# 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 04:19, 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