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Is x>y^2? (1) x>y+5 (2) x^2-y^2=0

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Is x>y^2? (1) x>y+5 (2) x^2-y^2=0 [#permalink] New post 11 Feb 2009, 03:30
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Is x>y^2?

(1) x>y+5
(2) x^2-y^2=0
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Re: DS [#permalink] New post 11 Feb 2009, 03:48
ritula wrote:
Is x>y^2?

(1) x>y+5
(2) x^2-y^2=0



1)
y=-15 x= -5 x>y^2? No
y=-5 x = 1 x>y^2? Yes
Not sufficient
2)

x=-y= 2 2>4 No
x=-y=1/2 1/2>1/4 Yes

Not sufficient

combined


x>y+5 --> x-y>5

x^2-Y^2 = (x-y)(x+y)=0

x-y>5 --> x=-y
-> 2x>5 -->x>2.5

clearly x>y^2 ---> No for all values.

Sufficient

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Re: DS [#permalink] New post 11 Feb 2009, 12:14
IMO C.

From clue 1 -

x > y + 5 ==> x - y > 5

Pick some intelligent numbers that satisfy this clue (In this case I try to pick the (x, y) combinations as (+,+), (+,-), (-,+), (-,-). Even if any two combinations result in the Question stem with two different ans, I ignore the clue. If all these combinations result in only one ans, then I check for the fractions between (-1,0) and (0,1) for x and y.

(x,y) = (7,1). Satifies clue 1. Substitue back in x > y^2 ==. 7 > 1 Yes.
(x,y) = (-1,-7). Satifies clue 1.Substitue back in x > y^2 ==. 1 > 49 No.

Two different ans. Hence the clue is insufficient.

From clue 2 -

x^2 - y^2 = 0 ==> x^2 = y^2. Coming back to the the question stem, x > y^2, it is same as "Is x > x^2". for any value of x where x< 0 or x > 1, x is always less than x^2. ie x > y^2 is No. But if we take a number between 0 and 1 for ex, x = 0.5, x > x^2 is true. Again we got two solutions. Hence the clue is insufficient.

Combine both the clues -

x - y > 5
x^2 - y^2 = 0 ==> (x+y)(x-y) = 0 From clue 1 we know that x-y <> 0 ==> x+y = 0 ==. y = -x.
Now substitute y back in x-y>5 ==> x-(-x) > 5 ==. 2x > 5 ==> x > 5/2 ==> x > 1.

We already know for x > 1, x <= x^2. Hence the question "Is x > y^2" can be answered as No. Sufficient.
Re: DS   [#permalink] 11 Feb 2009, 12:14
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