Question

Q12) Two years ago, a man was five times as old as his son. Two years later, his age will be 8 more than three times the age of his son. Find the present ages of the man and his * son.​

Answer 1

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Let the age of a man = x years

Let the age of a man = x years And the age of his son = y years

Two years ago,

Man’s age = (x – 2) years

Son’s age = (y – 2) years

According to the question,

(x – 2) = 5(y – 2)

⇒ x – 2 = 5y – 10

⇒ x = 5y – 10 + 2

⇒ x = 5y – 8 ...(i)

Two years later,

Father’s age = (x + 2) years

Son’s age = (y + 2) years

According to the question,

(x + 2) = 8 + 3(y + 2)

⇒ x + 2 = 8 + 3y + 6

⇒ x = 3y + 12 …(ii)

From Eq. (i) and (ii), we get

5y – 8 = 3y + 12

⇒ 5y – 3y = 12 + 8

⇒ 2y = 20

⇒ y = 10

On putting the value of y = 11 in Eq. (i), we get

x = 5(10) – 8

⇒ x = 50 – 8

⇒ x = 42

Hence, the age of man is 42 years and the age of his son is 10 years.