Difference between revisions of "2009 AMC 10A Problems/Problem 7"
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<math> | <math> | ||
\mathrm{(A)}\ \frac{12}{5} | \mathrm{(A)}\ \frac{12}{5} | ||
| − | \ | + | \qquad |
| + | \mathrm{(B)}\ 3 | ||
| + | \qquad | ||
| + | \mathrm{(C)}\ \frac{10}{3} | ||
| + | \qquad | ||
| + | \mathrm{(D)}\ 38 | ||
| + | \qquad | ||
| + | \mathrm{(E)}\ 42 | ||
</math> | </math> | ||
| − | = | + | ==Solution== |
| + | Rewording the question, we are being asked "<math>2</math> is <math>40\%</math> less than what number?" | ||
| + | If <math>x</math> represents the number we are looking for, then <math>40\%</math> less than the number would be represented by <math>x-0.4x</math> or <math>0.6x</math>. | ||
| + | Thus <math>0.6x=2</math>; solving for <math>x</math>, we get <math>x=\frac{10}{3}</math> <math>\longrightarrow \fbox{C}</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | If we change the question to 2% fat in the carton which is 50% less than whole milk, this means that there is 4% fat in the whole milk. The only answer closest and less than 4 is 10/3 <math>\implies \boxed{C}</math>. 38% fat is a lot of fat in milk :O | ||
| + | -RealityWrites | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC10 box|year=2009|ab=A|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:03, 21 August 2024
Contents
Problem
A carton contains milk that is 
% fat, an amount that is 
% less fat than the amount contained in a carton of whole milk. What is the percentage of fat in whole milk?
Solution
Rewording the question, we are being asked " is 
less than what number?"
If 
represents the number we are looking for, then less than the number would be represented by 
or 
.
Thus 
; solving for , we get 
Solution 2
If we change the question to 2% fat in the carton which is 50% less than whole milk, this means that there is 4% fat in the whole milk. The only answer closest and less than 4 is 10/3 
. 38% fat is a lot of fat in milk :O
-RealityWrites
See also
| 2009 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
