URDINESH: Aptitude
Showing posts with label Aptitude. Show all posts
Showing posts with label Aptitude. Show all posts

Sunday, May 18, 2014

27. CALENDAR


IMPORTANT FACTS AND FOAMULAE

Under this heading we mainly deal with finding the day of the week on a particular given date the  process of finding it lies on obtaining the number of odd days.

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.

LeapYear: Every year which is divisible by 4 is called a leap year.

Thus each one of the years 1992, 1996, 2004, 2008, 2012, etc. is a leap year. Every 4th century is a leap year but no other century is a leap year.thus each one of 400, 800, 1200,' 1600, 2000, etc. is a leap year.
None of 1900, 2010, 2020, 2100, etc. is a leap year.

An year which is not a leap year is called an ordinary year.

(I )An ordinary year has 365 days.   (II) A leap  year has 366 days.

Counting of Odd Days:

i)1 ordinary year = 365 days = (52 weeks + 1 day).
:. An ordinary year has 1 odd day.
ii)1 leap year = 366 days = (52 weeks + 2 days).
:. A leap year has 2 odd days.
_          iii)100 years = 76 ordinary years + 24 leap years
= [(76 x 52) weeks + 76 days) + [(24 x 52) weeks + 48 days]
= 5200 weeks + 124 days = (5217 weeks + 5 days).
:. 100 years contain 5 odd days.
200 years contain 10 and therefore 3 odd days.
300 years contain 15 and therefore 1 odd day.
400 years contain (20 + 1) and therefore 0 odd day.
Similarly, each one of 800, 1200, 1600, 2000, etc. contains 0 odd days.
Remark: (7n + m) odd days, where m < 7 is equivalent to m odd days.
Thus, 8 odd days ≡ 1 odd day etc.

No of odd days
0
1
2
3
4
5
6
Day
Sun.
Mon.
Tues.
Wed.
Thur.
Fri.
Sat.






SOLVED EXAMPLES

Ex: 1.Wbat was the day of the week on, 16th July, 1776?

Sol:  16th July, 1776 = (1775 years + Period from 1st Jan., 1776 to 16th July, 1776)
Counting of odd days :
1600 years have 0 odd day. 100 years have 5 odd days.
75 years = (18 leap years + 57 ordinary years)
= [(18 x 2) + (57 x 1)] odd days = 93 odd days
= (13 weeks + 2 days) = 2 odd days.

.. 1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day.
Jan.  Feb.  March   April   May   June   July
31   + 29  +   31    +  30   +  31  +  30  +16 = 198days
= (28 weeks + 2 days) =2days
:. . Total number of odd days = (0 + 2) = 2. Required day was 'Tuesday'.

Ex. 2. What was the day of the week on 16th August, 1947?

Sol. 15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th
Counting of odd days:
1600 years have 0 odd day. 300 years have 1 odd day.
47 years = (11 leap years + 36 ordinary years)
= [(11 x 2) + (36 x 1») odd days = 58 odd days = 2 odd days.
Jan. Feb. March April May June July Aug.
31 + 28 + 31 + 30 + 31 + 30 + 31 + 15
= 227 days = (32 weeks + 3 days) = 3,
Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.
Hence, the required day was 'Saturday'.


Ex. 3. What was the day of the week on 16th April, 2000 ?

Sol. 16th April, 2000 = (1999 years + Period from 1st Jan., 2000 to 16thA'
Counting of odd days:
1600 years have 0 odd day. 300 years have 1 odd day.
99 years = (24 leap years + 75 ordinary years)
= [(24 x 2) + (75 x 1)] odd days = 123 odd days
= (17 weeks + 4 days) = 4 odd days.
Jan. Feb. March April
31 + 29 + 31 + 16 = 107 days = (15 weeks + 2 days) = 2 odd,
Total number of odd days = (0 + 1 + 4 + 2) odd days = 7 odd days = 0 odd day.   Hence, the required day was 'Sunday'.


Ex. 4. On what dates of Jull.2004 did Monday fall?

Sol   . Let us find the day on 1st July, 2004.
2000 years have 0 odd day. 3 ordinary years have 3 odd days.

