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 = $$\ \frac{\ 5}{4}$$ 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.