2012 AMC 12A Problems/Problem 5

Revision as of 13:39, 12 February 2012 by Gina (talk | contribs)

Problem

A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of $280$ 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?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96$

Solution

So let the number of blueberries be $b,$ the number of raspberries be $r,$ the number of grapes be $g,$ and finally the number of cherries be $c.$

Observe that since there are $280$ pieces of fruit, \[b+r+g+c=280.\]

Since there are twice as many raspberries as blueberries, \[2b=r.\]

The fact that there are three times as many grapes as cherries implies, \[3c=g.\]

Because there are four times as many cherries as raspberries, we deduce the following: \[4r=c.\]

Note that we are looking for $c.$ So, we try to rewrite all of the other variables in terms of $c.$ The third equation gives us the value of $g$ in terms of $c$ already. We divide the fourth equation by $4$ to get that $r=\frac{c}{4}.$ Finally, substituting this value of $r$ into the first equation provides us with the equation $b=\frac{c}{8}$ and substituting yields: \[\frac{c}{4}+\frac{c}{8}+3c+c=280\] Multiply this equation by $8$ to get: \[2c+c+24c+8c=8\cdot 280,\] \[35c=8\cdot 280,\] \[c=64.\] \[\boxed{D}\]

See Also

2012 AMC 12A (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS