Question

Next door lives a man with his son. They both work in the same factory. I watch them going to work through my window. The father leaves for work 10 min earlier than his son. One day I asked him about it and he told me he takes 30 min to walk to his factory ,whereas his son is able to cover the distance in only 20 min. I wondered if the father were to leave the house 5 min earlier than his son, how soon would the son catch up with the father?

• A. 3.33 mins
• B. 8.3 mins
• C. 5.76 mins
• D. 8.25 mins

Answer 1

The correct option is B 8.3 mins

Let speed of son be s,and speed of father be f,then

$\frac{s}{f}=\frac{3}{2}$ (inverse ratio of the time) ( If D is the distance ,then $\frac{D}{f}=30$ and $\frac{D}{s}=20$ which implies $\frac{s}{f}=\frac{3}{2}$)

Let speed of dad be 20 and son be 30 m/s

Now in 5 mins, the father covers 100 m

And son covers 150 m

The son covers 100 m in 3.33 min,

Hence the son will be able to catch up with the father after 8.3 mins from the start. Option (b).