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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]  17 Mar 2011, 07:27
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27% (02:38) correct 72% (02:53) wrong based on 11 sessions
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?

(1) 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.
(2) More than half 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.
[Reveal] Spoiler: OA
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Re: ratios [#permalink]  17 Mar 2011, 07:48
1
KUDOS
I think no point solving this one. Use guesswork

In the weighted average if ratio (full time / part time) increases in division X, the ratio automatically decreases in division Y to keep the weighted average (ratio) same for company Z. This reasoning is for Statement 2)

The same can be applied intuitively to Statement 1) Hence both will be sufficient.

I will guess D after glancing for 60 secs.

Q))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?
(1) 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.
(2) More than half 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.
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Re: ratios [#permalink]  17 Mar 2011, 19:12
1
KUDOS
I dont understand why you have to solve in 60 seconds and guess. You can take 90 seconds and be sure.

Yes %Full-Z will be a weighted average of %Full-X and %Full-Y. So either X>Z>Y or X<Z<Y, Z will always be in the middle.

So (1) is enough, as if Y<Z, then 'X must >Z'

For 2 we see that X has more full time and less part time than Y, therefore X>Y, and by extension also X>Z.

So both are enough,(D)
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Re: ratios [#permalink]  17 Mar 2011, 22:05
1
KUDOS
I trudged through this for a while, and I can see now why it was not necessary:

PX + PY + FX + FY = Total

PX+PY = PZ

FX + FY = FZ

So the question is asking: FX/PX > (FZ)/(PZ)

From 1 : FY/(PY) < (FZ)/(PZ)

=> PZ * FY < FZ * PY

=> PZ ( FZ - FX) < FZ(PZ - PX)

=> PZ * FZ - PZ * FX < PZ*FZ - PX * FZ

=> PX * FZ < PZ * FX

=> FZ/PZ < FX/PX, so sufficient

From 2: FX > (FX + FY) /2 and PY > (PY + PX)/2

=> FX > FY and PY > PX

=> FX * PY > FY * PX

=> FX ( PZ - PX) > (FZ-FX)*PX

=> FX*PZ - FX*PX > FZ * PX - FX*PX

=> FX/PX > FZ/PZ, so sufficient

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Re: ratios [#permalink]  17 Mar 2011, 22:11
1
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@bostonrb

For 2 we see that X has more full time and less part time than Y, therefore X>Y, and by extension also X>Z.

jsu to calrify, are you saying that For x -> (Higher numerator)/(lower denominator) > For Y -> (Lower numerator)/(higher denominator) ?
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Re: ratios [#permalink]  17 Mar 2011, 22:29
1
KUDOS
Pretty much. What you wrote for denominators is accurate in terms of the ratio FT:PT, which is what the question is asking about.

In my post I referred to %Full Time, and although the relationship still stands, the lower/higher denominator would not be accurate for that because the denominator in that case would be FT+PT, not PT. Just ignore this caveat if its confusing.
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Re: ratios [#permalink]  18 Mar 2011, 12:29
@ subashghosh.....got it.....ur detailied explanation was helpful
regards
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Re: ratios [#permalink]  18 Mar 2011, 15:55
Lets say

Division X - Fulltime Employee's Fx
Partime Employee's Px

Division Y - Fulltime Employee's Fy
Partime Employee's Py

Total Full Time Employee's F = Fx+Fy
Total Part Time Emplyee's P = Px+Py

Rephrasing Given all they are asking is Fx/Px > (Fx+Fy)/(Px+Py) ? = > Fx/Px > Fy/Py?

1. Fy/Py < (Fx+Fy)/(Px+Py)

cross multiplying and solving this we get Fy/Py < Fx/Px , thats exactly what you had to find. So sufficient

2. Fx> Fy
Py> Px

Fx/Px Fy/Py
so division x has a bigger numerator than division y
and a small numerator than division y.

=> Fx/Px > Fy/Py

Sufficient

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Re: ratios [#permalink]  18 Mar 2011, 16:08
Lets say ratio was initially 1:1 for Z and for the divisions X 1:1 and for Y 1:1

Now I make the ratio for X as 1:4. Since X is 1:4 then Y will become 4:1 So that the weighted ratio is still 1:1. The same is true vice-versa.

