{"id":6018,"date":"2025-06-09T07:12:12","date_gmt":"2025-06-09T07:12:12","guid":{"rendered":"https:\/\/thecompanyboy.com\/?p=6018"},"modified":"2025-06-09T07:12:12","modified_gmt":"2025-06-09T07:12:12","slug":"rs-aggarwal-quantitative-aptitude-pdf-calendar-download","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/drive\/rs-aggarwal-quantitative-aptitude-pdf-calendar-download\/","title":{"rendered":"RS Aggarwal Quantitative Aptitude PDF Free Download: CALENDAR"},"content":{"rendered":"

CALENDAR\u00a0<\/strong><\/h1>\n

IMPORTANT FACTS AND FOAMULAE<\/h2>\n

Under this heading we mainly deal with finding the day of the week on a particular given date the\u00a0 process of finding it lies on obtaining the number of odd days.<\/p>\n

Odd Days<\/strong> : Number of days more than the complete number of weeks in a given<\/p>\n

Period <\/strong>., is the number of odd days during that period.<\/p>\n

LeapYear:<\/strong> Every year which is divisible by 4 is called a leap year.<\/p>\n

Thus each one of the years 1992, 1996, 2004, 2008, 2012, etc. is a leap year. Every 4th century is a leap year but no other century is a leap year.thus each one of 400, 800, 1200,’ 1600, 2000, etc. is a leap year.<\/p>\n

None of 1900, 2010, 2020, 2100, etc. is a leap year.<\/p>\n

An year which is not a leap year is called an ordinary year.<\/em><\/strong><\/p>\n

\u00a0<\/em>(I )An ordinary year has 365 days. (II) A leap\u00a0 year has 366 days.<\/p>\n

Counting of Odd Days<\/u>:<\/h2>\n

i)1 ordinary year = 365 days = (52 weeks + 1 day).<\/p>\n

:. An ordinary year has 1 odd day.<\/p>\n

ii)1 leap year = 366 days = (52 weeks + 2 days).<\/p>\n

:. A leap year has 2 odd days.<\/p>\n

_\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 iii)100 years = 76 ordinary years + 24 leap years<\/p>\n

= [(76 x 52) weeks + 76 days) + [(24 x 52) weeks + 48 days]<\/p>\n

= 5200 weeks + 124 days = (5217 weeks + 5 days).<\/p>\n

:. 100 years contain 5 odd days.<\/p>\n

200 years contain 10 and therefore 3 odd days.<\/p>\n

300 years contain 15 and therefore 1 odd day.<\/p>\n

400 years contain (20 + 1) and therefore 0 odd day.<\/p>\n

Similarly, each one of 800, 1200, 1600, 2000, etc. contains 0 odd days.<\/p>\n

Remark: (7n + m) odd days, where m < 7 is equivalent to m odd days.<\/p>\n

Thus, 8 odd days \u2261 1 odd day etc.<\/p>\n\n\n\n\n
\n

No of odd days<\/p>\n<\/td>\n

\n

0<\/p>\n<\/td>\n

\n

1<\/p>\n<\/td>\n

\n

2<\/p>\n<\/td>\n

\n

3<\/p>\n<\/td>\n

\n

4<\/p>\n<\/td>\n

\n

5<\/p>\n<\/td>\n

\n

6<\/p>\n<\/td>\n<\/tr>\n

\n

Day<\/p>\n<\/td>\n

\n

Sun.<\/p>\n<\/td>\n

\n

Mon.<\/p>\n<\/td>\n

\n

Tues.<\/p>\n<\/td>\n

\n

Wed.<\/p>\n<\/td>\n

\n

Thur.<\/p>\n<\/td>\n

\n

Fri.<\/p>\n<\/td>\n

\n

Sat.<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

SOLVED EXAMPLES<\/u><\/h2>\n

Ex: 1.Wbat was the day of the week on, 16th July, 1776?<\/strong><\/p>\n

Sol:<\/strong>\u00a0 16th July, 1776 = (1775 years + Period from 1st Jan., 1776 to 16th July, 1776)<\/p>\n

