Guess The Age

Four years ago the average age of A and B was $\color{#D61F06}{\dfrac{(A-4)+(B-4)}{2}=18}$ .Simplifying this we get $\color{#3D99F6}{A+B=44}$ .The average age of A,B and C today is $\color{#20A900}{\dfrac{A+B+C}{3}=24}$ .Substituting the value of $\color{#69047E}{A+B=44}$ and simplifying,we get $\color{rubinered}{C=28}$ .So the age of C 8 years from now will be $\color{#624F41}{28+8=\boxed{36}}$

well suppose the present ages of A, B and C be x, y and z respectively; now 4 years back the age of A and B will be x-4 and y-4 respectively; now according to the riddle find the average of their ages and put them equal to 18; {(x-4)+(y-4)}/2 =18) from here u will get x+y= 44; going through the question again we can calculate the age of C by following expression: (x+y+z)/3 = 24 and putting the value of (x+y) in the expression we will get the present age of C that is 28 and after 8 years C will be 36 years old

Four years ago: $A+B=36$

Today: $36+8+C=72$ $\implies$ $C=28$

Eight years from now: $C+8=28+8=$ $\large{\color{#D61F06}\boxed{36}}$