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# Each employee of Company Z is an employee of either division

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SVP
Joined: 29 Mar 2007
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Each employee of Company Z is an employee of either division  [#permalink]

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Updated on: 29 Oct 2013, 06:49
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95% (hard)

Question Stats:

43% (02:25) correct 57% (02:37) wrong based on 1503 sessions

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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?

(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 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

Originally posted by GMATBLACKBELT on 17 Oct 2007, 19:29.
Last edited by Bunuel on 29 Oct 2013, 06:49, edited 2 times in total.
Edited the question.
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19 Mar 2011, 05:59
4
4
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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Karishma
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18 Oct 2007, 03:00
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GMATBLACKBELT wrote:
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 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

D for me.

let's assume the following:
f1 = nb FT employee for div X
p1 = nb PT employee for div X
f2 = nb FT employee for div Y
p2 = nb PT employee for div Y

We are asked to show that:
f1/p1 > (f1+ f2)/(p1 + p2)
develop this inequation and we get:
f1p2 > f2p1

Stat.1 says: f2/p2 less then (f1+f2) divided by (p1+p2), develop it and we get f1p2 > f2p1 => stat.1 sufficient

stat.2 says: f1>f2 & p2>p1, multiply this and we get:
f1p2 > f2p1 => stat.2 is sufficient

Ans. is D

ps: i had problems posting this because of the inequations that are interpreted as CODE by the program
##### General Discussion
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18 Oct 2007, 09:32
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I think I understand A now. even w/o setting up equations, I think I can deduce that b/c the ratio of FT/PT for Y is less than Z then FT/PT for X is going to be greater than Y.

Essentially the question is asking is FT greater for X and is the PT greater for Y.

Ex/ X: FTx/PTx = 6/5 Y: FTy/PTy 1/3

FTx/PTx > FTx+FTy/PTx+PTy ---> process of elimination thru subtraction yields

Is (FTx)(PTy)> (FTy)(PTx)?

S1:
If we do set up equations then we get: FTx+FTy/PTx+PTy > FTy/PTy

which when worked out similar to the equation above yields:
(PTy)(FTx)> (FTy)(PTx) the same as the above so Suff.

S2: No need for equations here. We can answer the question from our rephrased question.

Since MORE than half of the employees at Z are FT under X and more than half are of the employees at Z are PT under Y. Then this answers the question that is the ratio for X:

FTx/PTx> FTx+FTy/PTx+PTy. B/c the FTx is > the the PTx.

Let me know if my logic is correct. I think the nature of this question just spooked me and I just guessed.
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17 Mar 2011, 07:48
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1
2
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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17 Mar 2011, 19:12
10
3
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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17 Mar 2011, 22:05
12
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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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17 Mar 2011, 22:11
1
@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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17 Mar 2011, 22:29
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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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18 Mar 2011, 15:55
2
1
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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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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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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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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18 Mar 2011, 19:38
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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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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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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

Posted from my mobile device

Posted from my mobile device
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30 May 2011, 22:05
Cant this be solved using fact and logic?

we know that company z, division x, division y each has both full time and part time employees.
There are no fractions , -ve number or 0 possibilities here.
Q is if FTX/PTX > FTY/PTY is true?

if it can be deduced that if FTZ/PTZ > FTY/PTY is true than Q is true.
Also for the Q condition to be true, company Z should have more full time employees in div X than div Y (i.e. in Q - num of left hand side should be higher than number of right hand side) and more part time employees in company Y than div X (i.e. in Q - denorminator of right hand side should be higher than denominatior of left hand side).

Both the statements gives very clear idea of the Q condition to be true.
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Re: Each employee of Company Z is an employee of either division  [#permalink]

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07 Aug 2013, 00:19
Is there an alternative explanation for this question? What would be the difficulty level of this question?
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Re: Each employee of Company Z is an employee of either division  [#permalink]

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07 Aug 2013, 04:25
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fozzzy wrote:
Is there an alternative explanation for this question? What would be the difficulty level of this question?

My solution: (From the diagram table)
we have to know, is a/c > (a+b)/(c+d) ?

from st(1), a+b/ c+d >b/d
or, ad>bc so st(1) alone is sufficient.

from st(2), we know a>b
and d>c .
Multiplying the both equation gives us, ad>bc. so st(2) alone is sufficient too.

Thus both statements individually are sufficient to solve the problem.
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Asif vai.....
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Re: Each employee of Company Z is an employee of either division  [#permalink]

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09 Aug 2013, 14:02
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Asifpirlo wrote:
fozzzy wrote:
Is there an alternative explanation for this question? What would be the difficulty level of this question?

My solution: (From the diagram table)
we have to know, is a/c > (a+b)/(c+d) ?

from st(1), a+b/ c+d >b/d
or, ad>bc so st(1) alone is sufficient.

from st(2), we know a>b
and d>c .
Multiplying the both equation gives us, ad>bc. so st(2) alone is sufficient too.

Thus both statements individually are sufficient to solve the problem.

this really helped at least 335 members..... glad to see it.
335 views 2 kudos..... not bad

thanks thanks.... love to see these
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Asif vai.....
Re: Each employee of Company Z is an employee of either division   [#permalink] 09 Aug 2013, 14:02

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