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Re: If XYZ>0 is X>0 [#permalink]
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DestinyChild wrote:
If XYZ>0 is X>0

XY > 0
XZ > 0

Obviously the answer is Yes or No;
Get your kudos for reasoning out why the answer is C and keep an eye on time too :)

Triple kudos for someone suggesting quicker technique for DS stems with triple variables...!


xy > 0 both are positive or both negative
xz > 0 both are positive or both negative

if x < 0 then z < 0 and y < 0 but xyz < 0 so this doesn't work
if x > 0 then y > 0 and z > 0 and xyz > 0 so this works and x > 0
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Re: If XYZ>0 is X>0 [#permalink]
cheers lagomez

will give you one...!
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Re: If XYZ>0 is X>0 [#permalink]
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1) x and y both are positive or negative, so x could be -ve, or +ve insuff.
2) x and z both are positive or negative, so x could be -ve, or +ve insuff.

For C
as xyz > 0, so if y and z both -ve, then x has to be +ve, and
if y and z become +ve then x has to be +ve to become xyz > 0.
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Re: If XYZ>0 is X>0 [#permalink]
C. For XYZ > 0 either all are positive or two of them negative. From St1 and St2 it is clear that both xy and xz has to be positive or both xz and xy has to be negative. in the latter case XYZ > 0 will be false.
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Re: If XYZ>0 is X>0 [#permalink]
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XYZ > 0 => X > 0 and YZ > 0 OR x < 0 and YZ < 0


(1)

XY > 0 means x > 0 and Y > 0 OR X < 0 and Y < 0

So (1) is not sufficient

(2)

XZ > 0 means X > 0 and Z > 0 OR X < 0 and Z < 0

So (2) is not sufficient

So if x < 0 then y < 0 and z < 0

=> xyz < 0 which is contrary to what is given in question

=> x > 0

Answer C.
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
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This one is pure fun!

xyz>0 this means that we need two number with same sign and the Notre number must be positive. otherwise we would get a negative number.

1-xy>0
Both x and y are positive or negative. NS

2-xz>0
Both x and y are positive or negative. NS

1 and 2 - xz>0 and xy>0 - SUFFICIENT
Lets assume that x is negative: if x is negative, then y must be negative and z must be negative. This would result in a negative number. This is not possible.
So we are sure that X is positive.
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If xyz > 0, is x > 0?

(1) xy > 0
(2) xz > 0


When you modify the original condition and the question, they become xyz>0 -> x>0? -> yz>0?. There are 3 variables(x,y,z) and 1 equation(xyz>0), which should match with the number of equations. So, you need 2 more equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer.
When 1) & 2), they become x(y^2)z>0. When dividing x(y^2)z>0 with y^2, it becomes yz>0, which is yes and sufficient.
Therefore, the answer is C.


l For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
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I solved this question "algebraically":

Question: xyz > 0 => x>0

(1) if xy > 0
dividing xyz > 0 by xy (inequality sign is not flipped)
=> z > 0 insufficient

(2) if xz > 0
dividing xyz > 0 by xz (inequality sign is not flipped)
=> y > 0 insufficient

(1&2) if z,y > 0
diving xyz > 0 by yz (inequality sign is not flipped)
=> x>0 sufficient

Is this a valid approach?
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
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benni089 wrote:
I solved this question "algebraically":

Question: xyz > 0 => x>0

(1) if xy > 0
dividing xyz > 0 by xy (inequality sign is not flipped)
=> z > 0 insufficient

(2) if xz > 0
dividing xyz > 0 by xz (inequality sign is not flipped)
=> y > 0 insufficient

(1&2) if z,y > 0
diving xyz > 0 by yz (inequality sign is not flipped)
=> x>0 sufficient

Is this a valid approach?


Yes, that's a valid approach.
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
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xyz > 0
x > 0?

Stem: yz ? 0

a) xz > 0 : Says nothing about y
b) xy > 0 : says nothing about z

Combining both L

a * b ( Since both are + ve)

x^2 * yz > 0

x^2 always positive. So yz > 0.

Answer C
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
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DestinyChild wrote:
If xyz > 0, is x > 0?

(1) xy > 0
(2) xz > 0


(1) xy > 0 --> If xy > 0, x can be both positive or negative but z has to be greater than 0
(2) xz > 0 --> If xz > 0, x can be both positive or negative but y has to be greater than 0

If z and y have to be greater than 0 then x has to be greater than 0 for xyz to be true.

