Difference between revisions of "2004 AMC 8 Problems/Problem 18"

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Alice=<math>10,6</math>
 
Alice=<math>10,6</math>
 
Ellen=<math>9,8</math>, therefore Alice scored the <math>6</math>.
 
Ellen=<math>9,8</math>, therefore Alice scored the <math>6</math>.
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This way, we don’t need to check anyways if anyone else got <math>6</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2004|num-b=17|num-a=19}}
 
{{AMC8 box|year=2004|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:53, 23 March 2020

Problem

Five friends compete in a dart-throwing contest. Each one has two darts to throw at the same circular target, and each individual's score is the sum of the scores in the target regions that are hit. The scores for the target regions are the whole numbers $1$ through $10$. Each throw hits the target in a region with a different value. The scores are: Alice $16$ points, Ben $4$ points, Cindy $7$ points, Dave $11$ points, and Ellen $17$ points. Who hits the region worth $6$ points?

$\textbf{(A)}\ \text{Alice}\qquad \textbf{(B)}\ \text{Ben}\qquad \textbf{(C)}\ \text{Cindy}\qquad \textbf{(D)}\ \text{Dave} \qquad \textbf{(E)}\ \text{Ellen}$

Solution 1

The only way to get Ben's score is with $1+3=4$. Cindy's score can be made of $3+4$ or $2+5$, but since Ben already hit the $3$, Cindy hit $2+5=7$. Similarly, Dave's darts were in the region $4+7=11$. Lastly, because there is no $7$ left, $\boxed{\textbf{(A)}\ \text{Alice}}$ must have hit the regions $6+10=16$ and Ellen $8+9=17$.

Solution 2

Taking the 2 largest scores to narrow out the points we get

Ellen with either $10,7$ or $9,8$ and Alice with either $10,6$ or $9,7$, the 2 pairs that work are Alice=$10,6$ Ellen=$9,8$, therefore Alice scored the $6$.

This way, we don’t need to check anyways if anyone else got $6$.

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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 AJHSME/AMC 8 Problems and Solutions

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