A pedestrian walks at a constant pace of 5 km/h. How many meters can he cover in exactly 18 minutes? MCQ with Answer and Explanation
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Question #1
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A train running at 72 km/h passes a tunnel 300 meters long in 25 seconds. Find the length of the train.
Correct Answer: Option A
Explanation:
Speed = 72 * (5/18) = 20 m/s. Total distance in 25 seconds = 20 * 25 = 500 meters. Length of train = 500 - 300 = 200 meters.
Speed = 72 * (5/18) = 20 m/s. Total distance in 25 seconds = 20 * 25 = 500 meters. Length of train = 500 - 300 = 200 meters.
This question belongs to:
Maths
Time Speed and Distance
Question #2
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Two points M and N are separated by 300 km. At 8:00 AM, a car leaves M for N at 60 km/h. At 9:00 AM, another car leaves N for M at 90 km/h. At what time will they pass each other?
Correct Answer: Option D
Explanation:
By 9:00 AM, the first car covers 60 km * 1 hour = 60 km. The remaining distance between them is 300 - 60 = 240 km. Relative speed = 60 + 90 = 150 km/h. Time to meet after 9:00 AM = 240 / 150 = 8/5 hours = 1 hour 36 minutes. Meeting time = 9:00 AM + 1 hour 36 minutes = 10:36 AM.
By 9:00 AM, the first car covers 60 km * 1 hour = 60 km. The remaining distance between them is 300 - 60 = 240 km. Relative speed = 60 + 90 = 150 km/h. Time to meet after 9:00 AM = 240 / 150 = 8/5 hours = 1 hour 36 minutes. Meeting time = 9:00 AM + 1 hour 36 minutes = 10:36 AM.
This question belongs to:
Maths
Time Speed and Distance
Question #3
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Walking at 6/7 of his usual speed, a man is 15 minutes late to his destination. What is his usual time to reach the destination?
Correct Answer: Option D
Explanation:
New speed = 6/7 of usual speed => New time = 7/6 of usual time. Difference = 1/6 of usual time = 15 minutes. Usual time = 15 * 6 = 90 minutes.
New speed = 6/7 of usual speed => New time = 7/6 of usual time. Difference = 1/6 of usual time = 15 minutes. Usual time = 15 * 6 = 90 minutes.
This question belongs to:
Maths
Time Speed and Distance
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