 9(a − b)².
2. x = [9(a + b) ± 3(a − b)]/18.
3. x = (12a + 6b)/18 = (2a + b)/3 or x = (6a + 12b)/18 = (a + 2b)/3.

Teacher's Note:
1. Compute D and notice it is a perfect square.
2. Simplify each root separately.

Question 9: The roots of \( 9x^2 - 3(a+b)x + ab = 0 \) are:
(a) \( \frac{a}{3},\ \frac{b}{3} \)
(b) \( a,\ b \)
(c) \( \frac{a}{9},\ \frac{b}{9} \)
(d) \( -\frac{a}{3},\ -\frac{b}{3} \)
Show Answer & Explanation

Answer: (a) \( \frac{a}{3},\ \frac{b}{3} \)

Explanation:
1. (3x − a)(3x − b) = 9x² − 3(a+b)x + ab, so x = a/3 or b/3.

Teacher's Note:
1. Factorise as a product of two linear factors.
2. Verify by expansion.

Question 10: If \( (1 + m^2)x^2 + 2mcx + c^2 - a^2 = 0 \) has equal roots, then:
(a) \( c^2 = a^2(1 - m^2) \)
(b) \( a^2 = c^2(1 + m^2) \)
(c) \( c = a(1 + m) \)
(d) \( c^2 = a^2(1 + m^2) \)
Show Answer & Explanation

Answer: (d) \( c^2 = a^2(1 + m^2) \)

Explanation:
1. D = 4m²c² − 4(1 + m²)(c² − a²) = 4[a²(1 + m²) − c²]. For D = 0, c² = a²(1 + m²).

Teacher's Note:
1. Equal roots mean D = 0.
2. Expand (1 + m²)(c² − a²).

Question 11: If the roots of \( (a-b)x^2 + (b-c)x + (c-a) = 0 \) are equal, then:
(a) \( 2b = a + c \)
(b) \( a = b + c \)
(c) \( 2a = b + c \)
(d) \( 2c = a + b \)
Show Answer & Explanation

Answer: (c) \( 2a = b + c \)

Explanation:
1. D = (b − c)² − 4(a − b)(c − a).
2. Expand: b² − 2bc + c² − 4(ac − a² − bc + ab) = (2a − b − c)².
3. D = 0 gives 2a = b + c.

Teacher's Note:
1. Compute the discriminant and look for a perfect square.
2. Alternatively note x = 1 is a root.

Question 12: Two consecutive positive integers have squares summing to 85. They are:
(a) 5 and 6
(b) 6 and 7
(c) 7 and 8
(d) 8 and 9
Show Answer & Explanation

Answer: (b) 6 and 7

Explanation:
1. x² + (x + 1)² = 85 gives x² + x − 42 = 0, so x = 6 (positive). The numbers are 6 and 7: 36 + 49 = 85.

Teacher's Note:
1. Let the integers be x and x + 1.
2. Reject the negative root.

Question 13: Two positive numbers differ by 2 and their product is 360. The numbers are:
(a) 18 and 20
(b) 16 and 18
(c) 15 and 24
(d) 20 and 22
Show Answer & Explanation

Answer: (a) 18 and 20

Explanation:
1. x(x + 2) = 360 gives x² + 2x − 360 = 0, so x = 18 (positive). The numbers are 18 and 20.

Teacher's Note:
1. Take the numbers as x and x + 2.
2. (x − 18)(x + 20) = 0.

Question 14: 300 apples are distributed equally among some students. If there had been 10 more students, each would have received one apple less. The number of students is:
(a) 40
(b) 60
(c) 30
(d) 50
Show Answer & Explanation

Answer: (d) 50

Explanation:
1. 300/x − 300/(x + 10) = 1.
2. 3000 = x(x + 10), so x² + 10x − 3000 = 0.
3. (x − 50)(x + 60) = 0, so x = 50.

Teacher's Note:
1. Share per student = 300 ÷ students.
2. Set up the difference equal to 1.

