BTGmoderatorLU wrote:Is |x+3| > |2x+y|?
(1) x > 0
(2) y > 0
The OA is E.
Let us say that:
x = 2 then 5 > |4 + y| then y can be 0.1 or y can be 10. We will get a Yes and a No.
Hence, E is the correct answer.
Has anyone another strategic approach to solve this DS question? Regards!
We are given that |x+3| > |2x+y|
=> x + 3 > 2x + y => x + y < 3 ---(1)
or
=> x + 3 < -(2x + y) => x + 3 < -2x - y => 3x + y < -3 ---(2)
If both the inequalities are either satisfied or not satisfied, then we have the unique answer; however, if there is a contradiction between (1) and (2), then we do not have a unique answer.
Let's take each statement one by one.
(1) x > 0
Clearly insufficient as we do not know the value of y.
(2) y > 0
Clearly insufficient as we do not know the value of x.
(1) and (2) together
Inequality (1): x + y < 3
Case 1: Say x = y = 1, then x + y = 2; thus, 2 < 3. The answer is Yes.
Case 2: Say x = y = 2, then x + y = 4; thus, 4 > 3. The answer is No. Insufficient.
The correct answer:
E
Hope this helps!
-Jay
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