Circular Path Radius

Let the distances between consecutive dustbins be $l_1, l_2, l_3, l_4, l_5$ in steps. Reaching every second dustbin means sum of 2 consecutive intervals, e.g. $l_i + l_{i+1} \ge 800$. Summing over all 5 dustbins: $\sum (l_i + l_{i+1}) = 2 \sum l_i \ge 5 \times 800 = 4000 \implies \sum l_i \ge 2000$. Reaching every third dustbin means sum of 3 consecutive intervals, e.g. $l_i + l_{i+1} + l_{i+2} \le 1260$. Summing over all 5 dustbins: $3 \sum l_i \le 5 \times 1260 = 6300 \implies \sum l_i \le 2100$. So total perimeter in steps is between 2000 and 2100 steps. Since 1 step = 0.77 m: $2000 \times 0.77 \le 2\pi r \le 2100 \times 0.77$ $1540 \le 2\pi r \le 1617$ $\frac{1540}{2\pi} \le r \le \frac{1617}{2\pi} \implies 245.1 \le r \le 257.3$. Among the given choices, 250 metres lies in this range.