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Is XY < Z? S1) xyz =1 S2) xy(z^2) =1

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Is XY < Z? S1) xyz =1 S2) xy(z^2) =1 [#permalink] New post 01 Dec 2005, 15:50
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A
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C
D
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Is XY < Z?

S1) xyz =1

S2) xy(z^2) =1
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Re: DS: number system [#permalink] New post 01 Dec 2005, 16:54
duttsit wrote:
Is XY < Z?

S1) xyz =1

S2) xy(z^2) =1


I would say C

1. x y z can be any combination for xyz=1
(-1,-1,1) (1/2,1/2,4) ...so insuff

2. Here too xyz can be a no. of combinations
(1,1,1)(1/2,1/2,2) .......insuff

1 & 2 together

..from 1. & 2. z=1 & xy=1 so sufficient
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 [#permalink] New post 01 Dec 2005, 18:00
(1). If x = 1/2, y = 2, z = 1, then xyz = 1, but xy = z
If x = 1/3, y = 1, z = 3, then xyz = 1, but xy < z

So not sufficient.

(2) If x = 1/2, y = 2, z = 1, then xy(z^2) = 1, but xy = z
If x = 1/3, y = 1, z = (3^1/2), then xy > z.

So not sufficient.

Using (1) and (2), xyz = xy(z^2) --> z = z^2 --> z(z-1) = 0 --> z = 0 or z = 1
For statement (1) and (2) to hold, z has to be 1. So x and y must cancel each other out to be 1. This will always result in xy=z. So sufficient.

Ans: C
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Re: DS: number system [#permalink] New post 01 Dec 2005, 20:41
duttsit wrote:
Is XY < Z?
S1) xyz =1
S2) xy(z^2) =1
C.

from i, we do not know anything about x, y and z.
from ii, nothing...

from i and ii, we know z=1. if z = 1, xy=1.
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 [#permalink] New post 01 Dec 2005, 20:58
Yes, both statements together isolates z to be only 1.

(C)
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 [#permalink] New post 01 Dec 2005, 23:07
andy_gr8 wrote:
How if Z is negative


combine 1 and 2, z=1 > 0
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 [#permalink] New post 02 Dec 2005, 09:03
Good job everyone. OA is C.
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  [#permalink] 02 Dec 2005, 09:03
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