A 10 km race is held on an 800 m circular track. Two runners P and Q compete, and their speeds are in the ratio 5 : 4. When the winner completes the full 10 km, how many complete times does the winner overtake (lap) the other on the track?

Introduction / Context:
This problem tests “relative pace” on a circular track. When two runners move at constant but different speeds, each full lap gained by the faster runner over the slower runner corresponds to the faster accumulating an extra one-track-length of distance relative to the slower. The question asks for how many complete overtakings (full laps gained) occur by the time the winner finishes the race distance.

Given Data / Assumptions:

• Total race distance for the winner = 10 km = 10,000 m.
• Track length (circumference) = 800 m.
• Speed ratio P:Q = 5:4 (constant speeds; no stops).

Concept / Approach:
The slower runner’s covered distance at the instant the faster finishes is proportional to the speed ratio. If the winner covers 10,000 m, the other covers 10,000 * (slower_speed / faster_speed). The number of complete overtakings equals (distance lead) / (track length), counting only full laps (integer part).

Step-by-Step Solution:

Verification / Alternative check:
If you simulate by laps: P’s laps = 10,000/800 = 12.5; Q’s laps = 8,000/800 = 10. The difference in completed laps is 2, confirming two full passes.

Why Other Options Are Wrong:
1 time or 3–4 times do not match the 2,000 m relative lead; only two full 800 m gains fit before the finish.

Common Pitfalls:
Counting the fractional (half) lap as a full overtake or mistakenly computing (5−4)/4 laps instead of using total lead divided by track length.

Final Answer:
2 times