Difference between revisions of "1988 AHSME Problems/Problem 25"
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| − | + | Let the variables <math>X</math>, <math>Y</math>, and <math>Z</math> represent the sums of the ages of the people in sets <math>X</math>, <math>Y</math>, and <math>Z</math> respectively. Let <math>x</math>, <math>y</math>, and <math>z</math> represent the numbers of people who are in sets <math>X</math>, <math>Y</math>, and <math>Z</math> respectively. Since the sets are disjoint, we know, for example, that the number of people in the set <math>X \cup Y</math> is <math>x+y</math> and the sum of their ages is <math>X+Y</math>, and similar results apply for the other unions of sets. Thus we have <math>X=37x</math>, <math>Y=23y</math>, <math>Z=41z</math>, <math>X+Y=29(x+y) \implies 37x + 23y = 29x + 29y \implies 8x = 6y \implies y = \frac{4}{3}x</math>, and <math>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</math>. Hence the answer is <math>\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</math>, which is <math>\boxed{\text{E}}</math>. 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. | |
Latest revision as of 13:35, 27 February 2018
Problem
and 
are pairwise disjoint sets of people. The average ages of people in the sets
and 
are 
and 
respectively.
Find the average age of the people in set 
.
Solution
Let the variables 
, 
, and represent the sums of the ages of the people in sets , , and respectively. Let 
, 
, and 
represent the numbers of people who are in sets , , and respectively. Since the sets are disjoint, we know, for example, that the number of people in the set 
is 
and the sum of their ages is 
, and similar results apply for the other unions of sets. Thus we have 
, 
, 
, 
, and 
. Hence the answer is 
, which is 
. 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.
See also
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