Counting of odd days :<\/p>\n

1600 years have 0 odd day. 100 years have 5 odd days.<\/p>\n

75 years = (18 leap years + 57 ordinary years)<\/p>\n

= [(18 x 2) + (57 x 1)] odd days = 93 odd days<\/p>\n

= (13 weeks + 2 days) = 2 odd days.<\/p>\n

\u00a0<\/strong>.. 1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day.<\/p>\n

Jan.\u00a0 Feb.\u00a0 March\u00a0\u00a0 April\u00a0\u00a0 May\u00a0\u00a0 June\u00a0\u00a0 July<\/p>\n

31\u00a0\u00a0 + 29\u00a0 +\u00a0\u00a0 31\u00a0\u00a0\u00a0 +\u00a0 30\u00a0\u00a0 +\u00a0 31\u00a0 +\u00a0 30\u00a0 +16 = 198days<\/p>\n

= (28 weeks + 2 days) =2days<\/p>\n

:. . Total number of odd days = (0 + 2) = 2. Required day was ‘Tuesday’.<\/p>\n

\u00a0<\/strong>Ex. 2. What was the day of the week on 16th August, 1947?<\/strong><\/p>\n

Sol.<\/strong> 15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th<\/p>\n

Counting of odd days:<\/p>\n

1600 years have 0 odd day. 300 years have 1 odd day.<\/p>\n

47 years = (11 leap years + 36 ordinary years)<\/p>\n

= [(11 x 2) + (36 x 1\u00bb) odd days = 58 odd days = 2 odd days.<\/p>\n

Jan. Feb. March April May June July Aug.<\/p>\n

31 + 28 + 31 + 30 + 31 + 30 + 31 + 15<\/p>\n

= 227 days = (32 weeks + 3 days) = 3,<\/p>\n

Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.<\/p>\n

Hence, the required day was ‘Saturday’.<\/p>\n

\u00a0<\/strong>Ex. 3. What was the day of the week on 16th April, 2000 ?<\/strong><\/p>\n

Sol<\/strong>. 16th April, 2000 = (1999 years + Period from 1st Jan., 2000 to 16thA’<\/p>\n

Counting of odd days:<\/p>\n

1600 years have 0 odd day. 300 years have 1 odd day.<\/p>\n

99 years = (24 leap years + 75 ordinary years)<\/p>\n

= [(24 x 2) + (75 x 1)] odd days = 123 odd days<\/p>\n

= (17 weeks + 4 days) = 4 odd days.<\/p>\n

Jan. Feb. March April<\/p>\n

31 + 29 + 31 + 16 = 107 days = (15 weeks + 2 days) = 2 odd,<\/p>\n

Total number of odd days = (0 + 1 + 4 + 2) odd days = 7 odd days = 0 odd day.\u00a0 \u00a0Hence, the required day was ‘Sunday’.<\/p>\n

Ex. 4. On what dates of Jull.2004 did Monday fall?<\/strong><\/p>\n

Sol <\/strong>\u00a0\u00a0. Let us find the day on 1st July, 2004.<\/p>\n

2000 years have 0 odd day. 3 ordinary years have 3 odd days.<\/p>\n

Jan. Feb. March April May June July<\/h3>\n

31 + 29 + 31 + 30 + 31 + 30 + 1<\/p>\n

= 183 days = (26 weeks + 1 day) = 1 t .<\/p>\n

Total number of odd days = (0 + 3 + 1) odd days = 4 odd days.\u00a0\u00a0\u00a0\u00a0\u00a0 ‘<\/p>\n

:. 1st July 2004 was ‘Thursday’,-,-\u00ad<\/p>\n

Thus, 1st Monday in July 2004 _as on 5th July.<\/p>\n

Hence, during July 2004, Monday fell on 5th, 12th, 19th and 26th. .<\/p>\n

Ex. 5. Prove that the calendar for the year 2008 will serve for the year 20ll <\/strong><\/p>\n

