2011 AIME II Problems/Problem 1: Difference between revisions
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== Problem == | == Problem == | ||
Gary purchased a large beverage, but only drank <math>m/n</math> of it, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only <math>2/9</math> as much beverage. | Gary purchased a large beverage, but only drank <math>m/n</math> of it, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only <math>2/9</math> as much beverage. Find <math>m+n</math>. | ||
== Solution == | == Solution == | ||
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>. | 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== | ||
Latest revision as of 05:12, 24 November 2024
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
