A student walks from his house at 5 , km/h an | vedclass

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Question

(B) Let the distance between the house and the school be $d \, km$.Let the scheduled time to reach the school be $t \, hours$.Case $1$: Speed $= 5 \,

Vedclass · Accepted Answer

$12.5$

Vedclass · Answer

(B) Let the distance between the house and the school be $d \, km$.Let the scheduled time to reach the school be $t \, hours$.Case $1$: Speed $= 5 \, km/h$,Time taken $= t + \frac{10}{60} \, hours$.Using the formula $	ext{Distance} = 	ext{Speed} 	imes 	ext{Time}$,we have $d = 5(t + \frac{1}{6})$.Case $2$: Speed $= 6 \, km/h$,Time taken $= t - \frac{15}{60} \, hours$.Using the formula,we have $d = 6(t - \frac{1}{4})$.Equating both expressions for $d$:$5(t + \frac{1}{6}) = 6(t - \frac{1}{4})$$5t + \frac{5}{6} = 6t - \frac{6}{4}$$t = \frac{5}{6} + \frac{3}{2} = \frac{5+9}{6} = \frac{14}{6} = \frac{7}{3} \, hours$.Now,substitute $t$ back into the distance equation:$d = 5(\frac{7}{3} + \frac{1}{6}) = 5(\frac{14+1}{6}) = 5(\frac{15}{6}) = 5(2.5) = 12.5 \, km$.

$A$ student walks from his house at $5 \, km/h$ and reaches his school $10 \, minutes$ late. If his speed had been $6 \, km/h$,he would have reached $15 \, minutes$ early. The distance of his school from his house is (in $km$):

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