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Logical Reasoning · Clocks _ Calenders

Basics- Formulas, Key Points and Examples

Explanation

🔵 CLOCKS AND CALENDARS 🟢 1. BASIC CLOCK CONCEPT A clock has 12 numbers and completes one full revolution in 12 hours.
360∘=12 hours 360^\circ=12\text{ hours}
The minute hand completes one revolution in 60 minutes.
360∘=60 minutes 360^\circ=60\text{ minutes}
Therefore:
1 minute=6∘ 1\text{ minute}=6^\circ
The hour hand moves 30° in one hour.
1 hour=30∘ 1\text{ hour}=30^\circ
The hour hand moves 0.5° in one minute.
1 minute=0.5∘ 1\text{ minute}=0.5^\circ
🟡 KEY POINT
Minute Hand Speed=6∘ per minute \text{Minute Hand Speed}=6^\circ\text{ per minute}
Hour Hand Speed=0.5∘ per minute \text{Hour Hand Speed}=0.5^\circ\text{ per minute}
🟢 2. MOVEMENT OF MINUTE HAND The minute hand moves 6° every minute.
Angle moved by Minute Hand=6M \text{Angle moved by Minute Hand} = 6M
where MM is the number of minutes. For example, in 20 minutes:
6×20=120∘ 6\times20=120^\circ
🟡 KEY POINT
Minute Hand Angle=6M \text{Minute Hand Angle}=6M
🟢 3. MOVEMENT OF HOUR HAND The hour hand moves 30° in one hour.
Angle moved in 1 hour=30∘ \text{Angle moved in 1 hour}=30^\circ
Since one hour contains 60 minutes:
Angle moved in 1 minute=3060 \text{Angle moved in 1 minute} = \frac{30}{60}
=0.5∘ =0.5^\circ
Therefore, at HH hours and MM minutes:
Hour Hand Angle=30H+0.5M \text{Hour Hand Angle} = 30H+0.5M
🟢 4. ANGLE BETWEEN THE HANDS OF A CLOCK At HH hours and MM minutes:
Hour Hand Angle=30H+0.5M \text{Hour Hand Angle} = 30H+0.5M
Minute Hand Angle=6M \text{Minute Hand Angle} = 6M
Therefore:
Angle=∣30H+0.5M−6M∣ \text{Angle} = \left| 30H+0.5M-6M \right|
Simplifying:
Angle=∣30H−5.5M∣ \text{Angle} = |30H-5.5M|
The smaller angle between the hands is:
Smaller Angle=min⁡(∣30H−5.5M∣,360−∣30H−5.5M∣) \text{Smaller Angle} = \min \left( |30H-5.5M|, 360-|30H-5.5M| \right)
🔴 IMPORTANT
If the calculated angle is greater than 180∘, subtract it from 360∘. \text{If the calculated angle is greater than }180^\circ, \text{ subtract it from }360^\circ.
