Difference between revisions of "2010 AMC 10A Problems/Problem 13"

(Problem)
(Problem)
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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  
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<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>
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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 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>
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== See Also ==
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{{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 $80$ kph and then stopped $20$ minutes for gas. After the stop, she drove at an average rate of $100$ kph. 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?

$\textbf{(A)}\ 80t+100(\frac{8}{3}-t)=250 \qquad \textbf{(B)}\ 80t=250 \qquad \textbf{(C)}\ 100t=250 \qquad  \textbf{(D)}\ 90t=250 \qquad \textbf{(E)}\ 80(\frac{8}{3}-t)+100t=250$


Solution

The answer is $A$ because she drove at $80$ kmh for $t$ hours (the amount of time before the stop), and 100 kmh for $\frac{8}{3}-t$ because she wasn't driving for $20$ minutes, or $\frac{1}{3}$ hours. Multiplying by $t$ gives the total distance, which is $250$ kms. Therefore, the answer is $80t+100(\frac{8}{3}-t)=250$ $\Rightarrow$ $\boxed{(A)}$


See Also

2010 AMC 10A (ProblemsAnswer KeyResources)
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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