Math, asked by yodares749, 9 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
24

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

Answered by Anonymous
0

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

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