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m02 #20

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Re: m02 #20 [#permalink] New post 07 Dec 2012, 05:32
Yep. Got it.

I guess the funda here is hidden in the Q stem.
Does not matter if the answer is a negative or positive as long as it is an integer.
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Re: m02 #20 [#permalink] New post 20 Dec 2012, 14:19
the question boils down to P*sqrt(q) is an integer only when p is an integer and q is a perfect sq

Statement 1 - both are satisfied so sufficient
statemnt 2 not sufficient

Answer A
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Re: m02 #20 [#permalink] New post 16 Oct 2013, 06:55
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If we had to add a constraint that is p\frac{q}{\sqrt{q}} a positive integer ? (or negative integer for that matter), will answer still be the same?
coz Sqroot Q = +-P as per statement 1. isn't it?
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Re: m02 #20 [#permalink] New post 16 Oct 2013, 10:54
Expert's post
nikhil007 wrote:
If we had to add a constraint that is p\frac{q}{\sqrt{q}} a positive integer ? (or negative integer for that matter), will answer still be the same?
coz Sqroot Q = +-P as per statement 1. isn't it?


Yes and in this case the answer would be C.
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Re: m02 #20 [#permalink] New post 05 Dec 2013, 06:02
Bunuel wrote:

(1) q = p^2 --> p=\sqrt{q} --> p\frac{q}{\sqrt{q}}=\sqrt{q}*\frac{q}{\sqrt{q}}=q=integer. Sufficient.

Answer: A.


Bunuel, Isn't p\frac{q}{\sqrt{q}} a mixed number. Equivalent to p + \frac{q}{\sqrt{q}}
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Re: m02 #20 [#permalink] New post 05 Dec 2013, 06:19
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sathishkumar434 wrote:
Bunuel wrote:

(1) q = p^2 --> p=\sqrt{q} --> p\frac{q}{\sqrt{q}}=\sqrt{q}*\frac{q}{\sqrt{q}}=q=integer. Sufficient.

Answer: A.


Bunuel, Isn't p\frac{q}{\sqrt{q}} a mixed number. Equivalent to p + \frac{q}{\sqrt{q}}


Nope. It's p*\frac{q}{\sqrt{q}}.
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Re: m02 #20 [#permalink] New post 06 Dec 2013, 06:49
IMO A...

Statement 1 is sufficient since q is +ve (given) and p can be +ve or -ve, the expression would always be an integer.
Statement 2 is not sufficient since we do not know whether (sqrt)q is integer or non-integer. So the output of the expression could be integer or non-integer.

So, A.
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Re: m02 #20   [#permalink] 06 Dec 2013, 06:49
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