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This one is strange, because if

If sqrt( 4a ) = integer

then

2*sqrt(a) must be integer -- and that is integer when sqrt(a) in integer
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I said the same thing than craky. Ans. D for me. But as previously mentioned the question is weirdly formulated.

Anyone cares to comment? :-D
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Don't know if I understand you guys but lets try with two numbers.

If a = 1.5^2 = 2.25
sqrt(4a) = sqrt(4 * 2.25) = sqrt ( 9 ) = 3 = integer
sqrt(a) = sqrt(2.25) = 1.5 = not integer

If a= 4
sqrt(4a) = sqrt(16) = 4 = integer
sqrt(4) = 2 = integer

The questions is: IF sqrt(4a) is an integer, will sqrt(a) be an integer as well? The converse is true but that is not the question here.
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craky
This one is strange, because if

If sqrt( 4a ) = integer

then

2*sqrt(a) must be integer -- and that is integer when sqrt(a) in integer

Not necessarily true. Even if \(\sqrt{a} = \frac{1}{2}, 2\sqrt{a} = 1\), an integer.

So if \(2\sqrt{a}\) is an integer, we cannot say whether \(\sqrt{a}\) is an integer.

Sneaky GMAT tricks!
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VeritasPrepKarishma,

Yes, but (1) tells us that A is a positive integer and therefore can't be 1/2 or 1/4? Therefore, wouldn't the answer be D?
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m990540
VeritasPrepKarishma,

Yes, but (1) tells us that A is a positive integer and therefore can't be 1/2 or 1/4? Therefore, wouldn't the answer be D?

Yes definitely. My reply was for the quoted comment above
'This one is strange, because if (I think he implied that you don't need the statements to answer the question)

If sqrt( 4a ) = integer

then

2*sqrt(a) must be integer and that is integer when sqrt(a) in integer


I had said that this is not true since 2*sqrt(a) can be an integer even if sqrt(a) is not. Implication: We need the statements to get to the answer, whatever the answer is. Think again and try to get the answer.
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rxs0005
If sqrt( 4a ) = integer is sqrt ( a ) an integer


(1) a is a positive integer
(2) a = n^6 where n is an integer


1. For \(sqrt(4a)\) to be an integer, a has to be a perfect square. therefore \(sqrt(a)\) will definitely be an integer. Sufficient
2. This statement says that a is a perfect square (of \(n^3\)). Sufficient.

Answer D.
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Wow very sneaky indeed.
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rxs0005
If \(\sqrt{4a}\) is an integer, is \(\sqrt{a}\) an integer

(1) a is a positive integer
(2) a = n^6, where n is an integer

if a= 1/4, then \sqrt{4a} = 1 (Integer), but \sqrt{a} is not an integer.

But if a is any +ve integer, then \sqrt{a} must be an integer in order to make \sqrt{4a} an integer.
\sqrt{4a} = 2 \sqrt{a}

(1) a is a positive integer. Sufficient (as mentioned above)

(2) a = n^6, where n is an integer. this means a is a +ve integer. Sufficient.

D is the answer
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Re: If sqrt(4a) is an integer, is sqrt (a) an integer

(1) a is a positive integer
(2) a = n^6, where n is an integer

Stem: Square root 4A = Integer, By squaring both sides 4A = (Integer)^2

(F1) A is a positive integer
4 * 1 = (Integer)^2 = (2)^2 Yes
4 *(2)^2 = (Integer)^2 Yes
Sufficient from stem it is implied that 4 * Any positive Integer is a perfect square.

(F2)a = n^6, where n is an integer
4 * (1)^6 = (2)^2 Yes
4 * (2)^6 = (8)^2 Yes
Sufficient

D
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