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Test: Calendar 4 - UPSC MCQ


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20 Questions MCQ Test UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making - Test: Calendar 4

Test: Calendar 4 for UPSC 2024 is part of UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making preparation. The Test: Calendar 4 questions and answers have been prepared according to the UPSC exam syllabus.The Test: Calendar 4 MCQs are made for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Calendar 4 below.
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Test: Calendar 4 - Question 1

What was the day of the week on 16th August, 1947?

Detailed Solution for Test: Calendar 4 - Question 1

15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th )
Counting of odd days:
1600 years have 0 odd day. 300 years have 1 odd day.
47 years = (11 leap years + 36 ordinary years)
= [(11 x 2) + (36 x 1) ] odd days
= 58 odd days
= 2 odd days
Jan   Feb   Mar   Apr   May   Jun   Jul   Aug
= 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15
= 227 days
= (32 weeks + 3 days)
= 3,
Total number of odd days
= (0 + 1 + 2 + 3) odd days
= 6 odd days.
Hence, the required day was 'Saturday'.

Test: Calendar 4 - Question 2

Prove that any date in March of a year is the same day of the week corresponding date in November that year.

Detailed Solution for Test: Calendar 4 - Question 2

We will show that the number of odd days between last day of February and last day of October is zero.
March   April   May   June   July   Aug.   Sept.   Oct.
31 + 30 + 31 + 30 + 31 + 31 + 30 + 31
= 241 days
= 35 weeks
= 0 odd day
Number of odd days during this period = 0.
Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows

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Test: Calendar 4 - Question 3

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

Detailed Solution for Test: Calendar 4 - Question 3

350 days = (350/7 = 50 weeks) i.e No odd days,
So it will be a Saturday.

Test: Calendar 4 - Question 4

The calendar for the year 1988 is same as which upcoming year ?

Detailed Solution for Test: Calendar 4 - Question 4

We already know that the calendar after a leap year repeats again after 28 years.
Here 1988 is a Leap year, then the same calendar will be in the year = 1988 + 28 = 2016.

Test: Calendar 4 - Question 5

Given that on 9th August 2016 is Saturday. What was the day on 9th August 1616 ?

Detailed Solution for Test: Calendar 4 - Question 5

We know that, After every 400 years, the same day occurs.
Thus, if 9th August 2016 is Saturday, before 400 years
i.e., on 9th August 1616 has to be Saturday.

Test: Calendar 4 - Question 6

Second & fourth Saturdays and every Sunday is a holiday. How many working days will be there in a month of 31 days beginning on a Friday ?

Detailed Solution for Test: Calendar 4 - Question 6

Given that the month begins on a Friday and has 31 days
Sundays = 3rd, 10th, 17th, 24th, 31st
⇒ Total Sundays = 5
Every second & fourth Saturday is holiday. 2nd & 4th Saturday in every month = 2
Total days in the month = 31
Total working days = 31 - (5 + 2) = 24 days.

Test: Calendar 4 - Question 7

On 17th March, 1997 Monday falls. What day of the week was it on 17th March, 1996?

Detailed Solution for Test: Calendar 4 - Question 7

The year 1996 is a leap year. So, it has 2 odd days. But 17 th March  comes after 29 th February. So, the day on 17 th March, 1997 will be 1 day  beyond the day on 17 th March, 1996.


Here 17 th March, 1997 is Monday. So, 17 th March, 1996 is a Sunday.

Test: Calendar 4 - Question 8

Which year has 366 days?

Detailed Solution for Test: Calendar 4 - Question 8

When a century year leaves a remainder 0, when divided by 400 then it is a leap year (366 days).
So, 1200 has 366 days.

Test: Calendar 4 - Question 9

What is 90 days from today?
(Hints : Today is 20th January 2017, Sunday)

Detailed Solution for Test: Calendar 4 - Question 9

Given Today is 20th January 2017, Sunday
In january, we have 31 days
February - 28 days (Non leap year)
March - 31 days
April - 30 days
⇒ Remaining days = 31 - 20 = 11 in Jan
11 in Jan + 28 in Feb + 31 in Mar = 11 + 28 + 31 = 70 days
More 20 days to complete 90 days ⇒ upto 20th April
Therefore, after 90 days from today i.e, 20th Jan 2017 is 20th Apr 2017.
Now, the day of the week will be
90/7 ⇒ Remainder '6'
As the day starts with '0' on sunday
6 ⇒ Saturday.
Required day is 20th April, Saturday.

Test: Calendar 4 - Question 10

On 24th Nov, 2007 Thursday falls. What day of the week was it on 10th Nov, 2006 ?

