AMC 10 · 2008 · #20

Grade 7 rate-ratio

Archetype Meeting Time with Piecewise Rates └ Sub-archetype Staggered Start or Schedule Table

ratecoordinate-geometrypattern-recognition physical-representationcasework ↑ Prerequisites: rate

📏 Long solution 💡 4 insights

Problem

A walker moves steadily while a truck drives faster but stops at regular intervals. The truck starts one interval ahead. Count how many times the two are at the same spot afterwards.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Draw a Diagram

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.

1STEP 1

Set the clock and the origin

A clock and an origin make both motions writable.

M(t) = 5t, truck starts at position 200

2STEP 2

Find the truck's 50-second rhythm

The truck repeats on a fixed cycle.

(200 ft)/(10 ft/s) = 20 s rolling, + 30 s parked = 50 s cycle

3STEP 3

Track one number: the gap

Tracking the gap replaces tracking both movers.

D(0) = 200; D rises +5 ft/s (truck rolling), D falls -5 ft/s (truck parked)

4STEP 4

Per-cycle swing walks to the first meeting

The per-cycle swing walks the gap down to the first meeting.

200 → 150 → 100 → 50 → 0 at t = 200 s

5STEP 5

Fifth cycle: one crossing

The next cycle gives one more crossing.

100 - 5(t - 220) = 0 → t = 240 s (meeting #2)

6STEP 6

Sixth cycle: two crossings

The following one gives two.

-50 + 5(t-250)=0 → t = 260; 50 - 5(t-270)=0 → t = 280

7STEP 7

Seventh cycle: last touch, then gone

After one last touch the gap stays negative forever.

-100 + 5(t-300)=0 → t = 320 (#5); after: D ≤ -50 ＜ 0 always

8STEP 8

Count the meetings

Counting the meetings gives 5, choice (B).

{200, 240, 260, 280, 320} → 5 meetings → (B)

Answer

Sanity-check by position, not just gap. Meeting #1 at t=200: Michael is at 5 · 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 · 320 = 1600 ft, and the truck's arrival times at pails 400,600,… 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.

💡Key takeaway

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.

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