Question

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​

Answer 1

Step-by-step explanation:

Firstly we will assume that present age of man is x years and present age of his son be equals to y years .

Now , 5 years ago,

Age of man = (x-5)years.

Age of son=(y-5) .

As given that "five years ago the man was 7 times as old as his son"

lt gives a linear equation in two variables in form of x and y.

Also,

five years hence ,

Age of man =(x+5)years

&

Age of son =(y+5)years .

Given that

"5 years hence the man will be 3 times as old as his son"

It will give second linear equation in two variables in form of x & y.

After solving these two equations simultaneously, we can calculate the values of x & y which are their present ages.

Answer 2

$\huge\sf{Solution}$

Let y be the age of son.

Let 7y be the age of father.

y + 5 , 7y + 5 are the present ages

= y + 10 future age of son.

7y + 10 is father.

= (7y + 10 = 3) (y + 10)

= (y = 5)

= (7y = 35)

(y + 5 = 5 + 5 = 10)

.•. 10 years is age of son and

(7y + 5 = 7) ×( 5 + 5 = 40) .•. 40 is age of father.

Hope it helps!