Question

five years ago a man was seven times old as his son. five years hence the father will be three times as old as his son find their present ages​

Answer 1

★${\underline{\underline{\bold{Given:-}}}}$

• 5 years ago,a man was 7 times as old as his son.
• 5 years hence, the father will be 3 times as old as his son.

★${\underline{\underline{\bold{To\:find:-}}}}$

• Their present ages.

★${\underline{\underline{\bold{Solution:-}}}}$

Consider,

• Present age of father = x years
• Present age of son = y years

According to the 1st condition :-

• 5 years ago,a man was 7 times as old as his son.

5 years ago,

• Age of father = (x-5) years
• Age of son = (y-5) years

$\to\sf{x-5=7(y-5)}$

$\to\sf{x-5=7y-35}$

$\to\sf{x=7y-35+5}$

$\to\sf{x=7y-30..............(i)}$

According to the 2nd condition :-

• 5 years hence, the father will be 3 times as old as his son.

After 5 years,

• Age of father = (x+5) years
• Age of son = (y+5) years

$\to\sf{x+5=3(y+5)}$

$\to\sf{x+5=3y+15}$

$\to\sf{7y-30+5=3y+15\:[put\:x=7y-30\: from\:eq(1)]}$

$\to\sf{7y-3y=15+30-5}$

$\to\sf{4y=40}$

$\to\sf{y=10}$

• Present age of son = 10 years.

Now , put y = 10 in eq(1) for getting the value of x.

$\to\sf{x=7y-30}$

$\to\sf{x=7\times\:10-30}$

$\to\sf{x=70-30}$

$\to\sf{x=40}$

• Present age of father = 40 years.