AMC 8 · 2025 · #19
Grade 6 rate-ratioProblem
Two towns, A and B, are connected by a straight road, 15 miles long. Traveling from town A to town B, the speed limit changes every 5 miles: from 25 to 40 to 20 miles per hour (mph). Two cars, one at town A and one at town B, start moving toward each other at the same time. They drive at exactly the speed limit in each portion of the road. How far from town A, in miles, will the two cars meet?
Pick an answer.
AMC 8 2025 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: 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?
Givens: Total road length $= 15$ miles, made of three $5$-mile segments; Speed limits from $A$ to $B$: $25$ mph, $40$ mph, $20$ mph; Both cars start at the same time and travel toward each other; Each car always drives at the posted speed of the segment it is currently in; Answer choices (miles from $A$): (A) $7.75$, (B) $8$, (C) $8.25$, (D) $8.5$, (E) $8.75$
Unknowns: The distance from town $A$ to the meeting point, in miles
Understand
Restated: 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?
Givens: Total road length $= 15$ miles, made of three $5$-mile segments; Speed limits from $A$ to $B$: $25$ mph, $40$ mph, $20$ mph; Both cars start at the same time and travel toward each other; Each car always drives at the posted speed of the segment it is currently in; Answer choices (miles from $A$): (A) $7.75$, (B) $8$, (C) $8.25$, (D) $8.5$, (E) $8.75$
Plan
Primary tool: #8 Analyze the Units
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
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 $\text{time} = \text{distance} \div \text{speed}$ 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.
Execute — Answer: D
4.MD.A.2 Step 1 - Draw the road and label the speed in each segment from Car A's point of view: $0$-$5$ mi at $25$ mph, $5$-$10$ mi at $40$ mph, $10$-$15$ mi at $20$ mph.
- Car A starts at mile $0$ moving right; Car B starts at mile $15$ moving left.
- Because the speed limits are set by the road (not by the car), Car B's first $5$ miles (from $15$ to $10$) are at $20$ mph, its middle $5$ (from $10$ to $5$) at $40$ mph, and its last $5$ (from $5$ to $0$) at $25$ mph.
💡 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.2 Step 2 - Find how long each car takes to finish its first $5$-mile segment, using $\text{time} = \text{distance} \div \text{speed}$.
- Track the units: miles divided by mph leaves hours, exactly what we want.
💡 Miles $\div$ (miles/hour) cancels "miles" and leaves "hours" — the units do the bookkeeping for us.
5.NF.A.1 Step 3 - Since $\tfrac{1}{5} < \tfrac{1}{4}$, Car A enters the middle segment first.
- So at the moment Car B finishes its own first segment (at time $t = \tfrac{1}{4}$ hr), Car A is already past mile $5$.
- The meeting must happen at or after $t = \tfrac{1}{4}$ hr, somewhere in the middle segment.
💡 Comparing $\tfrac{1}{5}$ and $\tfrac{1}{4}$ with unlike denominators is the standard Grade 5 fraction-comparison move.
5.NF.A.1 Step 4 - Find Car A's position at the moment $t = \tfrac{1}{4}$ hr.
- Car A used $\tfrac{1}{5}$ hr to cover the first $5$ miles, leaving $\tfrac{1}{4} - \tfrac{1}{5} = \tfrac{5}{20} - \tfrac{4}{20} = \tfrac{1}{20}$ hr in the middle segment at $40$ mph.
- That carries it an extra $40 \times \tfrac{1}{20} = 2$ miles past mile $5$.
💡 Subtracting $\tfrac{1}{4} - \tfrac{1}{5}$ uses Grade 5 unlike-denominator subtraction, then $40 \text{ mph} \times \text{hr}$ cleanly cancels to miles.
6.RP.A.3 Step 5 - Now both cars are in the middle segment, both moving at $40$ mph.
- Car A is at mile $7$ and Car B is at mile $10$, so the gap between them is $3$ miles.
- Two cars approaching each other close the gap at the SUM of their speeds, $40 + 40 = 80$ mph.
- So the remaining time until they meet is $\tfrac{3 \text{ mi}}{80 \text{ mph}} = \tfrac{3}{80}$ hr.
💡 Treating $80$ mph as the unit closing-rate (miles closed per hour) is Grade 6 rate reasoning at its cleanest.
5.NBT.B.7 Step 6 - During that last $\tfrac{3}{80}$ hr, Car A travels another $40 \times \tfrac{3}{80} = \tfrac{120}{80} = 1.5$ miles, ending at mile $7 + 1.5 = 8.5$.
- That is the meeting point, which sits inside the middle segment ($5 < 8.5 < 10$) as expected — matching choice (D).
💡 Adding the decimal $7 + 1.5 = 8.5$ is Grade 5 decimal arithmetic to hundredths.
4.MD.A.2 Draw the road and label the speed in each segment from Car A's point of view: $0 4.MD.A.2 Find how long each car takes to finish its first $5$-mile segment, using $\text{ 5.NF.A.1 Since $\tfrac{1}{5} < \tfrac{1}{4}$, Car A enters the middle segment first. So a 5.NF.A.1 Find Car A's position at the moment $t = \tfrac{1}{4}$ hr. Car A used $\tfrac{1} 6.RP.A.3 Now both cars are in the middle segment, both moving at $40$ mph. Car A is at mi 5.NBT.B.7 During that last $\tfrac{3}{80}$ hr, Car A travels another $40 \times \tfrac{3}{ Review
Reasonableness: 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 $\tfrac{5}{25} + \tfrac{3.5}{40} = \tfrac{1}{5} + \tfrac{7}{80} = \tfrac{16}{80} + \tfrac{7}{80} = \tfrac{23}{80}$ hr to reach mile $8.5$; Car B takes $\tfrac{5}{20} + \tfrac{1.5}{40} = \tfrac{1}{4} + \tfrac{3}{80} = \tfrac{20}{80} + \tfrac{3}{80} = \tfrac{23}{80}$ hr to reach mile $8.5$ from town $B$. Equal times — confirmed.
Alternative: Tool #13 (Convert to Algebra): let $d$ be the meeting distance from $A$. Because the meeting is in the middle segment, set Car A's time equal to Car B's time: $\tfrac{5}{25} + \tfrac{d-5}{40} = \tfrac{5}{20} + \tfrac{10-d}{40}$. Multiplying through by $40$ gives $8 + (d-5) = 10 + (10-d)$, i.e., $d + 3 = 20 - d$, so $2d = 17$ and $d = 8.5$ — the same answer, but the rate-reasoning path above avoids fractions until the very end.
CCSS standards used (min grade 6)
4.MD.A.2Solve word problems involving distances, time, liquid volumes, and money (Setting up the road picture and applying $\text{time} = \text{distance} \div \text{speed}$ for each car's first $5$-mile segment ($T_{A,1} = \tfrac{1}{5}$ hr, $T_{B,1} = \tfrac{1}{4}$ hr).)5.NF.A.1Add and subtract fractions with unlike denominators (Comparing $\tfrac{1}{5}$ vs $\tfrac{1}{4}$ and computing $\tfrac{1}{4} - \tfrac{1}{5} = \tfrac{1}{20}$ hr — the extra time Car A spends in the middle segment before Car B even leaves its first segment.)5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Adding the decimal distance $7 + 1.5 = 8.5$ miles to report the meeting point in the same format as the answer choices.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Adding the two cars' speeds to get an $80$ mph closing rate, then dividing the $3$-mile gap by $80$ mph to get the time until they meet — the core rate insight of the whole problem.)
⭐ 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!
⭐ 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!
More like this
Same archetype — closest grade level first.