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>\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 | + | 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>. |
==See also== | ==See also== | ||
Revision as of 15:33, 29 February 2020
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, 
.
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. 
