Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

It is currently 18 Jul 2019, 23:48

Close

GMAT Club Daily Prep

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.

Close

Request Expert Reply

Confirm Cancel

In a 200 member association consisting of men and women

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Intern
Intern
avatar
Joined: 18 Feb 2010
Posts: 22
In a 200 member association consisting of men and women  [#permalink]

Show Tags

New post 12 Dec 2010, 06:21
12
1
72
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

63% (02:23) correct 37% (02:31) wrong based on 1008 sessions

HideShow timer Statistics


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
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56270
Re: Least number of homeowners  [#permalink]

Show Tags

New post 12 Dec 2010, 06:59
13
19
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.
_________________
General Discussion
Intern
Intern
avatar
Joined: 18 Feb 2010
Posts: 22
Re: Least number of homeowners  [#permalink]

Show Tags

New post 12 Dec 2010, 11:10
Great! Thanks for the quick reply. Did not think it would be so straightforward

You rock!
NAD
Intern
Intern
User avatar
Joined: 30 Nov 2010
Posts: 23
Location: Boston
Schools: Boston College, MIT, BU, IIM, UCLA, Babson, Brown
Re: Least number of homeowners  [#permalink]

Show Tags

New post 12 Dec 2010, 13:57
awesome explanation. thanks!
Senior Manager
Senior Manager
User avatar
Status: Bring the Rain
Joined: 17 Aug 2010
Posts: 345
Location: United States (MD)
Concentration: Strategy, Marketing
Schools: Michigan (Ross) - Class of 2014
GMAT 1: 730 Q49 V39
GPA: 3.13
WE: Corporate Finance (Aerospace and Defense)
Re: Least number of homeowners  [#permalink]

Show Tags

New post 13 Dec 2010, 07:34
Manager
Manager
avatar
Joined: 20 Apr 2010
Posts: 174
Schools: ISB, HEC, Said
Re: Least number of homeowners  [#permalink]

Show Tags

New post 14 Dec 2010, 00:32
Nice Question and Great explaination by Bunuel .. Thanks a lot
Intern
Intern
avatar
Joined: 20 Jun 2011
Posts: 44
Re: Least number of homeowners  [#permalink]

Show Tags

New post 27 Jun 2012, 07:50
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 \(\frac{w}{20}\) ?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56270
Re: Least number of homeowners  [#permalink]

Show Tags

New post 27 Jun 2012, 08:00
2
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 \(\frac{w}{20}\) ?


\(0.05w=\frac{5}{100}*w=\frac{1}{20}*w=\frac{w}{20}\).

Hope it's clear.
_________________
Intern
Intern
avatar
Joined: 11 Jul 2012
Posts: 40
Re: In a 200 member association consisting of men and women  [#permalink]

Show Tags

New post 05 Nov 2012, 13:20
1
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
Manager
Manager
avatar
Joined: 26 Feb 2012
Posts: 97
Location: India
Concentration: General Management, Finance
WE: Engineering (Telecommunications)
Re: In a 200 member association consisting of men and women  [#permalink]

Show Tags

New post 09 Jun 2013, 05:42
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.

So ans is 41.

Rgds
Prasannajeet
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56270
Re: In a 200 member association consisting of men and women  [#permalink]

Show Tags

New post 09 Jun 2013, 08:06
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?
_________________
Director
Director
User avatar
S
Joined: 17 Dec 2012
Posts: 630
Location: India
Re: In a 200 member association consisting of men and women  [#permalink]

Show Tags

New post 09 Jun 2013, 19:32
7
1
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.
_________________
Srinivasan Vaidyaraman
Sravna Test Prep
http://www.sravnatestprep.com

Holistic and Systematic Approach
Intern
Intern
avatar
Joined: 28 Dec 2012
Posts: 9
GMAT ToolKit User
Re: In a 200 member association consisting of men and women  [#permalink]

Show Tags

New post 08 Aug 2013, 10:17
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.

E.
Intern
Intern
avatar
Joined: 29 Aug 2012
Posts: 29
WE: General Management (Consulting)
GMAT ToolKit User
Re: In a 200 member association consisting of men and women  [#permalink]

Show Tags

New post 27 Sep 2013, 01:25
1
I didn't set up any algebra equations.

Basically, we know that we have 1/5 of the men and 1/4 of the women. So both, men and women should be a multiple of 20 (lcm of 5 and 4).

Now, in order to minimize, we want the maximum multiple for men (because is the least fraction) and the minimun (dintinct to zero) for women.

So we have 180 men and 20 women. Now the solution is clear. 180*(1/5)+20(1/4)=41
Intern
Intern
avatar
Joined: 25 Nov 2013
Posts: 14
Concentration: Marketing, International Business
GMAT Date: 02-14-2014
GPA: 2.3
WE: Other (Internet and New Media)
Re: Least number of homeowners  [#permalink]

Show Tags

New post 05 Feb 2014, 06:34
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.

Where am I wrong please guide.

Thanks.
Manager
Manager
avatar
Joined: 15 Aug 2013
Posts: 239
Re: Least number of homeowners  [#permalink]

Show Tags

New post 25 May 2014, 18:43
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.

Am I using the right though process here?
Intern
Intern
avatar
Joined: 17 Jun 2015
Posts: 10
Re: In a 200 member association consisting of men and women  [#permalink]

Show Tags

New post 22 Jun 2015, 00:11
good easy solution
Intern
Intern
avatar
Joined: 08 Mar 2015
Posts: 4
Re: In a 200 member association consisting of men and women  [#permalink]

Show Tags

New post 14 Oct 2015, 06:05
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

brunel:

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.


brunel: If we want to minimize w, why cant we just assume that w =0. Ie. The number of women is 0. The question does not state anywhere that the number of women CANNOT be 0.
Intern
Intern
avatar
Joined: 26 Feb 2016
Posts: 12
Re: In a 200 member association consisting of men and women  [#permalink]

Show Tags

New post 13 Apr 2016, 23:07
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


I have a question about this method
i notice if i put w = 200 -m
i get
-m/20 + 50

then it would be completely different and il get min/max wrong way around?
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 7764
In a 200 member association consisting of men and women  [#permalink]

Show Tags

New post 13 Apr 2016, 23:39
3
sabxu1 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


I have a question about this method
i notice if i put w = 200 -m
i get
-m/20 + 50

then it would be completely different and il get min/max wrong way around?


Hi,
no you will still get the correct answer...
Two cases--
1) we maximize m, as the % of m is lower
2) If the answer is in w, we minimize w..


here you have taken your answer in m and you get it as \(50 - \frac{m}{20}\)..
so you have to maximize m..
m has to be a multiple of 20 to get an integer value for m/20..
m cannot be 200, as there are some w...


Next can be 200-20=180..
substitue m as 180, So number =\(50 - \frac{180}{20} = 50 - 9 = 41\)
_________________
GMAT Club Bot
In a 200 member association consisting of men and women   [#permalink] 13 Apr 2016, 23:39

Go to page    1   2    Next  [ 37 posts ] 

Display posts from previous: Sort by

In a 200 member association consisting of men and women

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne