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In a corporation, 50 percent of the male employees and 40
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26 Aug 2010, 07:40
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In a corporation, 50 percent of the male employees and 40 percent of the female employees are at least 35 years old. If 42 percent of all the employees are at least 35 years old, what fraction of the employees in the corporation are females? A. 3/5 B. 2/3 C. 3/4 D. 4/5 E. 5/6
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In a corporation, 50 percent of the male employees and 40
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26 Aug 2010, 08:12
udaymathapati wrote: In a corporation, 50 percent of the male employees and 40 percent of the female employees are at least 35 years old. If 42 percent of all the employees are at least 35 years old, what fraction of the employees in the corporation are females?
A. 3/5 B. 2/3 C. 3/4 D. 4/5 E. 5/6 Given weighted average employees who are at least 35 years old: \(\frac{0.5m+0.4f}{m+f}=0.42\). Question: \(\frac{f}{m+f}=?\) \(\frac{0.5m+0.4f}{m+f}=0.42\); \(50m+40f=42m+42f\); \(4m=f\); \(\frac{f}{m}=4\); \(\frac{f}{m+f}=\frac{4}{5}\). Answer: D.
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Re: Percent Problem
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26 Aug 2010, 09:46
I have a different approach that may be useful for you: Let's say that “x” is the % of male employees in the corporation. So, (1x) is the % of female employees in the same company. Therefore: \((50%)(x) + (40%)(1x) = 42%\) Expressed in fractions: \((1/2)(x) + (2/5)(1x) = 21/50\) We solve the equation and we have: \(x = 1/5\) \((1x) = 4/5\) Answer: 4/5 : D I think I deserve kudos
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Re: Percent Problem
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19 Oct 2010, 07:58
Hi Buneul 
I understood your explanation until the last step. How do you get from f/m= 4 to the 4/5?
Thanks!



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Re: Percent Problem
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19 Oct 2010, 14:33
Atrain13gm wrote: Hi Buneul 
I understood your explanation until the last step. How do you get from f/m= 4 to the 4/5?
Thanks! \(4m=f\) > \(\frac{f}{m+f}=\frac{4m}{m+4m}=\frac{4m}{5m}=\frac{4}{5}\). Hope it's clear.
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Re: GWD #1 Math 11
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05 Apr 2012, 20:34
eybrj2 wrote: In a corporation, 50 percent of the male employees and 40 percent of the female employeesa re at least 35 years old. If 42 percent of all the employees are at least 35 years old, what fraction of the employees in the corporation are females?
a) 3/5
b) 2/3
c) 3/4
d) 4/5
e) 5/6 You can use the weighted averages formula for a 10 sec solution. No of females/No of males = (50  42)/(42  40 ) = 4/1 No of females as a fraction of total employees = 4/(4+1) = 4/5 For details of this method, check: http://www.veritasprep.com/blog/2011/03 ... averages/
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Re: In a corporation, 50 percent of the male employees and 40
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02 Jun 2012, 19:06
your post on veritas blog is amazing Karishma
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Re: In a corporation, 50 percent of the male employees and 40
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29 Sep 2014, 00:20
Let Male employees = x Let Female employees = y Total Employees = x+y We require to find \(\frac{y}{x+y}\) Setting up the equation: \(\frac{50x}{100} + \frac{40y}{100} = \frac{42}{100} (x+y)\) \(\frac{y}{x} = 4\) \(\frac{y}{x+y} = \frac{4}{1+4} = \frac{4}{5}\) Answer = E
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Re: In a corporation, 50 percent of the male employees and 40
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05 Oct 2015, 21:42
x total number of employees f  all females xf  all males
we need f/x
50(xf)+40f=42x 50x50f+40f=42x 10f=8x f/x=8/10=4/5
D



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Re: In a corporation, 50 percent of the male employees and 40
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22 Aug 2016, 07:08
.5m+.40f=.42 m+f=1
solve these two equations, f=4/5



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Re: In a corporation, 50 percent of the male employees and 40
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06 Nov 2016, 20:47
I used the 2 way chart to help create the expression :
(50/100)x + (40/100)y = 42/100(x+y) .... where x is number of males, y is number of females and x+y is total.
Now when you simplify it, you will get 4x = y. Means for every 4 females there is 1 male... No of females is 80% or 4/5



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Re: In a corporation, 50 percent of the male employees and 40
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16 Dec 2017, 09:32
Males + Females =100  Equation 1
50 * Males + 40 * Females = 42 [ Males + Females ]  Equation 2
Solve for Females.
You will get 80 Answer ( Which is 80%)
therefore = 80/100=4/5



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In a corporation, 50 percent of the male employees and 40
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28 Mar 2018, 13:07
VeritasPrepKarishma wrote: eybrj2 wrote: In a corporation, 50 percent of the male employees and 40 percent of the female employeesa re at least 35 years old. If 42 percent of all the employees are at least 35 years old, what fraction of the employees in the corporation are females?
a) 3/5
b) 2/3
c) 3/4
d) 4/5
e) 5/6 You can use the weighted averages formula for a 10 sec solution. No of females/No of males = (50  42)/(42  40 ) = 4/1 No of females as a fraction of total employees = 4/(4+1) = 4/5 For details of this method, check: http://www.veritasprep.com/blog/2011/03 ... averages/Hi VeritasPrepKarishma, What if i use alligations method 5040 42 28 as far i remember the value that is closer to the mean has more weight right? so here # of female employees is more (its not surprising actually, they have always been smarter than men ) how to proceed further ? let me try ... so 2 here has more weight hence it should be in numerator, and 8 has less weight and should be in denominator hence \(\frac{2}{8}\) > \(\frac{1}{4}\) so ratio of females to men is 1 to 4 ? but the mean was closer to percentage of females, shouldnt females be more than men ?



