Moving 6/7 of its usual speed a train is 10 min late. Find its usual time to cover the journey.
Correct Answer: Description for Correct answer:
New speed = \( \Large \frac{6}{7} \) of usual speed
Now, time taken = \( \Large \frac{7}{6} \) of usual time (\( \Large \frac{7}{6} \) of the usual time)
- (usual time) = 10 min
=> \( \Large \frac{1}{6} \) of the usual time = 10 min
Therefore, Usual time = 60 min
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