AMC 10 Step by Step

2005 AMC 10B Problem 5

P1-10~1 minPrint

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?

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Key insight

If one third of the CDs costs one fifth of her money, then all the CDs cost three times as much: three fifths. Two fifths is left.

Solution

1. All the CDs cost the same, so buying three times as many CDs costs three times as much money.
2. One third of the CDs costs one fifth of her money. All of the CDs is three times one third, so they cost
3⋅15=35 of her money.3 \cdot \frac{1}{5} = \frac{3}{5} \text{ of her money}.
3. The question asks what is left. She started with all of it, 11 whole, so
1−35=25.1 - \frac{3}{5} = \frac{2}{5}.

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

Why this works

The money spent grows in step with the number of CDs bought, so tripling one amount triples the other. You never need her actual amount of money. Before choosing an answer, reread the question: here it asks for what is left, not what is spent.

Alternative approach

Make up numbers. Say she has 1515 dollars and wants 33 CDs. One fifth of her money is 33 dollars, and that buys one CD (one third of the CDs). So each CD costs 33 dollars, and all three cost 99 dollars. She has 15−9=615 - 9 = 6 dollars left, and 615=25\frac{6}{15} = \frac{2}{5}.

Concise solution

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

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}}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site. Also appeared as 2005 12B #3.