Difference between revisions of "2012 AMC 12A Problems/Problem 5"
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A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of <math>280</math> pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad? | A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of <math>280</math> pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad? | ||
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Latest revision as of 23:00, 3 July 2013
Problem
A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of 
pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad?
Solution
So let the number of blueberries be 
the number of raspberries be 
the number of grapes be 
and finally the number of cherries be 
Observe that since there are pieces of fruit,
![\[b+r+g+c=280.\]](http://latex.artofproblemsolving.com/3/8/e/38e13b54d1ccd5c52ea7d45d283ef7ccae426a5f.png)
Since there are twice as many raspberries as blueberries,
![\[2b=r.\]](http://latex.artofproblemsolving.com/a/3/a/a3ae4880cd0b104efc38649cb19fa55186cdbfe4.png)
The fact that there are three times as many grapes as cherries implies,
![\[3c=g.\]](http://latex.artofproblemsolving.com/c/8/e/c8e7d9bc5ee6212edbe561fa4c33a34cf9680e68.png)
Because there are four times as many cherries as raspberries, we deduce the following:
![\[4r=c.\]](http://latex.artofproblemsolving.com/5/4/c/54cea1f40d6de1b653d3e91349ea7525cf0ce26d.png)
Note that we are looking for So, we try to rewrite all of the other variables in terms of
The third equation gives us the value of 
in terms of 
already. We divide the fourth equation by 
to get that 
Finally, substituting this value of 
into the first equation provides us with the equation 
and substituting yields:
![\[\frac{c}{4}+\frac{c}{8}+3c+c=280\]](http://latex.artofproblemsolving.com/b/2/7/b276900e7b6be663c4d777224475d99e610ac30d.png)
Multiply this equation by 
to get:
![\[2c+c+24c+8c=8\cdot 280,\]](http://latex.artofproblemsolving.com/a/8/2/a822820bc1a07f28661af5fdf879635cca47b4c7.png)
![\[35c=8\cdot 280,\]](http://latex.artofproblemsolving.com/e/d/7/ed7ee4bf44aee65b1ea574ffcb64ddf72e73ccc4.png)
![\[c=64.\]](http://latex.artofproblemsolving.com/b/b/0/bb07c565cbc4e84c339061fc43debdf38c94d00b.png)
![\[\boxed{D}\]](http://latex.artofproblemsolving.com/d/a/d/dadc8e71363a80c53a137421994b9e02de6eeabd.png)
See Also
| 2012 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
