Set C is a set of positive integers where 4 is least. If the average

Question

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.

septwibowo · Answer

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

sandeep211986 · Answer

Let the last term in the series that is greatest term be x and number of terms be n.

Then range = x -4 and average of consecutive terms is sum of consecutive terms / number of terms .( n/2 (4 + x))/ n .
Since for an AP or consecutive terms with common difference of 1 the sum will be n/2 (first term + last term).
now x - 4 = (4 + x) / 2 solving we get x = 12.
So first term is 4 last term is 12 there fore we can find the number of terms.(sufficient).

statement 2 is not sufficient.
average is 8 so number of terms in the set with least number as 4 can be anything .Hence not sufficient.

Regards,
Sandeep
Kudos Please.

niks18 · Answer

Hi  septwibowo 
The 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   Sufficient

Approach 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  Insufficient

Option  A

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Set C is a set of positive integers where 4 is least. If the average

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Set C is a set of positive integers where 4 is least. If the average

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