Question

10 years ago father was 12 times as old as his son and 10 years hence he will be twice as old as his son find their present age

Answer 1

Step-by-step explanation:

Answer 2

$\huge{\bold{\underline{Solution:}}}$

Let the age of Father be x the age of Father's son be y.

A.T.Q,

• Case I

We subtract 10 years from both person's age because It is the case of 10 years in past. Father's age was 12 times son's ages.

$\Rightarrow x - 10 = 12(y - 10) \\ \Rightarrow x - 10 = 12y - 120 \\ \Rightarrow x - 12y = -110 \qquad \quad \bold{(1)}$

• Case II

Since It is the case of 10 years in future, So we add 10 years to both person's ages. Father's age will be two times son's age.

$\Rightarrow x + 10 = 2(y + 10) \\ \Rightarrow x + 10 = 2y + 20 \\ \Rightarrow x - 2y = 10 \qquad \quad \bold{(2)}$

On Solving the equations:

We get, y = 12

Putting y = 12 in (2):

$\Rightarrow x - 2y = 10 \\ \Rightarrow x - 2 \cdot 12 = 10 \\ \Rightarrow x = 10 + 24 = 34$

Father's age is x years and Son's age is y years.

Therefore,

Father's age is 34 years and Son's age is 12 years.