Jan. Feb. March April May June July

31 + 29 + 31 + 30 + 31 + 30 + 1
= 183 days = (26 weeks + 1 day) = 1 t .
Total number of odd days = (0 + 3 + 1) odd days = 4 odd days.      '
:. 1st July 2004 was 'Thursday',-,-­
Thus, 1st Monday in July 2004 _as on 5th July.
Hence, during July 2004, Monday fell on 5th, 12th, 19th and 26th. .


Ex. 5. Prove that the calendar for the year 2008 will serve for the year 20ll

Sol. In order that the calendar for the year 2003 and 2014 be the same, 1st  January of  both the years must be on the same day of the week.
For this, the number of odd days between 31st Dec., 2002 and 31st Dec.,2013 must be the same.
We know that an ordinary year has 1 odd day and a leap year has 2 odd During this period, there are 3 leap years, namely 2004, 2008 and 2012 and 8 ordinary years.
Total number of odd days = (6 + 8) days = 0 odd day.
Hence, the calendar for 2003 will serve for the year 2014.

Ex. 6. Prove that any date in March of a year is the same day of the week corresponding date in November that year.

We will show that the number of odd days between last day of February and last  day of October is zero.    .
March    April    May    June    July    Aug.    Sept.   Oct.
31     +   30  +  31    + 30   + 31 +   31  +    30   +  31
= 241 days = 35 weeks = 0 odd day. ,Number of odd days during this period = O.
Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows.



28. CLOCKS

IMPORTANT FACTS

The Face or dial of a watch is a circle whose circumference is divided into 60  equal parts, called  minute spaces.
A clock  has two hands, the smaller one is called the hour hand or short hand while the larger one is called the minute hand or long hand..
i) In 60 minutes, the minute hand gains 55 minutes on the hour hand.
ii) In every hour, both the hands coincide once.
iii) The hands are in the same straight line when they are coincident or opposite to each other.
iv) When the two hands are at right angles, they are 15 minute spaces apart.
v)When the hand's are in opposite directions, they are 30 minute spaces apart.
 vi)Angle traced by hour hand in 12 hrs = 360°.
vii)Angle traced by minute hand in 60 min. = 360°.

Too Fast and Too Slow: If a watch or a clock indicates 8.15, when the correct time , 8 is said to be 15 minutes too fast.
On the other hand, if it indicates 7.45, when the correct time is 8, it is said to be 15 minutes too slow.

SOLVED EXAMPLES


Ex 1:Find the angle between the hour hand and the minute hand of a clock when 3.25.

Solution:angle  traced by the hour hand in 12 hours = 360°
Angle traced by it in three hours 25 min (ie) 41/12 hrs=(360*41/12*12)° =102*1/2°
angle traced by minute hand in 60 min. = 360°.
Angle traced by it in 25 min. = (360 X 25 )/60= 150°
Required angle = 1500 – 102*1/2°= 47*1/2°

Ex 2:At what time between 2 and 3 o'clock will the hands of a clock be together?

Solution: At 2 o'clock, the hour hand is at 2 and the minute hand is at 12, i.e. they are 10 min  spaces apart.
To be together, the minute hand must gain 10 minutes over the hour hand.
Now, 55 minutes are gained by it in 60 min.
10 minutes will be  gained in (60 x 10)/55  min. = 120/11 min.
The hands will coincide at 120/11 min. past 2.

Ex. 3. At what time between 4 and 5 o'clock will the hands of a clock be at right angle?

     Sol: At 4 o'clock, the minute hand will be 20 min. spaces behind the hour hand, Now, when the two hands are at right angles, they are 15 min. spaces apart. So, they are at right angles in following two cases.
       Case I. When minute hand is 15 min. spaces behind the hour hand:
In this case min. hand will have to gain (20 - 15) = 5 minute spaces. 55 min. spaces are gained by it in 60 min.

 5 min spaces will be gained by it in 60*5/55  min=60/11min.

