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

Revision as of 14:17, 26 July 2017

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

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

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

Since there are $30$ grams of filler, and $30$ is $\frac{1}{4}$ of $120$ , the whole, we have $1 - \frac{3}{4} = \boxed{75\%}$ .

2003 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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 AJHSME/AMC 8 Problems and Solutions

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 ==Solution==
 There are <math> 30 </math> grams of filler, so there are <math> 120-30= 90 </math> grams that aren't filler. <math> \frac{90}{120}=\frac{3}{4}=\boxed{\mathrm{(D)}\ 75\%} </math>.
+==Solution2==
+Since there are <math> 30 </math> grams of filler, and <math> 30 </math> is <math>\frac{1}{4}</math> of <math> 120 </math>, the whole, we have <math>1 - \frac{3}{4} = \boxed{75\%}</math>.
 ==See Also==
 {{AMC8 box|year=2003|num-b=2|num-a=4}}
 {{MAA Notice}}