Smaller Angle=360∘−Larger Angle \text{Smaller Angle} = 360^\circ-\text{Larger Angle}
🟢 5. RIGHT ANGLE BETWEEN CLOCK HANDS A right angle is:
90∘ 90^\circ
Therefore, the clock hands are perpendicular when:
∣30H−5.5M∣=90 |30H-5.5M|=90
🟡 KEY POINT
Right Angle=90∘ \text{Right Angle}=90^\circ
🟢 6. STRAIGHT ANGLE BETWEEN CLOCK HANDS A straight angle is:
180∘ 180^\circ
Therefore, the clock hands are opposite when:
∣30H−5.5M∣=180 |30H-5.5M|=180
🟡 KEY POINT
Straight Angle=180∘ \text{Straight Angle}=180^\circ
🟢 7. COINCIDENCE OF CLOCK HANDS The hands coincide when both hands are at the same position. Therefore:
30H+0.5M=6M 30H+0.5M=6M
Simplifying:
30H=5.5M 30H=5.5M
Therefore:
M=30H5.5 M=\frac{30H}{5.5}
M=60H11 M=\frac{60H}{11}
The hands coincide approximately every:
72011 minutes \frac{720}{11}\text{ minutes}
Therefore:
72011=65511 minutes \frac{720}{11} = 65\frac{5}{11}\text{ minutes}
🔴 IMPORTANT
The hour and minute hands coincide 11 times in 12 hours. \text{The hour and minute hands coincide 11 times in 12 hours.}
🟢 8. OPPOSITE POSITIONS OF CLOCK HANDS The hands are opposite when the angle between them is:
180∘ 180^\circ
Therefore:
∣30H−5.5M∣=180 |30H-5.5M|=180
🟢 9. CLOCK HANDS AT A GIVEN ANGLE If the angle between the hands is θ\theta:
∣30H−5.5M∣=θ |30H-5.5M|=\theta
For a 60∘60^\circ angle:
∣30H−5.5M∣=60 |30H-5.5M|=60
For a 120∘120^\circ angle:
∣30H−5.5M∣=120 |30H-5.5M|=120
For a 90∘90^\circ angle:
∣30H−5.5M∣=90 |30H-5.5M|=90
🟢 10. ANGLE MOVED BY THE MINUTE HAND The minute hand completes:
360∘ 360^\circ
in 60 minutes. Therefore:
Angle per minute=36060 \text{Angle per minute} = \frac{360}{60}
=6∘ =6^\circ
Hence:
Angle in M minutes=6M \text{Angle in }M\text{ minutes} = 6M
🟢 11. ANGLE MOVED BY THE HOUR HAND The hour hand completes:
360∘ 360^\circ
in 12 hours. Therefore:
Angle per hour=36012 \text{Angle per hour} = \frac{360}{12}
=30∘ =30^\circ
Since one hour contains 60 minutes:
Angle per minute=3060 \text{Angle per minute} = \frac{30}{60}
=0.5∘ =0.5^\circ
🟢 12. CLOCK GAINING TIME If a clock gains time, it moves faster than the correct clock.
Gain per hour=Total GainTotal Time in Hours \text{Gain per hour} = \frac{\text{Total Gain}}{\text{Total Time in Hours}}
🟡 KEY POINT
A fast clock shows more time than the actual time. \text{A fast clock shows more time than the actual time.}
🟢 13. CLOCK LOSING TIME If a clock loses time, it moves slower than the correct clock.
Loss per hour=Total LossTotal Time in Hours \text{Loss per hour} = \frac{\text{Total Loss}}{\text{Total Time in Hours}}
🟡 KEY POINT
A slow clock shows less time than the actual time. \text{A slow clock shows less time than the actual time.}
🟢 14. RELATIVE GAIN AND LOSS OF CLOCKS If one clock gains time and another clock loses time, their relative difference increases.
