Question

4 yrs ago fathet was 6 time as old as his son . 10 yrs later the father will be 2 and halftime as old as his son
find the present age of both

Answer 1

Let,

Father's age = 'x' years

Son's age = 'y' years

$<u>4 years ago</u>$

Father's age = ( x - 4 ) years

Son's age = ( y - 4 ) years

A.T.Q.

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

x - 4 = 6y - 24

x - 6y = - 24 + 4

$<b>x - 6y = - 20</b>$

$<u>10 years later</u>$

Father's age = ( x + 10 ) years

Son's age = ( y + 10 ) years

A.T.Q.

( x + 10 ) = 2.5 ( y + 10 )

x + 10 = 2.5y + 25

x - 2.5y = 25 - 10

$<b>x - 2.5y = 15</b>$

$<i>Subtract both the equations.</i>$

( x - 6y ) - ( x - 2.5y ) = - 20 - 15

x - 6y - x + 2.5y = - 35

- 3.5y = - 35

y = 10

$<i>Put this in 1st equation.</i>$

x - 6 ( 10 ) = - 20

x - 60 = - 20

x = - 20 + 60

x = 40

Hence,
$<b>$
Father's age = 40 years

Son's age = 10 years