Question

In a family of four, the average age is 20. The parents’ age difference is 6 and the siblings’ age difference is 4. If the ages of the eldest and youngest member differ by 33, find the age of the youngest member of the family.

Solution:

Let the age of the parents be x and x+6.

Let the age of the siblings be y and y+4.



x+6-y=33

(x+x+6+y+y+4)/4 = 20



Solving the equations, y=4​

Answer 1

Answer:

Let the present age of father's be x years and present age of son's be y years.

According to the problem

x=6y

After 4 years

x+4=4(y+4)

Hence we get two equations

x=6y ......... (1)

x+4=4(y+4) ..... (2)

Simplifying eq (2)

x+4=4y+16

x−4y=12

Put x=6y in eq (2), we get

6y−4y=12

2y=12

y=6 years

and x=6y=36 years

Present age of son =6 years

and present age of father =36 years