Exams › IBPS PO › Quantitative Aptitude › Calendar
5 questions with worked solutions.
Q1. It was Sunday on January 1, 2006. What was the day of the week on January 1, 2010?
Answer: Friday
From Jan 1, 2006 to Jan 1, 2010, the odd days are: 2006 = 1, 2007 = 1, 2008 = 2 (leap year), 2009 = 1. Total odd days = 5, which means 5 days after Sunday, i.e. Friday.
Q2. What was the day of the week on 28th May, 2006?
Answer: Sunday
From 1 January 2006 to 28 May 2006, count the days elapsed: Jan 31, Feb 28, Mar 31, Apr 30, May 27 = 147 days. Since 147 is divisible by 7, the weekday remains the same as 1 January 2006, which was Sunday.
Q3. What was the day of the week on 17th June, 1998?
Answer: Wednesday
1998 is a non-leap year. From 1 January to 17 June, the number of days elapsed is 31 + 28 + 31 + 30 + 31 + 16 = 167 days, which gives 167 mod 7 = 6. Using the standard calendar calculation for 1998, 17 June 1998 falls on Wednesday.
Q4. What will be the day of the week on 15th August, 2010?
Answer: Sunday
2010 is a non-leap year. Counting the days from 1 January 2010 to 15 August 2010 gives 226 days elapsed, and 226 mod 7 = 2. Starting from the known weekday of 1 January 2010, this leads to Sunday on 15 August 2010.
Q5. Today is Monday. After 61 days, it will be:
Answer: Saturday
The day of the week repeats every 7 days. Since 61 mod 7 = 5, 61 days after Monday is 5 days ahead of Monday, which is Saturday.
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