AMC 10 · 2008 · #25
Grade 7 rate-ratioPick an answer.
AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 = truck - 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.
Set the clock and the origin
Start the clock at t = 0 when Michael passes a pail and call that pail position 0, so M(t) = 5t and the truck starts 200 ft ahead.
Pinning t=0 to a clean event (Michael at a pail, truck leaving the next) makes every later position a simple rate times time.
6.RP.A.2Draw A DiagramFind the truck's 50-second rhythm
The truck rolls 200 ft at 10 ft/s in 20 s, then parks 30 s: a fixed 50-second cycle per pail.
Distance divided by speed is time, so the pail spacing and truck speed fix a repeating 20-on / 30-off heartbeat.
7.RP.A.3Analyze The UnitsTrack one number: the gap
Follow one gap D = truck - Michael. D(0) = 200; it rises 5 ft/s while the truck rolls, falls 5 ft/s while it parks, and D = 0 is a meeting.
Watching the single signed gap turns a two-body chase into one number sliding up and down toward zero.
7.NS.A.3Introduce A VariablePer-cycle swing walks to the first meeting
Each cycle nets +100 then -150, so D drops 50 a cycle: 200, 150, 100, 50, 0, giving the first meeting at t = 200 s.
Each cycle nets a steady -50, so the head start of 200 is erased in exactly four cycles.
Each cycle nets the same steady change, so a fixed head start is erased in a whole number of cycles.
▸ Why?
The gap changes at the difference of the two paces, so a repeating pattern moves it by a fixed amount.
▸ Why?
Repeating the same fixed step makes an evenly spaced list, so the crossing point can be counted straight off.
Fifth cycle: one crossing
From D = 0 the gap climbs to 100 by t = 220, then falls back to 0 at t = 240 s: Michael passes the parked truck.
The truck jumps ahead while rolling, then sits still, so Michael's steady pace reels it back to one crossing.
7.EE.B.4Introduce A VariableSixth cycle: two crossings
Starting at -50, the gap crosses 0 at t = 260 (truck catches him) and again at t = 280 s (Michael retakes it): two meetings.
Because the gap swings from below zero up above zero and back, it crosses the meeting line twice in one cycle.
7.EE.B.4Introduce A VariableSeventh cycle: last touch, then gone
From -100 the gap just touches 0 at t = 320 s, both reaching the pail together; after that every cycle's peak stays below 0.
Once the gap's highest point in a cycle drops below zero, Michael is permanently ahead and no more meetings are possible.
7.NS.A.3Eliminate PossibilitiesCount the meetings
They meet at t = 200, 240, 260, 280, 320 seconds — five times in all, with none to follow. The answer is (B) 5.
Listing the exact crossing times and confirming the list is closed gives an airtight count.
4.NBT.B.4Eliminate PossibilitiesChase 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.
- Set the clock and the origin
- Find the truck's 50-second rhythm
- Track one number: the gap
- Per-cycle swing walks to the first meeting
- Fifth cycle: one crossing
- Sixth cycle: two crossings
- Seventh cycle: last touch, then gone
- Count the meetings