Quantitative Reasoning - CALENDAR Problems

CALENDAR Problems

① A calendar is a particular measure of time.

② The smallest unit of calendar is a day. It is an average time in which the earth completes onw rotation on its axis.

③ The time in which the earth travels round the sun is known as a <b>solar year</b>.

④ The solar year consists of 365 days 5 hours and 48 minutes and 48 seconds. An ordinary year consists of 365 days.

⑤ To adjust the differece between solar year and oridnary year, every fourth year has 366 days and is called leap year.

⑥ Every fourth century is a leap year but no other century is a leap year.



⑦ In a given period, the number of days more than the Complete weeks are called odd days.

⑧ One ordinary year = 365 days = 52 weeks + 1 day. So an ordinary year has 1 odd day.

⑨ One leap year =366 days = 52 weeks + 2 days. So a leap year has 2 odd days.

⑩ One Century = 100 years = 76 ordinary years + 24 leap years
= 76 Odd Days + 24 ✕ 2 = 124 Odd days
= 17 weeks + 5 days
= 5 Odd days
Thus 100 years contain 5 odd days.

⑪ 200 years contain 3 odd days. 300 years contain 1 odd day. 400 years contain 0 odd day.

⑫ No. Of Odd Days for Century Years are:

100 years – 5	500 -5	900-5	1300-5	1700-5	2100-5
200 years – 3	600-3	1000-3	1400-3	1800-3	2200-3
300 years – 1	700-1	1100-1	1500-1	1900-1	2300-1
400 years – 0	800-0	1200-0	1600-0	2000-0	2400-0

⑬ 1st century 1 A.D. was Monday (Code for Week days Monday =1 , Tuesday = 2... Sunday = 7)

⑭ Following months have 3 odd days: January, March, May, July, August, October, December

⑮ Following months have 2 Odd days: April, June, September, November

⑯ Feburary month has 0 odd day in an ordinary year and 1 odd day in a leap year.



Question: What was the day of the week on August 15, 1947

Answer:

August 15, 1947 means = 1946 years + 7 Months + 15 days

Odd days in 1600 years = 0
Odd days in  300 years = 1
46 years = 11 leap years + 35 ordinary years
= 11 ✕ 2 + 35 ✕ 3 = 22 + 35 = 57 Odd days
= 8 weeks + 1 Odd day
= 1 Odd day
∴ Odd days in 1946 =  0 + 1 + 1 = 2 Odd days

Number odd days from Jan 1- Aug 15 1947 = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 days
= 32 weeks + 3 Odd days = 3 Odd days

Total number of Odd days on Aug 15, 1947 = 2 + 3 = 5 Odd days
  = Friday



Question: Prove that the last day of the century cannot be either Tuesday, Thursday or Saturday

Answer: A century = 100 years = 76 Ordinary Years + 24 leap years
=  76 ✕ 1 + 24 ✕ 2 = 76 + 48 = 124
=  17 weeks + 5 days
=  5 Odd days
Last days of a century = Friday

   Similarly 200 years = 3 Odd days
   ∴ Last day of 2nd century = Wednesday

   Similarly 300 years = 1 Odd day = Monday

   400 years = 0 Odd day = Last day is Sunday.
   Thus last day of the century cannot be either Tuesday, Thursday or Saturday.


Question: Prove that the calendar for 2003 will serve for the year 2014.

Answer: No. of Odd days between Dec 31, 2002 to Dec 31, 2013 = 3 leap years + 8 Ordinary years
= 3 ✕ 2 + 8 ✕ 1 = 14 Odd days
= 2 weeks = 0 Odd day.
∴ the calendar for 2003 will serve for the year 2014.


Question: Today is Friday. What day will it be after 62 days?

Answer: Each day is repeated after every 7 days. After 63 days, it will be Friday.
⇒  After 62 days, it will be Thursday