The Ages 3 times Elder Problem is shown below
Question :
The ages of two brothers differ by 16 years. If 6 years ago, the elder one was 3 times as old as the younger one, find their present ages ?
Solution :
Consider the ages of the brothers as X and Y
Now the first statement gives us the equation
X – Y = 16 …………………………(1)
Now 6 years ago Brother X was 3 times as old as Brother Y, implies
(X-6) = 3(Y-6) ………………..(2)
So, solving this equation we get …
X -6 = 3Y – 18
X – 3Y = -12 …………………..(3)
Subtracting eqn (1) from (3), we get
X – Y – X +3Y = 16 – (- 12)
2Y = 16 + 12 = 28
So, Y = 28/2 = 14
So , from eqn (1), X = 16 +14 = 30
Hence the brothers current ages are 30 yrs & 14 yrs.
To understand this problem better, you can watch my video on Ages – Basics 