clocks and calendars

Calendars 9

Introduction

common in-various competitive exams. The method of solving such questions lies in the concept of obtaining the number of odd days. Before jumping to the topic, let us review some of the basic concepts.

(1) Whenever the number of year is exactly divisible by 4 (except the century years), then it is a Leap year.

(2) Whenever the number of year is not divisible by 4, then it is an Ordinary year.

(3) In case of the century years, if the number of year is exactly divisible by 400, then it is a Leap year.

(4) Whenever the number of year is not divisible by 400, then it is an Ordinary year.

Ordinary year: An ordinary year can be defined as the year having 365 days which is equal to 52 weeks and an extra day.

Century year: A year is a century year if it is divisible by 100.

Non-Century year: A year is a non-century year if it; not a century year.

Leap year: A year is a leap year if it is a non-century •ear that is divisible by 4, or a century year that is divisible by 400.

How to find the number of odd days?

The total number of days for a specific period of time is when divided by 7; the remainder obtained in such a case is termed as the odd day(s).

Counting of Odd days:

i. 1 ordinary year = 365 days = 52 weeks + 1 odd day

ii. 1 leap year = 366 days = 52 weeks + 2 odd days

iii. 1st century years = 100 years = 76 ordinary years + 24 leap years = 76 + 2 x 24 = 124 odd days = 5 odd days

Now, based on the above fact, we can conclude that the number of odd days in

(i) 100 years = 5(ii) 200 years = 3

(iii) 300 years = 1(iv) 400 years = 0

The following points have been observed:

The following table is based on the fact that 151 January, 1 A.D. was a Monday. This table is helpful in solving the question which assumes the given information.

No.of odd days

1 2 3 4 5 6 7

Dyas Mon. Tue. Wed. Thu Fri Sat. Sun

(2) In an Ordinary year, the calendar for the month of January is the same as the calendar for the month of

'The questions on this topic are very

October. In short, In an Ordinary year, January = October.

(3) In a Leap year, the calendar for the month of January is the same for the month July. In short, In a Leap year, January = July.

Example 1:

Find the day of the week on 16th July, 1776.

Solution:

16th July, 1776 means

(1775 years + 6 months + 16 days)

Now, 1600 years have 0 odd days.

100 years have 5 odd days.

75 years contain 18 leap and 57 ordinary years and therefore, (36 + 57) or 93 or 2 odd days.

1775 years give 0 + 5 + 2 = 7 and so 0 odd days. Also, number of days from 1st Jan, 1776 to 16th July, 1776

Jan. Feb. Mar. Apr. May Jun. Jul.

31 + 29 + 31 + 30 + 31 + 30 + 16

= 198 days = 28 weeks + 2 days = 2 odd days.

Total number of odd days = 0 + 2 = 2.

Hence, the day on 16th July, 1776 was Tuesday.

Example 2:

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

Solution:

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

Example 3:

On January 12, 1980, it was Saturday. The day of the week on January 12,1979 was: Solution:

The year 1979 being an ordinary year, it has 1 odd day. So, the day on 12th January 1980 is one day beyond the day on 12th January, 1979. But, January 12, 1980 being Saturday, January 12, 1979 was Friday.

Example 4:

February 20, 1999 was Saturday. What day of the week was on December 30, 1997?

Solution:

The year during this interval was 1998 and it was not a leap year. Now, we calculate the no. of odd days in 1999 up to February 19:

January 1999 gives 3 odd days

19 February 1999 give 5 odd days

1998, being ordinary year, gives 1 odd day

In 1997, December 30 and 31 give 2 odd days

Total number of odd days = 3 +5+1+2 = 11 days = 4 odd days.

Therefore, December 30, 1997 was 4 days before Saturday i.e., on Tuesday.

Example 5:

The year next to 1987 having the same calendar as that of 1987 is:

Solution:

Starting with 1987, we go on counting the number of odd days till the sum is divisible by 7. Number of odd days = 1 (1987) + 2(1988) +1 (1989] + 1(1990) + 1 (1991) + 2(1992) + 1(1993) + 1 (1994) + 1(1995) + 2(1996) + 1(1997) = 14/7 = 0 odd days.

So, the year next to 1987 having the same calendar as that of 1987 is 1998.

