Question

A man is twice as old his son.Twenty years ago,the father was 4 times as old as the son.Find the present age of each​

Answer 1

Let his son age be x years.

So , that man age would be 2x yrs .

Now , twenty years ago .

As per our equation,

son age would be ( x - 20 ) years

and the man age would be ( 2x - 20 ) years.

According to the question :-

$\color{red}\:2x - 20 = 4(x - 20)$

$2x - 20 = 4x - 80 \\ 4x - 2x = 80 - 2 0 \\ 2x = 60 \\ x = \frac{ \cancel{60}}{ \cancel{2}} = \: 30$

Therefore, the present age of each are as follow

son age = 30 years

Man age = 60 years

Answer 2

Answer:

Let the present age of son be x and of his father be 2x.

Before 20 year,

2x - 20 = 4(x - 20)

2x - 20 = 4x - 80

2x-4x=-80+20

- 2x = - 60

x = 30

So, age of son =30year and age of father = 60year.

Step-by-step explanation:

Hope you get the correct answer.