Question

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

Answer 1

Hint- Here we will proceed by assuming the present age of man and present age of son be x, y. Then we will convert conditions into linear equations and solve the linear equations to get the required answer.

Complete step-by-step solution-
Let the present age of man is $x$
And the present age of his son is $y$.
According to the question,
Case 1-
After 6 years:
$(x + 6) = 3(y + 6)$
$\Rightarrow x + 6 = 3y + 18 $
$\Rightarrow x + 6 – 3y – 18 = 0$
$\Rightarrow x – 12 = 3y $
$\Rightarrow x – 3y = 12 $ ………………. (1)

Case 2-
Three years ago:
$ x -3 = 9(y – 3) $
$\Rightarrow x – 3 = 9y – 27 $
$\Rightarrow x – 9y = -24 $ ……………. (2)
Now subtracting equation 2 from equation 1,
$(x – 3y = 12)$ - $ (x – 9y = -24)$
We get-
$\Rightarrow 6y = 36$
$\Rightarrow y = 6$
Substituting the value of y in equation 1,
$ x – 3xy = 12 $
We get-
$\Rightarrow x – 3x (6) = 12$
$\Rightarrow x = 12 + 18$
$\Rightarrow x = 30 $
Therefore, the age of man is 30 years.
Age of the son is 6 years.

Note- While solving this question, we can assume any variables instead of x and y. Also we can solve the linear equations by any method like substitution method, elimination method or cross-multiplication method.