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2009 AMC 10A Problem 4

2009 AMC 10A Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Linear Equations & Word Problems · difficulty 1 of 5 · about 2 min · position P1-10

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?

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

Answer

Key insight

Budget time, not speed: the swim takes 1/8 hour and the run 1/2 hour, leaving 11/8 hour for the 15-mile ride.

Solution

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

Why this works

Multi-leg trips are handled by converting every leg into the quantity that adds: time. Speeds do not add and cannot be averaged directly, but times do. Once the time budget for the unknown leg is known, one division gives the required speed.

The trap

Averaging the speeds or leg times instead of computing time = distance / speed for each leg and subtracting from 2 hours.

Common mistakes

Averaging the speeds or leg times instead of computing time = distance / speed for each leg and subtracting from 2 hours.
Computing the swim time as $\frac{1}{4} \cdot 2 = \frac{1}{2}$ hour (multiplying instead of dividing), which leads to a speed of $12$ , choice (E).

Techniques

Set up the equation/formula and compute; no special trick needed