Exercise

1. The first Republic Day of India was celebrated on 26th January, 1950. What was the day of the week on that date?

(1) Wednesday (2) Thursday (3) Friday (4) Saturday

2. Mahatma Gandhi was born on 2nd October, 1869. The day of the week was


3. India got Independence on 15th August 1947. What was the day of the week?


4. Smt Indira Gandhi died on 31 st October, 1984. The day of the week was:

(1) Monday (2) Tuesday (3) Wednesday (4) Friday

5. What day of the week was 20th June, 1837?

(1) Monday (2) Tuesday (3) Thursday (4) Friday

6. If 23rd April, 1984 was a Monday, which day of the week was 15th August in that year?

(1) Monday (2) Wednesday (3) Tuesday (4) Thursday

7. If 3rd March, 1984 was a Sunday, then which day of the week was 13th July, 1987?

(1) Monday (2) Sunday (3) Saturday (4) Tuesday

8. If 10th April, 1883 was a Wednesday, then which day of the week was 23rd August, 1879?

(1) Sunday (2) Tuesday (3) Monday (4) Friday

9. January 16, 1997 was a Thursday. What day of the week was January 4, 2000?

(1) Monday (2) Sunday (3) Tuesday (4) Wednesday

10. March 5, 1999 was on Friday, what day of the week was March 5, 2000?

(1) Friday (2) Tuesday (3) Monday (4) Sunday

I1. Monday falls on 4th April, "1988. What was the day on 3rd November, 1987?

(1) Tuesday (2) Sunday (3) Monday (4) Wednesday

12. The year after 1991 having the same calendar as that of 1991 is:

(1)1998 (2)2001 (3)2002 (4)2003

13. Which year will have the same calendar as that of 2004?

(1)2008 (2)2012 (3) 2032 (4) 2030

14. If a year starts and ends with Monday, then how many Mondays, are there in that year?

(1)51 (2)53 (3) 52 (4) Can't say

15. Which dates of April, 2012 will be a Sunday?

(1)1, 8,15,22,29 (2)3, 10,17,24,31 (3) 2, 9, 16, 23, 30 (4) can’t say

Clocks 10

Introduction

Many a times, questions appear on clocks in certain exams. Here, we discuss some concepts related to clocks covering all type of questions asked.

The dial of a clock is a circle whose circumference is divided into 12 parts, called hour spaces. Each hour space is further divided into 5 parts, called minute spaces. This way, the whole circumference is divided into 12 x 5 = 60 minute spaces.

The time taken by the hour hand (smaller hand) to cover a distance of an hour space is equal to the time taken by the minute hand (longer hand) to cover a distance of the whole circumference. Thus, we may conclude that in 60 minutes, the minute-hand gains 55 minutes over the hour-hand.

Note: The above statement (given in bold) is very much useful in solving the problems in this chapter, so it should be remembered. The above statement wants to say that:

"In an hour, the hour-hand moves a distance of 5 minute spaces whereas the minute-hand moves a distance of 60 minute spaces. Thus, the minute-hand remains 60 - 5 = 55 minute spaces ahead of the hour-hand."

Some other facts:

1. In every hour, both the hands coincide once.

2. When the two hands are at right angle, they are 15 minute spaces apart. This happens twice in every hour.

3. When the hands are in opposite directions, they are 30 minute spaces apart. This happens once in every hour.

4. The hands are in the same straight line when I they are opposite each other.

5. The hour hand moves around the whole circumference of clock once in 12 hours. Sc the minute-hand is twelve times faster than the hour-hand.

6. The clock is divided into 60 equal minute divisions.

7. 1 minute division = ��

�� = 6° apart.

8. The clock has 12 hours numbered frorr-1 to 12 serially arranged.

9. Each hour number is evenly and equally: separated by five minute divisions

(= 5 x 6°), = 30° apart.

10. In one minute, the minute-hand moves one' minute division or 6°.

11. In one minute, the hour hand moves ��

�

12. In one minute the minute-hand gains 5 ��

� more than the hour-hand.

13. When the hands are together, they are 0° apart hence,

� Formed, in 12 hours

Formed in 24 hours

0° or 180°

11 22

90° or any other angle

22 44

As per the required angle difference between the minute-hand and the hour-hand and the initial (or starting) position of the hour-hand, difference formulae are used to find out the required time. Nov, consider the Rules (Quicker Methods) given in the following pages.

