Question

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

Answer 1

Answer:

Step-by-step explanation:

Given :-

Ten year ago ,father was 12 times as old as his son 10 years hence, he will be twice as old as his son will be.

To Find :-

Their ages.

Solution :-

Let the age of the son before 10 years be x.

And age of the father before 10 years be 12x.

10 years hence,

2(x + 10 + 10) = (12x + 10 + 10)

⇒ 2x + 40 = 12x + 20

⇒ 10x = 20  

⇒ x = 20/10

⇒ x = 2  

Age of son at present = x + 10 = 2 + 10 = 12 years.

Age of the father at present = 12x + 10 = 24 + 10 = 34 years.

Answer 2

Step-by-step explanation:

Given:-

• Ten year ago ,father was 12 times as old as his son 10 years hence, he will be twice as old as his son will be.

To Find:-

• Their present ages

Solution:-

Suppose ten years ago, son's present age was x years and suppose father's age was 12x years.

➟ Present age of son = x + 10 years

➟Present age of Father = 12x + 10 years

➳ After 10 years,

• ➻ Son's age will be (x + 10 + 10) years = x + 20 years
• ➻ Father's age will be (12x + 10 + 10) years = 12x + 20 years

⇨ It is given that the age of father will be 2 times than that of son after 10 years . So, if we multiply 2 to the age of son after 10 years, it will be equal to the age of father after 10 years.

♦ ACQ,

⤐ 12x + 20 = 2(x + 20)

⤐ 12x + 20 = 2x + 40

⤐ 12 - 2x = 40 - 20

⤐ 10x = 20

⤐ x = 20/10

⤐ x = 2

So, Present age of son = x + 10 years

➨ 2 + 10

➨12 years

Present age of father = 12x + 10 years

⤕ 12(2) + 10 years

⤕ 24 + 10 years

⤕ 34 years