Important Formulas: Clock and Calendar

The number of questions related to the clock and calendar topic in competitive exams can vary greatly depending on the specific exam and its structure. Generally, these topics form a small portion of the reasoning or quantitative section and may contribute approximately 1-5 questions in a typical question paper, though the exact number changes with each exam format.

Clock

Minute spaces

• Minute spaces: The clock face is a circle divided into 60 equal parts; each part is one minute space.
• Degrees per minute space: Each minute space corresponds to 6° because the full circle is 360° and 360 ÷ 60 = 6.

Hour hand and minute hand

• Hour hand: The shorter hand; it completes one full rotation (360°) in 12 hours.
• Minute hand: The longer hand; it completes one full rotation (360°) in 60 minutes (1 hour).

Calendar

Ordinary year and leap year

• Ordinary year: A year with 365 days (52 weeks + 1 day).
• Leap year: A year with 366 days (52 weeks + 2 days).
• Leap year rule (Gregorian calendar): A year is a leap year if it is divisible by 4, except that years divisible by 100 are not leap years unless they are also divisible by 400. (Thus 2000 is a leap year, while 1900 is not.)

Odd days

Odd days are the days left over after grouping days into complete weeks. When calculating the day of the week after a given number of days, only the number of odd days (days modulo 7) matters.

Important observations and formulas for clocks

• Circle and hours: A clock is a complete circle of 360° divided into 12 equal hour parts.

• \( \dfrac{360}{12} = 30^\circ \)

• Minute hand speed: The minute hand covers 360° in 60 minutes.

• \( \dfrac{360^\circ}{60\ \text{min}} = 6^\circ\ \text{per minute} \)

  
• Hour hand movement: The hour hand moves 30° in 60 minutes, i.e. it moves 0.5° per minute.

• \( \dfrac{30^\circ}{60\ \text{min}} = 0.5^\circ\ \text{per minute} \)

• Relative speed of the minute hand w.r.t. the hour hand: The minute hand moves at 6°/min and the hour hand at 0.5°/min; the relative speed is 5.5°/min (or 11/2 °/min).

• \(6^\circ - 0.5^\circ = 5.5^\circ = \dfrac{11}{2}^\circ\ \text{per minute}\)

  
• Coincidence of hands: The hour and minute hands coincide 11 times in 12 hours and therefore 22 times in 24 hours.
• Opposite (180°) positions: The hands are opposite to each other 11 times in 12 hours and 22 times in 24 hours.
• Straight line instances (coincident or opposite): In 12 hours the hands are in a straight line (either coincident or opposite) 22 times; in 24 hours this happens 44 times.
• Right angles (90°): The hands form a right angle 22 times in 12 hours and 44 times in 24 hours.
• Minute and hour hand separation in minute spaces: Right angle corresponds to 15 minute spaces; opposite direction corresponds to 30 minute spaces.
• Fast / slow clock wording: If a clock shows 9:15 when the true time is 9:00, it is 15 minutes fast; if it shows 8:45 when the true time is 9:00, it is 15 minutes slow.
• Time between successive coincidences: The hands meet every \(\dfrac{720}{11}\) minutes (approximately 65.4545 minutes). This explains why there are 11 meetings in 12 hours.  
  \( \text{Time between successive coincidences} = \dfrac{360^\circ}{\text{relative speed}} = \dfrac{360}{5.5} = \dfrac{720}{11}\ \text{minutes}\)

Clocks - useful algebraic formulas

Use these forms depending on which hand is ahead or behind. Put each equation on its own line.

General single formula for the angle between the hands at H hours and M minutes

\( \theta = \left|30H - \dfrac{11}{2}M\right| \)

If the minute hand is ahead of the hour hand

\( \theta = \left|\dfrac{11}{2}M - 30H\right| \)

If the minute hand lags behind the hour hand

\( \theta = \left|30H - \dfrac{11}{2}M\right| \)

Special cases derived from the formula:

• When hands coincide: \( \theta = 0 \Rightarrow 30H = \dfrac{11}{2}M \Rightarrow M = \dfrac{60H}{11}.\)
• When hands are at right angle (90°): \( \left|30H - \dfrac{11}{2}M\right| = 90 \Rightarrow M = \dfrac{60H \pm 180}{11}.\)
• When hands are opposite (180°): \( \left|30H - \dfrac{11}{2}M\right| = 180 \Rightarrow M = \dfrac{60H \pm 360}{11}.\)

