Difference between revisions of "2011 AIME II Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | Let <math>x</math> be the [[fraction]] consumed, then <math>(1-x)</math> is the fraction wasted. We have <math>1 | + | Let <math>x</math> be the [[fraction]] consumed, then <math>(1-x)</math> is the fraction wasted. We have <math>\frac{1}{2} - 2x =\frac{2}{9} (1-x)</math>, or <math>9 - 36x = 4 - 4x</math>, or <math>32x = 5</math> or <math>x = 5/32</math>. Therefore, <math>m + n = 5 + 32 = \boxed{037}</math>. |
| + | |||
| + | == Solution 2 == | ||
| + | <cmath> | ||
| + | WLOG, Gary purchased \( n \) liters and consumed \( m \) liters. | ||
| + | After this, he purchased \( \frac{n}{2} \) liters, and consumed \( 2m \) liters. | ||
| + | He originally wasted \( n-m \) liters, but now he wasted \( \frac{n}{2} - 2m \). | ||
| + | \[ | ||
| + | \frac{n}{2} - 2m = \frac{4}{18} \cdot (n-m) | ||
| + | \] | ||
| + | \[ | ||
| + | 9n - 36m = 4n - 4m \implies 5n = 32m \implies \frac{m}{n} = \frac{5}{32}. | ||
| + | \] | ||
| + | </cmath> | ||
| + | |||
| + | Thus, the answer is <math>\boxed{37}</math> | ||
| + | ~idk123456 | ||
==See also== | ==See also== | ||
{{AIME box|year=2011|n=II|before=First Problem|num-a=2}} | {{AIME box|year=2011|n=II|before=First Problem|num-a=2}} | ||
| − | [[Category:Intermediate | + | [[Category:Intermediate Algebra Problems]] |
| + | {{MAA Notice}} | ||
Latest revision as of 05:12, 24 November 2024
Contents
Problem
Gary purchased a large beverage, but only drank 
of it, where 
and 
are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only 
as much beverage. Find 
.
Solution
Let 
be the fraction consumed, then 
is the fraction wasted. We have 
, or 
, or 
or 
. Therefore, 
.
Solution 2
![WLOG, Gary purchased \( n \) liters and consumed \( m \) liters. After this, he purchased \( \frac{n}{2} \) liters, and consumed \( 2m \) liters. He originally wasted \( n-m \) liters, but now he wasted \( \frac{n}{2} - 2m \). \[ \frac{n}{2} - 2m = \frac{4}{18} \cdot (n-m) \] \[ 9n - 36m = 4n - 4m \implies 5n = 32m \implies \frac{m}{n} = \frac{5}{32}. \]](http://latex.artofproblemsolving.com/0/a/e/0ae5bb5ad67d8892aa574179eb5ca0eae3dddd5f.png)
Thus, the answer is 
~idk123456
See also
| 2011 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
