Walking at 3 , km/h,Pintu reaches his school | vedclass

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

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$2$

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(B) Let the distance of the school from his house be $x \, km$.The time taken to reach school at $3 \, km/h$ is $t_1 = \frac{x}{3} \, hours$.The time taken to reach school at $4 \, km/h$ is $t_2 = \frac{x}{4} \, hours$.According to the problem,the difference between these two times is $5 \, minutes$ late and $5 \, minutes$ early,which totals $10 \, minutes$.Converting $10 \, minutes$ to hours: $10 \, minutes = \frac{10}{60} \, hours = \frac{1}{6} \, hours$.Therefore,the equation is:$\frac{x}{3} - \frac{x}{4} = \frac{1}{6}$Taking the least common multiple $(LCM)$ of $3$ and $4$,which is $12$:$\frac{4x - 3x}{12} = \frac{1}{6}$$\frac{x}{12} = \frac{1}{6}$$x = \frac{12}{6} = 2 \, km$.Thus,the distance of the school from his house is $2 \, km$.

Walking at $3 \, km/h$,Pintu reaches his school $5 \, minutes$ late. If he walks at $4 \, km/h$,he will be $5 \, minutes$ early. The distance of Pintu's school from his house is (in $km$):

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