Find all School-related info fast with the new School-Specific MBA Forum

It is currently 29 Jul 2016, 01:29
GMAT Club Tests

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

1 KUDOS received
Intern
Intern
avatar
Joined: 18 Feb 2010
Posts: 28
Followers: 1

Kudos [?]: 58 [1] , given: 5

In a 200 member association consisting of men and women [#permalink]

Show Tags

New post 12 Dec 2010, 06:21
1
This post received
KUDOS
24
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

60% (02:51) correct 40% (01:58) wrong based on 657 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
[Reveal] Spoiler: OA
Expert Post
6 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 34107
Followers: 6104

Kudos [?]: 76806 [6] , given: 9992

Re: Least number of homeowners [#permalink]

Show Tags

New post 12 Dec 2010, 06:59
6
This post received
KUDOS
Expert's post
12
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 \(w=20\).

Hope it's clear.
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 18 Feb 2010
Posts: 28
Followers: 1

Kudos [?]: 58 [0], given: 5

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: 31
Location: Boston
Schools: Boston College, MIT, BU, IIM, UCLA, Babson, Brown
Followers: 0

Kudos [?]: 9 [0], given: 10

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: 406
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)
Followers: 7

Kudos [?]: 43 [0], given: 46

Re: Least number of homeowners [#permalink]

Show Tags

New post 13 Dec 2010, 07:34
Great explanation.
_________________

Go Blue!

GMAT Club Premium Membership - big benefits and savings

Manager
Manager
avatar
Joined: 20 Apr 2010
Posts: 210
Schools: ISB, HEC, Said
Followers: 4

Kudos [?]: 66 [0], given: 28

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: 46
Followers: 1

Kudos [?]: 58 [0], given: 1

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 \(\frac{w}{20}\) ?
Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 34107
Followers: 6104

Kudos [?]: 76806 [1] , given: 9992

Re: Least number of homeowners [#permalink]

Show Tags

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


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

Hope it's clear.
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 11 Jul 2012
Posts: 46
Followers: 0

Kudos [?]: 2 [0], given: 0

Re: In a 200 member association consisting of men and women [#permalink]

Show Tags

New post 05 Nov 2012, 13:20
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
Manager
Manager
avatar
Joined: 26 Feb 2012
Posts: 116
Location: India
Concentration: General Management, Finance
WE: Engineering (Telecommunications)
Followers: 0

Kudos [?]: 18 [0], given: 56

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
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 34107
Followers: 6104

Kudos [?]: 76806 [0], given: 9992

Re: In a 200 member association consisting of men and women [#permalink]

Show Tags

New post 09 Jun 2013, 08:06
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?
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

2 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 17 Dec 2012
Posts: 442
Location: India
Followers: 22

Kudos [?]: 361 [2] , given: 14

Re: In a 200 member association consisting of men and women [#permalink]

Show Tags

New post 09 Jun 2013, 19:32
2
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.
_________________

Srinivasan Vaidyaraman
Sravna
http://www.sravnatestprep.com

Classroom and Online Coaching

Intern
Intern
avatar
Joined: 28 Dec 2012
Posts: 10
Followers: 0

Kudos [?]: 22 [0], given: 19

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.
Current Student
avatar
Joined: 29 Aug 2012
Posts: 31
WE: General Management (Consulting)
Followers: 1

Kudos [?]: 13 [0], given: 6

GMAT ToolKit User
Re: In a 200 member association consisting of men and women [#permalink]

Show Tags

New post 27 Sep 2013, 01:25
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
Manager
Manager
avatar
Joined: 07 Apr 2012
Posts: 126
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
GMAT 1: 590 Q48 V23
GPA: 3.9
WE: Operations (Manufacturing)
Followers: 0

Kudos [?]: 10 [0], given: 45

Re: Least number of homeowners [#permalink]

Show Tags

New post 13 Oct 2013, 07:47
How can we solve it by Calculus ? Since w is linear on RHS, how to differentiate it ?
Intern
Intern
avatar
Joined: 25 Nov 2013
Posts: 19
Concentration: Marketing, International Business
GMAT Date: 02-14-2014
GPA: 2.3
WE: Other (Internet and New Media)
Followers: 1

Kudos [?]: 5 [0], given: 11

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.
Senior Manager
Senior Manager
avatar
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 39 [0], given: 23

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?
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 10646
Followers: 496

Kudos [?]: 131 [0], given: 0

Premium Member
Re: In a 200 member association consisting of men and women [#permalink]

Show Tags

New post 21 Jun 2015, 11:02
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.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Intern
Intern
avatar
Joined: 17 Jun 2015
Posts: 10
Followers: 0

Kudos [?]: 0 [0], given: 0

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: 5
Followers: 0

Kudos [?]: 0 [0], given: 0

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.
Re: In a 200 member association consisting of men and women   [#permalink] 14 Oct 2015, 06:05

Go to page    1   2    Next  [ 24 posts ] 

    Similar topics Author Replies Last post
Similar
Topics:
1 Experts publish their posts in the topic The ratio of men to women in a class is 6 to 5. If 2 men and 1 woman Bunuel 11 03 Jan 2016, 12:15
3 Experts publish their posts in the topic The ratio of men to women at a banquet is 6:5. If ten men leave the pa shasadou 1 12 Sep 2015, 12:48
3 Experts publish their posts in the topic A division of a company consists of seven men and five women mikemcgarry 6 08 Jan 2013, 10:42
2 Experts publish their posts in the topic There are 4 more women than men on Centerville's board of Walkabout 4 07 Dec 2012, 03:20
5 If there are twice as many women as men in a group and an eq mehulsayani 8 08 Sep 2012, 06:01
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  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.