=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
The question may be modified as follows:
\(\frac{1}{x} >\frac{1}{y}\)
=> \(xy^2 > x^2y\) by multiplication by \(x^2y^2\)
=> \(xy^2 - x^2y > 0\)
=> \(xy(y-x) > 0\)
Since we have 2 variables (\(x\) and \(y\)) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Applying both conditions together yields \(y > x > 0\). So, \(x>0, y>0\) and \(y-x>0.\)
It follows that the product \(xy(y – x)\) is positive. Therefore, both conditions are sufficient when considered together.
Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
If \(x = 2\) and \(y = 3\), then \(\frac{1}{x} = \frac{1}{2}\), \(\frac{1}{y} = \frac{1}{3}\), and the answer is ‘yes’.
If \(x = -2\) and\(y = 3\), then \(\frac{1}{x} = -\frac{1}{2}\), \(\frac{1}{y} = \frac{1}{3}\), and the answer is ‘no’.
Condition 1) is not sufficient on its own.
Condition 2)
This condition provides us with no information about the variable y, so it is not sufficient.
Therefore, the answer is C.
Answer: C
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provides an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
_________________