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Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0
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12 Feb 2016, 05:57
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.