Difference between revisions of "2004 AMC 8 Problems/Problem 11"
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The numbers <math>-2, 4, 6, 9</math> and <math>12</math> are rearranged according to these rules: | The numbers <math>-2, 4, 6, 9</math> and <math>12</math> are rearranged according to these rules: | ||
| − | 1. The largest | + | 1. The largest isn't first, but it is in one of the first three places. |
| − | 2. The smallest | + | 2. The smallest isn't last, but it is in one of the last three places. |
| − | 3. The median | + | 3. The median isn't first or last. |
What is the average of the first and last numbers? | What is the average of the first and last numbers? | ||
| Line 16: | Line 16: | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2004|num-b=10|num-a=12}} | {{AMC8 box|year=2004|num-b=10|num-a=12}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 01:14, 16 January 2024
Problem
The numbers 
and 
are rearranged according to these rules:
1. The largest isn't first, but it is in one of the first three places.
2. The smallest isn't last, but it is in one of the last three places.
3. The median isn't first or last.
What is the average of the first and last numbers?
Solution
From rule 1, the largest number, , can be second or third. From rule 2, because there are five places, the smallest number 
can either be third or fourth. The median, 
can be second, third, or fourth. Because we know the middle three numbers, the first and last numbers are 
and 
, disregarding their order. Their average is 
.
See Also
| 2004 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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. 
