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GMATPrep Question - Math_DS

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GMATPrep Question - Math_DS [#permalink] New post 19 Aug 2006, 04:17
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Here a GMATPrep question. Kindly disregard the marked answer.
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 [#permalink] New post 19 Aug 2006, 09:53
I think it is D
Statement 1 tells us N(ft at y)/N(pt at y) < N(ft at Z)/N(pt at Z)
for 100 Ft and 100 Pt, i can have 20ft/30pt for y and then 80ft/70pt for X

thus x > Z
suff..

Statement 2
Take if Z has 100 FT and 100 parttime
X has more than half full time and Y has more than half parttime..
meaning
N(ft at X)/N(pt at X) will always be > 1 (say 51/49...51 full time and 49 partime at X )
Sufficient

Hence D
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 [#permalink] New post 20 Aug 2006, 00:08
Inituition says D here!

Let's see if I have the math to confirm it. Employees are either full time or part time. So we can simplify the question a little

If px, py and pz are the precentages of the employees of div. X, div. Y and Z that all full time, and nx, ny and nz the numbers of employees in X, Y and Z (nz=nx+ny) the question is equivalent to:

Is px > pz?

In general pz(nx+ny)=nypy+nxpx
ny(pz-py)+nxpz=nxpx

(1) says that pz-py > 0

So, nxpz<nxpx and pz<px SUFF

Really, when you think about it, it makes sense that either pz is between py and pz or px=py=pz ... (1) says py<pz , so it must be that pz<px

(2) Suppose pz ≥ px. Thus py ≥ pz and 1-px≥ 1-pz≥ 1-py

pxnx> 0.5nzpz So nx>0.5nz (a)

(1-py)ny>0.5(1-pz)nz. So ny>0.5nz (b)

Adding (a) and (b) we get, nx+ny>nz

However, as there is no overlap between divisions X and Y, this can't be true.

Thus px>pz SUFF


Answer D
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 [#permalink] New post 20 Aug 2006, 13:53
Yeah I would go with D on this, both seem to work for solutions.
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 [#permalink] New post 20 Aug 2006, 13:54
kevincan wrote:
Inituition says D here!

Let's see if I have the math to confirm it. Employees are either full time or part time. So we can simplify the question a little

If px, py and pz are the precentages of the employees of div. X, div. Y and Z that all full time, and nx, ny and nz the numbers of employees in X, Y and Z (nz=nx+ny) the question is equivalent to:

Is px > pz?

In general pz(nx+ny)=nypy+nxpx
ny(pz-py)+nxpz=nxpx

(1) says that pz-py > 0

So, nxpz<nxpx and pz<px SUFF

Really, when you think about it, it makes sense that either pz is between py and pz or px=py=pz ... (1) says py<pz , so it must be that pz<px

(2) Suppose pz ≥ px. Thus py ≥ pz and 1-px≥ 1-pz≥ 1-py

pxnx> 0.5nzpz So nx>0.5nz (a)

(1-py)ny>0.5(1-pz)nz. So ny>0.5nz (b)

Adding (a) and (b) we get, nx+ny>nz

However, as there is no overlap between divisions X and Y, this can't be true.

Thus px>pz SUFF


Answer D


Wao... simply amazing..thank you..
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  [#permalink] 20 Aug 2006, 13:54
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