Detailed Solution for Test: Calendar 4 - Question 10

The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 24th Nov, 2007 will be 1 day beyond the day on 24
th Nov, 2006.
But, 24th Nov, 2007 is Thursday.
24 - 10 = 14 days.
Therefore, 2 weeks ago it is same day.
Thus, 10th Nov, 2006 is one day before 10th Nov, 2007 i.e. it is Wednesday.

Test: Calendar 4 - Question 11

The year next to 2003 will have the same calendar as that of the year 2003?

Detailed Solution for Test: Calendar 4 - Question 11

Given year 2003, when divided by 4 leaves a remainder of 3.
NOTE: When remainder is 3, 11 is added to the given year to get the result.
So, 2003 + 11 = 2014

Test: Calendar 4 - Question 12

On 19th June, 1984 Monday falls. What day of the week was it on 19th June, 1985?

Detailed Solution for Test: Calendar 4 - Question 12

The year 1985 is an ordinary year. So, it has 1 odd day.
So, the day on 19th June, 1985 will be 1 day after the day on 19th June, 1984.
But, 19th June,1984 is Monday
So, 19th June, 1985 is Tuesday.

Test: Calendar 4 - Question 13

Suppose today is Friday, what day of the week will it be 65 days from now?

Detailed Solution for Test: Calendar 4 - Question 13

The day of the week repeats every 7 days.
Given today is Friday. Again Friday is repeated on the 7th day, 14th,... on 7 multiple days.
Hence, Friday is on the 63rd day, as 63 is multiple of 7.
Now, the required day of the week on the 65th day is Sunday.

Test: Calendar 4 - Question 14

How many seconds in 10 years?

Detailed Solution for Test: Calendar 4 - Question 14

We know that,
1 year = 365 days
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds.
Then, 1 year = 365 x 24 x 60 x 60 seconds. = 8760 x 3600
1 year = 31536000 seconds.
Hence, 10 years = 31536000 x 10 = 315360000 seconds.

Test: Calendar 4 - Question 15

The calendar of year 1939 is same as which year?

Detailed Solution for Test: Calendar 4 - Question 15

Given year 1939, when divided by 4 leaves a remainder of 3.
NOTE: When remainder is 3, 11 is added to the given year to get the result.
So, 1939 + 11 = 1950

Test: Calendar 4 - Question 16

What is two weeks from today?

Detailed Solution for Test: Calendar 4 - Question 16

We know that the day repeats every 7 days, 14 days, 21 days, .........
So if today is Monday, after 7 days it is again Monday, after 14 days again it is Monday.
Hence, after 2 weeks i.e, 14 days the day repeats and is the same day.

Test: Calendar 4 - Question 17

What will be your age in the year 2019 if you were born in 1995?

Detailed Solution for Test: Calendar 4 - Question 17

Calculating Age has 2 conditions. Let your Birthday is on January 1st.
1. If the month in which you are born is completed in the present year i.e, your birthday, then
Your Age = Present year - Year you are born
As of now, present year = 2018
i.e, Age = 2019 - 1995 = 24 years.
2. If the month in which you are born is not completed in the present year i.e, your birthday, then
Your Age = Last year - Year you are born
As of now, present year = 2019
i.e, Age = 2018 - 1995 = 23 years.

Test: Calendar 4 - Question 18

What was the day on 2nd Jan 1901?

Detailed Solution for Test: Calendar 4 - Question 18

2nd Jan 1901 means
(1900 years and 2 day)
Now, 1600 years have 0 odd day
300 years have 1 odd day
2 days has 2 odd day
Total no. of odd days = 0 + 1 + 2 = 3 days
Hence, the day on 2nd Jan 1901 was Wednesday.

Test: Calendar 4 - Question 19

What was the day on 16th June,1993?

Detailed Solution for Test: Calendar 4 - Question 19

16 June, 1993 = (1992 years + Period from 1.1.1993 to 16.6.1993)
Odd days in 1600 years = 0
Odd days in 300 years = 1
92 years = (69 ordinary years + 23 leap year) = (69 x 1 + 23 x 2)= 3 odd days
Jan. Feb. March April May June
(31 + 28 + 31 + 30 + 31 + 16 ) = 167 days
167 days = (23 weeks + 6 days) =6 odd days.
Total number of odd days = (0 + 1 + 3 + 6) = 3 odd days.
Given day is Wednesday.

Test: Calendar 4 - Question 20

Which century year is a leap year?

Detailed Solution for Test: Calendar 4 - Question 20

For an century year,it is divided by 400,when leaves a remainder 0 then it is a leap year. So, 1600

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