Difference between revisions of "2014 AMC 12B Problems/Problem 6"
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<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> | ||
| + | == See also == | ||
{{AMC12 box|year=2014|ab=B|num-b=5|num-a=7}} | {{AMC12 box|year=2014|ab=B|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:23, 22 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 
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 
ounces, and let the size of Ann's drink equal 
ounces. After both consume of their drinks, Ed and Ann have 
and 
ounces of their drinks remaining. Ann gives away 
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
![\[x + \frac{x}{8} + 2 = \frac{3x}{2} - \frac{x}{8} - 2\]](http://latex.artofproblemsolving.com/2/c/2/2c2c3436f4b65947526ec344f5f281a79790e74b.png)
![\[x = 16\]](http://latex.artofproblemsolving.com/6/6/4/664317c6d040b980c9763e609a05836a4e1d5f0f.png)
The total amount the two of them drank is simply
![\[x + \frac{3}{2}x = 16 + 24 = \boxed{\textbf{(D)}\ 40}\]](http://latex.artofproblemsolving.com/a/c/4/ac4bb21f88d45c3c0143c9a7daf1e06059180d73.png)
See also
| 2014 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
