D) Sunday

 It was Saturday on 31st December, 2005.

 Number of odd days from 2006 to 2009 = (1 + 1 + 2 + 1) = 5 days.

Hence on December 31, 2009, it was Thursday. Thus, on Jan 3, 2010 it was Sunday. 

D) Rs. Tuesday

 We know year 2004 is a leap year. Therefore, it has 2 odd days.

 Feb 2004 was not included as we are calculating from March 2004 to March 2005.

So it has 1 odd day. 

The day on March 13, 2005 will be 1 day ahead of the day on March 13, 2004.

Given that, March 13, 2005 is Wednesday.

So March 13, 2004 is Tuesday (1 day before to 13th March, 2005).

C) Saturday

100 years contain 5 odd days.

Therefore, Last day of 1st century is Friday

200 years contain (5 x2) is 3 odd days (approx.)

Last day of 2nd century is Wednesday

300 years contain ( 5x3) = 15 i.e. 1 odd day( approx.)

Last day of 3rd century is Monday 400 years contain 0 odd day

Last day of 4th century is Sunday.

Thus, the cycle is repeated.

Last day of a century cannot be Tuesday or Thursday or Saturday.

D) Tuesday

Step 1 : Take the last 2 digit of year : 76

Step 2 : Divide step 1 digit by 4 and take the quotient 76/4 we get 19

Step 3 : Take the day of the Date : 16

Step 4 : Code of the month : 6

Step 5 : Code of the year : 4

Step 6 : Add all of them and divide it by 7 remainder is the answer : 121/7 = 2 i.e Tuesday

[ It take only 20 - 30 sec to solve this sum using above trick given in the video ]

B) Sunday

Step 1 : Take the last 2 digit of year : 10

Step 2 : Divide step1 digit by 4 and take the quotient 10/4 we get 2

Step 3 : Take the day of the Date : 15

Step 4 : Code of the month : 2

Step 5 : Code of the year : 6

Step 6 : Add all of them and divide it by 7 remainder is the answer : 35/7 = 0 i.e Sunday

B) Thursday

The year 2006 was an ordinary year. So, it has 1 odd day.

The day on 7th Dec, 2007 will be 1 day beyond the day on 7th Dec, 2006.

But, 7th Dec, 2007 is Friday

So, 7th Dec, 2006 is Thursday.

C) Tuesday

The year 2007 is an ordinary year. So, it has 1 odd day.

1st day of the year 2007 was Monday

1st day of the year 2008 will be 1 day beyond Monday

Hence, it will be Tuesday.

B) 6th,13th,20th,27th

Let us find the day on 1st July, 2003.

Step 1 : Take the last 2 digit of year : 03

Step 2 : Divide step1 digit by 4 and take the quotient 03/4 we get 00

Step 3 : Take the day of the Date : 01

Step 4 : Code of the month : 6

Step 5 : Code of the year : 6

Step 6 : Add all of them and divide it by 7 remainder is the answer : 16/7 = 2 i.e Tuesday

Thus, 1st Sunday in July 2003 is 6th July.

So, during July 2003, Sunday fell on 6th, 13th, 20th and 27th.

C) Saturday

Step 1 : Take the last 2 digit of year : 47

Step 2 : Divide step1 digit by 4 and take the quotient 47/4 we get 11

Step 3 : Take the day of the Date : 16

Step 4 : Code of the month : 2

Step 5 : Code of the year : 0

Step 6 : Add all of them and divide it by 7 remainder is the answer : 76/7 = 6

i.e Saturday

A) 2018

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.

Sum = 14 odd days 0 odd days

Calendar for the year 2018 will be the same as for the year 2007

C) Wednesday

Step 1 : Take the last 2 digit of year : 99

Step 2 : Divide step1 digit by 4 and take the quotient 99/4 we get 24

Step 3 : Take the day of the Date : 16

Step 4 : Code of the month : 4

Step 5 : Code of the year : 0

Step 6 : Add all of them and divide it by 7 remainder is the answer : 143/7 = 3 i.e wednesday

A) Wednesday

First, we count the number of odd days for the left over days in the given period.
Here, given period is from 15.8.2012 to 11.6.2013
Aug Sept Oct Nov Dec Jan Feb Mar Apr May Jun
16 30 31 30 31 31 28 31 30 31 11(left days)
Therefore, 2 + 2 + 3 + 2 + 3 + 3 + 0 + 3 + 2 + 3 + 4 (odd days) = 6 odd days
So, the given day Thursday + 6 = Wednesday.

D) Sunday

The year 2004 was a leap year. So, it had 2 odd days.

The day on Feb 8, 2004 must be 2 days before the day on Feb 8, 2005.

