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

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If x < y < z, is xyz > 0? (1) xy > 0. (2) xz [#permalink]  09 Mar 2011, 14:01
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27% (02:26) correct 72% (00:58) wrong based on 11 sessions
If x < y < z, is xyz > 0?

(1) xy > 0.
(2) xz > 0.
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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. [#permalink]  09 Mar 2011, 14:35
Expert's post
banksy wrote:
254. If x < y < z, is xyz > 0?
(1) xy > 0.
(2) xz > 0.

If x < y < z, is xyz > 0?

(1) xy > 0 --> x and y have the same sign. Now, if both x and y are positive then 0<x<y<z so all three are positive thus xyz>0, but if both x and y are negative then z could be positive as well as negative thus xyz may or may not be positive. Not sufficient.

(2) xz > 0 --> x and z have the same sign and as x < y < z then all three have the same sign: if all of them are positive then xyz>0 but if all of them are negative then xyz<0. Not sufficient.

(1)+(2) It's still possible all three to be positive as well as negative. Not sufficient.

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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. [#permalink]  09 Mar 2011, 15:38
x,y,z need not be integers .

1. xy>0 , doesnt say anything about z. Not sufficient.

2. same thing here , we dont know anything about y . Not sufficient.

Together x>0 , y>0,z>0 xyz >0
x<0, y<0,z<0 xyz <0 , so not sufficient.

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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. [#permalink]  09 Mar 2011, 20:45
From (1) xy > 0 means both have same sign, hence if both are negative:

then xyz < 0 if z < 0, and xyz > 0 if z is positive

And if both are poitive, then xyz is always > 0. So (1) is not enough.

From (2) xz > 0, so here too both x and z have same sign, and hence y will also have same sign

But if x,y and z are negative, then xyz < 0 and if x,y and z are positive, then xyz > 0

From(1) and (2), x, y,z can be either all +ve or all -ve, so the answer is E.
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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. [#permalink]  18 Mar 2011, 04:30
This one is quite simple to be honest...think about it. The question is asking if either x, y or z are negative. If one of them is negative, then xyz = -ve. More importantly if z is negative then all three numbers are negative.
Statement 1 says xy>0 so they can be either positive or negative. INSUFF
Statement 2 says xz>0 ....same as above.
1+2 is the same deal. So E
Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0.   [#permalink] 18 Mar 2011, 04:30
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