## Clocks and Calendars

### Clock

A clock is an instrument having the numbers 1 to 12 or equivalent roman numerals around its face which display time divided into hours, minutes and seconds.

#### Hour Hand

The smaller or slower hand of a clock is called the hour hand. It takes two revolution in a day.

#### Minute Hand

The bigger or faster hand of a clock is called the minute hand.

It takes one revolution in every hour.

#### Second Hand

Second hand bigger or faster then minute hand. It makes one revolution per minute.

### Important Points Related to Clock

#### Angle makes by hour hand and minute hand

**Minute Hand**:- Minute hand complete one revolution i,e, 360° in 60 minutes.

360° = 60 minutes

180° = 30 minutes

90° = 15 minutes

30° = 5 minutes

6° = 1 minute

**∴ 1 minute = 6°**

Clearly, minute hand make 6° angle in 1 minute.

**Hour hand**:- Hour hand complete one revolution i.e, 360° in 12 hours.

12 h = 360°

1 h = 30°

60 min = 30°

∴ 1 min = | 1° |

2 |

Clearly, hour hand make | 1° | in 1 minute . |

2 |

Relative Speed between Minute hand and hour hand = 6 - | 1° | = 5 | 1° | degree per minute. |

2 | 2 |

**Note :**- In every minute, minute hand goes 5

^{1}/

_{2}degree more than hour hand.

#### Speed of Hour hand and Minute hand

Speed of hour hand = 5 min/hour

Speed of minute hand = 60 min/hour

Relative speed = 60 - 5 = 55 min/hour**Note**- In every hour, minute hand goes 55 min more than hour hand.

### Specific features of hour hand and minute hand

#### When they will make 90° angle

If two hands are at 90° they are 15 min apart.

Its happens twice in 1 hr.

1 h = 2 times

12 h = 22 times

**∴** 1 day = 24 h = 44 times

#### When they will make 180° angle

If two hands are at 180° they are 30 min apart.

direction opposite to each other.

It happens once in 1 hr.

**1 h = 1 times**

12 h = 11 times

∴ 1 day = 24 h = 22 times,

12 h = 11 times

∴ 1 day = 24 h = 22 times

#### When they will make a straight line

If angle between them 180°, then 30 min apart directly opposite.

If angle between them 0°, then no difference, overlap

It happens once in 1 hr.

**1 h = 1 times**

12 h = 11 times

1 day = 24 h = 22 times,

12 h = 11 times

1 day = 24 h = 22 times

#### Angle between minute hand and hour hand

Let we have to find angle between minute hand and hour hand at 'H hours and M minutes'.

Angle between two hands = | 11 | M - 30H |

2 |

**Note :-**If sign is negative, then ignore the negative sign.

**Ex-** Find the angle between minutes hand and hour hand at 4 : 20 A.M.

**Solution:-** Here, H = 4 and M = 20

According to formula,

Angle between two hands h hours and m minutes = | 11 | M - 30H |

2 |

= | 11 | × 20 - 30 × 4 |

2 |

Ignore the negative sign the required angle is 10°

### Important Formula :-

#### (1) Two hands at Together :-

Minute hand and hour hand of a clock will be together between h and ( h + 1 ) O' clock at **( 60/11 )h** minutes past h.

**Ex-** At what time between 2 and 3 O’clock are the two hands of the clock together? **Solution :-** Here, h = 2

According to formula ,

Two hands together at = | 60 | h minutes |

11 |

= | 60 | × 2 |

11 |

= | 120 | = 10 | 10 | minutes. |

11 | 11 |

Hence, they will be together at 2 : 10 | 10 | minutes. |

11 |

#### (2) To hands at Right angle :-

Minute hand and hour hand of a clock will right angle between h and (h+1) O' clock at** ( 5h ± 15 ) × 12/11** minutes past h.

