Calendar maths Problems

Calendar maths Problems
Leap Year Problems
Calendar trick
Calendar's Problems for Competitive Exam, 
Calendar’s questions, and answers,                          Part-II
Q1. If 9^th Feb 2013 is Tuesday, which would be the day of the week on 9^th Feb. 2012?
a. Wednesday
b. Monday
c. Sunday
d. Thursday

Ans: C.Sunday

Here, from 9^th Feb 2012 to 9^th Feb.2013 there are 366 days because Feb of 2012 has 29 days due to leap year. So, here the extra day is 2. Oppositely, if we back from 9^th Feb. 2013 to 9^th Feb 2012 we will get the same 2 extra days. So we must back 2 days from the given day. That is, Tuesday-2days=Sunday.

Q2. If 19^th March 2013 is Tuesday, which would be the day of the week on 19^th March 2012?

a.       Wednesday

b.       Monday

c.       Sunday

d.       Thursday

Ans: b. Monday

Here, from 19^th March 2012 to 19^th Feb.2013 there are 365 days because March of 2013 has 28 days due to Normal year. So, here the extra day is 1. Oppositely, if we back from 19^th March. 2013 to 19^th March 2012 we will get the same 1 extra day. So we must back 1 day from the given day. That is, Tuesday-1day=Monday.




Table-I

Same Month Code:

Name of Months	Same months Code of Normal Years( NY)	Same months Code of Leap Years( LY)
January	0	6
February	3	2
March	3	3
April	6	6
May	1	1
June	4	4
July	6	6
August	2	2
September	5	5
October	0	0
November	3	3
December	5	5

Table-II

Century Code:

Century	Century Code	Century	Century Code	Century	Century Code	Century	Century Code
⇧	6	⇧	4	⇧	2	⇧	0
800		900		1000		1100
1200		1300		1400		1500
1600 (⇧-400)		1700(⇧-400)		1800		1900
2000  (⇩-400)		2100(⇩-400)		2200		2300
2400		2500		2600		2700
2800		2900		3000		3100
⇩		⇩		⇩		⇩

Table-III

Day Code ( Day against Code)

Code→	0	1	2	3	4	5	6
Day→	Sunday	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday

NOTE: To find the answer like question type " What was the day of the week on 10^th  January 2012? " we must know/have the following 5 steps

1 1  Given date=

2 2  Given month code= (from table-I)

3 3 Last 2(two) digits of the given year=

4 4  Last 2(two) digits of the given year÷4= Take the only Quotient (if last 2 digits less than 4, then it will be Zero)

5 5 Century Code= (from table-II)

Then add points 1+2+3+4+5=Total÷7=remainder, then match the remainder with day code ( from Table-III), that day will be the result.

Q. What was the day of the week on 15^th August 1947?

• a.   Friday
• b.     Monday
• c.     Sunday
• d.     Thursday

Ans: A. Friday

1.    1.   G/D=15

2.     2.  M/Code=2 (As per Table-I, 1947 is a normal year and month code of Aug. is 2)

3.     3. L2D of G/Y=47

4.     4.  L2D of G/Y=47÷4=11 (Only Quotient)

5.     5.  Cent. /Code=0 (here 1947 is 1900 century so code is 0, as per table-II)

Now, 15+2+47+11+0=75÷7= remainder is 5. And 5 is the code of ‘FRIDAY’ (as per Table-III). So the result is Friday

Q. What was the day of the week on 15^th August 2017?

• a.   Friday
• b.     Monday
• c.     Sunday
• d.     Tuesday

Ans: D. Tuesday

1.    1.   G/D=15

2.     2.  M/Code=2 (As per Table-I, 2017 is a normal year and month code of Aug. is 2)

3.     3. L2D of G/Y=17

4.     4.  L2D of G/Y=17÷4=4 (Only Quotient)

5.     5.  Cent. /Code=6 (here 2017 is 2000 century so the code is 6, as per table-II)

Now, 15+2+17+4+6=44÷7= remainder is 2. And 2 is the code of ‘TUESDAY’ (as per Table-III). So the result is Tuesday

Q. What was the day of the week on 15^th March 2017?

• a.   Wednesday
• b.     Monday
• c.     Sunday
• d.     Thursday

Ans: A. Wednesday

1.    1.   G/D=15

2.     2.  M/Code=3 (As per Table-I, 2017 is a normal year and month code of March. is 3)

3.     3. L2D of G/Y=17

4.     4.  L2D of G/Y=17÷4=4 (Only Quotient)

5.     5.  Cent. /Code=0 (here 2017 is 2000 century so the code is 6, as per table-II)

Now, 15+3+17+4+6=45÷7= remainder is 3. And 3 is the code of ‘Wednesday’ (as per Table-III). So the result is Wednesday.

