Question

Six years hence a man's age will be three times his son's age, and three years ago he was nine times as old as his son. Find their present ages.

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

SOLUTION

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

6 years hence, their ages will be (x + 6) years and (y + 6) years.

According to the problem :

$x + 6 = 3(y + 6)$

$= > x + 6 = 3y + 18$

$= > x - 3y = 12$

Equation (i).

3 years ago, their ages were (x - 3) years and (y - 3) years.

According to the problem :

$x - 3 = 9(y - 3)$

$= > x - 3 = 9y - 27$

$= > x - 9y = - 24$

Equation (ii).

Subtracting (ii) from (i).

We get :

$= > 6y = 36$

$= > y = \frac{36}{6}$

$= > y = 6.$

Substituting this value of y in (i).

We get :

$x - 3 \times 6 = 12$

$= > x - 18 = 12$

$= > x = 12 + 18$

$= > x = 30.$

Hence :

The present age of the man is 30 years and that of his son is 6 years.

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