AMC 10 · 2012 · #16
Grade 6 arithmeticThree 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?
Pick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The track is a circle of length $500$ meters; All three runners start together, at the same point and same instant, all clockwise; Their constant speeds are $4.4$, $4.8$, and $5.0$ meters per second; Distance covered equals speed times time: $d = v\,t$; Answer choices: (A) $1{,}000$, (B) $1{,}250$, (C) $2{,}500$, (D) $5{,}000$, (E) $10{,}000$ seconds
Unknowns: The earliest time $t>0$, in seconds, at which all three runners occupy the same point again
Understand
Restated: 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.
Givens: The track is a circle of length $500$ meters; All three runners start together, at the same point and same instant, all clockwise; Their constant speeds are $4.4$, $4.8$, and $5.0$ meters per second; Distance covered equals speed times time: $d = v\,t$; Answer choices: (A) $1{,}000$, (B) $1{,}250$, (C) $2{,}500$, (D) $5{,}000$, (E) $10{,}000$ seconds
Plan
Primary tool: #7 Identify Subproblems
Secondary: #4 Introduce a Variable, #8 Analyze the Units, #3 Eliminate Possibilities
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.
Execute — Answer: C
6.RP.A.3 Step 1 When two runners meet on a loop
- On a circular track, two runners who left together are side by side again exactly when the faster one has pulled ahead by a whole number of full laps — that is, when the distance between them is a multiple of $500$ meters.
- After $t$ seconds each has gone (its speed) times $t$, so the gap is the speed difference times $t$.
- The meeting condition for a pair is therefore: (speed gap) $\times\, t$ is a multiple of $500$.
💡 On a loop, catching up by exactly one full lap drops you right back beside the other runner.
6.NS.B.3 Step 2 Slowest pair meets every 1250 s
- Take the $4.4$ and $4.8$ m/s runners.
- Their speed gap is $4.8-4.4=0.4$ m/s, so the faster one gains $0.4$ meter each second.
- To gain a full $500$-meter lap takes $500\div 0.4 = 1250$ seconds, and they line up again at every multiple of that.
- So this pair is together at $t = 1250,\ 2500,\ 3750,\ \ldots$
💡 Divide one lap by how fast the gap grows to get the time between meetings.
6.NS.B.3 Step 3 Fastest pair meets every 2500 s
- Now the $4.8$ and $5.0$ m/s runners.
- Their speed gap is $5.0-4.8=0.2$ m/s.
- Gaining a full $500$-meter lap at only $0.2$ meter per second takes $500\div 0.2 = 2500$ seconds.
- So this pair is together at $t = 2500,\ 5000,\ \ldots$
💡 A smaller speed gap closes a lap more slowly, so this pair meets less often.
6.NS.B.4 Step 4 Line up both pair rhythms
- All three are together only when both pairs are together at the same instant.
- The first pair meets at multiples of $1250$; the second at multiples of $2500$.
- The earliest time that is a multiple of both is the least common multiple, $\operatorname{lcm}(1250,2500)=2500$.
- The remaining pair, $4.4$ vs $5.0$, has gap $0.6$ and its own period $500\div 0.6 = 2500/3$ s; but at $t=2500$ its gap is $0.6\times 2500 = 1500 = 3\times 500$, a whole $3$ laps, so it is together too.
- Its gap is just the two smaller gaps added, so it is taken care of automatically.
💡 Both meeting rhythms first coincide at their least common multiple.
4.OA.B.4 Step 5 Confirm every runner hits a whole lap
- Check $t=2500$ seconds against the distances: $4.4\times 2500 = 11000 = 22$ laps, $4.8\times 2500 = 12000 = 24$ laps, and $5.0\times 2500 = 12500 = 25$ laps.
- Every distance is a multiple of $500$, so all three runners are back exactly at the starting point at the same time — and step 4 showed this is the earliest such moment.
- The answer is (C) $2500$.
💡 If every runner's total distance is a whole number of laps, they all sit on the same starting point.
6.RP.A.3 On a circular track, two runners who left together are side by side again exactl 6.NS.B.3 Take the $4.4$ and $4.8$ m/s runners. Their speed gap is $4.8-4.4=0.4$ m/s, so t 6.NS.B.3 Now the $4.8$ and $5.0$ m/s runners. Their speed gap is $5.0-4.8=0.2$ m/s. Gaini 6.NS.B.4 All three are together only when both pairs are together at the same instant. Th 4.OA.B.4 Check $t=2500$ seconds against the distances: $4.4\times 2500 = 11000 = 22$ laps Review
Reasonableness: The answer must be a common multiple of the pair periods $1250$ and $2500$, and $2500$ is exactly $\operatorname{lcm}(1250,2500)$, so nothing smaller can work — this rules out (A) $1000$ and (B) $1250$ (at $1250$ s the fastest pair has gap $0.2\times 1250 = 250$, 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.
Alternative: Work purely in relative distances (Tool #4, Introduce a Variable): the pairwise gaps after $t$ seconds are $0.4t$, $0.2t$, and $0.6t$ meters. All three coincide when each of these is a multiple of $500$ at the same time. The binding condition is $0.2t = 500m$ (the smallest gap grows slowest), giving $t = 2500m$; the smallest positive value is $t=2500$, and one checks $0.4(2500)=1000$ and $0.6(2500)=1500$ are also multiples of $500$. Same answer, (C).
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Turning speeds into distances with $d=v\,t$ and stating the on-track meeting condition as "the speed-gap times the time is a full number of laps.")6.NS.B.3Fluently add, subtract, multiply, and divide multi-digit decimals (Subtracting the decimal speeds to get the gaps $0.4$ and $0.2$ m/s and dividing $500$ by each to find the pair periods $1250$ and $2500$ seconds.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Combining the two pair periods with $\operatorname{lcm}(1250,2500)=2500$ to get the first time all runners meet.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Confirming that at $t=2500$ each runner's distance ($11000$, $12000$, $12500$) is a multiple of $500$, so all three land on the same point.)
⭐ 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 $\operatorname{lcm}(1250,2500)=2500$ seconds.
⭐ 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 $\operatorname{lcm}(1250,2500)=2500$ seconds.
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