Variants in a Clock: It is evident that the two hands: a clock will subtend an angle '6' between them. At any time, the same can be found out using the following formula:

� = �� m – 30 h ��

Or

� == 30 h - ��

(Here m = minutes and h = hours)

Example 1:

At what time between 3 O'clock and 4 O'clock, will e minute-hand and the hour-hand of a clock coincide with each other?

Solution:

When the two hands of a clock coincide with each other the angle between them is 0°.

� = �� m – 30 h

Here,

� = o° and h = 3

� O = �� m – 30h

� �� m = 30 � 3

� m = �� 16 ��

Therefore, the two hands of the clock are coincide

At 16 �� past 3.

Example 2:

At what time between 4 O'clock and 5 O'clock will the hands of a clock be in the same straight line but not together?

Solution:

When the two hands of the clock are in the same straight line but not together then the angle between them is 180°.

� = �� m – 30h

Here,

h = 4 and 9 = 180°

180 = �� - 30 � 4

� m = �� = ��

Therefore, the hands of the clock are on the same

Straight line at 4 hours ��

Example 3:

What is the angle between the minute-hand and the hour-hand of a clock at 3 hrs? 20 min.?

Solution:

� = �� m – 30 h

� = angle

m = minutes

h = hours

Here,

m = 20 and h = 3

� = �� x20 -30x3

= 110-90 = 20°

� 0 = 20°

Gain or Lose: In correct clock hands of a clock coincide every 65 ��min.

If hands of a clock coincide in less than

65 ��min then clock gains time and if hands of a

Clock coincides in more than 65 ��min then clock loses time.

Example 4:

The minute-hand of a clock overtakes the hour-hand at intervals of 65 minutes of the correct time. How much in a day does the clock gain or lose?

Solution:

In a correct clock, the hands of a clock coincide every 65 �� minutes. But in this case they are

together again after 65 minutes, hence clock gains time.

Gain in 65 minutes =!� ��" �!�� =

�� minutes.

Gain in one day (24x60 min.)

=�� #�� = �� min. = $� ��

��

Too Fast and Too Slow:

If a watch indicates 9.20, when the correct time is 9.10, it is said to be 10 minutes too fast. And if it indicates 9.00, when the correct time is 9.10, it is said to be 10 minutes too slow.

Example 5:

A watch, which gains uniform ally, was observed to be 5 minutes, slow at 10 a.m. on a Tuesday. On the next day at 11 a.m. it was noticed that watch was 5 minutes fast. When did the watch show the correct time? Solution:

Total hours from 10 a.m Tuesday to 11 a.m. on next day = 25 hours.

The watch gains (5 + 5) = 10 minutes in 25 hours. The watch gains 5 min. in

= � �� = �� %&' 12 �� hrs.

= 12 �� hrs. From 10 a.m. Tuesday = 10: 30 p.m. Tuesday

Example 6:

There are two clocks, both set to show the correct time at 10 p.m. One clock gains one minute in an hour while the other gains 2 minutes in one hour, then by how many minutes do the two clocks differ at 10 a.m. on the next day?

Solution:

Difference in minutes between the two clocks in one hour = 1 minute. Total number of hours (10 p.m to 10 a.m. on next day) = 12 hours. The two clocks differ by = 1 x 12 = 12 minutes

Example 7:

If the time in a clock is 8 hours 20 minutes, then what time does it show on the mirror?

Solution:

The time shown by the clock, when seen in the mirror is

= 12 hours - 8 hours 20 minutes

= 3 hours 40 minutes,

Exercise

1. A clock is started at noon. By 10 minutes past 5, the angle that the hour-hand has turned through is:

(1)145° (2)150° (3)155° (4)160°

2. An accurate clock shows 8 O'clock in the morning. Through how many degrees will the hour-hand rotate when the clock shows 2 O'clock in the afternoon?