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Re: In a corporation, 50 percent of the male employees and 40
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29 Mar 2018, 00:04
dave13 wrote: VeritasPrepKarishma wrote: eybrj2 wrote: In a corporation, 50 percent of the male employees and 40 percent of the female employeesa re at least 35 years old. If 42 percent of all the employees are at least 35 years old, what fraction of the employees in the corporation are females?
a) 3/5
b) 2/3
c) 3/4
d) 4/5
e) 5/6 You can use the weighted averages formula for a 10 sec solution. No of females/No of males = (50  42)/(42  40 ) = 4/1 No of females as a fraction of total employees = 4/(4+1) = 4/5 For details of this method, check: http://www.veritasprep.com/blog/2011/03 ... averages/Hi VeritasPrepKarishma, What if i use alligations method 5040 42 28 as far i remember the value that is closer to the mean has more weight right? so here # of female employees is more (its not surprising actually, they have always been smarter than men ) how to proceed further ? let me try ... so 2 here has more weight hence it should be in numerator, and 8 has less weight and should be in denominator hence \(\frac{2}{8}\) > \(\frac{1}{4}\) so ratio of females to men is 1 to 4 ? but the mean was closer to percentage of females, shouldnt females be more than men ? The method of allegation is the same as the weighted average formula I have used. To ensure that there is no confusion, use w1/w2 = (A2  Aavg)/(Aavg  A1) No diagram is required here. The details of this are provided in the link given in my post above.
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In a corporation, 50 percent of the male employees and 40
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29 Mar 2018, 09:48
VeritasPrepKarishma, thanks! I wish your website featured search box i wanted to find the link where it is explained whhe sometimes numeratot and denominato values switch places but couldnt find it...



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Re: In a corporation, 50 percent of the male employees and 40
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20 Apr 2018, 16:36
udaymathapati wrote: In a corporation, 50 percent of the male employees and 40 percent of the female employees are at least 35 years old. If 42 percent of all the employees are at least 35 years old, what fraction of the employees in the corporation are females?
A. 3/5 B. 2/3 C. 3/4 D. 4/5 E. 5/6 No. of male=m No. of Female=f (0.5)*m + (0.4)f = 0.42(m+f) 0.5m + 0.4f = 0.42m + 0.42f 0.08m = 0.02f 8m=2f (multiply both side by 100) 4m=f f/m=4 : 1 So, m+f = 5 f/(m+f)=4/5, which is option D.



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In a corporation, 50 percent of the male employees and 40
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23 Apr 2018, 16:08
Alternate way  you can set up two equations:
Males = M and Females = F Set the total to be 100 since we are dealing with percents and no value has been provided.
M+F = 100 (equation 1)
The question says 50% of males and 40% of females are at least 35. The question also says that 42% of total employees are at least 35. .5M + .4F = 42 > multiply each term by 100 50M+ 40F = 4200 > reduce down by dividing each by 10 5M+4F= 420 (equation 2)
Subtracting equation 2 from equation 1. You get M=20 and F=80.
F/M+F = 80/20+80 = 80/100 80/100 = 4/5
answer D



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Re: In a corporation, 50 percent of the male employees and 40
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01 Sep 2018, 05:54
VeritasKarishma wrote: eybrj2 wrote: In a corporation, 50 percent of the male employees and 40 percent of the female employeesa re at least 35 years old. If 42 percent of all the employees are at least 35 years old, what fraction of the employees in the corporation are females?
a) 3/5
b) 2/3
c) 3/4
d) 4/5
e) 5/6 You can use the weighted averages formula for a 10 sec solution. No of females/No of males = (50  42)/(42  40 ) = 4/1 No of females as a fraction of total employees = 4/(4+1) = 4/5 For details of this method, check: http://www.veritasprep.com/blog/2011/03 ... averages/ VeritasKarishma, you are perfect and your solutions are very well constructed and easy to follow. Thanks a lot for your valuable contribution!!!
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Re: In a corporation, 50 percent of the male employees and 40
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11 Sep 2018, 05:34
dave13 wrote: VeritasPrepKarishma wrote: eybrj2 wrote: In a corporation, 50 percent of the male employees and 40 percent of the female employeesa re at least 35 years old. If 42 percent of all the employees are at least 35 years old, what fraction of the employees in the corporation are females?
a) 3/5
b) 2/3
c) 3/4
d) 4/5
e) 5/6 You can use the weighted averages formula for a 10 sec solution. No of females/No of males = (50  42)/(42  40 ) = 4/1 No of females as a fraction of total employees = 4/(4+1) = 4/5 For details of this method, check: http://www.veritasprep.com/blog/2011/03 ... averages/Hi VeritasPrepKarishma, What if i use alligations method 5040 42 28 as far i remember the value that is closer to the mean has more weight right? so here # of female employees is more (its not surprising actually, they have always been smarter than men ) how to proceed further ? let me try ... so 2 here has more weight hence it should be in numerator, and 8 has less weight and should be in denominator hence \(\frac{2}{8}\) > \(\frac{1}{4}\) so ratio of females to men is 1 to 4 ? but the mean was closer to percentage of females, shouldnt females be more than men ? I think this is more cleaner
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Re: In a corporation, 50 percent of the male employees and 40
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11 Sep 2018, 22:35
Thanks for the solution guys, this is a really good question




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