Question

Five years ago, a man was seven time as old as his son.Five years hence, the father will be three time as old as his son. Find their present ages


​

Answer 1

Answer:

let father be x

let son be y

five year ago

(5-x) =7(5-y)

5-x=35-7y

-x+7y=30 equation 1

five year hence

(5+x)=3(5+y)

5+x=15+3y

x-3y=10 equation 2

from equation 1 and 2

-x+7y=30

x-3y=10

4y=20

y=5

x=25

Answer 2

Solution:-

let father's age = x

son's age = y

5 years hence, age of father = x+5                                  

age of son = y+5

So (x+5) = 3(y+5)

⇒ x =3y+10

5 years ago, age of father = x-5 

age of son = y-5

So x-5 = 7(y-5)

⇒3y + 10 - 5 = 7y - 35

⇒ 4y = 40

⇒y = 10  age of son

x = 3y +10 = 40

son age = 10 year

father's age= 40 year