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== |
Latest revision as of 16:33, 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 |
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