The averages of X and Y are 14 apart. The average of X ∪ Y is 6 above that of Y and 8 below that of X. So the sizes of X and Y must be in the ratio of 6:8.
The averages of Y and Z are 18 apart, with the average of Y ∪ Z 10 above that of Y and 8 below that of Z. So the sizes of Y and Z must be in the ratio of 8:10.
Thus X, Y, and Z are in the ratio of 6:8:10. For convenience, we can take them to in fact be of sizes 6, 8, and 10.
Then X has 6*37 = 222 years of age in it. Y has 8*23 = 184 years of age in it. And Z has 10*41 = 410 years of age in it. That's a total of 222 + 184 + 410 = 816 years of age. There are a total of 24 people in X, Y, and Z combined, so the average age is 816/24 = 34.
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