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# Percentages

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Intern
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Percentages [#permalink]  09 Feb 2018, 09:01
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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
CEO
Joined: 07 Jun 2014
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GRE 1: 323 Q167 V156
WE: Business Development (Energy and Utilities)
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Re: Percentages [#permalink]  09 Feb 2018, 15:23
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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!
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Sandy
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Re: Percentages [#permalink]  10 Feb 2018, 08:59
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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
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Re: Percentages [#permalink]  23 Feb 2018, 01:07
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
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Re: Percentages [#permalink]  23 Feb 2018, 08:32
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
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