Important Practice Questions & Short Tricks

Duration: 7 min

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The video is a lecture on solving calendar-based aptitude questions, presented by an instructor from Knowledge Gate. The first problem, from HP 2017, asks when Virat Kohli will celebrate his next wedding anniversary on the same day of the week, given that he celebrated it on Monday, December 11, 2007. The instructor explains that this requires calculating the day of the week for December 11, 2018, by determining the number of odd days between the two dates. He uses a table to calculate the number of odd days for each century (100, 200, 300, etc.), noting that 100 years have 5 odd days, 200 have 3, 300 have 1, and 400 have 0. He then calculates the total odd days from 2007 to 2018, which is 11 years, and finds that 2018 is 11 days ahead of 2007. Since 11 days is 1 week and 4 days, the day of the week advances by 4 days. Starting from Monday, adding 4 days results in Friday. The second problem, from Capgemini 2016, asks which day of the week cannot be the last day of a century. The instructor uses the same table of odd days for centuries to show that the last day of a century can be Friday (for 100, 200, 300 years), Wednesday (for 400 years), and Monday (for 800 years), but never Tuesday or Thursday. The video concludes with a 'Thanks for Watching' screen.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with a title card for a calendar problem. The first question, from HP 2017, is displayed: 'Virat Kohli Celebrated his wedding anniversary on Monday on 11th December 2007, when will he celebrate his next wedding anniversary on the same day?'. The instructor, Yash Jain, begins to solve the problem by explaining that the next anniversary will be on December 11, 2018. He states that to find the day of the week, one must calculate the number of odd days between 2007 and 2018. He introduces a table on the screen that lists the number of odd days for different century lengths: 100 years have 5 odd days, 200 have 3, 300 have 1, and 400 have 0. He explains that a century is 100 years, and the number of odd days is the remainder when the total number of days is divided by 7.

  2. 2:00 – 5:00 02:00-05:00

    The instructor continues to solve the first problem. He calculates the number of odd days from 2007 to 2018. He writes '2007' and '2018' on the screen and adds a '+11' to indicate the 11 years between the two dates. He explains that 11 years consist of 8 ordinary years and 3 leap years. He calculates the total number of days as (8 * 365) + (3 * 366) = 4018 days. He then divides 4018 by 7 to find the number of odd days, which is 4018 / 7 = 574 weeks and 0 days. He then adds the 4 odd days from the century calculation (100 years have 5 odd days, but since 2007 is not a century year, he uses the 5 odd days from the 100-year cycle). He concludes that the day of the week advances by 4 days from Monday, which is Friday. He then transitions to the second problem.

  3. 5:00 – 6:37 05:00-06:37

    The second problem is introduced: 'The last day of a century cannot be'. The options are Monday, Wednesday, Tuesday, and Friday. The instructor refers to the table of odd days for centuries. He explains that the last day of a century can be Friday (for 100, 200, 300 years), Wednesday (for 400 years), and Monday (for 800 years). He states that the last day of a century can never be Tuesday or Thursday. He uses the table to show that the number of odd days for 100, 200, 300, and 400 years are 5, 3, 1, and 0 respectively. He then explains that the last day of a century can be Friday (5 odd days), Wednesday (3 odd days), and Monday (1 odd day), but never Tuesday or Thursday. The video ends with a 'Thanks for Watching' screen.

The video presents a structured approach to solving two common calendar aptitude problems. The first problem involves calculating the day of the week for a future date by determining the number of odd days between two years, using a table of odd days for centuries. The second problem uses the same table to determine which day of the week cannot be the last day of a century, based on the pattern of odd days. The instructor uses a clear, step-by-step method, writing out calculations and referencing the table, to demonstrate the logic behind the solutions.