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Each employee of Company Z is an employee of either division

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Each employee of Company Z is an employee of either division [#permalink] New post 23 Apr 2008, 20:29
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Each employee of Company Z is an employee of either division X or division Y, but not both. If each division has some part-time employees, is the ratio of the number of full-time employees to the number of part-time employees greater for division X than for Company Z?

I. The ratio of the number of full-time employees to the number of part-time employees is less for division Y than for Company Z

II. More than half of the full-time employees of Company Z are employees of division X, and more than half of the part-time employees of Company Z are employees of division Y.

Please share your explanation
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Re: DS: tricky [#permalink] New post 23 Apr 2008, 22:00
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Lets say number of part time employees in Y = Py
Lets say number of part time employees in X = Px
Lets say number of full time employees in Y = Fy
Lets say number of full time employees in X = Fx

Total number of Part time employees = Px + Py
Total number of Full time employees = Fx + Fy

Question is asking Fx/Px > Fx+Fy/Px+Py

Statement 1:
Tells us that Fy/Py < Fx+Fy/Px+Py. This statement does not talks about division X so it is insufficient.

Statement 2:
Assume Number of Part Time employee 100x where x is any positive number, and number of full time employees 100y, where y is any positive number. As per this statement at least 51x are working in department y and at most 49x is working in department x. Similarly at least 51y are working for department x and at most 49y is working for department y.

Ratio for Department x= 51y/49x
Ratio for whole company = 100y/100x = y/x

So question is answered. Answer B.
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Re: DS: tricky [#permalink] New post 28 Apr 2008, 00:52
abhijit_sen wrote:
Lets say number of part time employees in Y = Py
Lets say number of part time employees in X = Px
Lets say number of full time employees in Y = Fy
Lets say number of full time employees in X = Fx

Total number of Part time employees = Px + Py
Total number of Full time employees = Fx + Fy

Question is asking Fx/Px > Fx+Fy/Px+Py

Statement 1:
Tells us that Fy/Py < Fx+Fy/Px+Py. This statement does not talks about division X so it is insufficient.

Statement 2:
Assume Number of Part Time employee 100x where x is any positive number, and number of full time employees 100y, where y is any positive number. As per this statement at least 51x are working in department y and at most 49x is working in department x. Similarly at least 51y are working for department x and at most 49y is working for department y.

Ratio for Department x= 51y/49x
Ratio for whole company = 100y/100x = y/x

So question is answered. Answer B.


Quote:
I think its D.
If Rz>Ry the x ratio (Rx) must be greater than Rz. Rz isn't the weighted average of Rx and Ry, but it must be between them (or equal to them).
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Re: DS: tricky [#permalink] New post 28 Apr 2008, 01:36
eschn3am wrote:
Answer D

Company average is the weighted average of the 2 divisions. Statement 1 tells us that Y is below the company average, so X must be above it to compensate.
Statement 2 tells us that X is more than half the full time employees and less than half of the part time, so it has a greater ratio than Y, which means it has a greater ratio than the company (since one must be greater and the other smaller, unless they're both equal to the co. average)
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Re: DS: tricky [#permalink] New post 29 Apr 2008, 01:25
Let me, a Vietnamese Engineer, solve it :)

Let Xf: Full-time employees of X
Let Xp: Part-time employees of X
Let Yf: Full-time employees of Y
Let Yp: Part-time employees of Y

Since Full-time employees of company Z: Xf+Yf
Part-time employees of company Z: Xp+Yp

The question is: Xf/Xp>(Xf+Yf)/(Xp+Yp)? {*}
We must be sure that all Xi and Yi are both greater than zero. So, simplifying {*} then we have

Xf/Xp>Yf/Yp (necessary and sufficient conditions of {*} to be true)

1. Yf/Yp<(Xf+Yf)/(Xp+Yp) ==> Xf/Xp>Yf/Yp ==> Sufficient
2. Xf>(Xf+Yf)/2 ==> Xf>Yf (a)
Yp>(Xp+Yp)/2 ==> Yp>Xp (b)

Since a and b are true we have Xf.Yp> Yf.Yp>Yf.Xp ==> Xf/Xp>Yf/Yp ==> Sufficient

So the answer is D.

If you have any troubles pls send your email to duongminhtan@hotmail.com

Last edited by lexis on 30 Apr 2008, 00:37, edited 2 times in total.
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Re: DS: tricky [#permalink] New post 29 Apr 2008, 17:56
lexis wrote:
Let me, an Vietnamese Engineer, solve it :)

Let Xf: Full-time employees of X
Let Xp: Part-time employees of X
Let Yf: Full-time employees of Y
Let Yp: Part-time employees of Y

Since Full-time employees of company Z: Xf+Yf
Part-time employees of company Z: Xp+Yp

The question is: Xf/Xp>(Xf+Yf)/(Xp+Yp)? *
We must be sure that all Xi and Yi are both greater than zero. So, simplifying * then we have

Xf/Xp>Yf/Yp (necessary and sufficient conditions of * to be true)

1. Yf/Yp<(Xf+Yf)/(Xp+Yp) ==> Xf/Xp>Yf/Yp ==> Sufficient
2. Xf>(Xf+Yf)/2 ==> Xf>Yf (a)
Yp>(Xp+Yp)/2 ==> Yp>Xp (b)

Since a and b are true we have Xf.Yp> Yf.Yp>Yf.Xp ==> Xf/Xp>Yf/Yp ==> Sufficient

So the answer is D.

If you have any troubles pls send your email to duongminhtan@hotmail.com


I still dint get how A...

I went for B..can someone explain in a simpler way please!!
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Re: DS: tricky [#permalink] New post 29 Apr 2008, 21:49
Yep, I got tricked and picked B too.
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Re: DS: tricky [#permalink] New post 30 Apr 2008, 03:32
Keep in mind that if we prove Xf/Xp>Yf/Yp from given information in one case, then the case must be true.

Hope you could get it.
Re: DS: tricky   [#permalink] 30 Apr 2008, 03:32
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