Question

Five years ago a man was seven times as old as his son. Five year's hence the father will be three times as old as his son . Find their present ages .

Answer 1

Let the past age of son = x

Let the past age of father = 7x

So,

Present ages are x+5 and 7x+5

Future age of son = x+10

Future age of father = 7x+10

so the equation will be

7x+10 = 3*(x+10)

7x+10 = 3x+30

7x-3x = 30-10

4x = 20

x = 20/4

x = 5

7x = 35

x+5 = 10

7x+5 = 40

Answer 2

$\huge\sf{Answer:-}$

Let y be the age of son.

Let 7y be the age of father.

y + 5 , 7y + 5 are the present ages

= y + 10 future age of son.

7y + 10 is father.

= (7y + 10 = 3) (y + 10)

= (y = 5)

= (7y = 35)

(y + 5 = 5 + 5 = 10)

.•. 10 years is age of son and

(7y + 5 = 7) ×( 5 + 5 = 40) .•. 40 is age of father.

Hope it helps!