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If x and y are integers such that x < y < 0, what is x

Expert replies
by zenithexe » Thu Jul 02, 2009 11:47 pm
If x and y are integers such that x < y < 0, what is x – y?

(1) (x + y)(x – y) = 5
(2) xy = 6


I have no idea how to solve this one, any help is appreciated. Thanks
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Source: — Data Sufficiency |

by tohellandback » Fri Jul 03, 2009 12:12 am
IMO A
1)( x + y)(x – y) = 5
possibilities:
(x+y)=1 ,(x-y)=5- rejected because x and y are negative
(x+y)=5 ,(x-y)=1, rejected for the same reason

(x+y)=-1 ,(x-y)=-5, if x+y=-1 x=-y-1, now x has to be negative. for any value of y<0, it is impossible. so rejected
(x+y)=-5 ,(x-y)=-1. SUFFICIENT

2)xy=6. x and y have to be negative. so one of them is -2 and other is -3 but we need to know which one is what. INSUFF
Last edited by tohellandback on Fri Jul 03, 2009 12:25 am, edited 1 time in total.
The powers of two are bloody impolite!!
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by zenithexe » Fri Jul 03, 2009 12:22 am
wow, never thought I'd see a guy/gal from Tokyo on this site. Its nice to know there is someone else studying for Gmat in Japan.

Thanks for the solution, I get it now. :D
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by tohellandback » Fri Jul 03, 2009 12:24 am
never thought I'd see a guy/gal from Tokyo on this site
umm, why not?? :D
The powers of two are bloody impolite!!
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by zenithexe » Fri Jul 03, 2009 12:29 am
Coz I did search for Tokyo (or was it Japan) on "Let's meet up" section and I found none.

I guess I should pm for this kind of stuff, it's getting off the topic.
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by benissleeping » Fri Jul 03, 2009 12:34 am
This was my thinking on the Q:

(1) Multiply out x^2 - y^2 = 5
Just by looking at this you can see it solves for x=-4 and y=-3 (16-9=5)
and x=-3 and y=-2 (9-4=5)

x-y = -1

(2) xy=6
x could be -6 and y -1 or x -3 and y -2. NS

So A
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by zenithexe » Fri Jul 03, 2009 12:54 am
benissleeping wrote:This was my thinking on the Q:

(1) Multiply out x^2 - y^2 = 5
Just by looking at this you can see it solves for x=-4 and y=-3 (16-9=5)
and x=-3 and y=-2 (9-4=5)
Thanks!
just out of curiosity how do you quickly come up with the numbers (X^2=16, Y^2=9)? and it there a way to be sure that those numbers are only numbers capable of satisfying the clause? I wish I could do that.
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by tryin700 » Fri Jul 03, 2009 1:10 am
Is it A

Given x<y<1

1) (x + y)(x – y) = 5

RHS = +5

so 2 combination are possible

-5X-1 or -1X-5

this means x=-3 and y=-2

SUFF

2) xy=6
possible cases -1 and -6 or -3 or -2

INSUFF

Hence A
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by ogbeni » Thu Jul 09, 2009 9:55 am
Why is statement 1 sufficient when (X-Y) could be either -1 or -5? I eliminated statement 1 based on those 2 possibilities.

Look forward to getting an explanation! Thnx
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by matanga » Thu Jul 09, 2009 1:53 pm
ogbeni wrote:Why is statement 1 sufficient when (X-Y) could be either -1 or -5? I eliminated statement 1 based on those 2 possibilities.

Look forward to getting an explanation! Thnx
Because you have 2 negative numbers, and, since y>x, when you subtract y from x, you are technically adding a value to x to make it closer to 0. Had the question not stated y was negative, then your logic would have been correct.


Hope this helps !
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by ogbeni » Thu Jul 09, 2009 2:13 pm
Oh i get it now. The fact that x<y<0 limits the possibilities of x and y to x = -3 and y = -2.

My goodness!! With the time pressure on the GMAT and without considering values for x and y, I just reasoned that the values of (x+y) x (x-y) could be -5,-1 or -1, -5 therefore eliminate - ARRGHHHH I hate that!!!!!!! lol

Thanks for clarifying
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