Difference between revisions of "2010 AMC 10A Problems/Problem 13"
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== Solution == | == 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 <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> | + | 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> |
== See Also == | == See Also == | ||
{{AMC10 box|year=2010|ab=A|num-b=12|num-a=14}} | {{AMC10 box|year=2010|ab=A|num-b=12|num-a=14}} | ||
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{{MAA Notice}} | {{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 |
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