2021 AMC 12B Spring Problem 12

Question

Suppose that

S

is a finite set of positive integers. If the greatest integer in

S

is removed from

S,

then the average value (arithmetic mean) of the integers remaining is

32.

If the least integer in

S

is also removed, then the average value of the integers remaining is

35.

If the greatest integer is then returned to the set, the average value of the integers rises to

40.

The greatest integer in the original set

S

is

72

greater than the least integer in

S.

What is the average value of all the integers in the set

S?

36.2

36.4

36.6

36.8

37

Answer: D

Concepts:mean system of equations

Difficulty rating: 1630

Small hint:

Let

n=|S|,

total

T,

greatest

M,

least

L;

write each average as an equation

Big hint:

Subtracting

\dfrac{T-M}{n-1}=32

from

\dfrac{T-L}{n-1}=40

gives

M-L=8(n-1)

Solution:

Let

n=|S|,

let

T

be the total,

M

the greatest, and

L

the least. Then

\dfrac{T-M}{n-1}=32,

\dfrac{T-M-L}{n-2}=35,

and

\dfrac{T-L}{n-1}=40.

Subtracting the first from the third:

\dfrac{M-L}{n-1}=8.

Since

M-L=72,

we get

n-1=9,

so

n=10.

Then

T-M=288

and

T-L=360.

The middle equation gives

T-M-L=35\cdot 8=280,

so

L=288-280=8

and

M=80.

Thus

T=288+80=368,

and the average is

\dfrac{368}{10}=36.8.

Thus, the correct answer is D.