To solve the exercise you can use a rule of three, like this
[tex]\begin{gathered} 6\text{ liters}\rightarrow3\text{ days} \\ 10\text{ liters}\rightarrow x\text{ days} \end{gathered}[/tex][tex]\begin{gathered} x=\frac{10\text{ liters }\cdot3\text{ days}}{6\text{ liters}} \\ x=\frac{10\cdot3\text{ days}}{6} \\ x=\frac{10\cdot3}{6}\text{days} \\ x=\frac{30}{6}\text{days} \\ x=5\text{ days} \end{gathered}[/tex]Therefore,
