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Each employee of Company Z is an employee of either division

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Re: Each employee of Company Z is an employee of either division  [#permalink]

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New post 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) ?
or, ac+ad>ac+bc
or, ad>bc ?

from st(1), a+b/ c+d >b/d
or, ad+bd>bc+bd
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.
Thus Answer is (D)
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File comment: use this chart to solve this problem in less than 2 min
chart.png
chart.png [ 11.03 KiB | Viewed 5552 times ]


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Re: Each employee of Company Z is an employee of either division  [#permalink]

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New post 29 Oct 2013, 05:35
2
Z is divided in to X and Y => Y <------- Z ------> X ( imagine its a chart not sure how to draw it out in here)

let the part time employees under division X be Px
Let the full time employees under division X be Fx

let the part time employees under division Y be Py
Let the full time employees under division Y be Fy

So the question is asking

Is \(\frac{Fx}{Px} > \frac{Fx+ Fy}{Px + Py}\) ?

We can cross multiply since all values positive

=> \(Fx.Px + Fx.Py > Fx.Px + Px.Fy\)

So the question \(is Fx.Py > Px.Fy?\)

Statement 1 says => 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

\(\frac{Fy}{Py} < \frac{Fy + Fx}{Px + Py}\)

Cross multiply we get

=> \(Fy.Px + Fy.Py < Fy.Py + Fx.Py\)
=> \(Fy.Px < Fx.Py\)

re-arrange\(Fx.Py > Fy.Px\) sufficient

Statement 2 => More than half of the full-time employees of Company Z are employees of Division X, and more than half of thepart-time employees of Company Z are employees of Division Y

\(\frac{Fx}{Fx + Fy}> \frac{1}{2}\) => \(2Fx > Fx + Fy\)=> \(Fx > Fy\)

And

\(\frac{Py}{Py + Px} > \frac{1}{2}\)

=> \(2Py > Py + Px\)

=>\(Py > Px\) => \(\frac{Px}{Py} < 1\) and \(Fx > Fy\) => \(\frac{Fx}{Fy}> 1\)

The question is\(Fx.Py > Px.Fy\)

Re-arrange the question \(Fx.Py > Px.Fy\)

\(\frac{Fx}{Fy}\) ( Greater than 1) > \(\frac{Px}{Py}\) ( less than 1) from statement above

sufficient

Answer is D
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Re: Each employee of Company Z is an employee of either division  [#permalink]

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New post 22 Nov 2013, 12:16
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


Nough said. This problem is easy to solve just conceptually. So drop your pens and lets start inmersing in the world of weighted average.
1) Ratio of FT/PT is less in Y than in total. This is STRAIGHT weighted average. It means then that this ratio in X will have to be higher to compensate and breakeven (Remember weighted average = 0). We could use differentials but its pretty clear to me.

Sufficient

2) We are being asked is FTX/PTX > FTY/PTY. From this the answer is clearly yes
Sufficient

Hence D

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Re: Each employee of Company Z is an employee of either division  [#permalink]

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New post 24 Nov 2013, 03:50
very hard questions. we can not use formular because it take a lot of time

just "feel" , draw 2 rantangular, each divided into 2 part, one part full time, the ther par time. we can do this question

this is very hard. I will redo this question may times
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Re: Each employee of Company Z is an employee of either division  [#permalink]

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New post 16 Dec 2013, 14:17
1
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


Can we have a short walkthrough for this problem. Would be more than happy to provide some nice Kudos as Christmas present
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Each employee of Company Z is an employee of either division  [#permalink]

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New post 06 Nov 2015, 07:16
Weighted average concept

Stmnt 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

ratio y < ratio z

since z is the overall average ratio, it has to definitely lie between the ratio for x and y. The question is which of the 2 ratios (ratiox or ratio y ) is higher..

stmt 1 says ratio y is less than ratio z which implies that ratio x is the greatest > ratio z> ratio y

SUFFICIENT

Stmnt 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

if you are assigning 51% or more of permanent employees to x then you assign 49% or less to y
if you are assigning 51% or more of temp employees to y then you assign 49% or less to x

thus numerator of x> numerator of y
denom x< denom y

this will always result is ratio x>ratio y [dividing a larger number into smaller parts>dividing a smaller number into larger parts]
this now implies something similar to statement 1

SUFFICIENT
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Re: Each employee of Company Z is an employee of either division  [#permalink]

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New post 10 Nov 2015, 10:49
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

This is a '2by2' question, most common type of question in GMAT math
Attachment:
GCDS GMATBLACKBELT Each employee of Company Z (20151107).jpg
GCDS GMATBLACKBELT Each employee of Company Z (20151107).jpg [ 23.17 KiB | Viewed 2281 times ]


If we modify the question, we get a/c>(a+b)/(c+d)?, or a(c+d)>c(a+b)?, or ac+ad>ca+cb?, or ad>cb, ultimately.
There are 4 variables (a,b,c,d) but only 2 equations are given from the 2 conditions, so there is high chance (E) will be our answer.
Looking at the conditions,
From condition 1, b/d<(a+b)/(c+d), or b(c+d)<d(a+b, or bc+bd<da+bd, or bc<ad. This answers the question 'yes' and this is sufficient.
For condition 2, a>(a+b)/2, or a>b and d>(c+d)/2, or d>c. This also answers the question 'yes' and this is sufficient as well.
Condition 1 = condition 2. The answer becomes (D).

