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