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2005 AMC 10B Problem 5

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2005 AMC 10B Problem 5AMC 10 Step by Step
Name:                             Date: September 13, 2026
Ratios, Percents & Averages · difficulty 1 of 5 · about 1 min · position P1-10

Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?

  1. A)

    15\frac15

  2. B)

    13\frac13

  3. C)

    25\frac25

  4. D)

    23\frac23

  5. E)

    45\frac45

Answer

C

Key insight

If one fifth of the money buys one third of the CDs, all the CDs cost three times that, three fifths, leaving two fifths.

Solution

One third of the CDs cost one fifth of her money. Since the CDs are equally priced, the full set costs three times as much:
3⋅15=35 of her money.3 \cdot \frac{1}{5} = \frac{3}{5} \text{ of her money}.

What remains is
1−35=25.1 - \frac{3}{5} = \frac{2}{5}.

The answer is (C) 25\boxed{\textbf{(C)}\ \frac{2}{5}}.

Why this works

Cost is proportional to the number of CDs, so scaling the fraction of CDs by 33 scales the fraction of money by 33 as well. No actual amounts are needed; proportion problems often close with a single multiplication and a "what is left" subtraction.

Alternative approach

Pick numbers: suppose she has 1515 dollars and there are 33 CDs. One fifth of the money, 33 dollars, buys one CD, so all three cost 99 dollars and 66 dollars remain, which is 615=25\frac{6}{15} = \frac{2}{5}.

The trap

Answering 3/5 (the fraction spent) instead of the fraction left over.

Common mistakes

  • Answering 3/5 (the fraction spent) instead of the fraction left over.
  • Multiplying 15\frac{1}{5} by 13\frac{1}{3} or subtracting 15\frac{1}{5} from 13\frac{1}{3} instead of scaling the cost up by 33.

Techniques

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