For this problem we are subtracting to find a previous time. 

We can subtract $\displaystyle 3$ hours from $\displaystyle 4$ and $\displaystyle 15$ minutes from $\displaystyle 30$. 

$\displaystyle \frac{\begin{array}[b]{r}4\\ -\ 3\end{array}}{ \ \ \ \space1}$   $\displaystyle \frac{\begin{array}[b]{r}30\\ -\ 15\end{array}}{ \ \ \ \space 15}$

He starts to get ready to leave at $\displaystyle 1:15$

For this problem we are subtracting to find a previous time. 

We can subtract $\displaystyle 2$ hours from $\displaystyle 7$ and $\displaystyle 10$ minutes from $\displaystyle 16$. 

$\displaystyle \frac{\begin{array}[b]{r}7\\ -\ 2\end{array}}{ \ \ \ \space5}$   $\displaystyle \frac{\begin{array}[b]{r}16\\ -\ 10\end{array}}{ \ \ \ \ \ \space 6}$

He starts to get ready to leave at $\displaystyle 5:06$

For this problem we are subtracting to find a previous time. 

We can subtract $\displaystyle 2$ hours from $\displaystyle 7$ and $\displaystyle 15$ minutes from $\displaystyle 25$. 

$\displaystyle \frac{\begin{array}[b]{r}7\\ -\ 2\end{array}}{ \ \ \ \space5}$   $\displaystyle \frac{\begin{array}[b]{r}25\\ -\ 15\end{array}}{ \ \ \ \space 10}$

He starts to get ready to leave at $\displaystyle 5:10$

For this problem we are subtracting to find a previous time. 

We can subtract $\displaystyle 2$ hours from $\displaystyle 7$ and $\displaystyle 30$ minutes from $\displaystyle 50$. 

$\displaystyle \frac{\begin{array}[b]{r}7\\ -\ 2\end{array}}{ \ \ \ \space5}$   $\displaystyle \frac{\begin{array}[b]{r}50\\ -\ 30\end{array}}{ \ \ \ \space 20}$

He starts to get ready to leave at $\displaystyle 5:20$

For this problem we are subtracting to find a previous time. 

We can subtract $\displaystyle 2$ hours from $\displaystyle 7$ and $\displaystyle 5$ minutes from $\displaystyle 5$. 

$\displaystyle \frac{\begin{array}[b]{r}7\\ -\ 2\end{array}}{ \ \ \ \space5}$   $\displaystyle \frac{\begin{array}[b]{r}5\\ -\ 5\end{array}}{ \ \ \ \ \space 0}$

He starts to get ready to leave at $\displaystyle 5:00$

For this problem we are subtracting to find a previous time. 

We can subtract $\displaystyle 2$ hours from $\displaystyle 7$ and $\displaystyle 20$ minutes from $\displaystyle 35$. 

$\displaystyle \frac{\begin{array}[b]{r}7\\ -\ 2\end{array}}{ \ \ \ \space5}$   $\displaystyle \frac{\begin{array}[b]{r}35\\ -\ 20\end{array}}{ \ \ \ \space 15}$

He starts to get ready to leave at $\displaystyle 5:15$

For this problem we are subtracting to find a previous time. 

We can subtract $\displaystyle 2$ hours from $\displaystyle 7$ and $\displaystyle 42$ minutes from $\displaystyle 55$. 

$\displaystyle \frac{\begin{array}[b]{r}7\\ -\ 2\end{array}}{ \ \ \ \space5}$   $\displaystyle \frac{\begin{array}[b]{r}55\\ -\ 42\end{array}}{ \ \ \ \space 13}$

He starts to get ready to leave at $\displaystyle 5:13$

For this problem we are subtracting to find a previous time. 

We can subtract $\displaystyle 2$ hours from $\displaystyle 8$ and $\displaystyle 50$ minutes from $\displaystyle 58$. 

$\displaystyle \frac{\begin{array}[b]{r}8\\ -\ 2\end{array}}{ \ \ \ \space6}$   $\displaystyle \frac{\begin{array}[b]{r}58\\ -\ 50\end{array}}{ \ \ \ \space 8}$

He started watching the television show at $\displaystyle 6:08$

Megan has $\displaystyle 11g$ of water and when she adds $\displaystyle 36g$ she will have $\displaystyle 47g$

$\displaystyle 11g+36g=47g$

Megan has $\displaystyle 15g$ of water and when she adds $\displaystyle 14g$ she will have $\displaystyle 29g$

$\displaystyle 15g+14g=29g$