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

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

Revision as of 14:12, 29 November 2020

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 (Problems • Answer Key • Resources)
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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