Introduction
11 Plus exams will always have a surprise element in their tests every year. These surprise elements are mainly derived from logical application of mathematical or literacy concpet from our day to day life. A teacher at Champs Learning explains this as an effort by schools to identify the students who can apply their learning analytically. After the concpet of time-speed-distance, concept of calender and problems based on it also fall into puzzle segment like age related problems. Most students face difficulty in solving Calendar math problems. So here we try to provide a simplified procedure to solve calendar reasoning problems.
Let us begin with the basics. We know that in an ordinary year there are 365 days, which means 52 × 7 + 1, or 52 weeks and one day. This additional day, we call an odd day. If 1st January of this year is on Sunday, then 1st January next year will be exactly 52 full weeks and a day after that – so on a Monday.
This is all right as long as the year is not a leap year. The Earth actually completes 1 orbit around the Sun in more than 365 days, i.e in 365 Days 5 Hours 48 minutes and 45 seconds or takes approximately 6 hours more. A leap year occurs every 4 years to adjust for the 1/4th day, 6 x 4= 24 hours, so every 4th year has 366 days (or 2 odd days). And as far as the few odd minutes of the orbit time are concerned, well every 100 years starting 1 AD, the year is declared to be a non–leap year, but every 4th century is a leap year. So any year divisible by 400 will be a leap year e.g. : 1200, 1600 and 2000. The years 1800, 1900 will be non leap years.
Concept of ‘odd-days’
The concept of odd days is very important in determining the days of the week. Let us look at how many odd days will there be in a century – i.e. 100 years. There will be 24 leap years and 76 non–leap years. As studied earlier each leap year and 2 odd days and each non-leap year has 1 odd day. Therefore, there will be 24 × 2 + 76 × 1 = 124 total odd days. Since 7 odd days make a week, to find out the net odd days, divide 124 by 7. The remainder is 5 – this is the number of odd days in a century.
Calendar Aptitude Tricks
You can remember the following points relating to the concepts of calendar:
100 years give us 5 odd days as calculated above.
200 years give us 5 x 2 = 10 – 7 (one week) 3 odd days.
300 years give us 5 x 3 = 15 – 14 (two weeks) 1 odd day.
400 years give us {5 x 4 + 1 (leap century)} – 21} (three weeks) 0 odd days.
Month of January gives us 31 – 28 = 3 odd days.
Month of February gives us 28 – 28 = 0 odd day in a normal year and 1 odd day in a leap year and so on for all the other months.
In total first six months i.e. January to June give us 6 odd days in a normal year and 7 – 7 = 0 odd days in a leap year. This is going to help, when you want to find a day, which is after 30th June. In total first nine months i.e. January to September give us 0 odd days in a normal year and 1 odd day in a leap year. Sometimes a reference date might be given to you. This makes your task easier, as you then start counting odd days only from that day.
When you count from the beginning i.e. 1st January, 0001
- 1 odd day mean – Monday
- 2 odd days mean – Tuesday
- 3 odd days mean – Wednesday and so on 6 odd days means Saturday.
Now let us solve an illustration:
Illustration: Any date in March is the same day of the week as corresponding date in of the same year.
A. OctoberB. NovemberC. JuneD. September
Answer: Option 2
Solution: 2 months have same corresponding days if number of odd days between these 2 months is 0. i.e. the total number of days are divisible by 7.
Now, between March and November, total number of days = 245.
Hence number of odd days =(245/7) 0.
So these 2 months have exactly same calendar.
Practice problems
Q.1. What was the day on 10th May, 1999?
A. Monday B. Friday C. Saturday D. Thursday
Sol : Option A
We know that in 1600 years there will be 0 odd days.
So start counting from here. In 300 years after that, there will be 1 odd day.
10 May, 1999 implies that starting from the end of 1900, 98 years, 4 months and 10 days have elapsed since then. 98 years have 24 leap years and 74 non leap years leading to 122 odd days.
Dividing by 7 and checking remainder, net odd days = 3.
In the 4 months and 10 days of 1999, there are 31 days in Jan, 28 in Feb, 31 in March, and 30 in April.
Total days elapsed in 1999 = 31+28+31+30+10 = 130.
So net odd days = 4.
Adding up all the odd days we have got so far we get a total of 1 + 3 + 4 = 8.
Net odd day = 1, so May 10, 1999 was a Monday.
(The rule is that 0 odd days means the day is a Sunday, 1 means Monday, 2 means Tuesday and so on.)
