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(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.

(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.

(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

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.
_________________

(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

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.....

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.

Please correct it. 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.

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.

Please correct it. 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.
_________________

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.

(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 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

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.