One clock gains, another clock loses time. When will both clocks next show the same time? Problem: A wall clock gains $2$ minutes in $12$ hours while a table clock loses $2$ minutes every $36$ hours. Both are set correctly at $12$ noon on Tuesday. When will both clocks next show the same time?
Choices:
$\text{a.)}\quad 12:30\quad  \text{night}\\
\text{b.)}\quad 12\quad  \large  \text{noon}\color{green}{\checkmark}\\
\text{c.)}\quad 01:30\quad  \text{night}\\
\text{d.)}\quad 12\quad   \text{night}$
My work: Time in which they meet is
\begin{align}
\dfrac{60}{\dfrac{2}{12\times 60}+\dfrac{2}{36\times 60}} &=16200\quad  \text{min}\\[1.5em]
&=270\quad  \text{hrs}
\end{align}
How to proceed?
 A: Your answer, (b), is correct, but the following may help clean up any problems with your reasoning:
You want to find out how many hours, $t$, elapse before the wall clock, $W_c$, and the table clock, $T_c$, show the same time again. 
Since the wall clock gains $2$ minutes every $12$ hours, it really gains $\frac{t}{360}$ hours every $t$ hours (e.g., $\frac{12}{360}=\frac{1}{30}$ and $\frac{1}{30}$ hours is equivalent to $2$ minutes). Thus, we can let the total time elapsed on $W_c$ for $t$ hours be
$$
W_c = t+\frac{t}{360}.\tag{1}
$$
For the table clock, $T_c$, we actually lose $2$ minutes every $36$ hours; that is, we lose $\frac{t}{1080}$ hours every $t$ hours (for example, $\frac{36}{1080}=\frac{1}{30}$ and $\frac{1}{30}$ hours is equivalent to $2$ minutes, as described above). Thus, we can let the total time elapsed on $T_c$ for $t$ hours be
$$
T_c = t-\frac{t}{1080}.\tag{2}
$$
Now what? This depends on the kind of clock you are using. If you are using a regular wall-clock that does not differentiate between AM and PM (i.e., a 12-hour clock), then we will need to figure out when 
$$
W_c-T_c=12.\tag{3}
$$
However, if you are using a clock that does differentiate between AM and PM, then you will need to figure out when
$$
W_c-T_c=24.\tag{4}
$$
Using a 12-hour clock: We substitute $(1)$ and $(2)$ into $(3)$ to get
$$
\frac{t}{360}+\frac{t}{1080}=12\Longleftrightarrow 4t=12960\Longleftrightarrow \color{red}{t=3240}.
$$
Thus, $3240$ hours will have elapsed. Note that
$$
\underbrace{3240}_{\text{hours}} = \underbrace{19}_{\text{weeks}}\cdot \underbrace{168}_{\text{hours/week}} + \underbrace{48}_{\text{hours}}.
$$
Thus, $19$ weeks and $48$ hours will have elapsed. Since your clocks began on a Tuesday at noon, they will next meet again on $\boxed{\color{red}{\text{Thursday at noon}}}$.
Using a 24-hour clock: We substitute $(1)$ and $(2)$ into $(4)$ to get
$$
\frac{t}{360}+\frac{t}{1080}=24\Longleftrightarrow 4t=25920\Longleftrightarrow \color{red}{t=6480}.
$$
Thus, $6480$ hours will have elapsed. Note that
$$
\underbrace{6480}_{\text{hours}} = \underbrace{38}_{\text{weeks}}\cdot \underbrace{168}_{\text{hours/week}} + \underbrace{96}_{\text{hours}}.
$$
Thus, $38$ weeks and $96$ hours (four days) will have elapsed. Since your clocks began on a Tuesday at noon, they will next meet again on $\boxed{\color{red}{\text{Saturday at noon}}}$.
