# Clock and Calendar Questions with Answers and Formulas

> Clock and calendar questions with answers: angle between the hands, when the hands meet, faulty clocks, odd days and the day of the week for any date.

Source: https://www.sansal.in/aptitude/clocks-and-calendars · Last updated: 2026-10-11 · How often it appears: Medium. Usually one or two questions, most often the angle between the hands or the day of the week for a date.

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.

## Concepts

### 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

| Formula | Rule | Note |
| --- | --- | --- |
| Speed of the hands | Minute hand 6° per minute; hour hand 0.5° per minute | The 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 H | θ = 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 minutes | If 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 0 | Leap 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 3 | 31-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 Saturday | Use 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. |

## Worked examples

### Example 1

**Question:** Find the angle between the hands of a clock at 3:40.

**Solution:**

1. Use θ = |30H − 5.5M| with H = 3 and M = 40.
2. 30 × 3 = 90 and 5.5 × 40 = 220.
3. θ = |90 − 220| = 130°. This is under 180°, so it is already the smaller angle.

**Answer:** 130°

**Check:** 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°.

### Example 2

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

**Solution:**

1. At 4:00 the minute hand is at 12 and the hour hand at 4, so the minute hand is 4 × 30 = 120° behind.
2. It gains 5.5° every minute, so it closes the gap in 120 ÷ 5.5 minutes.
3. 120 ÷ 5.5 = 240/11 = 21 9/11 minutes.

**Answer:** 21 9/11 minutes past 4

**Check:** Formula route: M = 60H/11 = 240/11, the same answer. At that moment both hands are 1440/11° (about 130.9°) from 12.

### Example 3

**Question:** What day of the week was 15 August 1947, India's first Independence Day?

**Solution:**

1. 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.
2. 1600 years = 4 × 400 years → 0 odd days. The next 300 years (1601–1900) → 1 odd day.
3. 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.
4. January to July 1947 (not a leap year): 3 + 0 + 3 + 2 + 3 + 2 + 3 = 16 odd days = 2 weeks + 2 → 2 odd days.
5. 15 days of August = 2 weeks + 1 → 1 odd day.
6. Total = 0 + 1 + 1 + 2 + 1 = 5. Odd day 5 is Friday.

**Answer:** Friday

**Check:** 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

1. What is the smaller angle between the hands of a clock at 2:50?
   - A. 215°
   - B. 155°
   - C. 145°
   - D. 140°
   - **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?
   - A. 5 5/11 minutes past 7
   - B. 38 2/11 minutes past 7
   - C. 5 minutes past 7
   - D. 54 6/11 minutes past 7
   - **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)?
   - A. 22
   - B. 44
   - C. 48
   - D. 24
   - **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?
   - A. Loses 32 8/11 minutes
   - B. Gains 34 10/11 minutes
   - C. Loses 34 10/11 minutes
   - D. Gains 32 8/11 minutes
   - **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?
   - A. 1900
   - B. 2100
   - C. 2000
   - D. 1800
   - **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?
   - A. Saturday
   - B. Sunday
   - C. Monday
   - D. Friday
   - **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?
   - A. Monday
   - B. Wednesday
   - C. Sunday
   - D. Thursday
   - **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?
   - A. 2027
   - B. 2026
   - C. 2032
   - D. 2028
   - **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.

## Related topics

- [Time, Speed & Distance](https://www.sansal.in/aptitude/time-speed-and-distance)
- [Number System](https://www.sansal.in/aptitude/number-system)
- [Puzzles](https://www.sansal.in/aptitude/puzzles)

Practise these on Sansal: https://www.sansal.in/practice/aptitude