What time are the minute and hour hands of a clock perpendicular? 
At noon the minute and hour hands of a clock coincide. Assuming the hands of the clock move continuously:
  a) What is the first time $T_1$ when they are perpendicular?
  b) What is the next time $T_2$ when they again coincide?

I believe I have solved this problem, but I would like someone to verify my answer.
I got $T_1$ to be 12:16 P.M. and 21.818181... seconds ($\frac{180}{11}$ minutes after noon).
I got $T_2$ to be 1:05 P.M. and 27.272727... seconds ($\frac{720}{11}$ minutes after noon).
If there are any questions relating to this problem, please feel free to ask them, as I will do my best to answer them.
 A: If the minute hand has made $x$ revolutions around the clock ($60x$ minutes), the hour hand has made $\frac1{12}x$ revolutions; the difference between these two values represents an angle. Both hands start at 0 revolutions.
For (1), the relevant angle is $\frac14$ of the circle:
$$x=\frac14+\frac1{12}x$$
Solving, we get $x=\frac3{11}$, which corresponds to a time of $\frac{180}{11}$ minutes after noon.
For (2) we replace $\frac14$ with 1, since the minute hand has lapped the hour hand. Then
$$x=1+\frac1{12}x$$
Solving, we get $x=\frac{12}{11}$, i.e. $\frac{720}{11}$ minutes after noon.
Therefore, both your solutions are correct.
A: At noon the hands coincide. In $15 + \delta$ minutes, the angle between them will be $90^\circ$.
The minute hand moves at $\dfrac{360^\circ}{60\ \text{min}} = 6$ degrees per minute. The hour hand moves at $\dfrac{360^\circ}{12\ \text{hr}} = \dfrac 12$ degrees per minute. So we need to solve
\begin{align}
   (15+\delta)\left(6 - \frac 12 \right) &= 90 \\
   15 + \delta &= \frac{180}{11} \\
   \delta &= \frac{15}{11} \text{min}
\end{align}
So $T_1 = 12:15 + 0:01\frac{4}{11} = 12:16\frac{4}{11}$.
The hands of the clock will be $90^\circ$ apart every $1 + \epsilon$ hours after that, and they will return to their original position after $11$ changes.
\begin{align}
   11(1 + \epsilon) &= 12 \\
   1 + \epsilon &= \frac{12}{11} \\
   \epsilon &= \frac{1}{11} \text{hr} \\
   \epsilon &= 5\frac{5}{11} \text{min}
\end{align}
So
$T_2 =  12:16\frac{4}{11} + 1:05\frac{5}{11} = 1:05\frac{9}{11}$
$T_3 =  1:05\frac{9}{11} + 1:05\frac{5}{11} = 2:11\frac{3}{11}$
$\vdots$