So basically the DS is based on truism.
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Re: ratios [#permalink]  18 Mar 2011, 16:25
I dont think you understand the meaning of 'weighted' average. You are assuming X and Y have the same size/weight
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Re: ratios [#permalink]  18 Mar 2011, 16:34
Pardon me. a) ratios are fractions and not certain numbers and b) there is no statement - saying one division is superior over the other. I dont know what you mean by weight.
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Re: ratios [#permalink]  18 Mar 2011, 16:49
Well if you are aiming for a high score you probably should find out
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Re: ratios [#permalink]  18 Mar 2011, 16:56
High score requires common sense not math
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Re: ratios [#permalink]  18 Mar 2011, 19:38
Actually, both of you are using exactly the same concept. You are just applying it differently.
bostonrb has depicted the use of weighted averages to solve it, gmat1220 is using it intuitively to solve the question and of course both of you are correct...

Quote gmat1220:
Lets say ratio was initially 1:1 for Z and for the divisions X 1:1 and for Y 1:1

Now I make the ratio for X as 1:4. Since X is 1:4 then Y will become 4:1 So that the weighted ratio is still 1:1.

This just means we are considering weights even if we are showing it in the form of ratios.

According to statement 1, Full:Part for Y is less than that for Z, it means fraction of Full time employees (Full/Total) is less in Y as compared to Z. Hence to get a higher fraction in Z, X must have an even higher fraction of Full time employees (Full/Total)
So if one group has a smaller Full:Part ratio, the other group must have a bigger Full:Part ratio.
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Re: ratios [#permalink]  18 Mar 2011, 20:10
Actually... I did not use weights, just that z will always be between x and y regardless of their weights.

Dont mean to be rude but you gotta go read up on weights too, because you also dont understand them. e.g. If you make the ratio of X as 1:4 as gmat1220 says, it is completely INCORRECT to say that Y's becomes 4:1. And this is because of weights.

z= weightX*AverageX+weightY*AverageY. NOT (AverageX+AverageY)/2

Belie' dat
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Re: ratios [#permalink]  18 Mar 2011, 20:21
Karishma,
I didn't do math in this question. So let me throw numbers
20:40 initially for company Z
Let's change ratio of div X to 10:10 then the ratio of div Y becomes 10:30. But yes the intuition was right

If 40:40 initially for company Z
Let's change ratio of div X to 10:30 then ratio of Y becomes 30:10

Pls verify. Thanks for confirming

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Re: ratios [#permalink]  19 Mar 2011, 05:59
1
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Don't worry. I understand weighted average. What is intriguing is that we use these concepts often without even meaning to use them. When I say 'he is using weights', it means the general concept of 'weighted averaging'

Let me give you an example:

I have two solutions X and Y, 1 lt each, of wine and water. Solution X has wine:water in the ratio 2:3. I mix these two solutions to get wine:water ratio in the mixture as 1:1. What is wine:water in Y?
Guess what, the ratio of wine:water in Y is 3:2.

You can do it in various ways
Say working with concentration on wine:
1/2 = (2/5 * 1 + x*1)/2
x = 3/5

or we can simply say that wine in X is 40%, so wine in Y must be 60% to get 50% in mixture. This is averaging (or more generically weighted averaging... ) Here the weights are the same....

But the thing to note is that it doesn't matter whether we have the weights or now because due to averaging, the mixture will have concentration of wine in between X and Y.
If one understands this, one may not need to use any math and do it intuitively. Others may use the equations above to arrive at the same conclusion.

gmat1220, I believe, was using a generic example to explain why his intuition said what it did. If you want to get a ratio of 1:1 in the combined mixture and one mixture has a ratio of 1:4, the other will have 4:1 provided they both have equal weights (but we don't have to worry about the weights since eventually they do not matter in our question) So basically, it is the same general concept of weighted averages.

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Re: ratios [#permalink]  19 Mar 2011, 06:33
Its an average weight problem .Have a look at an awesome article by Stacey (MGMT)
http://www.beatthegmat.com/mba/2011/01/26/breaking-down-gmatprep-average-arithmetic-problems
Re: ratios   [#permalink] 19 Mar 2011, 06:33
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