Difference between revisions of "2011 AMC 12B Problems/Problem 3"

(Problem)
(Solution)
 
Line 14: Line 14:
 
So each person should pay
 
So each person should pay
 
<cmath> \frac{A+B}{2} </cmath>
 
<cmath> \frac{A+B}{2} </cmath>
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 LeRoy has already paid <math>A</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} \textbf{(C)}} </cmath>
 
<cmath> = \boxed{\frac{B-A}{2} \textbf{(C)}} </cmath>

Latest revision as of 10:54, 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 $A < B.$ How many dollars must LeRoy give to Bernardo so that they share the costs equally?

$\textbf{(A)}\ \frac{A+B}{2} \qquad \textbf{(B)}\ \frac{A-B}{2} \qquad \textbf{(C)}\ \frac{B-A}{2} \qquad \textbf{(D)}\ B-A \qquad \textbf{(E)}\ A+B$

Solution

The total amount of money that was spent during the trip was \[A + B\] So each person should pay \[\frac{A+B}{2}\] if they were to share the costs equally. Because LeRoy has already paid $A$ dollars of his part, he still has to pay \[\frac{A+B}{2} - A =\] \[= \boxed{\frac{B-A}{2} \textbf{(C)}}\]

See also

2011 AMC 12B (ProblemsAnswer KeyResources)
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. AMC logo.png