# Problem involving time minute hand versus hour hand

I came across a problem in my textbook that goes as follows:

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?

I don't understand their approach to solving it:

Let x be the number of minutes past 4 o'clock. The minute hand moves 12 times faster than the hour hand. Therefore, the hour hand moves x/12 minutes when the minute hand moves x minutes. Now, for the minute hand to coincide with the hour hand, it must move 20 minute divisions. Then, finally, it will then move 13 minute divisions ahead of the clock. Thus, > it moves 33 minutes in total. Therefore, the required number of minutes past 4 o'clock is: $$x = \frac{x}{12} + 33$$

What I don't understand about the above equation is as follows:

(1) The hour hand moves, but I don't see where it was accounted for in the above equation.

(2) The minute hand does move 33 minutes in total, but why is that being added to the number of minutes the hour hand moves? Part of those 33 minutes is used to catch up to the hour hand, so why is it being added to the hour hands movement?

If anyone could intuitively explain the above equation, that would be appreciated.

Thanks.

Literally just after I posted this question, it came to me and I facepalmed hard. The minute hand, after 4 o'clock, moves $x$ minutes total to be 13 minutes ahead of the hour hand. So, the hour hand, moves $\frac{x}{12}$ minutes on the clock in total. To catch up with the hour hand, the minute hand has to move (if the hour hand is stationary) 33 minutes. Add that to the number of minutes the hour hand moves in total ($\frac{x}{12}$) and you get the equation, $$x = \frac{x}{12} + 33$$ or, in English: $$total_{minutes} = total_{hourhandminutes} + minutes_{assuminghourhandisstationary}$$