It is currently 10 Dec 2018, 12:50

### 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

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

# Over the past year, the number of men in Pleasantville

Author Message
TAGS:
Intern
Joined: 26 Jan 2018
Posts: 3
Followers: 0

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

Over the past year, the number of men in Pleasantville [#permalink]  09 Feb 2018, 09:01
1
KUDOS
00:00

Question Stats:

100% (02:31) correct 0% (00:00) wrong based on 21 sessions
Over the past year, the number of men in Pleasantville increased by 20% while the number of women in Pleasantville decreased by 20%. If the number of men and women in Pleasantville are now equal to each other, what was the percentage change in the total population of Pleasantville over the past year?

10% decrease
4% decrease
no change
4% increase
10% increase
[Reveal] Spoiler: OA
GMAT Club Legend
Joined: 07 Jun 2014
Posts: 4749
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 93

Kudos [?]: 1651 [4] , given: 396

Re: Percentages [#permalink]  09 Feb 2018, 15:23
4
KUDOS
Expert's post
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!
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

GRE Instructor
Joined: 10 Apr 2015
Posts: 1232
Followers: 45

Kudos [?]: 1110 [2] , given: 7

Re: Percentages [#permalink]  10 Feb 2018, 08:59
2
KUDOS
Expert's post
goutham v wrote:
Over the past year, the number of men in Pleasantville increased by 20% while the number of women in Pleasantville decreased by 20%. If the number of men and women in Pleasantville are now equal to each other, what was the percentage change in the total population of Pleasantville over the past year?

10% decrease
4% decrease
no change
4% increase
10% increase

Let's assign some nice values to the number of men and women AFTER the increases and decreases occurred.
Let's say that AFTER the increases and decreases occurred, there were 24 men and 24 women.
So, AFTER the increases and decreases occurred, the TOTAL population = 48

Let M = number of men BEFORE the increase
Let W = number of women BEFORE the decrease

The number of men in Pleasantville increased by 20%
So, 1.2M = 24
So, M = 24/1.2 = 20
There were 20 men BEFORE the increase

The number of women decreased by 20%
So, 0.8F = 24
So, F = 24/0.8 = 30
There were 30 women BEFORE the decrease

TOTAL population BEFORE the increases and decreases = 20 + 30 = 50

What was the percentage change in the total population of Pleasantville over the past year?
The population decreased from 50 to 48
ELIMINATE C, D and E

Percent change = 100(new - old)/old
= (100)(48 - 50)/50
= -4%
= B

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com

Manager
Joined: 15 Feb 2018
Posts: 53
Followers: 1

Kudos [?]: 17 [1] , given: 33

Re: Percentages [#permalink]  23 Feb 2018, 01:07
1
KUDOS
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!
Intern
Joined: 17 Jun 2017
Posts: 3
Followers: 0

Kudos [?]: 4 [1] , given: 8

Re: Percentages [#permalink]  23 Feb 2018, 08:32
1
KUDOS
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%

Re: Percentages   [#permalink] 23 Feb 2018, 08:32
Display posts from previous: Sort by