Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In a 200 member association consisting of men and women [#permalink]
12 Dec 2010, 05:21

1

This post received KUDOS

13

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

59% (02:53) correct
41% (01:50) wrong based on 489 sessions

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?

Re: Least number of homeowners [#permalink]
12 Dec 2010, 05:59

4

This post received KUDOS

Expert's post

9

This post was BOOKMARKED

nades09 wrote:

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 \(w=20\).

Re: Least number of homeowners [#permalink]
27 Jun 2012, 06:50

Bunuel wrote:

nades09 wrote:

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.

Could you explain this one; \(=0.05w+40=\frac{w}{20}+40\) ? How did you get \(\frac{w}{20}\) ?

Re: Least number of homeowners [#permalink]
27 Jun 2012, 07:00

1

This post received KUDOS

Expert's post

superpus07 wrote:

Bunuel wrote:

nades09 wrote:

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.

Could you explain this one; \(=0.05w+40=\frac{w}{20}+40\) ? How did you get \(\frac{w}{20}\) ?

Re: In a 200 member association consisting of men and women [#permalink]
05 Nov 2012, 12:20

I got E too (41) X: number of men and 200-X: number of women 0.20*X +0.25*(200-X) = 50 - 0.05*X = 50 - (1/20)*X To minimize this, we should maximize X (i.e X = 180, since there must be some women and X should be a multiple of 20) 50 - (1/20)*180 = 41 Brother Karamazov

Re: In a 200 member association consisting of men and women [#permalink]
09 Jun 2013, 04:42

Hi Bunuel

I have reached to the answer but the solution is little weird after I saw yours. Please guide me whether we can go by this

Question 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

Solution simple

Out of 200 20% are male i.e 40 and 25% are female i.e 50 , so total homeowner is 90.

Now min number homeowner is 40 and max is 90 so question ask us to find least and 41 has least value among all option.

Re: In a 200 member association consisting of men and women [#permalink]
09 Jun 2013, 07:06

Expert's post

prasannajeet wrote:

Hi Bunuel

I have reached to the answer but the solution is little weird after I saw yours. Please guide me whether we can go by this

Question 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

Solution simple

Out of 200 20% are male i.e 40 and 25% are female i.e 50 , so total homeowner is 90.

Now min number homeowner is 40 and max is 90 so question ask us to find least and 41 has least value among all option.

So ans is 41.

Rgds Prasannajeet

We are told that exactly 20% of men and exactly 25 % women are homeowners, not that 20% of 200 are males (homeowners)...

If there are 40 males and 50 females who are the remaining 200-(90)=110 people? _________________

Re: In a 200 member association consisting of men and women [#permalink]
09 Jun 2013, 18:32

1

This post received KUDOS

1. The percentage of women who are homemakers is higher than the percentage of men who are homemakers. So the least number of homemakers would be, when the number of women is the lowest possible 2. The number of women should be a multiple of 4 and the number of men should be a multiple of 5 3. Given our objective in (1) and the constraint in (2), the minimum number of women is 20 and the corresponding number of men is 180 4. So 25% of 20 is 5 and 20% of 180 is 36. The least number of homemakers is 5+36= 41.

Therefore the answer is choice E. _________________

Re: In a 200 member association consisting of men and women [#permalink]
08 Aug 2013, 09:17

Total numbers = 200. Men = X, Women = Y. We are told that exactly 20% of men are homeowner => X is a multiple of 5 => X=5a (a>=1). Similarly, Y = 4b (b>=1). Find min(a+b)

X + Y = 200 <=> 5a + 4b = 200. a and b are all integers. When a = 36, b=5 => (a+b)--->min.

Re: Least number of homeowners [#permalink]
05 Feb 2014, 05:34

Bunuel wrote:

nades09 wrote:

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.

Hi Bunuel,

I solved it with a similar approach but came out with the option A. Can you please explain where am I wrong:

Let males be = m , then women = 200-m

0.2m +0.25(200-m) we need to minimize m

so solving the equation gives us 50-m/20

For m to be an integer, the least value it can have is 20. So the ans: 50-1= 49.

Re: Least number of homeowners [#permalink]
25 May 2014, 17:43

Bunuel wrote:

nades09 wrote:

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.

Hi Bunuel, while solving this, I substituted for males instead of female and my equation looked like this:

.2m+.25(200-m) === 50 - 0.05m

Now, since it's asking me to minimize this value, would I choose the GREATEST possible value for m? How would I go about choose a value? At first, i thought it would be 196 but that doesn't lend itself well to being an integer.

Re: In a 200 member association consisting of men and women [#permalink]
21 Jun 2015, 10:02

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Hey, everyone. After a hectic orientation and a weeklong course, Managing Groups and Teams, I have finally settled into the core curriculum for Fall 1, and have thus found...

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

After I was accepted to Oxford I had an amazing opportunity to visit and meet a few fellow admitted students. We sat through a mock lecture, toured the business...