Calendar and Clock
Category : 8th Class
Introduction
Calendar and Clocks are essential part of our day-to-day lives. Every student must be confident about not only using them but also be able to apply their usage in our lives. This chapter will help you explore the various applications of Calendar and Clocks.
TO SOLVE PROBLEMS BASED ON CALENDER REMEMOER FOLLOWING POINTS:
We are supposed to find the day of the week on a given date.
For this, we use the concept of odd days.
(i) Odd Days: In a given period, the number of days more than the complete weeks are called odd days.
(ii) Leap Year:
(1) Every year divisible by 4 is a leap year, if it is not a century.
(2) Every 4th century (i.e., divisible by 400) is a leap year and no other century is a leap year.
Note: A leap year has 366 days.
Examples.
(iii) Ordinary Year:
The year which is not a leap year is called an ordinary year.
An ordinary year has 365 days.
(iv) Counting of odd days:
(1) 1 ordinary year =365 days = (52 weeks + 1 day)
\[\therefore \] 1 ordinary year has 1 odd day.
(2) 1 leap year = 366 days = (52 weeks + 2 days).
\[\therefore \] 1 leap year has 2 odd days.
(3) 100 years = 76 ordinary years + 24 leap years
\[=(76\times 1+24\times 2)\] odd days = 124 odd days.
= (17 weeks + 5 days) = 5 odd days.
\[\therefore \] Number of odd days in 100 years = 5
Number of odd days in 200 years\[=(5\times 2)=3\] odd days.
Number of odd days in 300 years \[=(5\times 3)=1\] odd day.
Number of odd days in 400 years \[=(5\times 4+1)=0\] odd days.
Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd days.
(v) First January 1 AD was Monday. Therefore, we must count days from Sunday, i.e. Sunday for 0 odd days, Monday for 1 odd day, Tuesday for 2 odd days and so on.
(vi) February in an ordinary year gives no odd day, but in a leap year gives one odd day.
EXAMPLE 1:
What was the day of the week on 15th August, 1947?
Sol. 15th August, 1947 = (1946 years + Period from 1.1.1947 to 15.8.1947)
Odd days in 1600 years = 0
Odd days in 300 years \[=(5\times 3)=15=1\]
46 years =(11 leap years +35 ordinary years)
\[=(11\times 2+35\times 1)\] odd days = 57 odd days
= (8 weeks + 1 day) = 1 odd day.
\[\therefore \] Odd days in 1946 years \[=(0+1+1)=2\].
Jan. Feb. March April May June July Aug.
\[\left( 31+28+\text{ }31\text{ }+\text{ }30\text{ }+\text{ }31+30+31+15 \right)=227\]days.
227 days = (32 weeks + 3 days) = 3 odd days.
Total number of odd days \[=\left( 2\text{ }+\text{ }3 \right)=5.\]
Hence, the required day is Friday.
EXAMPLE 2:
What day of the week was 20th June 1837?
Sol. 20th June 1837 means 1836 complete years + first 5 months of the year 1837 + 20 days of June.
1600 years give no odd days.
200 years give 3 odd days.
36 years give (27 + 9) or 3 odd days.
1836 years give 6 odd days.
From 1st January to 20th June there are 3 odd days.
Odd days:
|
January |
:3 |
|
February |
:0 |
|
March |
:3 |
|
April |
:2 |
|
May |
:3 |
|
June |
:6 |
|
|
17 |
Therefore, the total number of odd days \[=\left( 6+3 \right)\]or 2 odd days.
This means that the 20th of June fell on the 2nd day commencing from Monday. Therefore, the required day was Tuesday.
TO SOLVE PROBLEMS BASED ON CLOCKS, REMEMBER THE FOLLOWING POINTS:
The face of the 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) 1 minute space \[=\frac{{{360}^{o}}}{60}\] .
(As \[{{360}^{o}}\] of the circle is divided into 60 minutes).
(iii) In one minute, the hour hand moves \[=\frac{360}{12\times 60}=\frac{360}{720}=\frac{{{1}^{o}}}{2}\]
(As there are 12 hours of 60 minutes each)
Thus, in one minute the minute hand gains \[5\frac{{{1}^{o}}}{2}\] over the hour hand.
(iv) In every hour, both the hands coincide once.
(v) The hands are in the same straight line when they are coincident or opposite to each other.
(vi) When the two hands are at right angles, they are 15 minute spaces apart.
(vii) The hands coincide 11 times in every 12 hours (between 11 and 1 O'clock there is a common position at 12 O'clock).
Hence, the hands coincide 22 times in a day.
(viii) The hands of a clock are at right angles twice in every hour, but in 12 hours they are at right angles 22 times since there are two common positions in every 12 hours.