(1)144° (2)150° (3)168° (4)180°

3. At what time between 9 O'clock and 10 O'clock will the hands of a watch coincide?

(1)10 hrs. = �( �� ' (2) 9 hrs. = �( �

�� '

(3) 11 hrs. = �( �� ' (4) 9 hrs. = �( �

�� '

4. The angle between the minute-hand and the hour-hand of a clock when the time is 8 : 30, is:

(1)80° (2)75° (3)60° (4)105°

5. At what time between 5 and 6 O'clock are the hands of a clock 3 minutes apart?

(1) 24 min. past 5 (2) �� min. past 5 (3) 30

min. past 5 (4) Both (1) and (2)

6. How many times do the hands of a clock coincide in a day?

(1)20 (2)21 (3) 22 (4) 24

7. The minute-hand of a clock overtakes the hour-hand at interval of 64 minutes of the correct time. How much does the clock gain or lose in a day?

(1)�# �� (3) �! �

�� (3) 90 min. (4) 96 min.

8. The minute-hand of a clock overtakes the hour-hand at intervals of 68 minutes of a correct time. How much in a day does the clock gain or lose?

(1) 53��) �� *+�& (2) �� *+�& (3)

�� )��) �� *+�& (4) ��

��) �� *+�&

9. A watch which gains uniformly is 2 minutes slow at noon on Monday and is 4 min. 48 sec fast at 2 p.m. on the following Monday. When did it show the correct time?

(1)2p.m. on Tuesday (2)2 p.m. on Wednesday (3)3 p.m. on Thursday (4)1 p.m. on Friday

10. A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the afternoon of the same cay, when the watch indicated quarter past 4 O'clock, the true time is:

(1) �( )�� past 3 (2) 4 p.m.

(3) �, )�� min. past 3 (4) # �

�� . past 4

11. A watch, which gains uniformly is 6 minutes slow at 4 p.m. on a Sunday and

10�� *+�& minutes fast on the following Sunday at 8 a.m. When did it show the

correct time?

(1)2: 00 a.m. on Monday (2)1: 36 a.m. on Tuesday

(3)1: 36 a.m. on Wednesday (4)1: 36 a.m. on Thursday

12. A clock gains 10 minutes in every 24 hours. It is set right on Monday at 8 a.m. What will be the correct time on the following Wednesday, when the watch indicates 6 p.m.?

(1)5:30 p.m. (2) 5: 24 p.m. (3) 5: 36 p.m. (4) 5: 20 p.m.

13. A clock is set right at 5 a.m. The clock loses 16 min. in 24 hrs. What will be the true time when the clock indicates 10 p.m. on the 4th day?

(1)12 p.m. (2) 11 a.m. (3) 11 a.m. (4) 10 a.m.

14. There are two clocks, both set to show the correct time at 10 a.m. One clock gains two minutes in one hour while the other gain one minute in one hour. If the clock which gains 2 minute shows the time as 22 minute past 9 p. m. On the same day, then what time the other watch show?

(1)9 hrs. 33 min. (2) 9 hrs. 12 min. (3) 9 hrs. 11 min. (4) 9 hrs. 23 min.

15. If the time in a clock is 6 hours 45 minutes, then what time does it show on the mirror?

(1)4 hrs. 15 min. (2) 5 hrs. 45 min.

Answer key

Calendars

Clocks

1. 2 Total number of odd days = 1600 years have 0 odd

day + 300 years have 1 odd day + 49 Jan have 5 odd days = 0 + 1 + 5 + 5 = 4 odd days So, the day was Thursday.

2. 4 1600 years have 0 odd day

200 years have 2x5 = 10, i.e., 3 odd days.years. That is, 17 x 2 + 51 =85 days, i.e., 1 odd day.of odd days = 31 (Jan.) + 28(Feb.) + 31 (Mar.) + 30(Apr.) + 31 (May)+ 30(Jun.) + 31 (Jul.) + 31 (Aug.) + 30(Sep.) + 2(Oct.)

Total odd days = 0 + 3 + 1 + 2 = 6 odd days.

The day was Saturday.

3.3 15 Aug., 1947 = (1600 + 300 + 46) years + 1 Jan. I(1600 + 300 + 46) years + 365(1600 + 300 + 46) years + (365 yea and 35 ordinary years) + 3 = 5 odd days. .

4. 3 83 years contain 20 leap years and 63 ordinary5 odd days. 1983 years contain (0 + 1+5) i.e., 6 odd days. Number of days from Jan., 1984 to 31st Oct., 1984 = (31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31) = 305 days = 4 odd days. .-. Total number of odd days = 6 + 4 = 10 i.So, 31st Oct., 1984 was Wednesday.