:. They are at right angles at 60/11min. past 4.
Case II. When the minute hand is 15 min. spaces ahead of the hour hand:
To be in this position, the minute hand will have to gain (20 + 15) = 35 minute spa' 55 min. spaces are gained in 60 min.
35 min spaces are  gained in (60 x 35)/55 min =40/11
          
     :. They are at right angles at 40/11 min. past 4.

Ex. 4. Find at what time between 8 and 9 o'clock will the hands of a clock being  the same straight line but not together.
       
       Sol: At 8 o'clock, the hour hand is at 8 and the minute hand is at 12, i.e. the two hands_ are 20 min. spaces apart.
To be in the same straight line but not together they will be 30 minute spaces apart. So, the minute hand will have to gain (30 - 20) = 10 minute spaces over the hour hand.
55 minute spaces are gained. in 60 min.
10 minute spaces will be gained in (60 x 10)/55 min. = 120/11min.
:. The hands will be in the same straight line but not together at 120/11 min.

Ex. 5. At what time between 5 and 6 o'clock are the hands of a clock 3minapart?
   
. Sol. At 5 o'clock, the minute hand is 25 min. spaces behind the hour hand.
       Case I. Minute hand is 3 min. spaces behind the hour hand.
In this case, the minute hand has to gain' (25 - 3) = 22 minute spaces. 55 min. are gained in 60 min.
22 min. are gaineg in (60*22)/55min. = 24 min.
:. The hands will be 3 min. apart at 24 min. past 5.
     Case II. Minute hand is 3 min. spaces ahead of the hour hand.
In this case, the minute hand has to gain (25 + 3) = 28 minute spaces. 55 min. are gained in 60 min.
 28 min. are gained in  (60 x 28_)/55=346/11
The hands will be 3 min. apart at 346/11 min. past 5.


Ex 6. Tbe minute hand of a clock overtakes the hour hand at intervals of 65 minutes of the correct time. How much a day does the clock gain or lose?
   
   Sol: In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60 minutes.
To be together again, the minute hand must gain 60 minutes over the hour hand. 55 min. are gained in 60 min.
60 min are gained in  60 x 60 min =720/11 min.
                                   55

But, they are together after 65 min.
Gain in 65 min ­=720/11-65 =5/11min.

Gain in 24 hours =(5/11 * (60*24)/65)min =440/43
The clock gains 440/43  minutes in 24 hours.

Ex. 7. A watch which gains uniformly, is 6 min. slow at 8 o'clock in the morning Sunday and it is 6 min. 48 sec. fast at 8 p.m. on following Sunday. When was it correct?

Sol. Time from 8 a.m. on Sunday to 8 p.m. on following Sunday = 7 days 12 hours = 180 hours

The watch gains (5 + 29/5) min. or 54/5 min. in 180 hrs.
Now 54/5  min. are gained in 180 hrs.
5 min. are gained in (180 x 5/54 x 5) hrs. = 83 hrs 20 min. = 3 days 11 hrs 20 min.
Watch is correct 3 days 11 hrs 20 min. after 8 a.m. of Sunday.
It will be correct at 20 min. past 7 p.m. on Wednesday.


Ex 8. A clock is set right at 6 a.m. The clock loses 16 minutes in 24 hours. What will be the true time when the clock indicates 10 p.m. on 4th day?      

Sol. Time from 5 a.m. on a day to 10 p.m. on 4th day = 89 hours.
      Now 23 hrs 44 min. of this clock = 24 hours of correct clock.

356/15 hrs of this clock = 24 hours of correct clock.
       89 hrs of this clock = (24 x 31556 x 89) hrs of correct clock.
                              = 90 hrs of correct clock.
       So, the correct time is 11 p.m.

 
Ex. 9. A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours will be the true time when the clock indicates 1 p.m. on the following day?
 
Sol. Time from 8 a.m. on a day  1 p.m. on the following day = 29 hours.
       24 hours 10 min. of this clock = 24 hours of the correct clock.
145 /6  hrs of this clock = 24 hrs of the correct clock

29 hrs of this clock = (24 x  6/145 x 29) hrs of the correct clock
= 28 hrs 48 min. of correct clock
The correct time is 28 hrs 48 min. after 8 a.m.
This is 48 min. past 12.



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