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 350,000 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:

(2) \(p\) is a positive integer --> \(p\frac{q}{\sqrt{q}}=integer*\sqrt{q}\). This product may or may not be an integer depending on \(q\). Not sufficient.

If we had to add a constraint that is \(p\frac{q}{\sqrt{q}}\) a positive integer ? (or negative integer for that matter), will answer still be the same? coz Sqroot Q = +-P as per statement 1. isn't it? _________________

Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back!

so now, the question is: Is \((p+{sqrt q})\) = k, where k is an integer?

1) \(q = p^2\) \({sqrt q}\) = + or - p

i: if \({sqrt q}\) = p, \((p+{sqrt q})\) = 2p but we do not know whether p is an integer - may or may not be. ii: if \({sqrt q}\) = -p, \((p+{sqrt q})\) = 0 - yes.

so insuff.

2) p is a positive integer p is an integer alone is not suff as we do not know whether \({sqrt q}\) is an integer?

togather 1 and 2: yes. since p is a positive integer, \((p+{sqrt q})\) = 2p = k, where k is an integer.

Now the question as is assumed something different and said OA is A. Looks like this is an error. Isn't below the same as what I had interpreted before.

If \(q\) is a positive integer, is \(p\frac{q}{\sqrt{q}}\) an integer?

So in the GMAT, based on above explanation, the following two have different answers depending on how the expression is written though it is the same expression? I am confused...

20) If \(q\) is a positive integer, is \(p\frac{q}{\sqrt{q}}\) an integer?

1. \(q = p^2\) 2. \(p\) is a positive integer

DIFFERENT FROM

20) If \(q\) is a positive integer, is (p√q + q )/√q an integer?

Never mind. A slight misunderstanding on my part. The question is assuming NOT a mixed fraction rather it is a simple multiplication and GMAT_tiger's solution assumes the same.

My question now is does GMAT have mixed fraction questions or is it always simple multiplication. In above case, what should be interpreted?

It's my understanding (I've read this in many places) that when the gmat presents a root in the stimulus (i.e. a variable under the square root symbol) it always means the positive square root.

Whereas a variable squared could have a positive or negative root, a variable rooted will be the positive root.

If that's the case then the answer to this should be D. _________________

If you like my post, a kudos is always appreciated

It's my understanding (I've read this in many places) that when the gmat presents a root in the stimulus (i.e. a variable under the square root symbol) it always means the positive square root.

Whereas a variable squared could have a positive or negative root, a variable rooted will be the positive root.

If that's the case then the answer to this should be D.

woops I read the question wrong. Since it is asking if a something is an integer regardless if it is positive or negative. The answer is indeed A. _________________

If you like my post, a kudos is always appreciated

If q is a positive integer, is (p* (root q ) + q ) / (root q) an integer?

1.) q = p ^^ 2 2.) p is a positive integer.

note: I was not aware how to post the mathematical expression but basically its a mixed fraction which I translated to math equation.

I believe either the question or explanation might have a typo.

stmt1: q = p^2 => p = sqrt(q) so expression is q+q/root q = 2sqrt(q) not necessarily integer. stmt2: p is positive integer. Not suff since sqrt q can be real number.

combine both since p = sqrt q and p is postive integer expression= 2 sqrt q = 2p hence an integer. so, both the stmts are reqd. _________________

note: I was not aware how to post the mathematical expression but basically its a mixed fraction which I translated to math equation.

I believe either the question or explanation might have a typo.

Statement 1) I used the following variables respectively Q:1,4,1,4 P:1,2,-1,-2

With these numbers I got all YES so it is sufficient.

Statement 2) I used the following variables respectively P:1,2,3 Q:1,2,3

When you plug these numbers into the original statement you will get Yes, No, No. Insufficient.

A is the answer. _________________

I'm trying to not just answer the problem but to explain how I came up with my answer. If I am incorrect or you have a better method please PM me your thoughts. Thanks!

note: I was not aware how to post the mathematical expression but basically its a mixed fraction which I translated to math equation.

I believe either the question or explanation might have a typo.

Statement 1) I used the following variables respectively Q:1,4,1,4 P:1,2,-1,-2

With these numbers I got all YES so it is sufficient.

Statement 2) I used the following variables respectively P:1,2,3 Q:1,2,3

When you plug these numbers into the original statement you will get Yes, No, No. Insufficient.

A is the answer.

You are wrong there, my friend. For S1, you have already demonstrated that expression can be integer, now what if: Q = 0.09, P = 0.3? The expression does not solve as integer. Hence 1 alone is not sufficient.

Now the expression \(p\frac{q}{\sqrt{q}}\) can be reduced to \(p + {\sqrt{q}\), which will be an integer only if \(p\) is an integer, and \(q\) is a perfect square of an integer. Both 1 and 2 together are sufficient. _________________

note: I was not aware how to post the mathematical expression but basically its a mixed fraction which I translated to math equation.

I believe either the question or explanation might have a typo.

Statement 1) I used the following variables respectively Q:1,4,1,4 P:1,2,-1,-2

With these numbers I got all YES so it is sufficient.

Statement 2) I used the following variables respectively P:1,2,3 Q:1,2,3

When you plug these numbers into the original statement you will get Yes, No, No. Insufficient.

A is the answer.

You are wrong there, my friend. For S1, you have already demonstrated that expression can be integer, now what if: Q = 0.09, P = 0.3? The expression does not solve as integer. Hence 1 alone is not sufficient.

Now the expression \(p\frac{q}{\sqrt{q}}\) can be reduced to \(p + {\sqrt{q}\), which will be an integer only if \(p\) is an integer, and \(q\) is a perfect square of an integer. Both 1 and 2 together are sufficient.

It states Q is a positive integer. Q cannot be 0.09. Am I missing something? _________________

I'm trying to not just answer the problem but to explain how I came up with my answer. If I am incorrect or you have a better method please PM me your thoughts. Thanks!

Am a lil confused. The ques says q is a positive andnothing abt p int Consider option 1 whr q=p^2 so if q is 3 p=sqrt of 3 then how is only stmt1 sufficient? IMO c!!

The expression above will remain an integer even if \(p\) is a negative root. It will just be a negative integer. Let me know if I'm missing anything here.

Aminayak wrote:

Am a lil confused. The ques says q is a positive andnothing abt p int Consider option 1 whr q=p^2 so if q is 3 p=sqrt of 3 then how is only stmt1 sufficient? IMO c!!

Solving given expression we have p * q/sqrt(q) = p * sqrt(q)?

1. Sufficient q = p^2 => p= sqrt(q)

=> p * sqrt(q) = sqrt(q) * sqrt(q) = +q or -q

as q is an integer , +q , -q are both integers. 2. Not sufficient.

p is a positive integer, but we dont whether q is a perfect square or not. if q is a perfect square , given expression is an integer. if q is not a perfect square, given expression is not an integer.