Relative Difference=Gain of First Clock+Loss of Second Clock \text{Relative Difference} = \text{Gain of First Clock} + \text{Loss of Second Clock}
If both clocks gain or both clocks lose time:
Relative Difference=∣Rate of First Clock−Rate of Second Clock∣ \text{Relative Difference} = \left| \text{Rate of First Clock} - \text{Rate of Second Clock} \right|
🟡 KEY POINT
Always compare the rates of the two clocks when solving relative clock problems. \text{Always compare the rates of the two clocks when solving relative clock problems.}
🟢 15. BASIC CALENDAR CONCEPT A calendar is based on days, weeks, months and years. One week contains:
7 days 7\text{ days}
A normal year contains:
365 days 365\text{ days}
A leap year contains:
366 days 366\text{ days}
🟢 16. ODD DAYS Odd days are the number of days left after complete weeks are removed. Since:
1 week=7 days 1\text{ week}=7\text{ days}
The number of odd days is the remainder when the number of days is divided by 7. For 365 days:
365=52×7+1 365=52\times7+1
Therefore:
Odd Days=1 \text{Odd Days}=1
For 366 days:
366=52×7+2 366=52\times7+2
Therefore:
Odd Days=2 \text{Odd Days}=2
🟢 17. NORMAL YEAR A normal year has:
365 days 365\text{ days}
Therefore:
365=52×7+1 365=52\times7+1
Hence:
Normal Year=1 Odd Day \text{Normal Year}=1\text{ Odd Day}
🟢 18. LEAP YEAR A leap year has:
366 days 366\text{ days}
Therefore:
366=52×7+2 366=52\times7+2
Hence:
Leap Year=2 Odd Days \text{Leap Year}=2\text{ Odd Days}
🟢 19. LEAP YEAR RULE A year is generally a leap year if it is divisible by 4. For example:
2024÷4=506 2024\div4=506
Therefore:
2024 is a leap year. 2024\text{ is a leap year.}
However, century years must also be divisible by 400. For example:
2000÷400=5 2000\div400=5
Therefore:
2000 is a leap year. 2000\text{ is a leap year.}
But:
1900÷400=4.75 1900\div400=4.75
Therefore:
1900 is not a leap year. 1900\text{ is not a leap year.}
🔴 IMPORTANT
A century year is a leap year only if it is divisible by 400. \text{A century year is a leap year only if it is divisible by 400.}
🟢 20. DAYS IN DIFFERENT MONTHS The number of days in each month is:
January=31 \text{January}=31
February=28 \text{February}=28
March=31 \text{March}=31
April=30 \text{April}=30
May=31 \text{May}=31
June=30 \text{June}=30
July=31 \text{July}=31
August=31 \text{August}=31
September=30 \text{September}=30
October=31 \text{October}=31
November=30 \text{November}=30
December=31 \text{December}=31
In a leap year:
February=29 \text{February}=29
🟢 21. DAYS IN A WEEK The seven days of the week are:
Sunday \text{Sunday}
Monday \text{Monday}
Tuesday \text{Tuesday}
Wednesday \text{Wednesday}
Thursday \text{Thursday}
Friday \text{Friday}
Saturday \text{Saturday}
🟡 KEY POINT
1 week=7 days 1\text{ week}=7\text{ days}
🟢 22. DAY AFTER A GIVEN NUMBER OF DAYS If today is a particular day, the day after NN days depends on the remainder when NN is divided by 7.
N=7q+r N=7q+r
where rr is the number of odd days. Therefore:
Required Day=Starting Day+r \text{Required Day} = \text{Starting Day}+r
🟡 KEY POINT
For calendar problems, divide the number of days by 7 and use the remainder. \text{For calendar problems, divide the number of days by 7 and use the remainder.}
🟢 23. DAY BEFORE A GIVEN NUMBER OF DAYS If we need to find the day before a given date, subtract the odd days.
N=7q+r N=7q+r
Therefore:
Required Day=Starting Day−r \text{Required Day} = \text{Starting Day}-r
🟢 24. DAY OF THE WEEK To determine the day of the week for a particular date, calculate the total number of odd days from a known reference date. The calculation generally involves:
Odd Days=Odd Days in Complete Years+Odd Days in Complete Months+Odd Days in Remaining Days \text{Odd Days} = \text{Odd Days in Complete Years} + \text{Odd Days in Complete Months} + \text{Odd Days in Remaining Days}
Then:
Required Day=Reference Day+Total Odd Days \text{Required Day} = \text{Reference Day}+\text{Total Odd Days}
The final result is reduced using:
Total Odd Days mod 7 \text{Total Odd Days}\bmod7
🟢 25. ODD DAYS IN COMPLETE YEARS For a group of years:
Total Days=365(Number of Normal Years)+366(Number of Leap Years) \text{Total Days} = 365(\text{Number of Normal Years}) + 366(\text{Number of Leap Years})
Then:
Odd Days=Total Days mod 7 \text{Odd Days} = \text{Total Days}\bmod7
🟢 26. ODD DAYS IN COMPLETE MONTHS The number of days in complete months is added before the required date. For a normal year:
Days in February=28 \text{Days in February}=28
For a leap year:
Days in February=29 \text{Days in February}=29
Then:
Odd Days=Total Days mod 7 \text{Odd Days} = \text{Total Days}\bmod7
🟢 27. NUMBER OF ODD DAYS BETWEEN TWO DATES To find the number of days between two dates:
Total Days=Days in Complete Years+Days in Complete Months+Remaining Days \text{Total Days} = \text{Days in Complete Years} + \text{Days in Complete Months} + \text{Remaining Days}
Then:
Odd Days=Total Days mod 7 \text{Odd Days} = \text{Total Days}\bmod7
🟢 28. SAME CALENDAR YEAR Two years can have the same calendar when their starting day is the same and the total odd days between them is a multiple of 7.
