Math, asked by no140, 7 months ago

The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages, in years, of the son and the father are, respectively

(A) 4 and 24                                       

(B) 5 and 30

(C) 6 and 36                                       

(D) 3 and 24​

Answers

Answered by Anonymous
33

Answer: (C) 6 and 36

Explanation: Let the age of father be x and of son is y.

Then according to question,

x = 6y …..(i)

Four years hence age of son will be y + 4 and age of father will be x + 4

Then according to question,

x + 4 = 4 (y + 4)

x – 4y = 12 …..(ii)

Solving equations (i) and (ii) we get:

y = 6 and x = 36

hope it's helps you ❤️

Answered by Anonymous
3

C) 6 and 36     

__________________

      

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

Similar questions