Ten years ago B was twice of A in age. If the ratio of their present ages is 4:3, what is the sum of their present ages?

Options

A. 25 years
B. 30 years
C. 40 years
D. 35 years

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Define the Variables
Let the present age of A be \(3x\) and the present age of B be \(4x\), based on the given ratio of their ages (4:3).

### Step 2: Set Up the Equation for Ages 10 Years Ago
Ten years ago, the age of A would be \(3x - 10\) and the age of B would be \(4x - 10\). According to the problem, ten years ago, B was twice the age of A. Therefore, we can set up the equation:
\[ 4x - 10 = 2(3x - 10) \]

### Step 3: Simplify the Equation
Expanding the right side of the equation gives:
\[ 4x - 10 = 6x - 20 \]

### Step 4: Rearrange the Equation
Now, let's rearrange the equation to isolate \(x\):
\[ 4x - 6x = -20 + 10 \]
\[ -2x = -10 \]
\[ x = 5 \]

### Step 5: Calculate Present Ages
Now that we have \(x\), we can find the present ages of A and B:
- Present age of A: \(3x = 3(5) = 15\)
- Present age of B: \(4x = 4(5) = 20\)

### Step 6: Find the Sum of Their Present Ages
Now, we can find the sum of their present ages:
\[ 15 + 20 = 35 \]

### Final Answer
The sum of their present ages is **35**.

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