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# If xyz 0, is x (y + z) 0? (1) y + z = y + z (2) x + y = x +

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Manager
Joined: 06 Apr 2010
Posts: 144
Followers: 3

Kudos [?]: 63 [0], given: 15

If xyz 0, is x (y + z) 0? (1) y + z = y + z (2) x + y = x + [#permalink]  30 Aug 2010, 09:19
00:00

Difficulty:

(N/A)

Question Stats:

75% (03:05) correct 25% (04:39) wrong based on 4 sessions
If xyz ≠ 0, is x (y + z) ≥ 0?
(1) │y + z│ = │y│ + │z│
(2) │x + y│ = │x│ + │y│
[Reveal] Spoiler: OA
Manager
Joined: 25 Jun 2010
Posts: 94
Followers: 1

Kudos [?]: 23 [2] , given: 0

Re: Modulus DS [#permalink]  30 Aug 2010, 09:34
2
KUDOS
xyz != 0,
xy+xz>= 0 ?

i). |y+z| = |y|+|z|
=> y,z both > 0 or y,z both < 0
So, xy+xz can be either -ve or +ve depending on sign/value of x.
Insufficient.

ii). |x+y| = |x|+|y|
=> x,y both > 0 or x,y both < 0
Again, xy+xz can be either +ve or -ve depending on z's value/sign.
Insufficient.

Combining both of them :
Either , x,y and z all >0
Or, x, y and z all < 0

In both the cases: xy + xz = +ve + (+ve) = +ve
Hence, "C".
Senior Manager
Joined: 14 Jun 2010
Posts: 334
Followers: 2

Kudos [?]: 14 [0], given: 7

Re: Modulus DS [#permalink]  31 Aug 2010, 09:20
Has to be C, same explanation as Anshu
Re: Modulus DS   [#permalink] 31 Aug 2010, 09:20
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