## 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

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