**Type 1: Calendar Tips and Tricks- To find the day of given date**

**Question 1. What was the day on 27th May 2006?**

**Options:**

**A. Monday**

**B. Saturday**

**C. Sunday**

**D. Friday**

**Solution: **27^{th} May, 2006 = (2005 years + Period from 1.1.2006 to 27.5.2006)

To calculate number of odd days till 2000, we need

Number of odd days in 1600 years = 0

Number of odd days in 400 years = 0

Now, for calculating odd days in the next five years,

5 years = (4 ordinary years +1 leap year) = (4+2)

= 6 odd days

Now to calculate number of odd days from 1^{st} January 2006 to 27^{th} May 2006

January (31 days) + February (28 days because 2006 is an ordinary year) + March (31 days) + (April 30 days) + May 27 days = 147 days

Total number of odd days in 147 days = ((147/7) = 21 weeks+ 0 odd days

Total number of odd days in the entire period = 0 (1600 years) + 0 (400 years) + 6 (5 years) + 0 (Period from 1.1.2006 to 27.5.2006) = 6 odd days

As per the table, on 27th May 2006 the day was Saturday.

Days | Sunday | Monday | Tuesday | Wednesday | Thursday | Friday | Saturday |

Number of odd days | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

**Correct option: B**

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