# Difference between revisions of "1993 AJHSME Problems/Problem 15"

## Problem

The arithmetic mean (average) of four numbers is $85$. If the largest of these numbers is $97$, then the mean of the remaining three numbers is

$\text{(A)}\ 81.0 \qquad \text{(B)}\ 82.7 \qquad \text{(C)}\ 83.0 \qquad \text{(D)}\ 84.0 \qquad \text{(E)}\ 84.3$

## Solution

Say that the four numbers are $a, b, c,$ & $97$. Then $\frac{a+b+c+97}{4} = 85$. What we are trying to find is $\frac{a+b+c}{3}$. Solving, $$\frac{a+b+c+97}{4} = 85$$ $$a+b+c+97 = 340$$ $$a+b+c = 243$$ $$\frac{a+b+c}{3} = \boxed{\mathrm{(A)}\ 81}$$