AMC 10 · 2008 · #25
Grade 7 rate-ratioMichael 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 traveling at 10 feet per second in the same direction as Michael 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?
Pick an answer.
AMC 10 2008 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: Michael walks a straight path at $5$ ft/s. Pails sit every $200$ ft. A garbage truck drives the same way at $10$ ft/s but stops $30$ seconds at each pail. At the instant Michael passes a pail, the truck is one pail ahead ($200$ ft in front) and just starting to move again. Count how many times Michael and the truck are at the same spot afterward.
Givens: Michael's speed: $5$ ft/s, constant; Pails spaced every $200$ ft along the path; Truck's speed while moving: $10$ ft/s, same direction; Truck stops $30$ seconds at every pail; At $t = 0$: Michael is at a pail (call it position $0$); the truck is at the next pail ($200$ ft ahead) just leaving it; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Unknowns: The number of times Michael and the truck occupy the same position (meetings) after $t = 0$
Understand
Restated: Michael walks a straight path at $5$ ft/s. Pails sit every $200$ ft. A garbage truck drives the same way at $10$ ft/s but stops $30$ seconds at each pail. At the instant Michael passes a pail, the truck is one pail ahead ($200$ ft in front) and just starting to move again. Count how many times Michael and the truck are at the same spot afterward.
Givens: Michael's speed: $5$ ft/s, constant; Pails spaced every $200$ ft along the path; Truck's speed while moving: $10$ ft/s, same direction; Truck stops $30$ seconds at every pail; At $t = 0$: Michael is at a pail (call it position $0$); the truck is at the next pail ($200$ ft ahead) just leaving it; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #8 Analyze the Units, #5 Look for a Pattern, #3 Eliminate Possibilities
Tool #1 (Draw a Diagram) frames it as a position-versus-time picture: Michael is a straight line, the truck is a staircase (slanted while rolling, flat while parked), and a meeting is a crossing point. Tool #8 (Analyze the Units) turns the raw numbers into the truck's rhythm: $200$ ft at $10$ ft/s is $20$ s rolling, then $30$ s parked, a $50$-second cycle. Tool #4 (Introduce a Variable) replaces two moving objects with one number, the gap $D = \text{truck} - \text{Michael}$; a meeting is simply $D = 0$, and $D$ only rises $5$ ft/s (truck rolling) or falls $5$ ft/s (truck parked). Tool #5 (Look for a Pattern) exploits the repeating $+100$ then $-150$ swing of $D$ each cycle to march to the first meeting fast. Tool #3 (Eliminate Possibilities) closes it: once $D$ can no longer reach $0$, no meetings remain, so the count is final.
Execute — Answer: B
6.RP.A.2 Step 1 Set the clock and the origin
- Start the clock ($t = 0$) at the moment Michael passes a pail, and call that pail position $0$.
- Michael's position is then $M(t) = 5t$.
- At this same instant the truck is at the next pail, position $200$, and is just leaving it.
- So the truck begins $200$ ft ahead of Michael.
💡 Pinning $t=0$ to a clean event (Michael at a pail, truck leaving the next) makes every later position a simple rate times time.
7.RP.A.3 Step 2 Find the truck's 50-second rhythm
- The truck rolls one pail-gap, $200$ ft, at $10$ ft/s, which takes $200 \div 10 = 20$ seconds.
- Then it parks $30$ seconds.
- So the truck repeats a fixed cycle: $20$ s rolling, then $30$ s stopped, for a total of $50$ seconds per pail.
💡 Distance divided by speed is time, so the pail spacing and truck speed fix a repeating 20-on / 30-off heartbeat.
7.NS.A.3 Step 3 Track one number: the gap
- Instead of two positions, follow their gap $D = (\text{truck position}) - (\text{Michael position})$.
- At $t = 0$, $D = 200$.
- While the truck rolls it goes $10$ ft/s and Michael $5$ ft/s, so the gap grows at $10 - 5 = 5$ ft/s.
- While the truck is parked it goes $0$ and Michael still does $5$ ft/s, so the gap shrinks at $5$ ft/s.
- A meeting is exactly $D = 0$; if $D$ is negative, Michael is ahead.
💡 Watching the single signed gap turns a two-body chase into one number sliding up and down toward zero.
7.RP.A.3 Step 4 Per-cycle swing walks to the first meeting
- In each $50$-second cycle the gap first rises for $20$ s at $+5$ ft/s ($+100$), then falls for $30$ s at $-5$ ft/s ($-150$): a net change of $-50$ per cycle.
- Starting from $D = 200$, the gap at the end of successive cycles is $150,\,100,\,50,\,0$.
- So at $t = 4 \times 50 = 200$ s the gap is $0$ for the first time: meeting #1, with the truck parked at a pail and Michael catching up to it.
💡 Each cycle nets a steady $-50$, so the head start of $200$ is erased in exactly four cycles.
7.EE.B.4 Step 5 Fifth cycle: one crossing
- From $t = 200$ ($D = 0$) the truck starts rolling, so the gap climbs to $+100$ at $t = 220$ (truck now ahead).
- Then the truck parks and the gap falls.
- It returns to $0$ when $100 - 5(t - 220) = 0$, i.e.
