2014 AMC 10B Problem 3

P1-10~2 minPrint

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

Peter drove the first third of his trip on a gravel road, the next $20$ miles on pavement, and the remaining one-fifth on a dirt road. In miles how long was Peter's trip?

Show solution without answering

Key insight

Gravel and dirt together are 1/3 + 1/5 = 8/15 of the trip, so the 20 paved miles are the remaining 7/15 of it.

Solution

The trip has three parts. Two are given as fractions of the whole trip and one is given in miles. So first find what fraction of the trip the paved part is.

Put the two fractions over a common denominator of $15$ : the gravel part is $\frac13 = \frac{5}{15}$ of the trip and the dirt part is $\frac15 = \frac{3}{15}$ .
Together they are $\frac{5}{15} + \frac{3}{15} = \frac{8}{15}$ of the trip.
The paved part is what is left: $1 - \frac{8}{15} = \frac{7}{15}$ of the trip.
That paved part is $20$ miles. So $\frac{7}{15}$ of the trip is $20$ miles, which means $\frac{1}{15}$ of the trip is $\frac{20}{7}$ miles.
The whole trip is $15$ of those pieces: $15 \times \frac{20}{7} = \frac{300}{7}$ miles.

Check: a third of $\frac{300}{7}$ is $\frac{100}{7}$ , a fifth is $\frac{60}{7}$ , and $\frac{100}{7} + \frac{140}{7} + \frac{60}{7} = \frac{300}{7}$ , since $20 = \frac{140}{7}$ .

The answer is $\boxed{\textbf{(E)}\ \frac{300}{7}}$ .

Why this works

When some parts of a whole are given as fractions and one part is given as an actual amount, the actual amount is "whatever fraction is left over." Find that leftover fraction, then scale up to the whole. Do not be put off by an answer like $\frac{300}{7}$ : the choices themselves show that the answer need not be a whole number.

Concise solution

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

Let the whole trip be $T$ miles. The gravel and dirt portions together are
$\frac13 T + \frac15 T = \frac{5 + 3}{15} T = \frac{8}{15} T.$

So the paved stretch is the rest: $T - \frac{8}{15}T = \frac{7}{15}T$ . That stretch is $20$ miles, so
$\frac{7}{15} T = 20 \quad\Longrightarrow\quad T = 20 \cdot \frac{15}{7} = \frac{300}{7}.$

The answer is $\boxed{\textbf{(E)}\ \frac{300}{7}}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)
Prealgebra, Chapter 13: Data and Statistics (§13.1)