Total Odd Days≡0(mod7) \text{Total Odd Days}\equiv0\pmod7
Therefore:
Starting Day of Year 1=Starting Day of Year 2 \text{Starting Day of Year 1} = \text{Starting Day of Year 2}
🟡 KEY POINT
Leap years affect the starting day of the following year. \text{Leap years affect the starting day of the following year.}
🟢 29. CALENDAR AFTER A NORMAL YEAR A normal year has one odd day. Therefore, if a normal year starts on a particular day:
Next Year Start=Starting Day+1 \text{Next Year Start} = \text{Starting Day}+1
For example, if a normal year starts on Monday:
Next Year Start=Tuesday \text{Next Year Start}=Tuesday
🟢 30. CALENDAR AFTER A LEAP YEAR A leap year has two odd days. Therefore:
Next Year Start=Starting Day+2 \text{Next Year Start} = \text{Starting Day}+2
For example, if a leap year starts on Monday:
Next Year Start=Wednesday \text{Next Year Start}=Wednesday
🟢 31. IMPORTANT CLOCK FORMULAS FOR REVISION Minute hand angle:
Minute Hand Angle=6M \text{Minute Hand Angle}=6M
Hour hand angle:
Hour Hand Angle=30H+0.5M \text{Hour Hand Angle}=30H+0.5M
Angle between hands:
Angle=∣30H−5.5M∣ \text{Angle}=|30H-5.5M|
Smaller angle:
Smaller Angle=min⁡(∣30H−5.5M∣,360−∣30H−5.5M∣) \text{Smaller Angle} = \min \left( |30H-5.5M|, 360-|30H-5.5M| \right)
Right angle:
∣30H−5.5M∣=90 |30H-5.5M|=90
Straight angle:
∣30H−5.5M∣=180 |30H-5.5M|=180
Coincidence:
M=60H11 M=\frac{60H}{11}
🟢 32. IMPORTANT CALENDAR FORMULAS FOR REVISION Number of days in a normal year:
365=52×7+1 365=52\times7+1
Therefore:
Normal Year=1 Odd Day \text{Normal Year}=1\text{ Odd Day}
Number of days in a leap year:
366=52×7+2 366=52\times7+2
Therefore:
Leap Year=2 Odd Days \text{Leap Year}=2\text{ Odd Days}
Odd days:
Odd Days=Number of Days mod 7 \text{Odd Days} = \text{Number of Days}\bmod7
Day after NN days:
N=7q+r N=7q+r
Required Day=Starting Day+r \text{Required Day} = \text{Starting Day}+r
Day before NN days:
N=7q+r N=7q+r
Required Day=Starting Day−r \text{Required Day} = \text{Starting Day}-r
🟡 KEY POINTS
1 hour=60 minutes 1\text{ hour}=60\text{ minutes}
1 minute=60 seconds 1\text{ minute}=60\text{ seconds}
1 week=7 days 1\text{ week}=7\text{ days}
1 normal year=365 days 1\text{ normal year}=365\text{ days}
1 leap year=366 days 1\text{ leap year}=366\text{ days}
1 normal year=1 odd day 1\text{ normal year}=1\text{ odd day}
1 leap year=2 odd days 1\text{ leap year}=2\text{ odd days}
