Question

(i) One year ago, a man was 8 times as old as his son. Now, his age is equal to the square of his son's age. Find their present ages.
(ii) A man is $3\frac{1}{2}$ times as old as his son. If the sum of the squares of their ages is 1325, find the ages of the father and the son. (CBSE 2017]

Answer 1

(i)
Let the present age of the son be x years.

∴ Present age of the man = x² years

One year ago,

Age of the son = (x − 1) years

Age of the man = (x² − 1) years

According to the given condition,

Age of the man = 8 × Age of the son

$$\begin{gathered} \therefore x^2-1=8\left(x-1\right) \\ \Rightarrow x^2-1=8x-8 \\ \Rightarrow x^2-8x+7=0 \\ \Rightarrow x^2-7x-x+7=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x\left(x-7\right)-1\left(x-7\right)=0 \\ \Rightarrow \left(x-1\right)\left(x-7\right)=0 \\ \Rightarrow x-1=0\mathrm{or}x-7=0 \\ \Rightarrow x=1\mathrm{or}x=7 \end{gathered}$$
∴ x = 7 (Man's age cannot be 1 year)

Present age of the son = 7 years

Present age of the man = 7² years = 49 years

(ii)
Let the age of man be m and the age of son be s
It is given that man is $3\frac{1}{2}$ times as old as his son.
$$\begin{gathered} \Rightarrow m=3\frac{1}{2}s \\ \Rightarrow m=\frac{7}{2}s...\left(\mathrm{i}\right) \end{gathered}$$
Also given that
$m^2+s^2=1325.....\left(\mathrm{ii}\right)$
Put value of (i) in (ii), we get
$$\begin{gathered} {\left(\frac{{\displaystyle 7}}{{\displaystyle 2}}s\right)}^2+s^2=1325 \\ \frac{\Rightarrow 49s^2+4s^2}{4}=1325 \\ 53s^2=5300 \\ \Rightarrow s^2=100 \\ \Rightarrow s=\pm 10 \end{gathered}$$
Ignore the negative value
So, the age of son = s = 10 years
Also, from (i) we have
$$\begin{gathered} m=\frac{7}{2}s \\ \Rightarrow m=\frac{7}{2}\times 10 \\ \Rightarrow m=35 \end{gathered}$$
So, age of man = 35 years
Age of son = 10 years