Advanced Analytical Skills โ€” II

Unit 4: Advanced Calendar & Clocks

Master the art of finding any day in history, calculating clock angles in seconds, and cracking aptitude exams โ€” from SSC/Bank PO to CAT and placement tests.

โฑ๏ธ Time to Complete: 6โ€“8 hours  |  ๐Ÿ“ 30 MCQs (Bloom's Mapped)  |  15 Worked Examples

๐ŸŽฏ Exams this unlocks: SSC CGL  |  Bank PO/Clerk  |  CAT/MAT  |  Campus Placements (TCS, Infosys, Wipro)

Section 1

Opening Hook โ€” Why Calendar & Clocks Matter

๐Ÿ—“๏ธ Can You Tell What Day 15 August 1947 Was?

India gained independence on 15 August 1947. But was it a Monday? A Friday? The answer โ€” it was a Friday. And you don't need Google to figure this out. With the techniques in this chapter, you'll calculate the day for any date in history in under 30 seconds โ€” mentally.

This isn't a party trick. Calendar and clock reasoning is one of the most frequently tested topics in competitive exams across India. SSC CGL, Bank PO, IBPS Clerk, CAT, MAT, and nearly every IT company placement paper (TCS NQT, Infosys, Wipro, Cognizant) includes 2โ€“4 questions on this topic.

The best part? Unlike other aptitude topics that need heavy math, calendar & clock problems follow fixed formulas and shortcuts. Once you learn them, they become free marks โ€” you can solve them faster than reading the question!

๐Ÿ“ SSC CGL๐Ÿฆ Bank PO๐Ÿ“Š CAT/MAT๐Ÿ’ป TCS NQT๐Ÿ’ป Infosys๐Ÿ’ป Wipro
The Gregorian calendar, which we use today, was introduced by Pope Gregory XIII in 1582. Before that, the Julian calendar was used, which had a slight error โ€” it was 11 minutes and 14 seconds too long per year. By 1582, this had accumulated to a 10-day error! Gregory simply "deleted" 10 days โ€” October 4, 1582 was followed by October 15, 1582.
Section 2

Learning Outcomes โ€” Bloom's Taxonomy Mapped

Bloom's LevelLearning Outcome
๐Ÿ”ต RememberState the odd-day values for each month, leap year rules, and century codes
๐Ÿ”ต UnderstandExplain the concept of odd days and why 400-year cycles repeat to the same day
๐ŸŸข ApplyCalculate the exact day of the week for any given date using the odd-day method
๐ŸŸข AnalyzeDetermine clock angles, overlap times, and gain/loss of faulty clocks using formulas
๐ŸŸ  EvaluateSolve advanced multi-step calendar and clock problems from competitive exam papers
๐ŸŸ  CreateDesign and solve original calendar/clock problems; verify answers using multiple methods
Section 3

Calendar โ€” Basics & the Odd Days Concept

1. What Are Odd Days?

The concept of odd days is the backbone of all calendar problems. An "odd day" is the remainder when the total number of days is divided by 7. Since a week has 7 days, any complete set of 7 days brings us back to the same day. The leftover days are called odd days.

๐Ÿ“… Understanding Odd Days

Definition

Odd days = Total number of days mod 7 (remainder when divided by 7)

Example

If today is Monday and we go forward 10 days: 10 รท 7 = 1 remainder 3. So 3 odd days from Monday โ†’ Thursday.

Day-Number Mapping

0 = Sunday, 1 = Monday, 2 = Tuesday, 3 = Wednesday, 4 = Thursday, 5 = Friday, 6 = Saturday

2. Leap Year Rules

A year is a leap year if:

  • It is divisible by 4 โ€” e.g., 2024, 2028 โœ…
  • BUT if it's a century year (ends in 00), it must be divisible by 400
  • 1600, 2000 โ†’ Leap โœ… | 1700, 1800, 1900 โ†’ NOT Leap โŒ | 2100 โ†’ NOT Leap โŒ
Students often say "1900 is a leap year because 1900 รท 4 = 475." Wrong! Century years must pass the รท 400 test. Only 1600, 2000, 2400 etc. are leap centuries. This is the #1 trap in exam questions.

3. Odd Days in Standard Periods

PeriodTotal DaysOdd DaysHow?
Ordinary Year3651365 รท 7 = 52 weeks + 1 day
Leap Year3662366 รท 7 = 52 weeks + 2 days
100 yearsโ€”576 ordinary + 24 leap = 76 + 48 = 124 odd days โ†’ 124 mod 7 = 5
200 yearsโ€”35 + 5 = 10 โ†’ 10 mod 7 = 3
300 yearsโ€”15 + 5 + 5 = 15 โ†’ 15 mod 7 = 1
400 yearsโ€”05ร—4 + 1 (extra leap in 400th year) = 21 โ†’ 21 mod 7 = 0
Memorise: 100yr โ†’ 5, 200yr โ†’ 3, 300yr โ†’ 1, 400yr โ†’ 0. These are fixed values. Every 400 years, the calendar repeats exactly. That's why 1 Jan 2000 and 1 Jan 1600 were both Saturdays!

4. Odd Days for Each Month

MonthDaysOdd DaysMonthDaysOdd Days
January313July313
February (Ordinary)280August313
February (Leap)291September302
March313October313
April302November302
May313December313
June302
Memory trick โ€” "3-0-3-2-3-2-3 | 3-3-2-3-2-3" โ€” Read as: Jan(3), Feb(0), Mar(3), Apr(2), May(3), Jun(2), Jul(3) | Aug(3), Sep(2), Oct(3), Nov(2), Dec(3). For leap years, Feb becomes 1 instead of 0.
Section 4

Finding the Day โ€” Step-by-Step Algorithm

The Universal Day-Finding Algorithm

๐Ÿงฎ Steps to Find the Day of Any Date

Step 1: Count odd days from the century part (use the table: 100โ†’5, 200โ†’3, 300โ†’1, 400โ†’0)

Step 2: Count odd days from the remaining years (each ordinary year = 1, each leap year = 2)

Step 3: Count odd days from the completed months of the year

Step 4: Add the given date

Step 5: Total all odd days โ†’ divide by 7 โ†’ remainder gives the day

Day mapping: 0=Sun, 1=Mon, 2=Tue, 3=Wed, 4=Thu, 5=Fri, 6=Sat

โœ๏ธ Worked Example 1: What day was 15 August 1947?

๐Ÿ“Œ India's Independence DayIntermediate

Step 1 โ€” Century part: 1947 โ†’ 1900 years = 19 centuries. Breakdown: 400ร—4 = 1600 years (0 odd days) + 300 years (1 odd day) = 1 odd day.

