Important Practice Questions & Short Tricks
Duration: 6 min
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This educational video presents a series of logical reasoning problems related to calendar calculations, typical of competitive exams. The lecture begins with a question about determining the day of the week for June 11th, 2013, given that August 15th, 2012, was a Thursday. The instructor uses a step-by-step method, calculating the number of days between the two dates and then finding the remainder when divided by 7 to determine the day of the week. The process involves identifying the number of days in each month and accounting for leap years. The video then transitions to a second problem, asking for the day of the week for January 7th, 2013, given that July 9th, 2012, was a Sunday. The final segment presents a multiple-choice question about which year among the options (2004, 2005, 2008) has 365 days, which is a test of understanding leap years. The video concludes with a 'Thanks for Watching' screen.
Chapters
0:00 – 2:00 00:00-02:00
The video opens with a title card for a calendar problem. The first question is displayed: 'If Aug 15th, 2012 falls on Thursday then June 11th, 2013 falls on which day?'. The instructor, Yash Jain Sir, begins the solution by writing '15th Aug 2012 -> Thursday' on the blackboard. He then calculates the number of days from August 15, 2012, to August 15, 2013, which is 366 days because 2012 is a leap year. He notes that 366 mod 7 equals 2, meaning the day of the week advances by 2 days. He then calculates the number of days from June 11, 2013, to August 15, 2013, which is 65 days. He writes '65 days' and calculates 65 mod 7, which is 2. He then subtracts this from the total advance, resulting in a net advance of 0 days, concluding that June 11, 2013, is a Thursday.
2:00 – 5:00 02:00-05:00
The instructor continues the solution for the first problem. He writes '15th Aug 2013 -> Friday' on the board, indicating the day of the week for August 15, 2013. He then calculates the number of days from June 11, 2013, to August 15, 2013, which is 65 days. He writes '65 days' and calculates 65 mod 7, which is 2. He then subtracts this from the total advance, resulting in a net advance of 0 days, concluding that June 11, 2013, is a Thursday. He then moves to the second problem: 'If July 9th, 2012 falls on Sunday then Jan 7th, 2013 falls on which day?'. He writes '9th July 2012 -> Sunday' on the board. He calculates the number of days from July 9, 2012, to July 9, 2013, which is 366 days. He notes that 366 mod 7 equals 2, meaning the day of the week advances by 2 days. He then calculates the number of days from January 7, 2013, to July 9, 2013, which is 183 days. He writes '183 days' and calculates 183 mod 7, which is 5. He then subtracts this from the total advance, resulting in a net advance of 0 days, concluding that January 7, 2013, is a Sunday.
5:00 – 5:58 05:00-05:58
The video transitions to a new question: 'Which of the following year has 365 days?'. The options are a) 2004, b) 2005, c) 2008, d) All of the above. The instructor explains that a year with 365 days is a common year, not a leap year. He explains the rule for leap years: a year is a leap year if it is divisible by 4, but not by 100, unless it is also divisible by 400. He checks the options: 2004 is divisible by 4 and not by 100, so it is a leap year (366 days). 2005 is not divisible by 4, so it is a common year (365 days). 2008 is divisible by 4 and not by 100, so it is a leap year (366 days). Therefore, only 2005 has 365 days. The video ends with a 'Thanks for Watching' screen.
The video provides a comprehensive tutorial on solving calendar-based logical reasoning problems. It demonstrates a systematic approach: first, calculate the total number of days between two dates, then find the remainder when divided by 7 to determine the day of the week shift. The key concept is that a common year advances the day of the week by 1, while a leap year advances it by 2. The video also reinforces the rule for identifying leap years, which is essential for accurate calculations. The progression from one problem to the next builds a clear understanding of the underlying principles, making it a valuable resource for students preparing for competitive exams.