CBSE Class 10 Mathematics MCQs Set 04

Answer: (a) 33 km/h, 44 km/h

Explanation:
1. 1. Let the speed of the passenger train be \( x \) km/h; the express train goes at \( (x+11) \) km/h.
2. 2. Time difference: \( \dfrac{132}{x} - \dfrac{132}{x+11} = 1 \), which simplifies to \( x^2 + 11x - 1452 = 0 \).
3. 3. \( (x+44)(x-33) = 0 \), so \( x = 33 \) (the negative root is rejected).
4. 4. Speeds are 33 km/h and 44 km/h.

Teacher's Note:
1. Take the slower speed as x and write time = distance / speed for both.
2. Reject the negative root because a speed cannot be negative.

Answer: (a) 60 km/h

Explanation:
1. 1. Let the express train speed be \( v \) km/h; the other train goes at \( (v-12) \) km/h.
2. 2. \( \dfrac{240}{v-12} - \dfrac{240}{v} = 1 \) gives \( v^2 - 12v - 2880 = 0 \).
3. 3. \( (v-60)(v+48) = 0 \), so \( v = 60 \) km/h.

Teacher's Note:
1. The slower train takes more time, so its time minus the faster train's time equals 1 hour.
2. Check: 240/48 - 240/60 = 5 - 4 = 1.

Answer: (c) 40 km/h, 50 km/h

Explanation:
1. 1. Let the ordinary bus speed be \( x \) km/h; the deluxe bus goes at \( (x+10) \) km/h.
2. 2. \( \dfrac{600}{x} - \dfrac{600}{x+10} = 3 \) gives \( x^2 + 10x - 2000 = 0 \).
3. 3. \( (x+50)(x-40) = 0 \), so \( x = 40 \).
4. 4. The speeds are 40 km/h and 50 km/h.

Teacher's Note:
1. Form the equation from the difference of the two times.
2. Check: 600/40 - 600/50 = 15 - 12 = 3.

Answer: (d) 25 km/h

Explanation:
1. 1. Let the original speed be \( x \) km/h.
2. 2. \( \dfrac{300}{x} - \dfrac{300}{x+5} = 2 \) gives \( x^2 + 5x - 750 = 0 \).
3. 3. \( (x+30)(x-25) = 0 \), so \( x = 25 \) km/h.

Teacher's Note:
1. Original time minus new time is 2 hours.
2. Check: 300/25 - 300/30 = 12 - 10 = 2.

Answer: (c) 30 hours

Explanation:
1. 1. Let the slower tap take \( x \) hours; the faster takes \( (x-10) \) hours.
2. 2. \( \dfrac{1}{x-10} + \dfrac{1}{x} = \dfrac{1}{12} \) gives \( x^2 - 34x + 120 = 0 \).
3. 3. \( (x-30)(x-4) = 0 \), so \( x = 30 \) or \( x = 4 \). Since \( x - 10 \) must be positive, \( x = 30 \) hours.

Teacher's Note:
1. Reject the root that makes the faster tap's time negative.
2. Check: 1/20 + 1/30 = 5/60 = 1/12.

Answer: (c) \( d^2 = 3x^2 \)

Explanation:
1. 1. In right \( \triangle PAB \): \( \tan 30^\circ = \dfrac{x}{d} \), so \( d = \sqrt{3}\,x \).
2. 2. Squaring, \( d^2 = 3x^2 \).

Teacher's Note:
1. Angle of depression at P equals angle of elevation at B.
2. Square both sides to remove the root.

Answer: (b) \( d = y \)

Explanation:
1. 1. In right \( \triangle QAB \): \( \tan 45^\circ = \dfrac{y}{d} = 1 \).
2. 2. Hence \( y = d \).

Teacher's Note:
1. tan 45 degrees is 1.
2. Height and distance are equal for 45 degrees.

Answer: (d) \( 25(\sqrt{3}+3) \) m

Explanation:
1. 1. \( y = x + 50 \) and \( y = d = \sqrt{3}\,x \).
2. 2. \( \sqrt{3}x = x + 50 \), so \( x = \dfrac{50}{\sqrt{3}-1} = 25(\sqrt{3}+1) \).
3. 3. \( y = \sqrt{3}\times 25(\sqrt{3}+1) = 25(3+\sqrt{3}) \) m.

Teacher's Note:
1. Combine the two relations to find x first.
2. Rationalise 50/(root 3 - 1) by multiplying by (root 3 + 1).

Answer: (d) \( 25(\sqrt{3}+3) \) m

Explanation:
1. 1. \( AB = d = y = 25(3+\sqrt{3}) \) m = \( 25(\sqrt{3}+3) \) m.

Teacher's Note:
1. From the 45 degree reading, distance equals the new height.
2. Order of terms in the bracket does not matter.

Answer: (a) \( 75(1+\sqrt{3}) \) m

Explanation:
1. 1. In right \( \triangle QAC \): \( \tan 30^\circ = \dfrac{y}{AC} \), so \( AC = \sqrt{3}\,y \).
2. 2. \( AC = \sqrt{3}\times 25(3+\sqrt{3}) = 25(3\sqrt{3}+3) = 75(1+\sqrt{3}) \) m.

Teacher's Note:
1. Use the 30 degree reading at Q for the car at C.
2. Factor 25 x 3 = 75 out of the bracket.

Answer: (c) 150

Explanation:
1. 1. Class mark \( = \dfrac{140+160}{2} = 150 \).

Teacher's Note:
1. Class mark is the average of the two limits.
2. It is the midpoint of the class.

Answer: (a) 120-140

Explanation:
1. 1. \( n = 50 \), \( \dfrac{n}{2} = 25 \). Cumulative frequencies: 12, 26, 34, 40, 50.
2. 2. The first cumulative frequency above 25 is 26, so the median class is 120-140.

Teacher's Note:
1. Find n/2 first.
2. Pick the class whose cumulative frequency first exceeds it.

Answer: (d) Rs. 145.2

Explanation:
1. 1. Class marks 110, 130, 150, 170, 190; \( \sum f x = 1320+1820+1200+1020+1900 = 7260 \).
2. 2. Mean \( = \dfrac{7260}{50} = 145.2 \).

Teacher's Note:
1. Multiply class marks by frequencies and add.
2. Divide by 50.

Answer: (b) 140

Explanation:
1. 1. The highest frequency is 14, so the modal class is 120-140 and its upper limit is 140.

Teacher's Note:
1. The modal class has the highest frequency.
2. Read the larger limit.

Answer: (a) Rs. 125

Explanation:
1. 1. \( l=120, h=20, f_1=14, f_0=12, f_2=8 \).
2. 2. \( \text{Mode} = 120 + \dfrac{14-12}{28-12-8}\times 20 = 120 + 5 = 125 \).

Teacher's Note:
1. f0 is the frequency before the modal class and f2 the one after.
2. 2/8 x 20 = 5.