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If x, y, and z are nonzero numbers, is (x)(y + z) > 0? [#permalink]
 26 Jul 2008, 00:41
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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, 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]
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]
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]
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 don't know about z.. it can be +ve or -ve leads mutliple answers.. insuffciient 2) y and z should have same sign don't know about x.. it can be +ve or -ve leads mutliple answers.. 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, 14:48
good one 
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Re: DS absolute value 2 (n3.7) [#permalink]
06 Aug 2008, 18:09
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, 18:09
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