If x < y < z, is xyz > 0? : GMAT Data Sufficiency (DS)
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If x < y < z, is xyz > 0?

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

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New post 09 Mar 2011, 14:01
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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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New post 09 Mar 2011, 14:35
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.

Answer: 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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New post 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.


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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New post 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]

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

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New post 29 Apr 2014, 02:04
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.

Answer: E.



what is wrong with my approach

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

either x > 0 or y > z
y > z is discarded because it contradicts the stem
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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New post 29 Apr 2014, 03:01
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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.

Answer: E.



what is wrong with my approach

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


either x > 0 or y > z
y > z is discarded because it contradicts the stem
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.

Adding/subtracting/multiplying/dividing inequalities: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html

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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New post 29 Apr 2014, 03:13
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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.

Answer: E.



what is wrong with my approach

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

either x > 0 or y > z
y > z is discarded because it contradicts the stem
therefore x > 0 therefor y & z is also > 0

Hence xyz > 0 ,Sufficient.


Note From Bunuel:

ADDING/SUBTRACTING INEQUALITIES:


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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New post 30 Apr 2014, 15:33
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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New post 01 May 2014, 08:54
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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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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New post 13 Mar 2016, 07:09
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, 07:09
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