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:

Hello Fluke and Everyone in this forum - can you please explain the conceptual way of to solve this and improve my understanding ?

My approach to solve this:

Statement 1: Sufficient, because the average of the even number of consecutive integers can never be an integer.

Statement 2: I understand what the statement means, but I am struggling to understand the concept.

My understanding is

Average = Sum/Number of terms Therefore, Sum = Average * Number of Terms Therefore S=Average/n. But what can we conclude from this? The book says that this is sufficient as well.

So the answer is D. But I don't understand the Statement 2.

statement 2 says : S is positive and S>n.....(eq) now as you said sum = avg*n = S

subtitute value of S in equation we get => avg*n>n now since n cant be negative, we can cancel n out on both sides without a change of sign which gives me =>avg>1 so it succesfully answers NO! to is avg=1?

hope its clear... _________________

It matters not how strait the gate, How charged with punishments the scroll, I am the master of my fate : I am the captain of my soul. ~ William Ernest Henley

So if 0<s<n, then also the answer is average cannot be 1, because s has at least to be 1 and n hs to be greater than 1, so in either case the average cannot be 1. Is my thinking correct fluke and everyone please? _________________

So if 0<s<n, then also the answer is average cannot be 1, because s has at least to be 1 and n hs to be greater than 1, so in either case the average cannot be 1. Is my thinking correct fluke and everyone please?

s<n or s>n means \(s \ne n\). If s NOT EQUAL n, the average can't be 1. _________________

Helpful Geometry formula sheet:best-geometry-93676.html I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.

Hello Fluke and Everyone in this forum - can you please explain the conceptual way of to solve this and improve my understanding ?

My approach to solve this:

Statement 1: Sufficient, because the average of the even number of consecutive integers can never be an integer.

Statement 2: I understand what the statement means, but I am struggling to understand the concept.

My understanding is

Average = Sum/Number of terms Therefore, Sum = Average * Number of Terms Therefore S=Average/n. But what can we conclude from this? The book says that this is sufficient as well.

So the answer is D. But I don't understand the Statement 2.

Is the average of n consecutive integers = 1?

Statement 1

n is even

If n is even then the average of consecutive integers will not be an integer at all so it won't be 1 for sure

Just to illustrate

n = 2, the avarege of 1 and 2 is 1.5 n=4, the average of 1,2,3 and 4 is 2.5 etc...

Sufficient

Statement 2

We can manipulate this to 0<s/n<1

So we are told that the average is in fact less than 1

Re: Is the average of n consecutive integers equal to 1 ? [#permalink]

Show Tags

01 Feb 2014, 04:03

+1 D

Stmt1: n is even.

let n=4, and 4 consecutive integers can be a-1,a,a+1,a+2. sum=a-1+a+a+1+a+2 = 4a+2 their average is (4a+2)/4 a+1/2 since a is integer and adding a fraction to a. mean can never be 1. SUFF

Stmt2: 0<S<n.

for average to be 1 S must be equal to n. Hence Stmt2 is also SUFF. _________________

Re: Is the average of n consecutive integers equal to 1 ? [#permalink]

Show Tags

01 Oct 2015, 03:50

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. _________________

Re: Is the average of n consecutive integers equal to 1 ? [#permalink]

Show Tags

26 Apr 2016, 22:45

if average has to be = 1 , sum of elements must be equal to n . E G. 0,1,2 ; -1,0,1,2,3 .

if number of elements is odd in consecutive numbers, middle element= average. if number of elements is even , average = sum of middle two elements/2 ---> odd number/2 so average would not be an integer.

statement 1: n is even therefore n can not be equal to 1 so it gives a definite Ans statement 2: 0<S<n but for average to be 1, S must be equal to n so it gives a definite Ans

therefore D.

gmatclubot

Re: Is the average of n consecutive integers equal to 1 ?
[#permalink]
26 Apr 2016, 22:45

Last year when I attended a session of Chicago’s Booth Live , I felt pretty out of place. I was surrounded by professionals from all over the world from major...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...