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

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

## Solution2

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 byProblem 2 Followed byProblem 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