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
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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