Step 2 โ€” Remaining years: 47 years (from 1900 to 1947). Leap years in 47 years: 47 รท 4 = 11 leap years, 36 ordinary years. Odd days = 36ร—1 + 11ร—2 = 36 + 22 = 58. 58 mod 7 = 2 odd days.

Step 3 โ€” Months (Jan to Jul of 1947): Jan(3) + Feb(0) + Mar(3) + Apr(2) + May(3) + Jun(2) + Jul(3) = 16. 16 mod 7 = 2 odd days.

Step 4 โ€” Date: 15 โ†’ 15 mod 7 = 1 odd day.

Step 5 โ€” Total: 1 + 2 + 2 + 1 = 6. 6 mod 7 = 6 โ†’ Saturday.

Wait โ€” historical records say it was a Friday! Let's recheck. Actually, the count from 1 AD to 1900 gives us exactly 1 odd day, and 47 years (1900โ€“1946, i.e., years 00โ€“46 give 11 leap years: '04,'08,'12,'16,'20,'24,'28,'32,'36,'40,'44 = 11). So 36 + 22 = 58 โ†’ 58 mod 7 = 2. Janโ€“Jul = 3+0+3+2+3+2+3 = 16 โ†’ 2. Date = 15 โ†’ 1. Total = 1+2+2+1 = 6. But wait โ€” we need to check: 1900 is not a leap year!

Corrected calculation: 1600 years โ†’ 0 odd days. Next 300 years (1601โ€“1900) โ†’ 1 odd day. Years 1901โ€“1947 = 47 years. Leap years: '04,'08,'12,'16,'20,'24,'28,'32,'36,'40,'44 = 11 leaps. Ordinary = 36. Odd days = 36 + 22 = 58 โ†’ mod 7 = 2. Months Janโ€“Jul 1947 = 16 โ†’ mod 7 = 2. Date = 15 โ†’ mod 7 = 1. Total = 0 + 1 + 2 + 2 + 1 = 6 โ†’ Saturday?

Historical fact: 15 August 1947 was indeed a Friday. The discrepancy arises from the convention used. Using the Doomsday method or counting from 1 Jan 1 AD (which was a Monday, odd day = 1), we get: Total odd days = 5 โ†’ Friday โœ….

Using the shortcut method directly: From 1 Jan 0001 to 15 Aug 1947: (1600 yrs โ†’ 0) + (300 yrs โ†’ 1) + (47 yrs โ†’ 11 leaps ร— 2 + 36 ร— 1 = 58 โ†’ 2) + (Janโ€“Jul โ†’ 3+0+3+2+3+2+3 = 16 โ†’ 2) + (15 days โ†’ 1) = 0+1+2+2+1 = 6. Starting from Monday (Day 1 of 1 AD), 6 days later โ†’ Sunday if 0=Sun. But 1 Jan 1 AD = Monday. So our reference is: 0 odd days = Sunday for the "total odd days from day zero" method, giving 6 = Saturday. Under the corrected exam convention (0=Sun, 1=Monโ€ฆ6=Sat), 6 = Saturday. But many exam solutions count 1 Jan 0001 as Day 1 (Monday).

โœ… Standard Exam Method (most textbooks):

Odd days from 1 to 1946 years: 1600โ†’0, 300โ†’1, 46 yrs (11 leap + 35 ordinary = 22+35 = 57 โ†’ 57 mod 7 = 1). Total century + years = 0+1+1 = 2. Janโ€“Jul = 3+0+3+2+3+2+3 = 16 โ†’ 2. Date = 15 โ†’ 1. Grand total = 2+2+1 = 5 โ†’ 5 = Friday โœ…

Key: Count completed years (1 to 1946), NOT up to 1947.

The #1 error: Counting the given year itself in "remaining years." If the date is in 1947, count years 1901โ€“1946 (= 46 years), NOT 1901โ€“1947. You count only completed years before the given date.

โœ๏ธ Worked Example 2: What day was 26 January 1950? (Republic Day)

๐Ÿ“Œ Competitive Exam FavouriteIntermediate

Step 1: 1600 years โ†’ 0 odd days. 300 years (1601โ€“1900) โ†’ 1 odd day.

Step 2: 1901โ€“1949 = 49 years. Leap years: '04,'08,'12,'16,'20,'24,'28,'32,'36,'40,'44,'48 = 12. Ordinary = 37. Odd days = 37 + 24 = 61 โ†’ 61 mod 7 = 5.

Step 3: No completed months (date is in January). Odd days = 0.

Step 4: Date = 26. 26 mod 7 = 5.

Step 5: Total = 0 + 1 + 5 + 0 + 5 = 11. 11 mod 7 = 4 โ†’ Thursday โœ…

India's first Republic Day was on a Thursday.

โœ๏ธ Worked Example 3: What day will be 1 January 2030?

๐Ÿ“Œ Future Date CalculationBeginner

Step 1: 2000 years โ†’ 400ร—5 = 0 odd days (every 400 years = 0).

Step 2: 2001โ€“2029 = 29 years. Leap years: '04,'08,'12,'16,'20,'24,'28 = 7. Ordinary = 22. Odd days = 22 + 14 = 36. 36 mod 7 = 1.

Step 3: No completed months. = 0.

Step 4: Date = 1 โ†’ 1.

Total: 0 + 1 + 0 + 1 = 2 โ†’ 2 = Tuesday โœ…

Section 5

Century Code & Month Code Shortcuts

The Quick-Code Method (Zeller-Style Shortcut)

Instead of computing odd days from scratch, use pre-computed codes for centuries and months to find the day in seconds.

๐Ÿท๏ธ Century Codes (for Gregorian Calendar)

Century (first 2 digits of year)CodeExamples
17xx41776, 1799
18xx21857, 1899
19xx01947, 1999
20xx62000, 2024, 2099
21xx42100, 2199
22xx22200, 2299

Pattern: The codes cycle as 4 โ†’ 2 โ†’ 0 โ†’ 6 every 4 centuries. Easy to remember!

๐Ÿ“† Month Codes

MonthCodeMonthCode
January0July6
February3August2
March3September5
April6October0
May1November3
June4December5

Memory Mnemonic: "0-3-3-6-1-4-6-2-5-0-3-5" โ€” Memorise as: "Zero-Three-Three-Six-One-Four | Six-Two-Five-Zero-Three-Five"

The Quick Formula

Day-Finding Quick Formula

Day = (Century Code + Last 2 digits of year + Last 2 digits รท 4 + Month Code + Date) mod 7

โš ๏ธ For Jan/Feb of a leap year, subtract 1 from the total before taking mod 7.

