**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.

**Solution:**

(i)

Let the present age of the son be *x* years.

∴ Present age of the man = *x*2 years

One year ago,

Age of the son = (*x* − 1) years

Age of the man = (*x*2 − 1) years

According to the given condition,

Age of the man = 8 × Age of the son

$\therefore x^{2}-1=8(x-1)$

$\Rightarrow x^{2}-1=8 x-8$

$\Rightarrow x^{2}-8 x+7=0$

$\Rightarrow x^{2}-7 x-x+7=0$

$\Rightarrow x(x-7)-1(x-7)=0$

$\Rightarrow(x-1)(x-7)=0$

$\Rightarrow x-1=0$ or $x-7=0$

$\Rightarrow x=1$ or $x=7$

∴ *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.

$\Rightarrow m=3 \frac{1}{2} s$

$\Rightarrow m=\frac{7}{2} s \quad \ldots$ (i)

Also given that

$m^{2}+s^{2}=1325 \quad \ldots \ldots$ (ii)

Put value of (i) in (ii), we get

$\left(\frac{7}{2} s\right)^{2}+s^{2}=1325$

$\frac{\Rightarrow 49 s^{2}+4 s^{2}}{4}=1325$

$53 s^{2}=5300$

$\Rightarrow s^{2}=100$

$\Rightarrow s=\pm 10$

Ignore the negative value

So, the age of son = *s* = 10 years

Also, from (i)* *we have

$m=\frac{7}{2} s$

$\Rightarrow m=\frac{7}{2} \times 10$

$\Rightarrow m=35$

So, age of man = 35 years

Age of son = 10 years