1997 AMC 12 Problem 18

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

A total increase of

10

raises the mean by

2

, determining the list length

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

m

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.

Problem 18 in other years