Clocks and Calendars

Overview

Clocks and Calendars appears about 1 times per paper across the last four AFCAT solved papers, placing it in the high yield band of Reasoning and Aptitude.

Clocks and Calendars is a small but reliable scoring slot in AFCAT reasoning. Over the last four solved papers the topic delivers roughly one question per attempt, and every one of those questions is solvable from a tiny formula sheet. There is no figure work, no language ambiguity and no syllogism trap — the only risks are arithmetic slips and the century-leap-year exception. Treat this topic as an almost-guaranteed +3, not a thinking problem. The clock half rests on two facts: the minute hand sweeps 6° per minute and the hour hand 0.5° per minute, so the minute hand gains 5.5° on the hour hand every minute. The calendar half rests on one fact: every block of days contributes a fixed number of leftover days after removing whole weeks, and those leftover days (the odd days) tell you the day of the week. Master those two ideas and the topic is essentially closed.

Why this topic is pure formula work

AFCAT does not test clock reading or calendar memory. It tests whether you can convert a time or a date into an angle or an odd-day count and then plug into a one-line formula. Every stem in the four recent solved papers reduces to one of five templates:

• Find the angle between the hands at a given H:M.
• Find the time between H and H+1 when the hands coincide, are at 90°, or are opposite.
• A clock gains or loses a fixed number of seconds per day; when does it show the correct time again.
• A particular date is a given weekday; what weekday is the same or a different date some years later.
• What was the day on a far-back date such as 15 August 1947 or 26 January 1950.

Because the templates are fixed, the only way to fail is to misuse a sign or forget the century exception. The sections below give you each template plus the trap attached to it.

Clock basics — angular speed of each hand

A clock face is a 360° circle. The minute hand completes one full circle in 60 minutes; the hour hand completes one full circle in 12 hours (720 minutes). Convert each to degrees per minute and memorise the three numbers below.

The 5.5°/min relative speed is the single most useful number in clock problems. Every coincidence question, every right-angle question and every opposite-hands question is solved by dividing some target gap by 5.5.

Angle between the hands at H:M

At H hours and M minutes past the hour, the hour hand has swept (30H + 0.5M) degrees from the 12-o'clock mark, and the minute hand has swept 6M degrees. The signed difference between them is (30H − 5.5M). Take the magnitude, and if the answer exceeds 180° subtract from 360° to give the smaller angle.

Formula: Angle between hands = |30H − 5.5M|, then take min(angle, 360 − angle).

Hands coincide — every 65 and 5/11 minutes

The two hands point in the same direction at 12:00. After that, the minute hand has to gain a full 360° on the hour hand to catch up again. At a relative speed of 5.5°/min, the time taken is 360 ÷ 5.5 = 720/11 minutes, which is 65 minutes and 5/11 of a minute (about 65 minutes 27.27 seconds).

In 12 hours the hands therefore coincide 11 times (not 12), because the 12-hour interval contains only eleven gaps of 720/11 minutes. In 24 hours they coincide 22 times.

Coincidence time between H and H+1: at exactly H o'clock the hour hand is 30H degrees ahead of the 12 mark, and the minute hand is at zero, so the minute hand is 30H° behind. To erase this gap at 5.5°/min:

Minutes past H = 30H ÷ 5.5 = (60H) ÷ 11.

Hands at a right angle — 22 times in 12 hours

The hands form a 90° angle whenever the gap between them is exactly 90° on either side. In each hour this happens twice except in the bands around 3 o'clock and 9 o'clock, where one of the two events slides into the neighbouring hour and the count comes out to 22 right angles in 12 hours (44 in 24 hours).

Formula: set 30H − 5.5M = ±90 and solve for M.

Worked check between 2 and 3 o'clock: M = (60 − 90)/5.5 is negative, so the first right angle is actually before 2:00 (it sits between 1 and 2). The second right angle is at (60 + 90)/5.5 = 150/5.5 = 27 3/11 minutes past 2.

Hands in a straight line opposite — 11 times in 12 hours

The hands point in exactly opposite directions when the gap between them is 180°. The simplest equation is 30H − 5.5M = ±180.

Like coincidences, opposite events occur 11 times every 12 hours (22 times a day). Including the coincidences, the hands form a straight line — same direction or opposite — 22 times in 12 hours.

Quick example: between 4 and 5 o'clock the hands are opposite when 30(4) − 5.5M = −180, giving 5.5M = 300, so M = 300/5.5 = 54 6/11 minutes past 4.

Faulty clocks — gaining or losing time

A clock that gains or loses time can be modelled as one that runs at a fixed ratio to a true clock. If a clock gains G seconds per day, then in one true day it shows (86400 + G) seconds; the ratio of faulty to true is (86400 + G) / 86400.

Two question patterns dominate:

• When will the faulty clock show the correct time again? A clock that gains or loses time shows the right time again only when the total drift is a multiple of 12 hours (43,200 seconds) on a 12-hour dial, or 24 hours on a 24-hour digital display. Compute days needed = 43,200 / (drift per day in seconds).
• What is the true time when the faulty clock shows X? Use the ratio. If the faulty clock shows that T_f seconds have passed since the last sync, true seconds elapsed = T_f × 86400 / (86400 + G) for a gaining clock.