Result: 0=Sun, 1=Mon, 2=Tue, 3=Wed, 4=Thu, 5=Fri, 6=Sat

โœ๏ธ Worked Example 4: Quick Code โ€” 15 August 1947

๐Ÿ“Œ Verify with Quick FormulaBeginner

Century = 19 โ†’ Code = 0

Last 2 digits = 47. 47 รท 4 = 11 (integer part).

Month code for August = 2

Date = 15

Total = 0 + 47 + 11 + 2 + 15 = 75

75 mod 7 = 5 โ†’ Friday โœ…

โœ๏ธ Worked Example 5: Quick Code โ€” 26 January 1950

๐Ÿ“Œ Republic Day VerificationBeginner

Century = 19 โ†’ Code = 0

Last 2 digits = 50. 50 รท 4 = 12.

Month code for January = 0

Date = 26

Total = 0 + 50 + 12 + 0 + 26 = 88

88 mod 7 = 4 โ†’ Thursday โœ…

Speed technique for exams: With practice, you can find the day of ANY date in 10โ€“15 seconds using the Quick Code method. Many competitive exam toppers swear by this shortcut โ€” it saves 2โ€“3 minutes per paper.
Section 6

Advanced Calendar Problems

1. "What Day After N Days?" Type

โœ๏ธ Worked Example 6: If today is Wednesday, what day will it be after 1000 days?

๐Ÿ“Œ Exam ClassicBeginner

1000 รท 7 = 142 remainder 6.

So 6 odd days after Wednesday: Wed โ†’ Thu(1) โ†’ Fri(2) โ†’ Sat(3) โ†’ Sun(4) โ†’ Mon(5) โ†’ Tue(6)

Answer: Tuesday โœ…

โœ๏ธ Worked Example 7: If 1 Jan 2023 was Sunday, what day was 31 Dec 2023?

๐Ÿ“Œ Year-End CalculationBeginner

2023 is an ordinary year โ†’ 365 days โ†’ 1 Jan to 31 Dec = 364 days later (since 1 Jan is day 1, 31 Dec is day 365, so gap = 364).

364 รท 7 = 52 remainder 0. So 31 Dec is the same day as 1 Jan โ†’ Sunday โœ….

Alternatively: An ordinary year has 1 odd day, but that means 1 Jan of the next year shifts by 1 day. From 1 Jan to 31 Dec of the same year = 364 days = 0 odd days = same day.

2. "Same Calendar" Problems

๐Ÿ“… When Does a Calendar Repeat?

A calendar repeats when the total odd days from one year to another = 0 (i.e., divisible by 7) AND both years have the same leap/ordinary status.

Quick Rules

โ€ข After an ordinary year, the calendar shifts by 1 day

โ€ข After a leap year, the calendar shifts by 2 days

โ€ข A calendar can repeat in: 5, 6, 7, 11, 12, 28 years (depending on leap year positions)

โœ๏ธ Worked Example 8: The calendar of 2023 will be the same as which future year?

๐Ÿ“Œ Same Calendar ProblemIntermediate

2023 is an ordinary year starting on Sunday. We need to find the next ordinary year where cumulative odd days from 2023 = 0 mod 7.

YearLeap?Odd DaysCumulativeCum mod 7
2023No111
2024Yes233
2025No144
2026No155
2027No166
2028Yes281
2029No192
2030No1103
2031No1114
2032Yes2136
2033No1140 โœ…

Cumulative = 0 at 2034 (the year after accumulating 14 odd days). But we need 2034 to also be an ordinary year โ€” it is! Answer: 2034 โœ…

That's 11 years later. The calendar of 2023 repeats in 2034.

3. Counting Days Between Two Dates

โœ๏ธ Worked Example 9: How many days between 15 Mar 2024 and 20 Sep 2024?

๐Ÿ“Œ Date DifferenceBeginner

Remaining days in March: 31 โˆ’ 15 = 16

April: 30, May: 31, June: 30, July: 31, August: 31, September (up to 20): 20

Total = 16 + 30 + 31 + 30 + 31 + 31 + 20 = 189 days โœ…

Calendar reasoning appears in the General Intelligence section of SSC exams โ€” typically 2โ€“3 questions worth 4โ€“6 marks. In Bank PO Prelims, 1โ€“2 questions appear in the Reasoning section. Knowing shortcuts can save precious minutes in these time-pressured exams.
Section 7

Clocks โ€” Fundamentals

1. Basic Clock Mechanics

๐Ÿ• How Clock Hands Move

Hour Hand

โ€ข Completes one full rotation (360ยฐ) in 12 hours

โ€ข Speed = 360ยฐ รท 12 = 30ยฐ per hour = 0.5ยฐ per minute

Minute Hand

โ€ข Completes one full rotation (360ยฐ) in 60 minutes (1 hour)

โ€ข Speed = 360ยฐ รท 60 = 6ยฐ per minute

Relative Speed

โ€ข Minute hand gains on hour hand at 6ยฐ โˆ’ 0.5ยฐ = 5.5ยฐ per minute

โ€ข In 1 hour, the minute hand gains 330ยฐ over the hour hand (5.5ยฐ ร— 60)

2. The Angle Formula

Angle Between Clock Hands

At H hours and M minutes, the angle between the hour and minute hands:

ฮธ = |30H โˆ’ 5.5M|

If ฮธ > 180ยฐ, the actual (reflex-free) angle = 360ยฐ โˆ’ ฮธ

Where does this formula come from?

  • Hour hand position at H:M = 30H + 0.5M degrees (from 12 o'clock)
  • Minute hand position at H:M = 6M degrees (from 12 o'clock)
  • Angle between them = |30H + 0.5M โˆ’ 6M| = |30H โˆ’ 5.5M|

โœ๏ธ Worked Example 10: Find the angle between the hands at 3:20.

๐Ÿ“Œ Basic Angle ProblemBeginner

H = 3, M = 20

ฮธ = |30 ร— 3 โˆ’ 5.5 ร— 20| = |90 โˆ’ 110| = |โˆ’20| = 20ยฐ โœ…

โœ๏ธ Worked Example 11: Find the angle at 7:45.

๐Ÿ“Œ Reflex Angle CheckBeginner

H = 7, M = 45

ฮธ = |30 ร— 7 โˆ’ 5.5 ร— 45| = |210 โˆ’ 247.5| = 37.5ยฐ

Since 37.5ยฐ < 180ยฐ, the answer is 37.5ยฐ โœ…

Speed hack: For quick mental math, remember 30H gives you the hour hand position (at exact hour). Then subtract 5.5 ร— M. The absolute value is your answer. If it exceeds 180ยฐ, subtract from 360ยฐ.
Section 8

Angle-Based Clock Problems

1. "At What Time is the Angle Xยฐ?"

These problems are the reverse of finding the angle โ€” you're given the angle and asked to find the time.

Finding Time for a Given Angle

Set |30H โˆ’ 5.5M| = ฮธ and solve for M.

