Senior Manager
Joined: 30 Nov 2008
Posts: 495
Schools: Fuqua
Followers: 10
Kudos [?]:
114
[0], given: 15

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)(xy) = 0 From clue 1 we know that xy <> 0 ==> x+y = 0 ==. y = x. Now substitute y back in xy>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.
