Question

How many times from 4 pm to 10 pm, the hands of a clock are at right angles?

• A. 11
• B. 6
• C. 9
• D. 10

Hint:

Use the relative speed method: right-angle interval is \( \frac{90}{5.5} = 16\frac{4}{11} \) minutes between successive events in one direction; counting across a time span by dividing total minutes by this interval gives a fast and reliable total.

Answer 1

The Correct Option is A

Key facts: the minute hand moves 6° per minute, the hour hand moves 0.5° per minute, and their relative speed is \(6 - 0.5 = 5.5^\circ\) per minute.
Right angles occur whenever the absolute angular difference between the hands equals \(90^\circ\).
General equation for times after \(h\) o’clock (where \(h\) is an integer hour and \(m\) minutes after \(h\)) is \(|30h + 0.5m - 6m| = 90\).
This simplifies to \(|30h - 5.5m| = 90\).
Solving gives two families in each hour: \(5.5m = 30h + 90\) and \(5.5m = 30h - 90\), provided the resulting \(m\) lies in \([0,60)\).
Between any two successive hours, there are two valid solutions except where one falls exactly at a boundary duplicating the adjacent hour, hence the classic rule of thumb: right angles occur twice every hour.
Interval length from 4:00 to 10:00 is 6 hours — namely 4–5, 5–6, 6–7, 7–8, 8–9, 9–10.
Naively multiplying gives \(6 \times 2 = 12\) occurrences.
But check for boundary duplication at exactly 4:00 or 10:00 — do the hands make a \(90^\circ\) angle at either boundary time.
At 4:00, the hour hand is at \(120^\circ\) and the minute hand at \(0^\circ\), difference \(120^\circ\), not right angle — so no boundary count at 4:00.
At 10:00, the hour hand is at \(300^\circ\) and the minute hand at \(0^\circ\), difference \(300^\circ\) (or \(60^\circ\) the other way), not right angle — so no boundary count at 10:00.
Why then is the total not 12 — we must verify minute-range validity in each hour, because in one hour of any 6-hour stretch the two theoretical times may collapse into one valid time due to out-of-range minutes.
Let us compute explicitly for one pair of hours to see the pattern and confirm the count.
For 4–5: set \(h=4\).
Case A: \(5.5m = 30\cdot4 + 90 = 210 \Rightarrow m = 38\frac{2}{11}\) (valid).
Case B: \(5.5m = 30\cdot4 - 90 = 30 \Rightarrow m = 5\frac{5}{11}\) (valid).
For 5–6: \(h=5\).
Case A: \(5.5m = 240 \Rightarrow m = 43\frac{7}{11}\) (valid).
Case B: \(5.5m = 60 \Rightarrow m = 10\frac{10}{11}\) (valid).
Continuing similarly for 6–7, 7–8, 8–9, 9–10 gives two valid times each — twelve in total — but we must note that across any 12-hour span there are 22 right angles, which averages to \(22/12 \approx 1.833\) per hour, not exactly 2 everywhere, leading to one fewer in any 6-hour block than 12.
A well-known property is: in any continuous span of 6 hours, the hands are at right angles 11 times because the relative cycle completes every \(65\frac{5}{11}\) minutes, and \(6\) hours equals \(360\) minutes which fits \(360 / 32.727...\approx 11\) right-angle events exactly.
Hence, from 4 pm to 10 pm the hands are at right angles 11 times.
Therefore, the correct answer is 11.