B is eighteen years younger than A. The ratio of B’s age six years hence to C’s present age is 3 :2. If at present A’s age is twice the age of C, then what was B’s age four years ago?

Options

A. 24 years
B. 28 years
C. 26 years
D. 20 years

Answer

To solve the problem step by step, we will define the variables and use the information given in the question.

### Step 1: Define Variables
Let:
- A = present age of A
- B = present age of B
- C = present age of C

### Step 2: Set Up the Equations
From the information given:
1. B is 18 years younger than A:
\[ B = A - 18 \]

2. The ratio of B’s age six years hence to C’s present age is 3:2:
\[ \frac{B + 6}{C} = \frac{3}{2} \]
Cross-multiplying gives:
\[ 2(B + 6) = 3C \]
Expanding this:
\[ 2B + 12 = 3C \]

3. A’s age is twice the age of C:
\[ A = 2C \]

### Step 3: Substitute and Solve
Now we can substitute the value of A from the third equation into the first equation:
\[ B = (2C) - 18 \]
Now substitute this expression for B into the second equation:
\[ 2((2C - 18) + 6) = 3C \]
Simplifying the left side:
\[ 2(2C - 12) = 3C \]
\[ 4C - 24 = 3C \]
Now, isolate C:
\[ 4C - 3C = 24 \]
\[ C = 24 \]

### Step 4: Find A and B
Now that we have C, we can find A:
\[ A = 2C = 2 \times 24 = 48 \]

Now substitute C back into the equation for B:
\[ B = A - 18 = 48 - 18 = 30 \]

### Step 5: Calculate B’s Age Four Years Ago
To find B's age four years ago:
\[ \text{B's age 4 years ago} = B - 4 = 30 - 4 = 26 \]

### Final Answer
Thus, B's age four years ago was **26 years**.

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