2n years ago, the age of Raju was four times that of his son and n years ago, the age of Raju was thrice that of his son. If n years later, the sum of the ages of Raju and his son will be 80 years, then the difference in the ages of Raju and his son is20 years.40 years.24 years.30 years.

Let's denote the current age of Raju as R and the current age of his son as S.

From the problem, we have the following equations:

1. 2n years ago, Raju was four times the age of his son. This gives us the equation: R - 2n = 4*(S - 2n).

2. n years ago, Raju was thrice the age of his son. This gives us the equation: R - n = 3*(S - n).

3. n years later, the sum of their ages will be 80. This gives us the equation: R + n + S + n = 80.

4. The difference in their ages is unknown. This gives us the equation: R - S = D, where D is the difference we are trying to find.

We now have a system of four equations with four unknowns (R, S, n, D). We can solve this system to find the values of the unknowns.

Let's start by simplifying the first two equations:

1. R - 4S = -6n
2. R - 3S = -2n

Subtract the second equation from the first to get: -S = -4n, or S = 4n.

Substitute S = 4n into the third equation to get: R + 2n + 4n + n = 80, or R + 7n = 80.

Substitute S = 4n into the fourth equation to get: R - 4n = D.

We now have two equations with two unknowns (R and n):

1. R + 7n = 80
2. R - 4n = D

Subtract the second equation from the first to get: 11n = 80 - D, or n = (80 - D) / 11.

Substitute n = (80 - D) / 11 into the second equation to get: R = D + 4*(80 - D) / 11.

We now have a single equation with a single unknown (D). We can solve this equation to find the value of D.

Unfortunately, without numerical options for D, we cannot proceed further. If you provide numerical options for D, we can solve the equation to find the correct answer.