Question

40. 18 years ago, a man was three times as old as his son.
Now, the man is twice as old as his son. The sum of
the present ages of the man and his son is
(S.S.C., 2003)​

Answer 1

Solution :

Let the son's age be r years & the man's age be m years respectively;

A/q

$\underbrace{\bf{18\:years\:ago\::}}$

The age of son's was = (r-18) years .

The age of man's was = (m-18) years .

Now;

$\mapsto\tt{(m-18) = 3(r-18)}\\\\\mapsto\tt{m-18 =3r - 54}\\\\\mapsto\tt{m-3r = -54 + 18}\\\\\mapsto\tt{m-3r = -36..................(1)}$

&

$\mapsto\tt{m=2r......................(2)}$

∴Putting the value of m in equation (1),we get;

$\mapsto\tt{2r-3r=-36}\\\\\mapsto\tt{\cancel{-}r=\cancel{-}36}\\\\\mapsto\bf{r=36\:years}$

∴Putting the value of r in equation (2),we get;

$\mapsto\tt{m=2\times 36}\\\\\mapsto\bf{m=72\:years}$

Thus;

The present age of son's = r = 36 years .

The present age of man's = m = 72 years .

Now,sum of the present both person ages are = r + m = (36 + 72) years = 108 years .