For S, a set of five integers, is the mean of the set an integer in S?

Question

For S, a list of five integers, is the mean of the list an integer in S?

(1) The difference between the smallest integer in S and the median of S is the same as the difference between the median and the highest integer in S.

(2) The difference between the second smallest integer in S and the median of S is the same as the difference between the median and the second highest integer in S.

yashikaaggarwal · Answer

Let the five number be a1, a2, a3, a4, and a5

statement 1 = a3 - a1 = a5 - a3
=> 2a3 = a1+a5
so, a1, a3, and a5 are at consecutive distance

case 1 : a1, a2, a3, a4, and a5 = 1,2,3,4,5
mean = 3

case 2: a1, a2, a3, a4, and a5 = 4,6,12,16,20
mean = 58/5 = 11.6 (not in s list)
(insufficient)

statement 2 = a2 - a1 = a4 - a3
=> 2a3 = a2+a4
so, a2, a3, and a4 are at consecutive distance

case 1 : a1, a2, a3, a4, and a5 = 1,2,3,4,5
mean = 3

case 2: a1, a2, a3, a4, and a5 = 2,6,12,18,21
mean = 59/5 = 11.8 (not in s list)
(insufficient)

Statement 1&2 together: a1, a2, a3, a4, and a5 are at consecutive distance 
so their mean will always be equal to the middle term that is a3
(sufficient)

Answer should be C

IanStewart · Answer

Neither statement could be sufficient alone, because using either we can freely determine two values in the set, and can easily get a non-integer mean.

Both Statements together tell you the set is "symmetric" (if you have a value above the median, there is a corresponding value equally far below the median). For any symmetric set, the mean and median are equal, so the two Statements together are sufficient.

If you weren't familiar with symmetric sets, you could do the problem algebraically. If m is the median, we know our smallest and largest values are m - d and m + d (from Statement 1), and our second-largest and second-smallest values are m - c and m + c (from Statement 2). So our list is

m-d, m-c, m, m+c, m+d

and summing this list and dividing by 5, we find the mean is equal to m, the median, which is an integer in our list.

dimri10 · Answer

If we pick numbers:
Statement 1.:
6==N==12==n==18
If n are 9 and 15 then 12 (the mean) is in the list. If n are 8 and 13 or 7 and 16, i.e , then the mean is not in the list.
Same for statement 2 
Both means that the gap between each N is equal, so the median is the mean 
6 ---9--12---15---18

Posted from my mobile device

p05eidon · Answer

For S, a list of five integers, is the mean of the list an integer in S?

Statement-1 
(1) The difference between the smallest integer in S and the median of S is the same as the difference between the median and the highest integer in S.

Let us consider the list 1,2,3,4,5
this list satisfies statement-1. Also, we can say that the mean is an integer in the list.

Let us consider another list 1,1,3,4,5
This list also satisfies statement-1. But mean is not an integer in the list.

So, statement-1 is not sufficient.

Statement-2 
(2) The difference between the second smallest integer in S and the median of S is the same as the difference between the median and the second-highest integer in S.

Let us consider the list 1,2,3,4,5
this list satisfies statement-2. Also, we can say that the mean is an integer in the list.

Let us consider another list 1,2,3,4,6
This list also satisfies statement-2. But mean is not an integer in the list.

So, statement-2 is not sufficient.

Combining the 2 statements

upon combining the two statements, we can understand that the list contains integers which are evenly spaced (Like 1,2,3,4,5 or 1,3,5,7,9).
In an evenly spaced odd-numbered list, the mean always an integer in the list.

Combining two statements is sufficinet.

Answer - Option C

bumpbot · Answer

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For S, a set of five integers, is the mean of the set an integer in S?

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For S, a set of five integers, is the mean of the set an integer in S?

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