This gives two solutions: M = (30H โˆ’ ฮธ) / 5.5 and M = (30H + ฮธ) / 5.5

Accept only solutions where 0 โ‰ค M < 60

โœ๏ธ Worked Example 12: At what time between 4 and 5 will the hands make a 90ยฐ angle?

๐Ÿ“Œ Classic Right-Angle ProblemIntermediate

H = 4, ฮธ = 90ยฐ

Case 1: 30(4) โˆ’ 5.5M = 90 โ†’ 120 โˆ’ 5.5M = 90 โ†’ 5.5M = 30 โ†’ M = 30/5.5 = 5 5โ„11 minutes

Case 2: 30(4) โˆ’ 5.5M = โˆ’90 โ†’ 120 โˆ’ 5.5M = โˆ’90 โ†’ 5.5M = 210 โ†’ M = 210/5.5 = 38 2โ„11 minutes

Answer: The hands are at 90ยฐ at 4:05 5โ„11 and 4:38 2โ„11 โœ…

2. Mirror Image of Clock

If you see a clock in a mirror, the time appears reversed. To find the actual time from the mirror image:

Mirror Image Time

Actual Time = 11:60 โˆ’ Mirror Time (for hours 1โ€“11)

Actual Time = 23:60 โˆ’ Mirror Time (for hours past 12)

Shortcut: Subtract the mirror time from 12:00 (adjust if minutes > 0)

โœ๏ธ Worked Example 13: A clock shows 8:20 in a mirror. What is the actual time?

๐Ÿ“Œ Mirror ImageBeginner

Actual time = 11:60 โˆ’ 8:20 = 3:40 โœ…

(11 hrs 60 min โˆ’ 8 hrs 20 min = 3 hrs 40 min)

Section 9

Overlap, Right Angle & Straight Line Positions

1. Hand Overlap (0ยฐ โ€” Hands Coincide)

๐Ÿ”„ When Do the Hands Overlap?

The hands overlap when the angle between them = 0ยฐ.

Key Facts

โ€ข The hands overlap 11 times in 12 hours (NOT 12 times โ€” they don't overlap separately at 12 o'clock and between 11โ€“12 and 12โ€“1).

โ€ข The hands overlap 22 times in 24 hours.

โ€ข Time between consecutive overlaps = 12/11 hours = 1 hr 5 min 27 3โ„11 sec โ‰ˆ 65.45 minutes.

Overlap Formula Between H and (H+1)

Set 30H โˆ’ 5.5M = 0 โ†’ M = 60H/11

BetweenOverlap TimeM = 60H/11
12โ€“112:00:00H=0, M=0
1โ€“21:05 5โ„11H=1, M=60/11=5 5โ„11
2โ€“32:10 10โ„11H=2, M=120/11=10 10โ„11
3โ€“43:16 4โ„11H=3, M=180/11=16 4โ„11
4โ€“54:21 9โ„11H=4, M=240/11=21 9โ„11
5โ€“65:27 3โ„11H=5, M=300/11=27 3โ„11
6โ€“76:32 8โ„11H=6, M=360/11=32 8โ„11
7โ€“87:38 2โ„11H=7, M=420/11=38 2โ„11
8โ€“98:43 7โ„11H=8, M=480/11=43 7โ„11
9โ€“109:49 1โ„11H=9, M=540/11=49 1โ„11
10โ€“1110:54 6โ„11H=10, M=600/11=54 6โ„11
11โ€“1212:00:00(Next cycle)

2. Right Angle (90ยฐ) โ€” Hands Perpendicular

๐Ÿ“ When Are the Hands at Right Angles?

โ€ข The hands are at 90ยฐ 44 times in 24 hours (22 times in 12 hours).

โ€ข Between H and (H+1): solve |30H โˆ’ 5.5M| = 90 โ†’ two solutions usually.

โ€ข Between 2โ€“3 and 3โ€“4, one right angle is "shared," so only 1 right angle occurs between 2โ€“4 and 8โ€“10 (each).

3. Straight Line (180ยฐ) โ€” Hands Opposite

โ†”๏ธ When Are the Hands in a Straight Line?

โ€ข Hands are in a straight line when angle = 180ยฐ (opposite) or 0ยฐ (overlap).

โ€ข Hands are at 180ยฐ 11 times in 12 hours, 22 times in 24 hours.

โ€ข Between H and (H+1): solve |30H โˆ’ 5.5M| = 180 โ†’ M = (30H โˆ’ 180)/5.5 or M = (30H + 180)/5.5

โ€ข Hands in a straight line (0ยฐ or 180ยฐ) = 22 times in 12 hours, 44 times in 24 hours.

โœ๏ธ Worked Example 14: At what time between 5 and 6 do the hands overlap?

๐Ÿ“Œ Overlap ProblemBeginner

M = 60H/11 = 60 ร— 5 / 11 = 300/11 = 27 3โ„11 minutes

Answer: The hands overlap at 5:27 3โ„11 โœ…

At 12:00, both hands overlap AND form a straight line simultaneously. This is the only time when 0ยฐ and 180ยฐ coincide in meaning โ€” both hands point in the same direction. Between 5โ€“6 and 6โ€“7, the "opposite" position doesn't occur because the hour hand moves too fast relative to the 180ยฐ mark.
Section 10

Faulty Clocks โ€” Gain & Loss of Time

1. The Concept

A faulty clock runs faster (gains time) or slower (loses time) than a correct clock. These problems ask you to find the correct time given that a clock gains or loses a certain amount per hour/day.

โฐ Gain/Loss Framework

If a clock GAINS x minutes per hour

In 1 real hour, the faulty clock shows 60 + x minutes.

When faulty clock shows T minutes, real time = T ร— 60 / (60 + x) minutes.

If a clock LOSES x minutes per hour

In 1 real hour, the faulty clock shows 60 โˆ’ x minutes.

When faulty clock shows T minutes, real time = T ร— 60 / (60 โˆ’ x) minutes.

When do two faulty clocks show the correct time together?

A clock that gains/loses will show the correct time again when it has gained/lost exactly 12 hours (720 minutes).

Key Faulty Clock Formulas

Time gained/lost per day: If a clock gains x min/hour โ†’ gains 24x min/day

Correct time after T hours: Real time elapsed = T ร— 60/(60 ยฑ x) hours

Clock shows right time again: After gaining/losing 720 min โ†’ 720/gain_per_hour hours

โœ๏ธ Worked Example 15: A clock gains 5 minutes every hour. If set correctly at 12:00 noon, what will it show when the actual time is 4:00 PM?

๐Ÿ“Œ Gain of TimeIntermediate

Real time elapsed: 12:00 noon to 4:00 PM = 4 hours = 240 minutes.

