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Is xy > 0?

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Is xy > 0? [#permalink]

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New post 14 Aug 2017, 04:52
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Re: Is xy > 0? [#permalink]

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New post 14 Aug 2017, 05:02
(1) x is positive not sufficient
(2) y is negative not sufficient
Combine sufficient
C


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Re: Is xy > 0? [#permalink]

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New post 16 Aug 2017, 03:22
Stmnt 1) x= 5,2
Stmnt 2) y = -4,-1

So xyis always <0
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Re: Is xy > 0? [#permalink]

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New post 16 Aug 2017, 05:57
From statements 1 and 2; x=5 or x=2 and y=-1 or y=-4; xy has to be negative
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Re: Is xy > 0? [#permalink]

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New post 16 Aug 2017, 07:16
Statement 1: Quadratic equation with solutions: x= 5 or x=2
we do not know anything about y
Not sufficient

Statement 2: Quadratic equation with solutions: y= -1 or y=-4
we do not know anything about X
Not sufficient

Combining both we get X always +ve and Y always -ve

Therefore C
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Re: Is xy > 0? [#permalink]

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New post 16 Aug 2017, 18:12
answer is C,
statement 1 we got x is positive
statement 2 we got y is negative
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Re: Is xy > 0? [#permalink]

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New post 16 Aug 2017, 19:29
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St1- X is always positive. No Info about y. Eliminate.
St2- Y is always negative. No Info about x. Eliminate.

1+2- XY is always negative. Sufficient. Answer is C.
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Re: Is xy > 0? [#permalink]

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New post 24 Aug 2017, 12:40
Bunuel wrote:
Is xy > 0?

(1) x^2 – 7x + 10= 0
(2) y^2 + 5y + 4 = 0


We need to determine whether xy > 0. If xy > 0, either both x and y are positive or both are negative.

Statement One Alone:

x^2 – 7x + 10 = 0

x^2 – 7x + 10 = 0

(x - 5)(x - 2) = 0

x = 5 or x = 2

Since we know nothing about y, statement one alone is not sufficient to answer the question.

Statement Two Alone:

y^2 + 5y + 4 = 0

y^2 + 5y + 4 = 0

(y + 4)(y + 1) = 0

y = -4 or y = -1

Since we know nothing about x, statement two alone is not sufficient to answer the question.

Statements One and Two Together:

Using statements one and two, we know that x = 5 or x = 2 and y = -4 or y = -1. Thus, xy will always be less than zero.

Answer: C
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Re: Is xy > 0?   [#permalink] 24 Aug 2017, 12:40
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