A tiger is 50 of its own leaps behinds a deer. The tiger takes 5 leaps and per minutes to the deer's 4. If the tiger and the deer cover 8 m and 5 m per leap respectively, what distance will the tiger have to run before it caches the deer?

Answer: Option C

Speed of tiger = 40 m/min
Speed of deer = 20 m/min.
Relative speed = 40 - 20 = 20 m/min.
Initial difference in distance = 50 × 8 = 400 m
Time take to catch = $$\frac{400}{20}$$ = 20 min.
Distance traveled in 20 min,
= 20 × 40
= 800 m