# Calendar Quiz 1

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# Calendar Quiz 1

### Introduction

Calendar deals with odd days, leap year, ordinary year, counting of odd days, and day of the week related to odd days. To find the day of the week on a given date concept of odd days is used. The article Calendar Quiz 1 lists important Calendar practice questions for competitive exams like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and etc.

### Concepts

Odd days: Extra days, apart from the complete weeks in given periods are called odd days.
Leap year: A leap year is divisible by 4 except for a century. For a century to be a leap year, it must be divisible by 400.
Examples:
Years like 1988, 2008 are leap year (divisible by 4).
Centuries like 2000, 2400 are leap year ( divisible by 400).
Years like 1999, 2003 are not leap year (not divisible by 4).
Centuries like 1700, 1800 are not leap year ( not divisible by 400).
In a century, there is 76 ordinary year and 24 leap year.

Ordinary year: Ordinary year is other than leap years. A ordinary year has 365 days.
Counting of odd days: (i) 1 ordinary year = 365 days = (52 weeks + 1 day).
An ordinary year has one odd day.
(ii) 1 leap year = 366 days = (52 weeks + 2 days). A leap year has 2 odd days.
Day of the week related to odd days: Let the number of days be 0, 1, 2, 3, 4, 5 , 6 and their days are Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday respectively.

### Q1

It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
A. Sunday B. Saturday C. Friday D. Wednesday

C
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.

### Q2

What was the day of the week on 28th May, 2006?
A. Thursday B. Friday C. Saturday D. Sunday

D
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.

### Q3

What was the day of the week on 17th June, 1998?
A. Monday B. Tuesday C. Wednesday D. Thursday

C
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
168 days = 24 weeks = 0 odd day.
Total number of odd days = (0 + 1 + 2 + 0) = 3.
Given day is Wednesday.

### Q4

What will be the day of the week 15th August, 2010?
A. Sunday B. Monday C. Tuesday D. Friday

A
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 May 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.

### Q5

Today is Monday. After 61 days, it will be:
A. Wednesday B. Saturday C. Tuesday D. Thursday

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