2008 AMC 10B Problem 25

P21-25~10 minPrint

Linear Equations & Word Problems Games & Processes Organized listing / direct enumeration Compute small cases, spot the pattern, generalize

Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck travels at $10$ feet per second in the same direction as Michael and stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?

Show solution without answering

Key insight

Track the truck's lead over Michael. Each 50-second cycle it gains 100 feet while driving and loses 150 while stopped, so the lead shrinks by 50.

Solution

Michael and the truck meet whenever the truck's lead over Michael is exactly $0$ . So follow that lead through time.

One cycle of the truck.

The truck drives $200$ feet at $10$ feet per second, taking $20$ seconds, then stops for $30$ seconds. So its pattern repeats every $50$ seconds.
While the truck drives ( $20$ seconds), it goes $10$ feet per second and Michael goes $5$ . The truck pulls ahead by $5$ feet each second: $5 \times 20 = 100$ feet.
While the truck is stopped ( $30$ seconds), Michael closes in by $5$ feet each second: $5 \times 30 = 150$ feet.
At the start the truck is just leaving the next pail, so its lead is $200$ feet.

The lead, cycle by cycle. Record it at the end of each drive and each stop.

Cycle	After driving	After stopping
1	$300$	$150$
2	$250$	$100$
3	$200$	$50$
4	$150$	$0$
5	$100$	$-50$
6	$50$	$-100$
7	$0$	$-150$
8	$-50$	$-200$

A negative lead means Michael is ahead.

Counting the meetings. Each time the lead reaches or passes through $0$ is a meeting.

Cycle 4, stop: the lead falls from $150$ to exactly $0$ . Michael reaches the truck just as it starts again. Meeting 1. The truck then drives off ahead.
Cycle 5, stop: the lead falls from $100$ past $0$ to $-50$ . Michael passes the parked truck. Meeting 2.
Cycle 6, drive: the lead rises from $-50$ past $0$ to $50$ . The truck passes Michael. Meeting 3.
Cycle 6, stop: the lead falls from $50$ past $0$ to $-100$ . Michael passes it again. Meeting 4.
Cycle 7, drive: the lead rises from $-100$ to exactly $0$ . The truck catches Michael just as it reaches a pail and stops. Meeting 5. Then Michael walks on ahead.
After that, each drive raises the lead by only $100$ feet, but the lead starts each drive at $-150$ or lower. It never gets back to $0$ .

They meet $5$ times.

The answer is $\boxed{\textbf{(B)}\ 5}$ .

Why this works

Two moving objects meet exactly when the distance between them is zero, so follow that one number instead of both positions. The truck's lead goes up and down in a repeating pattern but drifts down by $50$ feet each cycle. Checking the lead at every turning point catches every meeting, including the two where the lead just touches $0$ at a turning point without crossing. Once even the highest point of a cycle stays below $0$ , there are no more meetings.

Alternative approach

Draw a graph of distance against time. Michael is a straight line rising $5$ feet per second. The truck is a staircase: steep pieces (rising $10$ feet per second for $20$ seconds) joined by flat pieces ( $30$ seconds each). Count the places where Michael's line meets the staircase, including two places where it just touches a corner. There are $5$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let $g(t)$ be the truck's lead over Michael, in feet. At $t = 0$ , $g = 200$ .

The truck's motion repeats every $50$ seconds: it drives $200$ feet in $20$ seconds, then stops for $30$ seconds. While driving, the truck gains $5$ ft/s on Michael, so $g$ rises by $100$ ; while stopped, Michael closes at $5$ ft/s, so $g$ falls by $150$ . Net change per cycle: $-50$ .

Record $g$ at the end of each drive and each stop:

cycle	after drive	after stop
1	$300$	$150$
2	$250$	$100$
3	$200$	$50$
4	$150$	$0$
5	$100$	$-50$
6	$50$	$-100$
7	$0$	$-150$
8	$-50$	$-200$

Meetings are the moments $g = 0$ . In cycle 4, $g$ reaches $0$ exactly as the stop ends (meeting 1). In cycle 5, $g$ goes from $100$ down through $0$ to $-50$ (meeting 2). In cycle 6, the drive takes $g$ from $-50$ up through $0$ to $50$ (meeting 3), and the stop takes it back down through $0$ to $-100$ (meeting 4). In cycle 7, the drive brings $g$ from $-100$ up to exactly $0$ (meeting 5), after which the truck stops and Michael pulls ahead. From then on, each drive raises $g$ by only $100$ while $g$ starts at $-150$ or lower, so $g$ stays negative forever.

They meet $5$ times. The answer is $\boxed{\textbf{(B)}\ 5}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Intermediate Counting & Probability, Chapter 13: Events With States (§13.4)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
Introduction to Algebra, Chapter 7: Proportion (§7.4)