Clock Math: Finding the current time when the minute hand and hour hand are opposite to each other

Question

Problem

It is now between 10:00 and 11:00 Six minutes from now, the minute hand of a watch will be exactly opposite the place where the hour hand was three minutes ago. What is the exact time now?

My attempt

If we say that each minute that passes there is a $12$ degrees increase then there are $720$ degrees in a full revolution of the clock. If we let the current angle of the hour hand be $x$ degrees then we know that we are at a degrees position after $600$ ( after $10$ means $600$ degrees and under $660$). Three minutes ago means we subtract an angle of $36$ and 6 minutes from now means we add an angle of $72$. I also subtracted $360$ from the left hand side because I know that would be on the opposite side of the hour hand.

$600 + \frac{720}{x-36} - 360 = 720 + \frac{720}{x+72}$

Obviously, my equation is not correct when I try to solve it, it doesent make much sense. Can anyone tell me where my logic is wrong, and give me some hints? I've been trying to make an equation for a while now. Thanks.

Answer 1

Denote the position of the minute and hour hand by $m$ and $h$ (in units), respectively, $0\leq m,h \leq 60$. Then, in units/minute, $$\frac{dm}{dt} = 1, \, \frac{dh}{dt} = \frac{1}{60} \\ \implies m(t)=t \,, h(t)= \frac{t}{60}$$

Suppose at $t=0$, the time was exactly $10:00$, and $t_0$ minutes have passed since then. Further, say the minute and hour hand moved by $m’$ and $h’$ units. Then, the position of the minute hand after $6$ minutes will be $t_0$ and the position of the hour hand three minutes ago was $50 + h’ - \frac{3}{60}$ units. We want the difference of these positions to be $30$ units. So, $$ m’-h’=17-\frac {1}{20} \\ \implies t-\frac{t}{60} = \frac{339}{20} \\ \implies t \approx 17.237$$ Hence, $(m’,h’)= (17.237, 0.287)$ and the time right now is $\color{purple}{10:17:14}$!

Answer 2

Let $m_0$, $h_0$ be the positions of the minute hand and hour hand right now, respectively. Let m(t), h(t) be the position on the watch as a function of time where $t$ is in minutes and the position is in ticks and is between $0$ and $59$, inclusive.

Then $$m(t)=m_0 + t$$ because the minute hand moves at one tick per minute.

Likewise, $$h(t)=h_0+\frac{t}{12}$$ because the hour hand moves at five ticks per sixty minutes which is $\frac{1}{12}$ ticks per minute.

We are given that $h(-3)+30=m(6)\implies h_0-\frac{1}{4}+30 = h_0+\frac{119}{4}=m(6) = m_0+6$.

Therefore, $$h_0=m_0-\frac{95}{4}=m_0-23\frac{3}{4}$$

We are also told that it is between $10$:$00$ and $11$:$00$ so $h_0$ is somewhere near $50$. Since we are restricting the positions to be between $0$ and $59$ we rewrite $$h_0=m_0-23\frac{3}{4}$$ as $$h_0=m_0+36\frac{1}{4}$$

We also know that at $10$:$00$, $$m(t)=0\implies t = -m_0$$ and $$h(t)=50 = h_0+\frac{t}{12}$$

Substituting $-m_0$ for $t$ and $m_0+36\frac{1}{4}$ for $h_0$ gives

$$50 = m_0+36\frac{1}{4}-\frac{m_0}{12}\implies 13\frac{3}{4}=\frac{11}{12}m_0\implies 165=11m_0\implies m_0=15$$

Therefore, $h_0=51\frac{1}{4}$ and it is $10$:$15$.