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If xyz 0, is x (y + z) 0?

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If xyz 0, is x (y + z) 0? [#permalink] New post 30 Aug 2010, 09:19
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A
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C
D
E

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77% (01:58) correct 23% (02:47) wrong based on 13 sessions
If xyz ≠ 0, is x (y + z) ≥ 0?

(1) │y + z│ = │y│ + │z│
(2) │x + y│ = │x│ + │y│

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-xyz-0-is-x-y-z-107551.html
[Reveal] Spoiler: OA
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Re: If xyz 0, is x (y + z) 0? [#permalink] New post 30 Aug 2010, 09:34
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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".
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Re: If xyz 0, is x (y + z) 0? [#permalink] New post 31 Aug 2010, 09:20
Has to be C, same explanation as Anshu
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Re: If xyz 0, is x (y + z) 0? [#permalink] New post 26 Apr 2015, 13:10
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Re: If xyz 0, is x (y + z) 0? [#permalink] New post 27 Apr 2015, 00:35
Expert's post
If xyz ≠ 0, is x (y + z) >= 0?

xyz ≠ 0 means that neither of unknowns is 0.

(1) |y + z| = |y| + |z| --> either both \(y\) and \(z\) are positive or both are negative, because if they have opposite signs then \(|y+z|\) will be less than \(|y|+|z|\) (|-3+1|<|-3|+1). Not sufficient, as no info about \(x\).

(2) |x + y| = |x| + |y| --> the same here: either both \(x\) and \(y\) are positive or both are negative. Not sufficient, as no info about \(z\).

(1)+(2) Either all three are positive or all three are negative --> but in both cases the product will be positive: \(x(y+z)=positive*(positive+positive)=positive>0\) and \(x(y+z)=negative*(negative+negative)=negative*negative=positive>0\). Sufficient.

Answer: C.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-xyz-0-is-x-y-z-107551.html
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Re: If xyz 0, is x (y + z) 0?   [#permalink] 27 Apr 2015, 00:35
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