When Can You Use a 2007 Calender Again
| 11. | The calendar for the yr 2007 will be the same for the year: | |||||||
Answer: Selection D Explanation: Count the number of odd days from the yr 2007 onwards to get the sum equal to 0 odd day. Yr : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 Odd twenty-four hour period : 1 2 ane 1 1 2 one 1 1 2 ane Sum = 14 odd days |
0 odd days.
Calendar for the yr 2018 will exist the same every bit for the yr 2007.| Bhuvana said: (Jun 4, 2010) | |
| Please explain me that how 14 odd days is like to 0 odd days | |
| Mani said: (Jun 27, 2010) | |
| How to observe odd days? and how information technology'south possible? | |
| Mani said: (Jun 27, 2010) | |
| Need a brief explanation for this problem? | |
| Mani` said: (Jun 27, 2010) | |
| I don't have much time to expect for your caption, It's very emergency ? so, delight respond to my questions ? | |
| Harru said: (Jul 2, 2010) | |
| Sum = 14 odd days 0 odd days. plz clerify this point | |
| Sundar said: (Aug eighteen, 2010) | |
| How-do-you-do Friends, Delight go through the basics terms and formulas given beneath before solving the problems. Link: http://world wide web.indiabix.com/aptitude/agenda/formulas Hope this will assist you to solve the problems in this section. | |
| Nitesh Nandwana said: (Oct 10, 2010) | |
| The number of days more than the complete weeks are called odd days. | |
| Prasad said: (December 23, 2010) | |
| For every ordinary year number of odd days = 1 odd mean solar day For leap year number of odd days = 2 odd days. If the odd days sum=7 implies odd days=0 then over again loop starts from one,ii,iii,iv,..7 . therefore 2013=1 (SUM=ix), 2017=1 (SUM=14) | |
| Manisha said: (Jan vi, 2011) | |
| You are right prasad | |
| Priya said: (Mar 22, 2011) | |
| This calendar method will take time to detect the mean solar 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) | |
| Hi All Pls explicate me why 2018 is non included in this footstep. | |
| Jati Northward Kamani said: (Jul 7, 2011) | |
| Please explicate me how to fourteen odd days to equal 0 odd day. | |
| Sadik said: (Jul 14, 2011) | |
| Prasad explained very cleary all the best prasad go along it up. | |
| Shahir said: (Jul 14, 2011) | |
| We are supposed to detect the day of the calendar week on a given date. For this, nosotros use the concept of 'odd days'. In a given menstruation, the number of days more than 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, even for 2012 two odd days are assigned not for 2011 equally the former is divisible by 4... | |
| Prem said: (Aug 3, 2011) | |
| Just elementary every 12 month once calender remain same. Am I correct ? | |
| Rehan said: (Aug 5, 2011) | |
| Prasad is admittedly right. | |
| Asmath said: (Aug 22, 2011) | |
| Prasad you are admittedly correct. | |
| Riya said: (Aug 24, 2011) | |
| Hii prasad. I have an confusion, Let suppose I accept an year x and want to know the adjacent same year, if I get sum of odd days 7 in this case => odd days=0, Now what is the answer? 1) is the next year will be the aforementioned equally x as you told in your comment? | |
| 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 corresponding years. 1, i, 2, 1, ane, 1, 2, i, ane, ane, 2. Till 2007 total odd days becomes 0 hence reply should be 2008 but information technology is 2013 why ? | |
| Sandy said: (Aug 29, 2011) | |
| The year 2002 calender and the yr 2007 take the same calender. | |
| Apbakshi said: (Sep two, 2011) | |
| @ sandy. Will you lot tell me the adding behind that ? | |
| Svetta said: (Jan 22, 2012) | |
| Wow prasad! nice. | |
| Rambabu said: (Mar 1, 2012) | |
| Please briefly explicate this problem. | |
| Mahesh said: (Mar 22, 2012) | |
| Please tell us how to findout a twenty-four hours when date, month, yr are given. | |
| Ramu said: (Aug 28, 2012) | |
| @anand ques is to notice a yr whose calender wil b aforementioned as 2007 so you should go for coming years where the sum of odd days will be multiples of 7(which makes exact calendar week) for a regular yr nosotros wil be having 365/7 which gives residue 1 will be having one odd day and for leap years we will exist havin 2 odd days Year ODD DAYS 2007 one 2008(Bound) 2 2009 one 2010 one 2011 i 2012(Spring) 2 2013 1 2014 1 2015 1 2016(Spring) 2 2017 ane here we volition become the sum of odd days equally 14 which once more makes verbal 2 weeks...so adjacent yr will get the same calender of 2007 | |
