Is |x + 3| |2x + y| ? (1) x 0 (2) y 0

Question

Is |x + 3| > |2x + y| ?

(1) x > 0
(2) y > 0

chetan2u · Answer

We require both x and y to answer the question.

Even when both combined, there are no restrictions on the upper end of x and y, so E should be a straight forward answer.

Combined 
Both x+3 and 2x+y are positive, so the question becomes Is x+3>2x+y? Or Is 3>x+y?
If x is 100 and y is 200, then no. 
If x is 1 and y is 2, then yes. 
Insufficient

E

ManyataM · Answer

IMO the answer is option E

State 1 : tells about only x and no information about y .. insuff
State 2 : tells about y and no information about x ... insuffi

By combining also, we know they are greater than 0 but if both are 1 then it will hold a definite no and if they are diff it will hold either yes or no . insufficient on combination too

TarunKumar1234 · Answer

Is |x + 3| > |2x + y| ?

Stat1: x > 0
we don't know, if y > 0 or y < 0. Not sufficient

Stat2: y > 0
we don't know, if x > 0 or x < 0. Not sufficient

Combining both, x> 0 and y >0, which means, x + 3 > 2x + y or, x +y <3. It can be Yes or No both, as X and Y both are positive Not sufficient

So, I think E.

Bunuel · Answer

The statements alone are clearly insufficient because each gives info only about one of the variables.

When taken together: we know that both x and y are positive, thus both x+3 and 2x+y are positive, hence we can get rid of the modulus. The question becomes is x + 3 > 2x + y --> is x + y < 3. We cannot answer this question.

Answer: E.

Rohit5678 · Answer

Stmt 1 alone: Not Suff since no info on y. Eliminate A and D
Stmt 2 alone: Not Suff since no info on x. Eliminate B

Stmts 1&2 together: (x+3) & (2x+y) are positive. That’s all it provides. It gives various results for various inputs of (x,y). So its Not Suff (Eliminate C) and E is the answer.

nsdl1985 · Answer

How can we be sure that squaring both the sides is not needed here? 
Is squaring both sides and then solving a lengthy process? 
In which circumstances squaring both sides is needed?

Is |x + 3| > |2x + y| ? (1) x > 0 (2) y > 0

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