Oh, let's do this the annoying way.
The minute hand moves at $360\frac {\text{degrees}}{\text{revolutions}}\frac {1 \text{revolution}}{60 \text{minutes}}=6\frac{\text{degrees}}{\text{minutes}}$
The hour hand moves at $\frac{360 degrees}{1revolution}\frac{1revolution}{12hours}\frac{1hour}{60 minute}= \frac 12 \frac{degree}{minute}$.
If a minute hand is $x$ degrees away from the hour hand how long will it take for the minute hand to become $x + 180$ degrees away from the hour hand?
Well that's a matter of solving $x + 6t = x + \frac 12t + 180$ or $t= \frac {180}{\frac {11}{2}}=\frac {360}{11}$. (Roughly $32$ minutes.)
At $3:00$ the hour hand and the minute hand are perpendicular. The next time that will happen will be when the minute hand moves for $90$ degrees before the hour had to $90$ degrees after the hour. That will happen in about $32$ minutes. (At $3:32$). Every $\frac {360}{11}$ theminute hand will move 180 degrees further than the hour hand and the hands will be perpendicular.
In a $12$ hour (or a $60*12 = 720$ minute period) this will happen $\frac {720}{\frac {360}{11}} = 22$ times.
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Here's a better way, although it was easy to make annoying sign and logic errors if you aren't careful.
The minute hand travels $6$ degrees a minute. ($\frac {360}{60} = 6$). The hour hand travels $.5$ degrees a minute. ($\frac {360}{12*60} = .5$).
In a 12-hour period there are $12*60=720$ minutes.
Let's say $\theta_t = 6t$ is the angle of the minute hand after $t; 0 \le t < 720$ minutes. Note: it's very possible that $\theta(t) > 360$.
Let's say $\phi_t = .5t$ is the angle of the hour hand after $t$ minutes..
Obviously $\theta_t \ge \phi_t$.
$\theta_t$ and $\phi_t$ are perpendicular if $\theta_t = \phi_t + 90 + k*180$ for some non-negative integer $k$.
So we need the find out how many solutions there are to:
$6t = .5t + 90 + k*180; 0 \le t < 720$ there are.
So $t = \frac {90 + k*180}{5.5}$ will have one solution for each integer $k$.
So how many $k$ are there so that $0 \le t = \frac {90 + k*180}{5.5} < 720$ are there?
Multiply everything by $11$: $0 \le 2(90 + k*180)< 11*720$
Divide everything by $180$: $0 \le 1 + 2k < 11*4$
$-1 \le 2k < 11*4 -1$
$\frac 12 \le k < 2*11 - \frac 12$
And as $k$ is a non-negative integer $0 \le k < 22$. There are $22$ possible $k$s which give a solution.
This also gives us the times this occurs:
$t = \frac {90 + k*180}{5.5} = \frac {180}11 + k *\frac {360}{11} \approx 16.36 + 32.72k$ (or a little more than every half-hour):
So at 12:16, 12:49, 1:22, 1:54, 2:27, 3:00, etc.
Best Answer
On a normal clock, the hand cross each other 11 times in 12 hours, which means that they cross once every $\dfrac{12}{11} = 1.090909…$ hours or every 1 hour, 5 minutes, and 27.272727… seconds. The incorrect clock gains about 27 seconds at every crossing, so each day it gains $22(27.272727… seconds)= 10 minutes every day.
Note: These answers are all similar enough that different methods of calculating the answer could be the reason for the varying results.