Question

Thirteen years before fathers age was four times of sons age. After seven years fathers age will be double that of sons age. What will be the age of the father now?

Answer 1

21 years ago of the father

Answer 2

Given:

Thirteen years before father's age was four times of son's age. After seven years father's age will be double that of the

$x - 4y + 39 = 0$

son's age.

To find:

The age of father now.

Solution:

The father's age at present is 53 years.

To answer this question, we will follow the following steps:

Let the present age of the father and son be x years and y years respectively.

Thirteen years before,

The age of father = (x - 13) years

The age of son = (y - 13) years

Now,

According to the question, we have,

$x - 13 = 4(y - 13)$

$x - 13 = 4y - 52$

$x - 4y + 39 = 0 \: \: (i)$

Also,

After 7 years,

father's age = (x + 7) years

Son's age = (y + 7) years

So,

$x + 7 = 2(y + 7)$

$x + 7 = 2y + 14$

$x - 2y - 7 = 0 \: \: (ii)$

After subtracting (i) from (ii), we get

$2y = 46$

$y = 23 \: years$

On putting the value of y in (ii), we get

$x - 46 - 7 = 0$

$x = 53 \: years$

Hence, the present age of the father is 53 years.