Quick template: a clock gains 3 minutes (180 s) in 24 hours. To regain correct time on a 12-hour dial it must drift exactly 12 hours = 43,200 s. Days needed = 43,200 / 180 = 240 days.

Calendar basics — leap year rule and odd days

The Gregorian calendar uses a three-line leap rule:

1. Years divisible by 4 are leap years (e.g., 2024, 2028).
2. Century years (1700, 1800, 1900, 2100) are leap years only if divisible by 400.
3. So 2000 is a leap year; 1900, 2100, 2200, 2300 are not.

The odd-day idea is the engine for every day-finding problem. Removing complete weeks from any block of days leaves a remainder of 0 to 6 days; that remainder is the number of odd days.

Odd days across whole centuries

For day-finding on dates from earlier centuries you need the odd-day total for 100, 200, 300 and 400 years. Apply the leap rule: in any 100-year block of the Gregorian calendar there are 76 ordinary years and 24 leap years (because one of the century years inside the block is not leap unless it is divisible by 400).

Odd days in 100 years = 76 × 1 + 24 × 2 = 124 days = 17 weeks + 5 odd days.

The cycle is locked: 5, 3, 1, 0 and then repeats. So 2000 years carry 0 odd days, 2100 years carry 5, 2200 carry 3, and so on.

Day-of-week shift from one year to the next

If a date is given as day D in year Y and you must find the same date in year Y+1, count the odd days in the intervening 365 or 366 days.

• If neither year is leap, the shift is +1 day of the week.
• If the intervening 12 months include 29 February (i.e., Y is a leap year and the date is on or before 29 February, or Y+1 is a leap year and the date is after 28 February), the shift is +2 days.

Multi-year shift: sum the odd days. From a date in 2018 to the same date in 2025 you cross 2020 and 2024 as leap years; 5 ordinary + 2 leap = 5 + 4 = 9 odd days = 2 days of the week (9 mod 7).

Day-of-week for any given date

The cleanest method without memorising Zeller's congruence is the odd-day chain. Pick a reference: 1 January of year 1 of the Gregorian calendar was a Monday. To find the weekday of a date d/m/y, count odd days in: (i) complete centuries before y, (ii) extra years of the present century, (iii) months of the present year up to but not including m, (iv) days in month m up to d.

Add the four totals, take modulo 7, and read the weekday from this table:

Month odd days within a normal year (used in step iii):

In a leap year add 1 to every cumulative value from March onward.

Common AFCAT traps

• Century-leap-year exception. 1900 was not a leap year, but a hasty answer using only the divisible-by-4 rule treats it as leap. Test divisibility by 400 for every century year in your odd-day chain.
• Direction error on faulty clocks. A gaining clock shows a future time; a losing clock shows a past time. Write G or L on the rough sheet before solving.
• Wrong hour for hands events. If your computed M is negative or ≥ 60, the event has slipped into the neighbouring hour. This is the source of most wrong answers on 'between 3 and 4 o'clock' or 'between 9 and 10 o'clock' items, where the 90° event naturally sits at the hour boundary.
• Same direction vs straight line. 'Hands form a straight line' is satisfied by both coincidence and opposite positions; that is 22 events in 12 hours. 'Hands in opposite direction' is 11 events only. Read the stem carefully.
• Day-shift sign. Shifts due to elapsed years are always forward (+). Do not subtract odd days when moving to a later year.
• 5/11 fractions. Coincidence times always carry a fraction with denominator 11. If your answer comes out with a clean 5 or 10 in the minute place, recheck.

Time budget on test day

Clocks and calendars should not eat more than 60 seconds. Run this sequence:

1. Identify the template — angle, coincidence, right-angle, opposite, faulty clock, day shift, day finder.
2. Write the controlling formula on the rough sheet (|30H − 5.5M|, 30H/5.5, 43,200/drift, odd-day sum).
3. Plug in, simplify, pick the option whose form (degree or fractional minute) matches your computed answer.

Skip only if the stem requires a far-back date and your odd-day table is shaky. In that case mark it for review and move on; one missed mark is cheaper than four minutes lost.

Worked AFCAT-style examples

Exam-day strategy

1. Memorise the six small numbers: 6°/min, 0.5°/min, 5.5°/min, 720/11 min, 11 coincidences per 12 hours, 22 right angles per 12 hours.
2. Solve every hand-event question by writing 30H − 5.5M = 0, ±90 or ±180 and dividing — no diagrams.
3. For faulty-clock items, fix the direction (gain vs lose) on the rough sheet first; compute drift in seconds and divide into 43,200 s (12-hour dial).
4. For calendar items, build a four-part odd-day sum: centuries, extra years, completed months, completed days. Add and take mod 7.
5. Test every leap year against both the ÷4 rule and the century ÷400 exception before assigning it 2 odd days.
6. Cap each item at 60 seconds; if the odd-day chain feels long, mark for review and move on.

Frequently asked questions

Related topics