Q.2. May 10, 1999 was a Monday. Then what was the day on 10–Dec–2001?
A. Monday B. Thursday C. Saturday D. Friday
Sol : Option A
May 10, 2001 will be a Thursday (1999 was a non-leap year so add one day and 2000 is a leap year, so add 2 odd days).
Now start counting the days from May 10, 2001 to 10–Dec–2001. Complete months in between are June, July, Aug, Sep, Oct, Nov – total days = 30 + 31 + 31 + 30 + 31 + 30 = 183 days. Plus 21 days in May and 10 days in Dec.
So total days = 214. Net odd days = 4. So 10–Dec–2001 will be 4 days after Thursday, which is Monday.
Q.3. What was the day on 15th August, 1947?
A. Saturday B. Thursday C. Tuesday D. Friday
Sol : Option D
Up to first 1600 years no odd day.
From 1601 – 1900 1 odd day
From 1901 – 1946: 46 + 11 (leap)
= 57 – 56 (complete weeks) 1 odd day
First six months give = 6 odd days
July 3 = 3 odd days
Up to August 15 1 odd days
Total up to 15th August 10 odd days
Total no. of odd days 12 – 7 (complete week) = 5
Now counting from the beginning, 5 odd days, so India’s first Independence Day was on a Friday.
Q.4. Jan 5, 1991 was a Saturday. What was the day of the week on March 3, 1992?
A. Friday B. Thursday C. Tuesday D. Saturday
Sol : Option C
5.1. 1991– Saturday
∴5.1. 1992 was Sunday.
From 5.1.1992 to 3.3.1992, number of days = 26 + 29 + 3 = 58.
Hence number of odd days = 58/7 ⇒2
Day on 3.3.1992 was 2 days ahead of Sunday i.e. Tuesday.
Q.5. If Sunday falls on 4th April, 1992. What was the day on 3rd Nov. 1991?
A. Tuesday B. Friday C. Wednesday D. Saturday
Sol : Option D
3.4.92 was Monday.
Number of days from 3.4.92 to 3.11.91
= 4 + 31 + 29 + 31 + 31 + 27 = 153.
Hence number of odd days = 153/7⇒6
Hence the day on 3.11.91 was 6 days behind Sunday i.e. Saturday.
Q.6. It was Thursday on 2nd Jan 1997. What day of the week will be on 15th March 1997?
A. Tuesday B. Saturday C. Sunday D. Thursday
Sol : Option B
2.1.97 was Thursday.
Number of days from 2.1.97 to 15.3.97
= 29 + 28 + 15 = 72.
Hence number of odd days = 72/7⇒2
Hence the day on 15.3.97 will be 2 days ahead of Thursday i.e. Saturday.
Q.7. What was the day on 29th January, 1950?
A. Friday B. Sunday C. Wednesday D. Saturday
Sol : Option B
29.1.1950. Upto 1900, number of odd days = 1. From 1900 to 1949, there are 49 years in which 12 are leap years and 37 are normal years
Total number of odd days = (12 2) + (37 1) = 24 + 37 = 61 5.
Also 29 days of January will have 1 odd day. Hence the total number of days = 1 + 5 + 1 = 7
0.
Hence the day on this date i.e. 29.1.1950 was Sunday.
Q8. Today is Tuesday. After 1 yr, 68 days it will be:
A. Friday B. Thursday C. Tuesday D. Monday
Sol : Option D
Today is Tuesday. 1 yr, 68 days means 365 + 68 = 433 days.
Hence number of odd days = 433/7⇒6
Hence the day would be 6 days ahead of Tuesday i.e. Monday.
Q9. The year next to 1986 having the same calendar as that of 1986 was:
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A. 2001 B. 1998 C. 1997 D. 2002
Sol : Option C
2 different years will have the same calendar if the total number of odd days from one year to the other is either 7 or multiple of 7.
So from 1986, if we go forward, odd days in 1987– 1, 1988 – 2, 1989 – 1, 1990 – 1, 1991 – 1, 1992 – 2, 1993 – 1, 1994 – 1, 1995 – 1, 1996 –2, 1997– 1. Total number of odd days up to 1997= 1 + 2 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 = 14.
Hence 1997 will have same calendar as 1986.
Q10. If today is Tuesday. After 68 days, it will be:
A. Monday B. Thursday C. Sunday D. Friday
Sol : Option C
Each day of the week is repeated after 7 days. So, after 70 days, it will be Tuesday.∴ After 68 days, it will be Sunday.