Difference between revisions of "2014 AMC 12B Problems/Problem 6"

(Answer was incorrectly marked as C, although the numerical value was correct)
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The total amount the two of them drank is simply  
 
The total amount the two of them drank is simply  
 
<cmath>x + \frac{3}{2}x = 16 + 24 = \boxed{\textbf{(D)}\ 40}</cmath>
 
<cmath>x + \frac{3}{2}x = 16 + 24 = \boxed{\textbf{(D)}\ 40}</cmath>
 
(Solution by kevin38017)
 

Revision as of 22:36, 20 February 2014

Problem

Ed and Ann both have lemonade with their lunch. Ed orders the regular size. Ann gets the large lemonade, which is 50% more than the regular. After both consume $\frac{3}{4}$ of their drinks, Ann gives Ed a third of what she has left, and 2 additional ounces. When they finish their lemonades they realize that they both drank the same amount. How many ounces of lemonade did they drink together?

$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}}\ 40\qquad\textbf{(E)}\ 50$ (Error compiling LaTeX. Unknown error_msg)

Solution

Let the size of Ed's drink equal $x$ ounces, and let the size of Ann's drink equal $\frac{3}{2}x$ ounces. After both consume $\frac{3}{4}$ of their drinks, Ed and Ann have $\frac{x}{4}$ and $\frac{3x}{8}$ ounces of their drinks remaining. Ann gives away $\frac{x}{8} + 2$ ounces to Ed.

In the end, Ed drank everything in his original lemonade plus what Ann gave him, and Ann drank everything in her original lemonade minus what she gave Ed. Thus we have \[x + \frac{x}{8} + 2 = \frac{3x}{2} - \frac{x}{8} - 2\] \[x = 16\] The total amount the two of them drank is simply \[x + \frac{3}{2}x = 16 + 24 = \boxed{\textbf{(D)}\ 40}\]