Question

Four years ago a father was nine times as old as his son, and 8 years hence the father's age will be three times the son's age. Find their present ages​

Answer 1

Answer:

Let father's age = x

son's age = y

Five years hence, age of father = x+5 

                                 age of son = y+5

So (x+5) = 3(y+5)

⇒ x =3y+10

Five years ago, age of father = x-5 

                                 age of son = y-5

So x-5 = 7(y-5)

⇒3y + 10 - 5 = 7y - 35

⇒ 4y = 40

⇒y = 10  <<<<<<<<son's age

   x = 3y +10 = 40 <<<<<<father's age.

Answer 2

Answer:-

$\sf{The \ father's \ and \ his \ son's \ ages \ are}$

$\sf{40 \ years \ and \ 8 \ years \ respectively.}$

Given:

• Four years ago a father was nine times as old as his son.

• 8 years hence the father's age will be three times the son's age.

To find:

• Present ages.

Solution:

$\sf{Let \ the \ father's \ age \ be \ x \ years \ and}$

$\sf{his \ son's \ age \ be \ y \ years.}$

$\sf{According \ to \ the \ first \ condition.}$

$\sf{x-4=9(y-4)}$

$\sf{\therefore{x-4=9y-36}}$

$\sf{\therefore{x-9y=-32...(1)}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{x+8=3(y+8)}$

$\sf{\therefore{x+8=3y+24}}$

$\sf{\therefore{x-3y=24-8}}$

$\sf{\therefore{x-3y=16...(2)}}$

$\sf{Subtract \ equation (1) \ from \ equation (2)}$

$\sf{x-3y=16}$

$\sf{-}$

$\sf{x-9y=-32}$

__________________

$\sf{6y=48}$

$\sf{\therefore{y=8}}$

$\sf{Substitute \ y=8 \ in \ equation (2)}$

$\sf{x-3(8)=16}$

$\sf{\therefore{x=16+24}}$

$\sf{\therefore{x=40}}$

$\sf\purple{\tt{\therefore{The \ father's \ and \ his \ son's \ ages \ are}}}$

$\sf\purple{\tt{40 \ years \ and \ 8 \ years \ respectively.}}$