Question

Ten years ago, father was 12 times as old as his son and 10 years hence, he will be twice as old as his son will be. Find their present ages

Answer 1

Answer:

Father : 34 years old

Son : 12 years old

Step-by-step explanation:

Let father be x, son be y.

12 (y-10) = x-10

2 (y+10) = x+10

12y -120 = x - 10

12y = x + 110

2y + 20 = x + 10

2y = x - 10

y = (x - 10)/2

12(x-10)/2 = x + 110

6x - 60 = x + 110

5x = 170

x = 34

y = (34-10)/2 = 12

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assumption,

Ten years ago age of the son be p

Also,

Ten years ago age of father be 12p

Now,

Situation,

Ten years hence,

2(p + 10 + 10) = (12p + 10 + 10)

⇒ 2p + 40 = 12p + 20

⇒ 10p = 20

⇒ p = 20/10

⇒ p = 2

So,

Present age of son

= p + 10

= 2 + 10

= 12 years

Present age of father

= 12p + 10

= 24 + 10

= 34 years