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# Test: Calendars- 1

## 10 Questions MCQ Test Quantitative Aptitude for Banking Preparation | Test: Calendars- 1

Description
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, long questions & short questions for UPSC on EduRev as well by searching above.
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 = 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
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.