Difference between revisions of "2011 AMC 10B Problems/Problem 4"
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(A) <math>\frac{A+B}{2}</math> (B) <math>\frac{A-B}{2}</math> (C) <math>\frac{B-A}{2}</math> (D) <math>B-A</math> (E) <math>A+B</math> | (A) <math>\frac{A+B}{2}</math> (B) <math>\frac{A-B}{2}</math> (C) <math>\frac{B-A}{2}</math> (D) <math>B-A</math> (E) <math>A+B</math> | ||
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| + | == Solution == | ||
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| + | The difference in how much LeRoy and Bernardo paid is <math>B-A</math>. The average of how much each paid is <math>\frac{A+B}{2}</math>. So LeRoy needs to pay Bernardo <math>\frac{B-A}{2}</math>. | ||
Revision as of 13:45, 27 March 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 
dollars and Bernardo had paid 
dollars, where 
. How many dollars must LeRoy give to Bernardo so that they share the costs equally?
(A) 
(B) 
(C) 
(D) 
(E) 
Solution
The difference in how much LeRoy and Bernardo paid is . The average of how much each paid is . So LeRoy needs to pay Bernardo .
