It's an Age Problem

Let $A,B,C$ be the present ages of Andrew, Bernard and Charlie, respectively.

$B=12+A$ $\color{#D61F06}(1)$

$\dfrac{B-5}{C-5}=\dfrac{2}{3}$ $\color{#D61F06}(2)$

$C+4=3(A+3)$ $\implies$ $C+4=A+9$ $\implies$ $3A-C=-5$ $\color{#D61F06}(3)$

Substitute $\color{#D61F06}(1)$ in $\color{#D61F06}(2)$ , we have

$\dfrac{12+A-5}{C-5}=\dfrac{2}{3}$ $\implies$ $\dfrac{7+A}{C-5}=\dfrac{2}{3}$ $\implies$ $21+3A=2C-10$ $\implies$ $3A-2C=-31$ $\color{#D61F06}(4)$

Subtract $\color{#D61F06}(4)$ from $\color{#D61F06}(3)$ , we get $C=26$ . It follows that $A=7$ and $B=19$ .

If $A+10=17$ , then $B+10=29$ and $C+10=36$ , so the desired answer is

$17+29+36=$ $\boxed{82}$

This is the algebra equation :

• $Bernard = Andrew + 12$

• $\frac{Bernard - 5}{Charlie - 5} = \frac{2}{3}$

• $Charlie + 4 = 3(Andrew + 3)$

Their ages now are :

• $Andrew = 7$

• $Bernard = 19$

• $Charlie = 26$

Solve it by your self :D