Chase problem — tiger vs deer (uniform leaps): A tiger is behind a deer by a head start equal to 50 of the tiger’s own leaps. The tiger makes 5 leaps per minute and covers 8 m per leap. The deer makes 4 leaps per minute and covers 5 m per leap. How many metres will the tiger run before it catches the deer?

Introduction / Context:
Relative speed chase problems compare how quickly a pursuer closes the gap on a target when both move simultaneously. Here, movement is in “leaps per minute,” but we convert each to metres per minute to compute the catch-up distance cleanly.

Given Data / Assumptions:

• Tiger: 5 leaps/min, 8 m/leap ⇒ 40 m/min
• Deer: 4 leaps/min, 5 m/leap ⇒ 20 m/min
• Initial head start = 50 tiger leaps = 50 * 8 = 400 m
• Both run in the same straight line with constant speeds.

Concept / Approach:
The time to catch equals initial lead / relative speed. Relative speed (tiger w.r.t. deer) = tiger speed − deer speed.

Step-by-Step Solution:

Verification / Alternative check:
If both run 20 min, the deer covers 20 * 20 = 400 m; adding to the tiger’s initial deficit of 400 m confirms equality at the catch point (tiger total 800 m vs deer total 400 m after the head start).

Why Other Options Are Wrong:

• 600 m, 700 m, 1000 m, 500 m: Each ignores either the correct relative speed or the 50-leap (400 m) head start.

Common Pitfalls:
Using leaps/min directly without converting to m/min, or misreading “50 leaps” as metres. Another mistake is adding speeds instead of taking the difference for same-direction chase problems.

Final Answer:
800 m