A man reduces his speed from 20 kmph to 18 kmph. So, he takes 10 minutes more than the normal time. what is the distance traveled by him.

Solution (By Examveda Team)

Usual time = 9 × 10 = 90 min = $$\frac{3}{2}$$ h
Thus,
Distance traveled = Speed × time = $$20 \times \frac{3}{2}$$  = 30 km.

Alternatively,
As the speed decreases from 20 kmph to 18 kmph i.e. 10 % increment in usual time.
10% = 10 min
100% = 100 min.
Now,
Distance traveled by him,
= $$\frac{{100}}{{60}} \times 18$$
= 30 km.