# Difference between revisions of "2009 AMC 10A Problems/Problem 18"

## Problem

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

$\mathrm{(A)}\ 30\% \qquad \mathrm{(B)}\ 40\% \qquad \mathrm{(C)}\ 49\% \qquad \mathrm{(D)}\ 51\% \qquad \mathrm{(E)}\ 70\%$

### Solution 1

Out of the soccer players, $40\%$ swim. As the soccer players are $60\%$ of the whole, the swimming soccer players are $0.4 \cdot 0.6 = 0.24 = 24\%$ of all children.

The non-swimming soccer players then form $60\% - 24\% = 36\%$ of all the children.

Out of all the children, $30\%$ swim. We know that $24\%$ of all the children swim and play soccer, hence $30\%-24\% = 6\%$ of all the children swim and don't play soccer.

Finally, we know that $70\%$ of all the children are non-swimmers. And as $36\%$ of all the children do not swim but play soccer, $70\% - 36\% = 34\%$ of all the children do not engage in any activity.

A quick summary of what we found out:

• $24\%$: swimming yes, soccer yes
• $36\%$: swimming no, soccer yes
• $6\%$: swimming yes, soccer no
• $34\%$: swimming no, soccer no

Now we can compute the answer. Out of all children, $70\%$ are non-swimmers, and again out of all children $36\%$ are non-swimmers that play soccer. Hence the part of non-swimmers that plays soccer is $\frac{36}{70} \approx \boxed{51\%}$.

### Solution 2

Let us set that the total number of children is $100$. So $60$ children play soccer, $30$ swim, and $0.4\times60=24$ play soccer and swim.

Thus, $60-24=36$ children only play soccer.

So our numerator is $36$.

Our denominator is simply $100-\text{Swimmers}=100-30=70$

And so we get $\frac{36}{70}$ which is roughly $51.4=\boxed{\text{D}}$