2000 AMC 8 Problems/Problem 23
Contents
Problem
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
Solution
Remember that if a list of numbers has an average of
, then the sum
of all the numbers on the list is
.
So if the average of the first numbers is
, then the first four numbers total
.
If the average of the last numbers is
, then the last four numbers total
.
If the average of all numbers is
, then the total of all seven numbers is
.
If the first four numbers are , and the last four numbers are
, then all "eight" numbers are
. But that's counting one number twice. Since the sum of all seven numbers is
, then the number that was counted twice is
, and the answer is
Algebraically, if , and
, you can add both equations to get
. You know that
, so you can subtract that from the last equation to get
, and
is the number that appeared twice.
Yay! :D
Video Solution
https://youtu.be/xdNSN7hYK6k Soo, DRMS, NM
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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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