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19 Aug 2006, 05:17
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N
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Here a GMATPrep question. Kindly disregard the marked answer.
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19 Aug 2006, 10:53
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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
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
20 Aug 2006, 01:08
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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
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
