AMC10 2008 B

Answer (B): Number the pails consecutively so that Michael is presently at pail 0 and the garbage truck is at pail 1. Michael takes $200/5 = 40$ seconds to walk between pails, so for $n \ge 0$ he passes pail $n$ after $40n$ seconds. The truck takes 20 seconds to travel between pails and stops for 30 seconds at each pail. Thus for $n \ge 1$ it leaves pail $n$ after $50(n-1)$ seconds, and for $n \ge 2$ it arrives at pail $n$ after $50(n-1) - 30$ seconds. Michael will meet the truck at pail $n$ if and only if $50(n-1) - 30 \le 40n \le 50(n-1)$ or, equivalently, $5 \le n \le 8$. Hence Michael first meets the truck at pail 5 after 200 seconds, just as the truck leaves the pail. He passes the truck at pail 6 after 240 seconds and at pail 7 after 280 seconds. Finally, Michael meets the truck just as it arrives at pail 8 after 320 seconds. These conditions imply that the truck is ahead of Michael between pails 5 and 6 and that Michael is ahead of the truck between pails 7 and 8. However, the truck must pass Michael at some point between pails 6 and 7, so they meet a total of five times.