Question

20 years ago the age of a father was 4 times the age of his son. After 4 years,
from now the age of the father will be double that of his son. The present age of
his son is.​

Answer 1

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

Scenario 1: 20 years ago,

→ x - 20 = 4(y - 20)

→x - 20 = 4y - 80

→ x - 4y = - 60

Or - x + 4y = 60 __(1)

Scenario 2: After 4 years,

→ x + 4 = 2(y + 4)

→ x + 4 = 2y + 8

→ x - 2y = 4 ___(2)

On adding (1) and (2) we get,

→ 2y = 64

→ y = 32

Then, age of father be

→ x - 2(32) = 4

→ x = 68

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

Answer 2

Answer:

68years,32years

Reason:

Let father's age be x

his sons age be y

according to the conditions

(x-20)=4(y-20)....(1)

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

From 1

x-20=4y-80

implies x=4y-80+20

implies x=4y-60.....(3)

From (2)

x+4=2y+8

put the value of x from equation 3 in equation 2 to get

4y-60+4=2y+8

implies 2y=60+8-4

y=64/2=32..

now put the value of y in 3 to get

x=4(32)-60

x=128-60=68.

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