1965 AMC 12 Problem 26

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

Express

m

in terms of the subgroup means

k

and

l

Compare

m=\frac{2k+3l}{5}

with

p=\frac{k+l}{2}

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.

Problem 26 in other years