Walking at $\frac{7}{8}$ th of his usual speed, a man reaches late at his office by $15$ min. Find his usual time (in minutes).

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Solution

Speed becomes $\frac{7}{8}$ of usual speed, it means time become $\frac{8}{7}$ of usual time taken.
[ As speed is inversely proportional to time taken when distance is constant. ST = D ]
So, $\frac{8}{7}$ of usual time = usual time $+ 15$ min
or, $\frac{8}{7}$ of usual time - usual time $= 15$ min
or, $\frac{1}{7}$ of usual time $= 15$ min
So, usual time $= 15 \times 7$
or, Usual time $= 105$ min $= 1 h o u r 45$ min.

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