Question

The age of a father 10 years ago was thrice
the age of his son. Ten years hence, the
father's age will be twice that of his son.
The ratio of their present ages is:​

Answer 1

Answer:

Step-by-step explanation:

Let their present ages be x ( father ) and y (son)

10 years ago ..their ages were

Father: x-10

Son: y-10

According to first condition..

Father's age = 3× sons ages

x-10= 3( y-10)

x-10= 3y-30

x-3y= -30+10

x-3y = -20 .......(1)

Ten years hence ..their ages will be ...

Father: x+10

Son : y+10

According to second condition...

Father's age=2×sons age

x+10= 2(y+10)

x+10= 2y+20

x-2y= 20-10

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

Solving equations (1) & (2) simultaneously...

x-3y= -20

x-2y= 10

-. +. -

----------------

-y= -30

y=30

Putting this in equation (1) to get value of x...

x-2y= 10

x-2(30)=10

x-60=10

x=10+60

x=70

Thus father's present age is 70 years ...and sons age is 30 years.