AMC 10 · 2025 · #1
Grade 6 rate-ratioAndy and Betsy both live in Mathville. Andy leaves Mathville on his bicycle at 1:30, traveling due north at a steady 8 miles per hour. Betsy leaves on her bicycle from the same point at 2:30, traveling due east at a steady 12 miles per hour. At what time will they be exactly the same distance from their common starting point?
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AMC 10 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: Andy leaves a starting point at $1{:}30$, biking due north at a steady $8$ mph. Betsy leaves the same point at $2{:}30$, biking due east at a steady $12$ mph. Find the clock time at which Andy and Betsy are the same distance from the starting point.
Givens: Andy departs at $1{:}30$ heading due north at a steady $8$ mph.; Betsy departs at $2{:}30$ from the same point heading due east at a steady $12$ mph.; Betsy leaves exactly $1$ hour after Andy.; Answer choices: (A) $3{:}30$, (B) $3{:}45$, (C) $4{:}00$, (D) $4{:}15$, (E) $4{:}30$.
Unknowns: The clock time when the two riders are equally far from the starting point.
Understand
Restated: Andy leaves a starting point at $1{:}30$, biking due north at a steady $8$ mph. Betsy leaves the same point at $2{:}30$, biking due east at a steady $12$ mph. Find the clock time at which Andy and Betsy are the same distance from the starting point.
Givens: Andy departs at $1{:}30$ heading due north at a steady $8$ mph.; Betsy departs at $2{:}30$ from the same point heading due east at a steady $12$ mph.; Betsy leaves exactly $1$ hour after Andy.; Answer choices: (A) $3{:}30$, (B) $3{:}45$, (C) $4{:}00$, (D) $4{:}15$, (E) $4{:}30$.
Plan
Primary tool: #8 Analyze the Units
Secondary: #16 Change Focus / Count the Complement, #4 Introduce a Variable
Perpendicular directions tempt you to reach for the Pythagorean theorem, but that would give the distance BETWEEN the riders. The question asks how far each is FROM the start, and since each travels a single straight line, the units tell the whole story: miles $=$ (miles per hour) $\times$ (hours). That is Tool #8 — let the units pick the model. Once distance is just speed $\times$ time, Tool #16 (change focus) makes it fast: instead of tracking two growing distances, watch the single gap between them and how quickly it closes. Tool #4 (introduce a variable) gives the same answer through a set-the-distances-equal equation, kept as a backup.
Execute — Answer: E
6.RP.A.3 Step 1 Pick the model from the units
- Each rider travels one straight line away from home, so distance from the start is speed $\times$ time.
- The units confirm the model: miles $=$ (miles/hour) $\times$ hours.
- Because we want each rider's distance FROM home, not the distance BETWEEN them, the perpendicular directions never enter and no Pythagorean theorem is needed.
💡 Going straight out from home means your distance from home is just how fast times how long.
6.RP.A.3 Step 2 Andy's one-hour head start
- Andy rides a full hour, from $1{:}30$ to $2{:}30$, before Betsy even starts.
- In that hour he covers $8 \times 1 = 8$ miles.
- So at $2{:}30$ Andy is $8$ miles from home while Betsy is still $0$ miles from home.
💡 The one-hour gap turns into an $8$-mile lead in distance-from-home.
6.RP.A.3 Step 3 Track the gap, not both distances
- Rather than follow two growing distances, watch the gap between them.
- After $2{:}30$, each hour Betsy's distance-from-home grows by $12$ miles while Andy's grows by only $8$, so the gap shrinks by $12 - 8 = 4$ miles every hour.
💡 Betsy is faster, so she eats into Andy's lead at the difference of their speeds.
6.NS.B.2 Step 4 When does the gap reach zero?
- The gap starts at $8$ miles (Andy's lead at $2{:}30$) and closes at $4$ miles per hour.
- It disappears — meaning the two distances are equal — after $8 \div 4 = 2$ hours of Betsy riding.
💡 An $8$-mile lead closing at $4$ miles per hour is gone in exactly $2$ hours.
3.MD.A.1 Step 5 Convert to clock time
- Betsy starts at $2{:}30$, so $2$ hours later the clock reads $4{:}30$.
- At that moment Andy has ridden $3$ hours and Betsy $2$ hours, and both are $24$ miles from home.
- The time is $4{:}30$, which is choice (E).
💡 Add the $2$-hour ride onto Betsy's $2{:}30$ start to land on the clock time.
6.RP.A.3 Each rider travels one straight line away from home, so distance from the start 6.RP.A.3 Andy rides a full hour, from $1{:}30$ to $2{:}30$, before Betsy even starts. In 6.RP.A.3 Rather than follow two growing distances, watch the gap between them. After $2{: 6.NS.B.2 The gap starts at $8$ miles (Andy's lead at $2{:}30$) and closes at $4$ miles pe 3.MD.A.1 Betsy starts at $2{:}30$, so $2$ hours later the clock reads $4{:}30$. At that m Review
Reasonableness: Check the tie directly: at $4{:}30$ Andy has ridden $3$ hours and Betsy $2$ hours, giving $8 \times 3 = 24$ miles and $12 \times 2 = 24$ miles — an exact match, so (E) is consistent. Earlier choices fail: at $4{:}00$ Andy is $8 \times 2.5 = 20$ miles out but Betsy only $12 \times 1.5 = 18$ miles, so Betsy is still behind. The gap first hits zero at $4{:}30$.
Alternative: Introduce a variable (Tool #4): let $t$ be the hours since Andy left, so Betsy has ridden $t - 1$ hours. Setting the distances equal gives $8t = 12(t - 1)$, hence $8t = 12t - 12$ and $t = 3$, so the clock reads $1{:}30 + 3 = 4{:}30$. This works, but the variable lands on both sides of the equation — formally a Grade 8 move — while the gap-closing view keeps everything at Grade 6 rate reasoning.
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Modeling each rider's distance from home as speed $\times$ time, and finding how fast the gap between the two distances closes ($12 - 8 = 4$ mph).)6.NS.B.2Fluently divide multi-digit numbers using the standard algorithm (Dividing the $8$-mile lead by the $4$-mph closing rate to get the $2$ hours Betsy needs to catch up in distance.)3.MD.A.1Tell and write time to the nearest minute and solve elapsed time problems (Adding the $2$-hour riding interval onto Betsy's $2{:}30$ start to convert the answer into the clock time $4{:}30$.)
⭐ When two people ride straight away from the same spot, don't juggle both distances — watch the gap between them and how fast it closes. Andy's $8$-mile head start vanishing at $4$ miles per hour means they tie $2$ hours after Betsy starts, at $4{:}30$.
⭐ When two people ride straight away from the same spot, don't juggle both distances — watch the gap between them and how fast it closes. Andy's $8$-mile head start vanishing at $4$ miles per hour means they tie $2$ hours after Betsy starts, at $4{:}30$.
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