The average age of Akhil and Sunita is 51 years of Sunita and Veena is 31 years, and that of Veena and Akhil is 33 years. The age of Veena is:

Options

A. 13 years
B. 49 years
C. 53 years
D. 17 years

Answer

To find the age of Veena, we will use the information given about the average ages of Akhil, Sunita, and Veena. Let's break it down step by step.

### Step 1: Set up the equations based on the average ages.

1. The average age of Akhil (A) and Sunita (S) is 51 years:
\[ \frac{A + S}{2} = 51 \]
Multiplying both sides by 2 gives:
\[ A + S = 102 \quad \text{(Equation 1)} \]

2. The average age of Sunita (S) and Veena (V) is 31 years:
\[ \frac{S + V}{2} = 31 \]
Multiplying both sides by 2 gives:
\[ S + V = 62 \quad \text{(Equation 2)} \]

3. The average age of Veena (V) and Akhil (A) is 33 years:
\[ \frac{V + A}{2} = 33 \]
Multiplying both sides by 2 gives:
\[ V + A = 66 \quad \text{(Equation 3)} \]

### Step 2: Solve the equations.

Now we have three equations:
1. \( A + S = 102 \) (Equation 1)
2. \( S + V = 62 \) (Equation 2)
3. \( V + A = 66 \) (Equation 3)

We can solve these equations step by step.

### Step 3: Express one variable in terms of another.

From Equation 1:
\[ S = 102 - A \quad \text{(Substituting into Equation 2)} \]

Substituting \( S \) into Equation 2:
\[ (102 - A) + V = 62 \]
This simplifies to:
\[ V = 62 - 102 + A \]
\[ V = A - 40 \quad \text{(Equation 4)} \]

### Step 4: Substitute \( V \) into Equation 3.

Now substitute Equation 4 into Equation 3:
\[ (A - 40) + A = 66 \]
This simplifies to:
\[ 2A - 40 = 66 \]
Adding 40 to both sides gives:
\[ 2A = 106 \]
Dividing by 2 gives:
\[ A = 53 \quad \text{(Age of Akhil)} \]

### Step 5: Find the age of Sunita using \( A \).

Substituting \( A \) back into Equation 1:
\[ 53 + S = 102 \]
Solving for \( S \):
\[ S = 102 - 53 = 49 \quad \text{(Age of Sunita)} \]

### Step 6: Find the age of Veena using \( A \).

Substituting \( A \) into Equation 4:
\[ V = 53 - 40 = 13 \quad \text{(Age of Veena)} \]

### Conclusion

The age of Veena is **13 years**.

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