A student goes to school at the rate of $\frac{5}{2}$ km/hr and reaches 6 minutes late. If he travels at the speed of 3 km/hr, he reaches 10 minutes earlier. The distance of the school is :
[A]4 km
[B]10 km
[C]20 km
[D]45 km

4 km
Distance of school = x km
Difference of time = 16 minutes $= \frac{16}{60}\ hour$
$\therefore \frac{x}{\frac{5}{2}}-\frac{x}{3} = \frac{16}{60}$
$=> \frac{2x}{5} - \frac{x}{3} = \frac{4}{15}$
$=> \frac{6x-5x}{15} = \frac{4}{15}$
$=> \frac{x}{15} = \frac{4}{15}$
$=> x = 4\ km$
Hence option [A] is correct answer.