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:

(1) 2x - 1 is odd --> \(2x-1=odd\) --> \(2x=even\) --> \(x=\frac{even}{2}\) --> \(x\) is an integer. Now, \(x=\frac{even}{2}\) could be even (consider \(x=\frac{even}{2}=\frac{4}{2}=2=even\)) as well as odd (consider \(x=\frac{even}{2}=\frac{2}{2}=1=odd\)). Not sufficient.

(2) x^3 is odd. If \(x=integer\) then in order \(x^3=odd\) to hold true, it must be odd (answer YES), but \(x\) could also be some irrational number, for example \(x=\sqrt[3]{5}\) (answer NO). Not sufficient.

(1)+(2) Since from (1) \(x=integer\) then from (2) we have that \(x=odd\). Sufficient.

I don't understand your approach for the first argument:

\(2x-1=odd\)

If the result muss be odd, so x must be even. It will be even if x is 2 or greater than 2. If x is odd the result won't be odd. If I say x = 1 so the result will be Zero. But Zero is neither even nor odd. So the satetement is sufficient ?

I don't understand your approach for the first argument:

\(2x-1=odd\)

If the result muss be odd, so x must be even. It will be even if x is 2 or greater than 2. If x is odd the result won't be odd. If I say x = 1 so the result will be Zero. But Zero is neither even nor odd. So the satetement is sufficient ?

Several things:

1. \(2x-1=odd\) --> \(2x=odd+1=odd+odd=even\) --> so \(2x=even\) --> \(x=\frac{even}{2}=integer\). Hence \(2x-1=odd\) just means that \(x\) is an integer (it can be even as well as odd).

2. If \(x=1\)the result wont be zero, it'l be 1, so odd: \(2*1-1=2-1=1=odd\).

3. Zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Or in another way: an even number is an integer of the form \(n=2k\), where \(k\) is an integer. So for \(k=0\) --> \(n=2*0=0\).

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.
_________________

I understand why A is wrong...for ex. x=0.5... I understand why B is wrong but can't see why C is ok...hm..

Hi, 1) Statement 1 gives you 2x-1 is odd .. so 2x is even .. x can be i) odd, ii) even or iii) a fraction with a denominator of 2 in its simplified form..

2) statement 2 gives you x^3 is odd.. this gives you two cases of x.. i) x is odd, or ii) x is third root of some odd number..

You have correctly realized that A and B are not sufficient alone..

But combined.. for both 1 and 2 to be true, the value of x has to be something that fits in both the cases.. look at the possible values of x, only X IS ODD is common to both.. so x is odd and C is the answer..

I understand why A is wrong...for ex. x=0.5... I understand why B is wrong but can't see why C is ok...hm..

Hi, 1) Statement 1 gives you 2x-1 is odd .. so 2x is even .. x can be i) odd, ii) even or iii) a fraction with a denominator of 2 in its simplified form..

2) statement 2 gives you x^3 is odd.. this gives you two cases of x.. i) x is odd, or ii) x is third root of some odd number..

You have correctly realized that A and B are not sufficient alone..

But combined.. for both 1 and 2 to be true, the value of x has to be something that fits in both the cases.. look at the possible values of x, only X IS ODD is common to both.. so x is odd and C is the answer..

Hope it helps you

Sorry, but I think that a fraction with a denominator of 2 in its simplified form is not an option, because then 2x would be odd, and 2x -1= even. The only thing that statement 1 gives us is that x is an integer.
_________________

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...