Test: Calendars- 1 - UPSC MCQ

# Test: Calendars- 1 - UPSC MCQ

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## 10 Questions MCQ Test CSAT Preparation - Test: Calendars- 1

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Test: Calendars- 1 - Question 1

### If 15 March 1816 was Friday, what day of the week would 15th April 1916 be?

Detailed Solution for Test: Calendars- 1 - Question 1

We are given that 15th March 1816 was a Friday.

Now we know that 100 years have 5 odd days. So till 15th March 1916, we will be having 5 odd days.

So if we move from 15th March 1816 to 15th March 1916, we will encounter 5 odd days.

Now from 15th March 1916 to 15th April 1916 there would be 3 odd days.

So total number of odd days = 5 + 3 = 8

8 mod 7 = 1

So 15th April 1916 would be Friday + 1 = Saturday

Test: Calendars- 1 - Question 2

### A year 1991 is having a same calendar as that of the year X. Which of the following is a possible valueof X.

Detailed Solution for Test: Calendars- 1 - Question 2

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Test: Calendars- 1 - Question 3

### If 28th August 1946 was a Wednesday, what day of the week was 31 August 1961?

Detailed Solution for Test: Calendars- 1 - Question 3

It is given that 28th August 1946 was Wednesday.

From 28th August 1946 to 28th August 1961, we have 4 leap years and 11 normal years.

So the number of odd days would be 11*1 + 4*2 = 19

Now the date which is asked is 31 Aug 1961. So if we move from 28th August to 31st August, we will have 3 more odd days.

So total number of odd days = 5 + 3 = 8

Now 8 mod 7 = 1 .

So 31st August 1961 would be Wednesday + 1 = Thursday.

Test: Calendars- 1 - Question 4

If 09/12/2001(DD/MM/YYYY) happens to be Sunday, then 09/12/1971 would have been a

Detailed Solution for Test: Calendars- 1 - Question 4

30 years. The number of leap years is 8 (1972,1976,1980,1984,1988,1992,1996,2000).

So, the total number of days = 22*365 + 8*366 = 10958

10958 mod 7 = 3

Since 9/12/2001 is a Sunday, 9/12/1971 should be a Thursday.

Test: Calendars- 1 - Question 5

In 2016, Mohan celebrated his birthday on Friday. Which will be the first year after 2016 when Mohan will celebrate his birthday on a Wednesday? (He was not born in January or February)

Detailed Solution for Test: Calendars- 1 - Question 5

Since it has been mentioned that Mohan was not born in February, so he can’t be born on 29th Feb.

Hence He will celebrate his next birthday on a Wednesday in the year for which the sum of the odd days becomes 5 or a multiple of 5.

By his birthday in 2017, there will be 1 odd day.

By his birthday in 2018, there will be 2 odd days.

By his birthday in 2019, there will be 3 odd days.

By his birthday in 2020, there will be 5 odd days, as 2020 is a leap year.

So in 2020 He will celebrate his birthday on Wednesday.

Test: Calendars- 1 - Question 6

If 10th May, 1997 was a Monday, what was the day on Oct 10, 2001?

Detailed Solution for Test: Calendars- 1 - Question 6

In this question, the reference point is May 10, 1997 and we need to find the number of odd days from May 10, 1997 up to Oct 10, 2001.

► Now, from May 11, 1997 - May 10, 1998 = 1 odd day
► May 11, 1998 - May 10, 1999 = 1 odd day
► May 11, 1999 - May 10, 2000 = 2 odd days (2000 was leap year)
► May 11, 2000 - May 10, 2001 = 1 odd day
► Thus, the total number of odd days up to May 10, 2001 = 5.
► The remaining 21 days of May will give 0 odd days.
► In June, we have 2 odd days; in July, 3 odd days; in August, 3 odd days; in September,2 odd days and up to 10th October, we have 3 odd days. Hence, total number of odd days = 18 i.e. 4 odd days.
Since, May 10, 1997 was a Monday, and then 4 days after Monday would be Friday. So, Oct 10, 2001 was Friday.

Test: Calendars- 1 - Question 7

Find the leap year?

Detailed Solution for Test: Calendars- 1 - Question 7

Remember the leap year rule:

• Every year divisible by 4 is a leap year, if it is not a century.
• Every 4th century is a leap year, but no other century is a leap year.
• 800,1200 and 2000 comes in the category of 4th century (such as 400,800,1200,1600,2000 etc).

Hence, 800,1200 and 2000 are leap years.

Test: Calendars- 1 - Question 8

The century can end with:

Detailed Solution for Test: Calendars- 1 - Question 8
• 100 years contain 5 odd days.
∴ Last day of 1st century is Friday.
• 200 years contain (5 x 2) ≡ 3 odd days.
∴ Last day of 2nd century is Wednesday.
• 300 years contain (5 x 3) = 15 ≡ 1 odd day.
∴ Last day of 3rd century is Monday.
• 400 years contain 0 odd day.
∴ Last day of 4th century is Sunday.

This cycle is repeated.
∴ Last day of a century cannot be Tuesday or Thursday or Saturday.

Test: Calendars- 1 - Question 9

What was the day on February 9, 1979?

Detailed Solution for Test: Calendars- 1 - Question 9
• We know that in 1600 years, there will be 0 odd days. And in the next 300 years, there will be 1 odd day.
• From 1901 to 1978 we have 19 leap years and 59 non-leap years.
• So, the total number of odd days up to 31st Dec. 1978 is 19 x 2 + 59 = 97. On dividing 97 by 7 we get 6 as the remainder, which is the total number of odd days in these years.
• So, till 31st Dec. 1978 we have 1 + 6 = 7 odd days, which forms one complete week. Now, in 1979, we have 3 odd days in January, and 2 odd days in the month of February (up to 9th Feb).
• So, the total odd days are 3 + 2 = 5.
• Hence, 9th February 1979 was a Friday.
Test: Calendars- 1 - Question 10

Which calendar year will be same as the year 2008?

Detailed Solution for Test: Calendars- 1 - Question 10

For every 28 years, the calendars will same,
so the years 2008,2036 have the same calendar as 1980.

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