If I walk at 5, km/hr,I miss a train by 7, mi | vedclass

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(A) Let the distance between the house and the station be $D\, km$ and the scheduled time of the train be $T\, hours$.When walking at $5\, km/hr$,the

Vedclass · Accepted Answer

$6$

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(A) Let the distance between the house and the station be $D\, km$ and the scheduled time of the train be $T\, hours$.When walking at $5\, km/hr$,the time taken is $\frac{D}{5}$ hours. Since $I$ miss the train by $7\, minutes$ $(7/60\, hours)$,the time taken is $T + 7/60$.So,$\frac{D}{5} = T + \frac{7}{60} \implies T = \frac{D}{5} - \frac{7}{60} \dots (1)$When walking at $6\, km/hr$,the time taken is $\frac{D}{6}$ hours. Since $I$ reach $5\, minutes$ $(5/60\, hours)$ early,the time taken is $T - 5/60$.So,$\frac{D}{6} = T - \frac{5}{60} \implies T = \frac{D}{6} + \frac{5}{60} \dots (2)$Equating $(1)$ and $(2)$:$\frac{D}{5} - \frac{7}{60} = \frac{D}{6} + \frac{5}{60}$$\frac{D}{5} - \frac{D}{6} = \frac{5}{60} + \frac{7}{60}$$\frac{6D - 5D}{30} = \frac{12}{60}$$\frac{D}{30} = \frac{1}{5}$$D = \frac{30}{5} = 6\, km$Thus,the distance is $6\, km$.

If $I$ walk at $5\, km/hr$,$I$ miss a train by $7\, minutes$. However,if $I$ walk at $6\, km/hr$,$I$ reach the station $5\, minutes$ before the departure of the train. The distance between my house and the station is (in $km$)

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