The question is from CAT Calendars. How many years it takes to have the birthday on one particular day in a week? CAT exam is known to test on basics rather than high funda ideas. CAT also tests multiple ideas in the same question, and 2IIMs CAT question bank provides you with CAT questions that can help you gear for CAT Exam CAT 2020. Set Theory (especially constructing venn diagrams) is a frequently tested topic. Make sure you know the basics from this chapter.
Question 4: John was born on Feb 29th of 2012 which happened to be a Wednesday. If he lives to be 101 years old, how many birthdays would he celebrate on a Wednesday?
Let us do this iteratively. Feb 29th 2012 = Wednesday => Feb 28th 2012 = Tuesday
Feb 28th 2013 = Thursday (because 2012 is a leap year, there will be 2 odd days)
Feb 28th 2014 = Friday, Feb 28th 2015 = Saturday, Feb 28th 2016 = Sunday, Feb 29th 2016 = Monday
Or, Feb 29th to Feb 29th after 4 years, we have 5 odd days.
So, every subsequent birthday, would come after 5 odd days.
2016 birthday – 5 odd days
2020 birthday – 10 odd days = 3 odd days
2024 birthday – 8 odd days = 1 odd day
2028 birthday – 6 odd days
2032 birthday – 11 odd days = 4 odd days
2036 birthday – 9 odd days = 2 odd days
2040 birthday – 7 odd days = 0 odd days. So, after 28 years he would have a birthday on Wednesday
The next birthday on Wednesday would be on 2068 (further 28 years later), the one after that would be on 2096. His 84th birthday would again be a leap year.
Now, there is a twist again, as 2100 is not a leap year. So, he does not have a birthday in 2100. His next birthday in 2104 would be after 9 odd days since 2096, or 2 odd days since 2096, or on a Thursday.
From now on the same pattern continues. 2108 would be 2 + 5 odd days later = 7 odd days later. Or, 2108 Feb 29th would be a Wednesday.
So, there are 4 occurrences of birthday falling on Wednesday – 2040, 2068, 2096 and 2108.
The question is "How many birthdays would he celebrate on a Wednesday?"
Choice B is the correct answer.
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