When a person walks at $\frac{4}{5}$ th of his usual speed, he is late by 5 hrs. The usual time is

17 hrs

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16 hrs

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20 hrs

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25 hrs

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Solution

The correct option is C
20 hrs

Let, usual speed = S

Usual time = T and usual distance = D

$∴$ New Speed is $\frac{4 S}{5}$ and new time is $\frac{5 T}{4}$

$\Rightarrow \frac{5}{4} T - T = 5$

$∴$ T = 20 hrs

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