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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.
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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Since we have 2 variables (x and y) and 0 equations, C is most likely 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.
Assume a = x – y and b = y + 3.
When we try to check both conditions together, we realize condition 1) alone is sufficient because condition 1) tells us 2x – y = -3 for the following reason.
(x-y)(y+3)/4(x-y)^2+(y+3)^2= - 1/4
⇔ ab/4a^2+b^2=-1/4 (a = (x – y) and b = (y + 3))
⇔ 4a^2+b^2=-4ab (cross multiplying)
⇔ 4a^2+4ab+b^2=0 (adding 4ab to both sides)
⇔(2a+b)^2=0 (factoring)
⇔2a+b=0
⇔ 2(x - y) + (y + 3) = 0 (substituting (x – y) and (y + 3) back in)
⇔ 2x – 2y + y + 3 = 0 (multiplying 2 through the first bracket)
⇔ 2x – y + 3 = 0 (adding like terms)
Thus, we have 2x – y = -3 from condition 1) alone.
Condition 2)
Condition 2) is not sufficient, obviously, since condition 2) does not yield a unique solution.
Therefore, A is the answer.
Answer: A
This question is an application of CMT 4(B): condition 2) is easy to work with, and condition 2) is difficult to work with. For CMT 4(B) questions, we may assume condition 1) is sufficient. Since we can figure out condition 2) is not sufficient, we should be able to choose A.
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) provide 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.
