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 | ||
+ | After this, he purchased | ||
+ | He originally wasted | ||
+ | \[ | ||
+ | \frac{n}{2} - 2m = \frac{4}{18} \cdot (n-m) | ||
+ | \] | ||
+ | \[ | ||
+ | 9n - 36m = 4n - 4m \implies 5n = 32m \implies \frac{m}{n} = \frac{5}{32}. | ||
+ | \] | ||
+ | </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
[hide]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
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 |
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