The sum of ages

Let $M$ and $A$ be the present ages of Marvin and Ana, respectively.

$M=0.25A$ $\Longleftrightarrow$ $\color{#D61F06}\boxed{1}$

$\dfrac{A+15}{M+15}=\dfrac{7}{3}$ $\Longleftrightarrow$ $\color{#D61F06}\boxed{2}$

Substitute $\color{#D61F06}\boxed{1}$ in $\color{#D61F06}\boxed{2}$

$\dfrac{A+15}{0.25A+15}=\dfrac{7}{3}$

$3A+45=1.75A+105$

$3A-1.75A=105-45$

$1.25A=60$

$A=48$

It follows that, $M=0.25(48)=12$

So the sum of their ages $10$ years ago was $48-10+12-10=\boxed{40}$