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Reasoning Calendar Best Tricks Solve а а аґ а а ѕа аґќа 30 Se

текстовые задачи задачи на проценты задание 11 егэ математика
текстовые задачи задачи на проценты задание 11 егэ математика

текстовые задачи задачи на проценты задание 11 егэ математика After 5 year ny – 1 jan monday (2 od of previous year) after 5 year – (the total of odd days is 7 means monday) alternate method. add the odd day of years till it become 7 or multiple of 7. 2 ly 1 ny 1 ny 1 ny 2 ly = 7 (after 5 years) q 4: ram celebrated his 4th birthday on 29 feb 2020. when he was born. How to solve calendar reasoning questions – tips and tricks candidates can find various tips and calendar reasoning tricks from below for solving the questions that may come in competitive exams. tip # 1: in an ordinary year, the first day of the months are same for january and october (1 10), april and july (4 7), february, march and.

Consider The Following Aes S Box Used In The Bytesub Chegg
Consider The Following Aes S Box Used In The Bytesub Chegg

Consider The Following Aes S Box Used In The Bytesub Chegg The topic “calendar” falls under the category of logical reasoning as it involves a lot of logical discussion and analysis. one can definitely expect 2 to 4 problems in the question papers of various govt and bank exams. the following list of entrance exams have calendar as a topic under the logical reasoning section:. These are the steps to solve: 1) pick up last two digits from the given year. 2) divide it by 4 and take its quotient. 3) use date that is already given in the question. 4) use month code of the given month specified in the question. 5) use century code. 6) add altogether from 1 to all the way 5th step. Number of days between two dates. count the number of days in each year, month, and remaining days to calculate the total duration between two dates. 7. finding the day on a given date. use the formula: day = (reference day number of days) % 7, where reference day is the known day and number of days is the duration to be added. Ordinary year has 1 odd day. 1 leap year= 366 days= (52 weeks 2 days). one leap year has 2 odd days. 100 years = 76 ordinary years 24 leap years. = (76 x 1 24 x 2) odd days. = 124 odd days. = (17 weeks 5 days)= 5 odd days. so, the number of odd days in 100 years= 5. number of odd days in 200 years = (5×2).

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