Clocks

Module 13 — Lesson 0013 · In Depth · 45 min

A clock is a circular race — minute hand at 6°/min, hour hand at 0.5°/min, relative speed 5.5°/min. Every angle problem is |30H − 5.5M| with a reflex check; every faulty-clock problem is a ratio of real vs indicated time.

1. The Geometry of a Clock

Clock geometry — 30° per hour, 6° per minute 12 1 2 3 4 5 6 7 8 9 10 11 6°/min 0.5°/min Hour gap 30°, minute gap 6°
12 hour marks at 30°, 60 minute marks at 6° — every angle is a multiple of these.

2. Speed of the Hands

Hand Speed Derivation
Minute hand 6°/min 360° /60 min
Hour hand 0.5°/min 360° /720 min (12h)
Relative (minute gains) 5.5°/min =11/2 6 −0.5 (same direction)

Relative speed is the lap speed — minute hand laps the hour hand every 720/11 ≈65.45 min (not 60 — because hour hand moves).

3. Finding the Angle Between Hands

Standard formula: at H hours and M minutes:

θ = |30H − 5.5M| = |30H − (11/2)M|

H is hour digit (12 → 0), M is minutes. θ is internal (smaller) angle. If θ >180°, smaller angle = 360 − θ (reflex is the larger).

Worked: At 3:15 → θ=|30×3 −5.5×15|=|90−82.5|=7.5°. At 3:00 → |90−0|=90°. At 6:00 → |180−0|=180° (straight line). At 12:00 → |0−0|=0° (overlap).

4. Finding the Time for a Specific Angle

Given angle (0° overlap, 180° straight, 90° right), find M between H and H+1:

  1. Start at M=0: initial angle = 30H
  2. Desired angle = θ_des (e.g., 0, 90, 180)
  3. Distance to cover = |θ_des − 30H| (handling wrap) or more generally 5.5M = |30H − θ_des|
  4. M = Distance / 5.5

Example: Between 3 and 4, when are hands at 90°? At 3:00 angle 90° already (so M=0 is one solution). Second solution: need angle 90° on the other side → distance = 180−90=90° beyond? Actually between 3–4, after 3:00 the minute gains: solve |90−5.5M|=90 → M=0 or M=180/5.5≈32.73 min → times 3:00 and ~3:32:43.

5. Standard Frequency Observations

Position Angle Times in 12h Times in 24h
Coincide (overlap) 11 22
Straight line (opposite) 180° 11 22
Right angle 90° 22 44

Missing instance: No overlap between 11:00 and 1:00 — they coincide at exactly 12:00, counted once, so 11 not 12.

6. Gain and Loss of Time (Faulty Clocks)

Method 1: Ratio Method

Correct time : faulty time → find real duration for a given indicated duration.

Method 2: True Time from Gain

If a clock gains x minutes per 24h of real time:

Faulty covers 24h + x in 24h real → True Time = Indicated Duration × 24 / (24 + Gain)

Loses x → denominator 24 − x (since faulty covers less).

Worked: Clock gains 5 min per 24h. After 2 days real, faulty shows 48h +10 min. If faulty shows 2 days have passed (48h indicated), true time = 48×24/(24+5/60)? Actually x in hours: 5 min =0.0833h → true =48×24/(24.0833)=47.83h ≈47h 50m. For quick MCQs, use minutes: true = indicated ×1440/(1440+gain_minutes).

7. Advanced / Custom Clocks

If hour hand speed is not 0.5°/min (e.g., 0.7°/min as in puzzle): redefine relative speed = |MinuteSpeed − HourSpeed|, then reuse overlap/angle logic: time to next overlap = 360 / RelativeSpeed.

8. Summary of Key Variables

Symbol Meaning
H Hour (1–12, 12→0)
M Minutes (0–60)
θ Angle in degrees
5.5 =11/2 Relative speed °/min
T Total real time elapsed
Error Indicated − True

Hour space in degrees?

Minute hand speed?

Relative speed minute vs hour?

Angle at 3:15?

If θ>180°, smaller angle is?

How many overlaps in 12h?

Clock gains 6 min per 24h → true time for 24h indicated?

Notes

Geometry :-

360° total ; hour 30° ; minute 6°

Speeds :-

Minute 6°/min ; Hour 0.5°/min ; Relative 5.5=11/2 ; lap ≈65.45 min

Angle :-

θ=|30H−5.5M| ; if >180 → 360−θ ; 3:15→7.5°, 3:00→90°, 6:00→180°

Time for angle: M=|30H−θ|/5.5

Frequency :-

Overlap/straight:11 in 12h (22 in 24h) ; right:22 in 12h (44 in 24h) ; none 11→1

Gain/Loss :-

True = Indicated×24/(24±gain) ; gains + , loses −

Custom: new relative = |M_speed − H_speed|

Primary source: none pinned yet — drop your preferred video/book resource into RESOURCES.md and it will be linked here. Ask me anything that's unclear.

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