2009 AMC 10A Problem 4

Below is the professionally curated solution for Problem 4 of the 2009 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10A solutions, or check the answer key.

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Concepts:distance rate and time fraction

Difficulty rating: 1030

Eric plans to compete in a triathlon. He can average $2$ miles per hour in the $\tfrac14$ -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?

$\dfrac{120}{11}$

$11$

$\dfrac{56}{5}$

$\dfrac{45}{4}$

$12$

Solution:

The swim takes $\dfrac{1/4}{2} = \dfrac18$ hour and the run takes $\dfrac{3}{6} = \dfrac12$ hour. This leaves $2 - \dfrac18 - \dfrac12 = \dfrac{11}{8}$ hours for the bicycle ride.

His average speed must be $\dfrac{15}{11/8} = \dfrac{120}{11}$ miles per hour.

Thus, the correct answer is A.

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