AMC 10 Step by Step
Log inSign up
Printable

2014 AMC 10B Problem 3

Problem onlyBack to the problem
2014 AMC 10B Problem 3AMC 10 Step by Step
Name:                             Date: September 13, 2026
Ratios, Percents & Averages · difficulty 1 of 5 · about 2 min · position P1-10

Peter drove the first third of his trip on a gravel road, the next 2020 miles on pavement, and the remaining one-fifth on a dirt road. In miles how long was Peter's trip?

  1. A)

    3030

  2. B)

    40011\frac{400}{11}

  3. C)

    752\frac{75}{2}

  4. D)

    4040

  5. E)

    3007\frac{300}{7}

Answer

E

Key insight

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

Solution

Let the whole trip be TT miles. The gravel and dirt portions together are
13T+15T=5+315T=815T.\frac13 T + \frac15 T = \frac{5 + 3}{15} T = \frac{8}{15} T.

So the paved stretch is the rest: T−815T=715TT - \frac{8}{15}T = \frac{7}{15}T. That stretch is 2020 miles, so
715T=20⟹T=20⋅157=3007.\frac{7}{15} T = 20 \quad\Longrightarrow\quad T = 20 \cdot \frac{15}{7} = \frac{300}{7}.

The answer is (E) 3007\boxed{\textbf{(E)}\ \frac{300}{7}}.

Why this works

When two parts of a whole are given as fractions and the third as an actual amount, the actual amount equals "one minus the fractions" times the whole. Add the fractions with a common denominator, subtract from 11, and divide. The ugly fraction in the answer is a sign the computation was done correctly, not a red flag.

The trap

Treating the 20 miles as 1/3 or 1/5 of the trip, or adding 1/3 + 1/5 as 2/8.

Common mistakes

  • Treating the 20 miles as 1/3 or 1/5 of the trip, or adding 1/3 + 1/5 as 2/8.
  • Solving 815T=20\frac{8}{15}T = 20 (using the wrong part) to get 752\frac{75}{2}, choice (C).

Techniques

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