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Bunuel
A total of 22 men and 26 women were at a party. The average (arithmetic mean) age of all of the people at the party was exactly 66.74 years. If the average age of the men was exactly 69.74 years, which of the following was closest to the average age, in years, of the women?

(A) 61.24
(B) 63.74
(C) 64.24
(D) 64.74
(E) 69.24
\((22+26)*66.74 = 22*69.74 + 26*w\)

Or, \(3203.52 = 1534.28 + 26w\)

Or, \(w = \frac{3203.52 - 1534.28}{26}\)

Or, \(w = 64.20\)

Thus, the correct answer will be (C)
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Bunuel
A total of 22 men and 26 women were at a party. The average (arithmetic mean) age of all of the people at the party was exactly 66.74 years. If the average age of the men was exactly 69.74 years, which of the following was closest to the average age, in years, of the women?

(A) 61.24
(B) 63.74
(C) 64.24
(D) 64.74
(E) 69.24
The average age of the men is exactly three years more than the average age of the whole group, which got me to thinking . . .

These numbers are horrid for calculating. And although prompt asks for "closest to," the answer choices aren't far enough apart to estimate, IMO.

1. Use the group mean and the men's mean to find the weight (total number of years' difference from mean) of the men as a group

Whole group mean age is 66.74, and male age mean is 69.74.

Men's individual deviation from the mean is 69.74 - 66.74 = +3

There are 22 men, each is +3 away from the mean.

The sum of each man's deviation is +3 * 22 = +66.

2. Women as a group must balance out men

By definition of group mean, the women as a group must "balance out" this total of + 66.

Or, the sum of the positive differences from the mean (the men) must equal the sum of the negative differences from the mean (which must be the women's).

If you think about a scale with two plates, where men as a group are all on one plate, that group of men has a total weight.

The mean is a balance point.

The women as a group, on the other plate, must equal the absolute value of the men's total weight in order for the scale plates to "level" or balance each other out, such that there is a mean, i.e., the scale's balancing point.

3. Women's individual difference from mean calculated by women's group difference from mean divided by number of women

The men are heavier than the mean; the sum of their deviations is, as noted, +66.

So the women's group total difference from the mean must be - 66.

There are 26 women. Each woman's individual difference from the group's mean is \(\frac{-66}{26}\) = - 2.538 -- > round up to - 2.54 to make next subtraction easier

4. Average age of each woman

Group mean of 66.74 - individual woman's difference from mean (-2.54) will yield average age of each woman.

66.74 - 2.54 = 64.20

Answer (C).

66/26 wasn't a whole lot of fun by hand, but it looked better than multiplication and addition among 22, 26, 66.74, and 69.74. :shock: I think I might have been right. Hope it helps.
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22*69.74 + 26w = 66.74*48
w = (66.74*48 - 69.74*22)/26
= (69.74*48 -3*48 - 69.74*22)/26
= (69.74(26) - 3*48)/26
= 69.74 - 5.53
= 64.21
Approx 64.24
Option C
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Why doesn't weighted average concept work here?

22/26=11/13

shouldn't the ratio of (difference between average age of men : 69.64 and Average age of all: 66.94) , and (Average age of Women :X and Average age of all: 66.94)
be in the ratio 11/13?


But 69.74- 66.74=+3.



Can someone pls help me understand why can't we use cross method of weighted averages here?
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Lads' Average               Women Average
      ­69.74                              w
         \                                  /
            \                             /
                      66.74
                  Total Average
            /                            \
         /                                  \
  66.74 - w                69.74 - 66.74 = 3

(Yes, graphics design is my passion. :P)

[66.74 - w][/3] = [22][/26]

66.74 - w = 33 / 13 which is almost 2.5

w = 66.74 -2.5 = 64.24.­
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