The clock gains 5 min per hour, so in 1 real hour it shows 65 minutes.

In 4 real hours, the clock shows: 4 ร— 65 = 260 minutes = 4 hours 20 minutes.

Starting from 12:00, the faulty clock will show 4:20 PM โœ…

It's 20 minutes ahead of the correct time.

2. Two Clocks Together

๐Ÿ•๐Ÿ• Two Faulty Clocks Problem

If Clock A gains a min/hr and Clock B loses b min/hr, they show the same time again when:

Combined drift = (a + b) min/hr. They'll show the same time after drifting 12 hours = 720 minutes apart.

Time = 720 / (a + b) hours.

Students often confuse "what the faulty clock shows" with "what the real time is." Read the question carefully: "What is the correct time when the faulty clock shows X?" is the reverse of "What does the faulty clock show when the correct time is X?" โ€” they need different calculations!
Mental Math Challenge: A clock loses 3 minutes every hour. If it was set correctly at 8:00 AM on Monday, what time will it show at 8:00 AM on Tuesday (real time)? Answer: In 24 hours, it loses 24 ร— 3 = 72 minutes = 1 hour 12 minutes. So it shows 8:00 โˆ’ 1:12 = 6:48 AM.
Section 11

MCQ Assessment Bank โ€” 30 Questions (Bloom's Mapped)

Remember / Identify (Q1โ€“Q6)

Q1

How many odd days are there in an ordinary (non-leap) year?

  1. 0
  2. 1
  3. 2
  4. 3
Remember
โœ… Answer: (B) 1 โ€” An ordinary year has 365 days. 365 รท 7 = 52 weeks + 1 day. So 1 odd day.
Q2

How many odd days are there in a leap year?

  1. 0
  2. 1
  3. 2
  4. 3
Remember
โœ… Answer: (C) 2 โ€” A leap year has 366 days. 366 รท 7 = 52 weeks + 2 days.
Q3

Which of the following is NOT a leap year?

  1. 1600
  2. 1700
  3. 2000
  4. 2400
Remember
โœ… Answer: (B) 1700 โ€” Century years must be divisible by 400 to be leap years. 1700 รท 400 โ‰  integer.
Q4

The minute hand of a clock moves at:

  1. 0.5ยฐ per minute
  2. 6ยฐ per minute
  3. 30ยฐ per minute
  4. 12ยฐ per minute
Remember
โœ… Answer: (B) 6ยฐ per minute โ€” The minute hand covers 360ยฐ in 60 minutes = 6ยฐ per minute.
Q5

How many odd days are there in 400 years?

  1. 1
  2. 5
  3. 0
  4. 3
Remember
โœ… Answer: (C) 0 โ€” 400 years have exactly 0 odd days. The calendar repeats every 400 years.
Q6

The hour hand of a clock moves at:

  1. 6ยฐ per minute
  2. 1ยฐ per minute
  3. 0.5ยฐ per minute
  4. 0.25ยฐ per minute
Remember
โœ… Answer: (C) 0.5ยฐ per minute โ€” The hour hand covers 360ยฐ in 12 hours = 30ยฐ per hour = 0.5ยฐ per minute.

Understand / Explain (Q7โ€“Q12)

Q7

Why does the calendar repeat every 400 years?

  1. Because 400 is divisible by 4
  2. Because 400 years contain exactly 0 odd days
  3. Because every century year is a leap year
  4. Because 400 รท 7 = 0
Understand
โœ… Answer: (B) โ€” 400 years have exactly 0 odd days (97 leap years and 303 ordinary years = 97ร—2 + 303ร—1 = 497 odd days; 497 mod 7 = 0). So the cycle returns to the same starting day.
Q8

The hands of a clock overlap 11 times in 12 hours instead of 12 times because:

  1. The clock skips one overlap at midnight
  2. One overlap between 11 and 12 coincides with the 12 o'clock overlap
  3. The minute hand moves too slowly
  4. The hour hand moves backward once
Understand
โœ… Answer: (B) โ€” The overlap that would occur between 11โ€“12 actually happens at exactly 12:00, which is the start of the next cycle. So only 11 distinct overlaps occur in 12 hours.
Q9

If a clock gains 5 minutes per hour, which statement is true?

  1. In 12 real hours, the clock gains 1 hour
  2. In 12 real hours, the clock shows it's been 13 hours
  3. The clock will show the correct time after 144 hours
  4. All of the above
Understand
โœ… Answer: (D) โ€” Gain per 12 hrs = 60 min = 1 hr. Clock shows 13 hrs in 12 real hrs. It shows correct time after gaining 12 hrs = 720 min. 720/5 = 144 hours.
Q10

In the formula ฮธ = |30H โˆ’ 5.5M|, the term 5.5M represents:

  1. The net gain of the minute hand over the hour hand
  2. The position of the hour hand
  3. The total degrees covered by both hands
  4. The speed of the second hand
Understand
โœ… Answer: (A) โ€” 5.5M = 6M โˆ’ 0.5M. Here, 6M is the minute hand's position and 0.5M is the additional movement of the hour hand due to minutes. The difference (5.5M) is the net displacement the minute hand gains.
Q11

The century code for years in the 19xx range (e.g., 1947) in the quick day-finding method is:

  1. 6
  2. 4
  3. 2
  4. 0
Understand
โœ… Answer: (D) 0 โ€” The century codes cycle 4โ†’2โ†’0โ†’6. For 17xxโ†’4, 18xxโ†’2, 19xxโ†’0, 20xxโ†’6.
Q12

If the mirror image of a clock shows 2:50, the actual time is:

  1. 9:10
  2. 10:10
  3. 9:50
  4. 10:50
Understand
โœ… Answer: (A) 9:10 โ€” Actual time = 11:60 โˆ’ 2:50 = 9:10.

Apply / Calculate (Q13โ€“Q18)

Q13

What day of the week was 26 January 1950?

  1. Monday
  2. Wednesday
  3. Thursday
  4. Friday
Apply
โœ… Answer: (C) Thursday โ€” Using the quick code: Century(19)=0, Year=50, 50รท4=12, Month(Jan)=0, Date=26. Total = 0+50+12+0+26 = 88. 88 mod 7 = 4 = Thursday.
Q14

If today is Monday, what day will it be after 63 days?

  1. Monday
  2. Tuesday
  3. Sunday
  4. Wednesday
Apply
โœ… Answer: (A) Monday โ€” 63 รท 7 = 9, remainder 0. So 0 odd days โ†’ same day = Monday.
Q15

What is the angle between the hands of a clock at 5:30?

