1997 AHSME Problems/Problem 18
Problem
A list of integers has mode and mean . The smallest number in the list is . The median of the list is a member of the list. If the list member were replaced by , the mean and median of the new list would be and , respectively. If were instead replaced by , the median of the new list would be . What is ?
Solution
Let there be integers on the list. The list of integers has mean , so the sum of the integers is .
Replacing with will increase the sum of the list from to .
The new mean of the list is , so the new sum of the list is also .
Thus, we get , leading to numbers on the list.
If there are numbers on the list with mode and smallest number , then the list is
Since replacing with gives a new median of , and must be on the list of integers since is odd, , and the list is now
The sum of the numbers on this list is , so we get:
, giving answer .
The original list is , with mean and median and mode .
The second list is , with mean and median .
The third list is with median .
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.