By walking at $$\frac{4}{5}th$$ of his usual speed, a man reaches office 10 minutes later than usual. What is his usual time?
Let the usual speed and the time taken by the man = 5v, 4t respectively.
If the man walks at $$\frac{4}{5}th$$ of his usual speed, a man reaches office 10 minutes later than usual
i.e the time taken by the man =Â $$\ \frac{\ 5}{4}$$ of the usual time (Speed and time are inversely proportional to each other)
=5t
5t-4t = 10 minutes.
t = 10 minutes.
Usual time taken by the man = 4t = 40 minutes
B is the correct answer.
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