Hence, this day was Sunday

C) Wednesday

Step 1 : Take the last 2 digit of year : 98

Step 2 : Divide step1 digit by 4 and take the quotient 98/4 we get 24

Step 3 : Take the day of the Date : 17

Step 4 : Code of the month : 4

Step 5 : Code of the year : 0

Step 6 : Add all of them and divide it by 7 remainder is the answer : 143/7 = 3 i.e wednesday

B) Sunday

Step 1 : Take the last 2 digit of year : 06

Step 2 : Divide step1 digit by 4 and take the quotient 06/4 we get 01

Step 3 : Take the day of the Date : 28

Step 4 : Code of the month : 1

Step 5 : Code of the year : 6

Step 6 : Add all of them and divide it by 7 remainder is the answer : 42/7 = 0 i.e Sunday

C) Friday

Step 1 : Take the last 2 digit of year : 79

Step 2 : Divide step1 digit by 4 and take the quotient 79/4 we get 19

Step 3 : Take the day of the Date : 09

Step 4 : Code of the month : 3

Step 5 : Code of the year : 0

Step 6 : Add all of them and divide it by 7 remainder is the answer : 110/7 = 5 i.e Friday

D) Wednesday

rather than finding odd days between both the dates i will use the trick and directly find the day on Oct 10, 2001 because it will save time.

Step 1 : Take the last 2 digit of year : 01

Step 2 : Divide step1 digit by 4 and take the quotient 01/4 we get 00

Step 3 : Take the day of the Date : 10

Step 4 : Code of the month : 0

Step 5 : Code of the year : 6

Step 6 : Add all of them and divide it by 7 remainder is the answer : 17/7 = 3 i.e Wednesday

A) Friday

rather than finding odd days between both the dates i will use the trick and directly find the day on September 17, 1915 because it will save time.

Step 1 : Take the last 2 digit of year : 15

Step 2 : Divide step1 digit by 4 and take the quotient 15/4 we get 03

Step 3 : Take the day of the Date : 17

Step 4 : Code of the month : 5

Step 5 : Code of the year : 0

Step 6 : Add all of them and divide it by 7 remainder is the answer : 40/7 = 5 i.e Friday

D) 2nd, 9th, 16th, 23rd, 30th

Let first find the day on 1st April, 2002.

Step 1 : Take the last 2 digit of year : 02

Step 2 : Divide step1 digit by 4 and take the quotient 02/4 we get 00

Step 3 : Take the day of the Date : 01

Step 4 : Code of the month : 6

Step 5 : Code of the year : 6

Step 6 : Add all of them and divide it by 7 remainder is the answer : 15/7 = 01 i.e Monday

On April 1, 2002 it was Monday.

So in April 2002, Tuesday falls on 2nd, 9th, 16th and 23rd & 30th.

D) Tuesday

Step 1 : Take the last 2 digit of year : 95

Step 2 : Divide step1 digit by 4 and take the quotient 95/4 we get 23

Step 3 : Take the day of the Date : 02

Step 4 : Code of the month : 01

Step 5 : Code of the year : 00

Step 6 : Add all of them and divide it by 7 remainder is the answer : 121/7 = 02

i.e Tuesday

C) Monday

Each day of a week is repeated after 7 days, so after 70 days, it will be Wednesday.

Therefore, after 68 days, it will be Monday.

A) Wednesday

1997 was an ordinary year, it had 1 odd day. So, the day on February 5, 1998, would be one day beyond the day on February 5, 1997

∴ Thursday on February 5, 1998, would be one day beyond the day on February 5, 1997, so the day on February 5, 1997, was Wednesday.

D) Monday

Step 1 : Take the last 2 digit of year : 91

Step 2 : Divide step1 digit by 4 and take the quotient 91/4 we get 22

Step 3 : Take the day of the Date : 17

Step 4 : Code of the month : 04

Step 5 : Code of the year : 00

Step 6 : Add all of them and divide it by 7 remainder is the answer : 134/7 = 01.

i.e Monday

B) Wednesday

The year 1996 is divisible by 4, so it is a leap year with 2 odd days.

As per the question, the first day of the year 1996 was Monday, so the first day of the year 1997 must be two days after Monday. So, it was Wednesday.

C) Thursday

Step 1 : Take the last 2 digit of year : 50

Step 2 : Divide step1 digit by 4 and take the quotient 50/4 we get 12

Step 3 : Take the day of the Date : 26

Step 4 : Code of the month : 00

Step 5 : Code of the year : 00

Step 6 : Add all of them and divide it by 7 remainder is the answer : 88/7 = 04 i.e Thursday

A) 8y

There are 7 days in a week, so y weeks will contain 7y days.

∴The required number of days = 7y + y days = 8y days

A) 5^th July

April & July for all years have the same calendar. So, a day on any date of April will be the same day on the corresponding date in July.

∴The same day will fall on 5th July of the same year.

B) Friday

On December 31, 2004, it was Friday.

Number of odd days from the year 2005 to the year 2009 = 1+1+1+2(leap year) +1 = 6 odd days

∴On December 31, 2009, it was Thursday, so on January 1, 2010, it was Friday.

B) Friday

Total number of odd days between the years 2002 and 2006 = (2006 - 2002) + 1 = 5 odd days. The year 2004 is a leap year, it has two odd days. So, one extra odd day is added.

So, if it was Wednesday on March 1, 2006, it would be (Wednesday - 5) Friday on March 1, 2002.

B) 700

A century divisible by 400 is a leap year. The years 800, 1600 and 2000 are divisible by 400, so they are leap years.

The year 700 is not divisible by 400, so it is not a leap year.