AMC 10 · 2012 · #16

Grade 6 arithmetic
Archetype Meeting Time with Piecewise Rates └ Sub-archetype Constant-Rate Closing Speed
lcmrate identify-subproblems ↑ Prerequisites: lcmmultiples
📏 Medium solution 💡 3 insights
Problem
Three runners leave the same spot on a 500-meter circular track at the same instant, all going clockwise at steady speeds of 4.4, 4.8, and 5.0 meters per second. They keep running until the first moment all three are side by side again at some point on the track. Find how many seconds that takes.

Pick an answer.

(A)
1,000
(B)
1,250
(C)
2,500
(D)
5,000
(E)
10,000

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Three runners meeting at once is hard to attack head-on, so break it into pairs (Tool #7, Identify Subproblems). The key fact about a loop is that two runners who started together are side by side again exactly when the faster one has gained a whole number of laps on the slower. That turns each pair into a simple "how long to gain one lap" question. Introduce the time t (Tool #4) and use rate reasoning d=v t (Tool #8, Analyze the Units) to find how often each pair meets. Then the whole-group answer is just the first time that is a multiple of both pair periods — a least-common-multiple combine. Finally, Tool #3 (Eliminate Possibilities) checks the winning time against the answer choices by confirming every runner lands on a whole lap.

1STEP 1

When two runners meet on a loop

On a loop, two runners are side by side again exactly when their gap, (speed difference) × t, is a whole multiple of 500 meters.

gap=(v_fast-v_slow) t = 500k (k=1,2,3,…)
2STEP 2

Slowest pair meets every 1250 s

The 4.4 and 4.8 pair closes 0.4 m/s, so one 500-meter lap takes 500 ÷ 0.4 = 1250 s; they meet at every multiple.

500/(4.8-4.4)=500/0.4=1250 s
3STEP 3

Fastest pair meets every 2500 s

The 4.8 and 5.0 pair closes only 0.2 m/s, so a full lap takes 500 ÷ 0.2 = 2500 s — they meet far less often.

500/(5.0-4.8)=500/0.2=2500 s
4STEP 4

Line up both pair rhythms

All three coincide only when both rhythms do: lcm(1250, 2500) = 2500 s, where the 4.4 vs 5.0 gap is 1500 meters, a whole 3 laps.

lcm(1250, 2500)=2500 s
5STEP 5

Confirm every runner hits a whole lap

At t = 2500 they cover 11000, 12000, 12500 meters — exactly 22, 24, 25 laps — so all three stand at the start: (C) 2500.

4.4(2500)=11000, 4.8(2500)=12000, 5.0(2500)=12500 → all multiples of 500 → (C)
Answer
2,500
The answer must be a common multiple of the pair periods 1250 and 2500, and 2500 is exactly lcm(1250,2500), so nothing smaller can work — this rules out (A) 1000 and (B) 1250 (at 1250 s the fastest pair has gap 0.2 × 1250 = 250, only half a lap, so they are on opposite sides, not together). Larger choices like (D) 5000 and (E) 10000 are also common meeting times, but they are not the first one. At t=2500 every runner completes a whole number of laps (22, 24, 25), so they reunite at the start — consistent and clean.
💡Key takeaway

On a loop, two runners meet again each time the faster one gains a full lap; find how often each pair meets, then take the least common multiple — here lcm(1250,2500)=2500 seconds.

  • When two runners meet on a loop
  • Slowest pair meets every 1250 s
  • Fastest pair meets every 2500 s
  • Line up both pair rhythms
  • Confirm every runner hits a whole lap

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