MTS/ POSTMAN / GDS TO PA / PA RECRUITMENT MATERIAL

This material prepared and compiled by Akula. Praveen Kumar, SPM, Papannapet Sub Office-502 303, Medak Division, Andhra Pradesh (9849636361, 8019549939)

Disclaimer:- All questions/Information provided in this post are Compiled by A. Praveen Kumar for in good faith of Departmental Employees. The types of questions, number of questions and standard of questions may be vary in actual examination. This is my predictions only. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of the contents.

   

                      Calendar :

Some important facts about calendar we must know :

Odd days: Number of days more than the complete number of weeks in a given period is the number of odd days during that period.

Ordinary year has 365 days ;It has 1odd days.

Leap year has 366 days ; It has 2 odd days.

100 year contain 5 odd days.

200 year contain 3 odd days.

300 year contain 1 odd day.

400 year contain 0 odd day.

Sunday -------> '0' odd day.
Monday -------> '1' odd day.

Leap year:

Every Year which is divisible by 4.

Every century year must be divisible by both 4 and 400.

3.One leap year contains '2' odd days.

The years which are mul of '4' are called leap years.

Leap year -------> 366 days (feb --> 29 days).
Ordinary year -------> 365 days.

 MONTHS DAYS

Jan 31
Feb 28 (or) 29
Mar 31
Apr 30
May 31
Jun 30
Jul 31
Aug 31
Sep 30
Nov 30
Dec 31

One week = '7' days.

Leap year ------> '52' weeks + '2' odd days.
Ordinary year ------> '52' weeks + '1' odd day.

Steps involved in determining the day for given date.

§  Split the completed years into multiple of 400 and remainder.

§  Split the remainder into multiples of 100 and remainder.

§  Compute the odd day for the remainder years.

§  Compute the odd day for the completed months.

§  Compute the odd day for the date.

§  Add up the number of odd days obtained in (3) to (5).

§  Use the below table to determine the required day.

Odd days		1	2	3	4	5	6
Day	Sun	Mon	Tues	Wed	Thurs	Fri	Sat

                                  Example:

1.What was the day of the week on 16^th November 1989?

1989 = 1600 + 389 [Step 1]

389 = 300+ 89 [Step 2]

89 = 22 leap years + 67 ordinary years [Step 3]

= 22*2 + 67* 1 = 111 days = R{111/7} = 6 odd days.

10 months completed i.e up-to October .

No of days in ten months = 365 – (30+31) = 304 days.

No of odd days in ten months = R{304/7} = 3 odd days. [Step 4]

No of odd days for given date = R{16/7} = 2 odd days. [Step 5]

Sum of odd days from step 3,4,5 = 6 + 3 + 2 = 11 days = 4 odd days.

From the above table,we can find the day, The required day is Thursday.

2.On what dates of November 1989 did Thursday fall ?

These type of problems we need to find the day on 1^st November, 1989.

For years,we have 6 odd days,

For months, we have 3 odd days,

For dates,we have 1 odd day i.e 1^st November.

Sum of odd days = 10 days = 3 odd days.

The day on 1^st November 1989 is Wednesday. So first Thursday on November month is 2 nd November, 9^th ,16^th,23^th and 30^th .

3.The calendar for the year 2012 is same as for the year ?

Count the number of odd days from 2012 onwards to get 0 odd day.

Year	2012	2013	2014	2015	2016
Odd Days	2	1	1	1	2

At the end of 2016, 7 odd days = 0 odd day.

So calendar for the year 2012 is same as that for the year 2017.

4.Today is Saturday,After 65 days , it will be :

R[65/7] = 2 odd days.

This table is for these type of problems alone. If today is Monday we have to take 0 odd day as Monday , 1 odd day as Tuesday and so on . But In this problem,Today is Saturday So we have to starts with Saturday .

Odd days		1	2	3	4	5	6
Day	Sat	Sun	Mon	Tues	Wed	Thur	Fri

So the required day is Monday .

In calendar Problems , Above four models Frequent. Be thorough with the concepts which given. If you have any doubts plz post here.

                                          CLOCKS :

In a clock, the minute hand travels 360° in one hour or 6° in one minute and the hour hand moves 360° in 12 hours or 0.5° in one minute. Since the two hands move in the same direction the relative speed is 5.5°.

The following results are very useful to answer questions fast.

o The two hands overlap each other every 65 5/11 minutes.

o In a 12-hour gap the two hands would overlap / coincide 11times and hence in a 24-hour gap 22 overlap would take place.

o In a 12-hour gap the two hands would be perpendicular to each other 22 times.

o In a 12-hour gap the two hands would be in a straight line  22 times.

o In a 12-hour gap the two hands would be in a straight line but opposite in direction is 11  times.

o In 60 minutes, the minute hand gains 55 minutes on the hour hand.

Another one shortcut formula for Clocks:

Angle = 30H – (11/2)M

H represents the hour   : M represents the minute

                                           Example :

1.What is the angle between the two hands of a clock at 3:27 p.m.?

As per formula,

H = 3 ; M = 27

Angle = 30*3 – (11/2)27

= 90^o – 148.5^o

= 58.5^o (here no need to represent the negative in that)

2.At what time between 2 a.m. and 3 a.m. is the angle between the two hands of a clock 28°?