(ix) When the hands are in opposite directions, they are 30 minute spaces apart.
(x) Angle traced by hour hand in 12 hrs =\[{{360}^{o}}\].
(xi) Angle traced by minute hand in 60 min.=\[{{360}^{o}}\]
(xii) Interchangeable positions of minute hand and hour hand occur when the original interval between the two hands is \[\frac{60}{13}\]minute spaces or a multiple of this.
\[\frac{\text{True time interval}}{\text{Time interval in incorrect clock}}\]
\[\text{=}\frac{\text{1}}{\text{1 }\!\!\pm\!\!\text{ hour gained/lost in1hour by incorrect clock}}\]
True time interval____
Time interval in incorrect clock
(+) when incorrect clock gains time
(-) when incorrect clock loses time
In a correct (true) clock, both hands coincide at a interval of \[65\frac{5}{11}\]minutes.
But, if both hands coincide at an interval of x minutes
\[\left( \ne 65\frac{5}{11} \right)\]of correct time? then the clock is incorrect and, total time gained / lost \[60T\times \frac{65\frac{5}{11}-x}{x}\] (in T hours of correct time).
Too Fast and Too Slow: If a watch or a clock indicates 8.15, when the correct time is 8, it 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.
Clock image: Sum of the actual time and time observed in image =\[={{23}^{H}}{{59}^{M}}{{60}^{S}}\] (railway timing) or \[{{11}^{5}}{{59}^{M}}{{60}^{S}}\]
Miscellaneous Solved Examples
EXAMPLE: 1
At what time between 4 and 5 will the hands of a watch point in opposite directions?
Sol. They will be opposite to each other when there is a space o 30 minutes between them.
This will happen when the minute hand gains \[\left( 30+20 \right)\] minutes = 50 minutes.
The minute hand gains 50 minutes in
\[\frac{50\times 60}{55}=54\frac{6}{11}\]minutes.
i.e., they will point in opposite directions at \[54\frac{6}{11}\]minutes past 4.
EXAMPLE: 2
A watch which gains uniformly, is 5 min. slow at 8 o'clock in the morning on Sunday and it is 5 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
\[\therefore \]The watch gains \[\left( 5+5\frac{4}{5} \right)\] mm. or \[\frac{54}{5}\]min. in 180 hrs.
Now \[\frac{54}{5}\] min. are gained in 180 hrs.
\[\therefore \] 5 min. are gained in \[\left( 180\times \frac{5}{54}\times 5 \right)\] hrs. = 83 hrs 20 min.
= 3 days 11 hrs 20 min.
\[\therefore \] Watch is correct 3 days 11 hrs 20 min. after 8 a.m. of Sunday
\[\therefore \]It will be correct at 20 min. past 7 p.m. on Wednesday.
EXAMPLE: 3
A clock is set right at 1 p.m. If it gains one minute in an hour, then what is the true time when the clock indicates 6 p.m. the same day?
Sol. Time interval indicated by incorrect clock
= 6 p.m. -1 p.m. =5 hours.
Time gained by incorrect clock in one hour
\[=+1\text{minute}=+\frac{1}{60}hour.\]
Using the formula,
\[\frac{\text{True time interval}}{\text{Time interval in incorrect clock}}\]
\[\text{=}\frac{\text{1}}{\text{1+hour gained in1hour by in correct clock}}\]
\[\Rightarrow \] \[\frac{\text{True time interval}}{\text{5}}=\frac{1}{1+\frac{1}{60}}\]
\[\Rightarrow \] True time interval \[=\frac{5\times 60}{61}=4\frac{56}{61}\]
\[\therefore \] True time = 1 p.m. \[+4\frac{56}{61}\] hrs.
= 5 p.m. \[+\frac{56}{61}\]hrs. 5 p.m. \[+\frac{56}{61}\times 60\]min.
\[=55\frac{5}{61}\] minutes past 5.
EXAMPLE: 4
The minute hand of a clock overtakes the hour hand at intervals of 65 minutes of correct time. How much a day does the clock gain or lose?
Sol. In a correct clock, the minute hand gains 55 minute spaces over the hour hand in 60 minutes. To be together again, the minute hand must gain 60 minutes over the hour hand.
Now, 55 min. are gained in 60 min.
\[\therefore \] 60 min. are gained in \[\left( \frac{60}{55}\times 60 \right)\]min \[=65\frac{5}{11}\]min
But, they are together after 65 minutes.
\[\therefore \] Gain in 65 minutes \[=\left( 65\frac{5}{11-65} \right)=\frac{5}{11}\]min.
Gain in 24 hrs. \[=\left( \frac{5}{11}\times \frac{60\times 24}{64} \right)\]min. \[=10\frac{10}{143}\] mm.
