Difference between revisions of "2011 AIME II Problems/Problem 1"

m (Problem 1)
(Solution 2)
 
(15 intermediate revisions by 9 users not shown)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
Let <math>x</math> be the [[fraction]] consumed, then <math>(1-x)</math> is the fraction wasted. We have <math>1/2 - 2x = 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 n2 liters, and consumed 2m liters. 
 +
He originally wasted nm liters, but now he wasted n22m
 +
\[
 +
\frac{n}{2} - 2m = \frac{4}{18} \cdot (n-m)
 +
\]
 +
\[
 +
9n - 36m = 4n - 4m \implies 5n = 32m \implies \frac{m}{n} = \frac{5}{32}.
 +
\]
 +
</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 Geometry Problems]]
+
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 05:12, 24 November 2024

Problem

Gary purchased a large beverage, but only drank $m/n$ of it, where $m$ and $n$ are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only $2/9$ as much beverage. Find $m+n$.

Solution

Let $x$ be the fraction consumed, then $(1-x)$ is the fraction wasted. We have $\frac{1}{2} - 2x =\frac{2}{9} (1-x)$, or $9 - 36x = 4 - 4x$, or $32x = 5$ or $x = 5/32$. Therefore, $m + n = 5 + 32 = \boxed{037}$.

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 \( \frac{n}{2} - 2m \).   \[ \frac{n}{2} - 2m = \frac{4}{18} \cdot (n-m) \] \[ 9n - 36m = 4n - 4m \implies 5n = 32m \implies \frac{m}{n} = \frac{5}{32}. \]

Thus, the answer is $\boxed{37}$ ~idk123456

See also

2011 AIME II (ProblemsAnswer KeyResources)
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. AMC logo.png