The job is to find the time say, t minutes after 2 a.m. when (y – x) = 28° . there are two possibilities – one in which the minute hand is ahead of the hour hand and the other in which the hour hand is ahead of the minute hand.

CASE 1

Now, y = the angular movement of the minute hand in t minutes = 6t°.

Also x = 60 + 0.5t. [since at 2 a.m., 60° is already covered]

So, 6t – 60 – 0.5t = 28 or

5.5t …………………… = 88 or

t ………………………… = 16.

CASE 2

In this case, (x – y) = 28°, y = 6t and x = 60 + 0.5t.

60 + 0.5t – 6t …… = 28 or

5.5t …………………… = 32 or

t ……………………….. = 32/5.5

= 5 9/11.

Thus, there are two points of time at which the angle between the two hands

is 28° and these two points of time are 16 minutes past 2 a.m. and 5 9/11

minutes past 2 a.m.

                                    EXAMPLES

1.Today is Tuesday. After 62 days it will be:

Each day of the week is repeated after 7 days. After 63 days, it will be Tuesday.
After 62 days, it will be Monday.

2.Monday falls on 20th March, 1995. What was the day on 3rd November, 1994?

Counting the number of days after 3rd November, 1994 we have: Nov. Dec. Jan. Feb. March
27+ 31 + 31+ 28 + 20 = 137days =19weeks + 4days . Number of odd days = 4.
The day on 3rd November, 1994 is (7 - 4) days beyond the day on 20th March, 1995.
So, the required day is Thursday.

3.What will be the day of the week on 1st Jan 2001 ?

1600 yeaas contain 0 odd day.
300 years contain 1 odd day.
100 years contain 5 odd days.
2000 years contain (0+ 1 +5) odd days =6 odd days.
1st Jan, 2001 has one odd day.
Total number of odd days upto 1st Jan, 2001 = 7 odd days =Q odd days.
The day will be Sunday.

4.Today is 1st April. The day of the week is Wednesday. This is a leap year. The day of the week on this day after 3 years will be:

This being a leap year, none of the next 3 years is a leap year.
So,the day of the week will be 3 days beyond Wednesday.
The day after 3 years will be Saturday.

5.How many days are there from 2nd January 1995 to 15th March, 1995?

Jan Feb March 30 +28 + 15 = 73 days.

6.January 1, 1995 was a Sunday. What day of the week lies on January 1, 1996?

1995 being on ordinary year, it has 1 odd day. So, the first day of 1996 will be one day beyond Sunday, i.e. it will be Monday

7.On 8th Feb, 1995 it was Wednesday. The day of the week on 8th Feb, 1994 was

1994 being an ordinary year, it has 1 odd day. So, the day on 8th Feb, 1995 is one day beyond the day on 8th Feb, 1994. But, 8th Feb, 1995 was Wednesday. 8th Feb, 1994 was Tuesday.

8.May 6, 1993 was Thursday. What day of the week was on May 6, 1992?

1992 being a leap year, it has 2 odd days.
So, the day on May, 1993 is 2 days beyond the day on May 6, 1992.
But, on May 6, 1993 it was Thursday.
So, on May 6, 1992 it was Tuesday.

9.The year next to 1996 having the same calender as that of 1996 is:

Starting with 1996, we go on counting the number of odd days till sum is divisible by 7.
Year 1996 1997 1998 1999 2000
Odd days 2 1 1 1 2 =7 odd days
0 odd day.
Calendar for 2001 will be the same as that of 1995.

10.January 1, 1992 was Wednesday. What day of the week was on January 1,1993?

1992 being a leap year, it has 2 odd days. So, the first day of the year
1993 must be two days beyond Wednesday.
So, it was Friday.

11.On what dates of April, 1994 did Sunday fall?

Find the day on 1st April, 1994. 1600 years contain 0 odd day 300 years contain 1 odd day 93 years = (23 leap years + 70 ordinary years) = (46 + 70) odd days =4 odd days. Number of days upto 1St April, 1994 Jan Feb March April 31 + 28 ÷ 31 + 1 =9ldays=Ooddday. Total number of odd days = (0÷ 1 +4+0) =5 Odd day.  Day on 1st April 1994 is ‘Friday’. Sunday was on 3rd April, 1994. Thus, Sunday fell on 3rd, 10th, 17th & 24th

12.The day on 5th March of a year is the same day on what date of thesame year?

Since any date in March is the same day of the week as the corresponding date in November of that year, so the same day falls on 5th November.

13.The calendar for 1990 is the same as for

count the number of days for 1990 onwards to get 0 odd day.
Year 1990 1991 1992 1993 1994 1995
Odd days 1 1 2 1 1 1 = 7 or 0 odd day.
Calendar for 1990 is the same as for the year 1996

14.If the first day of the year 1991 was Tuesday, what day of the week must have been on 1st January, 1998?

Total number of odd days from 1st Jan, 1991 to 1st Jan, 1998
Year 1991 1992 1993 1994 1995 1996 1997
Odd days I + 2 + 1 + 1 + 1 + 2 + 1 = 9 odd days
2 odd days
The day is 2 days beyond the day on 1st Jan, 1991.
i.e. The required day must be Thursday.