Difference between revisions of "2009 AMC 10A Problems/Problem 4"

Revision as of 10:55, 4 July 2013

Problem

Eric plans to compete in a triathlon. He can average $2$ miles per hour in the $\frac{1}{4}$ -mile swim and $6$ miles per hour in the $3$ -mile run. His goal is to finish the triathlon in $2$ hours. To accomplish his goal what must his average speed in miles per hour, be for the $15$ -mile bicycle ride?

$\mathrm{(A)}\ \frac{120}{11} \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ \frac{56}{5} \qquad \mathrm{(D)}\ \frac{45}{4} \qquad \mathrm{(E)}\ 12$

Solution

Since $d=rt$ , Eric takes $\frac{\frac{1}{4}}{2}=\frac{1}{8}$ hours for the swim. Then, he takes $\frac{3}{6}=\frac{1}{2}$ hours for the run. So he needs to take $2-\frac{5}{8}=\frac{11}{8}$ hours for the $15$ mile run. This is $\frac{15}{\frac{11}{8}}=\frac{120}{11} \frac{\text{miles}}{\text{hour}}$

$\longrightarrow \fbox{A}$

2009 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==See also==
 {{AMC10 box|year=2009|ab=A|num-b=3|num-a=5}}
+{{MAA Notice}}

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Difference between revisions of "2009 AMC 10A Problems/Problem 4"

Revision as of 10:55, 4 July 2013

Problem

Solution

See also