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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