Difference between revisions of "2011 AMC 12B Problems/Problem 3"
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<cmath> \frac{A+B}{2} - A = </cmath> | <cmath> \frac{A+B}{2} - A = </cmath> | ||
<cmath> = \boxed{\frac{B-A}{2}\ \((C)} </cmath> | <cmath> = \boxed{\frac{B-A}{2}\ \((C)} </cmath> | ||
| + | |||
== See also == | == See also == | ||
{{AMC12 box|year=2011|before=Problem 2|num-a=4|ab=B}} | {{AMC12 box|year=2011|before=Problem 2|num-a=4|ab=B}} | ||
Revision as of 16:08, 27 February 2011
Problem
LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid A dollars and Bernardo had paid B dollars, where 
How many dollars must LeRoy give to Bernardo so that they share the costs equally?
Solution
The total amount of money that was spent during the trip was
![\[A + B\]](http://latex.artofproblemsolving.com/d/6/2/d62b57eca33a7910224a3d3c0311598f808a06b0.png)
So each person should pay
![\[\frac{A+B}{2}\]](http://latex.artofproblemsolving.com/2/5/8/2582e48b89cce9e96302e33b01566fe34a450d0d.png)
if they were to share the costs equally. Because LeRoy has already paid 
dollars of his part, he still has to pay
![\[\frac{A+B}{2} - A =\]](http://latex.artofproblemsolving.com/0/7/7/077a77a211a52462acd409a7e5bcaed528d683eb.png)
\[= \boxed{\frac{B-A}{2}\ \((C)}\] (Error compiling LaTeX. Unknown error_msg)
See also
| 2011 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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 | |
