SSC CGL  >  CSAT Preparation for UPSC CSE  >  Test: Calendars- 2 Download as PDF

Test: Calendars- 2


Test Description

10 Questions MCQ Test CSAT Preparation for UPSC CSE | Test: Calendars- 2

Test: Calendars- 2 for SSC CGL 2022 is part of CSAT Preparation for UPSC CSE preparation. The Test: Calendars- 2 questions and answers have been prepared according to the SSC CGL exam syllabus.The Test: Calendars- 2 MCQs are made for SSC CGL 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Calendars- 2 below.
Solutions of Test: Calendars- 2 questions in English are available as part of our CSAT Preparation for UPSC CSE for SSC CGL & Test: Calendars- 2 solutions in Hindi for CSAT Preparation for UPSC CSE course. Download more important topics, notes, lectures and mock test series for SSC CGL Exam by signing up for free. Attempt Test: Calendars- 2 | 10 questions in 10 minutes | Mock test for SSC CGL preparation | Free important questions MCQ to study CSAT Preparation for UPSC CSE for SSC CGL Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you?
Test: Calendars- 2 - Question 1

What was the day of the Week on 17th June 1998?

Detailed Solution for Test: Calendars- 2 - Question 1

17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)
Odd days in 1600 years = 0
Odd days in 300 years = (5 x 3) ≡ 1
97 years has 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.
Jan.         Feb.       March       April         May         June
(31     +     28     +     31     +     30     +     31     +     17) = 168 days
Therefore 168 days = 24 weeks = 0 odd day.
Total number of odd days = (0 + 1 + 2 + 0) = 3.
Given day is Wednesday.

Test: Calendars- 2 - Question 2

On what dates of April 2001 did Wednesday fall?

Detailed Solution for Test: Calendars- 2 - Question 2

We shall find the day on 1st April, 2001.
1st April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)
Odd days in 1600 years = 0
Odd days in 400 years = 0
Jan. Feb. March April
(31 + 28 + 31 + 1)     = 91 days ≡ 0 odd days.
Total number of odd days = (0 + 0 + 0) = 0
On 1st April, 2001 it was Sunday.
In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th

Test: Calendars- 2 - Question 3

16th July 1776,the day of the week was?

Detailed Solution for Test: Calendars- 2 - Question 3

16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)
Counting of odd days :
1600 years have 0 odd day
100 years have 5 odd days
75 years = (18 leap years + 57 ordinary years)
= [(18 x 2) + (57 x 1)]
= 93 (13 weeks + 2 days)
= 2 odd days
1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day
Jan   Feb   Mar   Apr   May   Jun   Jul
31 + 29 + 31 + 30 + 31 + 30 + 16
= 198 days
= (28 weeks + 2 days)
Total number of odd days = (0 + 2) = 2
Required day was 'Tuesday'.
 

Test: Calendars- 2 - Question 4

How many days are there in x weeks x days

Detailed Solution for Test: Calendars- 2 - Question 4

x weeks x days = (7x + x) days 
= 8x days.

Test: Calendars- 2 - Question 5

What will be the day of the week 15th August 2010?

Detailed Solution for Test: Calendars- 2 - Question 5

15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)
Odd days in 1600 years = 0
Odd days in 400 years = 0
9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days ≡ 4 odd days.
Jan.     Feb.   March   April     Mayb   June     July       Aug.
(31   +   28   +   31   +   30   +   31   +   30   +   31   +   15) = 227 days
∴ 227 days = (32 weeks + 3 days) ≡ 3 odd days.
Total number of odd days = (0 + 0 + 4 + 3) = 7 ≡ 0 odd days.
Given day is Sunday
 

Test: Calendars- 2 - Question 6

It was Sunday on Jan 1, 2006. What was the day of the Week Jan 1, 2010

Detailed Solution for Test: Calendars- 2 - Question 6

On 31st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
∴ On 31st December 2009, it was Thursday.
Thus, on 1st Jan, 2010 it is Friday.

Test: Calendars- 2 - Question 7

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

Detailed Solution for Test: Calendars- 2 - Question 7

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- 2 - Question 8

The last day of a Century cannot be

Detailed Solution for Test: Calendars- 2 - 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- 2 - Question 9

On 8th Feb,2005 it was Tuesday.What was the day of the Week on 8th Feb,2004?

Detailed Solution for Test: Calendars- 2 - Question 9

The year 2004 is a leap year. It has 2 odd days.
∴ The day on 8th Feb, 2004 is 2 days before the day on 8th Feb, 2005.
Hence, this day is Sunday.

Test: Calendars- 2 - Question 10

What was the day of the week on 28th May 2006?

Detailed Solution for Test: Calendars- 2 - Question 10

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) ≡ 6 odd days
Jan.         Feb.       March       April         May
(31     +     28     +     31     +     30     +     28 ) = 148 days
∴ 148 days = (21 weeks + 1 day) ≡ 1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7 ≡ 0 odd day.
Given day is Sunday.
 

72 videos|64 docs|92 tests
Use Code STAYHOME200 and get INR 200 additional OFF
Use Coupon Code
Information about Test: Calendars- 2 Page
In this test you can find the Exam questions for Test: Calendars- 2 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Calendars- 2, EduRev gives you an ample number of Online tests for practice