When Is a 2007 Calendar Useful Again
| eleven. | The calendar for the year 2007 will be the same for the year: | |||||||
Answer: Option D Explanation: Count the number of odd days from the year 2007 onwards to go the sum equal to 0 odd day. Year : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 Odd twenty-four hour period : 1 2 1 1 one 2 1 1 one 2 i Sum = 14 odd days |
0 odd days.
Calendar for the twelvemonth 2018 will be the same as for the year 2007.| Bhuvana said: (Jun 4, 2010) | |
| Please explain me that how 14 odd days is similar to 0 odd days | |
| Mani said: (Jun 27, 2010) | |
| How to find odd days? and how it'southward possible? | |
| Mani said: (Jun 27, 2010) | |
| Need a cursory caption for this problem? | |
| Mani` said: (Jun 27, 2010) | |
| I don't accept much time to wait for your explanation, It's very emergency ? so, please respond to my questions ? | |
| Harru said: (Jul ii, 2010) | |
| Sum = 14 odd days 0 odd days. plz clerify this bespeak | |
| Sundar said: (Aug 18, 2010) | |
| Hi Friends, Please become through the basics terms and formulas given below before solving the problems. Link: http://www.indiabix.com/aptitude/calendar/formulas Promise this will assist you to solve the issues in this section. | |
| Nitesh Nandwana said: (Oct ten, 2010) | |
| The number of days more than than the complete weeks are called odd days. | |
| Prasad said: (Dec 23, 2010) | |
| For every ordinary year number of odd days = one odd solar day For leap yr number of odd days = two odd days. If the odd days sum=vii implies odd days=0 and so again loop starts from i,2,3,4,..7 . therefore 2013=1 (SUM=9), 2017=1 (SUM=fourteen) | |
| Manisha said: (Jan 6, 2011) | |
| You lot are right prasad | |
| Priya said: (Mar 22, 2011) | |
| This calendar method volition take time to find the day of the week.... I have seen a new piece of cake method in an quantitative aptitude book Beacon....if possible read that book | |
| Hariharan said: (Jun 15, 2011) | |
| Howdy All Pls explain me why 2018 is not included in this footstep. | |
| Jati N Kamani said: (Jul seven, 2011) | |
| Please explain me how to xiv odd days to equal 0 odd 24-hour interval. | |
| Sadik said: (Jul fourteen, 2011) | |
| Prasad explained very cleary all the best prasad keep it up. | |
| Shahir said: (Jul 14, 2011) | |
| Nosotros are supposed to find the day of the week on a given date. For this, nosotros use the concept of 'odd days'. In a given menstruum, the number of days more the complete weeks are called odd days. | |
| Akash said: (Jul 27, 2011) | |
| Can anyone explain how we are counting the days in respect to the odd days? | |
| Madhu said: (Jul 30, 2011) | |
| @Akash Actually 7 odd days = 0 odd days right?? And, fifty-fifty for 2012 2 odd days are assigned not for 2011 as the former is divisible past 4... | |
| Prem said: (Aug 3, 2011) | |
| Just unproblematic every 12 month once calender remain same. Am I right ? | |
| Rehan said: (Aug 5, 2011) | |
| Prasad is admittedly right. | |
| Asmath said: (Aug 22, 2011) | |
| Prasad yous are absolutely right. | |
| Riya said: (Aug 24, 2011) | |
| Hii prasad. I have an defoliation, Permit suppose I accept an year x and want to know the next same year, if I go sum of odd days 7 in this instance => odd days=0, Now what is the answer? 1) is the adjacent year will be the aforementioned as x as you told in your annotate? | |
| Anand said: (Aug 26, 2011) | |
| I have query regarding year 2002. Years- 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013. No of odd days for respective years. 1, one, ii, 1, 1, 1, ii, 1, 1, one, 2. Till 2007 total odd days becomes 0 hence respond should exist 2008 simply it is 2013 why ? | |
