2021 AMC 10B Problem 19

Problem 19 of 25HarderAlgebra

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

72

greater than the least integer in

S.

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

S?

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Solution

Let the sum of all the integers be

s,

the greatest number be

g,

the least number be

l,

and the size of

S

n.

The two averages of sets with

n-1

elements give

\begin{aligned} s-l&=40(n-1),\\ s-g&=32(n-1). \end{aligned}

Subtracting and using

g-l=72

gives

72=8(n-1),

n=10.

Removing both extremes leaves

8

numbers with sum

8\cdot35=280.

On the other hand, removing only the least number leaves a sum of

9\cdot40=360,

g=360-280=80.

Hence

l=80-72=8,

and the original sum is

s=280+80+8=368.

The required average is

\frac{s}{n}=\frac{368}{10}=36.8.

Thus, the answer is D .