The minute-hand and second-hand of a clock cross each other _______ times between 09:15:00 AM and 09:45:00 AM on a day.

The correct option is 30.

To understand why this is the correct answer, let's analyze the movement of the second-hand relative to the minute-hand step-by-step.

First, let's look at the time interval: from 09:15:00 AM to 09:45:00 AM. This is a duration of exactly 30 minutes.

During any standard 1-minute interval (for example, from 09:15:00 AM to 09:16:00 AM):
- The second-hand starts at the 12 o'clock position (0 seconds), completes one full rotation of 360°, and returns to the 12 o'clock position at the end of the minute.
- The minute-hand moves forward by exactly 1 minute space (which is 6° of rotation).
- Because the second-hand sweeps through the entire clock face (360°) while the minute-hand moves only a tiny fraction (6°), the second-hand must cross the minute-hand exactly once during this minute.

Let's verify the crossings at the boundaries of each minute interval:
- For the first interval from 09:15:00 AM to 09:16:00 AM:
At exactly 09:15:00 AM, the second-hand is at the 12 o'clock position (0 seconds) and the minute-hand is at the 3 o'clock position (15 minutes). They are not crossing at the start. As the second-hand sweeps around, it crosses the minute-hand shortly after it passes the 15-minute mark (around 15 seconds into the minute). It then reaches 09:16:00 AM without any further crossing in that interval.
- This pattern continues for each of the 30 individual 1-minute intervals. In each interval, the second-hand crosses the minute-hand exactly once.

Since there are exactly 30 one-minute intervals between 09:15:00 AM and 09:45:00 AM, and the hands cross exactly once in each interval, the total number of crossings is:
30 intervals × 1 crossing/interval = 30 crossings.