Question

A tiger is 50 leaps of its own behind a deer. The tiger takes 5 leaps per minute to the deer's 4. If the tiger and the deer cover 8 metre and 5 metre per leap respectively, what distance in metres will the tiger have to run before it catches the deer?

Answer 1

(800)
Tiger is 50 leap of its own behind deer.

Given tiger's one leap is 8m. So initial separation of Tiger and deer = 50$\times$8 =400m/min

Tiger goes 5 leap in a minute which is equal to 5 $\times$ 8 = 40m/min

Deer goes 4 leap in a minute which is equal to 4$\times$5 = 20m/min.

Now relative distance of 400m have to be covered with (40 - 20) = 20m/min

Required time =$\frac{400m}{(40-20)m/min}$

$\frac{Distance}{\mathrm{Relative}Speed}=20min$

So, distance travelled by Tiger

$\frac{40m}{min}\times 20min=800m/min$