Difference between revisions of "2003 AMC 8 Problems/Problem 3"

(Problem 3)
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==Solution==
 
==Solution==
 
There is <math> 30 </math> grams of filler, so there is <math> 120-30= 90 </math> grams that aren't filler. <math> \frac{90}{120}=\frac{3}{4}=\boxed{\mathrm{(D)}\ 75\%} </math>.
 
There is <math> 30 </math> grams of filler, so there is <math> 120-30= 90 </math> grams that aren't filler. <math> \frac{90}{120}=\frac{3}{4}=\boxed{\mathrm{(D)}\ 75\%} </math>.
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{{AMC8 box|year=2003|num-b=2|num-a=4}}

Revision as of 09:37, 25 November 2011

Problem

A burger at Ricky C's weighs $120$ grams, of which $30$ grams are filler. What percent of the burger is not filler?

$\mathrm{(A)}\ 60\% \qquad\mathrm{(B)}\ 65\% \qquad\mathrm{(C)}\ 70\% \qquad\mathrm{(D)}\ 75\% \qquad\mathrm{(E)}\ 90\%$

Solution

There is $30$ grams of filler, so there is $120-30= 90$ grams that aren't filler. $\frac{90}{120}=\frac{3}{4}=\boxed{\mathrm{(D)}\ 75\%}$.

2003 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AJHSME/AMC 8 Problems and Solutions
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