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Difference between revisions of "1987 AHSME Problems/Problem 17"

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\textbf{(D)}\ \text{Ann, Dick, Carol, Bill}\qquad
 
\textbf{(D)}\ \text{Ann, Dick, Carol, Bill}\qquad
 
\textbf{(E)}\ \text{Ann, Dick, Bill, Carol}</math>   
 
\textbf{(E)}\ \text{Ann, Dick, Bill, Carol}</math>   
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== Solution ==
 +
Let the scores be <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. Then we have <math>A + C = B + D</math>, <math>A + B > C + D</math>, and <math>D > B + C</math>. Call these equations <math>(1)</math>, <math>(2)</math>, and <math>(3)</math> respectively. Then adding <math>(1)</math> and <math>(2)</math> gives <math>2A + B + C > B + C + 2D \implies 2A > 2D \implies A > D</math>. Now subtracting <math>(1)</math> from <math>(2)</math> gives <math>A + B - A - C > C + D - B - D \implies B - C > C - B \implies 2B > 2C \implies B > C</math>. Finally from <math>(3)</math>, since  <math>D > B + C</math> and <math>C</math> is non-negative, we must have <math>D > B</math>, so putting our results together we get <math>A > D > B > C</math>, which is answer <math>\boxed{\text{E}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 13:38, 1 March 2018

Problem

In a mathematics competition, the sum of the scores of Bill and Dick equalled the sum of the scores of Ann and Carol. If the scores of Bill and Carol had been interchanged, then the sum of the scores of Ann and Carol would have exceeded the sum of the scores of the other two. Also, Dick's score exceeded the sum of the scores of Bill and Carol. Determine the order in which the four contestants finished, from highest to lowest. Assume all scores were nonnegative.

$\textbf{(A)}\ \text{Dick, Ann, Carol, Bill} \qquad \textbf{(B)}\ \text{Dick, Ann, Bill, Carol} \qquad \textbf{(C)}\ \text{Dick, Carol, Bill, Ann}\\ \qquad \textbf{(D)}\ \text{Ann, Dick, Carol, Bill}\qquad \textbf{(E)}\ \text{Ann, Dick, Bill, Carol}$

Solution

Let the scores be $A$, $B$, $C$, and $D$. Then we have $A + C = B + D$, $A + B > C + D$, and $D > B + C$. Call these equations $(1)$, $(2)$, and $(3)$ respectively. Then adding $(1)$ and $(2)$ gives $2A + B + C > B + C + 2D \implies 2A > 2D \implies A > D$. Now subtracting $(1)$ from $(2)$ gives $A + B - A - C > C + D - B - D \implies B - C > C - B \implies 2B > 2C \implies B > C$. Finally from $(3)$, since $D > B + C$ and $C$ is non-negative, we must have $D > B$, so putting our results together we get $A > D > B > C$, which is answer $\boxed{\text{E}}$.

See also

1987 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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