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Now P and q are non zero intergers; so if p=1 and q =2; then p/q = 1/2 p2/q2 = 1/4.. So it is not an integer.

So the answer is B.. any1 get my logic?

p=1 and q=2 are not valid choices for statement (1). It says that n^2 is an integer but for these values of p and q we have that n^2=(1/2)^2=1/4 which is not an integer.

Complete solution:

If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or an irrational number (for example: \(\sqrt{3}\)), (note that \(n\) can not be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> \(\frac{2n+4}{2}=n+2=integer\) --> \(n=integer\). Sufficient.

If n=p/q ( p and q are nonzero integers), is n an integer?

(1) n^2 is an integer (2) (2n+4)/2 is an integer

My doubt is if n^2 is an integer then n can be non integere also eg Sqrt2 Then how cum D is the answer

Good question. If I am correctly recalling, I faced it before.

The logic goes like this: If p and q are integers, then the square of the value from p/q never becomes an integer.

Let me ask you a counter question: can you get sqrt (2) dividing p by q such that p and q are integers? If yes, its B and vice versa.
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a) n=n^2: if n was a decimal,with for example one decimal digit, then in the tenths' place could be: 1: but n^2 would have in the tenths' place 1 -> n^2 not an integer 2: but n^2 would have in the tenths' place 4 -> n^2 not an integer 3: but n^2 would have in the tenths' place 9 -> n^2 not an integer 4: but n^2 would have in the tenths' place 6 -> n^2 not an integer 5: but n^2 would have in the tenths' place 5 -> n^2 not an integer 6: but n^2 would have in the tenths' place 6 -> n^2 not an integer 7: but n^2 would have in the tenths' place 9 -> n^2 not an integer 8: but n^2 would have in the tenths' place 4 -> n^2 not an integer 9: but n^2 would have in the tenths' place 1 -> n^2 not an integer So, given that n^2 is an integer means that n is an integer (the decimal unit is 0) b)(2n+4)/2 is an integer->n+2 is an integer-> n is an integer. So the best answer is d, every statement provides sufficient information by itself

You end up with a couple of pieces of information - q*q is a divisor of p*p. - q*q & p*p by definition are not prime numbers so you can manipulate as per below

If \(n = p/q\) is not an integer, then q is not a divisor p then q is not a divisor of p*p and (\(p*p/q\)) is some arbitrary non integer number then \((p*p/q)/ q = (p*p) / (q*q)\) can't be an integer either which breaks the whole thing

1) n^2 is an integer --> to be an integer must be either an integer or an irrational number (for example: ), (note that can not be reduced fraction, for example or as in this case won't be an integer). But as can be expressed as the ratio of 2 integers, , then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: is an integer. Sufficient.

1) n^2 is an integer --> to be an integer must be either an integer or an irrational number (for example: ), (note that can not be reduced fraction, for example or as in this case won't be an integer). But as can be expressed as the ratio of 2 integers, , then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> --> . Sufficient.

Answer: D.

what if q=1?

The same. If \(q=1\) then as given that \(n=\frac{p}{q}\) then \(n=p\) and as also given that \(p\) is an integer then \(n\) is an integer too.
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This is how I solved it. Let me know if I'm wrong.

before we start, we know that P & Q are integers that =/= 0. also \(n= \frac{p}{q}\)

1. statement 1: \(n^2=\frac{p^2}{q^2}\) we know:

\({p}\) is divisible by \({q}\) \(\frac{p^2}{q^2}\) is an integer

the only way for \({n}\)not to be true is for \({p}\) or \(q\) to be \(\sqrt{q}\) or \(\sqrt{p}\) but we know that these are integers so statement 1 is sufficient.

Guys the question states that p and q are non-zero integers. So neither can they be fractions nor they can be irrational numbers.

Hence p and q are only integers but cannot be zero.

Given that n = p/q; from statement n^2 is integer. Hence n^2 = (p/q)^2 is an integer so p/q has to be an integer as p and q can only be integers in such a way that p is a multiple of q and hence n is an integer.

Hence sufficient.

second statement is obviously sufficient for n to be integer. Hence D

Good one. The important point is that it is mentioned that p and q are non-zero integers. This helps resolve the statement(i) which says n*2 is an integer.

Concluded D in ~1 mins and felt I deduced the answer pretty quickly so rechecked.
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My attempt to capture my B-School Journey in a Blog : tranquilnomadgmat.blogspot.com

Okay, I am not a math genius by any means, but this is how it goes:

n = p/q, where we know that p and q are non-zero INTEGERS means that the result (n) will either be an integer or a fraction. It cannot be \sqrt{2} or \sqrt{3}.

So, once you've determined that much, when you go down to the first statement, you are told n^2 is an integer.

Remember, a squared fraction is smaller than the fraction you were squaring.

So: THE SQUARE OF A FRACTION CANNOT BE AN INTEGER!!! IT WILL BE A SMALLER FRACTION. Therefore, for the square of n to be an integer, p/q must produce a non-zero integer. It could be 1 or it could be 1000, but n would have to be an integer.