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In a 200 member association consisting of men and women, exactly 20% 17 Jan 2019, 19:10
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nades09
In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A. 49
B. 47
C. 45
D. 43
E. 41
Show more

Letting m = the number of men in the association and w = the number of women in the association, we know that:

m + w = 200

m = 200 - w

Thus:

0.2(200 - w) + 0.25w = homeowners

40 - 0.2w + 0.25w = homeowners

40 + 0.05w = homeowners

40 + w/20 = homeowners

We see that w must be a multiple of 20 and since we want the least number of homeowners, w = 20. So the least number of homeowners is 40 + 20/20 = 40 + 1 = 41.

Answer: E
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In a 200 member association consisting of men and women, exactly 20% 05 Dec 2019, 03:28
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Bunuel
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In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A) 49
B) 47
c) 45
D) 43
E) 41

Please explain

Thanks
NAD


Let the # of women be \(w\), then # of men will be \(200-w\). We want to minimize \(0.25w+0.2(200-w)\) --> \(0.25w+0.2(200-w)=0.05w+40=\frac{w}{20}+40\), so we should minimize \(w\), but also we should make sure that \(\frac{w}{20}+40\) remains an integer (as it represent # of people). Min value of \(w\) for which w/20 is an integer is for \(w=20\) --> \(\frac{w}{20}+40=1+40=41\).

Answer: E.

Or: as there are bigger percentage of homeowner women then we should minimize # of women, but we should ensure that \(\frac{1}{4}*w\) and \(\frac{1}{5}*(200-w)\) are integers. So \(w\) should be min multiple of 4 for which \(200-w\) is a multiple of 5 (basically w should be min positive multiple of 20), which turns out to be for \(w=20\).

Hope it's clear.
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Bunuel
why do not we took at first For example: Let the # of men be \(m\), then # of women will be \(200-m\). if so after calculating i got answer 49? why this process goes to wrong answer?
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In a 200 member association consisting of men and women, exactly 20% 05 Dec 2019, 16:05
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Hi hudacse6,

To start, there are additional pieces of information that you have to consider beyond that fact that there are 200 total people. We are also told that 20% of men and 25% of women are homeowners - and those are both specific fractions.

20% = 1/5, so we know that the number of men MUST be a multiple of 5
25% = 1/4, so we know that the number of women MUST be a multiple of 4

These facts tell us that M and W cannot be "any" numbers from 1-200, they can only be specific numbers AND the sum of those two numbers must add up to 200. In addition, we want the LEAST number of homeowners, meaning that we want to maximize the number of men (since a smaller percentage are homeowners) and minimize the number of women.

It's not clear whether you accounted for that additional information in your calculations. If you repost and include the 'steps' that you worked through, then we should be able to define why you did not get to the correct answer.

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In a 200 member association consisting of men and women, exactly 20% 01 Jun 2021, 05:34
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Concept: The question deal with Minima/Maxima problem


Solution:-


Let there be w women


=>No of men =200-w


25% of women =0.25 w are home owners


Also, 20% of men = 0.2(200-w) are home owners


Thus 0.25w + 0.2(200-w) = 0.05w + 40 =w/20 + 40


We have to minimize this equation =>w has to be minimized


Minimum value of w for which w/20 is an integer(as number of women as homeowners cannot be a non-integer) = 20


So, at w=20, w/20 +40 =41 (option e)

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In a 200 member association consisting of men and women, exactly 20% 23 Sep 2021, 23:41
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can anyone explain the Through Weighted Avg Method?


as per my knowledge Taking extremes of Homeowners of the ratio that they have given Homeowners 5:4 -TOTAL OF 9

Taking extremes 36-------x--------45

and avg comes around 40.5 so we can solve as above to get the answer to Option (E)- 41

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In a 200 member association consisting of men and women, exactly 20% 24 Sep 2021, 06:58
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kskumar
can anyone explain the Through Weighted Avg Method?


as per my knowledge Taking extremes of Homeowners of the ratio that they have given Homeowners 5:4 -TOTAL OF 9

Taking extremes 36-------x--------45

and avg comes around 40.5 so we can solve as above to get the answer to Option (E)- 41

avigutman VeritasKarishma
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You're right that we can solve this using a weighted average, kskumar!
Your approach isn't quite right, though. They have not given us a ratio of 5:4.
They have given us a ratio of 1:5 (where 5 ratio units represents the total number of men in the association) and a ratio of 1:4 (where 4 ratio units represent the total number of women in the association). Adding the 5 ratio units and the 4 ratio units is where you went wrong. These ratios may have different scale factors, so you can't add them. What you could do, though, if you wanted to, is infer that the number of men is some positive multiple of 5 and the number of women is some positive multiple of 4.

Now, as for solving this problem using weighted average thinking: if all 200 people were men, there would be 40 homeowners, and if all 200 people were women, there would be 50 homeowners. Therefore, the actual number of homeowners must be one of the integers inside that range, exclusive (why exclusive? because the question is nonsensical if there are only members of one gender). Now that we know the number of owners is at least 41 and at most 49, we can answer the question with ease.
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In a 200 member association consisting of men and women, exactly 20% 24 Jun 2024, 22:01
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Bunuel

nades09
In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A) 49
B) 47
c) 45
D) 43
E) 41

Please explain

Thanks
NAD

Let the # of women be \(w\), then # of men will be \(200-w\). We want to minimize \(0.25w+0.2(200-w)\) --> \(0.25w+0.2(200-w)=0.05w+40=\frac{w}{20}+40\), so we should minimize \(w\), but also we should make sure that \(\frac{w}{20}+40\) remains an integer (as it represent # of people). Min value of \(w\) for which w/20 is an integer is for \(w=20\) --> \(\frac{w}{20}+40=1+40=41\).

