AMC 10 · 2010 · #13
Grade 7 rate-ratioAngelina drove at an average rate of 80 kmh and then stopped 20 minutes for gas. After the stop, she drove at an average rate of 100 kmh. Altogether she drove 250 km in a total trip time of 3 hours including the stop. Which equation could be used to solve for the time t in hours that she drove before her stop?
Pick an answer.
AMC 10 2010 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: Angelina drove at $80$ km/h, stopped $20$ minutes for gas, then drove at $100$ km/h. The whole trip covered $250$ km and lasted $3$ hours including the stop. Let $t$ be the number of hours she drove before the stop. Which equation could be used to find $t$?
Givens: First leg speed is $80$ km/h; Gas stop lasts $20$ minutes (no driving); Second leg speed is $100$ km/h; Total distance driven is $250$ km; Total trip time, including the stop, is $3$ hours; Answer choices: (A) $80t+100\left(\frac{8}{3}-t\right)=250$, (B) $80t=250$, (C) $100t=250$, (D) $90t=250$, (E) $80\left(\frac{8}{3}-t\right)+100t=250$
Unknowns: The equation whose solution is $t$, the hours driven before the stop
Understand
Restated: Angelina drove at $80$ km/h, stopped $20$ minutes for gas, then drove at $100$ km/h. The whole trip covered $250$ km and lasted $3$ hours including the stop. Let $t$ be the number of hours she drove before the stop. Which equation could be used to find $t$?
Givens: First leg speed is $80$ km/h; Gas stop lasts $20$ minutes (no driving); Second leg speed is $100$ km/h; Total distance driven is $250$ km; Total trip time, including the stop, is $3$ hours; Answer choices: (A) $80t+100\left(\frac{8}{3}-t\right)=250$, (B) $80t=250$, (C) $100t=250$, (D) $90t=250$, (E) $80\left(\frac{8}{3}-t\right)+100t=250$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #8 Analyze the Units, #4 Introduce a Variable
The question does not ask for a number — it asks which equation models the trip. That is exactly Tool #13 (Convert to Algebra): translate each English phrase into a symbol and glue them into one equation. Tool #8 (Analyze the Units) is essential first, because the stop is given in minutes while everything else is in hours — the $20$ minutes must become $\frac{1}{3}$ hour before any time can be subtracted, and each distance is a rate (km/h) times a time (h). Tool #4 (Introduce a Variable) names the two driving times: $t$ before the stop and $\frac{8}{3}-t$ after it, since the two driving legs must share the $\frac{8}{3}$ hours of actual driving. Guessing (Tool #6) is unhelpful here because no leg time is a whole number and the answer is an equation, not a value.
Execute — Answer: A
6.RP.A.3 Step 1 Turn the stop into hours
- Everything is in kilometers and hours except the stop, which is in minutes.
- Convert $20$ minutes to hours, then subtract it from the $3$-hour total to find how long she was actually driving.
💡 The clock runs during the stop but the odometer does not, so only $\frac{8}{3}$ of the $3$ hours can create distance.
6.EE.A.2 Step 2 Name the two driving times
- Let $t$ be the hours driven before the stop.
- The two driving legs together must fill the $\frac{8}{3}$ driving hours, so the time after the stop is whatever is left.
💡 Once one piece of a fixed total is called $t$, the other piece is forced to be the total minus $t$.
6.RP.A.3 Step 3 Write each leg's distance
- Distance on each leg is its speed times its time.
- Use $80$ km/h for the first leg and $100$ km/h for the second, matching the times from the previous step.
💡 Rate times time gives distance, so km/h multiplied by hours cancels to kilometers for each leg.
7.EE.B.4 Step 4 Add the legs to $250$
- The two driving legs together cover the full $250$ km.
- Add the two distance expressions and set the sum equal to $250$ to get the equation for $t$.
- This matches choice (A).
💡 The whole $250$ km is just the first leg plus the second leg, so their expressions must add to $250$.
6.RP.A.3 Everything is in kilometers and hours except the stop, which is in minutes. Conv 6.EE.A.2 Let $t$ be the hours driven before the stop. The two driving legs together must 6.RP.A.3 Distance on each leg is its speed times its time. Use $80$ km/h for the first le 7.EE.B.4 The two driving legs together cover the full $250$ km. Add the two distance expr Review
Reasonableness: Solve the chosen equation to see it behaves sensibly: $80t + \frac{800}{3} - 100t = 250 \Rightarrow -20t = 250 - \frac{800}{3} = -\frac{50}{3} \Rightarrow t = \frac{5}{6}$ h. That is $50$ minutes before the stop, leaving $\frac{8}{3} - \frac{5}{6} = \frac{11}{6}$ h $\approx 110$ min after — both positive and inside the $3$-hour window, so the equation is physically meaningful. Choice (E) swaps the speeds onto the wrong legs, and (B), (C), (D) ignore the stop and the two-speed split, so only (A) fits.
Alternative: Eliminate possibilities (Tool #3): the true distance is a mix of $80$ and $100$ km/h over $\frac{8}{3}$ driving hours, so the equation must contain both $80$ and $100$ and the number $\frac{8}{3}$. Only (A) and (E) do; (E) attaches $100$ to the before-stop time $t$, which contradicts the story, leaving (A).
CCSS standards used (min grade 7)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Converting $20$ minutes to $\frac{1}{3}$ hour and using rate $\times$ time $=$ distance to write each leg as $80t$ and $100\left(\frac{8}{3}-t\right)$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming the before-stop time $t$ and writing the after-stop time as the expression $\frac{8}{3}-t$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Combining the two distance expressions into the single equation $80t+100\left(\frac{8}{3}-t\right)=250$ that models the whole trip.)
⭐ Subtract the stop time first, split the leftover driving time into before and after, then add the two speed-times-time distances to $250$.
⭐ Subtract the stop time first, split the leftover driving time into before and after, then add the two speed-times-time distances to $250$.
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