Study Notes for Part A: Mastering the Concept of Clock & Calendar (Download PDF)

By Renuka Miglani|Updated : June 21st, 2023

Hi Aspirants,

With only a few days left for CSIR NET 2023 exam, you should devote all your time in revising your concepts and topics. Practice as many mocks as you can to work on your speed and accuracy. At this high time, you should practice a lot of questions of those topics which have high chances of being asked in the exam. Clock & Calendar is one of the important and most scoring topics of the General Aptitude (Part A)  through which one can score full marks in this section.

To help you in your preparation, we have come up with the concept of the Clock & Calendar for the upcoming exam CSIR-NET 2021. Scroll down the article below and learn this concept to crack the exam.

Concept of Reasoning: Clock & Calender

CLOCK

 Important formula and equations:

  • A clock has two hands, hour hand and minute hand – the hour hand is the smaller one and the minute hand is the larger one.
  • A clock is a complete circle having 36°. It is divided into twelve equal parts that is each part is = 30°.
  • A minute hand takes one hour to complete a round and it covers 360° in 60 min. So, in 1 min. it covers = 6°/ min.
  • After every one hour both the hands coincide once. In 12 hours, both the hands coincide 11 times since between 1 1 and 1, they coincide only once i.e., at 12 o’clock.
  • When the two hands are at right angles, they are 15 min spaced apart. The hands are in the same straight line when they are coincident or opposite to each other.
  • In 1 hour, they will form two right angles and in 12 hours they will form 22 right angles.
  • The angle formed by both the minute and hour hand at 3 o’clock and 9 o’clock.
  • When the hands are in opposite directions, they are 30 min spaces apart.
  • If a clock indicates 7.15, when the correct time is 7, it is said to be 15 min too fast. Also, if a clock indicates 6.45, when the correct time is 7, it is said to be 15 min too slow. The minute hand gains 55 min over hour hand per hour.
  • 22 times in a day, the hands of a clock will be in a straight line but opposite in direction. 22 times in a day, the hands of a clock coincide.
  • 44 times in a day, the hands of a clock will be straight.
  • 44 times in a day, the hands of a clock are at right angles.

Ø Some common type of questions asked in the exam-

  1. To find the angle between the two hands (i.e., minute hand and hour hand) at a given time.
  2. To find the time when the angle between the two hands is given.
  3. Questions on losing or gaining time.
  4. Mirror image of clock –Just subtract the given time from 11:60 OR 23:60

Below we are going to share the types of questions with examples to explain it.

Q1. What is the angle between the two hands of a clock when the time shown by the clock is 8 p.m.? (in degrees).

  1. 120
  2. 150
  3. 60
  4. 50

Ans. A

Sol. 60 minutes made by minute hand complete the one round of 360°. So, 1 minute by minute hand makes = 360/60 = 6°.

At 08:00 PM, the minute hand will make 0 angle.

Now, 12 hours made by hour hand makes one round of 360°. So, 1 hour made by hour hand makes = 360/12 = 30°.

Now, the degree made by hour hand at 8 will be = 30 × 8 = 240°.

So, at 08:00 PM there will be two angles between minute and hour hand which are obtuse angle (240°) and acute angle (120°).

So, the correct option is (A).

Q2. Find the angle between the minute hand and the hour hand of a clock when the time is 5.10?

  1. 65°
  2. 45°
  3. 75°
  4. 95°

Ans. D

Sol. So, Angle by hour hand in 5.10 hrs. =  

Angle by minute hand in 10 min = × 10 = 60

Hence, angle between the minute hand and the hour hand of a clock when the time is; 5.10 = 155 – 60 = 95.

So, the correct option is (D).

Q3. The clock starts at noon. By 40 minutes past 4, the hour hand has turned through the angle is?

  1. 140°
  2. 145°
  3. 175°
  4. 155°

Ans. A

 Sol. Angle made by hour hand in 12 hours = 360°

Therefore, Angle made by hour hand in 4 hours 40 minutes i.e. 4 and 2/3 hours

= 360/12 × 14/3

= 30 × 14/3

= 10 × 14 = 140°

So, the correct option is (A).

Q4. If time in a watch is 09.20. What will be the mirror image of this time?

