Mathematics

The present age of a man is 3 years more than three times the age of his son. Three years hence, the man’s age will be 10 years more than twice the age of his son. Determine their present ages.

Linear Equations

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Answer

Let x be present age of the man and y be the present age of son.

Given,

The present age of a man is 3 years more than three times the age of his son,

⇒ x = 3y + 3     ……..(1)

Given,

Three years hence, the man’s age will be 10 years more than twice the age of his son.

⇒ x + 3 = 2(y + 3) + 10

⇒ x + 3 = 2y + 6 + 10

⇒ x = 2y + 16 - 3

⇒ x = 2y + 13     ……..(2)

From equation (1) and (2), we get :

⇒ 3y + 3 = 2y + 13

⇒ 3y - 2y = 13 - 3

⇒ y = 10.

Substituting value of y in equation (1), we get :

⇒ x = 3y + 3

⇒ x = 3(10) + 3

⇒ x = 30 + 3

⇒ x = 33.

Hence, the present age of son = 10 years and that of man = 33 years.

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