If a boy travels from his house to school at | vedclass

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(B) Let the distance of the school from the house be $x \, km$.Time taken at $4 \, km/h$ is $t_1 = \frac{x}{4} \, hours$.Time taken at $3 \, km/h$ is

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$4$

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(B) Let the distance of the school from the house be $x \, km$.Time taken at $4 \, km/h$ is $t_1 = \frac{x}{4} \, hours$.Time taken at $3 \, km/h$ is $t_2 = \frac{x}{3} \, hours$.The difference between the two times is $10 \, minutes$ early and $10 \, minutes$ late,which is $10 + 10 = 20 \, minutes$.Converting $20 \, minutes$ into hours: $\frac{20}{60} = \frac{1}{3} \, hours$.According to the problem:$\frac{x}{3} - \frac{x}{4} = \frac{1}{3}$Taking the least common multiple of $3$ and $4$,which is $12$:$\frac{4x - 3x}{12} = \frac{1}{3}$$\frac{x}{12} = \frac{1}{3}$$x = \frac{12}{3} = 4 \, km$.Therefore,the distance of the school from his house is $4 \, km$.

If a boy travels from his house to school at the rate of $4 \, km/h$,he reaches the school $10 \, minutes$ earlier than the scheduled time. However,if he walks at the rate of $3 \, km/h$,he reaches $10 \, minutes$ late. Find the distance of the school from his house (in $km$).

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