In a race of three horses, the first beat the second by 11 metres and the…

Concept: When racers run at constant speeds on the same track, the distance each has covered at a given instant is proportional to their speed. If X beats Y by d metres over a track of length L, then at the instant X finishes, Y has covered (L − d) metres, so the speed ratio vY/vX = (L − d)/L. Writing this ratio for the same pair from two different finish instants and equating the two expressions pins down the unknown track length.

1. When A finishes the race (covers L), B has covered L − 11, so vB/vA = (L − 11)/L.

2. At that same instant, C has covered L − 90 (A beats C by 90 m), so vC/vA = (L − 90)/L.

3. Dividing the two gives the speed ratio vB/vC = (L − 11)/(L − 90).

4. Separately, when B finishes the race (covers L), C has covered L − 80 (B beats C by 80 m), so vC/vB = (L − 80)/L, i.e. vB/vC = L/(L − 80).

5. Equating the two expressions for vB/vC: (L − 11)/(L − 90) = L/(L − 80).

6. Cross-multiplying: (L − 11)(L − 80) = L(L − 90).

7. Expanding: L² − 91L + 880 = L² − 90L.

8. Simplifying: 880 = L, so L = 880 metres.

Cross-check: with L = 880, when A finishes, B has covered 880 − 11 = 869 m and C has covered 880 − 90 = 790 m, giving vB/vC = 869/790 = 11/10. When B finishes (covers 880 m), C has covered 880 − 80 = 800 m, giving vC/vB = 800/880 = 10/11 — the reciprocal of the same ratio, confirming consistency.

So the racecourse is 880 metres long.