Question

Four years ago father’s age was 6 times that of his son. Twelve years from now, father’s age will

be twice that of the son. What is the ratio of father and son’s present ages?​

Answer 1

HELLO DEAR FRIEND

Let the age of age of father be x

and age of son be y

ATQ,

(x - 4) = 6(y - 4)

x - 4 = 6y - 24

x - 6y = - 24 + 4

x - 6y = - 20 ....(1)

also,

x + 12 = 2(y + 12)

x + 12 = 2y + 24

x - 2y = 24 - 12

x - 2y = 12 ....(1)

(1) - (2)

x - 6y - (x -2y) = - 20 - 12

x - 6y - x + 2y = - 32

- 4y = - 32

$\large{ \large{ \bf{y}} = \frac{ \large{ - 32}}{ \large{ - 4}}}$

$\large \boxed{y = 8}$

putting the value of on (2)

x - 2y = 12

x - 2(8) = 12

x - 16 = 12

x = 12 + 16

$\large \boxed{y = 28}$

So,

$\large{the \: present \: age \: of \: father = \boxed{28}}$

$\large{the \: present \: age \: of \: son \: = \boxed{8}}$

Ratio of ages of father and son

=

$\large{ \frac{ \large{ 28}}{ \large{ 8}}}$

$\large{ = \frac{ \large{ 7}}{ \large{ 2}}}$

$\huge\boxed {= \bf{7:2}}$

HOPE IT HELPS

Answer 2

Answer:

Step-by-step explanation:

take father's present age as x

take his son's age as y

four years ago:

x - 4 = (y-4)6

x-4 = 6y-24

x-6y = -20  -----------> eq 1

twelve years from now:

x+12 = (y+12)2

x+12 =  2y + 24

x -2y = 12 ---------------> eq 2

solving both the equations by elimination method; we get

x = 28

y = 8

now, the ratio between the present ages of father and son is :

28/8 = 7/2

i.e 7:2