# Clock problem in which the hour and minute hand will exchange positions

I've tried using the angle formula for clocks, but I'm not sure where I'm going wrong, i keep getting 60/11, but my logic is pretty sloppy so I'm not confident in that answer. Any help/hints would be nice, thanks!

(Picture: It is now sometime between 7:00 and 8:00. In less than one hour, the minute hand and hour hand will have exactly exchanged positions. Find the number of minutes after 7:00 that the time now is.)

• Your image file is not displaying on Firefox on android. Hence edit, and put your question in full please – unseen_rider Feb 27 '17 at 18:13
• It's there, sorry for the inconvenience! – Nick Brown Feb 27 '17 at 18:16
• It is now displaying fine. - Put the actual link where picture is at the end after question. – unseen_rider Feb 27 '17 at 18:22
• A related question is math.stackexchange.com/questions/336481, although that question asks for the difference beween the start and end times, not the starting time. – David K Feb 27 '17 at 20:04

Let $h_0$ be the initial time, expressed in hours.

Then $h_0 = 7 + t$ where $0 < t < 1$.

Let $m_0$ be number of minutes for the minute hand at time $h_0$.

Then $m_0=60t$.

Let $h_1$ be the time of the exchanged positions, expressed in hours.

Let $m_1$ be number of minutes for the minute hand at time $h_1$.

It's given that at time $h_1$, the hour and minute hand positions are the same as for time $h_0$, except that the hour and minute hands are exchanged.

\begin{align*} \text{Then }\;\;\frac{h1}{12}&=\frac{m0}{60}\\[6pt] \implies\; h1&=12\left(\frac{m0}{60}\right)=12\left(\frac{60t}{60}\right)=12t\\[10pt] \text{and }\;\;\frac{m1}{60}&=\frac{h0}{12}\\[6pt] \implies m1&=60\left(\frac{h0}{12}\right)=60\left(\frac{7+t}{12}\right)=35 + 5t \end{align*}

From given information, $7 < h_0 < 8$ and $h_0 < h_1 < h_0 + 1$, hence $7 < h_1 < 9$.

Suppose first that $7 < h1 < 8$. Then

\begin{align*} &h_1 = 7 + \frac{m_1}{60}\\[6pt] \implies\; &12t = 7 + \frac{35 + 5t}{60}\\[6pt] \implies\; &t = {\small{\frac{7}{11}}}\\[6pt] \implies\; & h_0 = h_1 = {\small{\frac{84}{11}}} \end{align*}

contrary to $h_0 < h_1$.

Thus, we must have $8 \le h_1 < 9$. Then

\begin{align*} &h_1 = 8 + \frac{m_1}{60}\\[6pt] \implies\; &12t = 8 + \frac{35 + 5t}{60}\\[6pt] \implies\; &t = {\small{\frac{103}{143}}}\\[6pt] \implies\; & h_0 = 7 + {\small{\frac{103}{143}}} \text{ and } h_1 = 8 + {\small{\frac{92}{143}}} \end{align*}

Thus, at time $h_0$, the number of minutes since $7$:$00$ is

$$m_0 = 60t = 60\left(\frac{103}{143}\right) = \frac{6180}{143}$$

You could start by drawing a picture in order to look for clues about how the solution should look.

The hour hand has to go somewhere between its 7:00 and the 8:00 positions. In less than an hour, it has to get to where the minute hand is now, so the minute hand can't be far away.

If the minute hand is "before" the hour hand (that is, on the same side of the hour hand as 7:00 position of the hour hand), it will take the hour hand much more than one hour to get there. So the minute hand must be "after" the hour hand. To get to where the hour hand is, the minute hand will have to pass the 12:00 position, so it will be after 8:00 by the time it gets to where the hour hand is now--which means the hour hand will be between 8:00 and 9:00 then, which means the minute hand must be between 8:00 and 9:00 now.

So you have an hour hand between 7:00 and 8:00 and a minute hand between 8:00 and 9:00, which is somewhere between $40$ and $45$ minutes past the hour--clearly not $\frac{60}{11}.$

But the angle moved by the minute hand (to get to the hour hand's starting point) plus the angle moved by the hour hand (to get to the minute hand's starting point) add up to a full circle. (Check this in the picture.) And the minute hand moves $12$ times as far as the hour hand during that period of time. From that you should be able to deduce how long the period of time is.

The simplest way to get the answer is probably to procede as follows. Since the two hands exchange their positions in less than one hour, in the initial position the minute hand must be more advanced than the hour hand. Let us call $x$ the initial angle (in radians) between the two hands. In the same time interval, the hour hand moves by an angle $x$, while the minute hand moves by an angle $2 \pi -x \,$. Because the angle covered by the minute hands must be $12$ times larger than that covered by the hour hand, we have

$$12 x= 2 \pi -x$$ $$x=\frac {2}{13} \pi$$

Now note that, in the initial position, if we call $y$ the angle between the minute hand and the position of the number 12:00, this angle must be $12$ times larger than that included between the hour hand and the number 7:00. So we must have

$$y=12 (y-x-\frac {7}{6} \pi) \\= 12 (y- \frac {2}{13} \pi -\frac {7}{6} \pi)$$

from which

$$y=\frac {206}{143} \pi$$

Converting in minutes, we have that the number of minutes after 7:00 of the initial time is

$$60 \cdot \frac {206/143 \, \,\pi }{2 \pi}=\frac {6180}{143} \approx 43.216$$

which corresponds to about 7:43 and $13$ seconds.