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# M02-20

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Math Expert
Joined: 02 Sep 2009
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15 Sep 2014, 23:18
Expert's post
11
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Difficulty:

45% (medium)

Question Stats:

58% (00:46) correct 42% (00:56) wrong based on 209 sessions

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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

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15 Sep 2014, 23:18
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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.

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14 Dec 2014, 10:45
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.

what if root of q is negative of p

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15 Dec 2014, 07:14
Arnav180 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.

what if root of q is negative of p

The result would still be an integer: $$p=-\sqrt{q}$$. Substitute $$p$$: $$p\frac{q}{\sqrt{q}}=-\sqrt{q}*\frac{q}{\sqrt{q}} = -q = \text{integer}$$. Sufficient.
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22 Jan 2015, 02:55
I think this question is poor and helpful.
Question appears to say

(p*underroot q + q)/underroot q

IT should be clearly mentioned that it is
(pq)/ underfoot q

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22 Jan 2015, 04:05
obishake wrote:
I think this question is poor and helpful.
Question appears to say

(p*underroot q + q)/underroot q

IT should be clearly mentioned that it is
(pq)/ underfoot q

Sorry but don't understand what you mean at all... What is confusing in $$p\frac{q}{\sqrt{q}}$$?
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09 Apr 2015, 19:48
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.

I had chosen C. The reason being, is that I substitute as is, that q=p^2 into the equation as that's where my brain goes... q is already isolated, why do extra work? When you actually use that though, you end up getting p^2 as your answer. So I thought, ok that is INSUFFICIENT because you don't know if p is an integer or not. Then voila, you look at the next constraint and it states that p is an integer....so I chose C.

I think a good takeaway from this, is just because it's given to you in a certain format and it can be used, doesn't mean it'll help you solve the question. The GMAT is full of tricks.....

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14 Apr 2015, 23:12
Bunuel wrote:
obishake wrote:
I think this question is poor and helpful.
Question appears to say

(p*underroot q + q)/underroot q

IT should be clearly mentioned that it is
(pq)/ underfoot q

Sorry but don't understand what you mean at all... What is confusing in $$p\frac{q}{\sqrt{q}}$$?

It looks like a mixed fraction, so that's a bit confusing.

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16 Aug 2015, 05:34
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I think this is a poor-quality question and I don't agree with the explanation. Hi

The question should have P*Q/.. rather than Pq/...

This creates confusion of mixed fraction and not multiplication.

Thanks
Rajesh

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20 Aug 2015, 08:41
mrk.iitm@gmail.com wrote:
I think this is a poor-quality question and I don't agree with the explanation. Hi

The question should have P*Q/.. rather than Pq/...

This creates confusion of mixed fraction and not multiplication.

Thanks
Rajesh

Edited as suggested. Thank you.
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20 Aug 2015, 10:56
Bunuel wrote:
mrk.iitm@gmail.com wrote:
I think this is a poor-quality question and I don't agree with the explanation. Hi

The question should have P*Q/.. rather than Pq/...

This creates confusion of mixed fraction and not multiplication.

Thanks
Rajesh

Edited as suggested. Thank you.

Hi Bunuel,
I think the answer to this should be C. According to statement 1, it should be p*\sqrt{q} where q=p^2. So now this becomes p^p=p^2.P can be a fraction or an integer so this is not sufficient.

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20 Aug 2015, 11:43
KS15 wrote:
Bunuel wrote:
mrk.iitm@gmail.com wrote:
I think this is a poor-quality question and I don't agree with the explanation. Hi

The question should have P*Q/.. rather than Pq/...

This creates confusion of mixed fraction and not multiplication.

Thanks
Rajesh

Edited as suggested. Thank you.

Hi Bunuel,
I think the answer to this should be C. According to statement 1, it should be p*\sqrt{q} where q=p^2. So now this becomes p^p=p^2.P can be a fraction or an integer so this is not sufficient.

No, since it's given that q (which equals to p^2) is an integer.
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20 Mar 2016, 17:50
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. I was wondering why "P" could not be considered a negative Integer, Which satisfies statement 1. Do the answer to this question change depending of wether " P " is negative or not ? I may be missing something Please clarify if possible.

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02 Jun 2016, 19:21
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 Bunuel,
I just wanted to re-confirm the key takeaway from this question;
1. if x=y^2 and given that x is integer, then y will always be and integer, right ?? alternatively we could plug in a few numbers and see.
2. if we square root on both sides of the equation x= y^2, we should get $$\sqrt{x} = |y|$$ ; where $$\sqrt{x}$$ and y are both integers and where the sign of y is not known

is there a gap in this understanding ? please assist

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12 Sep 2016, 06:00
Hello Bunuel,

but isn't √P^2=|P| and as we don't know the sign of p .??If that is the case then we get |P|=√q

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12 Sep 2016, 21:03
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Responding to a pm:

Ishanvs wrote:
1. if x=y^2 and given that x is integer, then y will always be and integer

No. Given y is an integer, x will be an integer but the other way around is not necessary.
Say x = 5 (an integer) but y is $$\sqrt{5}$$ or $$- \sqrt{5}$$.

Ishanvs wrote:
2. if we take square root on both sides of the equation x= y^2, we should get √x=|y| ; where √x and y are both integers and where the sign of y is not known

That depends on whether y is an integer. If, say, y = 5/3, x and y are not integers.

Ishanvs wrote:
isn't √P^2=|P| and as we don't know the sign of p .??If that is the case then we get |P|=√q

Yes it is. We don't know the sign of p using stmnt 1 alone. But absolute value of p is equal to $$\sqrt{q}$$. So the two will get cancelled off in either case. You might be left with a negative sign but you still get an integer.

Say p = 5
$$p^2 = q = 25$$
$$\sqrt{q} = 5$$
$$p * q / \sqrt{q} = 5 * 25 / 5 = 25$$

Say p = -5
$$p^2 = q = 25$$
$$\sqrt{q} = 5$$
$$p * q / \sqrt{q} = -5 * 25 / 5 = -25$$ (still an integer)
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03 Dec 2017, 06:10
I think Statement 2 is adequate as well.

Question says----> Q is an integer & Statement says P is an integer.

Then, P * √Q -----> will not be an integer.

Thus, B is sufficient. Please let me know if I am missing anything.

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03 Dec 2017, 07:09
quest800 wrote:
I think Statement 2 is adequate as well.

Question says----> Q is an integer & Statement says P is an integer.

Then, P * √Q -----> will not be an integer.

Thus, B is sufficient. Please let me know if I am missing anything.

$$integer*\sqrt{integer}$$ could be an integer as well as non-integer. For example, $$1*\sqrt{1}=1=integer$$ and $$1*\sqrt{2}=\sqrt{2}\neq integer$$.

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Re: M02-20   [#permalink] 03 Dec 2017, 07:09
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# M02-20

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