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# If x and y are integers, is k the square of an integer? (1)

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Manager
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If x and y are integers, is k the square of an integer? (1) [#permalink]  22 May 2004, 07:04
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If x and y are integers, is k the square of an integer?

(1) k = (x^2)(y^2)
(2) sqrt(k) = 4
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Manager
Joined: 10 Mar 2004
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Clearly each individual so (d)
Manager
Joined: 16 May 2004
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I agree with you but the answer key I have is "B"........
Maybe, the key is incorrect??
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Senior Manager
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k = (xy)^2 or (-xy)^2

The indefinite article an makes the answer D.
Senior Manager
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hallelujah1234 wrote:
k = (xy)^2 or (-xy)^2

The indefinite article an makes the answer D.

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
Manager
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hallelujah1234 wrote:
k = (xy)^2 or (-xy)^2

The indefinite article an makes the answer D.

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
Manager
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Re: DS_3_17 [#permalink]  28 May 2004, 13:12
becoolja wrote:
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.

Manager
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Re: DS_3_17 [#permalink]  28 May 2004, 13:18
sofere wrote:
becoolja wrote:
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.

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

lastochka wrote:
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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For (1): what if x=0? Could it be the factor that points that A is incorrect? Just asking....
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