Difference between revisions of "2009 AMC 10A Problems/Problem 18"
m (minor edit) |
m (→Solution 3) |
||
(5 intermediate revisions by 3 users not shown) | |||
Line 30: | Line 30: | ||
* <math>34\%</math>: swimming no, soccer no | * <math>34\%</math>: swimming no, soccer no | ||
− | 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 | + | 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 percent of non-swimmers that play soccer is <math>\frac{36}{70} \approx 51\% \Rightarrow \boxed{D}</math>. |
=== Solution 2 === | === Solution 2 === | ||
− | Let us set | + | Let us set the total number of children as <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. |
Thus, <math>60-24=36</math> children only play soccer. | Thus, <math>60-24=36</math> children only play soccer. | ||
Line 42: | Line 42: | ||
And so we get <math>\frac{36}{70}</math> which is roughly <math>51.4\% \Rightarrow \boxed{\text{D}}</math> | And so we get <math>\frac{36}{70}</math> which is roughly <math>51.4\% \Rightarrow \boxed{\text{D}}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | WLOG, let the total number of students be <math>100</math>. Draw a venn diagram with 2 circles encompassing these 4 regions: | ||
+ | |||
+ | |||
+ | Non-soccer players, non-swimmers: 34 people | ||
+ | |||
+ | Soccer players, non-swimmers: 36 people | ||
+ | |||
+ | Soccer players, swimmers: 24 people | ||
+ | |||
+ | Non-soccer players, swimmers: 6 people. | ||
+ | |||
+ | |||
+ | Hence the answer is <math>\frac{36}{70}=\frac{18}{35}</math>. We know this is a little bit larger than <math>\frac 12</math> because <math>\frac{17.5}{35}=\frac 12</math>. <math>\boxed{\textbf{(D) } 51\%}</math> | ||
+ | |||
+ | ~BakedPotato66 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/Ahei7_BKK9Q | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == |
Latest revision as of 10:29, 2 August 2021
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?
Solutions
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
Now 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 percent of non-swimmers that play soccer is
.
Solution 2
Let us set the total number of children as . 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
Solution 3
WLOG, let the total number of students be . Draw a venn diagram with 2 circles encompassing these 4 regions:
Non-soccer players, non-swimmers: 34 people
Soccer players, non-swimmers: 36 people
Soccer players, swimmers: 24 people
Non-soccer players, swimmers: 6 people.
Hence the answer is . We know this is a little bit larger than
because
.
~BakedPotato66
Video Solution
~savannahsolver
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.