Difference between revisions of "2010 AMC 10A Problems/Problem 13"
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<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> | ||
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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> | ||
Revision as of 03:19, 31 December 2010
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?
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 
