There are five dustbins along a circular path at different places. Ramesh takes multiple rounds of the path every morning, always at the same speed. He noticed that it took him a different number of steps to walk between any two consecutive dustbins. Ramesh also noticed that starting from any of the dustbins, it took a minimum 360 steps to reach every second dustbin, and a maximum 1260 steps to reach every third dustbin.
If Ramesh's step size is 0.77 meter, and the width of the path is negligible, which the following can be the radius of the circular path?
Answer: C
Asked: a possible radius of the circular path.
Let the five gaps between consecutive dustbins be \(g_1, \dots, g_5\) steps (all different), with circumference \(C = g_1 + \dots + g_5\) steps.
Reading the data as bounds: every walk to the second dustbin ahead (two consecutive gaps) takes at least 360 steps, and every walk to the third dustbin ahead (three consecutive gaps) takes at most 1260 steps.
- Adding all five two-gap sums counts each gap twice: \(2C \ge 5 \times 360\), so \(C \ge 900\).
- Adding all five three-gap sums counts each gap three times: \(3C \le 5 \times 1260\), so \(C \le 2100\). Equality would need all gaps equal to 420, which is not allowed because the gaps differ, so \(C < 2100\).
Convert to metres: \(C < 2100 \times 0.77 = 1617\) m, so \(r < \frac{1617}{2\pi} \approx 257.4\) m.
- 260 m would need about 2122 steps, which is impossible.
- 250 m needs \(2\pi \times 250 \approx 1570.8\) m, about 2040 steps. This works: for example, gaps of 400, 412, 404, 416, 408 have every adjacent pair at least 808 (so every three-gap sum is at most 2040 − 780 = 1260) and every two-gap sum above 360.
Answer: 250 meters (see the note below on the wording)