Difference between revisions of "2011 AMC 10B Problems/Problem 4"

Latest revision as of 12:05, 24 January 2024

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?

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

Solution 1

The difference in how much LeRoy and Bernardo paid is $B-A$ . To share the costs equally, LeRoy must give Bernardo half of the difference, which is $\boxed{\textbf{(C) } \;\frac{B-A}{2}}$

Solution 2 (Use an example)

Since there are no restrictions on cost paid besides $A<B$ , we can use an example where $A = 40$ and $B = 50$ . Quickly, we realize the only way they could pay the same amount of money is if they both pay 45 dollars. This means LeRoy must give Bernardo $50 - 45 = 5$ . Looking at the answer choices we see only $\frac{50-40}{2} = 5$ works. $\boxed{\textbf{(C) } \;\frac{B-A}{2}}$

~vsinghminhas

Video Solution

https://youtu.be/v8a5x98Xy_s

~savannahsolver

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 11: / Line 11: @@
 ==Solution 2 (Use an example)==
-Since there are no restrictions on cost paid besides <math>A<B</math>, we can use an example where <math>A = 40</math> and <math> B = 50</math>. Quickly, we realize the only way they could pay the same amount of money is if they both pay 45 dollars. This means LeRoy must give Bernardo <math>50 - 45 = 5</math>. Looking at the answer choices we see only <math>\frac{50-40}{2} = 5</math> works. <math>\boxed{(C) \;\frac{B-A}{2}}</math>
+Since there are no restrictions on cost paid besides <math>A<B</math>, we can use an example where <math>A = 40</math> and <math> B = 50</math>. Quickly, we realize the only way they could pay the same amount of money is if they both pay 45 dollars. This means LeRoy must give Bernardo <math>50 - 45 = 5</math>. Looking at the answer choices we see only <math>\frac{50-40}{2} = 5</math> works. <math>\boxed{\textbf{(C) } \;\frac{B-A}{2}}</math>
 ~vsinghminhas

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