Question

A father is 7 times as old as his son. Three
years
ago, the father was 13 times as old as his son.
What are their
present
ages?​

Answer 1

Answer:

present age of son x, father is 7x

3years ago

7x-3=13x-39

36=4x

x=9,son is 9,father is 45

Answer 2

Solution :

Let the present age of son's be r years & the present age of father's be 7r years respectively;

A/q

$\underbrace{\bf{3\:years\:ago\::}}}$

The son age was (r-3) years.

The father age was (7r-3) years.

$\longrightarrow\sf{(7r - 3) = 13(r-3)}\\\\\longrightarrow\sf{7r-3 = 13r - 39}\\\\\longrightarrow\sf{7r-13r = -39 + 3}\\\\\longrightarrow\sf{-6r = -36}\\\\\longrightarrow\sf{r = \cancel{-36/-6}}\\\\\longrightarrow\bf{r = 6\:years}$

Thus;

The present age of son's will be 6 years & the present age of father's will be 7r = 7 × 6 = 42 years.