ds with inequalities

I also think it's A.

But again, my DS hit rate is about 40%

ern5231 wrote:Is x^2+x> y^2?
(1)x^2+y^2=1
(2)x>0

OA is E but I feel it is C. When both conditions are taken into account the only possible values are :
x=1 and y=0 which can establish :

1^2+1>0^2 hence sufficient. Is my analogy correct?

Actually, you don't see the word "integer" in the problem, so just by considering the integer solutions of the problem you are missing some results. What if x^2 = 0.64 and y^2 = 0.36 or the other way around?

This is why the answer is indeed E.

I also think its C.

Somebody please explain..

Thanks,
V

Ohh!! thanks DanaJ..
I didnt read the q properly ...

In questions where the properties of the variables are not mentioned (integer, range, etc.), if you decide to pick numbers, use different scenarios like a) x is a positive integer b) x is a fraction between 0 and 1 c) x is negative and MAYBE even d )x is a negative fraction between 0 and -1.

I solved this algebraically:

in 1
x^2 + y^2 = 1
so y^2 = 1-x^2

plugging into question stem we get:

x^2+x> 1-x^2
rearranging all terms to 1 side we get:
2x^2+x-1 > 0
factoring we get:
(2x-1)(x+1)>0
now multiplication must either be +ve and +ve or -ve and -ve to be >0

so in +ve/+ve case we get
x>1/2 and x>-1 --> therefore x>1/2

in -ve/-ve case we get
x<1/2 and x<-1 --> therefore x<-1

so insuff

clearly 2 is insuff alone

combined x>0 can lie anywhere from 0 to 1/2 OR >1/2 so insuff as well

ans is E

Lets break it down.

Question: Is x^2 + x > y^2?

Statement 1:
x^2 + y^2 = 1
=> y^2 = 1 - x^2

Question: Is x^x + x > 1 - x^2 ?

x=positive integer (2), 2^2 + 2 > 1 - 2^2 = TRUE

x=positive proper fraction (1/3), 1/9 + 1/3 > 1 - 1/9 = FALSE

Hence, statement 1 is insufficient.

Statement 2: x>0
The statement doesn't tell us anything about y. Therefore, it is insufficient on its own.

Statement 1 + 2:
We already considered two cases for which x>0 and received inconsistent results. Hence, the answer is E.