1988 AHSME Problems/Problem 25

Problem

$X, Y$ 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 $37, 23, 41, 29, 39.5$ and $33$ respectively. Find the average age of the people in set $X \cup Y \cup Z$ .

$\textbf{(A)}\ 33\qquad \textbf{(B)}\ 33.5\qquad \textbf{(C)}\ 33.6\overline{6}\qquad \textbf{(D)}\ 33.83\overline{3}\qquad \textbf{(E)}\ 34$

Solution

Let the variables $X$ , $Y$ , and $Z$ represent the sums of the ages of the people in sets $X$ , $Y$ , and $Z$ respectively. Let $x$ , $y$ , and $z$ represent the numbers of people who are in sets $X$ , $Y$ , and $Z$ respectively. Since the sets are disjoint, we know, for example, that the number of people in the set $X \cup Y$ is $x+y$ and the sum of their ages is $X+Y$ , and similar results apply for the other unions of sets. Thus we have $X=37x$ , $Y=23y$ , $Z=41z$ , $X+Y=29(x+y) \implies 37x + 23y = 29x + 29y \implies 8x = 6y \implies y = \frac{4}{3}x$ , and $X+Z = 39.5(x+z) \implies 37x + 41z = 39.5x + 39.5z \implies 74x + 82z = 79x + 79z \implies 3z = 5x \implies z = \frac{5}{3}x$ . Hence the answer is $\frac{X+Y+Z}{x+y+z} = \frac{37x + 23\times\frac{4}{3}x + 41\times\frac{5}{3}x}{x+\frac{4}{3}x+\frac{5}{3}x} = \frac{136x}{4x} = 34$ , which is $\boxed{\text{E}}$ . Notice that we did not need all three of the means of the unions of the sets - any two of them would have been sufficient to determine the answer.

1988 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1988 AHSME Problems/Problem 25

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