2009 AMC 10A Problems/Problem 4

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 $\frac{15}{\frac{11}{8}}=\frac{120}{11} \frac{\text{miles}}{\text{hour}}$

$\longrightarrow \fbox{A}$

See also

2009 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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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