Question

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 $a,$ $b,$ $c,$ $d,$ $e$ are chosen, we shall always have:

$m=p$

$m\geq p$

$m\gt p$

$m\lt p$

none of these

Answer: E

Concepts:mean weighted mean counterexample

Difficulty rating: 1610

Small Hint:

Express $m$ in terms of the subgroup means $k$ and $l$

Big Hint:

Compare $m=\frac{2k+3l}{5}$ with $p=\frac{k+l}{2}$

Solution:

We have $m=\frac{2k+3l}{5}$ and $p=\frac{k+l}{2},$ so $m-p=\frac{l-k}{10}.$ This difference may be positive, zero, or negative depending on the chosen numbers. None of the first four relations always holds.

Therefore, the correct answer is E.

1965 AMC 12 Problem 26

Answer: E

Difficulty rating: 1610

Small Hint:

Big Hint:

Solution:

Problem 26 in Other Years