1986 AHSME Problems/Problem 15

Problem

A student attempted to compute the average $A$ of $x, y$ and $z$ by computing the average of $x$ and $y$ , and then computing the average of the result and $z$ . Whenever $x < y < z$ , the student's final result is

$\textbf{(A)}\ \text{correct}\quad \textbf{(B)}\ \text{always less than A}\quad \textbf{(C)}\ \text{always greater than A}\quad\\ \textbf{(D)}\ \text{sometimes less than A and sometimes equal to A}\quad\\ \textbf{(E)}\ \text{sometimes greater than A and sometimes equal to A} \quad$

Solution

The true average is $A=\frac{x+y+z}{3}$ , and the student computed $B=\frac{\frac{x+y}{2}+z}{2}=\frac{x+y+2z}{4}$ , so $B-A = \frac{2z-x-y}{12} = \frac{(z-x)+(z-y)}{12}$ , which is always positive as $z>x$ and $z>y$ . Thus $B$ is always greater than $A$ , i.e. $\boxed{C}$ .

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1986 AHSME Problems/Problem 15

Problem

Solution

See also