CBSE Class 10 Mathematics Quadratic Equations MCQs Set 19

Answer: (b) −9

Explanation:
1. Put x = 3: 2(9) − 3(3) + p = 0.
2. 18 − 9 + p = 0, so p = −9.

Teacher's Note:
1. A root satisfies the equation.
2. Substitute carefully with brackets.

Answer: (a) \( -\frac{3}{2} \)

Explanation:
1. Sum of the roots = \( -\frac{b}{a} = \frac{3}{2} \).
2. Other root = \( \frac{3}{2} - 3 = -\frac{3}{2} \). Check with p = −9: product = \( \frac{p}{2} = -\frac{9}{2} \) and 3 × (−3/2) = −9/2.

Teacher's Note:
1. Use the sum of roots when one root is known.
2. Check with the product of the roots.

Answer: (d) −1

Explanation:
1. Put x = 2: 8 + 2k − 6 = 0.
2. 2k = −2, so k = −1.

Teacher's Note:
1. A root makes the equation true.
2. Solve the linear equation in k.

Answer: (c) 14

Explanation:
1. D = 4(k − 12)² − 8(k − 12) = 4(k − 12)(k − 14).
2. D = 0 gives k = 12 or k = 14.
3. But k = 12 makes the coefficient of x² zero, so the equation would not be quadratic. Hence k = 14.

Teacher's Note:
1. Reject any value that makes a = 0.
2. Factorise D instead of expanding it.

Answer: (b) \( \pm\frac{4}{3} \)

Explanation:
1. D = 9k² − 16.
2. D = 0 gives k² = \( \frac{16}{9} \), so k = \( \pm\frac{4}{3} \).

Teacher's Note:
1. Equal roots need D = 0.
2. Take both signs when you take a square root.

Answer: (a) 4 km/h

Explanation:
1. Let the stream speed be x km/h: \( \frac{60}{16-x} - \frac{60}{16+x} = 2 \).
2. 120x = 2(256 − x²), so x² + 60x − 256 = 0, that is (x + 64)(x − 4) = 0.
3. x = 4 (the negative value is rejected).

Teacher's Note:
1. Upstream speed = boat − stream; downstream speed = boat + stream.
2. Check: 60/12 − 60/20 = 5 − 3 = 2 hours.

Answer: (d) 32 km/h

Explanation:
1. \( \frac{288}{x} - \frac{288}{x+4} = 1 \), so 288 × 4 = x(x + 4).
2. x² + 4x − 1152 = 0, that is (x + 36)(x − 32) = 0.
3. x = 32 km/h.

Teacher's Note:
1. Time = distance ÷ speed.
2. Check: 288/32 = 9 h and 288/36 = 8 h.

Answer: (c) 8 cm

Explanation:
1. Let the shortest side be s. Then the hypotenuse is 2s + 1 and the third side is 2s − 1.
2. s² + (2s − 1)² = (2s + 1)², so s² = 8s, giving s = 8.
3. The sides are 8, 15 and 17 cm.

Teacher's Note:
1. Use Pythagoras with the three expressions.
2. 8² + 15² = 64 + 225 = 289 = 17².

Answer: (b) Rs. 30

Explanation:
1. \( \frac{1200}{x} - \frac{1200}{x+10} = 10 \), so 12000 = 10x(x + 10), that is x² + 10x − 1200 = 0.
2. (x + 40)(x − 30) = 0, so x = 30.

Teacher's Note:
1. Number of books = total money ÷ price.
2. Check: 1200/30 = 40 books and 1200/40 = 30 books.

Answer: (a) 9

Explanation:
1. Let the numbers be x (larger) and y (smaller): x² − y² = 45 and y² = 4x.
2. x² − 4x − 45 = 0, so (x − 9)(x + 5) = 0 and x = 9.
3. Then y² = 36, so y = 6.

Teacher's Note:
1. Substitute y² to get one quadratic.
2. Check: 81 − 36 = 45.

Answer: (d) \( \frac{3}{4} \)

Explanation:
1. Let the fraction be \( \frac{x}{x+1} \). Then \( \frac{x+3}{x+4} - \frac{x}{x+1} = \frac{3}{28} \).
2. \( \frac{3}{(x+4)(x+1)} = \frac{3}{28} \), so (x + 4)(x + 1) = 28 and x² + 5x − 24 = 0.
3. (x + 8)(x − 3) = 0, so x = 3 and the fraction is 3/4.

Teacher's Note:
1. Simplify the difference of fractions first.
2. Check: 6/7 − 3/4 = 3/28.

Answer: (c) 5

Explanation:
1. Let the value be y. Then y = \( \sqrt{20 + y} \), so y² = 20 + y.
2. y² − y − 20 = 0, that is (y − 5)(y + 4) = 0.
3. y must be positive, so y = 5.

Teacher's Note:
1. The repeating part is the same as the whole value.
2. A square root is never negative.