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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!
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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

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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 %.
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Re: Over the past year, the number of men in Pleasantville [#permalink]
Hello from the GRE Prep Club BumpBot!

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