5. 2 20th June, 1837 means "1836

(1)4 hrs. 15 min. (2) 5 hrs. 45 min. (3) 5 hrs. 15 min. (4) 5 hrs. 30 min

1. 2 Total number of odd days = 1600 years have 0 odd

day + 300 years have 1 odd day + 49 years (12 leap + 37 ordinary) have 5 odd days + 26 days of Jan have 5 odd days = 0 + 1 + 5 + 5 = 4 odd days So, the day was Thursday.

1600 years have 0 odd day

200 years have 2x5 = 10, i.e., 3 odd days. 68 years contain 17 leap years and 51 ordinThat is, 17 x 2 + 51 =85 days, i.e., 1 odd day. In 1869, up to 2nd Oct., total number

= 31 (Jan.) + 28(Feb.) + 31 (Mar.) + 30(Apr.) + 31 (May)+ 30(Jun.) + 31 (Jul.) + 31 (Aug.) + 30(Sep.) + 2(Oct.) = 275 days = 2 odd days.

d days = 0 + 3 + 1 + 2 = 6 odd days.

15 Aug., 1947 = (1600 + 300 + 46) years + 1 Jan. I 15 Aug. of 1947(1600 + 300 + 46) years + 365- 16 Aug. to 31 Dec 1947

46) years + (365 - 138) days Number of odd days = 0 + 1 +1 (from 11 leap yea and 35 ordinary years) + 3 = 5 odd days. .-. The day was Friday.

83 years contain 20 leap years and 63 ordinary years and therefore (40 + 0) odd days i.e., 1983 years contain (0 + 1+5) i.e., 6 odd days. Number of days from Jan.,

1984 to 31st Oct., 1984 = (31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31) = 305 days mber of odd days = 6 + 4 = 10 i.e 3 odd days.

So, 31st Oct., 1984 was Wednesday.

20th June, 1837 means "1836 complete years + fit

years (12 leap + 37 ordinary) have 5 odd days + 26 days of

68 years contain 17 leap years and 51 ordinary to 2nd Oct., total number

= 31 (Jan.) + 28(Feb.) + 31 (Mar.) + 30(Apr.) + 31 (May)+ 30(Jun.) + 31

15 Aug. of 1947 = =

138) days Number of odd days = 0 + 1 +1 (from 11 leap

years and therefore (40 + 0) odd days i.e., 1983 years contain (0 + 1+5) i.e., 6 odd days. Number of days from Jan.,

1984 to 31st Oct., 1984 = (31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31) = 305 days

5 months of the year 1837 + 20 days of June" 1600 years give no odd day 200 years give 3 odd days 36 years give 3 odd days. [36 years contain 9 leap years and 27 ordinary year Therefore, (27 + 18 =) 45 odd days = 3 odd days.] :. 1836 years give (0 + 3 + 3) = 6 odd days Now, from first January to 20th, June, we have, Jan. Feb. Mar. Apr. May Jun. odd days: 3 + 0 + 3 + 2+ 3 + 6 = 17 i.e., 3 odd days. � Total number of odd days = (6 + 3) = 9 odd da) i.e., 2 odd days. This means that the 20th June fell on the 2nd d< commencing from Monday. Therefore the required da was Tuesday.

6. 2 Counting the number of days after 23rd April, 1984 have: April may June July August days: 7 + 31 + 30 + 31 + 15 =114 days Number

of odd days in 114 days = ��) = 16 weeks + 2 odd days 2nd day after Monday is

Wednesday.

7. 3 It is given that 3rd March, 1984 was a Sunday So, 3rd March, 1987, was three days after Sunday, i.e., o Wednesday. Number of days from 3rd March, 1987 to 13th July 1987: March April May June July Days: 28 + 30 + 31 +

30 + 31 = 150 = � �) = 3 odd days 3rd day

after Wednesday is Saturday.

8.1 It is given that 10th April, 1883 was a Wednesday. Number of days from 10th April, 1883 to 23rd August 1883 April May June July August :

Days: 20 + 31 + 30 + 31 + 23 = 135

Number of odd days in 135 days =�� ) = 2 odd days.

