I need help with solving a math problem that involves clocks Here's the question.
"It is now between 10:00 and 11:00. Six minutes from now, the minute hand of a watch will be exactly opposite the place where the hour hand was three minutes ago. What is the exact time now?"
It's from Art of Problem Solving Volume 1, Chapter 4 Proportions. I haven't solved clock problems in the past and I'm not sure how to do this. The solution didn't help and was extremely complicated, so I was hoping someone could provide a solution that I might be able to understand. Any tips for future problems that might be similar to this would also be greatly appreciated!
 A: I'd set $t=0$ to correspond to 10:00. Let's write formulas for the position of each hand $t$ minutes after that, where position is measured as the angle (in degrees) clockwise from the "12".
Minute hand: $m(t)=\frac{360\text{ deg}}{60\text{ min}}t=6t$.
Hour hand: $h(t)=\frac{360\text{ deg}}{12\cdot60\text{ min}}t+300=0.5t+300$.
Now let $T$ be the time now. Your information says:
$$
\begin{align}
m(T+6)&=h(T-3)-180&&\text{(Note $h(T)$ is between 300 and 330.)}\\
6(T+6)&=0.5(T-3)+300-180
\end{align}
$$
You can solve this linear equation to get $T=15$. So it's 10:15.
A: Consider 10:00. The angle formed at 10:00 is 60 degrees. So, for every minute past ten, the minute hand increases the angle by 6 degrees (360 deg/60min), but the hour hand decreases it by 1/2 degree since it moves up. This leaves us with a net gain of 5.5 deg. We can this write the angle at 10: min to be 60+5.5(min).
Six minutes from now, the minute hand moves 66=36 degrees.3 minutes ago, the hour hand moves 1/23=1.5 degrees. Since the minute and hour hand are opposite, they forms a straight line, and our time 10:min must form an angle of 180-36-3/2= 142.5 deg. So, 60+5.5x = 142.5 and x= 15
