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# If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

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If x, y, and z are nonzero numbers, is (x)(y + z) > 0? [#permalink]  25 Jul 2008, 23:41
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If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

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

(2) |z + y| = |y| + |z|
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Re: DS absolute value 2 (n3.7) [#permalink]  26 Jul 2008, 06:22
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]  26 Jul 2008, 07:19
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]  26 Jul 2008, 09:21
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]  06 Aug 2008, 12:05
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]  06 Aug 2008, 13:48
good one
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Re: DS absolute value 2 (n3.7) [#permalink]  06 Aug 2008, 17:09
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: DS absolute value 2 (n3.7)   [#permalink] 06 Aug 2008, 17:09
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