1988 AMC 12 Problem 25

Question

X,

Y,

and

Z

are pairwise disjoint sets of people. The average ages of people in the sets

X,

Y,

Z,

X\cup Y,

X\cup Z,

and

Y\cup Z

are given in the table below.

Set	$X$	$Y$	$Z$	$X\cup Y$	$X\cup Z$	$Y\cup Z$
Average age of people in the set	$37$	$23$	$41$	$29$	$39.5$	$33$

Find the average age of the people in the set

X\cup Y\cup Z.

33

33.5

33.66

33.833

34

Let

x,y,z

be the sizes of the three sets

Each union average gives a linear relation among

x,y,z

Let the set sizes be

x,y,z.

The three union averages give

\begin{aligned} 37x+23y&=29(x+y),\\ 37x+41z&=39.5(x+z), \end{aligned}

and

23y+41z=33(y+z).

Thus

y=\frac{4x}{3}

and

z=\frac{5x}{3}.

The total average is

\frac{\begin{gathered} 37x+23(\frac{4x}{3})\\ {}+41(\frac{5x}{3}) \end{gathered}} {x+\frac{4x}{3}+\frac{5x}{3}}=34.

Thus the correct answer is E.

Problem 25 in other years