Question

A list of integers has mode $32$ and mean $22.$ The smallest number in the list is $10.$ The median $m$ of the list is a member of the list. If the list member $m$ were replaced by $m+10,$ the mean and median of the new list would be $24$ and $m+10,$ respectively. If $m$ were instead replaced by $m-8,$ the median of the new list would be $m-4.$ What is $m?$

$16$

$17$

$18$

$19$

$20$

Answer: E

Concepts:mean median mode

Difficulty rating: 2010

Small Hint:

A total increase of $10$ raises the mean by $2$ , determining the list length

Big Hint:

Order the five entries and use the two stated median changes to identify the entries beside $m$

Solution:

The list has $\frac{10}{2}=5$ entries. Write them $10\le a\le m\le b\le c.$ Replacing $m$ by $m+10$ makes that value the median, so $b\ge m+10$ and $c\ge m+10;$ because the mode is $32,$ we must have $b=c=32.$ The original total is $110,$ giving $a+m=36.$ Replacing $m$ by $m-8$ makes the median $a=m-4,$ so $2m-4=36$ and $m=20.$ Thus E is correct.

1997 AMC 12 Problem 18

Answer: E

Difficulty rating: 2010

Small Hint:

Big Hint:

Solution:

Problem 18 in Other Years