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.

Answer 1

(i)

Let the age of the son one year ago be x years.
Age of the father one year ago = 8x years.

Present age of the father $=8x+1$
Present age of the son $=x+1$
Given that $(8x+1)=(x+1{)}^2(8x+1)=(x^2+2x+1)x^2-6x=0x(x-6)=0x=0orx=6$

Son's age cannot be zero.

So, the present age of son is (x+1) i.e. 7 years.
Present age of the father = (8x+1) i.e. 49 years.

(ii)

Let age of son be x

then age of father $=3\frac{1}{2}x=\frac{7}{2}x$

Given sum of square of age = 1325

$x^2+(\frac{7}{2}x{)}^2=1325x^2+\frac{49x^2}{4}=1325\frac{4x^2+49x^2}{4}=132553x^2=5300x^2=100x=10$

Age of the son = 10

Aged of father $=\frac{7}{2}\times 10=35$