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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]

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26 Jul 2008, 00:41
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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

Last edited by Bunuel on 08 Feb 2014, 02:10, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: DS absolute value 2 (n3.7) [#permalink]

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26 Jul 2008, 07:22
judokan wrote:
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

(1) |x + y| = |x| + |y|

(2) |z + y| = |y| + |z|

From 1)

Either both X and Y are positive or both are negative. Then only 1) can hold true.

If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative.

From 2)

Either both Z and Y are positive or both are negative. Then only 2) can hold true.

If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative.

Combining 1) and 2)

When both X and Y are positive then Z is positive. i.e (x)(y + z)>0
When both X and Y are negative then Z is negative i.e (x)(y + z) >0

Thus C
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Re: DS absolute value 2 (n3.7) [#permalink]

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26 Jul 2008, 08:19
daemnn bro @ rahulgoyal1986! you're oh a roll!! I agree with C

What's the OA?

rahulgoyal1986 wrote:
judokan wrote:
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

(1) |x + y| = |x| + |y|

(2) |z + y| = |y| + |z|

From 1)

Either both X and Y are positive or both are negative. Then only 1) can hold true.

If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative.

From 2)

Either both Z and Y are positive or both are negative. Then only 2) can hold true.

If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative.

Combining 1) and 2)

When both X and Y are positive then Z is positive. i.e (x)(y + z)>0
When both X and Y are negative then Z is negative i.e (x)(y + z) >0

Thus C
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Re: DS absolute value 2 (n3.7) [#permalink]

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26 Jul 2008, 10:21
haidzz wrote:
daemnn bro @ rahulgoyal1986! you're oh a roll!! I agree with C

What's the OA?

rahulgoyal1986 wrote:
judokan wrote:
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

(1) |x + y| = |x| + |y|

(2) |z + y| = |y| + |z|

From 1)

Either both X and Y are positive or both are negative. Then only 1) can hold true.

If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative.

From 2)

Either both Z and Y are positive or both are negative. Then only 2) can hold true.

If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative.

Combining 1) and 2)

When both X and Y are positive then Z is positive. i.e (x)(y + z)>0
When both X and Y are negative then Z is negative i.e (x)(y + z) >0

Thus C

You guys do well.

OA is C
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Re: DS absolute value 2 (n3.7) [#permalink]

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06 Aug 2008, 13:05
judokan wrote:
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

(1) |x + y| = |x| + |y|

(2) |z + y| = |y| + |z|

1) x and y should have same sign
insuffciient
2) y and z should have same sign

combine. x,y,z must have same signs.

(x)(y + z) > 0 always true..

combined sufficient.
C

good question.
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Re: DS absolute value 2 (n3.7) [#permalink]

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06 Aug 2008, 14:48
good one
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Re: DS absolute value 2 (n3.7) [#permalink]

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06 Aug 2008, 18:09
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judokan wrote:
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

(1) |x + y| = |x| + |y|

(2) |z + y| = |y| + |z|

simplest way to solve this :
whether we say |a|+|b|= |a+b|
a,b are of same sign either a,b are +ve or both are -ve

hence now consider :
(1) x,y should be same sign
but (x)(y + z) > 0 depends on z also hence INSUFFI info

(2)z,y are same sign but (x)(y + z) > 0 depends on x hence INSUFFI info

now (1) and (2) say that x,y,z are of same sign hence (x)(y+z)>0 always SUFFI
hence IMO C
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Re: If x, y, and z are nonzero numbers, is (x)(y + z) > 0? [#permalink]

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07 Feb 2014, 20:03
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

(1) |x + y| = |x| + |y| -> sq and simplifying, xy = |x||y| => x and y have same sign. NOT SUFFICIENT

(2) |z + y| = |y| + |z| -> Sq and simplifying, zy = |z||y| => y and z have same sign. NOT SUFFICIENT.

Combining, x,y, and z have same sign. when all +ve, (x)(y + z) > 0. when all -ve, (x)(y + z) > 0. SUFFICIENT.

C.
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Re: If xyz ≠ 0, is x(y + z) >= 0? [#permalink]

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08 Feb 2014, 02:11
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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.

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] 08 Feb 2014, 02:11
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