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 is

Options

A. A)20 years
B. B)40 years
C. C)24 years
D. D)30 years

Answer

To solve the problem step by step, let's denote Raju's current age as \( r \) and his son's current age as \( s \).

### Step 1: Set up the equations based on the information given.

1. **2n years ago**:
- Raju's age was \( r - 2n \)
- Son's age was \( s - 2n \)
- According to the problem, \( r - 2n = 4(s - 2n) \)

Rearranging gives us:
\[ r - 2n = 4s - 8n \]
\[ r = 4s - 6n \quad \text{(Equation 1)} \]

2. **n years ago**:
- Raju's age was \( r - n \)
- Son's age was \( s - n \)
- According to the problem, \( r - n = 3(s - n) \)

Rearranging gives us:
\[ r - n = 3s - 3n \]
\[ r = 3s - 2n \quad \text{(Equation 2)} \]

### Step 2: Solve the equations.

Now we have two equations:
- From Equation 1: \( r = 4s - 6n \)
- From Equation 2: \( r = 3s - 2n \)

Setting the two expressions for \( r \) equal to each other:
\[ 4s - 6n = 3s - 2n \]
Rearranging gives:
\[ 4s - 3s = 6n - 2n \]
\[ s = 4n \quad \text{(Equation 3)} \]

### Step 3: Substitute \( s \) back to find \( r \).

Now substitute \( s = 4n \) into Equation 1:
\[ r = 4(4n) - 6n \]
\[ r = 16n - 6n \]
\[ r = 10n \quad \text{(Equation 4)} \]

### Step 4: Use the future age condition.

According to the problem, \( n \) years later, the sum of their ages will be 80:
\[ (r + n) + (s + n) = 80 \]
Substituting \( r \) and \( s \) from Equations 3 and 4:
\[ (10n + n) + (4n + n) = 80 \]
\[ 11n + 5n = 80 \]
\[ 16n = 80 \]
\[ n = 5 \]

### Step 5: Find the current ages.

Now substitute \( n = 5 \) back into Equations 3 and 4:
\[ s = 4n = 4(5) = 20 \]
\[ r = 10n = 10(5) = 50 \]

### Step 6: Calculate the difference in ages.

The difference in ages is:
\[ r - s = 50 - 20 = 30 \]

Thus, the difference in the ages of Raju and his son is **30 years**.

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