Difference between revisions of "2003 AMC 8 Problems/Problem 3"
AlcumusGuy (talk | contribs) |
(→Solution2) |
||
(3 intermediate revisions by 3 users not shown) | |||
Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
− | There | + | 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}} | {{AMC8 box|year=2003|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Latest revision as of 15:18, 26 July 2017
Problem
A burger at Ricky C's weighs grams, of which grams are filler. What percent of the burger is not filler?
Solution
There are grams of filler, so there are grams that aren't filler. .
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.