A legal clerk walks from his residence to the courthouse at 5 km/h and arrives 15 minutes late. If he sets a pace of 7.5 km/h, he arrives 5 minutes early. Find the distance. MCQ with Answer and Explanation

A legal clerk walks from his residence to the courthouse at 5 km/h and arrives 15 minutes late. If he sets a pace of 7.5 km/h, he arrives 5 minutes early. Find the distance.
A. 6.0 km
B. 5.0 km
C. 5.5 km
D. 4.5 km
Answer: Option B
Solution (By JKExamLibrary)
Time difference = 15 - (-5) = 20 minutes = 20/60 = 1/3 hours. Let distance be d. d/5 - d/7.5 = 1/3 => (1.5d - d) / 7.5 = 1/3 => 0.5d / 7.5 = 1/3 => d / 15 = 1/3 => d = 5 km.

This question belongs to: Maths Time Speed and Distance

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A cyclist handles 600 meters in exactly 2 minutes. Find his moving velocity in km/h.
A. 18 km/h
B. 20 km/h
C. 15 km/h
D. 22 km/h

Correct Answer: Option A


Explanation:
Time = 2 minutes = 120 seconds. Speed = 600 / 120 = 5 m/s. In km/h = 5 * (18/5) = 18 km/h.

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Question #2 Report Error
The speed indices of two sprinters are in the ratio 3 : 4. If the first sprinter requires 24 seconds to wrap up a bend, how much time does the second sprinter require?
A. 20 seconds
B. 18 seconds
C. 22 seconds
D. 16 seconds

Correct Answer: Option B


Explanation:
Ratio of speeds = 3:4, so ratio of times required = 4:3. Given 4 parts = 24 seconds => 1 part = 6 seconds. Time for the second sprinter = 3 parts = 3 * 6 = 18 seconds.

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Question #3 Report Error
A train 110 meters long is traveling at 60 km/h. How much time will it take to pass a platform 240 meters long?
A. 27 seconds
B. 18 seconds
C. 21 seconds
D. 24 seconds

Correct Answer: Option C


Explanation:
Total distance = 110 + 240 = 350 meters. Speed = 60 * (5/18) = 50/3 m/s. Time = 350 / (50/3) = 350 * 3 / 50 = 7 * 3 = 21 seconds.

This question belongs to: Maths Time Speed and Distance