Important observations and formulas for calendars

• Odd days per year: An ordinary year contributes 1 odd day; a leap year contributes 2 odd days.
• Odd days in 100 years: In 100 years there are 76 ordinary years and 24 leap years (in the 1st 100-year block under Gregorian rules), giving \(76\times1 + 24\times2 = 124\) odd days, which is \(124 \equiv 5 \pmod{7}\). So 100 years = 5 odd days.
• Odd days in multiples of 100: 200 years = 10 odd days ≡ 3 (mod 7); 300 years = 15 odd days ≡ 1 (mod 7); 400 years = 20 odd days ≡ 0 (mod 7). Therefore, the calendar repeats every 400 years under the Gregorian rule.
• Counting odd days for a given interval: Find total odd days from years, months, and days separately, reduce modulo 7, then map the result to days of the week.

Sample clock and calendar questions with solutions

Q1: If your birthday falls on the third Monday of September 2023, on what date will your birthday fall in 2027?

Sol: Identify the date in 2023: if the third Monday of September 2023 is the birthday, find the date of the third Monday.
September 2023 has Mondays on 4, 11, 18, 25. So the third Monday is 18 September 2023.
Find how weekdays shift from 2023 to 2027. The years 2024, 2025, 2026, 2027 are four years ahead; among these there is one leap year (2024), so total odd days = 3 ordinary years ×1 + 1 leap year ×2 = 5 odd days.
Advance the weekday of 18 September 2023 by 5 days to get the weekday for 18 September 2027.
Monday + 5 days = Saturday. Hence 18 September 2027 is a Saturday.
The question asks for the date of the third Monday in September 2027. Since 1 September 2027 is a Wednesday, the Mondays in September 2027 fall on 6, 13, 20 and 27; thus the third Monday is 20 September 2027.

Q2: Sarah set her clock to show 3 o'clock in the afternoon. How many degrees will the hour hand of the clock rotate when the clock shows 8 o'clock at night?

Sol: Compute the time difference: from 3:00 p.m. to 8:00 p.m. is 5 hours.
The hour hand moves 360° in 12 hours, so per hour it moves 30°.
\(30^\circ \times 5 = 150^\circ\)
Therefore, the hour hand rotates 150°.

Q3: Determine the year in which the calendar will repeat exactly as it was in the year 2015.

Sol: 2015 is a non-leap year.
For a non-leap year, the calendar often repeats after 11 years unless an intervening leap year pattern alters the interval. Add 11 years to 2015.
2015 + 11 = 2026.
Therefore, the calendar for 2026 will be the same as that for 2015.

Q4: Thomas Miller, a curious individual, approaches his mathematics teacher with a question regarding the day on which the 1st of the month will occur, given that the 9th of the month falls on the day just before Sunday.

Sol: If the 9th is on the day just before Sunday, then the 9th is a Saturday.
The 1st is eight days before the 9th, and eight days ≡ 1 day (mod 7), so the weekday of the 1st is one day earlier than the weekday of the 9th.
Saturday minus one day = Friday.
Hence the 1st of the month will fall on a Friday.

Q5: A wager was made between two friends, Harry and Ronald, to determine the probability of randomly selecting a leap year that has 53 Fridays.

Sol: A leap year has 366 days = 52 complete weeks + 2 extra days, so there are already 52 Fridays for sure.
The year will have a 53rd Friday exactly when one of the two extra days is a Friday.
List the possible ordered pairs of extra consecutive weekdays (these correspond to the weekday of 1 January in a leap year):

• (Sunday, Monday)
• (Monday, Tuesday)
• (Tuesday, Wednesday)
• (Wednesday, Thursday)
• (Thursday, Friday)
• (Friday, Saturday)
• (Saturday, Sunday)

Among these 7 equally likely possibilities, Friday appears in the extra-days pair in 2 cases: (Thursday, Friday) and (Friday, Saturday).
Therefore the probability of 53 Fridays in a randomly selected leap year is 2/7.