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Mathematics

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

Linear Eqns One Variable

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Answer

Let the son's present age be x years.

Then, the man's present age = 2x years.

20 years ago:

Son's age = (x - 20) years.

Man's age = (2x - 20) years.

According to the question, 20 years ago, the age of the man was 12 times the age of the son:

∴ 2x - 20 = 12(x - 20)

⇒ 2x - 20 = 12x - 240

⇒ 2x - 12x = -240 + 20 \quad [Transposing -20 to RHS and +12x to LHS]

⇒ -10x = -220

⇒ x = 22010\dfrac{-220}{-10}

⇒ x = 22

Son's present age = x = 22 years.

Man's present age = 2x years = (2 × 22) years = 44 years.

Hence, the present age of the man is 44 years and his son's age is 22 years.

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