Calendar maths Problems

Calendar maths Problems
Leap Year Problems
Calendar trick
Calendar's Problems for Competitive Exam, 
Calendar’s questions, and answers,                          Part-II
Calendar Maths Problems

Q1. If 9th Feb 2013 is Tuesday, which would be the day of the week on 9th Feb. 2012?
a. Wednesday
b. Monday
c. Sunday
d. Thursday
Ans: C.Sunday
Here, from 9th Feb 2012 to 9th 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 9th Feb. 2013 to 9th 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 19th March 2013 is Tuesday, which would be the day of the week on 19th March 2012?
a.       Wednesday
b.       Monday
c.       Sunday
d.       Thursday
Ans: b. Monday
Here, from 19th March 2012 to 19th 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 19th March. 2013 to 19th 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 typeWhat was the day of the week on 10th  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 15th 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 15th 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)
Now15+2+17+4+6=44÷7= remainder is 2And 2 is the code of ‘TUESDAY’ (as per Table-III). So the result is Tuesday
Q. What was the day of the week on 15th 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)
Now15+3+17+4+6=45÷7= remainder is 3And 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 is the code of ‘FRIDAY’ (as per Table-III). So the result is Friday.
Q. What will be the day of the week on 11th 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)
Now11+3+23+5+0=42÷7= remainder is 0And 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)
Nowthe 01+2+28+7+2=40÷7= remainder is 5. And 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-I1776 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)
Nowthe 16+6+76+19+4=121÷7= remainder is 2. And 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-I2002 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)
Nowthe 4+4+02+0+6=16÷7= remainder is 2. And 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-I2000 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)
Nowthe 22+5+00+00+6=31÷7= remainder is 3. And 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

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