Difference between revisions of "2011 AMC 12B Problems/Problem 3"
(→Solution) |
(→Solution) |
||
| Line 16: | Line 16: | ||
if they were to share the costs equally. Because Bernardo has already paid <math>B</math> dollars of his part, he still has to pay | if they were to share the costs equally. Because Bernardo has already paid <math>B</math> dollars of his part, he still has to pay | ||
<cmath> \frac{A+B}{2} - A = </cmath> | <cmath> \frac{A+B}{2} - A = </cmath> | ||
| − | <cmath> = \boxed{\frac{B-A}{2} | + | <cmath> = \boxed{\frac{B-A}{2} \textbf{(C)}} </cmath> |
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=2|num-a=4|ab=B}} | {{AMC12 box|year=2011|num-b=2|num-a=4|ab=B}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 09:53, 16 May 2015
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 Bernardo give to LeRoy 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 Bernardo 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)
![\[= \boxed{\frac{B-A}{2} \textbf{(C)}}\]](http://latex.artofproblemsolving.com/9/d/1/9d197d58d1b25b08a2a6cf8566a821572e559e99.png)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
