equation

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equation

by beater » Thu Mar 26, 2009 5:39 pm
x and y are integers such that x<y<0. What is x?

1. (x+y) (x-y) = 7
2. xy = 12

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by cramya » Thu Mar 26, 2009 5:54 pm
I would go wiht A


Stmt II

xy = 12

x=-4 y=-3
x=-6 y=-2

INSUFF

Stmt I

x^2-y^2 = 7


Only possible value is x=-4 and y=-3(x<y<0) since the distance between the difference of squares keeps on increasing and is unique between any 2 successive integer's perfect squares.This being the case the difference between the squares that are not successive increase even more.

SUFF

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by goelmohit2002 » Fri Mar 27, 2009 12:18 am
cramya wrote:I would go wiht A


Stmt II

xy = 12

x=-4 y=-3
x=-6 y=-2

INSUFF

Stmt I

x^2-y^2 = 7


Only possible value is x=-4 and y=-3(x<y<0) since the distance between the difference of squares keeps on increasing and is unique between any 2 successive integer's perfect squares.This being the case the difference between the squares that are not successive increase even more.

SUFF
Another approach to reach "A" may be

Let's take x + y = A
x - y = B.

Thus AB = 7

Now 7 = (-1) *(-7)
or
7 = 1 * 7

since both x and y are -ve so, second possibility is ruled out.

thus we have two equations and two variables....

x + y = -7
x - y = -1

Infact, IMO we even we do not even need to bother(since it is a DS question) whether

x + y = -7 or x + y = -1

just knowing that two variables and two equations should be sufficient to pick "A".

Thanks
Mohit