Last visit was: 23 Nov 2024, 13:40 It is currently 23 Nov 2024, 13:40

Close

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

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 GRE 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
avatar
Intern
Intern
Joined: 26 Jan 2018
Posts: 3
Own Kudos [?]: 10 [8]
Given Kudos: 0
Send PM
Most Helpful Expert Reply
User avatar
Retired Moderator
Joined: 07 Jun 2014
Posts: 4813
Own Kudos [?]: 11196 [6]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
Most Helpful Community Reply
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12196 [5]
Given Kudos: 136
Send PM
General Discussion
avatar
Manager
Manager
Joined: 15 Feb 2018
Posts: 53
Own Kudos [?]: 34 [3]
Given Kudos: 0
Send PM
Re: Over the past year, the number of men in Pleasantville [#permalink]
3
This is solution is very easy to follow.

sandy wrote:
Let the number of men in the past be \(x\) and let number of women be \(y\). Also assume currently the number of both men and women are 100 each.

So with 20% increase for men we can say \(1.2 \times x = 100\) or \(x=83.33\).
Also with 20% decrease for women we can say \(0.8 \times y = 100\) or \(y=125\).

Population in the past\(=x+y=83.33+125=208.33\).
Present population =\(100+100=200\).

% decrease= \(\frac{208.33-200}{208.33} \approx 3.99\).

Hence option B is correct!
avatar
Intern
Intern
Joined: 17 Jun 2017
Posts: 3
Own Kudos [?]: 7 [2]
Given Kudos: 0
Send PM
Re: Over the past year, the number of men in Pleasantville [#permalink]
2
Let the original number of males be M and females be F:
after a year, the number of males=1.2M
the numbers of females=0.8F


According to question: 0.8F=1.2M
=>F=3/2M
=>F=1.5M

Thus, initial population=M+F
=M+1.5M
=2.5M

Population after a year=0.8F+1.2M
=0.8*(1.5)M+1.2M
=2.4M

Percentage change = [(2.4M-2.5M)/2.5M]*100 %
= -1/25*100 %
= -4%
Answer: 4% decrease
Answer: B

avatar
Manager
Manager
Joined: 08 Dec 2018
Posts: 94
Own Kudos [?]: 70 [0]
Given Kudos: 0
Send PM
Re: Over the past year, the number of men in Pleasantville [#permalink]
2
Let's say, total population at the beginning =100
So, if Male=x, Female= (100-x)
Now the change is: 1.2x and .8(100-x)
By condition, 1.2x=.8(100-x)
So, x=40 ( Male), Female =60.
Current population Male=1.2*40=48, Female .8*60=48. So, Total=96.
% Change ( decrease) : 100-96= 4 %.
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5043
Own Kudos [?]: 74 [0]
Given Kudos: 0
Send PM
Re: Over the past year, the number of men in Pleasantville [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep 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.
Prep Club for GRE Bot
Re: Over the past year, the number of men in Pleasantville [#permalink]
Moderators:
GRE Instructor
84 posts
GRE Forum Moderator
37 posts
Moderator
1111 posts
GRE Instructor
234 posts

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