Minute hand moves 6∘ per minute. \text{Minute hand moves }6^\circ\text{ per minute.}
Hour hand moves 0.5∘ per minute. \text{Hour hand moves }0.5^\circ\text{ per minute.}
Use ∣30H−5.5M∣ to calculate the angle between clock hands. \text{Use }|30H-5.5M|\text{ to calculate the angle between clock hands.}
For calendar problems, reduce the number of days using division by 7. \text{For calendar problems, reduce the number of days using division by 7.}
A leap year is divisible by 4, except century years. \text{A leap year is divisible by 4, except century years.}
A century year is a leap year only when it is divisible by 400. \text{A century year is a leap year only when it is divisible by 400.}
Always check whether February has 28 or 29 days. \text{Always check whether February has 28 or 29 days.}

Example

🔵 CLOCKS AND CALENDARS — SOLVED EXAMPLES 🟢 EXAMPLE 1: ANGLE MOVED BY MINUTE HAND QUESTION: Find the angle moved by the minute hand in 25 minutes. SOLUTION:
Angle moved by minute hand=6M \text{Angle moved by minute hand} = 6M
=6×25 = 6\times25
=150∘ =150^\circ
🔵 ANSWER:
150∘ 150^\circ
🟢 EXAMPLE 2: ANGLE MOVED BY HOUR HAND QUESTION: Find the angle moved by the hour hand in 4 hours. SOLUTION:
Angle moved by hour hand=30H \text{Angle moved by hour hand} = 30H
=30×4 = 30\times4
=120∘ =120^\circ
🔵 ANSWER:
120∘ 120^\circ
🟢 EXAMPLE 3: ANGLE BETWEEN CLOCK HANDS QUESTION: Find the angle between the hands of a clock at 3:20. SOLUTION:
Angle=∣30H−5.5M∣ \text{Angle} = |30H-5.5M|
Here:
H=3,M=20 H=3,\quad M=20
Therefore:
=∣30(3)−5.5(20)∣ = |30(3)-5.5(20)|
=∣90−110∣ = |90-110|
=20∘ =20^\circ
🔵 ANSWER:
20∘ 20^\circ
🟢 EXAMPLE 4: ANGLE BETWEEN CLOCK HANDS QUESTION: Find the angle between the hands of a clock at 5:30. SOLUTION:
Angle=∣30H−5.5M∣ \text{Angle} = |30H-5.5M|
Here:
H=5,M=30 H=5,\quad M=30
Therefore:
=∣30(5)−5.5(30)∣ = |30(5)-5.5(30)|
=∣150−165∣ = |150-165|
=15∘ =15^\circ
🔵 ANSWER:
15∘ 15^\circ
🟢 EXAMPLE 5: SMALLER ANGLE QUESTION: Find the smaller angle between the hands of a clock at 8:40. SOLUTION:
Angle=∣30H−5.5M∣ \text{Angle} = |30H-5.5M|
Here:
H=8,M=40 H=8,\quad M=40
Therefore:
=∣30(8)−5.5(40)∣ = |30(8)-5.5(40)|
=∣240−220∣ = |240-220|
=20∘ =20^\circ
🔵 ANSWER:
20∘ 20^\circ
🟢 EXAMPLE 6: RIGHT ANGLE QUESTION: At what time between 3 and 4 o'clock will the hands of a clock be at right angles? SOLUTION: For a right angle:
∣30H−5.5M∣=90 |30H-5.5M|=90
Here:
H=3 H=3
Therefore:
∣30(3)−5.5M∣=90 |30(3)-5.5M|=90
