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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?

David K
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R K
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1 Answers1

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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 $\boxed{\color{red}{\text{Saturday at noon}}}$.

Daniel W. Farlow
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