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A factory has 500 workers, 15 percent of whom are women. If

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A factory has 500 workers, 15 percent of whom are women. If [#permalink] New post 17 Dec 2012, 06:00
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A factory has 500 workers, 15 percent of whom are women. If 50 additional workers are to be hired and all of the present workers remain, how many of the additional workers must be women in order to raise the percent of women employees to 20 percent?

(A) 3
(B) 10
(C) 25
(D) 30
(E) 35
[Reveal] Spoiler: OA
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Re: A factory has 500 workers, 15 percent of whom are women. If [#permalink] New post 17 Dec 2012, 06:04
Expert's post
Walkabout wrote:
A factory has 500 workers, 15 percent of whom are women. If 50 additional workers are to be hired and all of the present workers remain, how many of the additional workers must be women in order to raise the percent of women employees to 20 percent?

(A) 3
(B) 10
(C) 25
(D) 30
(E) 35


Currently the factory has 0.15*500=75 women.

After 50 additional workers are hired, there will be total of 550 workers and we need 0.2*550=110 of them to be women, so we need 110-75=35 more women.

Answer: E.
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Re: A factory has 500 workers, 15 percent of whom are women. If [#permalink] New post 03 Jul 2013, 09:08
We also can use mixture method

Say we add to 15% females of 550 ppl x % females of 50 ppl to get 20% females
So we have (x-20)/5=500/50
X=70%
70%of 50=35
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Re: A factory has 500 workers, 15 percent of whom are women. If [#permalink] New post 03 Jul 2013, 09:22
Or

(75+x)/550=20%
X =35
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Re: A factory has 500 workers, 15 percent of whom are women. If [#permalink] New post 22 Aug 2014, 19:24
I also solved this as a mixture problem (see image attached) where w denotes the number of new women hired.

Once I had the table set up I solved (75+w) / 550 = 0.20 => 75+w = 110 => w=35.

Therefore, 35 of the 50 workers hired need to be women.
Attachments

Mixture.png
Mixture.png [ 2.83 KiB | Viewed 453 times ]

Re: A factory has 500 workers, 15 percent of whom are women. If   [#permalink] 22 Aug 2014, 19:24
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