| Shashi said: (Oct 2, 2012) | |
| The calender for the yr 2008 will be the aforementioned for the year. | |
| Upendra said: (Sep fifteen, 2013) | |
| Guys calender will be same if their number of odd days will be same 2007..6*i+2=8,i odd solar day. 2014-3 odd days. 2017-0 odd days. 2018-1 days. So answer volition be 2018. | |
| Poornimajagatha said: (Sep 20, 2013) | |
| Every bit in formulas we heard that for 1990 = 1600+300 => 0+i => 1(odd day) => is that sun 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+6 = 2011. 2006+eleven = 2017. 2007+xi = 2018. | |
| Prabhuchandu said: (January 22, 2014) | |
| Easy method: 2000+28 = 2028. 2001+6 = 2007. 2002+xi = 2013. 2003+11 = 2015. 2004+28 = 2032. 2005+6 = 2011. 2006+eleven = 2017. 2007+11 = 2018. | |
| Sundar Gs said: (Jan 31, 2014) | |
| @Prabhuchandu. Can y'all explicate why you add years with 6, 11, 28...and all? | |
| Jenny said: (Jun 19, 2014) | |
| Why we want to find the odd numbers but from 2007 to 2017. Why we not discover the odd numbers 2018? | |
| Chandu said: (Sep 4, 2014) | |
| How can you tell that nosotros should add vi, 11, 28? Can you explain it. | |
| Amol Jadhav said: (Sep 26, 2014) | |
| @Prabhuchandu. Can you explain why you lot add years with 28, 6, 11, 11..and all? | |
| Shreyas D.J. said: (Oct 14, 2014) | |
| The addition is coming fourteen odd days = 0 odd days for year 2017 non 2018. And so why take 2018 if it has ane odd twenty-four hour period? | |
| Gopi said: (Dec 24, 2014) | |
| Please explain about 2018 why nosotros are not considering 2018? Delight explain rapidly very urgent. | |
| Rya said: (Jan 21, 2015) | |
| Please explain clearly how that 14 odd days is equivalent to 0 days? | |
| Katy said: (Feb 28, 2015) | |
| X/4 (x=given year). For any twelvemonth divided past 4, the possibility of remainders are 0, 1, 2, 3. If remainder is 0---->x+28. If remainder is 1---->x+vi. If remainder is 2 (or) three------>x+11. So 2007 divide by 4 gives 3 as remainder. And then 2007+eleven = 2018. | |
| Rupali said: (Mar 9, 2015) | |
| Why we are not considered 2018? Respond please. | |
| Akriti said: (Apr 2, 2015) | |
| Why we are not considering 2018? According to me information technology should be 2017 as both 2007 and 2017 has 0 odd days. Please explicate. | |
| Murali.O said: (Jul 6, 2015) | |
| Hai friends one modest way is to add the number in 11 ok. | |
| Master Black said: (Oct 28, 2015) | |
| Can y'all just calculate the next same year for 2008? | |
| Paveek said: (Nov 15, 2015) | |
| My best answer is hear: 1. Odd days: 2007 take 365 days (2007/4 = three (not perfect divisible) and then information technology is non Leap year and having 365 days. At present dissever by 7 (number of weeks) to know how many weeks in 2007. 365/7 = 52.i so 52 weeks and one day. So the yr 2007 have 52 weeks and ane day. This 1 day is consider every bit odd day). 2. So 2007 having i odd day. 2008/4 = 0 so LEAP yr comprise 366 day i.e. 52 week two odd days. Similarly 2009 2010 2011 2012 2013 2014 2015 2016 2017. 1 2 ane 1 1 2 1 1 1 2 1. 3. At present add all odd years because the year is repeated when odd days = 0. four. one+2+1+2+1 = 14. 5. 14 means it is perfect ii weeks and then no odd days. 6. We not get vii in adding and then we consider fourteen (perfect week is needed). seven. 2018 is not included because upto 2017 it is fourteen so next year is repeated year. | |
| Paveek said: (Nov xv, 2015) | |
| @Hariharan. 2018 is not included because upto 2017 information technology is 14 so the next twelvemonth is repeated year. We just demand perfect week (7, 14, 21....) adjacent year of the perfect week is the repeated year. | |
| Awdhesh said: (December 1, 2015) | |
| Explanation is not considered conspicuously. Find easy method please? | |
| Sudarshan said: (Dec sixteen, 2015) | |
| Equally per my calculations (all pl refer calendar) I have separate calculations for jump years and non jump years. Eg: Have a leap twelvemonth 2012. 5 years before this leap year and vi years subsequently this leap yr you become the aforementioned calendar (except January and Feb equally 2012 was a leap year and 2007 and 2018 were non jump years). Hence you can make it 2007 and 2018 equally the same. Pl apply this logic to all leap years. I know this because I tin can say any days from years ane A.D till infinity within seconds. | |
