Friday, 13 January 2017

Quantitative Reasoning - CALENDAR Problems

CALENDAR Problems

compete4Exams - Quantitative Resoning - CALENDAR Problems

A calendar is a particular measure of time.

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

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

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

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

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



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

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

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

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.

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

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


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

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

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

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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