Each employee of Company Z is an employee of either division X or divi

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

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 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

Hence the answer is D.
_________________

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.
_________________

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.

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.

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)

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


Hence Answer is D.

I dont think you understand the meaning of 'weighted' average. You are assuming X and Y have the same size/weight

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.

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.
_________________

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

Is there an alternative explanation for this question? What would be the difficulty level of this question?

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

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


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.

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.

_________________

Let Xf/Yf is the full time in Division X/Y, and Xp/Yp is part time in X/Y, X,Y, and Z
are number of employees in X, Y, and Z.
X=Xf+Xp
Y=Yf+Yp
Z=X+Y
For 1, Yf/Yp<Zf/Zp, as a compensation, Xf/Xp should be greater than Zf/Zp
For2, More than ?Zf /less than ?of the Zp should be greater than Zf/Zp
Answer is D

The key here is realizing that the ratio of the number of full-time employees to the number of part-time employees for the entire company must be between the respective ratios of division X and Y. So the ordering of the ratios can be:

X > Z > Y
or
Y > Z > X

The questions asks us if the ratio of X is greater than the ratio of Z (the entire company). So really the question is: is the ratio of X greater than Y?

In statement 1 we're told Z > Y. We can thus conclude that X > Z > Y . Sufficient.

In statement 2 we're told 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.

Picture two fractions: one for X and one for Y. Statement 2 tells us that the numerator of X will be greater than the the denominator of Y, and the numerator of Y will be less than the denominator of Y.

Statement 2 is sufficient as well.

Answer is D.
_________________

GMATBusters can we do these beautiful question via allegation ? if us kindly let me know thanks in advance