Suppose you have a clock that is set at the twelve o' clock position. Given one rotation how many times will both the minute and hour hand coincide?
The answer is pretty simple, it's every 12/11 hours, which makes sense because the hour hand moves at 30 degrees/hr while the minute hand moves at 360 degrees/hr so 360/330 = 12/11
One can also determine the answer by taking the sum $\sum_{x=0}^{\infty}(1/12)^x$, which also gives 12/11. Why did we take the geometric series of 1/12?
Best Answer
From an infinite series point of view, as indicated in my comment, the minute arm travels at 1/12 the speed of the hour arm, so $\sum=\frac{1}{1-\frac{1}{12}}$=$\frac{12}{11}$hour. So this is 1 hour and 1/11 of an hour. You can figure out that this corresponds 5 minutes, 27 and 3/11 seconds. But there is another way to calculate this (hence my post), without infinite series and that is just using a basic linear equation, as time continues linearly. If it is 1 o clock, the hour arm is 30 degrees ahead of the minute arm. Let $t$ be the time in seconds. The amount of degrees the minute arm covers in 1 second is $t/10$ and for the hour arm $t/120$. From here you can set up the equation $\frac{t}{10}=\frac{t}{120}+30$. Solving this equation results in $\frac{11t}{120}=30$ or $t=3600/11$ from which the 5 minutes, 27 +3/11 seconds follow. Using this principle over and over again, you can see on the arms eclipse 11 (not 12) times on a 12 hour cycle.