This is the solution that I found. I do not find this explanation satisfactory.

Can any of the experts please explain the solution better? Thanks in advance

Let's start by simplifying the original by multiplying both sides by y² (we can safely do so because we know that y² is positive):
Is \(x^2y^3\) - \(y^3\) > 1?
(1) x < y
If x and y are big positive numbers, \(y^3\) will dominate x²y² and the left side will be negative; if x and y are big negative numbers, x²y² and (-\(y^3\)) will both be positive and the left side will be a big positive number: insufficient.
(2) x² < y²
Tells us that |x| < |y|, but nothing about the signs of x and y. If we pick really big values with different signs we can generate both "yes" and "no" answers.
Together: combining doesn't help us at all, since we have the same generalizations from both: choose E.

chetan2u please help with this question.
I guessed the answer right but not sure about the approach.

\(x^2\) - y > 1/\(y^2\)

\(x^2\) - y - 1/\(y^2\) > 0

Lets try 2 first
(2) x² < y² or y^2 > x^2,
9 > 4, y can be -3 or 3 and x can be -2 or 2
1/4 > 1/16, y can be 1/2 or -1/2 and x can be 1/4 or -1/4

4 +3 - 1/9 > 0, This answers a yes to the question
1/16 - 1/2 - 1/4 < 0, This answers a no to the question

from 1)
x < y or y>x, here y & x can be again a fraction, integer or a number
1/2 > 1/4 or 3 > 2
1/16 - 1/2 - 1/4 < 0, This answers a no to the question
4 -3 - 1/9 > 0, This answers a yes to the question

The same test cases can be used again to show that after combining the 2, we wont get an unique answer

E
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Hi arpitkansal,

What \(x^2<y^2\) means is \(|x|<|y|\)..
X<y and |x|<|y| when combined mean y>x and y>0..
By looking at the equation, \(x^2-y>\frac{1}{y^2}\), we should realise that \(\frac{1}{y^2}\) will be positive.
We have to see if x^2-y can be both negative and positive.
1) x=2 and y=3
So \(2^2-3>\frac{1}{3^2}....4-3>\frac{1}{9}...1>\frac{1}{9}....\)yes
2) x=1 and y=4
\(1^2-4>\frac{1}{4^2}.....1-4>\frac{1}{16}.....-3>\frac{1}{16}...\).no

Thus no definite answer..

E

Take (x,y) = (0,1) & (0,-1).

We get E.

Hi Rahul, be careful with numbers you pick up..
(o, -1) does not fit into statement 1, which is x<y.. as 0>-1, meaning x>y

Here's how I solved, feel free to critique!

Is \(x^{2}\) - y > \(\frac{1}{y^{2}}\)? If \(y\neq{0}\).

I simplified the expression as follows:

Because \(y^{2}\) is always positive I multiplied both sides by \(y^{2}\) to get \(x^{2}y^{2}\) - \(y^{3}\) > 1.

From here I factored out \(y^{2}\) to get \(y^{2}\)(\(x^{2}\) - y) > 1.

The question is now:

Is \(y^{2}\)(\(x^{2}\) - y) > 1? If Y \(x\neq{0}\).

Statement 1: x < y

I picked x as -1 and y as 1. Plugging these values into the question the result is 0 > 1.

I then picked x as -3 and y as -2. Plugging these values into the question the result is 44 > 1.

Clearly insufficient.

Statement 2: \(x^{3}\) < \(y^{3}\)

I picked x as -1 and y as 1. Since they are raised to an odd power the signs remain the same for x and y (1 and -1, respectively).
Plugging these values into the question the result is 0 > 1.

I then picked x as -3 and y as -2. Since they are raised to an odd power the signs remain the same for x and y (-3 and -2, respectively).
Plugging these values into the question the result is 44 > 1.

Clearly insufficient.

Statement 1&2

Clearly insufficient.

Therefore, the answer is E.
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(1) x < y Let x = -3, y =-2. Then we have (-3)^2 --2 > 1/(-2)^2 --> is 9 -2 > 1/4? Yes

Let x =-1/2 let y =-1/4 is (-1/2)^2 --1/4 > 1/(-1/4)^2 ? is 1/4 - 1/4 >4 NO NS

(2) x^2 <y^2 This is equivalent to |x| <|y| Let x =-2 y =-3

Is (-2)^2 --3 > 1/(-3)^2 ? is 4+3 > 1/9? Yes

let x = 3 y =4 is (3)^2 -4 > 1/4^2? is 9-4 > 1/16 No

NS

(1) and (2) x < y and |x|<|y| so values like x = -4, y =-3 won't work Let x=0 y =1 is 0^2 -1 >1/1^2 No

Let x =3, y=4 is 3^2 -4 > 1/4^2 ? is 9-4 >1/4^2 Yes

NS Answer is E.

Hey VeritasKarishma Bunuel, could this be solved algebraically?

\(x^2-y>1/y^2…x^2y^2-y^3>1?\)
\([I.]…x=2,y=3:(4)(9)-(27)>1?…36-27>1?…answer=yes\)
\([II.]…x=1,y=4:(1)(16)-(64)>1?…16-64>1?…answer=no\)

(1) x < y: case [I.] and [II.] are possible; insufic.
(2) x² < y²: case [I.] and [II.] are possible; insufic.
(1&2): case [I.] and [II.] are possible; insufic.

Answer (E)

You are still using number plugging (which is pretty much the way to go in this question).
Also, how come the two examples work for both statements? When x = -3 and y = -4, x > y, not x < y

x < y and x^2 < y^2 when x and y are both positive.
Try
x = 1, y = 2
x = 1/4, y = 1/2
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There I fixed it. Thanks
But can you solve this algebraically? Or is it a waste of time?

I wouldn't even try algebra here.
x < y and x^2 < y^2 should remind you of number properties. For what kind of numbers does this hold? For non negative numbers. We know that numbers behave differently 'between 0 and 1' and '1 onwards'. So I will try both type of numbers.
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