Difference between revisions of "2004 AMC 8 Problems/Problem 18"
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==Solution 1== | ==Solution 1== | ||
| − | The only way to get Ben's score is with <math>1+3=4</math>. Cindy's score can be made of <math>3+4</math> or <math>2+5</math>, but since Ben already hit the <math>3</math>, Cindy hit <math>2+5=7</math>. | + | The only way to get Ben's score is with <math>1+3=4</math>. Cindy's score can be made of <math>3+4</math> or <math>2+5</math>, but since Ben already hit the <math>3</math>, Cindy hit <math>2+5=7</math>. Similarly, Dave's darts were in the region <math>4+7=11</math>. Lastly, because there is no <math>7</math> left, <math>\boxed{\textbf{(A)}\ \text{Alice}}</math> must have hit the regions <math>6+10=16</math> and Ellen <math>8+9=17</math>. |
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==Solution 2== | ==Solution 2== | ||
Revision as of 11:55, 12 October 2019
Contents
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 
through 
. Each throw hits the target in a region with a different value. The scores are: Alice 
points, Ben 
points, Cindy 
points, Dave 
points, and Ellen 
points. Who hits the region worth 
points?
Solution 1
The only way to get Ben's score is with 
. Cindy's score can be made of 
or 
, but since Ben already hit the 
, Cindy hit 
. Similarly, Dave's darts were in the region 
. Lastly, because there is no left, 
must have hit the regions 
and Ellen 
.
Solution 2
Taking the 2 largest scores to narrow out the points we get
Ellen with either 
or 
and Alice with either 
or 
, the 2 pairs that work are
Alice=
Ellen=, therefore Alice scored the .
See Also
| 2004 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
