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Re: Is 2|xy^2|> 0? [#permalink]
05 May 2013, 19:35

3

This post received KUDOS

Zarrolou wrote:

Is \(2|xy^2|> 0\)?

A)\(x>0\)

B)\(xy>0\)

This is a DS I created, I'll post the OA and the OE after some discussion

If you want give it a try, Kudos to the first user(s) who got it right!

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 . _________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

Re: Is 2|xy^2|> 0? [#permalink]
08 May 2013, 12:37

Zarrolou wrote:

Is 2|xy^2|> 0?

(1) x > 0

(2) xy > 0

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.

Re: Is 2|xy^2|> 0? [#permalink]
16 May 2013, 08:22

Zarrolou wrote:

Is 2|xy^2|> 0?

(1) x > 0

(2) xy > 0

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

Re: Is 2|xy^2|> 0? [#permalink]
23 Jun 2013, 07:45

shivdeepmodi wrote:

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

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 _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Is 2|xy^2|> 0? [#permalink]
01 Nov 2014, 15:35

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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