| Sudarshan said: (Dec sixteen, 2015) | |
| Every bit per my calculations (all pl refer calendar) I have dissever calculations for bound years and non leap years. Eg: Take a leap yr 2012. v years before this leap year and six years after this leap yr you get the same agenda (except January and February as 2012 was a leap year and 2007 and 2018 were non leap years). Hence you tin go far 2007 and 2018 as the same. Pl apply this logic to all spring years. I know this because I tin can say whatsoever days from years ane A.D till infinity within seconds. | |
| Sudarshan said: (December 16, 2015) | |
| As per my calculation (Please refer calendar if you want to verify). For leap years there is one calculation. So having leap year as a base v years before the leap year and half-dozen years subsequently the jump year you get the same calendar (Except January and February) as the base of operations year considered hither is a leap year. Eg. 2012 was a bound year. 5 years before 2012 i.e. 2007 had the aforementioned calendar as 2012 and 6 years later on 2012 that s 2018 are the same. Since 2012 was a leap year Jan and Feb would accept differed from 2018 and 2007 from 2007 as we all know bound year effects are observed only in Feb where 1 day is added in Feb. So hereby we conclude that 2007 is the aforementioned as 2018 as given in the question. I instance any days from years 1 A.D till infinity. | |
| Shekharrao said: (Dec 31, 2015) | |
| What is mean of ordinary year? | |
| Rakesh said: (Jun 2, 2016) | |
| How tin can we make up one's mind that the yr is an ordinary year and spring year? Please assist me. | |
| Saravanakumar said: (Jul 4, 2016) | |
| Information technology is as same as 400th yr (jump yr) tin any one differenciate it? | |
| Jesu said: (Jul 12, 2016) | |
| To notice that the year is a leap year or an ordinary year. The year should be divided past 4 if y'all get the residuum as 0 then it must be a leap year or else if you become a remainder as 1 or 2 or 3 then it must exist a ordinary year. Promise y'all understand. | |
| Gurubalaji said: (Jul 15, 2016) | |
| If you go a equal calendar for x twelvemonth= (ten + eleven) year. Because equal calender's obtain in every afterward 11 year. | |
| Guru Bhaskar Reddy said: (Jul xvi, 2016) | |
| It is so simple to summate , no need to observe Jan 1, 2007. If the twelvemonth is completely divisible with four, i.east residual = 0 then simply add together 23 years to the question twelvemonth. Example: Which year is aforementioned as; 2004 ? answer : 2004 + 23 = 2027. 2004, 2008, 2012............ +23. Is in that location any mistakes excuse me. | |
| Pradeep Verma said: (Jul 17, 2016) | |
| Please tell me the solution for this problem. If, 24march 1992 is Friday then which day fall on 24september 1993? | |
| Rakesh said: (Aug 10, 2016) | |
| It'due south for the upcoming years. How to calculate the same year backward? | |
| Venkatesh said: (Sep 4, 2016) | |
| 2007 is a normal year and front year 2006 is as well a normal year then we have to add together 11years to the given yr. i.e; 2007 + 11years = 2018. | |
| Ramandeep said: (Oct eight, 2016) | |
| Simply split 14 by 7 if zero remainder and so odd day zilch. | |
| Pavan Kumar said: (Oct sixteen, 2016) | |
| Thanks for the explanation. I take a small doubt. Every bit per your method for agenda 1988 twelvemonth the calendar should be repeated on 1992. Simply it is repeated on 2016. How information technology is possible? Delight explain. | |
| Narayan said: (October 31, 2016) | |
| Still I can non understand. Please help me. | |
| Rakesh Parmar said: (Nov 12, 2016) | |
| @All, Simply call back this. Split up Whatever yr by 4. If y'all go 0, add together 28 to given year, If you lot get 1 add + 6. | |
| Sonali said: (Dec 9, 2016) | |
| Thanks @Prasad. | |
| Santhosh said: (Jan 3, 2017) | |
| I have a query regarding the year 2006. No of odd days for respective years. 1, 1, two, i, i, one. Till 2011, total odd days becomes 0 hence answer should be 2011 but it is 2017 why? | |
| Santhosh said: (Jan three, 2017) | |
