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# DS 6

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SVP
Joined: 16 Oct 2003
Posts: 1805
Followers: 5

Kudos [?]: 154 [0], given: 0

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26 May 2004, 20:20
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

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SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA
Followers: 6

Kudos [?]: 101 [0], given: 0

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26 May 2004, 20:59
A) xyz > 0 if y = -1 z = 1 and x = -1 then xyz > 0 but x(y+z) = 0 (Insufficient)
B) yz > 0 y = -1, z = -1 yz = 1 > 0 if x = 1 then xyz = 1 > 0 but if x = -1 then xyz < 0 (Insufficient)

Combine these make y = -1 z = -2 then x has to be +ve because xyz > 0
Let x = 1 1(-1+-2) = -3
if y = 1 x = 2 and x = 1 then 1(1+2) = 3
(Insufficient)

E is the answer.
Director
Joined: 05 May 2004
Posts: 575
Location: San Jose, CA
Followers: 2

Kudos [?]: 61 [0], given: 0

Re: DS 6 [#permalink]

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28 May 2004, 18:12
Bhai wrote:

Ans: E

yz>0
xyz>0
which means x is positive
but y,z can be either positive or negative!
Manager
Joined: 07 May 2004
Posts: 183
Location: Ukraine, Russia(part-time)
Followers: 2

Kudos [?]: 17 [0], given: 0

Re: DS 6 [#permalink]

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29 May 2004, 03:30
Bhai wrote:

1 and 2 together are not sufficient, because

x = 1, y = -2, z = -3 satisfy 1 and 2 but do not satisfy x(y+z)>0.
Intern
Joined: 26 Jan 2003
Posts: 28
Location: Reality
Followers: 1

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

Re: DS 6 [#permalink]

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29 May 2004, 12:36
Bhai wrote:

(2) y>0, z>0 or y<0, z<0 (insufficient)
(1) x>0, yz>0 or x<0, yz<0 (insufficient)
(1)and(2) x>0, yz>0, y?0, z?0 => (y+z)?0

Re: DS 6   [#permalink] 29 May 2004, 12:36
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# DS 6

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