| Sandy said: (Aug 29, 2011) | |
| The year 2002 calender and the yr 2007 take the same calender. | |
| Apbakshi said: (Sep 2, 2011) | |
| @ sandy. Volition y'all tell me the adding behind that ? | |
| Svetta said: (January 22, 2012) | |
| Wow prasad! nice. | |
| Rambabu said: (Mar 1, 2012) | |
| Delight briefly explain this problem. | |
| Mahesh said: (Mar 22, 2012) | |
| Please tell us how to findout a 24-hour interval when date, month, year are given. | |
| Ramu said: (Aug 28, 2012) | |
| @anand ques is to find a twelvemonth whose calender wil b same as 2007 so y'all should go for coming years where the sum of odd days will be multiples of vii(which makes exact week) for a regular year we wil exist having 365/seven which gives balance 1 will be having 1 odd mean solar day and for leap years we will exist havin 2 odd days YEAR ODD DAYS 2007 1 2008(LEAP) 2 2009 1 2010 i 2011 one 2012(Leap) two 2013 1 2014 1 2015 one 2016(LEAP) ii 2017 1 here we will get the sum of odd days as 14 which again makes exact ii weeks...so next yr will get the same calender of 2007 | |
| Shashi said: (October ii, 2012) | |
| The calender for the yr 2008 will be the same for the yr. | |
| Upendra said: (Sep 15, 2013) | |
| Guys calender will be same if their number of odd days volition be same 2007..six*1+2=viii,1 odd 24-hour interval. 2014-three odd days. 2017-0 odd days. 2018-1 days. Then answer will be 2018. | |
| Poornimajagatha said: (Sep 20, 2013) | |
| As in formulas we heard that for 1990 = 1600+300 => 0+1 => i(odd day) => is that sunday or monday? | |
| Prabhuchandu said: (Jan 22, 2014) | |
| Easy method: 2000+28 = 2028. 2001+6 = 2007. 2002+11 = 2013. 2003+11 = 2015. 2004+28 = 2032. 2005+six = 2011. 2006+11 = 2017. 2007+11 = 2018. | |
| Prabhuchandu said: (Jan 22, 2014) | |
| Easy method: 2000+28 = 2028. 2001+half-dozen = 2007. 2002+11 = 2013. 2003+eleven = 2015. 2004+28 = 2032. 2005+vi = 2011. 2006+xi = 2017. 2007+11 = 2018. | |
| Sundar Gs said: (Jan 31, 2014) | |
| @Prabhuchandu. Tin can y'all explicate why y'all add years with half-dozen, xi, 28...and all? | |
| Jenny said: (Jun 19, 2014) | |
| Why nosotros want to find the odd numbers only from 2007 to 2017. Why we not discover the odd numbers 2018? | |
| Chandu said: (Sep 4, 2014) | |
| How can you tell that we should add together 6, 11, 28? Can you explain it. | |
| Amol Jadhav said: (Sep 26, 2014) | |
| @Prabhuchandu. Can you explicate why you lot add together years with 28, 6, 11, eleven..and all? | |
| Shreyas D.J. said: (Oct 14, 2014) | |
| The addition is coming 14 odd days = 0 odd days for twelvemonth 2017 non 2018. So why take 2018 if it has 1 odd day? | |
| Gopi said: (Dec 24, 2014) | |
| Please explain nearly 2018 why we are not considering 2018? Delight explain speedily very urgent. | |
| Rya said: (Jan 21, 2015) | |
| Please explain conspicuously how that 14 odd days is equivalent to 0 days? | |
| Katy said: (February 28, 2015) | |
| X/four (x=given twelvemonth). For whatsoever yr divided by 4, the possibility of remainders are 0, 1, 2, iii. If remainder is 0---->x+28. If remainder is 1---->x+vi. If remainder is 2 (or) 3------>10+11. So 2007 divide by four gives 3 as remainder. So 2007+11 = 2018. | |
| Rupali said: (Mar 9, 2015) | |
| Why we are non considered 2018? Reply please. | |
| Akriti said: (April two, 2015) | |
| Why we are not considering 2018? According to me it should be 2017 every bit both 2007 and 2017 has 0 odd days. Delight explain. | |
| Murali.O said: (Jul 6, 2015) | |
| Hai friends i small mode is to add together the number in 11 ok. | |
| Primary Black said: (October 28, 2015) | |
| Tin can you just calculate the next aforementioned year for 2008? | |
