Question

Ten years ago the age of father is 12 times the age of son. After ten years the
age of father is twice the age of son. Find their present ages.

Answer 1

let fathers age=x

let sons age=y

fathers age before 10 yrs= x-10

sons age before 10 yrs=y-10

according to question,

x-10=12(y-10)

x=12y-120+10

x=12y-110-----(1)

age of father after 10yrs=x+10

age of son after 10 yrs=y+10

according to question,

x+10=2(y+10)

x+10=2y +20

subsitituting (1)

12y-110+10=2y+20

12y-100=2y+20

12y-2y=20+100

10y=120

y=120/10

y=12

therefore,x= 12y-110

x=12(12)-110

x=144-110

x= 34....

Answer 2

Answer:

Father is 34 years old and Son is 12 years old.

Step-by-step explanation:

Given :

Ten years ago the age of father is 12 times the age of son

After ten years the  age of father is twice the age of son.

To find :

Their present ages

Solution :

Let their ages 10 years ago be -

• Son = y
• Father = 12(y)

Ages presently -

• Son = (y + 10)
• Father = 12y + 10

Ages after 10 years -

• Son = (y + 10) + 10
• Father = (12y + 10) + 10

$\rule{300}{1.5}$

According to the question,

After ten years the  age of father is twice the age of son

⇒ 2[(y + 10) + 10] = (12y + 10) + 10

⇒ 2(y + 20) = 12y + 20

⇒ 2y + 40 = 12y + 20

⇒ 2y - 12y = 20 - 40

⇒ 10y = 20

⇒ y = 2

$\rule{300}{1.5}$

Present age -

Father's present age =

⇒ 12y + 10

⇒ 12(2) + 10

⇒ 24 + 10

⇒ 34

Father is presently 34 years old.

Son's present age =

⇒ y + 10

⇒ 2 + 10

⇒ 12

Son's present age is 12 years.

∴ Father is 34 years old and Son is 12 years old.