Ten years ago, A was four times as old as B and after 10 years A will be twice as old as B find their present ages
Answers
Solution :-
Let :-
Present age of A = x
Present age of B = y
Ten years ago :-
Age of A = x - 10
Age of B = y - 10
After ten years :-
Age of A = x + 10
Age of B = y + 10
According to the first condition :-
According to the second condition :-
Equation (2) - Equation (1) :-
Substitute the value of y in eq (1) :-
Present age of A = 50 years
Present age of B = 20 years
♡ Answer :
- Present age of A = 50 years
- Present age of B = 20 years
♡ Given:
- Ten years ago, A was four times as old as B
- After 10 years A will be twice as old as B
♡ To find:
- The ages of A and B
♡ Solution:
Let the Present age of "A" be "x" and present age of "B" be "y".
So, ten years ago :
Age of A = x - 10
Age of B = y - 10
According to this :
x - 10 = 4 ( y - 10 )
x - 10 = 4y - 40
x - 4y = (- 40) + 10
x - 4y = (-30) _____(i)
And After ten years :
Age of A = x + 10
Age of B = y + 10
According to this :
x + 10 = 2 (y + 10)
x + 10 = 2y + 20
x - 2y = 20 - 10
x − 2y = 10 ______(ii)
By subtracting (ii) from (i) :-
x - 2y - (x - 4y) = 10 - (-30)
x - 2y - x + 4y = 10 + 30
4y − 2y = 40
2y = 40
y = 20
By Substituting the value of y in (i), we get :
x - (4 × 20) = (- 30)
x - 80 = (- 30)
x = (-30) + 80
x = 50
Therefore,
- Present age of A = 50 years
- Present age of B = 20 years
♡ Concepts Used:
- Assumption of unknown values
- Making expressions
- Equating the expressions
- Transposition Method
- Subtraction of equations
- Substitution of values