It's About Time

Let $\theta$ be the (clockwise) angle (in radians) between the minute hand and the 12:00 vertical, and let $\alpha$ be the (clockwise) angle between the hour hand and the 8:00 radial line. Then $\theta$ will be the same fraction of a full circle as $\alpha$ is of one-twelfth of a circle, (i.e., the angle the hour hand sweeps through in one hour). Thus

$\dfrac{\theta}{2\pi} = \dfrac{\alpha}{\dfrac{\pi}{6}} \Longrightarrow \theta = 12\alpha$ .

For the hour hand to then make the same (counterclockwise) angle with the 12:00 vertical as the (clockwise) angle the minute hand makes with the 12:00 vertical, we require that

$\theta = \dfrac{\pi}{2} + \left(\dfrac{\pi}{6} - \alpha\right) \Longrightarrow \theta + \alpha = \dfrac{2\pi}{3}$

$\Longrightarrow \theta + \dfrac{\theta}{12} = \dfrac{2\pi}{3} \Longrightarrow \dfrac{13}{12}\theta = \dfrac{2\pi}{3} \Longrightarrow \theta = \dfrac{8\pi}{13}$ .

In minutes past 8:15, the angle $\theta$ translates to

$\dfrac{\theta}{2\pi} * 60 - 15 = \dfrac{4}{13} * 60 - 15 = \dfrac{240}{13} - 15 = \dfrac{45}{13}$ minutes,

which in turn translates to $\dfrac{45}{13}*60 = 207.6923...$ seconds, the nearest integer to which is $\boxed{208}$ .

Mt. hand moves a distance of 1 $\color{#3D99F6}{minute}$ in 60 s.
Hr. hand moves a distance of 1/12 $\color{#3D99F6}{minute}$ in 60 s.
So they move towards one another at (1+1/12)= 13/12 $\color{#3D99F6}{minute}$ per 60 s.
$\implies~ 13/(12\times60)~ \color{#3D99F6}{minute}~ per~ s.$
The distance between them is (5 - 5/4) = 15\4 $\color{#3D99F6}{minute}$
t=d/v. So (15/4) / (13/720)=207.69231 s. ANSWER 208.

Note. at 8.15, Mt. hand has moved towards 9 by one fourth of distance between 8 and 9 o'clock.
At the end they do not meet but make the same angle with vertical.

If you know that a minute hand moves 6°/ minute and the hour hand moves 0.5° per minute this problem would be quite easier.. First, compute for the angle of the minute hand from the vertical at exactly 8:15

$\Rightarrow$ 15 $\times$ 6 = 90°

Next is the angle of the hour hand from the vertical at exactly 8:15

$\Rightarrow$ 60 + 0.5(15) = 67.5°

(At 8:00, the angle formed from 12 to 8 is 240°. Subtracting 180° because the problem asks for the same angle at vertical which gives 60°)

Let x be the minutes passed after 8:15 so that the minute hand and the hour hand of an (ideal) analog clock will make the same angle with the vertical

$90 - 6x = 67.5 + 0.5x$

$90 - 67.5 = 6x + 0.5x$

$\frac { 22.5 }{ 6.5 } =\frac { 6.5x }{ 6.5 }$

$x\quad =\quad \frac { 45 }{ 13 } minutes$

Converting minutes to seconds $\frac { 45 }{ 13 } \quad \times \quad 60\quad =\quad 207.692\quad \approx \quad 208\quad seconds$