| I accept a query regarding the year 2006. No of odd days for respective years. 1, one, two, ane, 1, 1. Till 2011, total odd days becomes 0 hence answer should be 2011 but it is 2017 why? | |
| Santhosh said: (Jan iii, 2017) | |
| I have a query regarding the twelvemonth 2006. No of odd days for corresponding years. 1, 1, 2, one, 1, 1. Till 2011, total odd days becomes 0 hence answer should be 2011 only information technology is 2017 why? | |
| Navnath said: (Jan 6, 2017) | |
| The year next to 1973 having the same calendar as that of 1973 is ____. ane) 1977 Please solve this. | |
| Mani said: (Jan vii, 2017) | |
| Cheers very much. I understand now. | |
| Sumanth said: (Jan 22, 2017) | |
| It is 1979 @Navnath. | |
| Niks said: (Feb 9, 2017) | |
| Thank you for easy method @Prabhuchandu. | |
| Krishna Kittu said: (Feb 17, 2017) | |
| For finding repeated calendars nosotros used formulae of adding 28, 6, 11. for finding the past repeated calendars we have to decrease aforementioned 28, 6, 11. Am I right? | |
| Riya Khandelwal said: (Apr ten, 2017) | |
| I am not understanding this. Please explain how 2018 volition be similar to 2007? | |
| Siri said: (Apr 21, 2017) | |
| Whenever they ask for the same calender we should check whether it is a bound year or not. If it is a leap year add 28 to the given year and so you lot volition get the aforementioned calendar for spring year for non- bound years. 6 and eleven (half-dozen is added when given twelvemonth is subtracted from previous spring year u shoud get 1. i.e ex=2005 is not a leap year..previous leap year is 2004 then (04-05=i), so add vi considering we got 1 after subtracting non-leap yr with previous leap year then (2005+half-dozen=2011). (eleven is added when you decrease non-leap year to previous jump year you will become 2 nor three). | |
| Aishwarya said: (Jun 5, 2017) | |
| The number of extra days more than a complete week in a given period is called odd days. Hence in that location are 14 days which can be of 2 weeks. As 7 days is i week and other vii days are another calendar week then at that place are no more than days which are beyond a complete week. Therefore, the odd days are 0. | |
| Sowmya said: (Jun 8, 2017) | |
| Is this procedure is same for leap year? Delight, anyone explain. | |
| Balu Srinivas Reddy said: (Jul thirteen, 2017) | |
| For 4n => add 28 years. If 2004 For 4n+1 => add half-dozen years. | |
| Vijay said: (Aug 9, 2017) | |
| Please explain in detail. | |
| Anu said: (Aug xv, 2017) | |
| It applies to all years: Bound twelvemonth == add 28. Here 2007, nearest leap year before 2007, 2004. Therefore 2007 is 3 yrs after 2004. Therefore add 11 to 2007=2017. | |
| Manjunath N said: (Aug 27, 2017) | |
| Assume 2007 starts with Sunday. 2008 starts - mon. | |
| Pratik said: (Sep 19, 2017) | |
| How to detect respond this question? Seema remembers that her female parent's altogether falls on after 21st January just before 25th January while his blood brother remembers that it falls after 23rd Jan but before 28th January. When is Seema female parent birthday? | |
| Sharayu said: (Sep 24, 2017) | |
| How is the odd days for year 2017 1? I still didn't go this? | |
| Prudhviraj said: (Oct 17, 2017) | |
| @ALL. start with following year | |
| Jitendra Singh said: (Nov 22, 2017) | |
| Calander of 2016 is repeated in which yr? Delight explain. | |
| Dfeverx said: (April 11, 2018) | |
| For every ordinary year number of odd days = one odd 24-hour interval. For leap yr number of odd days = ii odd days. It implies next twelvemonth i.e 2021 will accept the same calendar equally 2016. | |
| Shivam Giri said: (May 4, 2018) | |
| The calendar of 1872 repeated on which twelvemonth and why? Please explain the answer. | |
| Rupa said: (Jul i, 2018) | |
| Why this play a joke on is not applicable to 2008. Can anyone explain? | |
| Vinod said: (Jul 5, 2018) | |
| Nice, Thanks @Prasand. | |
| Evlin said: (Aug 19, 2018) | |
| Skilful 1 @Katy. It'southward a time efficient method. Cheers. | |
| Jitendra said: (Sep 28, 2018) | |
| 2007 terminal ii digit sectionalization by 4. if the remainder is 1, add 6. remainder is 2 add 11, and iii add 11 get an answer. | |
| Akanksha said: (Sep 29, 2018) | |
| Why do we take till 2017? Please explain. | |
| Semran Sheikh said: (Oct 9, 2018) | |
| Thanks @Prasad. | |
