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# A total of 22 men and 26 women were at a party. The average (arithmeti

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A total of 22 men and 26 women were at a party. The average (arithmeti  [#permalink]

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30 Jun 2017, 02:44
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Difficulty:

65% (hard)

Question Stats:

56% (02:44) correct 44% (02:34) wrong based on 57 sessions

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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

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Re: A total of 22 men and 26 women were at a party. The average (arithmeti  [#permalink]

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30 Jun 2017, 02:46
Bunuel wrote:
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

This is a modified question of the the following OG question: https://gmatclub.com/forum/a-total-of-2 ... 07390.html
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A total of 22 men and 26 women were at a party. The average (arithmeti  [#permalink]

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30 Jun 2017, 03:03
Bunuel wrote:
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

Total number of people $$= 22 + 26 = 48$$

Average age of all the people $$= 66.74$$

Total sum of ages $$= 66.74 * 48 = 3203.52$$

Average age of the men $$= 69.74$$

Total number of men $$= 22$$

Sum of ages of men $$= 1534.28$$

Sum of ages of women $$= 3203.52 - 1534.28 = 1669.24$$

Total number of women $$= 26$$

Average age of women $$= \frac{1669.24}{26} = 64.20$$

Closest value $$= 64.24$$

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Re: A total of 22 men and 26 women were at a party. The average (arithmeti  [#permalink]

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30 Jun 2017, 11:11
Bunuel wrote:
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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A total of 22 men and 26 women were at a party. The average (arithmeti  [#permalink]

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30 Jun 2017, 11:45
1
Bunuel wrote:
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

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. I think I might have been right. Hope it helps.
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A total of 22 men and 26 women were at a party. The average (arithmeti  [#permalink]

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02 Jul 2017, 05:48
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
A total of 22 men and 26 women were at a party. The average (arithmeti   [#permalink] 02 Jul 2017, 05:48
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# A total of 22 men and 26 women were at a party. The average (arithmeti

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