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

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If x < y < z, is xyz > 0? [#permalink] 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] New post 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 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] 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] 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] 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: If x < y < z, is xyz > 0? [#permalink] New post 23 Feb 2014, 04:30
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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. [#permalink] 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] 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] 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] 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] New post 01 May 2014, 08:54
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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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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
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

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If x < y < z, is xyz > 0?   [#permalink] 01 May 2014, 08:54
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