2004 AMC 8 Problem 11
Attempt Problem 11 of the 2004 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
11.
The numbers and are rearranged according to these rules:
The largest isn’t first, but it is in one of the first three places.
The smallest isn’t last, but it is in one of the last three places.
The median isn’t first or last.
What is the average of the first and last numbers?
Answer: C
Small hint:
Use the rules to locate the largest, smallest, and median numbers.
Big hint:
The first and last positions are filled by the two remaining numbers.
Solution:
Note that the largest, smallest, and median numbers cannot be the first or last number.
This means that the first and last numbers are and in some order. Their average is
Thus, C is the correct answer.
Problem 11 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1990 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
- 1994 AMC 8
- 1995 AMC 8
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- 1997 AMC 8
- 1998 AMC 8
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