| Paveek said: (November 15, 2015) | |
| My best reply is hear: ane. Odd days: 2007 have 365 days (2007/4 = 3 (not perfect divisible) then it is not LEAP year and having 365 days. Now dissever by 7 (number of weeks) to know how many weeks in 2007. 365/7 = 52.1 then 52 weeks and 1 twenty-four hours. So the yr 2007 take 52 weeks and 1 24-hour interval. This 1 day is consider as odd day). two. So 2007 having 1 odd day. 2008/4 = 0 so LEAP year comprise 366 day i.e. 52 week 2 odd days. Similarly 2009 2010 2011 2012 2013 2014 2015 2016 2017. i 2 i 1 i 2 1 1 one two 1. 3. At present add all odd years because the yr is repeated when odd days = 0. 4. i+2+1+2+i = 14. 5. 14 means it is perfect 2 weeks and then no odd days. half dozen. We not become 7 in adding and then nosotros consider xiv (perfect week is needed). vii. 2018 is not included because upto 2017 information technology is 14 so next year is repeated twelvemonth. | |
| Paveek said: (Nov 15, 2015) | |
| @Hariharan. 2018 is not included because upto 2017 it is 14 so the next year is repeated yr. We just need perfect week (7, fourteen, 21....) next yr of the perfect week is the repeated year. | |
| Awdhesh said: (Dec 1, 2015) | |
| Explanation is not considered clearly. Observe like shooting fish in a barrel method please? | |
| Sudarshan said: (Dec 16, 2015) | |
| As per my calculations (all pl refer calendar) I have separate calculations for leap years and non leap years. Eg: Have a leap year 2012. five years before this leap year and half-dozen years after this bound year y'all go the same agenda (except January and February equally 2012 was a leap year and 2007 and 2018 were non leap years). Hence you can arrive 2007 and 2018 as the aforementioned. Pl apply this logic to all leap years. I know this because I tin say whatsoever days from years ane A.D till infinity within seconds. | |
| Sudarshan said: (Dec 16, 2015) | |
| Equally per my calculations (all pl refer calendar) I accept separate calculations for jump years and not jump years. Eg: Take a leap year 2012. 5 years earlier this spring yr and 6 years later on this leap yr yous get the same agenda (except January and February as 2012 was a leap twelvemonth and 2007 and 2018 were not spring years). Hence you tin can get in 2007 and 2018 every bit the same. Pl use this logic to all spring years. I know this because I can say whatsoever days from years one A.D till infinity within seconds. | |
| Sudarshan said: (Dec sixteen, 2015) | |
| As per my calculation (Please refer calendar if y'all want to verify). For bound years there is one adding. So having leap year as a base 5 years earlier the jump yr and 6 years later on the leap year you lot become the same calendar (Except January and February) as the base year considered here is a leap year. Eg. 2012 was a leap yr. v years before 2012 i.e. 2007 had the aforementioned agenda as 2012 and six years later 2012 that s 2018 are the same. Since 2012 was a spring year Jan and Feb would have differed from 2018 and 2007 from 2007 every bit we all know jump twelvemonth effects are observed only in Feb where i day is added in Feb. So hereby we conclude that 2007 is the aforementioned equally 2018 as given in the question. I case any days from years 1 A.D till infinity. | |
| Shekharrao said: (Dec 31, 2015) | |
| What is hateful of ordinary twelvemonth? | |
| Rakesh said: (Jun ii, 2016) | |
| How can we make up one's mind that the year is an ordinary yr and jump year? Please help me. | |
| Saravanakumar said: (Jul 4, 2016) | |
| It is as same as 400th twelvemonth (bound year) can any 1 differenciate information technology? | |
| Jesu said: (Jul 12, 2016) | |
