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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]  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
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Re: A factory has 500 workers, 15 percent of whom are women. If [#permalink]  17 Dec 2012, 06:04
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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

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.

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Re: A factory has 500 workers, 15 percent of whom are women. If [#permalink]  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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Current Student
Joined: 23 Oct 2010
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Location: Azerbaijan
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Kudos [?]: 240 [0], given: 73

Re: A factory has 500 workers, 15 percent of whom are women. If [#permalink]  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]  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 [ 2.83 KiB | Viewed 4137 times ]

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A factory has 500 workers, 15 percent of whom are women. If [#permalink]  30 Oct 2014, 23:26
LalaB wrote:
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

Hello, could anyone please explain how we came into that equation from the first line?

I am clear up to this:
15% of 500 + x% of 50 = 20% of 550 [correction: 15% of 500, not 550]

* Then how can I come to the highlighted equation. Am I missing anything?

Last edited by appleid on 30 Oct 2014, 23:34, edited 1 time in total.
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Re: A factory has 500 workers, 15 percent of whom are women. If [#permalink]  04 Nov 2014, 00:12
appleid wrote:
LalaB wrote:
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

Hello, could anyone please explain how we came into that equation from the first line?

I am clear up to this:
15% of 500 + x% of 50 = 20% of 550 [correction: 15% of 500, not 550]

* Then how can I come to the highlighted equation. Am I missing anything?

Its called allegation method:

15% .................. x%

............. 20% ...............

(20-15)% .............. (x-20)%

$$\frac{x-20}{5} = \frac{500}{50}$$

x = 70%

$$\frac{70}{100} * 50 = 35$$

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Re: A factory has 500 workers, 15 percent of whom are women. If [#permalink]  19 Jan 2015, 01:14
Or, setting up an equation:

First, find how many women we have: 500*0.15 = 75 women now.
We need the number of women to be 20% of the total number of employees.

We will have 50 new employees, making the total # 500+50=550. We need to know how many of the additional workers need to be women in order for the women to be 0.20(550).

In other words: 75+x=0.20(550). This means how many women should we add to the 75 we already have so that the total # of women is 20% of the total # of workers (550).

Solving the equation we get:
75 + x = 0.20(550)
75 + x = 110
x = 110-75
x = 35 ANS E
Re: A factory has 500 workers, 15 percent of whom are women. If   [#permalink] 19 Jan 2015, 01:14
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