Three friends A, B and C decide to run around a circular track. They start at…
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Three friends A, B and C decide to run around a circular track. They start at the same time and run in the same direction. A is the quickest and when A finishes a lap, it is seen that C is as much behind B as B is behind A. When A completes 3 laps, C is the exact same position on the circular track as B was when A finished 1 lap. Find the ratio of the speeds of A, B and C?
- A.
5 : 4 : 2
- B.
4 : 3 : 2
- C.
5 : 4 : 3
- D.
3 : 2 : 1
Attempted by 13 students.
Show answer & explanation
Correct answer: C
Let the track length be L and let the speeds of A, B and C be a, b and c respectively. Time for A to finish one lap is T = L/a.
When A completes one lap, positions (measured as distances along the track from the start) are:
A: L (one full lap)
B: bT = (b/a) L
C: cT = (c/a) L
The gap from B to A is L - (b/a)L = (1 - b/a)L, and the gap from C to B is ((b - c)/a)L. The condition "C is as much behind B as B is behind A" gives
a = 2b - c
When A completes three laps (time 3T), C's position equals B's position at time T. Writing this as an equality of distances up to whole laps gives an integer k such that
3c = b + k a (for some integer k ≥ 0)
Substitute c = 2b - a into the second equation:
3(2b - a) = b + k a ⇒ 6b - 3a = b + k a ⇒ 5b = (k + 3) a
Hence b = (k + 3)/5 · a and c = 2b - a = (2k + 1)/5 · a. Since b < a we need (k + 3)/5 < 1, so k < 2. Therefore k can be 0 or 1.
Case k = 0: b = 3/5 a, c = 1/5 a → ratio a : b : c = 5 : 3 : 1.
Case k = 1: b = 4/5 a, c = 3/5 a → ratio a : b : c = 5 : 4 : 3.
Both ratios satisfy the problem conditions. Among the given choices, 5 : 4 : 3 appears and satisfies both conditions, so the correct answer from the list is 5 : 4 : 3.