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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