Five years ago, A was thrice as old as B and ten years later, A shall be twice as old as B. What are the present ages of A and B?
Answers
Given:
Five years ago, A was thrice as old as B
Ten years later, A shall be twice as old as B
Find:
Present age of A and B
Solution:
Five years ago, A was thrice as old as B
A's age = (x - 5)
B's age = (y - 5)
=> (x - 5) = 3(y - 5)
=> x - 5 = 3y - 15
=> x - 3y = -15 + 5
=> x - 3y = -10
=> x = 3y - 10 ........(i).
Ten years later, A shall be twice as old as B
A's age = (x + 10)
B's age = (y + 10)
=> (x + 10) = 2(y + 10)
=> x + 10 = 2y + 20
=> x - 2y = 20 - 10
=> x - 2y = 10 ..........(ii).
Now, putting the value of 'x' from equation (ii) in equation (i).
=> x - 2y = 10
=> 3y - 10 - 2y = 10
=> 3y - 2y = 10 + 10
=> y = 20
Present age of B = y = 20 years
Putting the value of 'y' in equation (i).
=> x = 3y - 10
=> x = 3 × 20 - 10
=> x = 60 - 10
=> x = 50
Present age of A = x = 50 years
Hence, the present age of 'A' is 50 years and the present age of 'B' is 20 years.
I hope it will help you.
Regards.
Given :
. Five years ago a was three times as old as b Ten years later a shall be twice older than b.
Assume that present age of a as x and that of b as y.
Five years ago, a was thrice as old as b
i.e. age of a was x - 5 and age of b was 3(y-5)
x - 5 = 3 (y - 5)
x - 5 = 3y - 15
x - 3y = -15+5
x - 3y = -10 ---------(1)
Ten years later, a shall be twice as old as b
i.e. age of a will be x + 10 and age of b will be 2(y+10)
x + 10 = 2 (y + 10)
x + 10 = 2y + 20
x - 2y = 20-10
x - 2y = 10 ---------(2)
By elimination method, we get
x - 3y = -10
x - 2y = 10
- y = -20
y = 20 i.e. present age of b
Substituting y = 20 in equation 1, we get
x - 3y = -10
x - 3(20) = -10
x - 60 = -10
x = -10 + 60
x = 50 i.e. present age of a.