# Aptitude Question ID: 106308

A student goes to school at the rate of $latex \frac{5}{2}&s=1$ 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 $latex = \frac{16}{60}\ hour&s=1$

$latex \therefore \frac{x}{\frac{5}{2}}-\frac{x}{3} = \frac{16}{60}&s=1$

$latex => \frac{2x}{5} – \frac{x}{3} = \frac{4}{15}&s=1$

$latex => \frac{6x-5x}{15} = \frac{4}{15}&s=1$

$latex => \frac{x}{15} = \frac{4}{15}&s=1$

$latex => x = 4\ km$

Hence option [A] is correct answer.