# QUESTIONS ON CALENDAR

### QUESTIONS ON CALENDAR

Most of these questions are taken from the previous year’s examinations conducted by the Staff Selection Commission (SSC) on Calendar of the Quantitative Aptitude (Maths) section of the following exams as well as other similar exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I & Tier-II

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Paper-I & Paper-II

SHIV KISHOR

#### Some important rules regarding calendar problems

The year consists of 365 days, 5 hours, 48 minutes (52 weeks and 1 odd day). An extra day is added once in every fourth year which was called the leap year, which has 366 days (52 weeks and 2 odd days).

#### To find the day of any given date of the year, you need to understand the calendar calculations:

1. First thing to remember, first January 1 AD was Monday, therefore, we must count days from Sunday. This means the 0th day was Sunday, so the 7th day was Sunday again and so on and so forth.

2. The day gets repeated after every seventh day (concept of a week), if today is Monday, then 28th day from now will also be Monday as it a multiple of 7 (28/7 = 4, so four weeks). Here the 30 day will be calculated by 30/7, which is 4 weeks and 2 days, these two days are called odd days. With starting day as Monday and two odd days, the day will be Wednesday; this point is the most critical in calendars. The other of looking at it is since the 28th day is Monday, so the 30th day will be Wednesday. But you have to understand and use the concept of odd days as the question may be about thousands of years.

3. In a normal year there are 365 days so 52 weeks and 1 odd day, in a leap year there are 366 days so 52 weeks and 2 odd days.

4. In 100 years there are 24 leap years and 76 normal years, so the number of odd days are 24*2 + 76 = 124, which is 17 weeks + 5 odd days, so 100 years have 5 odd days. Here you must remember that, though every century is divisible by 4, it’s not a leap year unless divisible by 400. Thus none of 100, 200, 300, 500, 600, 700, etc. is a leap year.

5. In 200 years the number of odd days is twice the number in 100 years which is 10, which is one week and 3 odd days, so 200 years have 3 odd days.

6. In 300 years, the number of odd days is 15, which is two weeks and 1 odd day, so 300 years have one odd day.

7. 400 year is a leap year; similarly the multiples of 400 are also leap years.

8. In 400 years, the number of odd days become 20 + 1(from the leap year), so total days are 21, which is three weeks and 0 odd days. In 400 years there are 0 odd days

#### QUERY 1

What was the day on 25th January, 1975?

A) Sunday
B) Thursday
C) Saturday
D) Friday

SHIV KISHOR
First we’ll count odd days of completed year, which are 1974 here
So, from the above explanation they can be grouped like this: 1600 + 300 + 74

Counting of odd days
In 1600 years = 0
In 300 years = 1
In 74 years = 1 (there are 18 leap years and 56 normal years, so the odd days are: 18*2 + 56*1 = 36 + 56 = 92, which is 13 weeks and 1 odd day)
In 25 days = 4 (there are 3 weeks and 4 odd days)

So, Total of odd days = 0 + 1 + 1 + 4 = 6; means Saturday (option ‘C’)

#### QUERY 2

How many times does the 29th day of the month occur in 400 consecutive years?

A) 4497 times
B) 5500 times
C) 3024 times
D) 4400 times

SHIV KISHOR
Number of leap years in 400 = 24*3 +25 = 97 years.
Hence, in 400 consecutive years February has the 29th day 97 times, and the remaining eleven months have the 29th day = 400 × 11 = 400 times.

Thus the 29th day of the month occurs
= 97 + 4400 = 4497 times (option ‘A’)

#### QUERY 3

Today is 3rd November. The day of the week is Monday. This is a leap year. What will be the day of the week on this date after 3 years?

A) Monday
B) Sunday
C) Friday
D) Thursday

SHIV KISHOR
As this is a leap year, so none of the next 3 years will be leap years.

Now number of odd days in these three years = 1*3 = 3
So the day after 3 years = 3rd beyond Monday i.e. it will be Thursday (option ‘D’)

#### QUERY 4

December 9, 2001 is Sunday. What was the day on December 9, 1971?

A) Thursday
B) Wednesday
C) Saturday
D) Sunday

SHIV KISHOR
Number of years = 2001 – 1971 = 30
Number of leap years = As the first leap year is 1972, they will be 8
Therefore the number of normal years = 30 – 8 = 22

Thus, odd days = 22*1 + 8*2 = 38 = remainder of 38/7 = 3

Hence the required day will be 3 days before Sunday; means THURSDAY (option ‘A’)

#### QUERY 5

The calendar of year 1982 is same as which year?

A) 1987
B) 1989
C) 1991
D) 1993

SHIV KISHOR
Remember that the calendar repeats after either 6 years or 11 years from a non-leap year. If the year comes just after a leap year the calendar will repeat after 6 years, otherwise 11 years. The year before 1982 is not a leap year, the calendar of this year, therefore, will repeat after 11 years.

Now 11th year from 1982 = 1982 +11 = 1993, so it’s the answer (option ‘D’)

#### QUERY 6

If today is Monday, what will be the day 350 days from now?

A) Thursday
B) Wednesday
C) Monday
D) Sunday

SHIV KISHOR
Number of odd days in 350 days = remainder of 350/7 = 0, means no odd days

So it will be Monday itself after 350 days from now (option ‘C’)

#### QUERY 7

If today is Monday, what will be the day one year and 50 days from now?

A) Thursday
B) Wednesday
C) Monday
D) Cannot be determined

SHIV KISHOR
Cannot be determined, as we don’t know if it is a leap year or not (option ‘D’)

#### QUERY 8

The calendar for the year 1984 is same as which upcoming year?

A) 2002
B) 2012
C) 2020
D) 2022

SHIV KISHOR
Remember the calendar repeats after 28 years after a leap year
As 1984 is a leap year, the same calendar will be in the year = 1984 + 28 = 2012 (option ‘B’)

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