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# If x>y>z, is xz>0? 1) xy<0 2) yz>0

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SVP
Joined: 14 Dec 2004
Posts: 1681

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If x>y>z, is xz>0? 1) xy<0 2) yz>0 [#permalink]

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19 Mar 2006, 06:02
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If x>y>z, is xz>0?

1) xy<0
2) yz>0

Kudos [?]: 172 [0], given: 0

Manager
Joined: 22 Apr 2005
Posts: 129

Kudos [?]: 5 [0], given: 0

Location: Los Angeles

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19 Mar 2006, 06:13
1. xy < 0 => ((x < 0) && (y > 0)) || ((x > 0) && (y < 0)) with given x>y>z => ((x > 0) && (y < 0)) => z < 0 => xz < 0

2. ((y > 0) && (z > 0)) || ((y < 0) && (z < 0)) with ((y < 0) && (z < 0)) we can't say anything about x.

A

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VP
Joined: 29 Dec 2005
Posts: 1337

Kudos [?]: 70 [0], given: 0

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19 Mar 2006, 07:06
Tyr wrote:
1. xy < 0 => ((x < 0) && (y > 0)) || ((x > 0) && (y < 0)) with given x>y>z => ((x > 0) && (y < 0)) => z < 0 => xz < 0

2. ((y > 0) && (z > 0)) || ((y < 0) && (z < 0)) with ((y < 0) && (z < 0)) we can't say anything about x.

A

altho, i did not follow your explanation, agree with you answer as A.

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GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5032

Kudos [?]: 457 [0], given: 0

Location: Singapore

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19 Mar 2006, 07:07
1) xY < 0

Could be x positive, y negative and z negative then xz = negative (E.g x = 10, y = -1, z = -10)
Or x negative, y negative and z negative then xz = positive (E.g x = -1, y = -2, z = -3)

Insufficient.

2) yz > 0.

Could be y positive, z positive, then xz would be positive (E.g. y = 10, z = 5 and x > 10)
Or y negative, z negative, then xz would be negative/positive. (E.g. y = -3, z = -5 and x > -3)

Insufficient.

Using 1) and 2)

Could be y negative, z negative, then x must be positive and xz = negative
or
Could be y positive, z positive, then x must be negative (Does not hold since x > y)

Ans C

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Manager
Joined: 23 Jan 2006
Posts: 191

Kudos [?]: 41 [0], given: 0

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19 Mar 2006, 07:13
I say A.

ywilfred wrote:
1) xY < 0

Could be x positive, y negative and z negative then xz = negative (E.g x = 10, y = -1, z = -10)
Or x negative, y negative and z negative then xz = positive (E.g x = -1, y = -2, z = -3)

Insufficient.

2) yz > 0.

Could be y positive, z positive, then xz would be positive (E.g. y = 10, z = 5 and x > 10)
Or y negative, z negative, then xz would be negative/positive. (E.g. y = -3, z = -5 and x > -3)

Insufficient.

Using 1) and 2)

Could be y negative, z negative, then x must be positive and xz = negative
or
Could be y positive, z positive, then x must be negative (Does not hold since x > y)

Ans C

This can't be if xy < 0. either one could be negative, but not both. (but from given information, we know that x>y, so x is positive and y is negative)

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Manager
Joined: 22 Apr 2005
Posts: 129

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Location: Los Angeles

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20 Mar 2006, 02:51
ywilfred wrote:
1) xY < 0

Could be x positive, y negative and z negative then xz = negative (E.g x = 10, y = -1, z = -10)
Or x negative, y negative and z negative then xz = positive (E.g x = -1, y = -2, z = -3)

Insufficient.

"Or x negative, y negative" negative * negative = positive // flaw.

Kudos [?]: 5 [0], given: 0

20 Mar 2006, 02:51
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# If x>y>z, is xz>0? 1) xy<0 2) yz>0

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