2010 AMC 10A Problem 13

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Concepts:distance rate and timelinear equation

Difficulty rating: 1370

13.

Angelina drove at an average rate of 8080 kph and then stopped 2020 minutes for gas. After the stop, she drove at an average rate of 100100 kph. Altogether she drove 250250 km in a total trip time of 33 hours including the stop. Which equation could be used to solve for the time tt in hours that she drove before her stop?

80t+100(83t)=25080t + 100\left(\dfrac83 - t\right) = 250

80t=25080t = 250

100t=250100t = 250

90t=25090t = 250

80(83t)+100t=25080\left(\dfrac83 - t\right) + 100t = 250

Solution:

Before the stop, Angelina drove 80t80t km.

The stop takes 13\frac{1}{3} of an hour, so her total driving time is 313=833-\frac13=\frac83 hours. After the stop, she drives for 83t\frac83-t hours, covering 100(83t)100\left(\frac83-t\right) km.

The total distance equation is 80t+100(83t)=250.80t+100\left(\frac83-t\right)=250.

Thus, A is the correct answer.

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