Question

The present age of Harish is 8 times the sum of the ages of his two sons. After 8 years, his age will be twice the sum of the ages of his two sons. The present age of Harish (in years) is:

• A. 31
• B. 34
• C. 33
• D. 32

Answer 1

The Correct Option is D

Let the present ages of Harish's two sons be \( x \) and \( y \). 

Then, the sum of the ages of his two sons is \( x + y \).

According to the problem:

1. The present age of Harish is 8 times the sum of the ages of his two sons.

\( H = 8(x + y) \)

2. After 8 years, Harish's age will be twice the sum of the ages of his two sons.

\( H + 8 = 2(x + y + 16) \)

Solve these equations:

From Equation (1):

\( H = 8(x + y) \)

Substitute into Equation (2):

\( 8(x + y) + 8 = 2(x + y + 16) \)

Expand and simplify:

\( 8(x + y) + 8 = 2x + 2y + 32 \)

\( 8x + 8y + 8 = 2x + 2y + 32 \)

Rearrange terms:

\( 8x + 8y - 2x - 2y = 32 - 8 \)

\( 6x + 6y = 24 \)

Divide by 6:

\( x + y = 4 \)

Substitute \( x + y = 4 \) back into the first equation:

\( H = 8(x + y) = 8 \times 4 = 32 \)

Thus, the present age of Harish is 32 years.

Answer 2

Let Harish's present age be \(H\) and the sum of the present ages of his two sons be \(S\). We are given that \(H = 8S\).

After 8 years, Harish's age will be \(H + 8\), and the sum of the ages of his two sons will be \(S + 8 + 8 = S + 16\). 

We are given that \(H + 8 = 2(S + 16)\).

Substituting \(H = 8S\) into the second equation, we get \(8S + 8 = 2(S + 16)\).

\(8S + 8 = 2S + 32\)

\(6S = 24\)

\(S = 4\)

Therefore, \(H = 8 \times 4 = 32\).