Therefore, answer choice C
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
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DestinyChild wrote:
If xyz > 0, is x > 0?

(1) xy > 0
(2) xz > 0


If xyz > 0, is x > 0?
The only way for xyz to be positive all three are positive (x is +ve ; y is +ve and z is +ve)
Or Only two of them are negative and one is positive

(1) xy > 0
x can be negative ; y can be negative (or both x and y positive)
INSUFFICIENT

(2) xz > 0
x can be negative ; z can be negative (or both x and z positive)
INSUFFICIENT

MERGING BOTH
x, y, z are all positive {therefore x is positive}
x,y,z are all negative {therefore x is negative}; BUT ALL THREE CANNOT BE NEGATIVE; OTHERWISE INFO IN THE QUESTION STEM WILL BECOME INCORRECT. xyz>0
So, x y z can only be positive

ANSWER IS C
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
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xyz > 0 in two cases:
1. all the numbers are positive.
2. two of the numbers are negative and the other is positve.

st 1:
xy > 0
This statement is insufficient because x could be positive or negative to make it work.
It does give a useful piece of information though:
if x and y are both postive, xy > 0 and therefore z would have to also be positve to make xyz > 0.
or
if x and y are both negative, xy > 0 and therefore z would have to be positve to make xyz > 0.

Basically, the statement is saying that z must be positive.

st 2:
xz > 0
This statement is insufficient because x could be positive or negative to make it work.
It does give a useful piece of information though:
if x and z are both postive, xz > 0 and therefore y would have to also be positve to make xyz > 0.
or
if x and z are both negative, xz > 0 and therefore y would have to be positve to make xyz > 0.

Basically, the statement is saying that y must be positive.

Combined
From the two statements we know that z must be positive (st 1) and y must be postive (st 2).
If y and z are both positive, the only way that xyz will be > 0 is if x is positive or x > 0.

Therefore C.
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
Bunuel wrote:
benni089 wrote:
I solved this question "algebraically":

Question: xyz > 0 => x>0

(1) if xy > 0
dividing xyz > 0 by xy (inequality sign is not flipped)
=> z > 0 insufficient

(2) if xz > 0
dividing xyz > 0 by xz (inequality sign is not flipped)
=> y > 0 insufficient

(1&2) if z,y > 0
diving xyz > 0 by yz (inequality sign is not flipped)
=> x>0 sufficient

Is this a valid approach?


Yes, that's a valid approach.


Hi Bunuel,

I have a doubt over here (Sorry if it is lame but I somehow can't wrap my head around it)

We can only divide two inequalities if both sides(LHS and RHS) of the two inequalities are positive and signs of inequalities(if one has > other should have < and vice versa) are opposite.

With the above concept, how can we divide xyz>0 by xy>0 ?
i. The Inequality sign is not opposite in both the inequalities
ii. RHS in both the above inequalities is 0 (it's not positive)

I know somewhere I am wrong above, need your help to clarify my doubt. Thanks in advance
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
Expert Reply
pikolo2510 wrote:
Bunuel wrote:
benni089 wrote:
I solved this question "algebraically":

Question: xyz > 0 => x>0

(1) if xy > 0
dividing xyz > 0 by xy (inequality sign is not flipped)
=> z > 0 insufficient

(2) if xz > 0
dividing xyz > 0 by xz (inequality sign is not flipped)
=> y > 0 insufficient

(1&2) if z,y > 0
diving xyz > 0 by yz (inequality sign is not flipped)
=> x>0 sufficient

Is this a valid approach?


Yes, that's a valid approach.


Hi Bunuel,

I have a doubt over here (Sorry if it is lame but I somehow can't wrap my head around it)

We can only divide two inequalities if both sides(LHS and RHS) of the two inequalities are positive and signs of inequalities(if one has > other should have < and vice versa) are opposite.

With the above concept, how can we divide xyz>0 by xy>0 ?
i. The Inequality sign is not opposite in both the inequalities
ii. RHS in both the above inequalities is 0 (it's not positive)

I know somewhere I am wrong above, need your help to clarify my doubt. Thanks in advance



For (1) we are dividing xyz > 0 by xy (which we are given to be positive) to get z > 0.
For (2) we are dividing xyz > 0 by xz (which we are given to be positive) to get y > 0.
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
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