Sheela walks a certain distance at (16/17)th of the usual speed and takes 5 minutes more than the usual time. Find the usual time taken. (in minutes)
Let usual speed = $$17$$ m/min and usual time taken = $$t$$ min
=> New speed = $$16$$ m/min and new time = $$(t+5)$$ min
Also, speed is inversely proportional to time.
=> $$\frac{17}{16}=\frac{t+5}{t}$$
=> $$17t=16t+80$$
=> $$17t-16t=t=80$$
$$\therefore$$ Usual time taken = 80 minutes
=> Ans - (C)
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