# 1965 AHSME Problems/Problem 26

## Problem

For the numbers $a, b, c, d, e$ define $m$ to be the arithmetic mean of all five numbers; $k$ to be the arithmetic mean of $a$ and $b$; $l$ to be the arithmetic mean of $c, d$, and $e$; and $p$ to be the arithmetic mean of $k$ and $l$. Then, no matter how $a, b, c, d$, and $e$ are chosen, we shall always have:

$\textbf{(A)}\ m = p \qquad \textbf{(B) }\ m \ge p \qquad \textbf{(C) }\ m > p \qquad \\ \textbf{(D) }\ m < p\qquad \textbf{(E) }\ \text{none of these}$

## Solution

We shall begin by eliminating some options through counterexamples. If $a=b=c=d=e$, then $m=k=l=a$, so $p=a$, and $m=p$. Answers (C) and (D) do not allow for $m=p$, so they can be eliminated. If we set $a=60$ and $b=c=d=e=0$, then $m=12$, $k=30$, $l=0$, and $p=15$. Here, $m, so we can throw out options (A) and (B) as well. Now, we are left with only option $\fbox{\textbf{(E) }none of these}$.

## See Also

 1965 AHSC (Problems • Answer Key • Resources) Preceded byProblem 25 Followed byProblem 27 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions

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