2009 AMC 10A Problem 4

P1-10~2 minPrint

Linear Equations & Word Problems Set up the equation/formula and compute; no special trick needed

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?

Show solution without answering

Key insight

Find how long the swim and the run take, subtract from 2 hours, and use the time that is left for the 15-mile bike ride.

Solution

Speeds cannot be added, but times can. So turn each part of the race into a time using $\text{time} = \dfrac{\text{distance}}{\text{speed}}$ .

1. Swim: $\frac14$ mile at $2$ miles per hour takes $\dfrac{1/4}{2} = \dfrac{1}{8}$ hour.
2. Run: $3$ miles at $6$ miles per hour takes $\dfrac{3}{6} = \dfrac{1}{2}$ hour.
3. Together these take $\frac18 + \frac48 = \frac58$ hour.
4. The whole race must take $2$ hours, so the bike ride gets
$2 - \frac{5}{8} = \frac{16}{8} - \frac{5}{8} = \frac{11}{8}\ \text{hour}.$
5. Speed is distance divided by time. For $15$ miles in $\frac{11}{8}$ hour:
$\frac{15}{11/8} = 15 \times \frac{8}{11} = \frac{120}{11}\ \text{miles per hour}.$

The answer is $\boxed{\textbf{(A)}\ \frac{120}{11}}$ .

Why this works

In a trip with several parts, the times of the parts add up to the total time, but the speeds do not. So change every known part into a time, find the time left over for the unknown part, and divide once to get its speed.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Compute the time each known leg consumes using $\text{time} = \frac{\text{distance}}{\text{speed}}$ .

Swim: $\dfrac{1/4}{2} = \dfrac{1}{8}$ hour. Run: $\dfrac{3}{6} = \dfrac{1}{2}$ hour.

Together these take $\frac{1}{8} + \frac{4}{8} = \frac{5}{8}$ hour, so the bike ride must fit into
$2 - \frac{5}{8} = \frac{11}{8}\ \text{hour}.$

Covering $15$ miles in $\frac{11}{8}$ hour requires a speed of
$\frac{15}{11/8} = \frac{15 \cdot 8}{11} = \frac{120}{11}\ \text{mph}.$

The answer is $\boxed{\textbf{(A)}\ \frac{120}{11}}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 7: Proportion (§7.4)