2000 AMC 8 Problem 23
Attempt Problem 23 of the 2000 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 2000 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
23.
There is a list of seven numbers. The average of the first four numbers is and the average of the last four numbers is If the average of all seven numbers is then the number common to both sets of four numbers is
Answer: B
Small hint:
Convert each average into a sum
Big hint:
The shared number is counted twice in the two four-number sums
Solution:
The sum of the first four numbers is The sum of the last four numbers is
The sum of all seven numbers is We know that the number common to both sets is included in both of the first two sums.
This means that the sum of the first two sums includes every number once, except for the common number which is included twice.
The third sum, however, only includes every number once. This means that the sum of the first two sums minus the third sum yields our desired number.
Therefore, the common number is
Thus, B is the correct answer.
Problem 23 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
- 1996 AMC 8
- 1997 AMC 8
- 1998 AMC 8
- 1999 AMC 8
- 2001 AMC 8
- 2002 AMC 8
- 2003 AMC 8
- 2004 AMC 8
- 2005 AMC 8
- 2006 AMC 8
- 2007 AMC 8
- 2008 AMC 8
- 2009 AMC 8
- 2010 AMC 8
- 2011 AMC 8
- 2012 AMC 8
- 2013 AMC 8
- 2014 AMC 8
- 2015 AMC 8
- 2016 AMC 8
- 2017 AMC 8
- 2018 AMC 8
- 2019 AMC 8
- 2020 AMC 8
- 2022 AMC 8
- 2023 AMC 8
- 2024 AMC 8
- 2025 AMC 8
- 2026 AMC 8