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Kritisood
If x and y are non-zero integers, Is x < y ?
(1) \(x = \frac{1}{y^{2}}\)
(2) x + y = 0

Question: Is x < y?

Statement 1: \(x = 1/y^2\)

for any negative value of y, x will be positive hence x > y i.e. x is NOT less than y

For positive integer values of y
at y = 1, x = 1 i.e. x is NOT less than y
at y = 2, x = 1/4 i.e. x is NOT less than y

CONSISTENT Answer hence

SUFFICIENT

Statement 2: x + y = 0

i.e. one of x and y is [positive and other negative but which one is what is unknown hence

NOT SUFFICIENT

Answer: Option A
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Kritisood
If x and y are non-zero integers, Is x < y ?

For positive integer values of y
at y = 1, x = 1 i.e. x is NOT less than y
at y = 2, x = 1/4 i.e. x is NOT less than y

In your second example, x is not an integer. That example shouldn't be considered.
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Kritisood
If x and y are non-zero integers, Is x < y ?
(1) \(x = \frac{1}{y^{2}}\)
(2) x + y = 0

x and y are nonzero integers.

We need to answer the question:

Is x < y ?

Statement One Alone:

=> x = 1/y^2

(x)(y^2) = 1

The product of two integers is equal to 1 if and only if both integers are equal to 1 or both are equal to -1.

Since the integer y^2 cannot be negative, it must be true that:

x = 1

y^2 = 1 => y = 1 OR y = -1

If x = 1 and y = 1, then the answer is No. If x = 1 and y = -1, then the answer is still No. Since we have a definite No answer to the question, statement one is sufficient. Eliminate answer choices B, C, and E.

Statement Two Alone:

=> x + y = 0

If x = 1 and y = -1, then the answer is No. Whereas, if x = -1 and y = 1, then the answer is Yes. Statement two is not sufficient.

Answer: A
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