Question

5 years ago, the age of a sister was twice the age of the other sister. 5 years hence, their ages will be in the ratio 2:3. Find their present ages.

Answer 1

Solution:-
Let the present age of elder sister be 'x' years.
And let the present age of younger sister be 'y' years.
5 years ago :
Age of elder sister = (x - 5) years
and age of younger sister = (y - 5) years
According to the first condition:-
(x - 5) = 2(y - 5)
x - 5 = 2y - 10
x - 2y = - 10 + 5
x = 2y - 5 .....................................(1)
5 years hence :
Elder sister's age = (x + 5) years
Younger sister's age = (y + 5) Years
Now, according to second condition:-
(x + 5)/(y +5) = 2/3
⇒ 2x + 10 = 3y + 15
⇒ 2x - 3y = 15 - 10
⇒ 2x - 3y = 5 ................................(2)
Now, substituting the value of x = 2y - 5 in the equation (2), we get
⇒ 2(2y - 5) - 3y = 5
⇒ 4y - 10 - 3y = 5
⇒ y = 5 + 10
⇒ y = 15
Substituting the value of y = 15 in equation (1), we get.
x = (2 × 15) - 5
x = 30 - 5
x = 25
So, the present age of younger sister is 15 years and present age of elder sister is 25 years.