Q. What was the day on 23rd Feb 2018?

• a.   Friday
• b.     Monday
• c.     Sunday
• d.     Thursday

Ans: A. Friday

1.    1.   G/D=23

2.     2.  M/Code=3 (As per Table-I, 2018 is a Normal year and month code of Feb. is 3)

3.     3. L2D of G/Y=18

4.     4.  L2D of G/Y=18÷4=4 (Only Quotient)

5.     5.  Cent. /Code=6 (here 2018 is 2000 century so the code is 6, as per table-II)

Now, the 23+3+18+4+6=54÷7= remainder is 5. And 5 is the code of ‘FRIDAY’ (as per Table-III). So the result is Friday.

Q. What will be the day of the week on 11^th March 2323?

• a.   Friday
• b.     Monday
• c.     Sunday
• d.     Thursday

Ans: C. Sunday

1.    1.   G/D=11

2.     2.  M/Code=3 (As per Table-I, 2023 is a normal year and month code of March. is 3)

3.     3. L2D of G/Y=23

4.     4.  L2D of G/Y=23÷4=5 (Only Quotient)

5.     5.  Cent. /Code=0 (here 2323 is 2300 century so the code is 0, as per table-II)

Now, 11+3+23+5+0=42÷7= remainder is 0. And 0 is the code of ‘Sunday’ (as per Table-III). So the result is Sunday.

Q. What was the day on 01st Feb 1028?

• a.   Friday
• b.     Monday
• c.     Sunday
• d.     Thursday

Ans: A. Friday

1.    1.   G/D=01

2.     2.  M/Code= 2(As per Table-I, 1028 is a Leap Year and Month code of Feb. is 2 )

3.     3. L2D of G/Y=28

4.     4.  L2D of G/Y=28÷4=7 (Only Quotient)

5.     5.  Cent. /Code= 2 (here 1028 is 1000 century so the code is 2, as per table-II)

Now, the 01+2+28+7+2=40÷7= remainder is 5. And 5 is the code of ‘FRIDAY’ (as per Table-III). So the result is Friday.

Q. What was the day of the week on 16th January 1776?

• a.   Tuesday
• b.     Monday
• c.     Sunday
• d.     Thursday

Ans: A. Tuesday

1.    1.   G/D=16

2.     2.  M/Code= 6 (As per Table-I, 1776 is a Leap Year and month code of Jan. is 6)

3.     3. L2D of G/Y=76

4.     4.  L2D of G/Y=76÷4=19 (Only Quotient)

5.     5.  Cent. /Code= 4 (here 1776 is 1700 century so the  cent. code is 4, as per table-II)

Now, the 16+6+76+19+4=121÷7= remainder is 2. And 2 is the code of ‘Tuesday’ (as per Table-III). So the result is Tuesday.

Q. What was the day of the week on 04th June 2002?

• a.   Tuesday
• b.     Monday
• c.     Sunday
• d.     Thursday

Ans: A. Tuesday

1.    1.   G/D=4

2.     2.  M/Code= 4 (As per Table-I, 2002 is a Normal Year and month code of June. is 4)

3.     3. L2D of G/Y=02

4.     4.  L2D of G/Y=02÷4=0 (Only Quotient)

5.     5.  Cent. /Code= 6 (here 2002 is 2000 century, so the  cent. code is 6, as per table-II)

Now, the 4+4+02+0+6=16÷7= remainder is 2. And 2 is the code of ‘Tuesday’ (as per Table-III). So the result is Tuesday.

Q. What was the day of the week on 22nd Dec. 2000?

• a.   Wednesday
• b.     Monday
• c.     Sunday
• d.     Thursday

Ans: A. Wednesday

1.    1.   G/D=22

2.     2.  M/Code= 5 (As per Table-I, 2000 is a Leap Year and month code of Dec. is 5)

3.     3. L2D of G/Y=00

4.     4.  L2D of G/Y=00÷4=0(Only Quotient)

5.     5.  Cent. /Code= 6 (here  2000 century year, so the  cent. code is 6, as per table-II)

Now, the 22+5+00+00+6=31÷7= remainder is 3. And 3 is the code of ‘Wednesday’ (as per Table-III). So the result is Wednesday.


NOTE: 
G/D=Given date.
M/Code=Month Code
L2D=Last two digits 
G/Y=Given Year


                                                                                Continue to Part-III