Question

5 year ago a man was seven time as old as his son. Five years hence, the father will be three times as old as his son. Find their present age. ​

Answer 1

Answer :-

40 years and 10 years are present ages of man and son respectively.

Solution :-

Let the present age of a man be 'x' years and the present age of son be 'y' years

5 years ago

Age of man = (x - 5) years

Age of son = (y - 5) years

Age of man 5 years ago = 7 times the age of sone 5 years ago

⇒ (x - 5) = 7(y - 5)

⇒ x - 5 = 7y - 35

⇒ x - 7y = - 35 + 5

⇒ x - 7y = - 30 ---(1)

5 years hence

Age of man = (x + 5) years

Age of son = (y + 5) years

Age of man three years hence = 3 times the age of son three years hence

⇒ (x + 5) = 3(y + 5)

⇒ x + 5 = 3y + 15

⇒ x - 3y = 15 - 5

⇒ x - 3y = 10 ---(2)

Subtract (2) from (1)

⇒ x - 7y - (x - 3y) = - 30 - 10

⇒ x - 7y - x + 3y = - 40

⇒ - 4y = - 40

⇒ 4y = 40

⇒ y = 40/4

⇒ y = 10

Present age of of son = y = 4 years

Substitute y = 10 in (2)

⇒ x - 3y = 10

⇒ x - 3(10) = 10

⇒ x - 30 = 10

⇒ x = 30 + 10

⇒ x = 40

Present age of man = x = 40 years

Therefore 40 years and 10 years are present ages of man and son respectively.

Answer 2

Step-by-step explanation:

$\huge{Your\:Answer}$

Sons age 5 years ago=x father's age = 7x

their present age = x+5 and 7x+5 resp

Sons age 5 years hence = x+10(since x is sons age 5 years ago from present age and we are talking of his age 5 years hence from his present age so 5+5=10)

 

fathers age 5 years hence =3(x+10)

hence the fathers present age will be 3(x+10)-5

also fathers present age =7x+5

therefore 7x+5=3(x+10)-5

7x+5=3x+30-5

7x-3x=25-5

4x=20

x=5

 

So, son's present age = 10years and fathers, present age = 40 years