Is 2|xy^2| 0?

Question

Is 2|xy^2|> 0?

(1) x > 0

(2) xy > 0

Archit143 · Answer

Hi
I think the question is flawed....U need to check it again...I may be wrong, If i am than pls correct me!

Consider kudos if my post helps!!!!!

Archit

anish123ster · Answer

B..
so as to confirm that it equation is not equal to zero..

navkaran · Answer

B
2nd option is necessary
Just need to see if the expression is not 0.
1 alone cant assure that, but 2 alone can

Bluelagoon · Answer

The given condition will be true if both x and y are not zero. B clearly states that. Hence B.

manishuol · Answer

This Q . asks Is  2|xy^2|> 0 ? or  |xy^2|> 0 ?

Statement :: 1 --- >  x>0  .. Dont't Know the value of Y ,.... It can be integer or 0 as well. Therefore, Insufficient.

Statement :: 2--- >  xy>0  ........ This means that either of them has +ve signs or -ve signs. Because for both the cases the condition ... xy>0  must be true. & if we use these values in the Q, stem... The result must be same for both the signs. Therefore, Sufficient.

Hence ................ B .

Archit143 · Answer

missed the point that y could be equal to 0....Of course anopther vote for B

Archit

mau5 · Answer

The question is asking , IS a mod quantity >0? This will always be true, unless there is a possibility that this mod quantity CAN be zero.

From F.S 1, we know that only x>0. y can be 0, in which case we get a NO and y can be a non-zero number, in which case we get a YES. Insufficient.

From F.S 2, we know that xy>0. Thus, both x and y are non-zero numbers. Sufficient.

B.

AbuRashid · Answer

Answer is B. for S1, the left side cannot be less than zero but certainly it can be zero if y = 0.

Zarrolou · Answer

Good job guys!

The OA in indeed B.

|abs|  can be  \geq{0} , so to make sure that is  >  we need to enstablish that both x,y are  
eq{0}

B does just that

Bluelagoon · Answer

Absolute values are always greater than equal to zero. So, the question is asking if both x and y are not zero. 
1)No mention of y .
2)Both are positive or negative but not zero. That's what we want to know. SUFFICIENT.

krrish · Answer

Stmt 1 is not sufficient, 
consider stmt2: sy is positive so no matter what ever may be the value of x, its in mod so its positive :: sufficient. 
B. is the answer

Reaper · Answer

Not clear about B. If we open the modulus we arrive at different values for xy > 0 x < 0, y < 0 then xy^2 < 0 x > 0, y > 0 then xy^2 > 0

Zarrolou · Answer

No matter what sign  xy^2  has, when we apply the mod it will become positive.

There is only one case in which this does not happen, and it's the case where  xy=0  =>  |xy^2|=0 .
With option B we are sure that neither x nor y is zero, so the expression  2|xy^2|  will be positive.

There is no need to open the abs value, all we need to check is whether x and y are not zero numbers => B does that and it's sufficient.

Hope it's clear

tchang · Answer

Since there's an absolute value symbol in the question, the ONLY time the left side ISN'T > 0 is when it EQUALS 0

So, if x or y = 0, then the answer is NO
If neither x nor y = 0, then the answer is YES

1) x > 0

x is positive, but we don't know about y
If y = 0, then NO
If y = anything else, then YES
Insufficient

2) xy > 0

x and y are either both positive or both negative. Neither can be 0 so the answer is always YES
Sufficient

Choose  B

t1000

srinjoy1990 · Answer

I somehow don't find the question convincing.

2|xy^2|> 0

note that y^2 > 0 for any real value of y. We do not consider imaginary values in gmat. If so,
then,

2|xy^2|> 0 => |x| > 0. [Note that |x| > 0 for any value except 0, x belongs to {(-inf, inf) - 0}]

If so, then the answer cannot be B.

B states xy > 0 This means x > 0 and y >0 or x< 0 and y <0. If y <0 then x <0. So in any case |x| > 0 since |x| != 0.

Then inequality does not depend on the value of y in anyway. IMO D.

Mo2men · Answer

y^2 could equal to zero too. i.e. y = 0. Hence the answer is to question is No. If y is any number then answer would be yes.

srinjoy1990 · Answer

Missed it you are right!

bumpbot · Answer

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Is 2|xy^2|> 0?

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Is 2|xy^2|> 0?

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