Difference between revisions of "2010 AMC 10A Problems/Problem 13"
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| − | ==Problem== | + | == Problem == |
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| − | <math>\textbf{(A)}\ 80t+100(\frac{8}{3}-t)=250 \qquad \textbf{(B)}\ 80t=250 \qquad \textbf{(C)}\ 100t=250 | + | Angelina drove at an average rate of <math>80</math> kmh and then stopped <math>20</math> minutes for gas. After the stop, she drove at an average rate of <math>100</math> kmh. 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? |
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| + | <math>\textbf{(A)}\ 80t+100\left(\frac{8}{3}-t\right)=250 \qquad \textbf{(B)}\ 80t=250 \qquad \textbf{(C)}\ 100t=250 \qquad \textbf{(D)}\ 90t=250 \qquad \textbf{(E)}\ 80\left(\frac{8}{3}-t\right)+100t=250</math> | ||
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| + | == Solution == | ||
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| + | 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 <math>100</math> 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> km. Therefore, the answer is <math>80t+100\left(\frac{8}{3}-t\right)=250</math> <math>\Rightarrow</math> <math>\boxed{(A)}</math> | ||
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| + | == See Also == | ||
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| + | {{AMC10 box|year=2010|ab=A|num-b=12|num-a=14}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:01, 29 October 2024
Problem
Angelina drove at an average rate of 
kmh and then stopped 
minutes for gas. After the stop, she drove at an average rate of 
kmh. 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 kmh for 
because she wasn't driving for minutes, or 
hours. Multiplying by gives the total distance, which is km. 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