  1. 15ยฐ
  2. 20ยฐ
  3. 12ยฐ
  4. 25ยฐ
Apply
โœ… Answer: (A) 15ยฐ โ€” ฮธ = |30ร—5 โˆ’ 5.5ร—30| = |150 โˆ’ 165| = 15ยฐ.
Q16

At what time between 3 and 4 o'clock will the hands of a clock overlap?

  1. 3:15
  2. 3:16 4โ„11
  3. 3:18
  4. 3:20
Apply
โœ… Answer: (B) 3:16 4โ„11 โ€” M = 60ร—3/11 = 180/11 = 16 4โ„11 minutes past 3.
Q17

How many times do the hands of a clock make a right angle in 24 hours?

  1. 22
  2. 24
  3. 44
  4. 48
Apply
โœ… Answer: (C) 44 โ€” The hands make right angles 22 times in 12 hours, so 44 times in 24 hours.
Q18

A clock gains 3 minutes every hour. If set right at 8 AM, what will it show at 8 PM (real time)?

  1. 8:36 PM
  2. 8:24 PM
  3. 9:00 PM
  4. 8:30 PM
Apply
โœ… Answer: (A) 8:36 PM โ€” 12 hours elapsed. Gain = 3 ร— 12 = 36 minutes. Clock shows 8:00 + 12:36 = 8:36 PM.

Analyze / Compare (Q19โ€“Q24)

Q19

If 1 January 2024 is Monday, on which day will 1 January 2025 fall?

  1. Monday
  2. Tuesday
  3. Wednesday
  4. Thursday
Analyze
โœ… Answer: (C) Wednesday โ€” 2024 is a leap year (2 odd days). Monday + 2 = Wednesday.
Q20

The angle between the hands at 12:30 is:

  1. 180ยฐ
  2. 175ยฐ
  3. 165ยฐ
  4. 150ยฐ
Analyze
โœ… Answer: (C) 165ยฐ โ€” ฮธ = |30ร—12 โˆ’ 5.5ร—30| = |360 โˆ’ 165| = 195ยฐ. Since 195 > 180, actual = 360 โˆ’ 195 = 165ยฐ.
Q21

A clock loses 2 minutes per hour. It was set right at noon on Monday. When it shows 12 noon on Wednesday, the correct time is:

  1. Wednesday 1:39 PM approx
  2. Wednesday 2:00 PM
  3. Wednesday 12:48 PM
  4. Wednesday 1:00 PM
Analyze
โœ… Answer: (A) โ€” The faulty clock shows 48 hours. In 1 real hour, it shows 58 min. So faulty 48 hrs = 48 ร— 60 = 2880 faulty minutes. Real time = 2880 ร— 60/58 = 2979.3 min โ‰ˆ 49 hrs 39 min. Noon Monday + 49h39m โ‰ˆ Wednesday 1:39 PM.
Q22

Between 6 and 7, at what time are the hands of a clock at 180ยฐ (opposite)?

  1. 6:00 only
  2. 6:00 and 6:32 8โ„11
  3. There is no 180ยฐ position between 6 and 7
  4. 6:49 1โ„11
Analyze
โœ… Answer: (C) โ€” At 6:00 the hands are already at 180ยฐ. To find another: |30ร—6 โˆ’ 5.5M| = 180 โ†’ M = 0 or M = 360/5.5 = 65.45 (>60, invalid). So only at 6:00, but that's the start, so strictly "between" 6 and 7, it occurs at 6:00 (boundary) only. Many texts say there's NO 180ยฐ between 6 and 7 other than exactly 6:00.
Q23

What was the day on 2 October 1869 (birth of Mahatma Gandhi)?

  1. Saturday
  2. Sunday
  3. Monday
  4. Friday
Analyze
โœ… Answer: (A) Saturday โ€” Century(18)=2, Year=69, 69รท4=17, Month(Oct)=0, Date=2. Total = 2+69+17+0+2 = 90. 90 mod 7 = 6 = Saturday.
Q24

If today is Friday, what day was it 100 days ago?

  1. Monday
  2. Wednesday
  3. Tuesday
  4. Thursday
Analyze
โœ… Answer: (B) Wednesday โ€” 100 รท 7 = 14 remainder 2. Going back 2 days from Friday: Fri โ†’ Thu โ†’ Wed.

Evaluate / Judge (Q25โ€“Q27)

Q25

A student claims: "The year 2100 is a leap year because 2100 รท 4 = 525." What is wrong with this reasoning?

  1. Nothing โ€” 2100 is indeed a leap year
  2. The รท 4 rule only applies to non-century years; century years must be divisible by 400
  3. 2100 is divisible by 400
  4. Leap years don't exist after 2000
Evaluate
โœ… Answer: (B) โ€” Century years (ending in 00) must be divisible by 400 to be leap years. 2100 รท 400 = 5.25 โ‰  integer, so 2100 is NOT a leap year.
Q26

Two clocks are set to the correct time at noon. Clock A gains 2 min/hr and Clock B loses 3 min/hr. When will they show the same time again?

  1. After 144 hours
  2. After 120 hours
  3. After 100 hours
  4. After 288 hours
Evaluate
โœ… Answer: (A) โ€” They drift apart at (2+3) = 5 min/hr. They show the same time when drift = 720 min. Time = 720/5 = 144 hours = 6 days.
Q27

A student says "the hands of a clock are at 90ยฐ exactly 24 times in 12 hours." Why is this incorrect?

  1. The correct count is 22 because at two intervals (2โ€“4 and 8โ€“10), only 3 right angles occur instead of 4
  2. The correct count is 11
  3. The correct count is 44
  4. It is actually correct
Evaluate
โœ… Answer: (A) โ€” Between 2โ€“3 and 3โ€“4, one right angle position is "shared" (similarly 8โ€“9 and 9โ€“10). So we get 22 right angles in 12 hours, not 24.

Create / Design (Q28โ€“Q30)

Q28

Which of the following dates will fall on the SAME day as 1 Jan 2024 (Monday)?

  1. 1 Jan 2029
  2. 1 Jan 2030
  3. 1 Jan 2035
  4. 1 Jan 2052
Create
โœ… Answer: (D) โ€” From 2024 to 2052: count odd days over 28 years (includes 7 leap years). Total odd days = 21 + 14 = 35 โ†’ 35 mod 7 = 0. So 1 Jan 2052 is also Monday. (The 28-year cycle applies within a century.)
Q29

You need to set a digital alarm that rings every time the clock hands overlap. How many times will it ring between 6 AM and 6 PM?

  1. 10
  2. 11
  3. 12
  4. 22
Create
โœ… Answer: (B) 11 โ€” In any 12-hour window, the hands overlap 11 times. 6 AM to 6 PM = exactly 12 hours.
Q30

If you could design a clock where the minute hand moves at 7ยฐ per minute (instead of 6ยฐ), how many times would the hands overlap in 12 hours?

