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If \(q\) is a positive integer, is \(p\frac{q}{\sqrt{q}}\) an integer? 1. \(q = p^2\) 2. \(p\) is a positive integer Source: GMAT Club Tests  hardest GMAT questions note: I was not aware how to post the mathematical expression but basically its a mixed fraction which I translated to math equation. I believe either the question or explanation might have a typo.



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Re: m02 #20 [#permalink]
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04 Dec 2010, 05:49
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A.
s1 : since p^2= q it implies that p is an integer. If p was not an iteger it's square can never be an integer
s2: does not provide any extra info
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Re: m02 #20 [#permalink]
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07 Dec 2012, 06:14
ConkergMat wrote: If \(q\) is a positive integer, is \(p\frac{q}{\sqrt{q}}\) an integer? 1. \(q = p^2\) 2. \(p\) is a positive integer Source: GMAT Club Tests  hardest GMAT questions note: I was not aware how to post the mathematical expression but basically its a mixed fraction which I translated to math equation. I believe either the question or explanation might have a typo. If \(q\) is a positive integer, is \(p\frac{q}{\sqrt{q}}\) an integer?(1) \(q = p^2\) > \(p=\sqrt{q}\) > \(p\frac{q}{\sqrt{q}}=\sqrt{q}*\frac{q}{\sqrt{q}}=q=integer\). Sufficient. (2) \(p\) is a positive integer > \(p\frac{q}{\sqrt{q}}=integer*\sqrt{q}\). This product may or may not be an integer depending on \(q\). Not sufficient. Answer: A.
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Re: m02 #20 [#permalink]
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16 Oct 2013, 07:55
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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]
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18 Jan 2009, 09:42
Quote: If q is a positive integer, is \((p{sqrt q}+q)/{sqrt q}\)an integer?
1) \(q = p^2\) 2) p is a positive integer = \((p{sqrt q}+q)/{sqrt q}\) = \({sqrt q}(p+{sqrt q})/{sqrt q}\) = \((p+{sqrt q})\) so now, the question is: Is \((p+{sqrt q})\) = k, where k is an integer? 1) \(q = p^2\) \({sqrt q}\) = + or  p i: if \({sqrt q}\) = p, \((p+{sqrt q})\) = 2p but we do not know whether p is an integer  may or may not be. ii: if \({sqrt q}\) = p, \((p+{sqrt q})\) = 0  yes. so insuff. 2) p is a positive integer p is an integer alone is not suff as we do not know whether \({sqrt q}\) is an integer? togather 1 and 2: yes. since p is a positive integer, \((p+{sqrt q})\) = 2p = k, where k is an integer. C.
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Re: m02 #20 [#permalink]
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18 Jan 2009, 10:05
I picked C too.
Now the question as is assumed something different and said OA is A. Looks like this is an error. Isn't below the same as what I had interpreted before.
If \(q\) is a positive integer, is \(p\frac{q}{\sqrt{q}}\) an integer?
1. \(q = p^2\) 2. \(p\) is a positive integer



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Re: m02 #20 [#permalink]
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18 Jan 2009, 11:56
ConkergMat wrote: Isn't below the same as what I had interpreted before.
If \(q\) is a positive integer, is \(p\frac{q}{\sqrt{q}}\) an integer?
1. \(q = p^2\) 2. \(p\) is a positive integer Obviously not but the answer could be the same. Given that \(q\) is a positive integer, Is \(p\frac{q}{\sqrt{q}}\) = k? 1. \(q = p^2\) \(p\frac{q}{\sqrt{q}}\) \(p\frac{q}{lpl\}\) \({lql}\). it could be ve or +ve but is an integer. so suff. 2. \(p\) is a positive integer[/quote] not suff q could be 4 or 5. for this, A should be OA.
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Re: m02 #20 [#permalink]
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18 Jan 2009, 12:21
So in the GMAT, based on above explanation, the following two have different answers depending on how the expression is written though it is the same expression? I am confused...
20) If \(q\) is a positive integer, is \(p\frac{q}{\sqrt{q}}\) an integer?
1. \(q = p^2\) 2. \(p\) is a positive integer
DIFFERENT FROM
20) If \(q\) is a positive integer, is (p√q + q )/√q an integer?
1. \(q = p^2\) 2. \(p\) is a positive integer



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Re: m02 #20 [#permalink]
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18 Jan 2009, 21:57
Never mind. A slight misunderstanding on my part. The question is assuming NOT a mixed fraction rather it is a simple multiplication and GMAT_tiger's solution assumes the same.
My question now is does GMAT have mixed fraction questions or is it always simple multiplication. In above case, what should be interpreted?



