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 x#-y is (x-y)/(x+y)>1?
(1) x>0
(2) y<0
If we modify the original condition, as squared numbers are always positive, the inequality sign does not change even if we multiply certain positive integers.
So if we multiply (x+y)^2, (x-y)(x+y)>(x+y)^2, --> (x-y)(x+y)-(x+y)^2>0, (x+y)(x-y-x-y)>0, (x+y)(-2y)>0, and if we divide both sides by -2,
the inequality sign changes, and ultimately we want to know whether (x+y)y<0.
There are 2 variables, so we need 2 equations, which are provided by the 2 conditions. If we look at the conditions together, the answer to what we want to know becomes 'yes' for x=2, y=-1, but 'no' for x=2,y=-3. Hence, the conditions are insufficient and the answer becomes (E).
For cases where we need 2 more equation, 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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