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# m02 #20

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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]

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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]

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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]

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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]

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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]

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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]

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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]

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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]

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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]

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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.

A is the answer.
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Re: m02 #20 [#permalink]

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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.

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, 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.

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, 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]

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04 Dec 2010, 04:49
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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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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]

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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]

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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]

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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]

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08 Dec 2011, 06:34
I read the question as p^(q/sqrt(q)) and that was leading to a C answer.
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Re: m02 #20 [#permalink]

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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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