2 days after Wednesday is Friday. Number of odd days from 23rd August 1879 to 23rd August 1883 are five. So, 23rd August, 1879 is five days back to Friday is Sunday.

9.3 First we look for the leap years during this period. 1997, 1998, 1999 are not leap years. 1998 and 1999 together have net 2 odd days. Number of days remaining in 1997 = 365 - 16 = 349 days = 49 weeks 6 odd days. �Total number of odd days = 2 + 6 + 4=12 days = 7 days (1 week) + 5 odd days Hence, January 4,2000 will be 5 days beyond Thursday i.e., it was Tuesday.

10.4 Year 2000 is a leap year. Therefore, March 5, 2000 will be two days beyond Friday, i.e., on Sunday.

11.1 Counting the number of days after 3rd November, 1987 we have:

Nov. Dec. Jan. Feb. Mar. Apr.

Days: 27 + 31 + 31 + 29 + 31 + 4 = 153 days, containing 6 odd days, i.e., (7 - 6) = 1 day beyond the day on 4th April, 1988. So, the day was Tuesday.

12.3 We go on counting the odd days from 1991 onwards till the sum is divisible by 7. The number of such days are 14 up to the year 2032. So, the calendar for 1991 will be repeated in the year 2002.

13.3 The year 2004 is a leap year and a leap year repeats itself after 28 years 2004 + 28 = 2032 So, 2032 will have the same calendar as that of 2004.

14.2 As the given year starts and ends with Monday means the next year will start with Tuesday. Hence, the given year is a non-leap year. There will be 53 Mondays in the year.

15.1 1st April, 2012: 2000 + 11 + Number of days from 1st January 2012 to 1st April, 2012. Number of odd days in 2000 years = 0 Number of odd days in 11 years = 13

January February March April

Odd days: 3 + 1 + 3 + 1 =8

Total number of odd days = 8 + 13 + 0 = 21 = 0 odd days. Hence, 3st April, 2012 is a Sunday. 1st, 8th, 15th, 22nd and 29th of April, 2012 are Sunday's.

Clocks

1. 3 Angle traced by hour-hand in 12 hrs. = 360° Angle traced by

hour-hand in 5 hrs 10 min. i.e. �� hrs. = ( �� ) ° = 155°

2.4 Angle traced by hour-hand in 6 hours = ( �� 6) ° = 180°

3.2 When the two hands of the clock coincide, then the angle between them is 0°.

49 min.

4.2 The angle between the two hands of a clock at 8:30 is

5. 2 At 5 o'clock, the minute-

Case (i) : Minute-hand is 3 min. spaces behind the hourminute-hand has to gain (25

22 min. are gained in (

3 min. apart at 24 min. past 5. Case (ii): Minutehand. In this case, the minuteare gained in 60 min.

28 min. are gained in (

The hands will be 3 minutes apart

6. 3 The hands of a clock coincide 11 times icoincide only once, i.e. at 12 O'clock).in a day.

7.1 In a correct clock, the minute60 minutes. To be together again, the minutehand. 55 min. spaces are gained in 60 min.

When the two hands of the clock coincide, then the angle between them is 0°.

Therefore, the hands of the clock are together at 9 hrs.

the two hands of a clock at 8:30 is

-hand and the hour-hand are 25 min. spaces apart.

hand is 3 min. spaces behind the hour-hand. In this case, the hand has to gain (25-3) = 22 min. spaces. Now, 55 min. are gained in 60 min.

22)min = 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 hourIn this case, the minute-hand has to gain (25 + 3) = 28 min. spaces

28) =30 min.

be 3 minutes apart at 30 min. past 5. 11

The hands of a clock coincide 11 times in every 12 hours (Since between 11 and 1, they coincide only once, i.e. at 12 O'clock). The hands of a clock coincide 22 times

In a correct clock, the minute-hand gains 55 minute spaces over the hour-be together again, the minute-hand must gain 60 minutes over the hour

hand. 55 min. spaces are gained in 60 min.

When the two hands of the clock coincide, then the angle between them is 0°.

Therefore, the hands of the clock are together at 9 hrs.

hand are 25 min. spaces apart.

In this case, the Now, 55 min. are gained in 60 min.

