A tiger is 50 leaps of its own behind a deer. The tiger takes 5 leaps per minute to the deers 4. If the tiger and the deer cover 8 metre and 5metre per leap respectively, what distance in metres will the tiger have to run before it catches the deer?
Answers
Answer:
800 meters
Step-by-step explanation:
It is given that:
- A tiger is 50 leaps of its own from a deer.
- Tiger takes 5 leaps per minute.
- Deer takes 4 leaps per minute.
- Tiger covers 8 meters per leap.
- Deer covers 5 meters per leap.
Therefore, the tiger covers 5*8 = 40 meters/minute.
The deer covers 4*5 = 20 meters/minute.
Hence, the relative speed is 20 meters/minute.
The tiger is behind the deer by 50*8 = 400 meters.
Therefore, the time taken = distance/speed = 400/20 = 20 minutes.
Therefore, in the 20 minutes, the tiger will cover 20*40 = 800 meters.
Answer:
800 m
Step-by-step explanation:
Find the distance the tiger will cover in 1 min:
1 leap = 8 m
5 leaps = 8 x 5 = 40 m
Find the distance the deer will cover in 1 min:
1 leap= 5 m
4 leaps = 5 x 4 = 20 m
Define x:
Let x be the number of minutes need the tiger to catch up with the deer
Solve x:
The deer is 50 "tiger leaps" ahead
40x = 20x + (50 x 8)
40x + 20x + 400
20x = 400
x = 20 minutes
Find the distance the tiger will run in 20 minutes:
1 min = 40 m
20 mins = 40 x 20 = 800 m
Answer: The tiger needs to run 800m to catch the deer.