Difference between revisions of "2010 AMC 10A Problems/Problem 13"
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Angelina drove at an average rate of <math>80</math> kph and then stopped <math>20</math> minutes for gas. After the stop, she drove at an average rate of <math>100</math> kph. Altogether she drove <math>250</math> km in a total trip time of <math>3</math> hours including the stop. Which equation could be used to solve for the time <math>t</math> in hours that she drove before her stop? | Angelina drove at an average rate of <math>80</math> kph and then stopped <math>20</math> minutes for gas. After the stop, she drove at an average rate of <math>100</math> kph. Altogether she drove <math>250</math> km in a total trip time of <math>3</math> hours including the stop. Which equation could be used to solve for the time <math>t</math> in hours that she drove before her stop? | ||
− | <math>\textbf{(A)}\ 80t+100(\frac{8}{3}-t)=250 \qquad \textbf{(B)}\ 80t=250 \qquad \textbf{(C)}\ 100t=250 | + | <math>\textbf{(A)}\ 80t+100(\frac{8}{3}-t)=250 \qquad \textbf{(B)}\ 80t=250 \qquad \textbf{(C)}\ 100t=250 \qquad \textbf{(D)}\ 90t=250 \qquad \textbf{(E)}\ 80(\frac{8}{3}-t)+100t=250</math> |
− | \qquad \textbf{(D)}\ 90t=250 \qquad \textbf{(E)}\ 80(\frac{8}{3}-t)+100t=250</math> | ||
− | The answer is <math>A</math> because she drove at <math>80</math> kmh for <math>t</math> hours (the amount of time before the stop), and 100 kmh for <math>\frac{8}{3}-t</math> because she wasn't driving for <math>20</math> minutes, or <math>\frac{1}{3}</math> hours. Multiplying by <math>t</math> gives the total distance, which is <math>250</math> kms. Therefore, the answer is <math>80t+100(\frac{8}{3}-t)=250</math> | + | |
+ | ==Solution== | ||
+ | |||
+ | The answer is <math>A</math> because she drove at <math>80</math> kmh for <math>t</math> hours (the amount of time before the stop), and 100 kmh for <math>\frac{8}{3}-t</math> because she wasn't driving for <math>20</math> minutes, or <math>\frac{1}{3}</math> hours. Multiplying by <math>t</math> gives the total distance, which is <math>250</math> kms. Therefore, the answer is <math>80t+100(\frac{8}{3}-t)=250</math> <math>\Rightarrow</math> <math>\boxed{(A)}</math> | ||
+ | |||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2010|ab=A|num-b=12|num-a=14}} |
Revision as of 09:58, 2 January 2012
Problem
Angelina drove at an average rate of kph and then stopped minutes for gas. After the stop, she drove at an average rate of kph. Altogether she drove km in a total trip time of hours including the stop. Which equation could be used to solve for the time in hours that she drove before her stop?
Solution
The answer is because she drove at kmh for hours (the amount of time before the stop), and 100 kmh for because she wasn't driving for minutes, or hours. Multiplying by gives the total distance, which is kms. Therefore, the answer is
See Also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |