AMC 10 Step by Step

2009 AMC 10A Problem 18

P16-20~3 minPrint

At Jefferson Summer Camp, 60%60\% of the children play soccer, 30%30\% of the children swim, and 40%40\% of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer?

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Key insight

Pretend there are 100 children: 60 play soccer, 24 of those swim, so 36 soccer players do not swim, out of 70 children who do not swim.

Solution

Only percents are given, so pick a handy number of children: 100100.

1. Soccer players: 60%60\% of 100=60100 = 60.
2. Swimmers: 3030. So the children who do not swim number 100−30=70100 - 30 = 70.
3. Soccer players who swim: 40%40\% of the 6060 soccer players, which is 2424.
4. Soccer players who do not swim: 60−24=3660 - 24 = 36.
5. The question asks about the 7070 non-swimmers: what part of them play soccer?
3670≈0.514,\frac{36}{70} \approx 0.514,
which is 51%51\% to the nearest whole percent.

The answer is (D) 51%\boxed{\textbf{(D)}\ 51\%}.

Why this works

Using 100100 people turns every percent into a plain count. The words "percent of the non-swimmers" tell you what to divide by: the size of that group (7070), not the whole camp. Read each "of" carefully: "40%40\% of the soccer players" is a part of the 6060, not of the 3030 swimmers.

Concise solution

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

Since only percentages are involved, assume there are 100100 children.

  • Soccer players: 6060.
  • Swimmers: 3030, so non-swimmers: 7070.
  • Soccer players who swim: 40%40\% of 60=2460 = 24.
  • Soccer players who do not swim: 60−24=3660 - 24 = 36.

The question asks what fraction of the 7070 non-swimmers play soccer:
3670=1835≈0.514,\frac{36}{70} = \frac{18}{35} \approx 0.514,
which rounds to 51%51\%.

The answer is (D) 51%\boxed{\textbf{(D)}\ 51\%}.

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.