Minute hand and hour hand interchange

Question

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.

Answer 1

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.

Answer 2

HINT:

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 $$ x= 4.5454\ldots_{(12)} \\ y = 5.4545\ldots_{(12)}$$