[Math] Analog Clock Problem; Alignment of the Hour Hand and Minute Hand.

algebra-precalculusgeometrypuzzle

The question goes like this:

It is now between 9 and 10 o’clock. In 4 minutes, the hour hand will be directly opposite the position occupied by the minute hand 3 minutes ago. What time is it now?

The answer to this question is 9:20, but cannot figure out the solution. I've tried treating this as a distance problem. I used degrees(°) as the distance unit, so the hour hand has a speed of $\frac12°$ per minute and the minute hand $6°$ per minute. It did not help me at all.

Maybe I just needed to be pointed to the right direction. I appreciate any help. Thank you.

Best Answer

We will measure angles in degrees from $12:00$ in a clockwise direction.

Let $h(t)$ be the position of the hour hand, in degrees, at $t$ minutes after $9$:$00$ and let $m(t)$ be the position of the minute hand, in degrees, at $t$ minutes after $9$:$00$.

Then $$h(t) = 270 + \frac 12t$$

and $$m(t) = 6t$$

We need to solve

\begin{align} h(t+4) &= m(t-3) + 180 \\ 270 + \frac 12(t+4) &= 6(t-3) + 180 \\ t &= 20\; \text{minutes} \end{align}