Question

20 years ago a father was four times as old as his son and after 4 years he will be twice as old as his son what are their present ages​

Answer 1

Answer:

Father's present age = 68 years, Son's present age = 32 years

Step-by-step explanation:

Let father's present age be x and son's present age be y.

According to the problem,

(i) (x-20) = 4(y-20)

4y-x = 60

(ii) x+4 = 2(y+4)

x-2y = 4

Solving the two equations, we get:

x=68

y=32

Answer 2

Given:

20 years ago a father was four times as old as his son and after 4 years he will be twice as old as his son.

To find:

The present ages of father and son.

Solution:

The present ages of father and son are 68 years and 32 years respectively.

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

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

So,

20 years ago,

the age of son = (x - 20) years

the age of father = (y - 20) years

Now,

According to the question, we have

$y - 20 = 4(x - 20) \:$

This can be written as

$4x - y - 60 = 0 \: \:( i)$

Also,

After 4 years, the age of the son will be = (x + 4) years

After 4 years, the age of the father will be = (y + 4) years

So,

$y + 4 = 2(x + 4)$

This can be written as

$2x - y + 4= 0 \: \: (ii)$

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

$2x = 64$

$x = 32 \: years$

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

$y = 68 \: years$

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