Walking frac{6}{7} of his usual speed, a man | vedclass

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(B) Let the usual speed be $v$ and the usual time taken be $t$.Distance $d = v 	imes t$.When the man walks at $\frac{6}{7}$ of his usual speed, the n

Vedclass · Accepted Answer

$1 \, 	ext{hour} \, 12 \, 	ext{minutes}$

Vedclass · Answer

(B) Let the usual speed be $v$ and the usual time taken be $t$.Distance $d = v 	imes t$.When the man walks at $\frac{6}{7}$ of his usual speed, the new speed is $v' = \frac{6}{7}v$.The new time taken is $t' = \frac{d}{v'} = \frac{d}{\frac{6}{7}v} = \frac{7}{6} 	imes \frac{d}{v} = \frac{7}{6}t$.Given that the man is $12 \, 	ext{minutes}$ late, the difference in time is $t' - t = 12 \, 	ext{minutes}$.$\frac{7}{6}t - t = 12 \, 	ext{minutes}$.$\frac{1}{6}t = 12 \, 	ext{minutes}$.$t = 12 	imes 6 = 72 \, 	ext{minutes}$.$72 \, 	ext{minutes} = 1 \, 	ext{hour} \, 12 \, 	ext{minutes}$.

Walking $\frac{6}{7}$ of his usual speed, a man is $12 \, \text{minutes}$ late. The usual time taken by him to cover that distance is

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