Clocks & Calendars
Clocks and calendars look like two topics but use one idea: count how far something has moved, then throw away the full cycles. For a clock the cycle is 360° and the minute hand gains 5.5° a minute on the hour hand. For a calendar the cycle is 7 days, and what is left over is the number of odd days.
Updated 11 October 2026 · Usually one or two questions, most often the angle between the hands or the day of the week for a date.
How to think about clocks & calendars
Each hand has a fixed speed
The minute hand turns 360° in 60 minutes, so 6° a minute. The hour hand turns 30° in an hour, so 0.5° a minute — it does not sit on the hour mark, it creeps towards the next one. The minute hand therefore gains 5.5° on the hour hand every minute, and every clock question is a relative-speed question at that rate.
The hands meet every 65 5/11 minutes, not every hour
To catch the hour hand again, the minute hand must gain a full 360°, which takes 360 ÷ 5.5 = 720/11 = 65 5/11 minutes. So in 12 hours the hands are together 11 times, not 12, and opposite 11 times; they are at right angles twice in each cycle, 22 times. Double these for a day. The same number exposes a faulty clock: if its hands meet more often than every 65 5/11 minutes of true time, it is running fast.
Odd days are what is left after whole weeks
Days of the week repeat every 7 days, so only the remainder after dividing a number of days by 7 matters. That remainder is the number of odd days. An ordinary year has 365 days = 52 weeks + 1 odd day, and a leap year has 2. The same date a year later falls one weekday later, or two weekdays later if a 29 February lies in between.
Leap years and centuries
A year is a leap year if it is divisible by 4, except that a century year must be divisible by 400. So 2000 and 2024 are leap years; 1900 and 2100 are not. A century such as 1801–1900 has 24 leap years and 76 ordinary years: 48 + 76 = 124 odd days, which leaves 5. So 100 years give 5 odd days, 200 give 3, 300 give 1 and 400 give 0.
Formulas and shortcuts
- Speed of the hands
Minute hand 6° per minute; hour hand 0.5° per minuteThe minute hand gains 5.5° per minute on the hour hand.
- Angle at H:M
θ = |30H − 5.5M|Use 12-hour time. If θ is more than 180°, the smaller angle is 360° − θ.
- Hands θ apart between H and H+1 o'clock
M = (2/11)(30H ± θ) minutes past Hθ = 0 gives 60H/11 (together); θ = 180 for opposite; θ = 90 for a right angle. If a bracket is negative or 330 or more, add or subtract 360 to bring it into the range 0 to 330; if it cannot be brought into that range, that sign gives no time in this hour.
- Correct clock
Hands meet every 720/11 = 65 5/11 minutesIf a clock's hands meet every T minutes of true time, it gains when T is less than 65 5/11 and loses when T is more.
- Odd days in years
Ordinary year 1, leap year 2; 100 years 5, 200 years 3, 300 years 1, 400 years 0Leap year: divisible by 4; a century year only if divisible by 400 (2000 yes, 1900 no).
- Odd days in months
Jan 3, Feb 0 (1 in a leap year), Mar 3, Apr 2, May 3, Jun 2, Jul 3, Aug 3, Sep 2, Oct 3, Nov 2, Dec 331-day months give 3, 30-day months give 2.
- Odd days to weekday
0 Sunday, 1 Monday, 2 Tuesday, 3 Wednesday, 4 Thursday, 5 Friday, 6 SaturdayUse it on the total odd days from 1 January of year 1 up to and including the date. From a date whose day you already know, move forward by the remainder of (days elapsed ÷ 7) instead.
Solved examples
Q1. Find the angle between the hands of a clock at 3:40.
- Use θ = |30H − 5.5M| with H = 3 and M = 40.
- 30 × 3 = 90 and 5.5 × 40 = 220.
- θ = |90 − 220| = 130°. This is under 180°, so it is already the smaller angle.
Answer: 130°
From 12 o'clock, the minute hand is at 40 × 6 = 240° and the hour hand at 3 × 30 + 40 × 0.5 = 110°. The difference is 130°.
Q2. At what time between 4 and 5 o'clock will the hands of a clock be together?
- At 4:00 the minute hand is at 12 and the hour hand at 4, so the minute hand is 4 × 30 = 120° behind.
- It gains 5.5° every minute, so it closes the gap in 120 ÷ 5.5 minutes.
- 120 ÷ 5.5 = 240/11 = 21 9/11 minutes.
