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:

what it means is | r | != 1 i meant | r | not equal to 1 != ( not equal) in programming world i am sorry for this confusion

This means that I wrote new question:

If r is an integer and \(|r|!=1\) is \(r\) even?

(1) \(r\) is not positive.

(2) \(5r>-2\).

Answer: C.

As for original question:

If \(|r|\neq{1}\) is \(r=even\)?

\(|r|\neq{1}\) --> \(r\neq{1}\) and \(r\neq{-1}\).

(1) \(r\) is not positive --> Clearly insufficient, \(r\) can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).

(2) \(2r>-5\) --> \(r>-\frac{5}{2}=-2.5\) --> again \(r\) can be even or odd (except -1 and 1): -2, 0, 2, 3, 4, 5, ... Not sufficient.

(1)+(2) \(r\) is not positive and \(r>-2.5\) --> \(r\) can be -2, -1, or 0. But as given that \(r\neq{-1}\) then only valid solutions for \(r\) are -2 and 0, both are even. Sufficient.

Bunuel, If condition (2) above is correct = 5r>-2, it would mean r =0,1,2,.. I think you have the condition flipped? and if r =0,1,2, and from (1) we have r <> 1, <> -1, I believe that cannot let us confirm that r is even. So E

I just saw, (2) was a typo. Very rare you make a mistake!

In the solution above, is 0 considered positive. If not, then we are left with -2 and that is even, so the answer still works. _________________

Bunuel, If condition (2) above is correct = 5r>-2, it would mean r =0,1,2,.. I think you have the condition flipped? and if r =0,1,2, and from (1) we have r <> 1, <> -1, I believe that cannot let us confirm that r is even. So E

I just saw, (2) was a typo. Very rare you make a mistake!

In the solution above, is 0 considered positive. If not, then we are left with -2 and that is even, so the answer still works.

No mistake there. Nothing was flipped. Look carefully, I consider in my previous post two questions:

MINE: If r is an integer and \(|r|!=1\) is \(r\) even?

(1) \(r\) is not positive.

(2) \(5r>-2\).

Answer: C.

AND ORIGINAL:

If \(|r|\neq{1}\) is \(r=even\)?

(1) \(r\) is not positive.

(2) \(2r>-5\).

Answer: C.

Totally in my previous posts we have 3 questions. First 2 I "invented" because rxs0005 meant by | r | != 1 \(|r|\neq{1}\) and I understood it as it was written \(|r|!={1}\). So two questions I consider in my first post are those I "invented".

Question 1 (mine):

If r is an integer and \(|r|!=1\) is \(r\) even?

First of all factorial of a number equals to 1 in two cases: number equals to 0 --> 0!=1 or number equals to 1 --> 1!=1.

\(|r|!=1\) --> \(|r|=0\) or \(|r|=1\), so \(r\) can take 3 values: 0, 1, and -1, out of which only 0 is even. So basically questions asks: is \(r=0\)?

(1) \(r\) is not positive --> \(r\) can be 0 or -1. Not sufficient.

(2) \(2r>-5\) --> \(r>-\frac{5}{2}=-2.5\) --> \(r\) can be 0, 1 or -1. Not sufficient.

(1)+(2) still two values of \(r\) are possible 0 and -1. Not sufficient.

Answer: E.

Question #2 (mine):

If r is an integer and \(|r|!=1\) is \(r\) even?

(1) \(r\) is not positive --> \(r\) can be 0 or -1. Not sufficient.

(2) \(5r>-2\) --> \(r>-\frac{2}{5}=-0.4\) --> \(r\) can be 0 or 1. Not sufficient.

(1)+(2) only one value of \(r\) is possible: 0, hence \(r\) is even. Sufficient.

Answer: C.

Question #3 (original):

If \(|r|\neq{1}\) is \(r=even\)?

\(|r|\neq{1}\) --> \(r\neq{1}\) and \(r\neq{-1}\).

(1) \(r\) is not positive --> Clearly insufficient, \(r\) can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).

(2) \(2r>-5\) --> \(r>-\frac{5}{2}=-2.5\) --> again \(r\) can be even or odd (except -1 and 1): -2, 0, 2, 3, 4, 5, ... Not sufficient.

(1)+(2) \(r\) is not positive and \(r>-2.5\) --> \(r\) can be -2, -1, or 0. But as given that \(r\neq{-1}\) then only valid solutions for \(r\) are -2 and 0, both are even. Sufficient.

Answer: C.

BUT: if the second statement were as you wrote answer still would be C.

Question #4 (your's):

If \(|r|\neq{1}\) is \(r=even\)?

\(|r|\neq{1}\) --> \(r\neq{1}\) and \(r\neq{-1}\).

(1) \(r\) is not positive --> Clearly insufficient, \(r\) can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).

(2) \(5r>-2\) --> \(r>-\frac{2}{5}=-0.4\) --> \(r\) can be 0 or 2, 3, 4, ... Not sufficient.

(1)+(2) Only one solution is possible for \(r\) is 0, which is even. Sufficient.

what it means is | r | != 1 i meant | r | not equal to 1 != ( not equal) in programming world i am sorry for this confusion

This means that I wrote new question:

If r is an integer and \(|r|!=1\) is \(r\) even?

(1) \(r\) is not positive.

(2) \(5r>-2\).

Answer: C.

As for original question:

If \(|r|\neq{1}\) is \(r=even\)?

\(|r|\neq{1}\) --> \(r\neq{1}\) and \(r\neq{-1}\).

(1) \(r\) is not positive --> Clearly insufficient, \(r\) can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).

(2) \(2r>-5\) --> \(r>-\frac{5}{2}=-2.5\) --> again \(r\) can be even or odd (except -1 and 1): -2, 0, 2, 3, 4, 5, ... Not sufficient.

(1)+(2) \(r\) is not positive and \(r>-2.5\) --> \(r\) can be -2, -1, or 0. But as given that \(r\neq{-1}\) then only valid solutions for \(r\) are -2 and 0, both are even. Sufficient.

Answer: C.

Bunuel - I see that there was an original question and you had modified it. I got it. Thanks for the clarification. As I said, very rare you make a mistake _________________

Re: if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink]

Show Tags

22 Apr 2014, 06:57

jlgdr wrote:

OA on original question should be E. Please correct

Thanks

Cheers! J

Hi jlgdr, The answer for the original question is indeed E , if "!" meant as factorial. However if the same is mentioned as "not equal to", then the answer should be C. I believe the question is originally posted for "!" as nt equal to. In that case "C" is the correct answer as mentioned in the OA. Hope that helps.

Thanks.

gmatclubot

Re: if | r | != 1 is integer r even 1 r is not positive 2 2r
[#permalink]
22 Apr 2014, 06:57

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...

This highly influential bestseller was first published over 25 years ago. I had wanted to read this book for a long time and I finally got around to it...