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Re: m02 #20 [#permalink]
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17 Mar 2010, 17:16
It's my understanding (I've read this in many places) that when the gmat presents a root in the stimulus (i.e. a variable under the square root symbol) it always means the positive square root. Whereas a variable squared could have a positive or negative root, a variable rooted will be the positive root. If that's the case then the answer to this should be D.
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Re: m02 #20 [#permalink]
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17 Mar 2010, 21:51
firasath wrote: It's my understanding (I've read this in many places) that when the gmat presents a root in the stimulus (i.e. a variable under the square root symbol) it always means the positive square root.
Whereas a variable squared could have a positive or negative root, a variable rooted will be the positive root.
If that's the case then the answer to this should be D. woops I read the question wrong. Since it is asking if a something is an integer regardless if it is positive or negative. The answer is indeed A.
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Re: m02 #20 [#permalink]
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19 Mar 2010, 07:44
ConkergMat wrote: If q is a positive integer, is (p* (root q ) + q ) / (root q) an integer?
1.) q = p ^^ 2 2.) p is a positive integer.
note: I was not aware how to post the mathematical expression but basically its a mixed fraction which I translated to math equation.
I believe either the question or explanation might have a typo. stmt1: q = p^2 => p = sqrt(q) so expression is q+q/root q = 2sqrt(q) not necessarily integer. stmt2: p is positive integer. Not suff since sqrt q can be real number. combine both since p = sqrt q and p is postive integer expression= 2 sqrt q = 2p hence an integer. so, both the stmts are reqd.
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Re: m02 #20 [#permalink]
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03 Dec 2010, 11:16
ConkergMat wrote: If \(q\) is a positive integer, is \(p\frac{q}{\sqrt{q}}\) an integer? 1. \(q = p^2\) 2. \(p\) is a positive integer Source: GMAT Club Tests  hardest GMAT questions note: I was not aware how to post the mathematical expression but basically its a mixed fraction which I translated to math equation. I believe either the question or explanation might have a typo. Statement 1) I used the following variables respectively Q:1,4,1,4 P:1,2,1,2 With these numbers I got all YES so it is sufficient. Statement 2) I used the following variables respectively P:1,2,3 Q:1,2,3 When you plug these numbers into the original statement you will get Yes, No, No. Insufficient. A is the answer.
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Re: m02 #20 [#permalink]
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03 Dec 2010, 12:53
sonnco wrote: ConkergMat wrote: If \(q\) is a positive integer, is \(p\frac{q}{\sqrt{q}}\) an integer? 1. \(q = p^2\) 2. \(p\) is a positive integer Source: GMAT Club Tests  hardest GMAT questions note: I was not aware how to post the mathematical expression but basically its a mixed fraction which I translated to math equation. I believe either the question or explanation might have a typo. Statement 1) I used the following variables respectively Q:1,4,1,4 P:1,2,1,2 With these numbers I got all YES so it is sufficient. Statement 2) I used the following variables respectively P:1,2,3 Q:1,2,3 When you plug these numbers into the original statement you will get Yes, No, No. Insufficient. A is the answer. You are wrong there, my friend. For S1, you have already demonstrated that expression can be integer, now what if: Q = 0.09, P = 0.3? The expression does not solve as integer. Hence 1 alone is not sufficient. Now the expression \(p\frac{q}{\sqrt{q}}\) can be reduced to \(p + {\sqrt{q}\), which will be an integer only if \(p\) is an integer, and \(q\) is a perfect square of an integer. Both 1 and 2 together are sufficient.
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Re: m02 #20 [#permalink]
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03 Dec 2010, 13:13
vaibhavtripathi wrote: sonnco wrote: ConkergMat wrote: If \(q\) is a positive integer, is \(p\frac{q}{\sqrt{q}}\) an integer? 1. \(q = p^2\) 2. \(p\) is a positive integer Source: GMAT Club Tests  hardest GMAT questions note: I was not aware how to post the mathematical expression but basically its a mixed fraction which I translated to math equation. I believe either the question or explanation might have a typo. Statement 1) I used the following variables respectively Q:1,4,1,4 P:1,2,1,2 With these numbers I got all YES so it is sufficient. Statement 2) I used the following variables respectively P:1,2,3 Q:1,2,3 When you plug these numbers into the original statement you will get Yes, No, No. Insufficient. A is the answer. You are wrong there, my friend. For S1, you have already demonstrated that expression can be integer, now what if: Q = 0.09, P = 0.3? The expression does not solve as integer. Hence 1 alone is not sufficient. Now the expression \(p\frac{q}{\sqrt{q}}\) can be reduced to \(p + {\sqrt{q}\), which will be an integer only if \(p\) is an integer, and \(q\) is a perfect square of an integer. Both 1 and 2 together are sufficient.It states Q is a positive integer. Q cannot be 0.09. Am I missing something?
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Re: m02 #20 [#permalink]
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03 Dec 2010, 20:55
sonnco wrote: It states Q is a positive integer. Q cannot be 0.09. Am I missing something? Uh oh... Actually, I missed something here... My bad...
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Re: m02 #20 [#permalink]
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04 Dec 2010, 10:41
sleekmover wrote: A.
s1 : since p^2= q it implies that p is an integer. If p was not an iteger it's square can never be an integer
s2: does not provide any extra info
Posted from my mobile device Not necessarily. E.g. if P is \({\sqrt{2}}\) or \({\sqrt{3}}\) (just 2 of many possible examples) its square would be an integer.
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Re: m02 #20 [#permalink]
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08 Dec 2010, 06:47
Am a lil confused. The ques says q is a positive andnothing abt p int Consider option 1 whr q=p^2 so if q is 3 p=sqrt of 3 then how is only stmt1 sufficient? IMO c!!
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Re: m02 #20 [#permalink]
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08 Dec 2010, 08:43
Let me put what you said into a math statement: \(q = 3\) \(p = \sqrt{3}\) Then: \(p\frac{q}{\sqrt{q}} = \sqrt{3}\frac{3}{\sqrt{3}} = 1\) (integer) The expression above will remain an integer even if \(p\) is a negative root. It will just be a negative integer. Let me know if I'm missing anything here. Aminayak wrote: Am a lil confused. The ques says q is a positive andnothing abt p int Consider option 1 whr q=p^2 so if q is 3 p=sqrt of 3 then how is only stmt1 sufficient? IMO c!!
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Re: m02 #20 [#permalink]
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29 Apr 2011, 19:00
Solving given expression we have p * q/sqrt(q) = p * sqrt(q)?
1. Sufficient q = p^2 => p= sqrt(q)
=> p * sqrt(q) = sqrt(q) * sqrt(q) = +q or q
as q is an integer , +q , q are both integers. 2. Not sufficient.
p is a positive integer, but we dont whether q is a perfect square or not. if q is a perfect square , given expression is an integer. if q is not a perfect square, given expression is not an integer.
Answer is A.







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