septwibowo wrote:

Bunuel wrote:

Set C is a set of positive integers where 4 is least. If the average (arithmetic mean) of C equals its range, then how many positive integers does set C have?

(1) The elements in C are consecutive integers.

(2) The average (arithmetic mean) of C is 8.

When I read the question, I thought that we MUST HAVE all positive integers, since the least is 4 (and also the Q is stated that Set C is a set of POSITIVE integers)

Maybe this is the question that tests is the Set C consist of ALL Integers or NOT ALL integers?

1. Automatically sufficient.

2. Average 8 is not always resulted from of one set of ALL integers. Not sufficient.

A.

But I am not sure with my answer

Hi

septwibowoThe question is asking

How many integers i.e. count of no of elements in set C. Hence it is not about whether the elements are integer or not. Even if they are not, then also we will get a count.

First term is \(4\), let the last term be \(T_n\) and let the number of element be \(n\). Given \(T_n-4=Average=\frac{Sum}{n}, or Sum = n(T_n-4)\)

Statement 1: Approach 1:this implies that the set is an AP with common difference of \(1\). Sum of an AP series \(= \frac{n}{2}(first term+Last term)\)

or \(n(T_n-4)=\frac{n}{2}(T_n+4)\) or \(T_n=12\)

So the element will be from \(4\) to \(12\) i.e \(9\)

SufficientApproach 2: as the elements are consecutive, the average will be the middle number and as average is an integer, hence number of elements will be ODD. so you can simply list down numbers starting from 4 to arrive at a set 4,5,6,7,

8,9,10,11,12, with 8 as average = range (12-4)

Statement 2: implies \(T_n-4=8\) or \(T_n=12\), so Set C can have only two element {4,12} or can be {4,5,6,7,8,...12}. Hence

InsufficientOption

A