2021 AMC 10B Problem 19

P16-20~5 minPrint

Statistics & Data Systems of Equations Set up the equation/formula and compute; no special trick needed

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 $S$ ?

Show solution without answering

Key insight

Removing the greatest or the least leaves n − 1 numbers either way. Those two totals differ by 72 and the averages by 8, so n − 1 = 9.

Solution

An average hides a total, so turn every average into a total: total $=$ average $\times$ how many.

Let $S$ have $n$ numbers with total $T$ , greatest number $M$ and least number $m$ .

Remove the greatest. There are $n - 1$ numbers left: $T - M = 32(n-1)$ .
Remove the least too. There are $n - 2$ left: $T - M - m = 35(n-2)$ .
Put the greatest back. Only the least is missing, so there are $n - 1$ numbers: $T - m = 40(n-1)$ .
Also, $M - m = 72$ .

Step 1: find $n$ . Equations 1 and 3 both have $n - 1$ numbers, so compare them. Subtract 1 from 3:
$(T - m) - (T - M) = 40(n-1) - 32(n-1), \qquad M - m = 8(n-1).$
Since $M - m = 72$ , we get $8(n - 1) = 72$ , so $n - 1 = 9$ and $n = 10$ .

Step 2: find the rest.

Equation 1: $T - M = 32 \times 9 = 288$ .
Equation 2: $T - M - m = 35 \times 8 = 280$ . Comparing with the line above, $m = 288 - 280 = 8$ .
Then $M = 8 + 72 = 80$ , and $T = 288 + 80 = 368$ .
Check with equation 3: $T - m = 360 = 40 \times 9$ . Correct.

Step 3: the average. $\frac{368}{10} = 36.8$ .

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

Why this works

Averages are hard to combine, but totals just add and subtract. Once every fact is written as a total, look for two facts that use the same number of items. Subtracting them leaves only the difference between the greatest and least, which the problem gives you.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(D)}\ 36.8}$ .

Review in the AoPS books

Prealgebra, Chapter 13: Data and Statistics
Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Intermediate Algebra, Chapter 1: Basic Techniques for Solving Equations (§1.4)