Median of these 3 numbers a, b, and c is ‘b’ because a ≤ b ≤ c
Therefore, b = a + 11
Theoretically, the least value of c is when c = b.
Therefore, a + (a + 11) + (a + 11) = 60 (b and c are equal and b, the median, is a + 11)
Or 3a = 38 or a = 12.66
So, b = c = 12.66 + 11 = 23.66
However, we know that these numbers are all integers.
Therefore, a, b, and c cannot take these values.
So, the least value for c with this constraint is NOT likely to be when c = b
Let us increment c by 1. Let c = (b + 1)
In this scenario, a + (a + 11) + (a + 12) = 60
Or 3a = 37. The value of the numbers is not an integer in this scenario as well.
Let us increment c again by 1. i.e., c = b + 2
Now, a + (a + 11) + (a + 13) = 60
Or 3a = 36 or a = 12.
If a = 12, b = 23 and c = 25.
The least value for c that satisfies all these conditions is 25.
Correct answer is choice C.