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:

Re: GmatClub Test: Number Properies - I (DS) [#permalink]

Show Tags

07 May 2010, 11:21

ykaiim wrote:

Is 2(a+b-c) an odd integer?

1. a, b and c are consecutive numbers 2. b=a+c

The OA is A. But, I think it is incorrect. Case1: 0,1,2 the a+b-c = -1 (Yes) Case2: 1,2,3 the a+b-c = 0 (No)

Can someone tell me where I am wrong?

You forgot to multiply the -1 by 2

2(a+b-c) 2(0+1-2) 2(-1) = -2

or 2a + 2b - 2c 2(0) + 2(1) - 2(2) 2-4 = -2

Is the question correct? Multiplying anything by 2 will never give an odd so the question states the answer. In which case you wouldn't need to test 1 or 2

Re: GmatClub Test: Number Properies - I (DS) [#permalink]

Show Tags

07 May 2010, 17:07

lagomez wrote:

ykaiim wrote:

Is 2(a+b-c) an odd integer?

1. a, b and c are consecutive numbers 2. b=a+c

The OA is A. But, I think it is incorrect. Case1: 0,1,2 the a+b-c = -1 (Yes) Case2: 1,2,3 the a+b-c = 0 (No)

Can someone tell me where I am wrong?

You forgot to multiply the -1 by 2

2(a+b-c) 2(0+1-2) 2(-1) = -2

or 2a + 2b - 2c 2(0) + 2(1) - 2(2) 2-4 = -2

Is the question correct? Multiplying anything by 2 will never give an odd so the question states the answer. In which case you wouldn't need to test 1 or 2

I agree with lagomez. any number x 2 = Even Something must be wrong

Re: GmatClub Test: Number Properies - I (DS) [#permalink]

Show Tags

07 May 2010, 18:11

2

This post received KUDOS

Expert's post

ykaiim wrote:

Is 2(a+b-c) an odd integer?

1. a, b and c are consecutive numbers 2. b=a+c

The OA is A. But, I think it is incorrect. Case1: 0,1,2 the a+b-c = -1 (Yes) Case2: 1,2,3 the a+b-c = 0 (No)

Can someone tell me where I am wrong?

\(2(a+b-c)\) can be: odd, in case \(a+b-c=\frac{odd}{2}\); even, in case \(a+b-c=integer\) not an integer at all, in case \(a+b-c\) does not equal to any above. For example: \(a+b-c=\sqrt{2}\) or \(a+b-c=0.3\).

Guess statement (1) is saying: "a, b and c are consecutive integers". The word "consecutive" is redundant here. Just knowing that "a, b and c are integers" is enough to say that this statement is sufficient to answer the question. And the answer would be NO: a, b and c are integers --> \(a+b-c=integer\) --> \(2(a+b-c)=even\).

Statement (2) is clearly not sufficient. _________________

Re: GmatClub Test: Number Properies - I (DS) [#permalink]

Show Tags

07 May 2010, 18:31

Bunuel wrote:

ykaiim wrote:

Is 2(a+b-c) an odd integer?

1. a, b and c are consecutive numbers 2. b=a+c

The OA is A. But, I think it is incorrect. Case1: 0,1,2 the a+b-c = -1 (Yes) Case2: 1,2,3 the a+b-c = 0 (No)

Can someone tell me where I am wrong?

\(2(a+b-c)\) can be: odd, in case \(a+b-c=\frac{odd}{2}\); even, in case \(a+b-c=integer\) not an integer at all, in case \(a+b-c\) does not equal to any above. For example: \(a+b-c=\sqrt{2}\) or \(a+b-c=0.3\).

Guess statement (1) is saying: "a, b and c are consecutive integers". The word "consecutive" is redundant here. Just knowing that "a, b and c are integers" is enough to say that this statement is sufficient to answer the question. And the answer would be NO: a, b and c are integers --> \(a+b-c=integer\) --> \(2(a+b-c)=even\).