| Bapmo said: (Nov xi, 2018) | |
| Thanks @Prasad and @Manjunath.N. | |
| Samruddhia said: (Dec 12, 2018) | |
| @All. A leap yr calendar repeats itself in 28 years and an Ordinary yr Calender repeats itself in 6 or 11 years. Here, we can simply add together 2007+6=2013 (which is not there in the option) and 2007+11= 2018 which is the reply. | |
| Sheetal said: (Dec 28, 2018) | |
| In these types of questions, divide terminal 2digits of the given year from four, after that if divisible is-- 2 and iii add 11in information technology, i add 6 in it, 0 add 28 in it. Eg.2007(07/iv) divisible is three then add 11 in 2007= 2018. | |
| Manish said: (Jun 14, 2019) | |
| 2007 is a general year so either it will repeat subsequently twelvemonth or 11 twelvemonth let's come across; Twelvemonth : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017. If the addition is seven then the calendar will alter just 2012 is bound year so general year can't exist a bound twelvemonth so now we get till the fourteen. | |
| Pradyumna Kumar Tiwari said: (Jul 1, 2019) | |
| If the year is leap then add 28 to get the same agenda provided that no century yr which does not leap should fall in betwixt. If the yr is non leap, add 11 and if the result is non spring; the result has the aforementioned calender. If the result is leap then add together 6 to go same calender. It should be taken that no century twelvemonth which is not leap should not autumn in between. 2007 is not leap and then 2007+11=2018, 2018 is not leap; hence 2018 is the answer. | |
| 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 16, 2019) | |
| How answer is 2018? Please, anyone explain me. | |
| Fathima Safwana said: (October 30, 2019) | |
| Is this Holds expert for for twelvemonth where for which year 2009 will be same ? | |
| Ank said: (Dec 8, 2019) | |
| Step1- calculate odd days from given yr to that twelvemonth at which improver is divisible farewell 7. | |
| Mandakranta said: (Feb 24, 2020) | |
| 2009 was repeated in 2015. 2009 - 1 odd twenty-four hour period Again 2009 being not-bound year if we divide by four nosotros get a remainder 1. So 2009 + 11= 2020 which is not possible 2020 being a leap yr. So 2009 +vi =2015 is the repeated the year of 2009. | |
| Meghana said: (Mar 21, 2020) | |
| Ordinary yr = 1odd day. Spring year = 2odd days. They told to fïnd oud aforementioned calendar as 2007, then we should start counting an odd mean solar day from 2008, Add the above odd days. To get same calendar odd days should be multiple of vii. vii divided by vii gives reminder 0. Hence 2012 is the respond. | |
| Kunal said: (May 8, 2020) | |
| How the odd days taken here? | |
| Gireesh Kumar said: (May 17, 2020) | |
| @Kunal. See, what is the remainder after dividing the year by 4. If the residual is i add 5 years to the given twelvemonth. If the remainder is 2 add together 11 to the given year. If the remainder is 3 adding 11 to the given year. And so in the higher up question, they gave 2007. Subsequently dividing by four. Information technology gives the remainder as three. So simply add together 11 years to 2007. That will be 2018. Hope information technology helps y'all. | |
| Aniket said: (Jul 6, 2020) | |
| And then, What is the solution for 2008? Anyone explain to me. | |
| Shivam Padmani said: (Jul 27, 2020) | |
| Short play tricks for solving. One has to separate the given year past 4. | |
| Suraj said: (Nov 29, 2020) | |
| Why 2014 cannot be the answer. In that location are 7 odd days. Which sum up again to 0 odd days? | |
| Vigneshwaran said: (Apr 7, 2021) | |
| Thanks for explaining @Prasad. | |
| Mithilesh Mahatha said: (Apr 19, 2021) | |
| @All. Then, doubt here is how xiv = 0, | |
| Rama said: (Jun 23, 2021) | |
| Thank you for explaining @Prasad. | |
| Mesfin said: (Oct 2, 2021) | |
| Why get-go count odd number fron 2008 & why not starting time from 2007? Please explain me. | |
| Sonu Kumar said: (Apr 17, 2022) | |
| @All. Divide the given no past four if balance is 0 add 28 to the yr if rest is 1 add half dozen or if residual is 2or3 add together eleven, in this example, the remainder is 3. And so, 2007 + 11 = 2018. | |
| Ravi Maurya said: (Apr 30, 2022) | |
| This solution is not correct. Rule:- 1. Rule:- ii. Rule:-iii. | |
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