Question 15: The hypotenuse of a right triangle is 2 cm more than the base and 1 cm more than twice the altitude. The sides (base, altitude, hypotenuse) are:
(a) 12 cm, 5 cm, 13 cm
(b) 9 cm, 12 cm, 15 cm
(c) 15 cm, 8 cm, 17 cm
(d) 20 cm, 21 cm, 29 cm
Show Answer & Explanation

Answer: (c) 15 cm, 8 cm, 17 cm

Explanation:
1. Hypotenuse = 2h + 1; base = hypotenuse − 2 = 2h − 1.
2. (2h − 1)² + h² = (2h + 1)²: 5h² − 4h + 1 = 4h² + 4h + 1.
3. h² − 8h = 0, so h = 8; base 15, hypotenuse 17.

Teacher's Note:
1. Express all sides in terms of the altitude.
2. Check 15² + 8² = 17².

Question 16: The hypotenuse of a right-triangular field is 1 m more than twice the shortest side, and the third side is 7 m more than the shortest side. The sides are:
(a) 6 m, 8 m, 10 m
(b) 8 m, 15 m, 17 m
(c) 9 m, 12 m, 15 m
(d) 7 m, 14 m, 15 m
Show Answer & Explanation

Answer: (b) 8 m, 15 m, 17 m

Explanation:
1. s² + (s + 7)² = (2s + 1)².
2. 2s² − 10s − 48 = 0, i.e. s² − 5s − 24 = 0.
3. (s − 8)(s + 3) = 0, so s = 8.

Teacher's Note:
1. Take the shortest side as s.
2. Use Pythagoras.

Question 17: A motorboat with speed 15 km/h in still water goes 30 km downstream and returns in a total of 4 hours 30 minutes. The speed of the stream is:
(a) 5 km/h
(b) 3 km/h
(c) 4 km/h
(d) 6 km/h
Show Answer & Explanation

Answer: (a) 5 km/h

Explanation:
1. 30(15 − s) + 30(15 + s) = 900.
2. 900/(225 − s²) = 9/2.
3. 225 − s² = 200, so s = 5.

Teacher's Note:
1. Downstream speed 15 + s, upstream 15 − s.
2. Add the two times.

Question 18: A train takes 2 hours less for a 300 km journey if its speed is increased by 5 km/h. Its usual speed is:
(a) 20 km/h
(b) 30 km/h
(c) 15 km/h
(d) 25 km/h
Show Answer & Explanation

Answer: (d) 25 km/h

Explanation:
1. 300/v − 300/(v + 5) = 2.
2. 1500 = 2v(v + 5), so v² + 5v − 750 = 0.
3. v = 25.

Teacher's Note:
1. Time = distance ÷ speed.
2. Reject the negative root.

Question 19: A plane left 30 minutes late and, to reach a destination 1500 km away on time, increased its speed by 250 km/h. Its usual speed is:
(a) 500 km/h
(b) 1000 km/h
(c) 750 km/h
(d) 600 km/h
Show Answer & Explanation

Answer: (c) 750 km/h

Explanation:
1. 1500/v − 1500/(v + 250) = 1/2.
2. 1500 × 250 × 2 = v(v + 250), so v² + 250v − 750000 = 0.
3. v = 750 km/h.

Teacher's Note:
1. 30 minutes = 1/2 hour.
2. Equate the time saved to 1/2.

Question 20: In a two-digit number the product of the digits is 35, and adding 18 interchanges the digits. The number is:
(a) 75
(b) 57
(c) 53
(d) 37
Show Answer & Explanation

Answer: (b) 57

Explanation:
1. Digits a, b with ab = 35, so they are 5 and 7. 10a + b + 18 = 10b + a gives b − a = 2, so a = 5, b = 7 and the number is 57. Check: 57 + 18 = 75.

Teacher's Note:
1. Write the digits as a and b.
2. Use b − a = 2 together with ab = 35.

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Download Chapter MCQs: Class 10 Mathematics Chapter 04 Quadratic Equations

Class 10 Mathematics Chapter 04 Quadratic Equations Objective Test Questions

Test your conceptual understanding of Chapter 04 Quadratic Equations with these targeted multiple-choice questions. Designed in alignment with the latest CBSE curriculum for Class 10 Mathematics, these problem sets build accuracy and prepare students for objective exams.

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