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2021 AMC 10B Problem 19

2021 AMC 10B Problem 19AMC 10 Step by Step

Name: Date: September 13, 2026

Statistics & Data · difficulty 3 of 5 · about 5 min · position P16-20

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 $S$ ?

A)
$36.2$
B)
$36.4$
C)
$36.6$
D)
$36.8$
E)
$37$

Answer

Key insight

Both 'remove the greatest' and 'remove the least' leave n − 1 numbers, and their sums differ by 72, so 72/(n − 1) = 40 − 32.

Solution

Let $S$ have $n$ elements with total $T$ , greatest element $M$ and least element $m$ . Turn each average into a sum:

Remove $M$ ( $n - 1$ elements): $T - M = 32(n-1)$ .
Remove $M$ and $m$ ( $n - 2$ elements): $T - M - m = 35(n-2)$ .
Put $M$ back, so only $m$ is missing ( $n - 1$ elements): $T - m = 40(n-1)$ .
$M - m = 72$ .

Subtract (1) from (3): $M - m = 8(n - 1)$ . With (4), $8(n-1) = 72$ , so $n = 10$ .

Now (1) gives $T - M = 288$ and (2) gives $T - M - m = 280$ , so $m = 8$ , hence $M = 80$ and $T = 288 + 80 = 368$ .

The average of $S$ is $\frac{368}{10} = 36.8$ .

The answer is $\boxed{\textbf{(D)}\ 36.8}$ .

Why this works

Averages hide sums; once each statement is written as "sum $=$ average $\times$ count," the problem is a small linear system. The efficient move is to compare the two statements with the same count ( $n-1$ ): their sums differ by exactly $M - m$ , which the problem hands you. Always look for two equations that share a count and subtract.

The trap

Setting up the third average with n − 2 elements, forgetting that the greatest integer was put back and the set again has n − 1 elements.

Common mistakes

Setting up the third average with n − 2 elements, forgetting that the greatest integer was put back and the set again has n − 1 elements.
Solving for $n$ correctly but then averaging $32, 35, 40$ or reporting $T - M$ over $n$ instead of $T$ over $n$ .

Techniques

Set up the equation/formula and compute; no special trick needed