# 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

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