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 and k are positive integers, is \(\sqrt{n+k}>2\sqrt{n}\)?

Both parts of the inequality are positive, thus we can square it, to get "is \(n+k>4n\)?" --> is \(k>3n\)?

(1) k > 3n. Sufficient.

(2) n + k > 3n --> \(k>2n\). Not sufficient.

Answer: A.

Hi Bunuel,

I don't understand why n & k are positive then two parts of the inequility \(\sqrt{n+k}>2\sqrt{n}\) are positive.

What I understand is that: n, k>0 => n+k>0 => \(\sqrt{n+k}\) might be positive or negative. e.g: x=9>0 --> \(\sqrt{x}\) = 3 or -3 Same thought or n!

Please help me to clarify, thank you so much!

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{9}=3\), NOT +3 or -3. In contrast, the equation \(x^2=9\) has TWO solutions, +3 and -3. Even roots have only non-negative value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

Re: If n and k are positive integers, is (n+k)^1/2>2n^1/2? [#permalink]

Show Tags

27 Jun 2014, 04:06

Bunuel wrote:

LucyDang wrote:

Bunuel wrote:

If n and k are positive integers, is \(\sqrt{n+k}>2\sqrt{n}\)?

Both parts of the inequality are positive, thus we can square it, to get "is \(n+k>4n\)?" --> is \(k>3n\)?

(1) k > 3n. Sufficient.

(2) n + k > 3n --> \(k>2n\). Not sufficient.

Answer: A.

Hi Bunuel,

I don't understand why n & k are positive then two parts of the inequility \(\sqrt{n+k}>2\sqrt{n}\) are positive.

What I understand is that: n, k>0 => n+k>0 => \(\sqrt{n+k}\) might be positive or negative. e.g: x=9>0 --> \(\sqrt{x}\) = 3 or -3 Same thought or n!

Please help me to clarify, thank you so much!

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{9}=3\), NOT +3 or -3. In contrast, the equation \(x^2=9\) has TWO solutions, +3 and -3. Even roots have only non-negative value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

Re: If n and k are positive integers, is (n+k)^1/2>2n^1/2? [#permalink]

Show Tags

19 Aug 2014, 06:01

Bunuel wrote:

If n and k are positive integers, is \(\sqrt{n+k}>2\sqrt{n}\)?

Both parts of the inequality are positive, thus we can square it, to get "is \(n+k>4n\)?" --> is \(k>3n\)?

(1) k > 3n. Sufficient.

(2) n + k > 3n --> \(k>2n\). Not sufficient.

Answer: A.

How do you get to the conclusion that \(\sqrt{n+k}>2\sqrt{n}\)? Even if k > 3n it can be either n+k > 4n or n+k < 4n, since we don't know more about n.

If n and k are positive integers, is \(\sqrt{n+k}>2\sqrt{n}\)?

Both parts of the inequality are positive, thus we can square it, to get "is \(n+k>4n\)?" --> is \(k>3n\)?

(1) k > 3n. Sufficient.

(2) n + k > 3n --> \(k>2n\). Not sufficient.

Answer: A.

How do you get to the conclusion that \(\sqrt{n+k}>2\sqrt{n}\)? Even if k > 3n it can be either n+k > 4n or n+k < 4n, since we don't know more about n.

Thanks .

Not sure how you get the above...

Anyway, the question asks whether \(\sqrt{n+k}>2\sqrt{n}\)? After algebraic manipulations shown in my solution the question becomes: is \(k>3n\)? The first statement answers this question, which makes it sufficient.
_________________

Re: If n and k are positive integers, is (n+k)^1/2>2n^1/2? [#permalink]

Show Tags

07 Oct 2015, 02:14

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Problem: If n and k are positive integers, is \(\sqrt{n+k}\) > 2 \(\sqrt{n}\)?

(1) k > 3n (2) n + k > 3n

------ By squaring both sides of the equation, we are left with:

n + k > 4n

In its simplest form the equation is:

k > 3n ----

According to Manhattan GMAT 12th edition, the second statement is insufficient. Perhaps I am overlooking some mathematical principle, but if one compares the following two equations one should be able to conclude if the statement is true or not.

statement (2) from problem: n + k > 3n original equation squared: n + k > 4n

Problem: If n and k are positive integers, is \(\sqrt{n+k}\) > 2 \(\sqrt{n}\)?

(1) k > 3n (2) n + k > 3n

------ By squaring both sides of the equation, we are left with:

n + k > 4n

In its simplest form the equation is:

k > 3n ----

According to Manhattan GMAT 12th edition, the second statement is insufficient. Perhaps I am overlooking some mathematical principle, but if one compares the following two equations one should be able to conclude if the statement is true or not.

statement (2) from problem: n + k > 3n original equation squared: n + k > 4n

Any help or guidance would be much appreciated.

Merging topics. Please refer to the discussion above.

Re: If n and k are positive integers, is (n+k)^1/2>2n^1/2? [#permalink]

Show Tags

09 Dec 2016, 13:39

Hi,

If it were not given that n and k are positive integers, then taking the underroot of this expression (Underoort n+k > 2 underroot n) would have resulted in absolute value form as l n + k l > 4 l n l. Am I right?

Just wanted to make sure that I get the concept right.

Thanks for your help.

Regards, H

gmatclubot

Re: If n and k are positive integers, is (n+k)^1/2>2n^1/2?
[#permalink]
09 Dec 2016, 13:39

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...