Question

Two years ago a father has five times as old as his son. two years later, his age will be 8 year more than three times the age of son. find the present age of father and son

Answer 1

Let son's present age be x years

father's present age be y years

2 years ago ;

son's age = x - 2

father's age = y - 2

but a/q ;

y - 2 = 5 ( x - 2 ) = 5x - 10

or, 5x - y = 8 _______________(i)

Two years later ;

son's age = x + 2

father's age = y + 2

but a/q ;

y + 2 = 8 + 3 ( x + 2 )

or, y + 2 = 8 + 3x + 6 = 3x + 14

or, 3x - y = -12 ________________(ii)

Subtracting eqn (ii) from eqn (i) ; we get

2x = 20

or, x = 20 / 2 = 10 years (son's age)

y = 3x + 12 = 3×10 + 12 = 42 yrs (father's age)

Answer 2

Answer:

Given:

Two years ago a father was five times as old as his son. Two years later, his age will be 8 year more than three times the age of son.

To Find:

We need to find the present age of father and son.

Solution:

Let the present age of son be x years and that of father be y years.

Now, 2 years ago,

Son's age = x - 2

Father's age = y - 2

According to the question we have,

y - 2 = 5(x - 2)

=> y - 2 = 5x - 10

or y = 5x - 10 + 2

=> y = 5x - 8

=> y + 8 = 5x

=> 8 = 5x - y

or 5x - y = 8________(1)

Now, 2 years later,

Son's age = x + 2

Father's age = y + 2

According to the question, we have,

y + 2 = 8 + 3(x + 2)

=> y + 2 = 8 + 3x + 6

=> y + 2 = 14 + 3x

or 3x - y = -12_____(2)

On subracting equation 2 from equation 1 we get,

2x = 20

=> x = 20/2

=> x = 10

Son's age = x years = 10 years

Father's age = 3x + 12 = 3(10) + 12 = 30 + 12 = 42years.

Therefore Son's age is 10 years and father's age is 42 years.