= ( 5h - 15 ) × | 12 | When, h > 6 |

11 |

= ( 5h + 15 ) × | 12 | When, h < 6 |

11 |

**Ex-** At what time between 8 and 9 o' clock will the hands of a clock be at right angle?**Solution:-** Here, h = 8

According to formula,

Two hands at right angle = ( 5h - 15 ) × | 12 |

11 |

= ( 5 × 8 - 15 ) × | 12 | = 25 × | 12 |

11 | 11 |

Two hands at right angle= | 300 | = 27 | 3 |

11 | 11 |

Hence, they will be right angle at 8 : 27 | 3 | minutes. |

11 |

**Ex-** At what time between 4 and 5 o' clock will the hands of a clock be at right angle?**Solution:**- Here, h = 4

According to formula,

Two hands at right angle = ( 5h + 15 ) × | 12 |

11 |

= ( 5 × 4 + 15 ) × | 12 | = 35 × | 12 |

11 | 11 |

Two hands at right angle= | 420 | = 38 | 2 |

11 | 11 |

Hence, they will be right angle at 4 : 38 | 2 | minutes. |

11 |

#### (3) Two hands at Straight angle :-

Minute hand and hour hand of a clock will straight line at 180° between h and ( h + 1 ) o' clock at

( 5h ± 30 ) × 12/11 minutes past h.

= ( 5h - 30) × | 12 | When, h > 6 |

11 |

= ( 5h + 30) × | 12 | When, h < 6 |

11 |

**Ex-** At what time between 9 and 10 o' clock will the hands of a clock be in the same straight but not together?**Solution:-** Here, h = 9

According to formula,

Two hands at straight angle = ( 5h - 30 ) × | 12 |

11 |

= ( 5 × 9 - 30 ) × | 12 | = 15 × | 12 |

11 | 11 |

Two hands at straight angle= | 180 | = 16 | 4 |

11 | 11 |

Hence, they will be straight angle at 9 : 16 | 4 | minutes. |

11 |

**(4)**The hands of a clock are m minutes apart between h and ( h + 1 )

o' clock at ( 5h ± m ) × | 12 | minutes past h. |

11 |

**Ex- **Find the time between 8 and 9 O’clock when the two hands of a clock are 4 minutes apart. **Solution**:- Here, h = 8 and m = 4

According to formula,

( 5h ± m ) × | 12 |

11 |

= ( 5 × 8 ± 4 ) × | 12 | = ( 40 ± 4 ) × | 12 |

11 | 11 |

= 44 × | 12 | and 36 × | 12 |

11 | 11 |

= 48 and | 432 | |

11 |

apart at 39 | 3 | and 48 minutes past 8 O’clock |

11 |

**(5)** If the minute hand of a clock overtakes the hour hand at intervals of m minutes of the correct time,

then the clock losses or gains | 720 | - m | ( 60 × 24 ) | minutes. | ||||

11 | m |

**Ex- **The minutes hand of a clock overtakes the hour hand at intervals of 70 min of the correct time. how much in a day does the clock gain or loss?**Solution:-** Here, m = 60 minutes

According to formula,

720 | - m | ( 60 × 24 ) | ||||

11 | m |

720 | - 70 | ( 60 × 24 ) | ||||

11 | 70 |

= - | (720 - 770) | × | ( 6 × 24 ) |

11 | 7 |

= - | 50 | × | ( 6 × 24 ) |

11 | 7 |

= | -7200 | minutes |

77 |

= | 7200 | minutes loss . |

77 |

## Calendar :-

A calendar is a chart which show the day, week and months of a particular year. A calendar consist of 365 or 366 days divide into 12 months.

#### Ordinary Year :-

A year in which having 365 days is called an ordinary year. For example- 2011, 2015, 2019 etc.

#### Leap Year

A year in which having 366 days is called leap year. For example- 2000, 2012, 2016 etc.

If a normal year is divisible by four is called leap year but in the case of century it must be divisible by 400.**For example**- 2000, 1600, 2400, are leap year ( divisible by 400 )

1700, 1800, 1900 are not leap year ( not divisible by 400 )

In a century, there are 76 ordinary year and 24 leap years.

**Ex**- Right now how many century leap year crossed?**Solution**:- 400, 800, 1200, 1600, 2000 = 5.

**Ex**- In 100 years how many leap year are there?

Solution:- |
100 | - 1 = 25 - 1 = 24 years . |

4 |

#### Leap years in century years :-

In 100 years = 24 leap years

In 200 Years = 24 × 2 = 48

In 300 years = 24 × 3 = 72

In 400 years = 24 × 4 = 96 + 1 = 97

In 500 years = 24 × 5 = 120 + 1 = 121

**Ex**- In 400 years how many times we will get the date of 29th?**Solution**:- In ordinary year 29th comes 11 times

11 × 400 = 4400

In 400 years 97 leap years

∴ Total 29th in 400 years = 4400 + 97 = 4497

**Ex**- In 400 years how many times we will get the date of 29th feb?**Solution**:- We know that, In 400 years 97 leap years

∴ In 400 years 97 29th feb get.

