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Re: If XYZ>0 is X>0 [#permalink]
11 Apr 2011, 08:51
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. _________________
Re: If XYZ>0 is X>0 [#permalink]
11 Apr 2011, 10:54
2
This post received KUDOS
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Following chart lists down the possible scenarios in which XYZ > 0
X Y Z 1) - - + 2) - + - 3) + - - 4) + + +
As per statement I --> 1 & 4 have to be true for XY > 0. So I alone is insufficient as X can be +ve or -ve As per statement II --> 2& 4 have to be true for XZ > 0 So II alone is insufficient as X can be +ve or -ve If we combine both statements then only 4 is true and X has to be +ve
Re: If XYZ>0 is X>0 [#permalink]
11 Apr 2011, 12:43
2
This post received KUDOS
Expert's post
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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...!
Certainly each statement is not sufficient alone (the logic is identical for each: for S1, x and y can both be negative or both positive, and for S2, x and z can both be negative or both positive). You might notice that S1 tells us z is positive, and S2 tells us y is positive, but that's not what we're looking for.
Now, together, the inequality xy > 0 means 'x and y have the same sign'. So from the two statements we know x and y have the same sign, and x and z have the same sign, so x, y and z all have the same sign. If xyz > 0, they must all be positive, and the answer is C. _________________
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Re: If XYZ>0 is X>0 [#permalink]
11 Apr 2011, 17:41
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.
Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
18 Nov 2013, 07:29
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
26 Dec 2014, 09:21
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
09 Jan 2016, 20:53
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
11 Feb 2016, 09:38
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.
Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]
12 Feb 2016, 04:57
Expert's post
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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