  1. 11
  2. 12
  3. 13
  4. 14
Create
โœ… Answer: (C) 13 โ€” Relative speed = 7 โˆ’ 0.5 = 6.5ยฐ/min. Time per overlap = 360/6.5 โ‰ˆ 55.38 min. In 720 min: 720/55.38 โ‰ˆ 13 overlaps.
Section 12

Short Answer & Long Answer Questions

Short Answer Questions (8)

SA 1

Define "odd days" in the context of calendar problems. How many odd days does an ordinary year have?

Answer: Odd days are the remainder obtained when the total number of days is divided by 7. Since a week = 7 days, the "extra" days beyond complete weeks are called odd days. An ordinary year has 365 days. 365 รท 7 = 52 weeks + 1 odd day. So an ordinary year has 1 odd day, meaning if January 1 falls on a Monday, then January 1 of the next year will fall on Tuesday (one day ahead).
SA 2

State the leap year rule for century years with examples.

Answer: A century year (ending in 00) is a leap year only if it is divisible by 400. The normal "divisible by 4" rule does NOT apply to century years. Examples: 1600 โ†’ 1600 รท 400 = 4 โ†’ Leap โœ…. 2000 โ†’ 2000 รท 400 = 5 โ†’ Leap โœ…. 1700 โ†’ 1700 รท 400 = 4.25 โ†’ NOT Leap โŒ. 1800, 1900, 2100 โ†’ NOT Leap โŒ.
SA 3

Write down the number of odd days in 100, 200, 300, and 400 years.

Answer: 100 years โ†’ 5 odd days (76 ordinary years ร— 1 + 24 leap years ร— 2 = 124; 124 mod 7 = 5). 200 years โ†’ 3 odd days (10 mod 7 = 3). 300 years โ†’ 1 odd day (15 mod 7 = 1). 400 years โ†’ 0 odd days (21 mod 7 = 0, with the extra leap day for the 400th year). The calendar repeats exactly every 400 years.
SA 4

Derive the formula for the angle between clock hands: ฮธ = |30H โˆ’ 5.5M|.

Answer: Hour hand position at H hours M minutes = (H ร— 30) + (M ร— 0.5) = 30H + 0.5M degrees from 12 o'clock. Minute hand position = M ร— 6 = 6M degrees from 12 o'clock. Angle between them = |Hour position โˆ’ Minute position| = |30H + 0.5M โˆ’ 6M| = |30H โˆ’ 5.5M|. If this value exceeds 180ยฐ, subtract from 360ยฐ to get the smaller (non-reflex) angle.
SA 5

How many times do the hands of a clock overlap in 24 hours? Explain briefly.

Answer: The hands overlap 22 times in 24 hours (11 times in every 12-hour cycle ร— 2). It's 11 (not 12) because the overlap that would occur between 11 and 12 coincides with the 12 o'clock position, which begins the next cycle. The interval between consecutive overlaps is 12/11 hours โ‰ˆ 65 minutes 27 3โ„11 seconds.
SA 6

What is the century code for years in the 20xx range (2000โ€“2099)? Give the complete cycle pattern.

Answer: The century code for 20xx is 6. The complete cycle pattern repeats every 400 years: 17xx โ†’ 4, 18xx โ†’ 2, 19xx โ†’ 0, 20xx โ†’ 6, then 21xx โ†’ 4, 22xx โ†’ 2, 23xx โ†’ 0, 24xx โ†’ 6 (repeats). The pattern is 4-2-0-6.
SA 7

A clock gains 4 minutes every hour. How long (in real time) will it take to show the correct time again?

Answer: The clock gains 4 minutes per hour. It will show the correct time again when it has gained exactly 12 hours = 720 minutes (a full cycle of the 12-hour clock face). Time = 720 รท 4 = 180 hours = 7.5 days. After 180 real hours, the faulty clock will have gained exactly 12 hours and will show the correct time again.
SA 8

Find the mirror image time if a clock shows 5:25 in a mirror.

Answer: Actual time = 11:60 โˆ’ mirror time = 11:60 โˆ’ 5:25 = 6:35. So if the mirror shows 5:25, the actual time is 6:35.

Long Answer Questions (3)

LA 1

Using the odd-day method, find the day of the week for 15 August 1947 (India's Independence Day). Show all steps clearly.

Step-by-Step Solution:

Step 1 โ€” Odd days in the century part:
1947 โ†’ We need odd days from 1 AD to 1900 (19 centuries).
1600 years = 400 ร— 4 โ†’ 0 odd days.
Remaining 300 years (1601โ€“1900) โ†’ 1 odd day.
Century odd days = 0 + 1 = 1

Step 2 โ€” Odd days in remaining years (1901โ€“1946 = 46 years):
Note: We count completed years, so 1901 to 1946 = 46 years.
Leap years among these: 1904, 1908, 1912, 1916, 1920, 1924, 1928, 1932, 1936, 1940, 1944 = 11 leap years.
Ordinary years = 46 โˆ’ 11 = 35.
Odd days = 35 ร— 1 + 11 ร— 2 = 35 + 22 = 57. 57 mod 7 = 1 odd day.

Step 3 โ€” Odd days in months (Janโ€“Jul 1947):
1947 is not a leap year, so Feb = 28 days.
Jan(3) + Feb(0) + Mar(3) + Apr(2) + May(3) + Jun(2) + Jul(3) = 16. 16 mod 7 = 2 odd days.

Step 4 โ€” Date: 15 days. 15 mod 7 = 1 odd day.

Step 5 โ€” Total: 1 + 1 + 2 + 1 = 5.
5 โ†’ Friday (using 0=Sun, 1=Mon, ..., 5=Fri, 6=Sat).

โœ… 15 August 1947 was a Friday.

Verification with Quick Code Method: Century(19)=0, Year=47, 47รท4=11, Month(Aug)=2, Date=15. Total = 0+47+11+2+15 = 75. 75 mod 7 = 5 = Friday โœ…
LA 2

At what times between 4 o'clock and 5 o'clock will the hands of a clock be: (a) at right angles, (b) in a straight line (180ยฐ), and (c) overlapping? Calculate each to the exact fraction.

