A set of 13 integers has a median of 10 and an average of 15. If the highest term in the set is a unique, non-repeating integer, what is the lowest possible value of the highest integer in the set?
A. 81
B. 25
C. 22
D. 21
E. 20

Source:ExpertsGlobal

To make X- the last member of the set, we have to maximize other values in the set.

We know that median is the middle value. 7th term is 10. We need to minimize the 13th term.

1) 6 terms below Median can be not more than 10. So, 10 is the maximum value.
2) 7th term is 10. Given in question.
3) From 8th to 12th term can take maximum value of - One less than 13th term.

Numbers in the set are integers, Let X will be unique min 13th term, and 8th -12th numbers be X-1 each
Sum of the set = average*number of set members = 15*13

10*7 + 5*(x-1) + x = 13*15
5*(x-1) + x = 195 - 70
5*(x-1) + x = 125---------------------------> 1
5x-5+x = 195 - 70
6x = 125+5

x = 130/6 = 21.7

Since x- is integer we can say it is 21 or 22,

Let's assume x =21,

8th term - 12th term is 20. And 13th term is 21. Sum is 20*5 + 21= 121. But it should add up to 125. See eq 1

4 less than the value it should add up to. This 4 can be adjusted by increasing the values of last four terms ( 10th, 11th, 12th and 13th term).

So now, 10th term is 21, 11th term is 21, 12th term is 21 and 13th term is 22.

So Ans is - 22. C

_________________

Here's my method of finding it via concepts:
13 integers are present, and the median is 10.
In order to minimize the value of X, the largest integer, we have to maximize the value of all integers till 10.
The max value for each integer till the median is 10. So, our list has 7 integers, with value, '10'.
The sum of our list is 195.
10*number of integers till median = 10*7=70. Apart from these, we have 6 remaining integers.
Substract 70 from 195, we get 125.
So, the sum of the remaining 6 integers must be 125.
Now, in order to minimize the highest unique value, we must maximize the integers.
If we assume these to be 20, we have 6*20=120 which is less than 125.
Now assume them to be 21, we have 6*21=126, which is greater than 125.
Now write,
20 20 20 20 20 25(X) ...Matches our answer choice, but let's see if we can lessen the value of X, somehow.
20 20 20 20 21 24
20 20 20 20 22 23
20 20 20 21 21 22
After this, we will not be able to minimize the value of X, because then X would not be a UNIQUE integer in the list.
Hence, Option C is our answer.
_________________