Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.
_________________

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

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.

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.
_________________

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.
_________________

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.