1965 AHSME Problems/Problem 26

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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, $m<p$, 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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