Question

The age of a father eight years ago was eight times the age of his son After ten years the age of the father will be twice the age of the son find their present ages

Answer 1

Step-by-step explanation:

The age of a father is twice the square of the age of his son. Eight years hence, the age of his father will be 4 years more than 3 times the age of ...

Answer 2

Solution:-

Let, the present age of father be 'x' years

And, the present age of his son be 'y' years.

According to the Question

Eight Years Ago,

→ x-8 = 8 (y-8)

→ x-8 = 8y -64

→ x- 8y = -64 + 8

→ x-8y = 56 ---------------------(i)

After Ten Years ,

→ x +10 = 2(y+10)

→ x+10 = 2y +20

→ x-2y = 20-10

→ x-2y = 10 ----------------(ii)

Now, From equation (ii) we get

x = 10 +2y ------------(iii)

Substitute the Equation (iii) into (i) we get

→ 10 + 2y -8y = -56

→ -6y = -56-10

→ -6y = -66

→ 6y = 66

→ y = 66/6

→ y = 11 -------(iv)

Now Putting the value of equation (iv) into (iii) we get

→ x = 10 + 2(11)

→ x = 10 +22

→ x = 32

Therefore,

The Present age of Father is 32 Years.

The Present age of his son is 11 Years .