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This mock test of Test: Calendars- 1 for UPSC helps you for every UPSC entrance exam.
This contains 10 Multiple Choice Questions for UPSC Test: Calendars- 1 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Calendars- 1 quiz give you a good mix of easy questions and tough questions. UPSC
students definitely take this Test: Calendars- 1 exercise for a better result in the exam. You can find other Test: Calendars- 1 extra questions,
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QUESTION: 1

The century can end with:

Solution:

- 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.**

QUESTION: 2

Find the leap year?

Solution:

__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.

QUESTION: 3

What was the day on February 9, 1979?

Solution:

- 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**.

QUESTION: 4

What is the day on July 2 1985?

Solution:

**Every year has one odd day and a leap year has 2 odd days.**- Though 1984 is a leap year, we don't have Feb 29 in the required period.
- So, we get only one odd day and as we are moving back we get
**Tuesday**as the answer.

QUESTION: 5

India got independence on 15th August 1947. What was the day on that date?

Solution:

**We shall first calculate the number of odd days till 31 December 1946:**

- Number of odd days in the first 1600 years = 0 odd day
- Number of odd days in the next 300 years = 1 odd day
- Now, 46 years had 11 leap years and 35 ordinary years.
- The number of odd days in 46 years = (2 × 11) + (1 × 35) = 22 + 35 = 57 = 8 weeks and 1 odd day.

**Now, we shall calculate the number of odd days in 1947 till 15 August:
Month (Days):**

- January(31)
- February(28)
- March(31)
- April(30)
- May(31)
- June(30)
- July(31)
- August(15)

Days = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 **i.e.** 32 weeks and 3 odd days

So, the total number of odd days till 15 August 1947 = 0 + 1 + 1 + 3 = 5

On counting five days from Monday, we get Friday.

Therefore, **15 August 1947** was a** Friday**.

QUESTION: 6

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

Solution:

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.**

QUESTION: 7

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 ?

Solution:

Given that the month begins on a Friday and has 31 days

- Sundays = 3
^{rd}, 10^{th}, 17^{th}, 24^{th}, 31^{st}

⇒ Total Sundays = 5

Every second & fourth Saturday is holiday. - 2
^{nd}& 4^{th}Saturday in every month = 2 - Total days in the month = 31
- Total working days = 31 - (5 + 2) =
**24 days**

QUESTION: 8

Which calendar year will be same as the year 2008?

Solution:

For every **28 years, the calendars will same**,

so the years **2008,2036** have the same calendar as 1980.

QUESTION: 9

Today is Monday. After 61 days, it will be:

Solution:

- Each day of the week is repeated after
**7 days.** - So, after 63 days, it will be Monday.
- After
**61 days**, it will be**Saturday.**

QUESTION: 10

The day of the 5th november is equal to the day of the date in the same year?

Solution:

We will show that the number of odd days between the last day of February and the last day of October is zero.

- March, April, May, June, July, Aug, Sept, Oct ,
**i.e.**31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 = 241 days = 35 weeks = 0 odd day. - Number of odd days during this period = 0.

Thus, **5th March** of a year will be the same day as 5th November of that year.

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