AMC 8 · 2025 · #19

Grade 6 rate-ratio
Archetype Meeting Time with Piecewise Rates └ Sub-archetype Piecewise-Leg Time Balance
ratefraction-arithmeticlinear-equations-one-varinterval-arithmetic dimensional-analysisidentify-subproblemscasework ↑ Prerequisites: ratefraction-arithmetic
📏 Long solution 💡 3 insights 📊 Diagram
Problem
A straight 15-mile road from town A to town B is split into three 5-mile segments with posted speed limits 25, 40, and 20 mph (in that order from A to B). Two cars start at the same instant, one from A heading toward B and one from B heading toward A, each driving exactly the posted speed in whichever segment it is in. How many miles from town A do they meet?

Pick an answer.

(A)
7.75
(B)
8
(C)
8.25
(D)
8.5
(E)
8.75

AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Analyze the Units

This is a classic rate / distance / time problem, so Tool #8 (Analyze the Units) keeps us honest — every time we write a number we mark whether it is in miles, hours, or mph, and we use time = distance ÷ speed to move between them. Tool #1 (Draw a Diagram) lets us draw the road as three 5-mile segments with the posted speeds and the two cars' starting arrows, which makes the speed-by-segment bookkeeping concrete. Tool #7 (Identify Subproblems) breaks the trip into two clean phases: Phase 1 ends when one car finishes its first 5-mile segment, and Phase 2 is the remaining race in the middle segment where both cars travel at 40 mph — that lets us avoid setting up a single messy algebraic equation.

1STEP 1

Draw the road as three 5-mile segments. Speeds belong to the road, so Car A meets 25→40→20 while Car B meets them reversed: 20→40→25.

A: 0 → 5 → 10 → 15 ; B: 15 → 10 → 5 → 0
2STEP 2

Use time = distance ÷ speed on each first 5-mile segment: Car A takes 15\frac{1}{5} hr, Car B takes 14\frac{1}{4} hr.

T_A,1 = (5 mi)/(25 mph) = 15\frac{1}{5} hr T_B,1 = (5 mi)/(20 mph) = 14\frac{1}{4} hr
3STEP 3

Since 15\frac{1}{5} hr ＜ 14\frac{1}{4} hr, Car A enters the middle segment first, so they meet there (both at 40 mph) after t = 14\frac{1}{4} hr.

15\frac{1}{5} = 0.20 hr ＜ 14\frac{1}{4} = 0.25 hr
4STEP 4

At t = 14\frac{1}{4} hr, Car A has spent 14\frac{1}{4} − 15\frac{1}{5} = 120\frac{1}{20} hr in the middle at 40 mph, moving 40 × 120\frac{1}{20} = 2 miles past mile 5 — now at mile 7.

Car A at t = 14\frac{1}{4} hr: 5 + 40 × 120\frac{1}{20} = 5 + 2 = 7 mi
5STEP 5

Both cars are in the middle at 40 mph, 3 miles apart. Facing each other they close the gap at 40 + 40 = 80 mph, lasting 380\frac{3}{80} hr.

t_meet = (3 mi)/((40 + 40) mph) = 380\frac{3}{80} hr
6STEP 6

In that 380\frac{3}{80} hr Car A covers 40 × 380\frac{3}{80} = 1.5 more miles, so they meet at 7 + 1.5 = 8.5 — inside middle segment (5 ＜ 8.5 ＜ 10), choice (D).

Meeting point = 7 + 40 × 380\frac{3}{80} = 7 + 1.5 = 8.5 mi → (D)
Answer
8.5
Sanity check: by symmetry, if both cars went the same speeds in the same order they would meet exactly at the middle (mile 7.5). But Car A's first segment is faster (25 mph) than Car B's first segment (20 mph), so Car A should cover more than half the road before they meet — the answer should be greater than 7.5. 8.5 mi is greater than 7.5 and still inside the middle 5-10 segment, so it lines up with the picture. We can also verify with total times: Car A takes 525\frac{5}{25} + 3.540\frac{3.5}{40} = 15\frac{1}{5} + 780\frac{7}{80} = 1680\frac{16}{80} + 780\frac{7}{80} = 2380\frac{23}{80} hr to reach mile 8.5; Car B takes 520\frac{5}{20} + 1.540\frac{1.5}{40} = 14\frac{1}{4} + 380\frac{3}{80} = 2080\frac{20}{80} + 380\frac{3}{80} = 2380\frac{23}{80} hr to reach mile 8.5 from town B. Equal times — confirmed.
💡Key takeaway

This AMC 8 problem only needs Grade 6 rate reasoning — that two cars heading toward each other close the gap at the SUM of their speeds — that you already know!

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