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2010 AMC 10A Problem 13

2010 AMC 10A Problem 13AMC 10 Step by Step

Name: Date: September 13, 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

Driving time is 3 hours minus the 20-minute stop, i.e. 8/3 hours; distance is 80 times the first stretch plus 100 times the rest.

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

Why this works

Two-rate trips are always "rate one times time one plus rate two times time two equals total distance." The only bookkeeping is that the times must add to the driving time, not the total elapsed time, so the stop is subtracted first.

The trap

Using 3 hours instead of 8/3 as the driving time, or attaching 80 to the second stretch (choice E).

Common mistakes

Using 3 hours instead of 8/3 as the driving time, or attaching 80 to the second stretch (choice E).
Averaging the two speeds to $90$ km/h (choice D), which is only valid if the two stretches took equal time, something the problem does not say.

Techniques

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