Since mean and median are different, so we know that this is NOT an evenly spaced set.

There are 10 people

_ _ _ _ _ _ _ _ _ _

Now, we know that the range is 12 so the smallest term is X and the largest term is X+12.

Median = \(\frac{a_5 + a_6}{2} = 70\)

=\(a_5 + a_6 = 140\)

We take \(a_5\) and \(a_6\) as close as possible so we get \(a_5 = a_6 = 70\)

Our set as of now looks like

_ _ _ _ 70 70 _ _ _ _

When 74 is added the median changes from \(\frac{a_5 + a_6}{2}\) to \(a_6 = 70\)

The median remains the same, but since a new term is added , which is far away from 70.5 , the average will change. We need to check if the range changes or not because if (smallest integer < 74 < greatest integer), then the range remains same else the range changes as well.


We are given that the average = 70.5 so Sum = 705 \((70.5 \times 10)\)

Now we need to take 2 integers such that there is a difference of 12 between the two. Let's take the two integers as 64 and 76.

Till here, our set looks like

64 _ _ _ 70 70 _ _ _ 76

Let's maximize the value of 76 so that our set looks like

64 _ _ _ 70 70 76 76 76 76

We need to fill the 3 gaps and the remaining sum is \(705 - 76 \times 4 - 140 - 64 = 197\)
We will distribute 197 in the remaining three gaps such that each number is less than 69.

64 65 65 67 70 70 76 76 76 76

Now, if we add 74 to this set, the updated set becomes

64 65 65 67 70 70 74 76 76 76 76

The median, range remains same but the average changes.

Now, is it possible to take a number outside the range.

What if our set is

61 _ _ _ 70 70 _ _ _ 73

In this case is it possible to reach the sum 705? Let's check, we maximize the values so that our set looks like

61 70 70 70 70 70 73 73 73 73

The sum of this set becomes 703 so we need 2 extra and the way to do that is changing our smallest integer to 62 and largest integer as 74.
Now, when we add another 74 to the list the range will not change.

Hence, only the average changes.

OA, A