Question 41

Walking at $$\frac{3}{4}$$ of his usual pace, a man reaches his office 20 minutes late. Find his usual time.

Solution

Let usual speed = $$4$$ m/min and usual time taken be $$t$$ min

=> New speed = $$\frac{3}{4}\times4=3$$ m/min and new time = $$(t+20)$$ min

Speed is inversely proportional to time

=> $$\frac{4}{3}=\frac{t+20}{t}$$

=> $$4t=3t+60$$

=> $$t=60$$

$$\therefore$$ Usual time = 60 min = 1 hour

=> Ans - (B)

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