∣90−5.5M∣=90 |90-5.5M|=90
Considering the required position:
90−5.5M=−90 90-5.5M=-90
5.5M=180 5.5M=180
M=1805.5 M=\frac{180}{5.5}
M=36011 M=\frac{360}{11}
M=32811 M=32\frac{8}{11}
🔵 ANSWER:
3:32811 3:32\frac{8}{11}
🟢 EXAMPLE 7: STRAIGHT ANGLE QUESTION: At what time between 5 and 6 o'clock will the hands of a clock be opposite to each other? SOLUTION: For opposite hands:
∣30H−5.5M∣=180 |30H-5.5M|=180
Here:
H=5 H=5
Therefore:
∣150−5.5M∣=180 |150-5.5M|=180
Considering the required position:
150−5.5M=−180 150-5.5M=-180
5.5M=330 5.5M=330
M=3305.5 M=\frac{330}{5.5}
M=60 M=60
🔵 ANSWER:
6:00 6:00
🟢 EXAMPLE 8: COINCIDENCE OF HANDS QUESTION: At what time between 2 and 3 o'clock will the hands of a clock coincide? SOLUTION: For coincidence:
M=60H11 M=\frac{60H}{11}
Here:
H=2 H=2
Therefore:
M=60×211 M=\frac{60\times2}{11}
=12011 =\frac{120}{11}
=101011 =10\frac{10}{11}
🔵 ANSWER:
2:101011 2:10\frac{10}{11}
🟢 EXAMPLE 9: COINCIDENCE OF HANDS QUESTION: At what time between 7 and 8 o'clock will the hands of a clock coincide? SOLUTION:
M=60H11 M=\frac{60H}{11}
Here:
H=7 H=7
Therefore:
M=60×711 M=\frac{60\times7}{11}
=42011 =\frac{420}{11}
=38211 =38\frac{2}{11}
🔵 ANSWER:
7:38211 7:38\frac{2}{11}
🟢 EXAMPLE 10: CLOCK GAINS TIME QUESTION: A clock gains 5 minutes in 10 hours. How many minutes will it gain in 24 hours? SOLUTION:
Gain per hour=510 \text{Gain per hour} = \frac{5}{10}
=0.5 minutes =0.5\text{ minutes}
Therefore, gain in 24 hours:
=0.5×24 =0.5\times24
=12 minutes =12\text{ minutes}
🔵 ANSWER:
12 minutes 12\text{ minutes}
🟢 EXAMPLE 11: CLOCK LOSES TIME QUESTION: A clock loses 4 minutes in 8 hours. How many minutes will it lose in 24 hours? SOLUTION:
Loss per hour=48 \text{Loss per hour} = \frac{4}{8}
=0.5 minutes =0.5\text{ minutes}
Therefore:
Loss in 24 hours=0.5×24 \text{Loss in 24 hours} = 0.5\times24
=12 minutes =12\text{ minutes}
🔵 ANSWER:
12 minutes 12\text{ minutes}
🟢 EXAMPLE 12: FINDING ODD DAYS QUESTION: Find the number of odd days in 365 days. SOLUTION:
365=52×7+1 365=52\times7+1
Therefore:
Odd Days=1 \text{Odd Days}=1
🔵 ANSWER:
1 odd day 1\text{ odd day}
🟢 EXAMPLE 13: ODD DAYS IN A LEAP YEAR QUESTION: Find the number of odd days in 366 days. SOLUTION:
366=52×7+2 366=52\times7+2
Therefore:
Odd Days=2 \text{Odd Days}=2
🔵 ANSWER:
2 odd days 2\text{ odd days}
🟢 EXAMPLE 14: IDENTIFYING A LEAP YEAR QUESTION: Is 2024 a leap year? SOLUTION: A year divisible by 4 is generally a leap year.
