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

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

@@ Line 9: / Line 9: @@
 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 originally, but now he wasted \( \frac{n}{2} - 2m \).
+He originally wasted \( n-m \) liters, but now he wasted \( \frac{n}{2} - 2m \).
 \[
 \frac{n}{2} - 2m = \frac{4}{18} \cdot (n-m)
@@ Line 17: / Line 17: @@
 \]
 </cmath>
+Thus, the answer is <math>\boxed{37}</math>
 ~idk123456

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

Latest revision as of 05:12, 24 November 2024

Contents

Problem

Solution

Solution 2

See also