A man with frac{3}{5} of his usual speed reac | vedclass

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(C) Let the usual speed be $s$ and the usual time be $t$. The distance $d$ is constant.$d = s 	imes t$When the speed becomes $\frac{3}{5}s$,the time

Vedclass · Accepted Answer

$3 \frac{3}{4}$

Vedclass · Answer

(C) Let the usual speed be $s$ and the usual time be $t$. The distance $d$ is constant.$d = s 	imes t$When the speed becomes $\frac{3}{5}s$,the time taken is $t + 2 \frac{1}{2} = t + \frac{5}{2}$.Since distance is constant,$s 	imes t = \frac{3}{5}s 	imes (t + \frac{5}{2})$.Dividing both sides by $s$,we get $t = \frac{3}{5}(t + \frac{5}{2})$.$t = \frac{3}{5}t + \frac{3}{5} 	imes \frac{5}{2}$.$t - \frac{3}{5}t = \frac{3}{2}$.$\frac{2}{5}t = \frac{3}{2}$.$t = \frac{3}{2} 	imes \frac{5}{2} = \frac{15}{4} = 3 \frac{3}{4} 	ext{ hours}$.

$A$ man with $\frac{3}{5}$ of his usual speed reaches the destination $2 \frac{1}{2}$ hours late. Find his usual time to reach the destination? (in $hours$)

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