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# m02 #20

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m02 #20 [#permalink]  17 Jan 2009, 21:17
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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

[Reveal] Spoiler: OA
A

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]  18 Jan 2009, 08:42
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]  18 Jan 2009, 09:05
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]  18 Jan 2009, 10:56
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]  18 Jan 2009, 11:21
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]  18 Jan 2009, 20:57
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]  17 Mar 2010, 16:16
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]  17 Mar 2010, 20:51
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]  19 Mar 2010, 06:44
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]  03 Dec 2010, 10:16
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

[Reveal] Spoiler: OA
A

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.

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Re: m02 #20 [#permalink]  03 Dec 2010, 11:53
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

[Reveal] Spoiler: OA
A

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.

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]  03 Dec 2010, 12:13
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

[Reveal] Spoiler: OA
A

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.

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]  03 Dec 2010, 19:55
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]  04 Dec 2010, 04:49
3
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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]  04 Dec 2010, 09:41
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]  08 Dec 2010, 05:47
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]  08 Dec 2010, 07:43
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!!

Posted from my mobile device

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Re: m02 #20 [#permalink]  29 Apr 2011, 18:00
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.

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Re: m02 #20 [#permalink]  08 Dec 2011, 06:34
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Re: m02 #20 [#permalink]  07 Dec 2012, 05:14
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Expert's post
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

[Reveal] Spoiler: OA
A

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.

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Re: m02 #20   [#permalink] 07 Dec 2012, 05:14

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