For cases where we need 3 more equation, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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Re: Each employee of Company Z is an employee of either division  [#permalink]

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New post 11 Jun 2016, 02:16
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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


hard one.
pick numbers. suppose, Z have 50 full time employees and 50 part time employees.
and suppose the following:
50 full time employees of Z= 25 full time employee from X and 25 full time employee from Y
50 part time employee of Z= 25 partime one from X and 25 part time one from Y.

choice 1,
from choice 1, we have 24 full time one from Y, this mean 50-24=26 full time from X,
we have 26 part time from Y this mean 24 full time from X
enough
choice 2.
similar to choice 1.

answer D.

takeaway: PICK NUMBERS.
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Re: Each employee of Company Z is an employee of either division  [#permalink]

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New post 18 Jun 2016, 09:46
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) ?
or, ac+ad>ac+bc
or, ad>bc ?

from st(1), a+b/ c+d >b/d
or, ad+bd>bc+bd
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.
Thus Answer is (D)


I did not understand how below equations are deduced from statement2 :( .. Please explain!

from st(2), we know a>b
and d>c .
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Re: Each employee of Company Z is an employee of either division  [#permalink]

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New post 26 Jun 2016, 03:36
smartguy595 wrote:
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) ?
or, ac+ad>ac+bc
or, ad>bc ?

from st(1), a+b/ c+d >b/d
or, ad+bd>bc+bd
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.
Thus Answer is (D)


I did not understand how below equations are deduced from statement2 :( .. Please explain!

from st(2), we know a>b
and d>c .


Can someone explain how below equations are deduced from statement2 for above explanation!

from st(2), we know a>b
and d>c .
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Each employee of Company Z is an employee of either division  [#permalink]

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New post 05 Aug 2016, 10:58
Asifpirlo you absolutely nailed it. All the other concepts are rather abstract to me but you made it a lot easier with a chart. Thanks!

One question though, can't you just go directly to \(a/c > b/d\) without having to go through the \(a/c > (a+b)/(c+d)\) stage or am I missing something from taking that step? I believe it makes it even easier to setup intuitively
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Re: Each employee of Company Z is an employee of either division  [#permalink]

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New post 06 Apr 2017, 02:14
VeritasPrepKarishma wrote:
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.

gmat1220: I hope this answers your question too.



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Each employee of Company Z is an employee of either division  [#permalink]

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New post 12 Aug 2017, 21:32
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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


There are enough explanations for this question of weighted average. Somehow I wasn't convinced. I might have lacked the concept at the first place. So spent time and came up with a pictorial explanation for this conceptual question. Hope this helps those who ,like me, have faced difficulty in comprehending the essence of this question.


Refer the attachment for clearer visualization of the problem.

statement 1- consider the comparison only based on the element - "Ratio of F/P" for X & Y to that of Z.

statement 2 - consider the element - "Number of F" and "Number of Y" in Z.

statement 1 tells you about the blue portion in LHS - Ratio of F/P of Y in Z and question stem asks you about Yellow portion - Ratio of F/P in of X in Z. Definitely this is greater than the yellow portion.
statement 2 tells you about NX and DY and question stem asks you about (NX/DX) and (NY/DY). Note that smaller denominator of DX has given a greater value of the ratio NX/DY than the value of NY/DY(with a larger DY)

This is how both are sufficient and ans is D

Hope this was of help to those who got intimidated by a number of equations and distributions.
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Re: Each employee of Company Z is an employee of either division  [#permalink]

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New post 19 Oct 2017, 09:42
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


we dont need formulae here.

1)
if this ratio of full time to part time for Y is less than for Z, what make this happen. clearly, only because this ratio for x is greater than for Y, we have this ratio for y is less than for Z

2) similar to 1)
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Each employee of Company Z is an employee of either division  [#permalink]

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New post 18 Jun 2018, 05:59
fozzzy wrote:
Is there an alternative explanation for this question? What would be the difficulty level of this question?

I think the quickest and expected way to solve this is to:
Statement1:Realize that we've been given the answer, even though it seems too good to be true.

Statement2:use values
Where:
F is for full time
P is for Part time
Z is for Company Z
X is for division X, and
Y is for division Y

Say Fz/Pz ( which is Fx+Fy/Px+Py) = 3/3
Then according to statement2,

DivisionX= 2/unknown
DivisionY = unknown/2

Since an employee cannot work in both divisions then replace the 'unknown' with '1' and you have your answer.


Answer: D

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Re: Each employee of Company Z is an employee of either division  [#permalink]

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New post 01 Sep 2018, 20:18
in the test room when we are tired and nervous, this problem can kill us.

pick the specific numbers and get the answer
suppose company z has 20 employees, 10 are full time and 10 are part time.
condition 1
full time in Y is 5 and 5 is part time, so x must be 6 full time and 4 part time, at least, so. it is clear

condition 2
company z has 10 full time employee and 10 part time emplyees.

x has 6 full time, and y has 4 partime employee. it is good.
Re: Each employee of Company Z is an employee of either division &nbs [#permalink] 01 Sep 2018, 20:18

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