Walking at 6/7 of his usual speed, a man reaches his destination 10 minutes late. What is his usual time to reach the destination? MCQ with Answer and Explanation

Walking at 6/7 of his usual speed, a man reaches his destination 10 minutes late. What is his usual time to reach the destination?
A. 50 minutes
B. 70 minutes
C. 80 minutes
D. 60 minutes
Answer: Option D
Solution (By JKSSB Mock Tests)
New speed = 6/7 of usual speed => New time = 7/6 of usual time. Difference = 1/6 of usual time = 10 minutes. Usual time = 10 * 6 = 60 minutes.

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Practice More Time Speed and Distance Questions

Question #1
A train 160 meters long crosses a pole in 8 seconds. What is its speed in km/h?
A. 72 km/h
B. 64 km/h
C. 84 km/h
D. 80 km/h

Correct Answer: Option A


Explanation:
Speed of train = 160 / 8 = 20 m/s. In km/h = 20 * (18/5) = 72 km/h.

This question belongs to: Maths Time Speed and Distance
Question #2
Two outposts M and N are 320 km apart. Train A leaves M for N at 40 km/h at 6:00 AM. Train B leaves N for M at 60 km/h at 8:00 AM. At what time will they cross?
A. 10:36 AM
B. 10:24 AM
C. 10:48 AM
D. 10:12 AM

Correct Answer: Option B


Explanation:
By 8:00 AM, Train A has run for 2 hours, covering 40 * 2 = 80 km. Remaining separation distance = 320 - 80 = 240 km. Relative speed = 40 + 60 = 100 km/h. Time after 8:00 AM = 240 / 100 = 2.4 hours = 2 hours 24 minutes. Meeting time = 8:00 AM + 2 hours 24 minutes = 10:24 AM.

This question belongs to: Maths Time Speed and Distance
Question #3
Two points P and Q are 240 km apart. Train A starts from P towards Q at 60 km/h at 9:00 AM. Train B starts from Q towards P at 80 km/h at 10:00 AM. At what time will they meet?
A. 11:17 AM
B. 11:30 AM
C. 11:45 AM
D. 11:00 AM

Correct Answer: Option A


Explanation:
By 10:00 AM, Train A covers 60 km. Remaining distance = 240 - 60 = 180 km. Relative speed = 60 + 80 = 140 km/h. Time to meet after 10 AM = 180 / 140 = 9/7 hours = 1 hour and 17 minutes. Meeting time is 11:17 AM.

This question belongs to: Maths Time Speed and Distance