Question 74

Merlin walks certain distance at (9/10)th of the usual speed and takes 15 minutes more than the usual time. Find the usual time taken.

Solution

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Let usual speed = $$10$$ m/min and usual time taken = $$t$$ min

=> New speed = $$9$$ m/min and new time = $$(t+15)$$ min

Also, speed is inversely proportional to time.

=> $$\frac{10}{9}=\frac{t+15}{t}$$

=> $$10t=9t+135$$

=> $$10t-9t=t=135$$

$$\therefore$$ Usual time taken = 135 minutes

=> Ans - (B)

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