Q.What would be the time four and a half hour before 2.15 pm?

Clock and Calendar — From Intuition to Precision

Imagine you have a circular track. You start running, and after some time you come back to the same point. That's a cycle. A clock is just a cycle of 12 hours. A calendar is a cycle of days, months, and years. The entire topic is about counting how many times something repeats, and finding where you end up after a given number of steps.

The Intuition: Remainders Are Everything

Suppose today is Monday. What day will it be after 10 days? You don't count from Monday to Tuesday to Wednesday all the way — you notice that 7 days later it's Monday again. So 10 days = 7 days (back to Monday) + 3 more days. Monday + 3 = Thursday. The answer is Thursday.

That's the core idea: every cycle resets after a fixed period. For a clock, the cycle is 12 hours. For days of the week, the cycle is 7 days. For months, the cycle is 12 months. The only thing you ever need to compute is the remainder when the total time is divided by the cycle length.

The Precise Statement

Clock problems deal with the positions of hour and minute hands on a 12-hour dial. The key facts:

• The minute hand moves $360^\circ$ in 60 minutes → $6^\circ$ per minute.
• The hour hand moves $360^\circ$ in 12 hours = 720 minutes → $0.5^\circ$ per minute.
• At any time $t$ minutes past $H$ o'clock (where $H$ is from 1 to 12), the angle between the hands is:

(Take the smaller angle, i.e., $\min(\theta, 360-\theta)$.)

Calendar problems deal with days, weeks, months, and years. The key facts:

• A normal year has 365 days = 52 weeks + 1 odd day.
• A leap year has 366 days = 52 weeks + 2 odd days.
• A century year (like 1900) is a leap year only if divisible by 400.
• The day of the week advances by the number of odd days.

How to Think, Not Just What to Memorize

Don't memorize the formula $30H - 5.5t$ blindly. Derive it: at $H$ o'clock exactly, the hour hand is at $30H$ degrees (since $360/12 = 30$ degrees per hour). In $t$ minutes, the hour hand moves $0.5t$ degrees further, so its position is $30H + 0.5t$. The minute hand in $t$ minutes is at $6t$ degrees. The difference is $|30H + 0.5t - 6t| = |30H - 5.5t|$.

Similarly, for calendars: instead of memorizing "1900 is not a leap year", understand that the Earth's orbit is about 365.2422 days. The Gregorian rule (divisible by 400 for century years) corrects this drift. Every 400 years, there are exactly 97 leap years, giving $400 \times 365 + 97 = 146097$ days, which is exactly divisible by 7 — so the calendar repeats every 400 years.

The Only Two Types of Problems

Type 1: Find the angle or time on a clock.

Given a time, compute the angle. Or given an angle (like "hands coincide"), solve for $t$.

Type 2: Find the day of the week after N days/years.

Count the total number of days, find odd days modulo 7, and shift the starting day.

A Worked Example (Clock)

Q: At what time between 4 and 5 o'clock will the hands of a clock be together?

Solution: Let the time be $4: t$ minutes. For the hands to coincide, the angle must be $0^\circ$:

So the hands meet at $4:21\frac{9}{11}$.

A Worked Example (Calendar)

Q: What day of the week was 15 August 1947?

Solution:

• 1947 is not a leap year (not divisible by 4).
• Count odd days from a reference. A common reference: 1 Jan 0001 was Monday (by convention). But easier: 1600 was a Sunday (since the Gregorian cycle repeats every 400 years).
• From 1600 to 1946: 346 years. In 400 years, odd days = 0. So 1600 to 2000 has 0 odd days. But we only go to 1946.
• 1600 to 1900: 300 years. 300 years = 75 leap years (every 4th) but century years 1700, 1800, 1900 are not leap → 75 - 3 = 72 leap years. So odd days = $300 \times 1 + 72 = 372$ days → $372 \mod 7 = 1$ odd day.
• 1901 to 1946: 46 years. Leap years in this range: 1904, 1908, ..., 1944 → 11 leap years. So odd days = $46 + 11 = 57$ → $57 \mod 7 = 1$ odd day.
• Total odd days from 1600 to end of 1946 = $1 + 1 = 2$ odd days.
• 1600 Jan 1 was Sunday. So Jan 1, 1947 = Sunday + 2 = Tuesday.
• Now count days to 15 August 1947: Jan (31), Feb (28, not leap), Mar (31), Apr (30), May (31), Jun (30), Jul (31), Aug 15. Total = $31+28+31+30+31+30+31+15 = 227$ days. Odd days = $227 \mod 7 = 3$ (since $7 \times 32 = 224$, remainder 3).
• Jan 1 was Tuesday. Tuesday + 3 = Friday.

The Big Picture

Clock and Calendar is not about memorizing a dozen formulas. It's about one idea: cycles and remainders. Every problem reduces to:

1. Identify the cycle length (12, 60, 7, 365, 366, 400).
2. Compute the remainder when the total steps are divided by the cycle.
3. Use the remainder to find the final position.

Master that, and you can solve any clock or calendar problem on the spot.