**Ex**- In between two consecutive leap year how many normal year?**Solution**:- Consider two consecutive leap year 1984 and 1988.

1984, 85, 86, 87, 1988 Means 3 normal year.

Now consider two consecutive leap year 1896 and 1904.

1896, 97,98,99, 1900, 1901,02, 03, 1904 Means 7 normal leap year.

∴ Between two consecutive leap year 3 and 7 normal year.

### Odd days :-

When we divide the number of days by 7, if remainder left that remainder is called odd days.

For example- How many odd days in 1 normal year.

In normal year 365 days,

odd days = | 365 | = 1 odd days |

7 |

odd days = | 366 | = 2 odd days |

7 |

**Ex**- In 100 years, how many odd days are there?**Solution**:- We know that

In 100 years 76 normal year = 76 × 1 = 76

In 100 years 24 leap year = 24 × 2 = 48

Odd days = 76 + 48

odd days = | 124 | = 5 odd days |

7 |

#### Fast Trick to Find number of odd days in given year :-

First find number of leap years in given year, then Odd days = | Given year | + | Leap year |

7 | 7 |

**Ex**- In 50 years, how many odd days are there?**Solution**:- In 50 years 12 leap years

odd days = | 50 | + | 12 | = 1 + 5 = 6 odd days |

7 | 7 |

**Ex**- In 17 years, how many odd days ?**Solution**:- In 17 years 4 leap year

odd days = | 17 | + 4 = 3 + 4 = | 7 | = 0 odd days. |

7 | 7 |

#### Number of odd days in century years :-

We know that,

In 100 years = 5 odd days

In 200 years = 2 × 5 = 10 = | 10 | = 3 odd days |

7 |

In 300 years = 3 × 5 = 15 = | 15 | = 1 odd days |

7 |

In 400 years = 4 × 5 = 20 + 1 = | 21 | = 0 odd days |

7 |

**Note :-**As 400th is a leap year, therefor 1 more day has been taken

5 3 1 0

100 200 300 400

500 600 700 800

900 1000 1100 1200

1300 1400 1500 1600

Number of odd days in 100, 400, 900, 1300 = 5

Number of odd days in 200, 600, 1000, 1400 = 3

Number of odd days in 300, 700, 1100, 1500 = 1

Number of odd days in 400, 800, 1200, 1600 = 0

**Ex**- Find odd days in 2015.**Solution**:- 2015 is a ordinary year.

We know that, in 1 ordinary year 1 odd day.

∴ In 2015 = 1 odd day

**Ex**- Find odd days till 2015.**Solution**:- 2015 = 2000 + 15

In 2000 years, odd days = 0

In 15 years, 3 leap year = | 15 | + 3 = 1 + 3 = 4 |

7 |

### Month Code :-

Months | Code |
---|---|

January | 1 |

February | 4 |

March | 4 |

April | 0 |

May | 2 |

June | 5 |

July | 0 |

August | 3 |

September | 6 |

October | 1 |

November | 4 |

December | 6 |

### Day's Code :-

Days | Code |
---|---|

Monday | 1 |

Tuesday | 2 |

Wednesday | 3 |

Thursday | 4 |

Friday | 5 |

Saturday | 6 |

Sunday | 7/0 |

### To Find a Particular Day Without Given Date and Day :-

#### Year between 1900 - 1999 :-

1.**If the given year is Normal year**

**Step -1** - Consider last 2 digits of the given year

Step-2 :- |
Last 2 digits | = Quotient |

4 |

**Step-3**- Month code

**Step-4**- Date

**Step-5**- Add step 1 + 2 + 3 + 4

Step-6 :- |
Step-5 | = Remainder will be Day's Code |

7 |

**Ex**- Find the day of the week on 15 August 1947?**Solution**:- **Step-1** :- 47 = 47

Step-2 :- |
47 | = 11 |

4 |

**Step-3**; - Month code of August = 3

**Step-4**:- Date = 15

**Step-5**;- Sum = 47 + 11 + 3 + 15 = 76

**Step-6**;- 76/7 = 6 Friday

∴ It was Friday on 15th August 1947.

