How to Solve Clock Problems
Solve clock aptitude problems using angular speed and the relative-speed formula, with a worked example and practice questions with answers.
Expected Interview Answer
Clock problems are solved by treating the minute hand and hour hand as two objects moving at fixed angular speeds — 6° per minute and 0.5° per minute respectively — and finding the angle between them or the times they coincide.
The minute hand covers 360° in 60 minutes (6°/min); the hour hand covers 360° in 720 minutes (0.5°/min). Their relative speed is 6 − 0.5 = 5.5°/min, so the angle between them at m minutes past any hour H is |30H − 5.5m| degrees (taking the smaller of the result and 360° minus it). The hands coincide, are exactly opposite (180°), or perpendicular (90°) at times found by solving for m in that equation. In 12 hours, the hands coincide 11 times and are opposite 11 times, not 12, because the faster hand only "catches up" that many times.
- One relative-speed idea (5.5°/min) covers coincide/opposite/right-angle problems
- The formula |30H − 5.5m| generalizes to any target angle
- Explains the classic "11 times in 12 hours" result cleanly
AI Mentor Explanation
Picture two runners circling the boundary: the minute hand is a sprinter completing a lap every 60 minutes, the hour hand a jogger completing a lap every 720 minutes. The sprinter gains on the jogger at 5.5° per minute (their speed difference), so the sprinter laps the jogger only 11 times in 12 hours, not 12, because it needs that extra bit of lead each time. Clock problems solve exactly this: relative angular speed, applied to catch-up or angle questions.
Worked example (angle at a given time)
Hour hand
- 30×3 + 0.5×40 = 110°
Minute hand
- 6×40 = 240°
Angle
- |240−110| = 130°
Step-by-Step Explanation
Step 1
Compute hour hand angle
Hour angle = 30H + 0.5m degrees (H = hour, m = minutes past).
Step 2
Compute minute hand angle
Minute angle = 6m degrees.
Step 3
Take the absolute difference
Angle = |Hour angle − Minute angle|; if > 180°, use 360° minus it.
Step 4
Solve for m when angle is fixed
Set |30H − 5.5m| = target angle and solve for m to find coincide/opposite/right-angle times.
What Interviewer Expects
- Correct angular speeds: 6°/min (minute), 0.5°/min (hour)
- Relative speed of 5.5°/min and its role in every clock formula
- The 11-times-in-12-hours result for coincide and opposite cases
- Ability to set up and solve |30H − 5.5m| = target angle
Common Mistakes
- Assuming the hour hand stays fixed at the hour mark instead of moving continuously
- Using 12 instead of 11 for coincide/opposite counts in 12 hours
- Forgetting to take 360° minus the result when the raw angle exceeds 180°
- Mixing up which hand’s angle to subtract from which
Best Answer (HR Friendly)
“I treat the minute and hour hands as two hands moving at fixed speeds — 6 degrees a minute and half a degree a minute. Their relative speed is 5.5 degrees a minute, and that single number solves coincide, opposite, and right-angle problems by setting up the angle formula and solving for the time. The classic gotcha is that hands align only 11 times in 12 hours, not 12.”
Follow-up Questions
- Why do the clock hands coincide only 11 times in 12 hours, not 12?
- At what time between 3 and 4 o’clock do the hands form a right angle?
- How would you find when the hands are exactly opposite each other?
- How does a faulty clock that gains or loses time per hour affect these calculations?
MCQ Practice
1. The angle between the hands at 3:40 is?
Hour angle = 30×3+0.5×40 = 110°; minute angle = 6×40 = 240°; difference = 130°.
2. The relative speed of the minute hand with respect to the hour hand is?
Minute hand 6°/min minus hour hand 0.5°/min = 5.5°/min.
3. In a 12-hour period, the hands of a clock coincide how many times?
The minute hand gains a full 360° relative lap 11 times in 12 hours, so they coincide 11 times.
Flash Cards
Minute hand speed? — 6° per minute (360° / 60 min).
Hour hand speed? — 0.5° per minute (360° / 720 min).
Relative speed? — 6 − 0.5 = 5.5° per minute.
Coincide/opposite count in 12 hours? — 11 times each, not 12.