Prove that the calender for 1990 will be same for 2001 also.
Sol. It is clear that the calender for 1990 will be serve for 2001 if first January of both the years is the same weekdays. For that the number of odd days between 31 st December 1989 and 31st December 2000 must be zero. Odd days are as given below.
|
Year |
1990 |
1991 |
1992 |
1993 |
1994 |
1995 |
1996 |
1997 |
1998 |
1999 |
2000 |
|
Odd days |
1 |
1 |
(Leap) 1 |
1 |
1 |
1 |
(Leap) 2 |
1 |
1 |
1 |
(Leap) 2 |
Total number of odd days =14 days = 2 weeks + odd days.
EXAMPLE: 6
The minute hand of a clock overtakes (or coincides) the hour hand at intervals of 65 minutes of correct time. How much does the clock gain or lose in 12 hours?
Sol. In a correct clock, both the hands coincide (or overtake) at an interval of \[65\frac{5}{11}\]minutes.
But, here it coincides at an interval of 65 minutes.
Since \[65\frac{5}{11}>65,\]so, the clock is gaining time.
Using the formula,
Total time gained in T hours.
\[=(T\times 60)\frac{\left( 65\frac{5}{11}-x \right)}{x}\]minutes.
Where T = 12 hours, x = 65
\[\therefore \] Total time gained in 12 hours.
\[=(12\times 60)\frac{\left( 65\frac{5}{11}-65 \right)}{65}\]minutes.
\[=12\times 60\times \frac{5}{11}\times \frac{1}{65}\]minutes \[=5\frac{5}{143}\] minutes.
EXAMPLE: 7
A clock gains 5 minutes, in 24 hours. It was set right at 10 a.m. on Monday. What will be the true time when the clock indicates 10:30 a.m. on the next Sunday?
(1) 10a.m.
(2) 11a.m.
(3) 25 minutes past 10 a.m.
(4) 5 minutes to 11 a.m.
Sol. (1)
Time between 10 a.m. on Monday to 10:30 a.m. on Sunday \[=144\frac{1}{2}\]hours.
\[24\frac{1}{2}\]hours of incorrect clock = 24 hours of correct time.
\[\therefore \] \[144\frac{1}{2}\] hours of incorrect clock = x hours of correct time.
\[\therefore \] \[x=\frac{144\frac{1}{2}\times 24}{24\frac{1}{2}}=144\] hours i.e.,
The true time is 10 a.m. on Sunday.
EXAMPLE 8:
.At what time between 3 and 4 O'clock are the hands of a clock together —
(1) \[16\frac{4}{11}\]min. past3
(2) \[16\frac{6}{11}\]mm. past 3
(3) \[10\frac{10}{11}\]min. past3
(4) \[10\frac{6}{11}\]min. past3
Sol. (1)
At 3 O'clock, the hour hand is at 3 and the minute hand is at 12. i.e., they are 15 min. spaces apart. To be together, the minute hand must gain 15 min. over the hour hand.
Now 55 min. are gained in 60 min.
\[\therefore \] 15 min. will be gained in
\[\left( \frac{60}{55}\times 15 \right)\]min. \[=16\frac{4}{11}\]min.
So, the hands will coincide at \[16\frac{4}{11}\]min. past 3.
EXAMPLE 9:
Find at what time between 8 and 9 O'clock will the hands of a clock be in the same straight line but not together.
(1) \[10\frac{5}{11}\]min. past 8
(2) \[10\frac{8}{11}\]min. past 8
(3) \[10\frac{10}{11}\]min. past 8
(4) \[10\frac{6}{11}\]min. past 8
Sol. (3)
At 8 O'clock, the hour hand is at 8 and the min. 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 min. spaces apart. So, the min. hand will have to gain \[\left( 30-20 \right)\text{= }10\]min. spaces over the hour hand.
Now, 55 min. are gained in 60 min.
10 min. will be gained in \[\left( \frac{60}{55}\times 10 \right)\]min \[=10\frac{10}{11}\]min.
\[\therefore \] The hand will be in the same line but not together at \[10\frac{10}{11}\]min. past 8.
EXAMPLE 10:
Two men run a race of 200 meters. Man A wins by 10 meters. Because of this, they decide to make things fairer for the next race by making Man A stand 10 meters behind the line, thereby giving Man B a 10-meter head start. They both run the second race at exactly the same speed as before. What is the result?
Sol. Man A again wins. We know from the first race that Man A runs 200 meters at the same time that Man B runs 90 meters. Therefore, it follows that as Man A starts 10 meters behind the line, the men will be dead even at 10 meters short of the winning line. As man A is the faster runner, he goes on to overtake man B in the last 10 meters and win the race.
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