Competition · AMC preparation · step 4 of 4
AMC 8 · 2025 · #19
Grade 6 rate-ratio
Pick an answer.
AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Label each segment's speed
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.
Drawing the road with labeled speeds turns the word problem into a picture you can point at — Grade 4 distance/time word-problem territory.
4.MD.A.2Draw A DiagramFind each first segment time
Use time = distance ÷ speed on each first 5-mile segment: Car A takes hr, Car B takes hr.
Miles ÷ (miles/hour) cancels "miles" and leaves "hours" — the units do the bookkeeping for us.
4.MD.A.2Analyze The UnitsCompare the two times
Since hr ＜ hr, Car A enters the middle segment first, so they meet there (both at 40 mph) after t = hr.
Comparing 1/5 and 1/4 with unlike denominators is the standard Grade 5 fraction-comparison move.
5.NF.A.1Identify SubproblemsLocate Car A at that moment
At t = hr, Car A has spent − = hr in the middle at 40 mph, moving 40 × = 2 miles past mile 5 — now at mile 7.
Subtracting 1/4 - 1/5 uses Grade 5 unlike-denominator subtraction, then 40 mph × hr cleanly cancels to miles.
5.NF.A.1Analyze The UnitsFind the closing time
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 hr.
Treating 80 mph as the unit closing-rate (miles closed per hour) is Grade 6 rate reasoning at its cleanest.
Once both cars are together in the middle segment, 3 miles apart and each moving 40 mph, the time until they meet is the 3-mile gap divided by their combined 80 mph closing rate, which is 3/80 hour.
▸ Why?
The distance between the cars is a single 3-mile stretch of road, so each mile either car drives comes straight off that gap; in one hour Car A erases 40 miles of it and Car B erases another 40, a combined 80 miles per hour.
▸ Why?
The gap splits with no overlap into the part still ahead of Car A and the part still ahead of Car B, so the two cars' distances add back to the whole gap and their speeds add to the rate the gap shrinks.
▸ Why?
Shrinking at a steady 80 miles per hour, the gap loses 80 miles each hour, so the time to erase the 3-mile gap is 3 divided by 80.
▸ Why?
Distance equals rate times time, so 80 times the meeting time must equal 3 miles; dividing the 3 miles by 80 undoes that multiplication to release the time.
Find the meeting point
In that hr Car A covers 40 × = 1.5 more miles, so they meet at 7 + 1.5 = 8.5 — inside middle segment (5 ＜ 8.5 ＜ 10), choice (D).
Adding the decimal 7 + 1.5 = 8.5 is Grade 5 decimal arithmetic to hundredths.
5.NBT.B.7Analyze The UnitsThis 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!
- Label each segment's speed
- Find each first segment time
- Compare the two times
- Locate Car A at that moment
- Find the closing time
- Find the meeting point
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