1986 AMC 12 Problem 15

Question

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\lt y\lt z,

the student’s final result is

correct

always less than

A

always greater than

A

sometimes less than

A

and sometimes equal to

A

sometimes greater than

A

and sometimes equal to

A

Write both the true average and the student’s result as algebraic expressions

Subtract the true average and use the order

x\lt y\lt z

to determine the sign

The student’s result is

\frac{x+y+2z}{4},

while the true average is

\frac{x+y+z}{3}.

Their difference is

\begin{aligned} &\frac{x+y+2z}{4}-\frac{x+y+z}{3}\\ &\qquad=\frac{2z-x-y}{12}. \end{aligned}

Since

z\gt x

and

z\gt y,

the numerator is positive. The student’s result is therefore always greater than

A.

Thus the correct answer is C.

Problem 15 in other years