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2010 AMC 10A Problem 13

2010 AMC 10A Problem 13AMC 10 Step by Step

Name: Date: October 28, 2026

Linear Equations & Word Problems · difficulty 2 of 5 · about 2 min · position P11-15

Angelina drove at an average rate of $80$ km/h and then stopped $20$ minutes for gas. After the stop, she drove at an average rate of $100$ km/h. 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?

A) $80t + 100\left(\frac{8}{3} -t\right) = 250$
B) $80t = 250$
C) $100t = 250$
D) $90t = 250$
E) $80\left(\frac{8}{3} -t\right) + 100t = 250$

Answer

Key insight

Take the 20-minute stop out of the 3 hours first: she drove for 8/3 hours. Then distance = 80 times the first part plus 100 times the rest.

Solution

Distance equals speed times time, for each part of the trip. First find how long she was actually driving.

1. Driving time. The trip took $3$ hours, but $20$ minutes of that was the stop. $20$ minutes is $\tfrac{20}{60} = \tfrac13$ of an hour. So she drove for
$3 - \frac13 = \frac83 \text{ hours}.$
2. Time for each part. She drove $t$ hours before the stop. The rest of the driving time came after the stop: $\tfrac83 - t$ hours.
3. Distance for each part.
- Before the stop, at $80$ km/h: $80t$ km.
- After the stop, at $100$ km/h: $100\left(\tfrac83 - t\right)$ km.
4. Total distance. The two parts add to $250$ km:
$80t + 100\left(\frac{8}{3} - t\right) = 250.$

This is choice (A). Choice (E) has the speeds swapped: it uses $80$ km/h after the stop.

The answer is $\boxed{\textbf{(A)}\ 80t + 100\left(\frac{8}{3} -t\right) = 250}$ .

Why this works

A trip at two speeds is "speed times time for part one, plus speed times time for part two, equals total distance." The times must add up to the time spent moving, so subtract any stops first. Then check that each speed is paired with the right part of the trip.

The trap

Using 3 hours as the driving time, or matching 80 km/h with the time after the stop, which gives choice (E).

Concise solution

The $20$ -minute stop is $\dfrac{1}{3}$ hour, so the time actually spent driving is $3 - \dfrac13 = \dfrac{8}{3}$ hours.

If she drove $t$ hours before the stop, she drove $\dfrac{8}{3} - t$ hours after it. Distance is rate times time, so the first stretch covers $80t$ km and the second covers $100\left(\dfrac83 - t\right)$ km. Their sum is the total distance:
$80t + 100\left(\frac{8}{3} - t\right) = 250.$

The answer is $\boxed{\textbf{(A)}\ 80t + 100\left(\frac{8}{3} -t\right) = 250}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 7: Proportion (§7.4)

Techniques

Set up the equation/formula and compute; no special trick needed