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

Latest revision as of 14:18, 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. $\frac{90}{120}=\frac{3}{4}=\boxed{\mathrm{(D)}\ 75\%}$ .

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

@@ Line 1: / Line 1: @@
-==Problem 3==
+==Problem==
 A burger at Ricky C's weighs <math>120</math> grams, of which <math>30</math> grams are filler.
 What percent of the burger is not filler?
@@ Line 6: / Line 6: @@
 ==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 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>.
+==See Also==
+{{AMC8 box|year=2003|num-b=2|num-a=4}}
+{{MAA Notice}}