Question

Twenty years ago the age of a father was four times the age of his son . After four years from now the age of the father will be double that of the son . The present ages of the son is

Answer 1

Solution

$let \: the \: present \: age \: of \: the \: son \: be \: y$

$and \: age \: of \: father \: be \: x$

$20 \: years \: ago$

Age of son = y - 20.

And,

Age of father = x - 20

According to question

y - 20 = 4( x - 20)

Or,

x - 20 = 4x - 80

=> 4y - x = - 60......... (i)

4 years later..

Son's age = y + 4

Father's age = x + 4

But according to the question

According to the question;

⇒ y + 4 = 2(y + 4)

⇒ x + 4 = 2x + 8

⇒ x - 2y = 8 - 4

⇒ x - 2y = 4 .... (ii)

Subtracting eq (i) from (ii)

⇒ x - 4y - (x - 2y) = - 60 - 4

⇒ x - 4y - x + 2y = - 64

⇒ - 2y = - 64

⇒ y = \sf \dfrac{64}{2}

⇒ y = 32 (Son's age)

Therefore, father's age =

2( y + 4)

Put the value of y

2 ( 32 + 4)

= 68

Answer 2

Answer:

$\large{\underline{\boxed{\sf Present \: age \: of \: son = 32 \: years}}}$

Step-by-step explanation:

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

20 years ago,

Father's age = (x - 20) years

Son's age = (y - 20) years.

According to the question;

⇒ (x - 20) = 4(y - 20)

⇒ x - 20 = 4y - 80

⇒ x - 4y = - 80 + 20

⇒ x - 4y = - 60 .... (i)

_______________________

Four years later,

Father's age = (x + 4) years

Son's age = (y + 4) years

According to the question;

⇒ x + 4 = 2(y + 4)

⇒ x + 4 = 2y + 8

⇒ x - 2y = 8 - 4

⇒ x - 2y = 4 .... (ii)

_______________________

On Subtracting (i) from (ii) -

⇒ x - 4y - (x - 2y) = - 60 - 4

⇒ x - 4y - x + 2y = - 64

⇒ - 2y = - 64

⇒ y = $\sf \dfrac{64}{2}$

⇒ y = 32

On Substituting the value of y in (ii),

⇒ x - 2y = 4

⇒ x - 2 * 32 = 4

⇒ x - 64 = 4

⇒ x = 4 + 64

⇒ x = 68

_______________________

Hence, the present age of father is 68 years and present age of son is 32 years.