Question 62
Walking 5/6 th of his usual speed, a man is 10 min late. The usual time taken by him to cover that distance is:
Solution
Let us assume that his original speed and time are S and T mins respectively.
Now, walking at $$\frac{5S}{6}$$ speed, he takes T + 10 mins.
We know that for a constant distance, ratio of speeds is inverse to the ratio of tme taken.
$$\frac{S}{\frac{5S}{6}}=\frac{T+10}{T}$$
This gives T = 50 mins.
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