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If xy = – 6, what is the value of xy(x + y)? (1) x – y = 5

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If xy = – 6, what is the value of xy(x + y)? (1) x – y = 5 [#permalink]

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New post 27 Apr 2012, 10:39
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If xy = – 6, what is the value of xy(x + y)?

(1) x – y = 5
(2) xy^2= 18
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Apr 2012, 10:43, edited 1 time in total.
Edited the question and added the OA
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Re: If xy = – 6, what is the value of xy(x + y)? (1) x – y = 5 [#permalink]

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New post 27 Apr 2012, 10:48
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If xy = – 6, what is the value of xy(x + y)?

(1) \(x-y=5\) --> \(x=y+5\) --> \(xy=(y+5)y=-6\) --> \(y^2+5y+6=0\) --> \(y=-3\) or \(y=-2\). If \(y=-3\) then \(x=y+5=2\) so \(xy(x+y)=-6(x+y)=6\). But if \(y=-2\) then \(x=y+5=3\) so \(xy(x+y)=-6(x+y)=-6\). Two different answers, hence not sufficient.

(2) \(x*y^2=18\) --> \(xy*y=-6*y=18\) --> \(y=-3\) --> \(x=2\) --> \(xy(x+y)=-6(x+y)=6\). Sufficient.

Answer: B.
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Re: If xy = – 6, what is the value of xy(x + y)? (1) x – y = 5 [#permalink]

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New post 27 Apr 2012, 10:52
Bunuel wrote:
If xy = – 6, what is the value of xy(x + y)?

(1) \(x-y=5\) --> \(x=y+5\) --> \(xy=(y+5)y=-6\) --> \(y^2+5y+6=0\) --> \(y=-3\) or \(y=-2\). If \(y=-3\) then \(x=y+5=2\) so \(xy(x+y)=-6(x+y)=6\). But if \(y=-2\) then \(x=y+5=3\) so \(xy(x+y)=-6(x+y)=-6\). Two different answers, hence not sufficient.

(2) \(x*y^2=18\) --> \(xy*y=-6*y=18\) --> \(y=-3\) --> \(x=2\) --> \(xy(x+y)=-6(x+y)=6\). Sufficient.

Answer: B.


B is OA, thank you!!!
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Re: If xy = -6 , what is the value of xy(x+y ) [#permalink]

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New post 14 Feb 2013, 09:39
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roopika2990 wrote:
If xy = -6 , what is the value of xy(x+y ) ?
(1) x – y = 5
(2)xy^2 = 48


xy=-6 ----> x=-6/y

S1) x-y=5 ---------------> x=y+5 -------------> -6/y=y+5 --------> y^2 + 5y + 6 = 0 -------> y = -3 or -2 two values for y so not sufficiesnt

S2) xy^2 = 48 -------> We know x=-6/y -------> therefore (-6/y)(y^2)=48 -------> -6y = 48 -------> y = -8 and x = 6/8 so now we can calculate the value of xy(x+y) hence statement is sufficient

Regards,

Abhijit
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Re: If xy = -6 , what is the value of xy(x+y ) [#permalink]

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New post 14 Feb 2013, 11:36
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\((x+y)^2 = (x-y)^2+4xy\)

= 5^2+4(-6) = 25-24 = 1

From F.S 1, we know that (x+y)^2 will have two values 1 and -1. Hence, insufficient.

But, from F.S 2, we know that y will have only one value. Thus sufficient. No need to solve for either of the options.

B.
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Re: If xy = – 6, what is the value of xy(x + y)? (1) x – y = 5 [#permalink]

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Re: If xy = – 6, what is the value of xy(x + y)? (1) x – y = 5 [#permalink]

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New post 09 Feb 2016, 19:05
Looking for advice on this question.

I got this question right but could easily have gotten it wrong. For stmt II I figured out that I could plug in -6 for xy and then calculate the equation of -6y=18, which allowed me to solve for y. However for stmt I I eliminated it because although I recognized that from the stmt I could derive that both x and y are not decimals, one is positive and one is negative, and either x or y could equal -6 and the other 1 - I was suspicious that I could count on those values so I eliminated it.

Looking at the OA I see that the proper approach would have been to use factorization.

My question is - how was I supposed to intuitively know to use factorization with stmt I?

I am decent with algebra when I know the typical rules and common properties... Trouble is I've taught myself algebra from scratch using various gmat guides and so my knowledge is spotty. If anyone has any recommendations for any bare bones algebra books I'd appreciate it. I'm rewriting in 1.5 weeks.

Right now I'm trying to improve my algebra through drilling and examining my right and wrong answers.
Re: If xy = – 6, what is the value of xy(x + y)? (1) x – y = 5   [#permalink] 09 Feb 2016, 19:05
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If xy = – 6, what is the value of xy(x + y)? (1) x – y = 5

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