The hands will be

. spaces ahead of the hour- Now, 55 min.

n every 12 hours (Since between 11 and 1, they The hands of a clock coincide 22 times

-hand in every hand must gain 60 minutes over the hour-

60 min. spaces are gained in

But they are together at an interval of

8. 1 In a correct clock, the hands of a clock coincide every

9. 2 Time from 12 p.m. on Monday to 2 p.m. on the following Monday = 7 days 2 hours = 170 hours

Watch is correct 2 days 2 hrs. after 12 p.m. 0 Monday i.e., it will be correct at 2 p.m. on Wednesday

10. 2 Time from 7 a.m. to 4.15p.m. = 9hrs15min.=

3 min. 5 sec. of this clock = 3 min. of the correct clock

this clock = hrs. of the correct clock.

60 min. spaces are gained in x 65min. =65 min.

But they are together at an interval of 64 minutes. Gain in every 64 minutes.

In a correct clock, the hands of a clock coincide every

Time from 12 p.m. on Monday to 2 p.m. on the following Monday = 7 days 2 hours =

170 hrs. Now, min. are gained in 170 hrs.

t 2 days 2 hrs. after 12 p.m. 0 Monday i.e., it will be correct at 2 p.m. on

Time from 7 a.m. to 4.15p.m. = 9hrs15min.= hrs

3 min. 5 sec. of this clock = 3 min. of the correct clock

hrs. of the correct clock.

ain in every 64 minutes.

Time from 12 p.m. on Monday to 2 p.m. on the following Monday = 7 days 2 hours =

min. are gained in 170 hrs.

t 2 days 2 hrs. after 12 p.m. 0 Monday i.e., it will be correct at 2 p.m. on

hrs. of

correct clock. . The correct time is 9 hrs. after 7 a.m. i.e. 4 p.m

11. 3 Total time in hours from Sunday at 4 p.m. to the followin

16 = 160 hrs.Wednesday at 1.36 a.m

12. 3 Total number of hours from Monday at 8 a.m= 58 ft: 24 hrs. 10 min. of this clock are same as 24 hrs. o; correct clock.

Wednesday be 5 : 36 p.m.

13.2 Total number of hours from 5 a.m. on first day to 10 p.m. on 4th day is 89 hrs. 23 hrs. 44 min. of this clock are same as 24 hrs. of a correct clock.

hrs. of this clock = 24 hrs. of correct clock

89 hrs. of this = (

time is 11 p.m.

14.3 Difference in minute between the two clocks in one hour = 1 minute. Number of hours = 11 hours. In 11 hours, one of the clock gains 22 minutes The other clock which gains 1 minute per hour shows the time as 9:11 p.m.

. the correct clock = 9 hrs. of the The correct time is 9 hrs. after 7 a.m. i.e. 4 p.m

Total time in hours from Sunday at 4 p.m. to the following Sunday at 8 a.m. = 6 x 24 +

The watch was correct on

Total number of hours from Monday at 8 a.m following Wednesday at 6 p.m. 24 x 2 + 10 = 58 ft: 24 hrs. 10 min. of this clock are same as 24 hrs. o; correct clock.

The correct time on the Wednesday be 5 : 36 p.m.

Total number of hours from 5 a.m. on first day to 10 p.m. on 4th day is 89 hrs. 23 hrs. 44 min. of this clock are same as 24 hrs. of a correct clock.

hrs. of this clock = 24 hrs. of correct clock

89) hrs. of correct clock.

Difference in minute between the two clocks in one hour = 1 minute. Number of hours = 11 hours. In 11 hours, one of the clock gains 22 minutes and shows the time as 9:22 p.m. The other clock which gains 1 minute per hour shows the time as 9:11 p.m.

the correct clock = 9 hrs. of the

Sunday at 8 a.m. = 6 x 24 +

The watch was correct on

following Wednesday at 6 p.m. 24 x 2 + 10

The correct time on the following

Total number of hours from 5 a.m. on first day to 10 p.m. on 4th day is 89 hrs. 23 hrs. 44

The correct

Difference in minute between the two clocks in one hour = 1 minute. Number of hours = and shows the time as 9:22 p.m.

The other clock which gains 1 minute per hour shows the time as 9:11 p.m.

15. 3 The time shown by the clock when seen in the mirror = 12 hrs. - 6 hrs. 45 min. = 5 hrs. 15 min.

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