2024÷4=506 2024\div4=506
Since the division is exact:
2024 is a leap year 2024\text{ is a leap year}
🔵 ANSWER:
2024 is a leap year 2024\text{ is a leap year}
🟢 EXAMPLE 15: CENTURY YEAR QUESTION: Is 1900 a leap year? SOLUTION: 1900 is divisible by 100, so it is a century year. For a century year to be a leap year, it must be divisible by 400.
1900÷400=4.75 1900\div400=4.75
The division is not exact. Therefore:
1900 is not a leap year 1900\text{ is not a leap year}
🔵 ANSWER:
1900 is not a leap year 1900\text{ is not a leap year}
🟢 EXAMPLE 16: CENTURY LEAP YEAR QUESTION: Is 2000 a leap year? SOLUTION: Since 2000 is a century year, check divisibility by 400.
2000÷400=5 2000\div400=5
The division is exact. Therefore:
2000 is a leap year 2000\text{ is a leap year}
🔵 ANSWER:
2000 is a leap year 2000\text{ is a leap year}
🟢 EXAMPLE 17: DAYS AFTER A GIVEN DAY QUESTION: If today is Monday, what day will it be after 45 days? SOLUTION: Divide 45 by 7:
45=6×7+3 45=6\times7+3
Therefore:
Odd Days=3 \text{Odd Days}=3
Starting from Monday and moving 3 days:
Monday→Tuesday→Wednesday→Thursday \text{Monday}\rightarrow\text{Tuesday}\rightarrow\text{Wednesday}\rightarrow\text{Thursday}
🔵 ANSWER:
Thursday \text{Thursday}
🟢 EXAMPLE 18: DAYS BEFORE A GIVEN DAY QUESTION: If today is Friday, what day was it 20 days ago? SOLUTION:
20=2×7+6 20=2\times7+6
Therefore:
Odd Days=6 \text{Odd Days}=6
Moving 6 days backward from Friday:
Friday→Thursday→Wednesday→Tuesday \text{Friday}\rightarrow\text{Thursday}\rightarrow\text{Wednesday}\rightarrow\text{Tuesday}
→Monday→Sunday→Saturday \rightarrow\text{Monday}\rightarrow\text{Sunday}\rightarrow\text{Saturday}
🔵 ANSWER:
Saturday \text{Saturday}
🟢 EXAMPLE 19: DAYS IN A NORMAL YEAR QUESTION: If January 1 of a normal year is Monday, what day will January 1 of the next year be? SOLUTION: A normal year has:
365 days 365\text{ days}
Therefore:
365=52×7+1 365=52\times7+1
So there is:
1 odd day 1\text{ odd day}
Starting day:
Monday \text{Monday}
Move one day forward:
Monday→Tuesday \text{Monday}\rightarrow\text{Tuesday}
🔵 ANSWER:
Tuesday \text{Tuesday}
🟢 EXAMPLE 20: DAYS IN A LEAP YEAR QUESTION: If January 1 of a leap year is Wednesday, what day will January 1 of the next year be? SOLUTION: A leap year has:
366 days 366\text{ days}
Therefore:
366=52×7+2 366=52\times7+2
So there are:
2 odd days 2\text{ odd days}
Starting day:
Wednesday \text{Wednesday}
Move two days forward:
Wednesday→Thursday→Friday \text{Wednesday}\rightarrow\text{Thursday}\rightarrow\text{Friday}
🔵 ANSWER:
Friday \text{Friday}
🟢 EXAMPLE 21: DAYS IN A MONTH QUESTION: How many odd days are there in a 31-day month? SOLUTION:
31=4×7+3 31=4\times7+3
Therefore:
Odd Days=3 \text{Odd Days}=3
🔵 ANSWER:
3 odd days 3\text{ odd days}
🟢 EXAMPLE 22: DAYS IN FEBRUARY QUESTION: How many odd days are there in February of a normal year? SOLUTION: February has:
