Difference between revisions of "2009 AMC 10A Problems/Problem 18"
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Our denominator is simply <math>100-\text{Swimmers}=100-30=70</math> | Our denominator is simply <math>100-\text{Swimmers}=100-30=70</math> | ||
| − | And so we get <math>\frac{36}{70}</math> which is roughly <math>51.4=\boxed{\text{D}}</math> | + | And so we get <math>\frac{36}{70}</math> which is roughly <math>51.4\% = \boxed{\text{D}}</math> |
== See Also == | == See Also == | ||
Revision as of 13:17, 15 January 2018
Contents
Problem
At Jefferson Summer Camp, 
of the children play soccer, 
of the children swim, and 
of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer?
Solution 1
Out of the soccer players, swim. As the soccer players are of the whole, the swimming soccer players are 
of all children.
The non-swimming soccer players then form 
of all the children.
Out of all the children, swim. We know that 
of all the children swim and play soccer, hence 
of all the children swim and don't play soccer.
Finally, we know that 
of all the children are non-swimmers. And as 
of all the children do not swim but play soccer, 
of all the children do not engage in any activity.
A quick summary of what we found out:
: swimming yes, soccer yes
: swimming no, soccer yes
: swimming yes, soccer no
: swimming no, soccer no
: swimming yes, soccer no
: swimming no, soccer noNow we can compute the answer. Out of all children, are non-swimmers, and again out of all children are non-swimmers that play soccer. Hence the part of non-swimmers that plays soccer is 
.
Solution 2
Let us set that the total number of children is 
. So 
children play soccer, 
swim, and 
play soccer and swim.
Thus, 
children only play soccer.
So our numerator is 
.
Our denominator is simply 
And so we get 
which is roughly 
See Also
| 2009 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