| To find that the year is a leap yr or an ordinary twelvemonth. The yr should be divided past 4 if you get the remainder as 0 then information technology must exist a spring year or else if you lot go a remainder as ane or 2 or 3 then information technology must exist a ordinary year. Promise you lot understand. | |
| Gurubalaji said: (Jul fifteen, 2016) | |
| If you get a equal agenda for x year= (x + 11) yr. Because equal calender'due south obtain in every after 11 year. | |
| Guru Bhaskar Reddy said: (Jul 16, 2016) | |
| It is so unproblematic to calculate , no need to find Jan 1, 2007. If the year is completely divisible with 4, i.e residue = 0 and so simply add 23 years to the question year. Instance: Which year is aforementioned as; 2004 ? respond : 2004 + 23 = 2027. 2004, 2008, 2012............ +23. Is there any mistakes excuse me. | |
| Pradeep Verma said: (Jul 17, 2016) | |
| Please tell me the solution for this problem. If, 24march 1992 is Fri so which solar day fall on 24september 1993? | |
| Rakesh said: (Aug 10, 2016) | |
| It's for the upcoming years. How to calculate the same yr backward? | |
| Venkatesh said: (Sep iv, 2016) | |
| 2007 is a normal yr and front year 2006 is also a normal twelvemonth and then nosotros have to add 11years to the given yr. i.east; 2007 + 11years = 2018. | |
| Ramandeep said: (October viii, 2016) | |
| Only divide 14 past vii if zero remainder and so odd day zero. | |
| Pavan Kumar said: (Oct 16, 2016) | |
| Thanks for the caption. I take a pocket-size dubiousness. Every bit per your method for calendar 1988 year the calendar should be repeated on 1992. But it is repeated on 2016. How it is possible? Delight explain. | |
| Narayan said: (Oct 31, 2016) | |
| Still I can non understand. Please assist me. | |
| Rakesh Parmar said: (Nov 12, 2016) | |
| @All, Merely think this. Dissever Whatever year by 4. If you become 0, add 28 to given year, If you get 1 add together + 6. | |
| Sonali said: (Dec 9, 2016) | |
| Thank you @Prasad. | |
| Santhosh said: (Jan iii, 2017) | |
| I have a query regarding the year 2006. No of odd days for corresponding years. 1, 1, 2, 1, 1, ane. Till 2011, total odd days becomes 0 hence answer should be 2011 but it is 2017 why? | |
| Santhosh said: (Jan 3, 2017) | |
| I have a query regarding the year 2006. No of odd days for corresponding years. 1, i, 2, 1, 1, ane. Till 2011, full odd days becomes 0 hence answer should be 2011 simply information technology is 2017 why? | |
| Santhosh said: (January iii, 2017) | |
| I take a query regarding the year 2006. No of odd days for respective years. one, 1, 2, 1, 1, 1. Till 2011, total odd days becomes 0 hence answer should be 2011 but information technology is 2017 why? | |
| Navnath said: (Jan half dozen, 2017) | |
| The twelvemonth adjacent to 1973 having the aforementioned calendar as that of 1973 is ____. 1) 1977 Delight solve this. | |
| Mani said: (Jan 7, 2017) | |
| Give thanks you very much. I understand now. | |
| Sumanth said: (Jan 22, 2017) | |
| It is 1979 @Navnath. | |
| Niks said: (Feb 9, 2017) | |
| Thanks for easy method @Prabhuchandu. | |
| Krishna Kittu said: (Feb 17, 2017) | |
| For finding repeated calendars nosotros used formulae of calculation 28, 6, 11. for finding the by repeated calendars we accept to subtract same 28, half-dozen, eleven. Am I correct? | |
| Riya Khandelwal said: (Apr ten, 2017) | |
| I am not agreement this. Please explicate how 2018 will be similar to 2007? | |
| Siri said: (Apr 21, 2017) | |
