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A college reunion gathering had 22 males and 26 females. The average [#permalink]
I know this may sound weird but I would like to turn this problem into a chemical mixture one:

There are 2 solutions:
Solution Man: 22ml - 3%
Solution Woman: 26ml - x%
Mixture: 48 ml - 0%

(22.3+26.x)/48=0 ==> x=-2.5
Average age of woman = 66.74 - 2.5 = 64.24

Hope you find it interesting!
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Re: A college reunion gathering had 22 males and 26 females. The average [#permalink]
Beqa wrote:
I know this may sound weird but I would like to turn this problem into a chemical mixture one:

There are 2 solutions:
Solution Man: 22ml - 3%
Solution Woman: 26ml - x%
Mixture: 48 ml - 0%

(22.3+26.x)/48=0 ==> x=-2.5
Average age of woman = 66.74 - 2.5 = 64.24

Hope you find it interesting!


Very interesting approach!

But I cannot really follow the steps... Can you eleborate a bit more? Why the 3 and why the 0?
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Re: A college reunion gathering had 22 males and 26 females. The average [#permalink]
3
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T1101 wrote:
Beqa wrote:
I know this may sound weird but I would like to turn this problem into a chemical mixture one:

There are 2 solutions:
Solution Man: 22ml - 3%
Solution Woman: 26ml - x%
Mixture: 48 ml - 0%

(22.3+26.x)/48=0 ==> x=-2.5
Average age of woman = 66.74 - 2.5 = 64.24

Hope you find it interesting!


Very interesting approach!

But I cannot really follow the steps... Can you eleborate a bit more? Why the 3 and why the 0?


I would like to start with a familiar Weighted average problem:
How much alcohol 2% should be poured into 10 ml alcohol 10% to make alcohol 8%?

Using weighted average chart, we have:
2%................2% ------1
...........8%
10%...............6% ------3

The ratio between alcohol 2% and alcohol 10% is 1:3 --> You need 3.33ml alcohol 2%.
You may also notice that what really affects the ratio is the difference among figures given (2 and 6), NOT the figures themselves. For example, the answer is the same for the below 2 problems:
How much alcohol 12% should be poured into 10 ml alcohol 20% to make alcohol 18%?
How much alcohol 22,9876% should be poured into 10 ml alcohol 30,9876% to make alcohol 28,9876%?


Come back to our problem, as I found that it is actually a weighted average problem, I do not care the decimals but care the difference among figures given. Then I turn 66.74 and 69.74 into 0 and 3 for easier calculations.

If you find it helpful, please give me Kudo ^^
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Re: A college reunion gathering had 22 males and 26 females. The average [#permalink]
taniad wrote:
Selected the weighted average method.
(a+b)/sum of weights=0--- where a and b are the product of the total difference to the average with its' respective weights, in this case, men=22, women=26

Difference of Average age from Men' age= 66.74-69.74= -3
Difference of Average age from Women' age= x

Therefore, ((-3*22)+26x)/48=0
x=66/26= 2.5

avg age of women= total avg age-x=66.74-2.5= 64.24



Can you elaborate with this weighted average method? I've never seen this before.
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Re: A college reunion gathering had 22 males and 26 females. The average [#permalink]
blueblac92 wrote:
taniad wrote:
Selected the weighted average method.
(a+b)/sum of weights=0--- where a and b are the product of the total difference to the average with its' respective weights, in this case, men=22, women=26

Difference of Average age from Men' age= 66.74-69.74= -3
Difference of Average age from Women' age= x

Therefore, ((-3*22)+26x)/48=0
x=66/26= 2.5

avg age of women= total avg age-x=66.74-2.5= 64.24



Can you elaborate with this weighted average method? I've never seen this before.


Check out this link: https://gmatclub.com/forum/ultimate-gma ... 0bc931c9bb

Under point 18. Mixture Problems!
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Re: A college reunion gathering had 22 males and 26 females. The average [#permalink]
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Bunuel wrote:
A college reunion gathering had 22 males and 26 females. The average (arithmetic mean) age of all of the people at the party was exactly 66.74 years. If the average age of the males was exactly 69.74 years, which of the following was closest to the average age, in years, of the females?

(A) 61.24
(B) 63.74
(C) 64.24
(D) 64.74
(E) 69.24


We can create the weighted average equation:

66.74 = (69.74 x 22 + f x 26)/48

48 x 66.74 = 69.74 x 22 + f x 26

3203.52 = 1534.28 + 26f

1669.24 = 26f

64.2 ≈ f

Answer: C
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A college reunion gathering had 22 males and 26 females. The average [#permalink]
An alternate approach (estimation)

Female to Male ratio is 26:22 -> 13:11; In other words, women constitute slightly more than 50% in the total gathering.


By estimation, to bring the average down from 69.74 (if there were only men) to 66.74 (to the average of the group consisting of both men and women),women need to bring the average down by around 3 - However, since women are more than men in the group they can relax a bit and contribute slightly less than 3.

Now, that means --> 66.74 -3 = 63.74. So, the closest answer choice that is less than 3 units away from average but not too far away is (C) 64.24.

The key here is to realize that - If women contribute equally or if women's average age is exactly 3 units away (63.74), because of women's majority in the group, the average will be slightly less than 66.74.
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Re: A college reunion gathering had 22 males and 26 females. The average [#permalink]
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Re: A college reunion gathering had 22 males and 26 females. The average [#permalink]
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