## Tag Archives: clock puzzles

### More Clock Problems

Example.

At what time between 4 and 5 o’clock will the minute-hand of a watch be 13 minutes in advance of the hour hand?

Solution.

Let x denote the required number of minutes after 4 o’clock; then, as the minute hand travels twelve times as fast as the hour hand, the hout hand will move over x/12 minute divisions in x minutes. At 4 o’clock, the minute hand is 20 divisions behind the hour hand, and the finally minute hand is 13 divisions in advance; therefore the minute hand moves over $20+13$, that is,, 33 divisions more than the hour hand.

Hence, $x=\frac{x}{12}+33$ which implies $\frac{11x}{12}=33$ and hence, $x=36$.

Thus, the time is 36 minutes past 4.

If the question be asked as follows: “At what times between 4 and 5 o’clock will there be 13 minutes between the two hands, then we must also take into consideration, the case when the minute hand is 13 divisions behind the hour hand. In this case, the minute hand gains $20-13$ or 7 divisions.

Hence,, $x=\frac{x}{12}+7$ which gives $x=7 \frac{7}{11}$

Therefore, the times are $7\frac{7}{11}$ past 4, and $36^{'}$ past 4.

Homework for fun:

1. At what time between one and two o’clock are the hands of a watch first at right angles?
2. At what time between 3 and 4 o’clock is the minute hand one minute ahead of the hour hand?
3. When are the hands of a clock together between the hours of 6 and 7?
4. It is between 2 and 3 o’clock, and in 10 minutes the minute hand will be as much before the hour hand as it is not behind it; what is the time?
5. At what times between 7 and 8 o’clock will the hands of a watch be at right angles to each other? When will they be in the same straight line?

Hope you had enough fun! 🙂

More fun later,

Nalin Pithwa

### Interchanging the Hands of a Clock

Problem.

The biographer and friend of the immortal physicist Albert Einstein, A. Moszkowski, wished to distract his friend during an illness and suggested the following problem:

The problem he posed was this: “Take the position of the hands of a clock at 12 noon. If the hour hand and the minute hand were interchanged in this position the time would would still be correct. But, at other times (say at 6 o”clock) the interchange would be absurd, giving a position that never occurs in ordinary clocks: the minute cannot be on 6 when the hour hand points to 12. The question that arises is when and how often do the hands of a clock occupy positions in which interchanging the hands yields a new position that is correct for an ordinary clock?

“Yes,” replied Einstein, “this is just the type of problem for a person kept to his bed by illness; it is interesting enough and not so very easy. I am afraid thought that the amusement won’t last long because I already have my fingers on a solution.”

“Getting up in bed, he took a piece of paper and sketched the hypothesis of the problem. And, he solved it in no more time than it took me to state it.”

How is the problem tackled?

Solution:

We measure the distance of the hands around the dial from the point 12 in sixtieths of a circle.

Suppose one of the required positions of the hands was observed when the hour hand moved x divisions, from 12, and the minute hand moved q divisions. Since the hour hand passes over 60 divisions in 12 hours, or 5 divisions every hour, it covered the x divisions in x/5 hours. In other words, x/5 hours passed after the clock indicated 12 o’clock. The minute hand, covered y divisions in y minutes, that is, in y/60 hours. In other words, the minute hand passed the figure 12 a total of y/60 hours ago, or

$\frac{x}{5}-\frac{y}{60}$

hours after both hands stood at twelve. This number is whole(from 0 to 11) since it shows how many whole hours have passed since twelve.

When the hands are interchanged, we similarly find that $\frac{y}{5}-\frac{x}{60}$

whole hours have passed from 12 o’clock to the time indicated by the hands. This is a whole number from 0 to 11.

And, so we have the following system of equations:

$\frac{x}{5}-\frac{y}{60}=m$

$\frac{y}{5}-\frac{x}{60}=n$

where m and n are integers (whole numbers) that can vary between 0 and 11. From this system, we find

$x=\frac{60(12m+n)}{143}$

$y=\frac{60(12n+m)}{143}$

By assigning m and n the values from 0 to 11, we can determine all the required positions of the hands. Since each of the 12 values of m can be correlated with each of the 12 values of n, it would appear that the total number of solutions is equal to 12 times 12, that is, 144. Actually, however, it is 143 because when $m=0, n=0$ and also when $m=11, n=11$ we obtain the same position of the hands.

Put $m=11, n=11$, we have $x=60, y=60$

and the clock shows 12, as in the case of $m=0, n=0$.

We will not discuss all possible positions, but only two.

First example:

$m=1, n=1$

$x=\frac{60.13}{143}=5\frac{5}{11}$ and $y=5\frac{5}{11}$

and the clock reads 1 hour $5 \frac{5}{11}$ minutes; the hands, have merged by this time and they can of course be interchanged (as in all other cases of coincidence of the hands).

Second Example.

$m=8, n=5$.

$x=\frac{60(5+12.8)}{143} \approx 42.38$ and $y=\frac{90(8+12.5)}{143} \approx 28.53$.

The respective times are 8 hours 28.53 minutes and 5 hours 42.38 minutes.

More clock problems later,

Nalin Pithwa