Using ฮธ = |30H โˆ’ 5.5M| with H = 4:

(a) Right angles (ฮธ = 90ยฐ):
Case 1: 30(4) โˆ’ 5.5M = 90 โ†’ 120 โˆ’ 5.5M = 90 โ†’ 5.5M = 30 โ†’ M = 30/5.5 = 60/11 = 5 5โ„11 minutes.
Time: 4:05 5โ„11

Case 2: 30(4) โˆ’ 5.5M = โˆ’90 โ†’ 120 โˆ’ 5.5M = โˆ’90 โ†’ 5.5M = 210 โ†’ M = 210/5.5 = 420/11 = 38 2โ„11 minutes.
Time: 4:38 2โ„11

(b) Straight line, 180ยฐ (ฮธ = 180ยฐ):
Case 1: 30(4) โˆ’ 5.5M = 180 โ†’ 120 โˆ’ 5.5M = 180 โ†’ 5.5M = โˆ’60 โ†’ M = โˆ’60/5.5 (negative, invalid).
Case 2: 30(4) โˆ’ 5.5M = โˆ’180 โ†’ 120 โˆ’ 5.5M = โˆ’180 โ†’ 5.5M = 300 โ†’ M = 300/5.5 = 600/11 = 54 6โ„11 minutes.
Time: 4:54 6โ„11 (valid, since < 60)

(c) Overlapping (ฮธ = 0ยฐ):
30(4) โˆ’ 5.5M = 0 โ†’ 120 = 5.5M โ†’ M = 120/5.5 = 240/11 = 21 9โ„11 minutes.
Time: 4:21 9โ„11

Summary: Between 4 and 5, the hands are at 90ยฐ at 4:05 5โ„11 and 4:38 2โ„11, at 180ยฐ at 4:54 6โ„11, and overlapping at 4:21 9โ„11.
LA 3

Clock A gains 2 minutes per hour and Clock B loses 3 minutes per hour. Both are set correctly at 12:00 noon on Monday. (a) What will each clock show at 12:00 noon on Tuesday (real time)? (b) When will both clocks show the same time again? (c) When will each clock independently show the correct time again?

(a) Time shown at 12:00 noon Tuesday (24 real hours later):

Clock A (gains 2 min/hr): In 24 hours, it gains 2 ร— 24 = 48 minutes extra.
Clock A shows: 12:00 + 24:48 = 12:48 PM on Tuesday (48 minutes fast).

Clock B (loses 3 min/hr): In 24 hours, it loses 3 ร— 24 = 72 minutes = 1 hour 12 minutes.
Clock B shows: 12:00 + 22:48 = 10:48 AM on Tuesday (1 hr 12 min slow).

(b) When will both show the same time again?
They drift apart at (2 + 3) = 5 minutes per hour.
They show the same time when their total drift = 12 hours = 720 minutes.
Time = 720 รท 5 = 144 hours = 6 days.
Both clocks will show the same time again at 12:00 noon on the following Sunday.

(c) When will each show the correct time again?
Each clock shows the correct time when it has gained/lost exactly 12 hours (720 min).
Clock A: 720 รท 2 = 360 hours = 15 days.
Clock B: 720 รท 3 = 240 hours = 10 days.
Clock A shows correct time after 15 days (Saturday noon).
Clock B shows correct time after 10 days (Thursday noon).
Section 13

Formula Sheet & Chapter Summary

๐Ÿ“‹ Complete Formula Sheet โ€” Calendar

Calendar Formulas

Odd Days: Total days mod 7

Ordinary Year: 1 odd day | Leap Year: 2 odd days

Century Odd Days: 100yr โ†’ 5, 200yr โ†’ 3, 300yr โ†’ 1, 400yr โ†’ 0

Leap Year Test: Divisible by 4 (non-century) OR divisible by 400 (century)

Day Mapping: 0=Sun, 1=Mon, 2=Tue, 3=Wed, 4=Thu, 5=Fri, 6=Sat

Quick Code: Day = (Century Code + Year's last 2 digits + last2รท4 + Month Code + Date) mod 7

Century Codes: 17xxโ†’4, 18xxโ†’2, 19xxโ†’0, 20xxโ†’6 (cycle: 4-2-0-6)

Month Codes: J(0) F(3) M(3) A(6) M(1) J(4) J(6) A(2) S(5) O(0) N(3) D(5)

Leap year adj.: Subtract 1 from total for Jan/Feb of leap years

๐Ÿ“‹ Complete Formula Sheet โ€” Clocks

Clock Formulas

Hour hand speed: 0.5ยฐ per minute = 30ยฐ per hour

Minute hand speed: 6ยฐ per minute = 360ยฐ per hour

Relative speed: 5.5ยฐ per minute

Angle formula: ฮธ = |30H โˆ’ 5.5M| (if > 180ยฐ, use 360ยฐ โˆ’ ฮธ)

Overlap time (H to H+1): M = 60H / 11

Overlaps in 12 hrs: 11 | in 24 hrs: 22

Right angles in 12 hrs: 22 | in 24 hrs: 44

Straight lines (180ยฐ) in 12 hrs: 11 | in 24 hrs: 22

Time between overlaps: 12/11 hours โ‰ˆ 65 min 27 3โ„11 sec

Mirror time: 11:60 โˆ’ displayed time

Faulty clock (gains x min/hr): Shows 60+x min per real hour

Correct time after gaining: 720 / x hours to show right time again

๐Ÿ“– Chapter Summary

Key Takeaways โ€” Calendar & Clocks

โœ… Odd days = remainder when total days รท 7. This single concept solves all calendar day-finding problems.

โœ… Leap year rule for centuries: Must be divisible by 400, not just 4. (1900 โŒ, 2000 โœ…)

โœ… 400 years = 0 odd days โ€” the calendar repeats perfectly every 400 years.

โœ… The Quick Code method (Century + Year + Yearรท4 + Month + Date) mod 7 finds any day in ~15 seconds.

โœ… Clock angle formula: ฮธ = |30H โˆ’ 5.5M| โ€” memorise this one formula to solve 80% of clock problems.

โœ… The hands overlap 22 times/day, are at right angles 44 times/day, and are opposite 22 times/day.

โœ… Faulty clocks: Calculate gain/loss per hour, scale to find drift, use 720 min for "correct time again."

โœ… Mirror time: Subtract from 11:60 for a quick answer.

๐Ÿ“Š Learning Checkpoint

ConceptTool/MethodKey DeliverableExam Readiness
Odd DaysMental MathDay-finding for any dateโœ… SSC, Bank PO, CAT
Leap Year Rulesรท4 and รท400 testsIdentify leap/ordinary yearsโœ… All competitive exams
Quick Code MethodCentury + Month codes15-second day calculationโœ… Time-critical exams
Clock Angle Formulaฮธ = |30H โˆ’ 5.5M|Instant angle calculationโœ… SSC, Placements
Overlap/Right AngleM = 60H/11Exact overlap timesโœ… Bank PO, CAT
Faulty ClocksGain/Loss per hourCorrect time calculationโœ… Advanced aptitude
Mirror Image11:60 โˆ’ timeActual time from mirrorโœ… Reasoning section

โœ… Unit 4 complete. Ready for Unit 5!

[QR: Link to EduArtha video tutorial โ€” Calendar & Clocks]