Question 98

Dorlin walks certain distance at (5/6)th of the usual speed and takes 16 minutes more than the usual time. Find the usual time taken.

Solution

Let usual speed = $$6$$ m/min and usual time taken = $$t$$ min

=> New speed = $$5$$ m/min and new time = $$(t+16)$$ min

Also, speed is inversely proportional to time.

=> $$\frac{6}{5}=\frac{t+16}{t}$$

=> $$6t=5t+80$$

=> $$6t-5t=t=80$$

$$\therefore$$ Usual time taken = 1 hr 20 mins

=> Ans - (D)

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