Question

5 years ago a man was 7 times as old as
his son. After 5 years he will be the rice as old
as his son. Find their present ages.
​

Answer 1

Solution:-

Let , The Present Age of a Father = x years .

Age of the father 5 years ago = ( x - 5) years .

By the Problem ,

Age of his Son 5 years Ago

$= > \frac{1}{7} (x - 5)$

Age of father 5 years Hence = x +5

Age Of his Son 5 years hence = Age of his Son 5 years ago +10

$= > \frac{1}{7} (x - 5) + 10 = \frac{x}{7} - \frac{5}{7} + 1 \\ = > \frac{x - 5 + 70}{7} = \frac{x + 65}{7}$

Five years , hence , the father will be there time as old as his son then ,

$= > 3( \frac{x + 65}{7} ) = x + 5$

$3 \: (x + 65) = 7(x + 5)$

$3x + 195 = 7x + 35$

$7x - 3x = 195 - 35$

$4x = 160$

$x = 40$

Therefore, Present age of a father = 40 years.

Age of his son 5 years ago

$= > \frac{1}{7} (x - 5) = \frac{1}{7} (40 - 5) = \frac{1}{7} \times 35 = 5 \: years$

Hence, Present age of his son = 5 + 5 = 10 years .