Question

Five years ago a man was seven times old as his son. Five years hence, The father will be three times as old as his son. Find their present ages.

Answer 1

Hint: Start by assuming the present age as some variable and try to form relations as per the statement given in the question. Solve the relationships formed and find out the value of the variable assumed, ignore negative value if found as age can never be negative.

Complete step-by-step solution:
Let us take the present age of the person to be ‘x’ years and the son’s age be ‘y’ years.
Now, let us form the required relation of ages.
The age of the person 5 years from now will be=$x + 5$ years
The age of the person 5 years ago was =$x - 5$ years
The age of the son 5 years from now will be=$y + 5$ years
The age of the son 5 years ago was =$y - 5$ years
Now, according to the statement given in the question, we have
$x - 5 = 7(y - 5)$
Solving this, we get
$x - 5 = 7y - 35 \\
   \Rightarrow x - 7y = - 30 \to eqn.1 $
Similarly , we have one more condition,
$x + 5 = 3(y + 5)$
Solving this, we get
$ x + 5 = 3y + 15 \\
 \Rightarrow x - 3y = 10 \to eqn.2 $
Solving, eqn.1 and Eqn. 2 simultaneously, we get
$ - 4y = - 40 \\
 \Rightarrow y = 10 $
On substituting the value of y in equation 1 we get,
\[x = 40\]
Therefore, the present age of the person and son is 40 years and 10 years respectively.

Note: Such similar problems can be solved by following the approach of part by part work, read a statement and then form its equation or relation and repeat the same for the next statement. Attention is required while solving such questions as they can have quadratic equations, needed to be solved by either splitting middle term or discriminant rule and any negative values are to be neglected as age can never be negative.