Question

(ii) Nine years ago, father was 10 times as old as
his son. Fifteen years hence, the father will be
twice as old as his son. Find their present
ages.
Solution : (i) Let 9 years ago, son's age was x years
and father's age was y years. (ii) Form an equation
from the given conditions. (iii) Find their present
ages. (iv) Find their ages 15 years hence. (v) Form
another equation from the given condition.
(vi) Solve the simultaneous equations.​

Answer 1

Let the present of father be x years

Let the present age of son be y years

ATQ

Nine years ago

(x - 9) = 10( y - 9)

x - 9 =10y - 90

x - 10y = -90 +9

x - 10y = - 81 ________(1)

And

15 years hence

x +15 = 2(y +15)

x + 15 =2y +30

x - 2y = 30 - 15

x - 2y = 15__________(2)

Subtract eq( 2) from eq(1)

x - 10y - (x - 2y) = -81 - 15

y =12

Put y =12 in eq (1)

x - 10(12) =-81

x = - 81 +120

x =39

Therefore, age of father is 39 years

And age of son be 12 years

_____________________________

Hope it helps!!!

Keep calm and study hard

☘ℙ☘