Question

The sum of the ages of Anup and his father is $100$ years. When Anup is as old as his father now, he will be five times as old as his son Anuj is now. Anuj will be eight years older than Anup is now, when Anup is as old as his father. What are their ages now?

Answer 1

Let the present ages of Anup, his father and his son Anuj be $x,y$ and $z$, respectively.

Given that, the sum of ages of Anup and his father is $100$ years

$\Rightarrow x+y=100$ ..........(1)

Let their new ages after few years be $x^{\prime },y^{\prime }$ and $z^{\prime }$, respectively.

Given that, after those few years,

New age of Anup $=$ present age of father $=5\times$ present age of Anuj

i.e.,$x=y=5z$.........(2)

Also, given that, New age of Anuj $=8+$ present age of Anup

$\Rightarrow z=8+x$ .....(3)

Since, Number of years $=$ New age $-$ present age for each person

$\Rightarrow$ Number of years $x^{\prime }-x=y^{\prime }-y=z^{\prime }-z$

Taking first and last term we have,

$x^{\prime }-x=z^{\prime }-z$.....(4)

Using equations (1), (2), and (3), we get,

$x^{\prime }=y$

$x=100-y$

$z^{\prime }=8+x$

$\Rightarrow z^{\prime }=8+100-y$

$\Rightarrow z^{\prime }=108-y$

and $z=\frac{y}{5}$

Substituting these values in equation (4), we get:

$x^{\prime }-x=z^{\prime }-z$

$\Rightarrow y-(100-y)=108-y-\frac{y}{5}$

$\Rightarrow 2y-100-108=-y-\frac{y}{5}$

$\Rightarrow 2y-208=\frac{-5y-y}{5}$

$\Rightarrow 2y-208=-\frac{6y}{5}$

$\Rightarrow 10y-(5\times 208)=-6y$

$\Rightarrow 10y+6y=(5\times 208)$

$\Rightarrow 16y=(5\times 208)$

$\Rightarrow y=\frac{(5\times 208)}{16}$

$\Rightarrow y=65$ ---(i)

From (1),

$x+y=100$

$\Rightarrow x+65=100$

$\Rightarrow x=100-65$

$\Rightarrow x=35$---(ii)

From (2),

$y=5z$

$\Rightarrow z=\frac{y}{5}$

$\Rightarrow z=\frac{65}{5}$

$\Rightarrow z=13$---(iii)

Hence, from (i), (ii) and (iii) the present age of Anup is $35$ years, that of his father is $65$ years and that of his son, Anuj is $13$ years.