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M02-20

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Re: M02-20  [#permalink]

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New post 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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Re: M02-20  [#permalink]

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New post 22 May 2019, 10:14
Hello,
I was unable to answer the question correctly.
I answered D.

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.


Please help me understand why is my reasoning wrong for statement 2.

Thanks.
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Re: M02-20  [#permalink]

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

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.


Please help me understand why is my reasoning wrong for statement 2.

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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Re: M02-20  [#permalink]

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New post 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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Re: M02-20  [#permalink]

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New post 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.


Answer: A


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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Re: M02-20  [#permalink]

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New post 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.


Answer: A


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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