# Difference between revisions of "2021 AMC 10B Problems/Problem 19"

## Problem

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 is $S$ is [i]also[/i] 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 of $40$. 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 $S ?$ $\textbf{(A)} ~36.2 \qquad\textbf{(B)} ~36.4 \qquad\textbf{(C)} ~36.6 \qquad\textbf{(D)} ~36.8 \qquad\textbf{(E)} ~37$

## Solution

Let the lowest value be L and the highest G, and let the sum be Z and the amount of numbers n. We have $\frac{Z-G}{n-1}=32$, $\frac{Z-L-G}{n-2}=35$, $\frac{Z-L}{n-1}=40$, and $G=L+72$. Clearing denominators gives $Z-G=32n-32$, $Z-L-G=35n-70$, and $Z-L=40n-40$. We use $G=L+72$ to turn the first equation into $Z-L=32n+40$, which gives $n=10$. Turning the second into $Z-2L=35n+2$ we see $L=8$ and $Z=368$ so the average is $\frac{Z}{n}=\boxed{(D)36.8}$ ~aop2014

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