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Each employee of Company Z is an employee of either Division

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Intern
Joined: 01 Aug 2007
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Each employee of Company Z is an employee of either Division [#permalink]

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31 Aug 2007, 10:28
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 FT employees to the number of PT employees greatre for Division X than for Company Z?

1) The ratio of the number of FT employees to the number of PT employees is less for Division Y than for Company Z

2) More than half of the FT employees of Company Z are employees of Division X and more than half of the PT employees of Company Z are employees of Division Y

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Intern
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01 Sep 2007, 01:02
New to the GMAT here, will give this one a try:

1) Sufficient
2) Insufficient

Details:

1) Company Z: total ratio

- Division X: more (must be higher to balance out the division Y ratio)

- Division Y: less (given)

2) No specific information about the actual number of FT and PT employees, there could be an extreme difference in the number of FT versus PT employees in the company, one way or the other, that can throw the ratios anywhere.

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Manager
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01 Sep 2007, 07:36
I get B.

From stem, we have:

Company Z = Division X + Division Y = (FTx + PTx) + (FTy + PTy)

Where:
FTx=Full Time Employee of Div X
PTx=Part Time Employee of Div Y
FTy=Full Time Div Y
PTy=Part Time Div Y

We're looking to find if FTx/PTx is > or < (FTx+FTy)/(PTx+PTy)

1) FTy/PTy < (FTx+FTy)/(PTx+PTy)
Insufficient.
This tells us nothing about the number or ratio of part time/full time workers in X. It could be that Y has fewer total works than X or that Y has many more part time workers than X.

2) 1/2(FTx + FTy)< are in Div X and 1/2(PTx + PTy)< are in Div Y
Sufficient
This tells us that the FTx/PTx is at least greater than 1 and that (PTx + PTy) > than PTx. Therefore FTx/PTx has to be greater than (FTx + FTy)/(PTx + PTy) b/c the numerator of FTx/PTx is at least 2(FTx + FTy) while the denominator of FTx/PTx is at least 1/2(PTx+PTy)

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Intern
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03 Sep 2007, 09:55
OA is D.

Anyone else?

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Senior Manager
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03 Sep 2007, 11:32
Need to be found Fx/Px <> Fz/Pz

A) in first case info given is enough to see which one is higher:

Fy/Py<Fz/Pz

Fz=Fx+Fy Pz=Px+Py

(Fz-Fx)/(Pz-Px)<Fz/Pz

Pz(Fz-Fx)<Fz>FxPz or Fx/Px>Fz/Pz

which is the desired answer to the problem

SO a is SUFFICIENT

B) This case is more straightforward:

Fx>Fy and Px<Py>Fz/Pz because Fx is higher than 50% and Px is less than 50% making overal division higher than Fz/Pz.

Note that Fz/Pz = 1/1 = 1 Fx/Px = 0.6/0.4 = 1.5

SO b is also SUFFICIENT

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Manager
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03 Sep 2007, 16:04
I cannot understand Ferihere's logic.

Can you explain how did you arrive at the expression.

(Fz-Fx)/(Pz-Px)<Fz/Pz

implies Pz(Fz-Fx)<Fz>FxPz or Fx/Px>Fz/Pz
_________________

Regards

Subhen

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Manager
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03 Sep 2007, 22:49
Totally confused with this one - any OE?

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Intern
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04 Sep 2007, 00:35
Please see the attachment. I am unable to post the text properly.
Attachments

Ratio.doc [4.31 KiB]

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04 Sep 2007, 00:35
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