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Question

(C) Let the distance from the house to the school be $x \, km$.Let the scheduled time to reach the school be $t \, hours$.In the first case,the speed

Vedclass · Accepted Answer

$1.75$

Vedclass · Answer

(C) Let the distance from the house to the school be $x \, km$.Let the scheduled time to reach the school be $t \, hours$.In the first case,the speed is $v_1 = 2.5 \, km/hr$ and the time taken is $t_1 = t + \frac{6}{60} = t + 0.1 \, hours$.Using the formula $	ext{Distance} = 	ext{Speed} 	imes 	ext{Time}$,we have $x = 2.5(t + 0.1)$.In the second case,the speed is $v_2 = 2.5 + 1 = 3.5 \, km/hr$ and the time taken is $t_2 = t - \frac{6}{60} = t - 0.1 \, hours$.Thus,$x = 3.5(t - 0.1)$.Equating the two expressions for $x$: $2.5(t + 0.1) = 3.5(t - 0.1)$.$2.5t + 0.25 = 3.5t - 0.35$.$1.0t = 0.6$,so $t = 0.6 \, hours$.Substituting $t$ back into the first equation: $x = 2.5(0.6 + 0.1) = 2.5 	imes 0.7 = 1.75 \, km$.

Starting from his house,a student walks at a speed of $2.5 \, km/hr$ and reaches his school $6 \, minutes$ late. Next day he increases his speed by $1 \, km/hr$ and reaches the school $6 \, minutes$ early. How far is the school from his house? (in $km$)

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