Statement (2) is clearly not sufficient.

What numbers can be substituted for A, B, and C and when multiplied by 2 gives you an odd integer?

Re: GmatClub Test: Number Properies - I (DS) [#permalink]

Show Tags

07 May 2010, 18:49

1

This post received KUDOS

Expert's post

lagomez wrote:

Bunuel wrote:

ykaiim wrote:

Is 2(a+b-c) an odd integer?

1. a, b and c are consecutive numbers 2. b=a+c

The OA is A. But, I think it is incorrect. Case1: 0,1,2 the a+b-c = -1 (Yes) Case2: 1,2,3 the a+b-c = 0 (No)

Can someone tell me where I am wrong?

\(2(a+b-c)\) can be: odd, in case \(a+b-c=\frac{odd}{2}\); even, in case \(a+b-c=integer\) not an integer at all, in case \(a+b-c\) does not equal to any above. For example: \(a+b-c=\sqrt{2}\) or \(a+b-c=0.3\).

Guess statement (1) is saying: "a, b and c are consecutive integers". The word "consecutive" is redundant here. Just knowing that "a, b and c are integers" is enough to say that this statement is sufficient to answer the question. And the answer would be NO: a, b and c are integers --> \(a+b-c=integer\) --> \(2(a+b-c)=even\).

Statement (2) is clearly not sufficient.

What numbers can be substituted for A, B, and C and when multiplied by 2 gives you an odd integer?

Note that stem does not say that a, b and c are integers.

So for example: a=3/2, b=0, c=0 --> \(2(a+b-c)=3\). As I said if \(a+b-c=\frac{odd}{2}\), then \(2(a+b-c)=2*\frac{odd}{2}=odd\). This option is ruled out by statement (1), which says that a, b and c are integers.

Re: GmatClub Test: Number Properies - I (DS) [#permalink]

Show Tags

08 May 2010, 04:56

Bunuel wrote:

ykaiim wrote:

Is 2(a+b-c) an odd integer?

1. a, b and c are consecutive numbers 2. b=a+c

The OA is A. But, I think it is incorrect. Case1: 0,1,2 the a+b-c = -1 (Yes) Case2: 1,2,3 the a+b-c = 0 (No)

Can someone tell me where I am wrong?

\(2(a+b-c)\) can be: odd, in case \(a+b-c=\frac{odd}{2}\); even, in case \(a+b-c=integer\) not an integer at all, in case \(a+b-c\) does not equal to any above. For example: \(a+b-c=\sqrt{2}\) or \(a+b-c=0.3\).

Guess statement (1) is saying: "a, b and c are consecutive integers". The word "consecutive" is redundant here. Just knowing that "a, b and c are integers" is enough to say that this statement is sufficient to answer the question. And the answer would be NO: a, b and c are integers --> \(a+b-c=integer\) --> \(2(a+b-c)=even\).

Statement (2) is clearly not sufficient.

totally right I assumed that a, b and c were integers

Re: Is 2(a+b-c) an odd integer? 1. a, b and c are consecutive [#permalink]

Show Tags

08 Nov 2013, 12:38

If we substitute b = a + c in 2(a+b-c), we get 4a. Now, whatever the value of a, the outcome will always be even. This statement should be sufficient too. In my opinion, the answer should be D.

I am not sure that whether what I have done is correct but, this is what came to my mind when I was attempting this question; I marked D. _________________

--------------------------------------------------------------- Consider to give me kudos if my post helped you.

Re: Is 2(a+b-c) an odd integer? 1. a, b and c are consecutive [#permalink]

Show Tags

09 Nov 2013, 02:47

Expert's post

AasaanHai wrote:

If we substitute b = a + c in 2(a+b-c), we get 4a. Now, whatever the value of a, the outcome will always be even. This statement should be sufficient too. In my opinion, the answer should be D.

I am not sure that whether what I have done is correct but, this is what came to my mind when I was attempting this question; I marked D.

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...