Question 119

By walking at 3/4 of his usual speed, a man reaches his office 20 minutes later than usual. His usual time is

Solution

Let the usual time(T1) be T minutes .

and here we can see that Distance between office and home remains same.

So this is the case when distance traveled is constant

let the normal speed (S1) = S m/min

So the changed speed (S2) = $$\frac{3}{4}$$ S

T2 = (T + 20) minutes

hence , $$\frac{S1}{S2}$$ = $$\frac{T2}{T1}$$

$$\frac{4 S}{3 S}$$ = $$\frac{T + 20 }{T}$$

T = 60 minutes

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