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
https://www.youtube.com/watch?v=DZ_AzlW3qqM ~David
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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All AJHSME/AMC 8 Problems and Solutions |
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