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If xy<0, is x^2y<0?

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If xy<0, is x^2y<0?  [#permalink]

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New post 25 Jul 2016, 05:55
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If xy<0, is (x^2)y<0?
1) x>0
2) (x^3)y<0

*An answer will be posted in 2 days.

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Re: If xy<0, is x^2y<0?  [#permalink]

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New post 25 Jul 2016, 07:11
1
1) x>0
It gives that y should be <0 to have xy<0. So statement (x^2)y<0 will be true.

Sufficient!

2) (x^3)y<0
Either x<0 or y<0.

Insufficient!

Answer will be A.

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If xy<0, is x^2y<0?  [#permalink]

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New post 25 Jul 2016, 09:01
MathRevolution wrote:
If xy<0, is (x^2)y<0?
1) x>0
2) (x^3)y<0

*An answer will be posted in 2 days.



Given xy<0

Then x or y can +ve or -ve to be xy < 0 ..Finally asking to find (x^2)y<0 ? Yes or No question.

Stat 1: x > 0..i.e. X has to be +ve then y has to be -ve only then xy < 0.

(x^2)y<0 ..now sub x and y value in this equation.... +ve^2 * -ve will be < 0 ...Sufficient..

Stat 2: (x^3)y<0...Here we can take x or y can be +ve or -ve....multiple sets...no unique value...Insufficient...

Option A is correct answer.
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Re: If xy<0, is x^2y<0?  [#permalink]

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New post 25 Jul 2016, 15:54
1
Top Contributor
MathRevolution wrote:
If xy < 0, is x²y < 0?

1) x > 0
2) x³y < 0



Target question: Is x²y < 0 ?

Given: xy < 0
If the product xy is NEGATIVE, then there are only 2 possibilities
Possibility #1: x is positive and y is negative
Possibility #2: x is negative and y is positive


Statement 1: x > 0
If x is positive, then we can eliminate Possibility #2, leaving only Possibility #1
This means that x is POSITIVE and y is NEGATIVE
So, x²y = (POSITIVE²)(NEGATIVE) = NEGATIVE
In other words, we can conclude that x²y < 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x³y < 0
What can we conclude from this?
Well, we can conclude that EITHER possibility #1 is true, OR EITHER possibility #2 is true

If possibility #1 is true, then x is POSITIVE and y is NEGATIVE, which means x²y = (POSITIVE²)(NEGATIVE) = NEGATIVE. So, x²y < 0
If possibility #2 is true, then x is NEGATIVE and y is POSITIVE, which means x²y = (NEGATIVE²)(POSITIVE) = POSITIVE. So, x²y > 0

Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer =

Cheers,
Brent
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Re: If xy<0, is x^2y<0?  [#permalink]

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New post 26 Jul 2016, 18:23
Answer is A

- Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
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Re: If xy<0, is x^2y<0?  [#permalink]

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Re: If xy<0, is x^2y<0?   [#permalink] 16 Feb 2019, 05:24
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