Question

A man is twice as old as his son. 20 years ago, the age of the man was 12 tinmes the age of the son. Find their present ages.
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Answer 1

Answer:

Let the age of son be x years and father be 2x years.

20 years back, their ages were x−20 and 2x−20

As per the given condition, we get

2x−20=12(x−20)

⇒2x−20=12x−240

⇒2x−12x=20−240

⇒−10x=−220

⇒x=-220/-10

⇒x=22

∴ son's age is 22 years and father's age is (2×22)=44 years.

Answer 2

Let son's age be x years

Then, father's age=2x years

Before 20 years,

Son's age=x-20 years

And, Father's age=2x-20 years

ATQ,

12 (x-20)=2x-20

12x-240=2x-20

12x-2x=240-20

10x=220

x=22

Hence,present age of son-x=22 years And, present age of father=2x=2×22=44 years

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