Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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}\  \(\textbf{(C)}} </cmath>
+
<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 $A < B.$ How many dollars must Bernardo give to LeRoy 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 Bernardo has already paid $B$ 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 (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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_3&oldid=70238"