- $t = 240$: meeting #2, Michael overtaking the parked truck.
- By $t = 250$ the gap is $-50$.
💡 The truck jumps ahead while rolling, then sits still, so Michael's steady pace reels it back to one crossing.
7.EE.B.4 Step 6 Sixth cycle: two crossings
- Now the gap starts at $-50$ (Michael ahead).
- While the truck rolls, the gap rises and hits $0$ at $-50 + 5(t - 250) = 0$, i.e.
- $t = 260$: meeting #3, the truck catching Michael on the road.
- It keeps rising to $+50$ at $t = 270$, then the truck parks and the gap falls back to $0$ at $50 - 5(t - 270) = 0$, i.e.
- $t = 280$: meeting #4, Michael retaking the parked truck.
- The gap ends the cycle at $-100$.
💡 Because the gap swings from below zero up above zero and back, it crosses the meeting line twice in one cycle.
7.NS.A.3 Step 7 Seventh cycle: last touch, then gone
- The gap starts at $-100$.
- While the truck rolls it rises by $100$, just reaching $0$ exactly at $t = 320$: meeting #5, the truck pulling into the pail at the same instant Michael arrives.
- Then the truck parks and the gap falls to $-150$.
- From here every cycle's net $-50$ keeps the gap negative forever (peak of the next cycle is only $-50$), so it never returns to $0$.
💡 Once the gap's highest point in a cycle drops below zero, Michael is permanently ahead and no more meetings are possible.
4.NBT.B.4 Step 8 Count the meetings
- Collect the meeting times: $t = 200,\,240,\,260,\,280,\,320$ seconds.
- That is $5$ meetings in all, and none can follow.
- The answer is $\textbf{(B)}\ 5$.
💡 Listing the exact crossing times and confirming the list is closed gives an airtight count.
6.RP.A.2 Start the clock ($t = 0$) at the moment Michael passes a pail, and call that pai 7.RP.A.3 The truck rolls one pail-gap, $200$ ft, at $10$ ft/s, which takes $200 \div 10 = 7.NS.A.3 Instead of two positions, follow their gap $D = (\text{truck position}) - (\text 7.RP.A.3 In each $50$-second cycle the gap first rises for $20$ s at $+5$ ft/s ($+100$), 7.EE.B.4 From $t = 200$ ($D = 0$) the truck starts rolling, so the gap climbs to $+100$ a 7.EE.B.4 Now the gap starts at $-50$ (Michael ahead). While the truck rolls, the gap rise 7.NS.A.3 The gap starts at $-100$. While the truck rolls it rises by $100$, just reaching 4.NBT.B.4 Collect the meeting times: $t = 200,\,240,\,260,\,280,\,320$ seconds. That is $5 Review
Reasonableness: Sanity-check by position, not just gap. Meeting #1 at $t=200$: Michael is at $5 \cdot 200 = 1000$ ft; the truck reached pail $1000$ at $t=170$ and is parked there through $t=200$ — they coincide. Meeting #5 at $t=320$: Michael is at $5 \cdot 320 = 1600$ ft, and the truck's arrival times at pails $400,600,\dots$ are $20,70,120,170,220,270,320$, so it pulls into pail $1600$ exactly at $t=320$ — coincide. The five times $200,240,260,280,320$ split naturally into one late catch-up, a lone crossing, a double crossing, and a final tie, which is just what the rising-then-falling gap predicts. The count $5$ is choice (B); getting $4$ means missing the double-crossing cycle, and $6$ means wrongly counting the boundary touch at $t=320$ twice.
Alternative: Tool #1 (Draw a Diagram) done literally: plot position versus time. Michael is the straight line $y = 5t$. The truck is a staircase of slope-$10$ rolling segments joined by flat parked segments, starting at $(0, 200)$. Draw both and count intersection points of the line with the staircase — they cross five times (once on a flat step, once mid-slope, one slope-then-step double, and one exactly at a corner) before the staircase falls permanently below the line. Same answer, read straight off the graph.
CCSS standards used (min grade 7)
6.RP.A.2Understand the concept of a unit rate and use rate language (Reading the speeds $5$ ft/s and $10$ ft/s as unit rates to write positions as speed times time.)7.RP.A.3Use proportional relationships to solve multi-step ratio and percent problems (Turning $200$ ft at $10$ ft/s into a $20$-second roll plus $30$-second stop ($50$-second cycle), and tracking the repeating $+100/-150$ gap swing across cycles.)7.NS.A.3Solve real-world and mathematical problems involving the four operations with rational numbers (Following the signed gap $D$ up ($+5$ ft/s) and down ($-5$ ft/s), and recognizing when it turns permanently negative so meetings stop.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities to solve problems (Solving crossing-time equations like $100 - 5(t-220) = 0$ to pin the exact meeting times $240$, $260$, $280$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Accumulating cycle changes ($200,150,100,50,0$) and counting the final list of five meeting times.)
⭐ Chase problems get easy when you track just the gap between the two: it climbs while the truck rolls and shrinks while it's parked, and every time the gap hits zero they meet — here that happens five times.
⭐ Chase problems get easy when you track just the gap between the two: it climbs while the truck rolls and shrinks while it's parked, and every time the gap hits zero they meet — here that happens five times.
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