Question

2. Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages


please show step by step answer neatly.

Answer 1

$\huge\fcolorbox{black}{aqua}{Solution:-}$

Let Five years ago the age of son be x years and age of father be 7x years

Present age of son =x+5

Present age of father =7x+5

5 years later their age will (x+10) and (7x+10)

∴7x+10=3(x+10)

7x−3x=20

4x=20

x=5

So, the present age of son=x+5=5+5=10years

and the present age of father=7x+5=35+5=40years

Answer 2

Step-by-step explanation:

Age of the son=x

Age of the father = 7x

Present ages =x+5 and 7x +5

Future age of the son= x+ 10

Father =7x +10

7x+10 = 3(x+10)

X=5

7x=35

X+5=5+5=10

Son age =10

Father age = 7x+5=7×5+5=40 years