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

(New page: == Problem == At Jefferson Summer Camp, <math>60\%</math> of the children play soccer, <math>30\%</math> of the children swim, and <math>40\%</math> of the soccer players swim. To the nea...)
 
(Solution: Added 2nd Solution)
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</math>
 
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== Solution ==
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=== Solution 1 ===
  
 
Out of the soccer players, <math>40\%</math> swim. As the soccer players are <math>60\%</math> of the whole, the swimming soccer players are <math>0.4 \cdot 0.6 = 0.24 = 24\%</math> of all children.
 
Out of the soccer players, <math>40\%</math> swim. As the soccer players are <math>60\%</math> of the whole, the swimming soccer players are <math>0.4 \cdot 0.6 = 0.24 = 24\%</math> of all children.
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Now we can compute the answer. Out of all children, <math>70\%</math> are non-swimmers, and again out of all children <math>36\%</math> are non-swimmers that play soccer. Hence the part of non-swimmers that plays soccer is <math>\frac{36}{70} \simeq \boxed{51\%}</math>.
 
Now we can compute the answer. Out of all children, <math>70\%</math> are non-swimmers, and again out of all children <math>36\%</math> are non-swimmers that play soccer. Hence the part of non-swimmers that plays soccer is <math>\frac{36}{70} \simeq \boxed{51\%}</math>.
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=== Solution 2 ===
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Let us set that the total number of children is <math>100</math>. So <math>60</math> children play soccer, <math>30</math> swim, and <math>0.4\times60=24</math> play soccer and swim.
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Thus, <math>60-24=36</math> children only play soccer.
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So our numerator is <math>36</math>.
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Our denominator is simply <math>100-\text{Swimmers}=100-30=70</math>
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And so we get <math>\frac{36}{70}</math> which is roughly <math>51.4=\boxed{\text{D}}</math>
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2009|ab=A|num-b=17|num-a=19}}
 
{{AMC10 box|year=2009|ab=A|num-b=17|num-a=19}}

Revision as of 22:04, 6 February 2012

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 $\frac{36}{70} \simeq \boxed{51\%}$.

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 $51.4=\boxed{\text{D}}$

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 10 Problems and Solutions