28 days 28\text{ days}
Therefore:
28=4×7 28=4\times7
Hence:
Odd Days=0 \text{Odd Days}=0
🔵 ANSWER:
0 odd days 0\text{ odd days}
🟢 EXAMPLE 23: DAYS IN FEBRUARY OF A LEAP YEAR QUESTION: How many odd days are there in February of a leap year? SOLUTION: February has:
29 days 29\text{ days}
Therefore:
29=4×7+1 29=4\times7+1
Hence:
Odd Days=1 \text{Odd Days}=1
🔵 ANSWER:
1 odd day 1\text{ odd day}
🟢 EXAMPLE 24: FINDING THE DAY AFTER 100 DAYS QUESTION: If today is Tuesday, what day will it be after 100 days? SOLUTION:
100=14×7+2 100=14\times7+2
Therefore:
Odd Days=2 \text{Odd Days}=2
Starting from Tuesday:
Tuesday→Wednesday→Thursday \text{Tuesday}\rightarrow\text{Wednesday}\rightarrow\text{Thursday}
🔵 ANSWER:
Thursday \text{Thursday}
🟢 EXAMPLE 25: FINDING THE DAY BEFORE 100 DAYS QUESTION: If today is Sunday, what day was it 100 days ago? SOLUTION:
100=14×7+2 100=14\times7+2
Therefore:
Odd Days=2 \text{Odd Days}=2
Moving two days backward:
Sunday→Saturday→Friday \text{Sunday}\rightarrow\text{Saturday}\rightarrow\text{Friday}
🔵 ANSWER:
Friday \text{Friday}
🟢 EXAMPLE 26: NUMBER OF DAYS BETWEEN TWO DATES QUESTION: How many days are there from January 1 to January 31, excluding January 1? SOLUTION: January has:
31 days 31\text{ days}
After excluding January 1:
31−1=30 31-1=30
🔵 ANSWER:
30 days 30\text{ days}
🟢 EXAMPLE 27: SAME CALENDAR QUESTION: A normal year has 1 odd day. If January 1 of one year is Monday and the next year is also a normal year, what day will January 1 of the following year be? SOLUTION: First year contributes:
1 odd day 1\text{ odd day}
Second year also contributes:
1 odd day 1\text{ odd day}
Total:
1+1=2 odd days 1+1=2\text{ odd days}
Starting from Monday:
Monday→Tuesday→Wednesday \text{Monday}\rightarrow\text{Tuesday}\rightarrow\text{Wednesday}
🔵 ANSWER:
Wednesday \text{Wednesday}
🟢 EXAMPLE 28: MIXED CALENDAR PROBLEM QUESTION: If January 1 is Monday, what day will January 15 be? SOLUTION: The number of days after January 1 is:
15−1=14 15-1=14
Now:
14=2×7 14=2\times7
Therefore:
Odd Days=0 \text{Odd Days}=0
So the day remains unchanged. 🔵 ANSWER:
Monday \text{Monday}
🟢 EXAMPLE 29: LEAP YEAR FEBRUARY QUESTION: If February 1 of a leap year is Thursday, what day will March 1 be? SOLUTION: February in a leap year has:
29 days 29\text{ days}
Therefore:
29=4×7+1 29=4\times7+1
So:
Odd Days=1 \text{Odd Days}=1
Starting from Thursday:
Thursday→Friday \text{Thursday}\rightarrow\text{Friday}
🔵 ANSWER:
Friday \text{Friday}
🟢 EXAMPLE 30: MIXED CLOCK PROBLEM QUESTION: Find the angle between the hands of a clock at 7:25. SOLUTION:
Angle=∣30H−5.5M∣ \text{Angle} = |30H-5.5M|
Here:
H=7,M=25 H=7,\quad M=25
Therefore:
=∣30(7)−5.5(25)∣ = |30(7)-5.5(25)|
=∣210−137.5∣ = |210-137.5|
=72.5∘ =72.5^\circ
🔵 ANSWER:
72.5∘ 72.5^\circ