Answer: 21 9/11 minutes past 4
Formula route: M = 60H/11 = 240/11, the same answer. At that moment both hands are 1440/11° (about 130.9°) from 12.
Q3. What day of the week was 15 August 1947, India's first Independence Day?
- Count the odd days before the date in blocks: 1600 years, then 300 years, then 46 years, then the months of 1947, then the days of August.
- 1600 years = 4 × 400 years → 0 odd days. The next 300 years (1601–1900) → 1 odd day.
- 1901–1946 is 46 years with 11 leap years (1904 to 1944) and 35 ordinary years: 11 × 2 + 35 = 57 odd days = 8 weeks + 1 → 1 odd day.
- January to July 1947 (not a leap year): 3 + 0 + 3 + 2 + 3 + 2 + 3 = 16 odd days = 2 weeks + 2 → 2 odd days.
- 15 days of August = 2 weeks + 1 → 1 odd day.
- Total = 0 + 1 + 1 + 2 + 1 = 5. Odd day 5 is Friday.
Answer: Friday
February 1947 adds 0 because 1947 is not divisible by 4. Counting it as 1 gives Saturday — the most common slip in this method.
Practice questions with answers
1. What is the smaller angle between the hands of a clock at 2:50?
Answer: C. 145° — θ = |30 × 2 − 5.5 × 50| = |60 − 275| = 215°. That is more than 180°, so the smaller angle is 360° − 215° = 145°.
2. At what time between 7 and 8 o'clock are the hands of a clock in a straight line, pointing in opposite directions?
Answer: A. 5 5/11 minutes past 7 — M = (2/11)(30H ± θ) with H = 7 and θ = 180: (2/11)(210 − 180) = 60/11 = 5 5/11 minutes. The plus sign gives 390, and 390 − 360 = 30 is the same bracket, so there is only this one time. (38 2/11 minutes past 7 is when the hands are together.)
3. How many times are the hands of a clock at right angles in a day (24 hours)?
Answer: B. 44 — In 12 hours the minute hand gains 5.5 × 720 = 3960° = 11 full turns on the hour hand, and each turn passes through a right angle twice. That is 22 times in 12 hours, so 44 in a day.
4. The hands of a faulty clock coincide every 64 minutes of true time. How much does the clock gain or lose in a day?
Answer: D. Gains 32 8/11 minutes — In a correct clock the hands meet every 720/11 = 65 5/11 minutes. Here they meet sooner, so the clock is fast. In 64 true minutes it shows 720/11 minutes, a gain of 720/11 − 64 = 16/11 minutes. A day holds 1440 ÷ 64 = 22.5 such intervals, so the gain is 22.5 × 16/11 = 360/11 = 32 8/11 minutes.
5. Which of these years is a leap year?
Answer: C. 2000 — A century year is a leap year only if it is divisible by 400. Of these, only 2000 qualifies; 1800, 1900 and 2100 are divisible by 4 but not by 400.
6. Republic Day, 26 January 2024, fell on a Friday. On which day did 26 January 2025 fall?
Answer: B. Sunday — The year between the two dates includes 29 February 2024, so it has 366 days = 52 weeks + 2 odd days. Friday + 2 = Sunday. Forgetting the leap day gives Saturday.
7. On what day of the week did 26 January 1950, the first Republic Day, fall?
Answer: D. Thursday — 1900 years → 1 odd day (1600 years give 0, the next 300 give 1). 1901–1949 is 49 years with 12 leap years: 12 × 2 + 37 = 61 → 5 odd days. 26 days of January → 5. Total 1 + 5 + 5 = 11 → 4, which is Thursday.
8. The calendar for 2021 will be repeated in which year?
Answer: A. 2027 — A calendar repeats when a year starts on the same day and is also not a leap year. Add odd days from 2021: 2021 (1), 2022 (1), 2023 (1), 2024 (2), 2025 (1), 2026 (1) = 7, which is 0. So 2027 starts on the same day as 2021 and, like 2021, is not a leap year.
Where marks get lost
- Leaving the hour hand on the hour mark. At 3:40 it has moved 20° past the 3.
- Giving the reflex angle. If the formula gives more than 180°, subtract it from 360°.
- Treating every century year as a leap year. 1900 and 2100 are not; 2000 is.
- Missing a 29 February between two dates, which puts the answer one day short.
Reading is not practising.
Run a timed set on clocks & calendars and see which step you actually lose time on. Free, no card needed.
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