# Two boys start running straight toward each other from two points that are 100m apart.One runs with a speed of 5m/s,while other moves at 7m/s.How close are they to the slower one's starting point when they meet?

Let the two boys meet after $t$ sec their start.
$5 t + 7 t = 100$
$\implies t = \frac{100}{12} = \frac{25}{3}$ s
The distance of meeting point is $\frac{25}{3} \times 5 = \frac{125}{3}$ m apart from the starting point of slower boy.