Algebra clock problem An absent-minded watch repairman connected the hour hand to the minute hand pinion and the minute hand to the hour hand pinion and set the clock at 6AM which was the correct time then. How soon after 6AM (at what time) will this clock give the correct time again?
 A: Hint: Try writing out formulae for the positions of the hands on a correct clock and one the wrong clock in terms of time.
Correct clock:
the long hand is at $(\text{time in mins})\times (6^{\circ}), \mod 360^{\circ}$, since it starts straight up and moves by 360 degrees each hour;
the short hand is at $180^{\circ} +(\text{time in mins})\times \frac{1}{2}^{\circ}, \mod 360^{\circ}$, since it starts straight down and moves by 360 degrees every 12 hours.
Now write similar equations for the wrong clock, and look for values of time from which the long-hand equations give the same angle and so do the short hand equations.
A: correct answer is 7:06 AM, in 1 hour and 6 minutes the  minute hand will rotate 360+30 plus degrees and just cross 7 o'clock and the hour hand will rotate 30 degrees and come to 6 minutes (right next to 1 o'clock) 
A: Imagine a phantom hour hand that starts at $6$ (just like the actual hour hand)
but is connected to the correct pinion so that it moves at the speed an hour hand is supposed to move.
The clock cannot be showing the correct time unless the actual hour hand is where it is supposed to be, namely, where the phantom hour hand is.
So we have two "hands" that start at one position on the clock (pointing to $6$);
one hand rotates at a rate of one revolution per hour, the other at a rate of
one revolution every $12$ hours. They are next at the same position when 
$\frac{12}{11}$ hours have passed, the same amount of time as it takes between noon and when the hands meet again on a normal clock.
So far, we know only that the hour hand is in the correct position at $\frac{1}{11}$ hour after $7$ AM. But what about the minute hand?
If the minute hand does not end up where it is supposed to be at $\frac{1}{11}$ hour after $7$ AM, you will have to look for a later time when the clock is correct.
So the next step is to find out whether or not the minute hand will be where it should be at that time.
There are various ways of doing this, including direct computation (since you know exactly when to look at the hand's position); I leave that step "as an exercise."
