A person who left home between $4$ p.m. and $5$ p.m. returned between $5$ p.m. and $6$ p.m. and found that the hands of his watch has exactly changed places. When did he go out?

My attempt: The dial of a clock is divided into $60$ equal divisions. In one hour, the minute hand makes one complete revolution, i.e., it moves through $60$ divisions and the hour hand moves through $5$ divisions. Suppose, when the man went out, the hour hand was $x$ divisions ahead of the point labeled $12$ on the dial, where $20 < x < 25$ (as he went out between $4$ p.m. and $5$ p.m.). Also suppose, when the man returned, the hour hand was $y$ divisions ahead of zero mark and $25 < y < 30$. Since the minute hand and hour hand exactly interchanged places during the interval that the man was out, the minute hand was at y when he went out and at $x$ when he returned. Since the minute hand moves $12$ times as fast as the hour hand, we can say that the angle that the minute hand sweeps will be $12$ times that swept by the hour hand. But, I am not able to express this in terms of $x$ and $y$.

Any constructive hint is appreciated.

  • $\begingroup$ I think I would formulate the problem as $h_1=4+x$, $m_1=12x$, $h_2=5+y$, $m_2=12y$. $\endgroup$
    – Joe
    Sep 1, 2021 at 3:04
  • $\begingroup$ Can you please tell me what should be done after that. $\endgroup$
    – Ilovemath
    Sep 1, 2021 at 3:12
  • $\begingroup$ $h_1$ is the position of the hour hand when he left home, and $m_2$ is the position of the minute hand when he came back. What do we know about these? $\endgroup$
    – Joe
    Sep 1, 2021 at 3:22
  • $\begingroup$ We know that as the time for which minute hand and hour hand move is same , the angle they traverse will be in ratio of their speeds... $\endgroup$
    – Ilovemath
    Sep 1, 2021 at 3:28
  • $\begingroup$ I just meant that the position of the hour hand when he left is the same as the position of the minute hand when he came back. $\endgroup$
    – Joe
    Sep 1, 2021 at 3:46

2 Answers 2


Between the time you go out and the time you return, the hour hand describes the minor arc between the original positions of the hands, and the minute hand describes the major arc. So the total angle of rotation of the two hands together is one full revolution.

At the same time, we know the minute hand must have turned 12 times more than the hour hand. So the gap between the two hands when you go out is $1/13$ of a revolution.

Now assume the time you leave is $4 + x$ hours, where $0 < x < 1$. At that time the position of the minute hand, in relation to the top of the dial, is $x$ revolutions. The position of the hour hand is $(4 + x)/12$ revolutions. So we have $$x - (4 + x)/12 = 1/13.$$

The solution of this equation is $x = 64/143$.

So the time you leave is $26\frac{122}{143}$ minutes past four o'clock.



Let's consider a time as a number between $0$ and $12$. At time $x=$ the position of the hour hand, the position of the minute hand equals $12 \{x\}$, where $\{x\}$ is the fractional part of $x$. If $x = a_0. a_1 a_2 \ldots_{(12)}$ written in base $12$, then $12\{x\}= a_1.a_2 a_3 \ldots_{(12)}$.

Let $x$, $y$ be the departure, and arrival times. Then we have the system $$ y= 12\{x\}\\ x = 12\{y\} $$ and the conditions $4\le x\le 5$, $5\le y \le 6$. We conclude that in base $12$ we have $$ x= 4.5454\ldots_{(12)} \\ y = 5.4545\ldots_{(12)}$$

  • 2
    $\begingroup$ This answer is amazing! $\endgroup$
    – Mike
    Sep 1, 2021 at 4:51
  • $\begingroup$ @Mike, amazing answers deserve upvoted. $\endgroup$
    – Joe
    Sep 1, 2021 at 10:52
  • $\begingroup$ @Joe When I try to vote, I get this message: "Join Mathematics Stack Exchange to start earning reputation and unlocking new privileges like voting and commenting." So it's up to the registered members to vote. I haven't "unlocked" that "privilege." For some reason, I'm being allowed to comment, but when the site creators realize their mistake, I'm sure I'll get in trouble for usurping this privilege. $\endgroup$
    – Mike
    Sep 1, 2021 at 12:51
  • $\begingroup$ @Mike, I don't know why you received that message. You are a user: user 963216. You should only need a reputation $\ge 15$ to upvote, and your reputation is currently 126. $\endgroup$
    – Joe
    Sep 1, 2021 at 12:59
  • $\begingroup$ @Joe It's all right. Joe, have a look at the map here. I'm from one of the light areas. Are you from one of the dark areas? $\endgroup$
    – Mike
    Sep 1, 2021 at 13:50

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