Answer: E.

Or: as there are bigger percentage of homeowner women then we should minimize # of women, but we should ensure that \(\frac{1}{4}*w\) and \(\frac{1}{5}*(200-w)\) are integers. So \(w\) should be min multiple of 4 for which \(200-w\) is a multiple of 5 (basically w should be min positive multiple of 20), which turns out to be for \(w=20\).

Hope it's clear.
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­Bunuel what if you took women to be 200 - m? I don't get the same answer
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In a 200 member association consisting of men and women, exactly 20% 25 Jun 2024, 01:15
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Bunuel

nades09
In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A) 49
B) 47
c) 45
D) 43
E) 41

Please explain

Thanks
NAD
Let the # of women be \(w\), then # of men will be \(200-w\). We want to minimize \(0.25w+0.2(200-w)\) --> \(0.25w+0.2(200-w)=0.05w+40=\frac{w}{20}+40\), so we should minimize \(w\), but also we should make sure that \(\frac{w}{20}+40\) remains an integer (as it represent # of people). Min value of \(w\) for which w/20 is an integer is for \(w=20\) --> \(\frac{w}{20}+40=1+40=41\).

Answer: E.

Or: as there are bigger percentage of homeowner women then we should minimize # of women, but we should ensure that \(\frac{1}{4}*w\) and \(\frac{1}{5}*(200-w)\) are integers. So \(w\) should be min multiple of 4 for which \(200-w\) is a multiple of 5 (basically w should be min positive multiple of 20), which turns out to be for \(w=20\).

Hope it's clear.
­Bunuel what if you took women to be 200 - m? I don't get the same answer
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\(0.25(200-m)+0.2m=50 - \frac{ m}{20}\).

To minimize the above, we should maximize m. The maximum value of m, for which m/20 is an integer is m =180. In this case:

\(50 - \frac{180}{20} = 41\).




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In a 200 member association consisting of men and women, exactly 20% 02 Jan 2025, 21:30
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nades09
In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A. 49
B. 47
C. 45
D. 43
E. 41
Show more

There are many ways to solve this question. One of the most efficient would be to use integer solutions method since number of men and women have to be integers.

Method 1:

To minimize the number of homeowners, we must make the number of men as large as possible since only 20% of those are homeowners. Of the rest, 25% will be homeowners.

Since 20% of men are homeowners, it means the total number of men is a multiple of 5 (because 1/5 of men are homeowners which must be an integer). So say number of men = 5a
Since 25% of women are homeowners, it means the total number of women is a multiple of 4 (because 1/4 of women are homeowners which must be an integer). So say number of women = 4b

5a + 4b = 200

To minimize the number of homeowners, we must make the 5a group (number of men) as large as possible.

The first and simplest solution would be b = 0, a = 40. But this means there are no women and we must have men and women.
So next possible solution is b = 5, a = 36.

This minimizes the number of homeowners as 36 + 5 = 41

Answer (E)

Integer solutions to equations in 2 variables are discussed here: https://anaprep.com/algebra-integer-sol ... variables/

Method 2:

You can do the same thing orally also. To minimize the number of homeowners, we must make the number of men as large as possible since only 20% of those are homeowners. Of the rest, 25% will be homeowners.
So we start with taking all 200 as men. But we must have some women too so we check whether 4 women are possible. No because that gives us 196 men which is not divisible by 5. Then check whether 8 women are possible. No.
Since number of men must be divisible by 5, we can have at most 180 men which gives us 20 women.

Number of homeowners = 180/4 + 20/5 = 41
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In a 200 member association consisting of men and women, exactly 20% 13 Aug 2025, 13:12
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out of 200
20 percent men = 40
25 percent women = 50
90 Homeowners
So for least use weighed balance 40 Least 41 50
Hence 41 ( Also overlapping sets concept can be used)
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In a 200 member association consisting of men and women, exactly 20% 02 Sep 2025, 19:36
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sir, I took the opposite values for men and women. Men as X and women as 200-x, and I got 49. Where did I go wrong?
Bunuel



Let the # of women be \(w\), then # of men will be \(200-w\). We want to minimize \(0.25w+0.2(200-w)\) --> \(0.25w+0.2(200-w)=0.05w+40=\frac{w}{20}+40\), so we should minimize \(w\), but also we should make sure that \(\frac{w}{20}+40\) remains an integer (as it represent # of people). Min value of \(w\) for which w/20 is an integer is for \(w=20\) --> \(\frac{w}{20}+40=1+40=41\).

Answer: E.

Or: as there are bigger percentage of homeowner women then we should minimize # of women, but we should ensure that \(\frac{1}{4}*w\) and \(\frac{1}{5}*(200-w)\) are integers. So \(w\) should be min multiple of 4 for which \(200-w\) is a multiple of 5 (basically w should be min positive multiple of 20), which turns out to be for \(w=20\).

Hope it's clear.
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In a 200 member association consisting of men and women, exactly 20% 02 Sep 2025, 23:49
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Ilanchezhiyan
sir, I took the opposite values for men and women. Men as X and women as 200-x, and I got 49. Where did I go wrong?

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Please read the whole thread. Your doubt has already been addressed there!
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