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# M02-20

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GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5479
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

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06 Apr 2019, 03:26
Bunuel 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

$$p*\frac{q}{\sqrt{q}}$$
solve
we get p*√q

#1
$$q = p^2$$ ; p=√q
so p*√q ; √q*√q ; q
sufficient
#2
$$p$$ is a positive integer[/quote]
no relation with q..
insufficient
IMO A
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Joined: 20 Nov 2017
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22 May 2019, 10:14
Hello,
I was unable to answer the question correctly.

1) its self explanatory

2) Let p*(q/ √q) = z
squaring on both sides
p^2 * (q^2 / q) = z^2
i.e p^2 * q = z^2
We know q is +ve integer.
So the only way z^2 will be an integer, is if p is an integer. And statement 2 tells us exactly that.

Thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 59722

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22 May 2019, 21:08
yashraj1011 wrote:
Hello,
I was unable to answer the question correctly.

1) its self explanatory

2) Let p*(q/ √q) = z
squaring on both sides
p^2 * (q^2 / q) = z^2
i.e p^2 * q = z^2
We know q is +ve integer.
So the only way z^2 will be an integer, is if p is an integer. And statement 2 tells us exactly that.

Thanks.

p being an integer and q being an integer ensures that p^2*q is an integer but we need to ensure that p*√q is an integer. For example, if p=1 and q = 2, then p^2*q is an integer but p*√q is not.
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Joined: 03 Sep 2018
Posts: 252
Location: Netherlands
GMAT 1: 710 Q48 V40
GMAT 2: 780 Q50 V49
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11 Nov 2019, 12:03
$$p*\frac{q}{\sqrt{q}} \implies p*\sqrt{q}$$

$$q=p^2 \implies p*\sqrt{p^2} \implies p^2=q=integer$$
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Good luck to you.
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Joined: 11 Jun 2018
Posts: 19
Location: India
GMAT 1: 660 Q44 V36

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22 Nov 2019, 02:46
Bunuel wrote:
Official Solution:

(1) $$q = p^2$$. Take the square root from botth sides: $$p=\sqrt{q}$$. Substitute $$p$$: $$p*\frac{q}{\sqrt{q}}=\sqrt{q}*\frac{q}{\sqrt{q}} = q = \text{integer}$$. Sufficient.

(2) $$p$$ is a positive integer. So, we have that $$p*\frac{q}{\sqrt{q}}=\text{integer}*\sqrt{q}$$. This product may or may not be an integer depending on $$q$$. Not sufficient.

hi Bunnel,

stmt(A) can easily be made out to be p * |p|, in which case , we really cant say whether the expression is an integer or not.
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Joined: 02 Sep 2009
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22 Nov 2019, 02:49
dasoisheretorule wrote:
Bunuel wrote:
Official Solution:

(1) $$q = p^2$$. Take the square root from botth sides: $$p=\sqrt{q}$$. Substitute $$p$$: $$p*\frac{q}{\sqrt{q}}=\sqrt{q}*\frac{q}{\sqrt{q}} = q = \text{integer}$$. Sufficient.

(2) $$p$$ is a positive integer. So, we have that $$p*\frac{q}{\sqrt{q}}=\text{integer}*\sqrt{q}$$. This product may or may not be an integer depending on $$q$$. Not sufficient.

hi Bunnel,

stmt(A) can easily be made out to be p * |p|, in which case , we really cant say whether the expression is an integer or not.

You should do the way, so you should manipulate the way shown in the solution.
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Re: M02-20   [#permalink] 22 Nov 2019, 02:49

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# M02-20

Moderators: chetan2u, Bunuel