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I got a different answer,
K=x^2 * y^2
if x is 3 and y is 2, then we have k = 36, which means it's a square of 6.
pick another pair, 3 and 1 for example, and you get k = sqe. root of 10
hence, it's not A

B works for obvious reasons,
hence the asnwer i got is B

I got a different answer,
K=x^2 * y^2
if x is 3 and y is 2, then we have k = 36, which means it's a square of 6.
pick another pair, 3 and 1 for example, and you get k = sqe. root of 10 - Not correct - it is 9 hence, it's not A

B works for obvious reasons,
hence the asnwer i got is B

If x and y are integers, is k the square of an integer?

(1) k = (x^2)(y^2) (2) sqrt(k) = 4

(1) is not sufficient because the only way to know for sure that k is a perfect square is if x^2=y^2. Since this information cannot be derived by the statement then there is no way of knowing that k is a perfect square.

(2) sqrt(k) = 4
square both sides...k = 16,
Statement (2) is sufficient.

If x and y are integers, is k the square of an integer?

(1) k = (x^2)(y^2) (2) sqrt(k) = 4

(1) is not sufficient because the only way to know for sure that k is a perfect square is if x^2=y^2. Since this information cannot be derived by the statement then there is no way of knowing that k is a perfect square.

(2) sqrt(k) = 4 square both sides...k = 16, Statement (2) is sufficient.

The answer is B

Ignore this...I am very very wrong. Sorry. As usual Halle is right.

lastochka wrote:

I got a different answer, K=x^2 * y^2 if x is 3 and y is 2, then we have k = 36, which means it's a square of 6. pick another pair, 3 and 1 for example, and you get k = sqe. root of 10 hence, it's not A

B works for obvious reasons, hence the asnwer i got is B

3 and 1 does work...k = 3^2 * 1^1 = 9 * 1 = 9 NOT 9 +1 = 10

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