Clock Problems — Lesson + Practice

The two facts (memorize)

Minute hand: $6^\circ$ per minute. Hour hand: $0.5^\circ$ per minute.
So the minute hand gains on the hour hand at $5.5^\circ$/min. That one number does everything.

Let $t$ = minutes after $H$ o'clock.

Minute hand position: $6t$ (degrees from the 12).
Hour hand position: $30H + 0.5t$ (it starts at $30H^\circ$ — e.g. 3:00 $\to 90^\circ$).
Gap (hour $-$ minute) $= 30H - 5.5t$.

$$30H - 5.5t = \pm\,\theta \quad\Longrightarrow\quad t = \frac{30H \mp \theta}{5.5}$$

Coincide: $\theta = 0$. Right angle: $\theta = 90$. Opposite: $\theta = 180$.

Between 4:00 and 5:00, when do the hands coincide? Here $H = 4,\ \theta = 0$:

$$t = \frac{30(4)}{5.5} = \frac{120}{5.5} = 21\tfrac{9}{11}\ \text{min} \;\;\Rightarrow\;\; \mathbf{4\!:\!21\tfrac{9}{11}}$$

Check: minute $= 6(21.8) \approx 130.9^\circ$; hour $= 120 + 0.5(21.8) \approx 130.9^\circ$. Same spot. ✓

Now you — solo, no peeking. Narrate your steps; if you stall, name the step.

Between 7:00 and 8:00, when do the hands coincide?
Between 2:00 and 3:00, when are the hands at a right angle ($90^\circ$ apart)?

Answers — open only after you've committed

$t = \dfrac{30(7)}{5.5} = \dfrac{210}{5.5} = 38\tfrac{2}{11} \;\Rightarrow\; \mathbf{7\!:\!38\tfrac{2}{11}}$
The minute hand must get $90^\circ$ ahead: $\;t = \dfrac{30(2) + 90}{5.5} = \dfrac{150}{5.5} = 27\tfrac{3}{11} \;\Rightarrow\; \mathbf{2\!:\!27\tfrac{3}{11}}$. (The "$90^\circ$ behind" case gives $t < 0$ — it already happened before 2:00.)