2012 AMC 10A Problem 16

P16-20~4 minPrint

Linear Equations & Word Problems GCD & LCM Set up the equation/formula and compute; no special trick needed

Three runners start running simultaneously from the same point on a $500$ -meter circular track. They each run clockwise around the course maintaining constant speeds of $4.4$ , $4.8$ , and $5.0$ meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?

Show solution without answering

Key insight

Two runners are together whenever the faster has gained a whole lap, 500 m. Find when each pair meets, then the first time on both lists.

Solution

The runners do not need to be at the starting line, just at the same spot. So instead of tracking where each runner is, track how far ahead one runner is of another.

Two runners are at the same spot exactly when the faster one has gained a whole number of laps on the slower one: $500$ m, $1000$ m, $1500$ m, and so on.
Runners at $4.4$ and $4.8$ m/s. The faster one gains $0.4$ m every second. To gain $500$ m takes $500 \div 0.4 = 1250$ seconds. So these two are together at $1250, 2500, 3750, \dots$ seconds.
Runners at $4.8$ and $5.0$ m/s. The faster one gains $0.2$ m every second. To gain $500$ m takes $500 \div 0.2 = 2500$ seconds. So these two are together at $2500, 5000, \dots$ seconds.
If both pairs are together, all three are together, because the middle runner is with each of the others. So we need a time on both lists. The first one is $2500$ seconds.
Check at $2500$ seconds: the runners have gone $11000$ , $12000$ and $12500$ m. The gaps are $1000$ m and $500$ m, both whole laps, so they are all at the same spot. At $1250$ seconds the second gap is only $250$ m, so the runners are not all together yet.

The answer is $\boxed{\textbf{(C)}\ 2,500}$ .

Why this works

On a loop, "together" is about how far apart the runners are, not where they are. So subtract speeds and ask how long the faster runner takes to gain a full lap. Each pair meets on a regular schedule, and the whole group meets at the first time that is on every schedule. Two pairs are enough, because the third gap is just the other two gaps added together.

Alternative approach

Test the choices from the smallest. At $t = 1000$ the gaps are $0.4 \times 1000 = 400$ m and $0.2 \times 1000 = 200$ m, not whole laps. At $t = 1250$ they are $500$ m and $250$ m, and the second one fails. At $t = 2500$ they are $1000$ m and $500$ m, both whole laps. So the answer is $2500$ .

Concise solution

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

Two runners on a circular track are at the same point exactly when the distance between them is a whole number of laps. After $t$ seconds the gap between two runners is (difference of speeds) $\times\ t$ .

Speeds $4.4$ and $4.8$ differ by $0.4$ m/s, so they coincide when $0.4t$ is a multiple of $500$ , i.e. $t$ is a multiple of $1250$ .

Speeds $4.8$ and $5.0$ differ by $0.2$ m/s, so they coincide when $0.2t$ is a multiple of $500$ , i.e. $t$ is a multiple of $2500$ .

(The pair $4.4$ and $5.0$ differ by $0.6$ , giving multiples of $\frac{2500}{3}$ , but this is automatically satisfied once the other two conditions hold.)

All three are together when $t$ is a common multiple of $1250$ and $2500$ ; the first such time is $\operatorname{lcm}(1250, 2500) = 2500$ seconds.

The answer is $\boxed{\textbf{(C)}\ 2,500}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Number Theory, Chapter 3: Multiples and Divisors (§3.3, §3.4, §3.7)
Introduction to Algebra, Chapter 7: Proportion (§7.4)
Introduction to Number Theory, Chapter 4: Prime Factorization (§4.7)