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

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Manager
Joined: 10 Feb 2011
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If x < y < z, is xyz > 0? [#permalink]

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09 Mar 2011, 15:01
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69% (01:07) correct 31% (00:52) wrong based on 102 sessions

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

(1) xy > 0.
(2) xz > 0.
[Reveal] Spoiler: OA

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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. [#permalink]

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09 Mar 2011, 15:35
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 we would have that $$0<x<y<z$$, so in this case all three would be positive, which would mean that $$xyz>0$$, 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]

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09 Mar 2011, 16:38
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]

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09 Mar 2011, 21:45
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]

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18 Mar 2011, 05:30
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

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Re: If x < y < z, is xyz > 0? [#permalink]

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23 Feb 2014, 05:30
Bumping for review and further discussion.
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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. [#permalink]

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29 Apr 2014, 03:04
Bunuel wrote:
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 we would have that $$0<x<y<z$$, so in this case all three would be positive, which would mean that $$xyz>0$$, 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.

what is wrong with my approach

Taking 1 & 2
xy -xz > 0
x(y-z) > 0

either x > 0 or y > z
therefore x > 0 therefor y & z is also > 0

Hence xyz > 0 ,Sufficient.

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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. [#permalink]

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29 Apr 2014, 04:01
1
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Expert's post
abid1986 wrote:
Bunuel wrote:
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 we would have that $$0<x<y<z$$, so in this case all three would be positive, which would mean that $$xyz>0$$, 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.

what is wrong with my approach

Taking 1 & 2
xy -xz > 0
x(y-z) > 0

either x > 0 or y > z
therefore x > 0 therefor y & z is also > 0

Hence xyz > 0 ,Sufficient.

First of all you cannot subtract xz > 0 from xy > 0, because the signs of the inequalities are in the same direction, you can only add them.

Next, even if we had x(y-z) > 0, then it would mean that either both multiples are positive or both multiples are negative:

$$x>0$$ and $$y-z>0$$ ($$y>z$$).

$$x<0$$ and $$y-z<0$$ ($$y<z$$).

Hope it helps.
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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. [#permalink]

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29 Apr 2014, 04:13
1
KUDOS
abid1986 wrote:
Bunuel wrote:
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 we would have that $$0<x<y<z$$, so in this case all three would be positive, which would mean that $$xyz>0$$, 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.

what is wrong with my approach

Taking 1 & 2
xy -xz > 0
x(y-z) > 0

either x > 0 or y > z
therefore x > 0 therefor y & z is also > 0

Hence xyz > 0 ,Sufficient.

Note From Bunuel:

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

The 2 inequalities in question can be added but not subtracted.
For more on Inequalities refer: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html#p1242251

inequality-and-absolute-value-questions-from-my-collection-86939.html#p652806
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Re: If x < y < z, is xyz > 0? [#permalink]

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30 Apr 2014, 16:33
The way I approached this question is:

(1) xy > 0 - This tells us that xy are both positive or both negative but doesn't tell us anything about z; hence insufficient.

(2) xz > 0 - Same as (1) doesn't tell us anything about y.

Combined x,y,z might all be positive in which case xyz will be > 0 or all be negative is which case xyz < 0; hence insufficient.

Is this a correct approach or am I missing something?

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Re: If x < y < z, is xyz > 0? [#permalink]

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01 May 2014, 09:54
MarcusFenix wrote:
The way I approached this question is:

(1) xy > 0 - This tells us that xy are both positive or both negative but doesn't tell us anything about z; hence insufficient.

(2) xz > 0 - Same as (1) doesn't tell us anything about y.

Combined x,y,z might all be positive in which case xyz will be > 0 or all be negative is which case xyz < 0; hence insufficient.

Is this a correct approach or am I missing something?

Yes, your approach is correct. It's basically the same as in my post here: if-x-y-z-is-xyz-110615.html#p888498
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Re: If x < y < z, is xyz > 0? [#permalink]

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13 Mar 2016, 08:09
This Question is similar to the case of is PQR even
here we can make test cases for P,Q,R which shows us => PQR>0 or PQR <0
hence E
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Re: If x < y < z, is xyz > 0?   [#permalink] 13 Mar 2016, 08:09
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