Four different positive numbers are written in ascending order. One-third of the average of all the four numbers is 19 less than the greatest of these numbers. If the average of the first three numbers is 12, the greatest number among the given numbers is:

Let the four numbers are a, b, c, d where 'd' is the greatest number.

The average of the first three numbers is 12.

$$\Rightarrow$$  $$\frac{a+b+c}{3}$$ = 12

$$\Rightarrow$$  a + b + c = 36 ..............(1)

One-third of the average of all the four numbers is 19 less than the greatest of these numbers.

$$\Rightarrow$$  $$\frac{1}{3}\left[\frac{a+b+c+d}{4}\right]=d-19$$

$$\Rightarrow$$  a + b + c + d = 12d - 228

$$\Rightarrow$$  36 + d = 12d - 228 [From (1)]

$$\Rightarrow$$  11d = 264

$$\Rightarrow$$  d = 24

$$\therefore\ $$The greatest number among the given numbers = 24

Hence, the correct answer is Option B