GRE Prep 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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
In a 200 member club consisting of men and women, exactly 20% of men
[#permalink]
09 Sep 2021, 09:56
09 Sep 2021, 09:56
2
19
Bookmarks
00:00
Question Stats:
24% (02:18) correct
75% (02:02) wrong based on 61 sessions
Hide Show timer Statistics
In a 200 member club consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Least number of members who are homeowners |
41 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
In a 200 member club consisting of men and women, exactly 20% of men
[#permalink]
09 Sep 2021, 12:09
09 Sep 2021, 12:09
1
2
Bookmarks
edited:
Let men = \(M\) and women = \(W\), then the least number of homeowners is defined as the minimum of \(M/5+W/4\), where \(M+W=200\).
The denominator of \(20\) in \(\frac{4M+5W}{20}\) implies that for every 20 club members there can be either 5 men or 4 women owning homes. Thus, for 200 club members, the least number of homeowners would be 5 women out of 20 club members and \(180/5=36\) men as the total minimum required equal to 41.
Quantity A: 41
Quantity B: 41
answer is C
Let men = \(M\) and women = \(W\), then the least number of homeowners is defined as the minimum of \(M/5+W/4\), where \(M+W=200\).
The denominator of \(20\) in \(\frac{4M+5W}{20}\) implies that for every 20 club members there can be either 5 men or 4 women owning homes. Thus, for 200 club members, the least number of homeowners would be 5 women out of 20 club members and \(180/5=36\) men as the total minimum required equal to 41.
Quantity A: 41
Quantity B: 41
answer is C
KarunMendiratta wrote:
In a 200 member club consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Least number of members who are homeowners |
41 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Re: In a 200 member club consisting of men and women, exactly 20% of men
[#permalink]
30 Dec 2024, 08:19
30 Dec 2024, 08:19
Carcass pls provide an explanation for the above problem, quite unique question I find this one
