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:

Can someone please explain why point 2 is not correct? What I did is that 28 is factored into => 2*2*7 so therefore I concluded x must have at least one 7. so if x has one seven then \sqrt{7x} must be an integer. Am I doing something wrong here?

(1) \(\sqrt{\frac{x}{7}}\) is an integer. Given that \(\sqrt{\frac{x}{7}}=integer\) --> square it: \(\frac{x}{7}=integer^2\) --> \(x=7*integer^2\). So, \(\sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer\). Sufficient.

(2) \(\sqrt{28x}\) is an integer. If \(x=\frac{1}{28}\), then \(\sqrt{7x}=\frac{1}{2}\neq{integer}\) BUT if \(x=0\), then \(\sqrt{7x}=0=integer\). Not sufficient.

(1) \(\sqrt{\frac{x}{7}}\) is an integer. Given that \(\sqrt{\frac{x}{7}}=integer\) --> square it: \(\frac{x}{7}=integer^2\) --> \(x=7*integer^2\). So, \(\sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer\). Sufficient.

(2) \(\sqrt{28x}\) is an integer. If \(x=\frac{1}{28}\), then \(\sqrt{7x}=\frac{1}{2}\neq{integer}\) BUT if \(x=0\), then \(\sqrt{7x}=0=integer\). Not sufficient.

Answer: A.

Thanks sorry for asking again... But could you please explain if my method/ logic was completely wrong? ie if we break out 28 int primes we see that it has 2,2,7 so x must have at least one 7? therefore we can assume that \sqrt{7x} should be integer?

(1) \(\sqrt{\frac{x}{7}}\) is an integer. Given that \(\sqrt{\frac{x}{7}}=integer\) --> square it: \(\frac{x}{7}=integer^2\) --> \(x=7*integer^2\). So, \(\sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer\). Sufficient.

(2) \(\sqrt{28x}\) is an integer. If \(x=\frac{1}{28}\), then \(\sqrt{7x}=\frac{1}{2}\neq{integer}\) BUT if \(x=0\), then \(\sqrt{7x}=0=integer\). Not sufficient.

Answer: A.

Thanks sorry for asking again... But could you please explain if my method/ logic was completely wrong? ie if we break out 28 int primes we see that it has 2,2,7 so x must have at least one 7? therefore we can assume that \sqrt{7x} should be integer?

Thanks again! Alex

If x is NOT an integer you cannot make its prime factorization.

If we were told that x IS an integer, then \(\sqrt{28x}=2\sqrt{7x}=integer\). Thus x must be of the form \(7^{odd}*integer^2\), so in this case \(\sqrt{7x}=\sqrt{7*(7^{odd}*integer^2)}=\sqrt{7^{even}*integer^2}=integer\).

Can someone please explain why point 2 is not correct? What I did is that 28 is factored into => 2*2*7 so therefore I concluded x must have at least one 7. so if x has one seven then \sqrt{7x} must be an integer. Am I doing something wrong here?

Thanks! Alex

Quote:

\(\sqrt{7x}\) = integer ? or 7x= int^2 ?

(1) \(\sqrt{\frac{x}{7}}\) is an integer ---- lets say b x/7 = b^2 --> where b is an integer x= 7 *b^2 7x = (7*b) ^2 hence 7x=int^2--- sufficient

(2) \(\sqrt{28x}\) is an integer ---- lets say c

28x= c^2 7x=c^2/4 = (c/2)^2 If C is odd , then ans is No and if C is even then yes . ( If you knew that x was an integer , then this would have sufficed) Insufficient.
_________________

From F.S 1 , we know that \(\sqrt{\frac{x}{7}}\) = Integer(I)--> \(\sqrt{\frac{x*7}{7*7}}\) = \(\sqrt{7x}*\frac{1}{7}\) = I--> \(\sqrt{7x}\) = 7*I. Thus, an integer. Sufficient.

From F.S 2, we know that \(\sqrt{28x}\) = Integer(I) --> \(\sqrt{7*4x}\) = 2*\(\sqrt{7x}\) = I. Thus, \(\sqrt{7x}\) =\(\frac{I}{2}\). Depending on I being odd/even, the given expression will be a non-integer/integer. Insufficient.

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.
_________________

This question can be solved by TESTing VALUES, but you have to really be thorough with your TESTs (and think about what X COULD be given the information in the Facts).

We're asked if Root(7X) is an integer. This is a YES/NO question.

Fact 1: Root(X/7) is an integer.

This Fact tells us that X has to be a specific type of multiple of 7....

IF.... X = 7 the Root(7/7) = 1 and is an integer Root(49) = 7 and the answer to the question is YES

IF.... X = 28 the Root(28/7) = 2 and is an integer Root(196) = 14 and the answer to the question is YES

This hints at the pattern that X will be a "perfect square times a multiple of 7', which means that the answer to the question is ALWAYS YES. You can also use prime-factorization to prove that this is the case. Fact 1 is SUFFICIENT

Fact 2: Root(28X) is an integer.

This Fact tells us that X has to be a specific type of multiple of 7....

IF.... X = 7 the Root(196) = 14 and is an integer Root(49) = 7 and the answer to the question is YES

IF.... X = 1/28 the Root(1) = 1 and is an integer Root(7/28) = 1/2 and the answer to the question is NO Fact 2 is INSUFFICIENT

Is [m][square_root](7x)[/square_root][/m] an integer? [#permalink]

Show Tags

15 May 2016, 09:36

Is \(\sqrt{(7x)}\) an integer?

(1) \(\sqrt{(x/7)}\) is an integer.

(2) \(\sqrt{(28x)}\)is an integer.

Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient. EITHER statement BY ITSELF is sufficient to answer the question. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.

Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient. EITHER statement BY ITSELF is sufficient to answer the question. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.

Merging topics. Please refer to the discussion above.
_________________

Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient. EITHER statement BY ITSELF is sufficient to answer the question. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.

Hi lets see the what info Q has.. Is \(\sqrt{(7x)}\) an integer?.. possible - a) if x ia an integer and of type y^2*7.. b) if x is a fraction of type y^2/7..

lets see the statements (1) \(\sqrt{(x/7)}\) is an integer. this tells us that x = y^2*7, where y is an integer, SAME as case (a) above Suff

(2) \(\sqrt{(28x)}\)is an integer. this tells us that \(\sqrt{(4*7*x)}\)is an integer. so x = y^2/28.. if y is multiple of 2, ans is YES if y is not a multiple of 2, ans is No Insuff

Can someone please explain why point 2 is not correct? What I did is that 28 is factored into => 2*2*7 so therefore I concluded x must have at least one 7. so if x has one seven then \sqrt{7x} must be an integer. Am I doing something wrong here?

Thanks! Alex

A: let's say \(\sqrt{\frac{x}{7}}\) = m

\(\sqrt{\frac{x*7}{7*7}}\) = m

\(\sqrt{7x}\)/7 = m

=> \(\sqrt{7x}\) = 7*m => \(\sqrt{7x}\) is an integer. (m is given to be an integer)

A is sufficient.

B: let's say \(\sqrt{28x}\) = k

2*\(\sqrt{7x}\) = k

\(\sqrt{7x}\) = k/2

Here the answer depends on the parity (evenness or oddness) of k and parity of k is unknown. Hence B is not sufficient.

Final answer: A

gmatclubot

Re: Is (7x)^1/2 an integer?
[#permalink]
16 May 2016, 23:51

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...