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A college reunion gathering had 22 males and 26 females. The average

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A college reunion gathering had 22 males and 26 females. The average  [#permalink]

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New post 25 Sep 2018, 04:59
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  65% (hard)

Question Stats:

52% (02:26) correct 48% (01:40) wrong based on 66 sessions

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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

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Re: A college reunion gathering had 22 males and 26 females. The average  [#permalink]

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New post 26 Sep 2018, 02:34
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




Given,
Avg. of 48 total students (22M and 26F): 66.74 yrs
Total age must be: 66.74 * 48= 3203.52 yrs (total age of 22M+26F students)
Also can write: 22M+26F= 3203.52

Avg. age of 22M: 69.74 yrs
Total age of 22M students: 69.74 * 22= 1534.28 yrs
Also can write: 22M= 1534.28

Avg. of 26F : X yrs
Total age of 26F students= 26X
Also can write: 26F=26X

22M+26F= 3203.52 (replacing the 22M and 26F with respective values)
1534.28 + 26X = 3203.52
26X = 3203.52 - 1534.28 = 1669.24
X = 1669.24/26= 64.2015… approx. 64.24

Ans. is C

Bunuel Is there any easier way (without calculator) to do this?
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A college reunion gathering had 22 males and 26 females. The average  [#permalink]

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New post 26 Sep 2018, 03:27
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]

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New post 26 Sep 2018, 07:59
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]

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New post 26 Sep 2018, 20:18
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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]

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New post 26 Sep 2018, 20:58
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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
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Re: A college reunion gathering had 22 males and 26 females. The average  [#permalink]

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New post 27 Sep 2018, 00:27
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]

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New post 27 Sep 2018, 00:31
1
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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New post 29 Sep 2018, 17:56
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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Re: A college reunion gathering had 22 males and 26 females. The average &nbs [#permalink] 29 Sep 2018, 17:56
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