x and y are integers such that x<y<0. What is x?
1. (x+y) (x-y) = 7
2. xy = 12
equation
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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
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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Another approach to reach "A" may becramya 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
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