| Whenever they ask for the same calender we should bank check whether information technology is a bound year or not. If it is a leap yr add 28 to the given year then you lot volition get the same calendar for jump year for non- bound years. six and xi (6 is added when given twelvemonth is subtracted from previous leap year u shoud get 1. i.e ex=2005 is not a jump year..previous spring year is 2004 so (04-05=i), then add together 6 considering nosotros got 1 later subtracting non-leap year with previous spring yr and so (2005+half dozen=2011). (eleven is added when you decrease not-jump year to previous spring twelvemonth you lot will get 2 nor 3). | |
| Aishwarya said: (Jun 5, 2017) | |
| The number of extra days more a complete week in a given menses is called odd days. Hence there are 14 days which tin can exist of ii weeks. Equally 7 days is ane calendar week and other vii days are another week so there are no more than days which are across a complete week. Therefore, the odd days are 0. | |
| Sowmya said: (Jun eight, 2017) | |
| Is this procedure is same for jump yr? Please, anyone explain. | |
| Balu Srinivas Reddy said: (Jul thirteen, 2017) | |
| For 4n => add together 28 years. If 2004 For 4n+1 => add together half-dozen years. | |
| Vijay said: (Aug 9, 2017) | |
| Please explain in particular. | |
| Anu said: (Aug fifteen, 2017) | |
| It applies to all years: Jump year == add 28. Here 2007, nearest leap yr earlier 2007, 2004. Therefore 2007 is 3 yrs after 2004. Therefore add together 11 to 2007=2017. | |
| Manjunath North said: (Aug 27, 2017) | |
| Assume 2007 starts with Sunday. 2008 starts - monday. | |
| Pratik said: (Sep 19, 2017) | |
| How to find answer this question? Seema remembers that her female parent's altogether falls on later 21st January only before 25th January while his brother remembers that it falls afterwards 23rd January but before 28th January. When is Seema mother birthday? | |
| Sharayu said: (Sep 24, 2017) | |
| How is the odd days for year 2017 ane? I still didn't get this? | |
| Prudhviraj said: (Oct 17, 2017) | |
| @ALL. start with following yr | |
| Jitendra Singh said: (Nov 22, 2017) | |
| Calander of 2016 is repeated in which year? Delight explicate. | |
| Dfeverx said: (Apr 11, 2018) | |
| For every ordinary twelvemonth number of odd days = i odd day. For jump year number of odd days = 2 odd days. It implies next year i.e 2021 will have the same agenda equally 2016. | |
| Shivam Giri said: (May 4, 2018) | |
| The calendar of 1872 repeated on which year and why? Please explain the respond. | |
| Rupa said: (Jul 1, 2018) | |
| Why this trick is not applicative to 2008. Can anyone explain? | |
| Vinod said: (Jul 5, 2018) | |
| Dainty, Cheers @Prasand. | |
| Evlin said: (Aug 19, 2018) | |
| Proficient i @Katy. It's a fourth dimension efficient method. Cheers. | |
| Jitendra said: (Sep 28, 2018) | |
| 2007 last ii digit division by 4. if the remainder is 1, add half dozen. remainder is 2 add 11, and 3 add 11 get an answer. | |
| Akanksha said: (Sep 29, 2018) | |
| Why exercise we have till 2017? Please explicate. | |
| Semran Sheikh said: (October 9, 2018) | |
| Thanks @Prasad. | |
| Bapmo said: (Nov xi, 2018) | |
| Thanks @Prasad and @Manjunath.N. | |
| Samruddhia said: (Dec 12, 2018) | |
| @All. A leap twelvemonth calendar repeats itself in 28 years and an Ordinary yr Calender repeats itself in 6 or 11 years. Here, we can simply add 2007+vi=2013 (which is not there in the option) and 2007+eleven= 2018 which is the answer. | |
| Sheetal said: (Dec 28, 2018) | |
| In these types of questions, carve up last 2digits of the given year from 4, after that if divisible is-- 2 and three add together 11in it, 1 add together vi in it, 0 add 28 in it. Eg.2007(07/4) divisible is iii then add together 11 in 2007= 2018. | |
| Manish said: (Jun xiv, 2019) | |
