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Math Expert V
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A total of 22 men and 26 women were at a party, and the average  [#permalink]

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Question Stats: 72% (02:36) correct 28% (02:46) wrong based on 788 sessions

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A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?

(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33

Kudos for a correct solution.

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Re: A total of 22 men and 26 women were at a party, and the average  [#permalink]

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Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?

(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33

Kudos for a correct solution.

One can use the weighted average method:

$$\frac{Weight 1}{Weight 2} = \frac{Average2-Average}{Average-Average1}$$

For or example:

$$\frac{26}{22} = \frac{38-35}{35-Average1}$$

Solving for Average1 you will get exactly 32.46154. Answer D.
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Re: A total of 22 men and 26 women were at a party, and the average  [#permalink]

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Let Aw = average age of women
Average = (Number of men X average age of men + Number of women X average age of women)/total number of people

35 * 48 = 22 * 38 + 26 * Aw
=>1680 = 836 + 26 Aw
=> 26 Aw = 844
=>Aw = 32.5

Scale method -
If number of men and women were equal , the average would have been at the center of the average age of the individual groups
Average age of women = 32 as 35 will be equidistant from 32 and 38 (average age of men)
At this point we can eliminate A , B and C

But since , number of women are more , average age will be more than 32
Ratio of women to men = 26:22 = 13:11
If Aw= 32.5
Distance from the average
35- 32.5 :38-35
2.5 :3
5:6
which is rougly equal to inverse of 13:11

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Re: A total of 22 men and 26 women were at a party, and the average  [#permalink]

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Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?

(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33

Kudos for a correct solution.

You can also think of it like this on a scale method base:

x(Average female)..........(35-x)............Average(35)..........(3)............(38)Average male
#26 (No. of females)..........................................................................#22 (No. of males)

So again this shows that we have to groups which mixed together have an a weighted average of 35. The "weighs" in this example are the number of people. For men 22 and for woman 26. You can now equate the known weighs with the differences to average:

$$\frac{26}{22}=\frac{3}{(35-x)}$$

Then solve for x to get x= 32.4615

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Re: A total of 22 men and 26 women were at a party, and the average  [#permalink]

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Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?

(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33

Kudos for a correct solution.

38*22 + W*26 = 35*(22+26)
i.e. W = 844/26 = 32.46 = 32.5

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A total of 22 men and 26 women were at a party, and the average  [#permalink]

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Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?

(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33

Kudos for a correct solution.

Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

There are 48 people in total.
So, 22/48 = the proportion of women
And 26/48 = the proportion of men

Let x = average age of the women

So, plugging the information into the formula, we get: (22/48)(38) + (26/48)(x) = 35
Simplify: (11/24)(38) + (13/24)(x) = 35
Multiply both sides by 24 to get: (11)(38) + 13x = (35)(24)
Simplify: 418 + 13x = 840
Rearrange: 13x = 422
Solve: x = 422/13 ≈ 32.5

Cheers,
Brent
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Originally posted by GMATPrepNow on 20 Oct 2015, 08:03.
Last edited by GMATPrepNow on 06 Mar 2018, 10:49, edited 1 time in total.
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Re: A total of 22 men and 26 women were at a party, and the average  [#permalink]

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Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?

(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33

Kudos for a correct solution.

let average age of women be x

then, we have

$$22 * 38 + 26 * x = 48 * 35$$

=> $$x = 32.46$$

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Posts: 1
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Re: A total of 22 men and 26 women were at a party, and the average  [#permalink]

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1
I followed a similar approach to the scale. Given that m:w is 11:13 and we know that the AVG(m) = 38 and AVG(m+w)=35. We already know the AVG(w) has to be less than 35 given the higher weighting to w. I ended up multiplying the (11/13) * 38. To simplify, I rounded to redo the formula as (11/13)*39 = 33. Therefore, the answer must be just under 33 or 32.5. Also, keep in mind the key word of "closest", which implies that you do not need an EXACT answer (no crazy math).
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Re: A total of 22 men and 26 women were at a party, and the average  [#permalink]

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Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?

(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33

Kudos for a correct solution.

1) Since there are 48 people (22 men and 26 women) and the average age of all of them is 35, the total ages of all the people at the party is 48(35).

2) Since there are 22 men and their average age is 38, the sum of all the men's ages is 22(38).

