Test: Calendars- 1

# Test: Calendars- 1

Test Description

## 10 Questions MCQ Test CSAT Preparation for UPSC CSE | Test: Calendars- 1

Test: Calendars- 1 for Railways 2023 is part of CSAT Preparation for UPSC CSE preparation. The Test: Calendars- 1 questions and answers have been prepared according to the Railways exam syllabus.The Test: Calendars- 1 MCQs are made for Railways 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Calendars- 1 below.
Solutions of Test: Calendars- 1 questions in English are available as part of our CSAT Preparation for UPSC CSE for Railways & Test: Calendars- 1 solutions in Hindi for CSAT Preparation for UPSC CSE course. Download more important topics, notes, lectures and mock test series for Railways Exam by signing up for free. Attempt Test: Calendars- 1 | 10 questions in 10 minutes | Mock test for Railways preparation | Free important questions MCQ to study CSAT Preparation for UPSC CSE for Railways Exam | Download free PDF with solutions
 1 Crore+ students have signed up on EduRev. Have you?
Test: Calendars- 1 - Question 1

### Find the leap year?

Detailed Solution for Test: Calendars- 1 - Question 1

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 2

### The century can end with:

Detailed Solution for Test: Calendars- 1 - Question 2
• 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 3

### What is the day on July 2 1985?

Detailed Solution for Test: Calendars- 1 - Question 3
• 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.
Test: Calendars- 1 - Question 4

What was the day on February 9, 1979?

Detailed Solution for Test: Calendars- 1 - Question 4
• 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 5

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

Detailed Solution for Test: Calendars- 1 - Question 5

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 6

Which calendar year will be same as the year 2008?

Detailed Solution for Test: Calendars- 1 - Question 6

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

Test: Calendars- 1 - Question 7

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

Detailed Solution for Test: Calendars- 1 - Question 7

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.

Test: Calendars- 1 - Question 8

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: Calendars- 1 - Question 8

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: Calendars- 1 - Question 9

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

Detailed Solution for Test: Calendars- 1 - Question 9
• 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.
Test: Calendars- 1 - Question 10

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

Detailed Solution for Test: Calendars- 1 - Question 10

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.

## CSAT Preparation for UPSC CSE

72 videos|69 docs|92 tests
 Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code
Information about Test: Calendars- 1 Page
In this test you can find the Exam questions for Test: Calendars- 1 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Calendars- 1, EduRev gives you an ample number of Online tests for practice

## CSAT Preparation for UPSC CSE

72 videos|69 docs|92 tests