| 2007 is a general year so either it will repeat after year or xi year let's run into; Year : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017. If the addition is 7 then the calendar will change just 2012 is leap year then full general year can't be a jump year so at present we become till the fourteen. | |
| Pradyumna Kumar Tiwari said: (Jul i, 2019) | |
| If the year is leap then add together 28 to get the aforementioned calendar provided that no century year which does not leap should fall in between. If the year is not leap, add together xi and if the result is not jump; the upshot has the same calender. If the result is leap then add 6 to get same calender. It should be taken that no century year which is not spring should non fall in betwixt. 2007 is not leap so 2007+xi=2018, 2018 is not leap; hence 2018 is the respond. | |
| Pradyumna Kumar Tiwari said: (Jul 1, 2019) | |
| The two necessary conditions for the same calendar is; 1) Both years must either be normal or both leap. | |
| Ashwini said: (Aug xvi, 2019) | |
| How answer is 2018? Please, anyone explicate me. | |
| Fathima Safwana said: (Oct xxx, 2019) | |
| Is this Holds good for for year where for which year 2009 will be same ? | |
| Ank said: (Dec eight, 2019) | |
| Step1- calculate odd days from given year to that year at which addition is divisible bye 7. | |
| Mandakranta said: (Feb 24, 2020) | |
| 2009 was repeated in 2015. 2009 - one odd day Again 2009 beingness non-leap year if nosotros split up by 4 we get a balance one. And so 2009 + eleven= 2020 which is not possible 2020 being a leap year. And so 2009 +six =2015 is the repeated the year of 2009. | |
| Meghana said: (Mar 21, 2020) | |
| Ordinary year = 1odd mean solar day. Spring year = 2odd days. They told to fïnd oud same calendar as 2007, then we should first counting an odd 24-hour interval from 2008, Add the above odd days. To get aforementioned calendar odd days should be multiple of vii. 7 divided by 7 gives reminder 0. Hence 2012 is the answer. | |
| Kunal said: (May viii, 2020) | |
| How the odd days taken here? | |
| Gireesh Kumar said: (May 17, 2020) | |
| @Kunal. See, what is the remainder later on dividing the twelvemonth by four. If the remainder is 1 add 5 years to the given year. If the residual is 2 add xi to the given year. If the rest is 3 adding eleven to the given yr. Then in the above question, they gave 2007. After dividing by 4. It gives the remainder every bit 3. And so simply add 11 years to 2007. That will be 2018. Hope it helps you. | |
| Aniket said: (Jul 6, 2020) | |
| So, What is the solution for 2008? Anyone explain to me. | |
| Shivam Padmani said: (Jul 27, 2020) | |
| Short play a trick on for solving. One has to divide the given yr by 4. | |
| Suraj said: (Nov 29, 2020) | |
| Why 2014 cannot be the answer. In that location are 7 odd days. Which sum up once more to 0 odd days? | |
| Vigneshwaran said: (Apr 7, 2021) | |
| Thank you for explaining @Prasad. | |
| Mithilesh Mahatha said: (Apr nineteen, 2021) | |
| @All. So, doubt hither is how fourteen = 0, | |
| Rama said: (Jun 23, 2021) | |
| Thanks for explaining @Prasad. | |
| Mesfin said: (October two, 2021) | |
| Why start count odd number fron 2008 & why not start from 2007? Please explain me. | |
| Sonu Kumar said: (Apr 17, 2022) | |
| @All. Dissever the given no by iv if residuum is 0 add 28 to the year if residue is ane add 6 or if residual is 2or3 add 11, in this case, the remainder is three. So, 2007 + 11 = 2018. | |
| Ravi Maurya said: (Apr 30, 2022) | |
| This solution is not correct. Rule:- 1. Rule:- ii. Dominion:-3. | |
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