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By walking at 3/4 of his usual speed, a man reaches his office 20 minutes later than his usual time. The usual time taken by him to reach his office is
As distance is constant and we know s = \frac{d}{t} (where s is distance and t is time)
hence st = constant
or $$s_{1}t_{1}$$= $$s_{2}t_{2}$$
or $$s_{1}t_{1}$$=$$ \frac{3}{4}s_{1}(t_{1}+ 20)$$
hence $$t_{1}$$ = 60
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