\u00a0<\/strong>Sol<\/strong>. In order that the calendar for the year 2003 and 2014 be the same, 1st<\/sup>\u00a0 January of\u00a0 both the years must be on the same day of the week.<\/p>\n

For this, the number of odd days between 31st Dec., 2002 and 31st Dec.,2013 must be the same.<\/p>\n

We know that an ordinary year has 1 odd day and a leap year has 2 odd During this period, there are 3 leap years, namely 2004, 2008 and 2012 and 8 ordinary years.<\/p>\n

Total number of odd days = (6 + 8) days = 0 odd day.<\/p>\n

Hence, the calendar for 2003 will serve for the year 2014.<\/p>\n

Ex. 6. Prove that any date in March of a year is the same day of the <\/strong>week<\/strong> corresponding date in November that year.<\/strong><\/p>\n

\u00a0<\/strong>We will show that the number of odd days between last day of February and last\u00a0 <\/strong>day of October is zero.\u00a0\u00a0\u00a0 .<\/p>\n

March\u00a0\u00a0\u00a0 April\u00a0\u00a0\u00a0 May\u00a0\u00a0\u00a0 June\u00a0\u00a0\u00a0 July\u00a0\u00a0\u00a0 Aug.\u00a0\u00a0\u00a0 Sept.\u00a0\u00a0 Oct.<\/p>\n

31\u00a0\u00a0\u00a0\u00a0 +\u00a0\u00a0 30\u00a0 +\u00a0 31\u00a0\u00a0\u00a0 + 30\u00a0\u00a0 + 31 +\u00a0\u00a0 31\u00a0 +\u00a0\u00a0\u00a0 30\u00a0\u00a0 +\u00a0 31<\/p>\n

= 241 days = 35 weeks = 0 odd day. ,Number of odd days during this period = O.<\/p>\n

Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows.<\/p>\n

Accessing specific chapters, such as “Calendar,” from R.S. Aggarwal’s “Quantitative Aptitude” in PDF format may be challenging due to copyright restrictions. However, there are reputable resources that provide practice questions and explanations on calendar-related problems:<\/p>\n

    \n
  1. \n

    UPSC Fever<\/strong>: Offers a selection of calendar-related questions inspired by R.S. Aggarwal’s work, complete with solutions and explanations. This resource is beneficial for understanding the types of questions that appear in competitive exams.<\/p>\n<\/li>\n

  2. \n

    Google Books<\/strong>: Provides a limited preview of “Quantitative Aptitude” by R.S. Aggarwal, allowing you to access certain sections and topics, including parts of the “Calendar” chapter. This can be useful for reviewing specific concepts and sample problems.<\/p>\n<\/li>\n<\/ol>\n

    For comprehensive study, it’s advisable to obtain the complete book through authorized channels, such as purchasing a copy or accessing it via a library. This ensures you have full access to all topics and practice exercises, which is crucial for thorough preparation.<\/p>\n

    Remember, using official and reputable sources guarantees that you’re accessing accurate and high-quality study materials, essential for effective learning and success in competitive examinations.<\/p>\n

    RS Aggarwal Quantitative Aptitude PDF Free Download: CALENDAR<\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

    CALENDAR\u00a0 IMPORTANT FACTS AND FOAMULAE Under this heading we mainly deal with finding the day of the week on a particular given date the\u00a0 process of finding it lies on obtaining the number of odd days. Odd Days : Number of days more than the complete number of weeks in a given Period ., is […]<\/p>\n","protected":false},"author":41,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[126,127],"tags":[],"class_list":["post-6018","post","type-post","status-publish","format-standard","hentry","category-rs-aggarwal-quantitative-aptitude","category-rs-aggarwal-quantitative-aptitude-pdf"],"_links":{"self":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/6018","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/users\/41"}],"replies":[{"embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/comments?post=6018"}],"version-history":[{"count":0,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/6018\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/media?parent=6018"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/categories?post=6018"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/tags?post=6018"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}