2. **If the given year is Leap year**

In the case of | Jan | - ( -1 day ) Rest of the months i.e, March onward same as normal year. |

Feb |

**Ex**- Find the day of the week on 7 January 1904?**Solution**:- **Step-1**- 04

Step-2 :- |
04 | = 1 |

4 |

**Step-3**- Month code of January = 1

**Step-4**- Date = 7

**Step-5**- Sum = 4 + 1 + 1 + 7 = 13

Step-6 :- |
13 | = 6 |

7 |

**Step-7**- 6 - 1 = 5 Thursday

∴ It was Thursday on 7th January 1904.

#### Year between 2000-2099

1.**If the given year is Normal year**

**Step-1** - Consider last 2 digits of the given year

Step-2 :- |
Last 2 digits | = Quotient |

4 |

**Step-3**- Month code

**Step-4**- Date

**Step-5**- Add step 1 + 2 + 3 + 4

Step-6 :- |
Step-5 | = Remainder |

7 |

**Step-7**- Remainder - 1 = will be Day's Code

**Ex**-Find the day of the week on 20 January 2018?**Solution**:- **Step-1**- 18

Step-2 :- |
18 | = 4 |

4 |

**Step-3**- Month code of January = 1

**Step-4**- Date = 20

**Step-5**- Sum = 18 + 4 + 1 + 20 = 43

Step-6 :- |
43 | = 1 |

7 |

**Step-7**- 1 - 1 = 0 Saturday

∴ It was Saturday on 20 January 2018.

2.If the given year is Leap year In the case of |
Jan | - ( -2 days ) . |

Feb |

**Ex**- Find the day of the week on 8 February 2008?**Solution**:- **Step-1**- 08

Step-2 :- |
08 | = 2 |

4 |

**Step-3**- Month code of January = 4

**Step-4**- Date = 8

**Step-5**- Sum = 08 + 2 + 4 + 8 = 22

Step-6 :- |
22 | = 1 = 8 |

7 |

**Step-7**- 8 - 2 = 6 Friday

∴ It was Friday on 08 February 2008.

#### Year between 1800-1899

1.I**f the given year is Normal year**

**Step-1** - Consider last 2 digits of the given year

Step-2 :- |
Last 2 digits | = Quotient |

4 |

**Step-3**- Month code

**Step-4**- Date

**Step-5**- Add step 1 + 2 + 3 + 4

Step-6 :- |
Step-5 | = Remainder |

7 |

**Step-7**- Remainder + 2 = will be Day's Code

**Ex**- Find the day of the week on 7 October 1807?**Solution**:- **Step-1**- 07

Step-2 :- |
07 | = 1 |

4 |

**Step-3**- Month code of January = 1

**Step-4**- Date = 7

**Step-5**- Sum = 07 + 1 + 1 + 7 = 16

Step-6 :- |
16 | = 2 |

7 |

**Step-7**- 2 + 2 = 4 Wednesday

∴ It was Wednesday on 07 October 1807.

2.If the given year is Leap year In the case of |
Jan | - ( + 1 day ) . |

Feb |

**Ex**- Find the day of the week on 8 July 1806?**Solution**:- **Step-1**- 06

Step-2 :- |
06 | = 2 |

4 |

**Step-3**- Month code of July = 0

**Step-4**- Date = 8

**Step-5**- Sum = 06 + 2 + 0 + 8 = 16

Step-6 :- |
16 | = 2 |

7 |

**Step-7**- 2 + 1 = 3 Tuesday

∴ It was Tuesday on 08 July 1806.

### Day Gain/Loss :-

#### Ordinary Year ( ± 1 day )

1.When we proceed forward by 1 yr, then 1 day is gained.

For example- If 11th August 2013 is Sunday, then 11th August 2014 has to be Sunday + 1 = Monday.

2. When we move backward by 1 yr, then 1 day is lost.

For example- If 27th December 2013 is Friday, then 24th December 2012 has to be Friday - 1 = Thursday.

#### Leap Year ( ± 2 day )

1.When we proceed forward by 1 leap year, then 2 days are gained.

For example- If 29th December 2011 is Sunday, then 29th December 2012 has to be Sunday + 2 days = Tuesday.

2. When we move backward by 1 leap year, then 2 days are lost.

For example- If 22nd December 2012 is Sunday, then 22nd December 2011 has to be Sunday - 2 days = Friday.

### Special Case

2 days after Monday = + 2 = Wednesday

3 days after Monday = + 3 = Thursday

2 days before Monday = -2 = Saturday

3 days before Monday = -3 = Friday