2010 AMC 10A Problems/Problem 13

Revision as of 04:19, 31 December 2010 by DVA6102 (talk | contribs) (Problem)

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?

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

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 $80t+100(\frac{8}{3}-t)=250$

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