3) There are 26 women; letting their average age be x, the sum of all the women's ages is 26x.

Since the sum of the ages of all 48 people must be equal to the sum of the ages of the 22 men plus the sum of the ages of the 26 women, we have

48(35) = 22(38) + 26x

1680 = 836 + 26x

26x = 844

x = 844/26

x = 32 12/26 ≈ 32.5

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A total of 22 men and 26 women were at a party, and the average  [#permalink]

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Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?

(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33

Kudos for a correct solution.

Total age of men and women = 48*35 => 1,680

Total age of men is = 22*38 => 836

So, total age of women in = 1680 - 836 => 844

Average age of women is 844/26 => 32.46

Hence answer will be (D) 32.5 _________________
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Re: A total of 22 men and 26 women were at a party, and the average  [#permalink]

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7
Here is the MGMAT Solution to it.

Avg. age of men is +3 than the avg. age of the group.

Now,

3(22)+w(26)=0....(Make this differential cancel out,hence equal to 0)

22=no. of men
26=no. of women
3=differential from mean.

Hence w=-2.5

Thus group mean (35)+Differential (-2.5)=32.5

Hope its simple than all the long calculations.
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Re: A total of 22 men and 26 women were at a party, and the average  [#permalink]

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3
Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?

(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33

Kudos for a correct solution.

I uesd the Std Deviation method :
Ave of Man = 38 (which is 38-35 = 3 more than the ave of Man & Women together)
and No of man = 22
Therefore Deviation More than the Mean = 3 * 22 = 66

Also, we know that No of women = 26
Now, For any given set of numbers, Deviation More than the Mean is ALWAYS = Deviation Less than the Mean
Therefore, each women can have a deviation of = 66/26 ~= 2.5 from the Average Value
Therefore, Ave age of Women = 35-2.5 = 32.5

Kudos if u like the method....
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Re: A total of 22 men and 26 women were at a party, and the average  [#permalink]

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1
Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?

(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33

Kudos for a correct solution.

It initially took mme ~2mins to solve this, but tried a couple of ways and think weighted avg would be fastest as lesser big numbers to deal with:

[Women-avg: (w)]----------------35--------------------38

$$\frac{26}{22} = \frac{3}{(35-w)}$$
$$w = 35 - \frac{33}{13}$$
13 * 2 = 26; hence $$\frac{33}{13} = 2.x$$ (I wont even bother getting the decimal)
w = 35 - 2.x = 32.x (~32.5)
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Re: A total of 22 men and 26 women were at a party, and the average  [#permalink]

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Hi there Could you explain the logic behind this ratio formula please? Thanks so much!!
reto wrote:
Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?

(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33

Kudos for a correct solution.

One can use the weighted average method:

$$\frac{Weight 1}{Weight 2} = \frac{Average2-Average}{Average-Average1}$$

For or example:

$$\frac{26}{22} = \frac{38-35}{35-Average1}$$

Solving for Average1 you will get exactly 32.46154. Answer D.

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Re: A total of 22 men and 26 women were at a party, and the average  [#permalink]

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average age of man * number of men = summation of ages of men
average age of woman * number of women = summation of ages of women

summation of ages of men + summation of ages of women = summation of ages of men and women

Thus, (38*22) + (average age of woman*26) = 35*(22+26)

simplifying,
average age of woman = {[35*(22+26)]-(38*22)}/26 = a little greater than 32 => 32.5

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Posts: 60
Re: A total of 22 men and 26 women were at a party, and the average  [#permalink]

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GMATPrepNow wrote:
Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?

(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33

Kudos for a correct solution.

Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

There are 48 people in total.
So, 22/48 = the proportion of women
And 26/48 = the proportion of men

Let x = average age of the women

So, plugging the information into the formula, we get: (22/48)(38) + (26/48)(x) = 35
Simplify: (11/24)(38) + (13/24)(x) = 35
Multiply both sides by 24 to get: (11)(38) + 13x = (35)(24)
Simplify: 418 + 13x = 840
Rearrange: 13x = 422
Solve: x = 422/13 ≈ 32.5

Cheers,
Brent

How can we assume that 35 is the weighted average and not a simple avg of 48 people in total ? Re: A total of 22 men and 26 women were at a party, and the average   [#permalink] 11 Oct 2019, 03:46
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