Question

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

Answer 1

The ages of Man and son 5 years ago:

Let son's age be x
Man's age = 7x

The present ages of Man and Son,

Son's age = x + 5 years
Man's age = 7x + 5 years

The ages of Man and Son 5 years hence,

Son's age = x + 5 + 5 = x + 10 years
Man's age = 7x + 5 + 5 = 7x + 10 years

Given,
five yers hence,the father will be three times as old as his son.

The equation formed is as follows:


$3(x + 10) = 7x + 10$

$3x + 30 = 7x + 10$

$30 - 10 = 7x - 3x$
$20 = 4x$

$x = \frac{20}{4}$

$x = 5$

So,

Son's present age = x + 5 = 5 + 5 = 10 years

Man's age = 7x + 5 = 40 years.


So,

The present ages of Man and son are 10 years and 40 years respectively.

Answer 2

let the age of man be X years old
and his son be Y years old
Before five years.
x-5=(y-5)7
x-5=5y-35
x-5y=-30 ..........eq (i)
Five years hence, they will be
X+5=3(y+5)
x+5=3y+15
x-3y=10............eq(ii)
From eq(i) and (ii)

x-7y=-30
X-3y=10
- + -
------------ on subtracting
-4y=-40
y=10
from eq (i)
x-70=-30
x=-30+70
x=40