  1. 2.60
  2. 1.50
  3. 2.40
  4. 3.40

Sol. 11.60-9.20=2.40

 

CALENDAR

Important formula and equations:

  • As we know, a Calendar measures a day, a week, a month and a year. In an ordinary year, there are 365 days. The year which is not a leap year is an ordinary year.
  • Leap year: a year which is divisible by 4, if it is not a century year.
  • Odd days: In a given period, the number of days is more than the complete weeks. Counting of odd days -In an ordinary year, there are 365 days and in that there are (52 weeks + 1 day). This additional day is called an odd day.

So, in 1 Leap year there are 366 days which means (52 weeks + 2 days) this shows that there are 2 odd days in 1 leap year.

  • In 100 years = 76 ordinary years (non- leap years) + 24 leap years

= 76 × 1 + 24 × 2 = 124 odd days.

Since, 7 odd days make a week, so by dividing 124 by 7 we get the number of odd days in 100 years that will be 5 (which is the remainder).

Thus, calculating as above; the number of odd days in 200 years is 5 × 2 = 10 – 7 (1 week)

= 3 odd days.

  • The number of odd days in 300 years is 5 × 3 = 15 – 14 (2 weeks) = 1 odd day.
  • The number of odd days in 400 years is {5 × 4 + 1 (leap century) – 21} = 0 odd days. Likewise, each one of 800 years, 1200 years, 1600 years, 2000 years and so on has 0 odd days.

Given below is the table showing days of the week related to odd days.

No. of days

Days

0 or 7

Sunday

1

Monday

2

Tuesday

3

Wednesday

4

Thursday

5

Friday

6

Saturday

 Given below is the table showing months and their code.

Month

Code

Jan

1

Feb

4

Mar

4

Apr

0

May

2

Jun

5

Jul

0

Aug

3

Sep

6

Oct

1

Now

4

Dec

6

  • In a simple year, months have the same first day; Jan– Oct, Feb – Mar- Nov, Apr- Jul. In a leap year months those have the same first day; Jan-Apr-Jul, Feb-Aug, Mar-Nov.
  • The 1st and the last day of a simple year remain the same as First Jan is on Monday. Thus, the 31st Dec of that year will be on Monday as well.

Below we are going to share the types of questions with examples to explain it-

Q1. Mahatma Gandhi died on 30th January 1948. What was the day on which he died?

  1. Monday
  2. Thursday
  3. Saturday
  4. Friday

Ans. D

Sol. Up to 1600 AD we have 0 odd days; up to 1900 AD we have 1 odd day. Now in 47 years we have 11 leap years and 36 normal years.

Odd days from 1901 to 1947 = (11 × 2 + 36 × 1) = (22 + 36) = 58 odd days = 8 weeks +

2 odd days.

Total odd days up to 31st December 1947 is (1 + 2) = 3 odd days. 30 days of January contain only 4 weeks + 2 odd days.

So, 30th January 1948 has a total of 5 odd days. Day on 30th January 1948 = Friday.

So, the correct option is (D).

Q2. Aniruddh's birthday is on Sunday 2nd April. On what day of the week will be Priyansh's Birthday in the same year if Priyansh was born on 28th October?

  1. Saturday
  2. Wednesday
  3. Thursday
  4. Friday

Ans. A

Sol.

Month

Number of days

April

30

May

31

June

30

July

31

August

31

September

30

 Days left in April = 30 – 2 = 28 Days in October to count = 28

Thus total number of days from Aniruddh's birthday to Priyansh’s birthday = 28 + 31 + 30

+ 31 + 31 + 30 + 28 = 209

Thus number of odd days = 6

As, Aniruddh's birthday is on Sunday, Priyansh’s birthday will on Sunday + 6 days = Saturday.

So, the correct option is (A).

Q3. If 23 January 2005 was a Thursday, then what day was 22 January 2000?

  1. Wednesday
  2. Thursday
  3. Friday
  4. Sunday

Ans. A

Sol- Number of days between 22 January 2000 & 23 January 2005

365 × 3 + 366 × 2 – 22 + 23 = 1828

261 weeks and 1 day.

Since 23 January 2005 was given to be Thursday. 22 January 2000 will be Thursday - 1 = Wednesday. So, the correct option is (A).

 Download the PDF on Clock and Calendar - Click Here to Download

 

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