Three positive integers, a, b, and c, are such that their average is 28 and a ≤ b ≤ c. If the median is (a + 16), then what is the least possible value of c?

Correct option is D

Given:

a, b, c are positive integers such that a ≤ b ≤ c

Average = 28

Median = a + 16

Since a ≤ b ≤ c, median is b

So, b = a + 16

Solution:

Average = 28

​$$\frac{a + b + c}{3} = 28$$​​

a + b + c = 84
Substitute b = a + 16:

a + (a + 16) + c = 84

2a + 16 + c = 84

c = 84 - 2a - 16 = 68 - 2a

To find the least possible value of c, maximize a such that a ≤ b ≤ c

Also, since b = a + 16, and c ≥ b, so:

c = 68−2a  ≥ a+16 

68 − 2a ≥ a + 16

68 − 16 ≥ 3a

52 ≥ 3a

a ≤ 17.33

max integer a = 17

Now, put a = 17:

b = 33, c = 68 − 34 = 34

Conditions: a $$= 17 \leq b = 33 \leq c = 34$$​, all integers, median = 33 = a + 16, total = 84

Thus, option (d) is right.