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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.

Re: Average of Consecutive Integers [#permalink]
02 Jul 2011, 13:52

statement 1 is fairly clear as you said

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

Re: Average of Consecutive Integers [#permalink]
03 Jul 2011, 00:44

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? _________________

Re: Average of Consecutive Integers [#permalink]
03 Jul 2011, 01:27

1

This post received KUDOS

enigma123 wrote:

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]
01 Feb 2014, 03:03

+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. _________________

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gmatclubot

Re: Is the average of n consecutive integers equal to 1 ?
[#permalink]
01 Feb 2014, 03:03

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...