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27 Jun 2017, 04:48
Kudos
Bookmarks
GMATinsight
Bunuel
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?
(1) |xy| is NOT a square of an integer
(2) Point (x, y) is above x-axis
SOLUTION (OE) IS HERE.
(1) |xy| is NOT a square of an integer
(2) Point (x, y) is above x-axis
SOLUTION (OE) IS HERE.
√(x^2-y^2) = (x+y)
Only If y=0
Each statement says that y#0
Hence sufficient
Answer option D
There is another possibility, when x = -y. For example, x = 1 and y = -1.
27 Jun 2017, 05:05
Kudos
Bookmarks
Bunuel
GMATinsight
Bunuel
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?
(1) |xy| is NOT a square of an integer
(2) Point (x, y) is above x-axis
SOLUTION (OE) IS HERE.
(1) |xy| is NOT a square of an integer
(2) Point (x, y) is above x-axis
SOLUTION (OE) IS HERE.
√(x^2-y^2) = (x+y)
Only If y=0
Each statement says that y#0
Hence sufficient
Answer option D
There is another possibility, when x = -y. For example, x = 1 and y = -1.
Good point...
+1Kudos 
27 Jun 2017, 05:09
Kudos
Bookmarks
Bunuel
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?
(1) |xy| is NOT a square of an integer
(2) Point (x, y) is above x-axis
SOLUTION (OE) IS HERE.
(1) |xy| is NOT a square of an integer
(2) Point (x, y) is above x-axis
SOLUTION (OE) IS HERE.
\(\sqrt{x^2 - y^2} = x + y\)
-> x^2-y^2 = (x+y)^2
-> (x+y)(x-y ) = (x+y)^2
(x+y )(y) = 0
-> Either y =0 or x = -y which we require .............................................................DS
Option 1 : |xy| is NOT a square of an integer
so x =/= y,-y,0 .. Since these values will make the value |xy| as a square of an integer.
Hence SUFFICIENT.
Option 2 : Point (x, y) is above x-axis, So y>0
Also y=/= 0,
Now since -x<=y <= x....
Since y>0, y =/=-x
Hence SUFFICIENT....
