**Problem**

A car leaves city A for city B, 40 km distant, traveling 50 kph. Thirty minutes later, another car leaves city B for city A at 40 kph. At what point of their path will the first car pass 6 minutes before the second car?

**Solution**

Let C = point of the path where the first car pass 6 minutes before the second

t_{1} = time for the first car to travel from city A to C

$t_1 = \dfrac{x}{50}$

t_{2} = time for the second car to travel from city B to C

$t_2 = \dfrac{40 - x}{40} = 1 - \dfrac{x}{40}$

$t_1 = t_2 + \dfrac{30}{60} - \dfrac{6}{60}$

$\dfrac{x}{50} = \left( 1 - \dfrac{x}{40} \right) + \dfrac{2}{5}$

$\dfrac{9}{200}x = \dfrac{7}{5}$

$x = 31.11 ~ \text{km from city A}$ *answer*