13.CLOCK, CALANDER AND ARITHMETICAL REASONING
13.1 CLOCKS
This chapter of Mental Ability deals with different type of questions on clock which are invariably
asked in various competitive examination. Though we all know how to read time on a clock, yet solution
of questions on clock requires some level of intelligence which is above on common level. We have laid
down properties of clock with illustrative examples so that students may not find any difficulty while
solving questions given in the exercise.
The dial of the clock is numbered from 1 to 12 in such a way that each subsequent number is equidistant
(5 minutes space apart) from the preceding number. It has two needles known as minute hand and hour hand.
The distance between two consecutive numbers is 5 minutes. It means that the circumference of the clock measures
60 minutes space. When hour hand travels 5 minutes space, the minute hand travels 60 minutes space. In this way,
by traveling 60 minutes space, the minute hand gains 55 minutes on the hour hand.
Properties of Clock
1. In 60 minutes the minute hand gains 55 minutes on the hour hand.
2. A clock is said to be too fast when it shows time more than that of shown by a correct clock.
Likewise, a clock is said to be too slow when it shows time less than that of shown by a correct clock.
3. Both the hands of clock are at right angle (when the distance between two hand measures 90°)
when they are 15 minutes space apart. This situation occurs twice in one hour.
4. Both the hands of a clock are opposite to each other when they are 30 minutes space apart.
This situation occurs once in a hour.
5. The hands of the clock are in the same straight line when they are coincident or opposite to each other.
Illustration 1
At what time between 7 and 8 will the hands of a clock be in the same straight line, but not together?
(a) 5 minutes (b) minutes past 7
(c) minutes past 7 (d) minutes past 7
Solution
Figure (1) shows the positions of hands of the clock at 7 and figure (2) shows the positions of hands of clock when
both the hands are opposite in the straight line. From figure (1), it is clear that both the hands are 25 minute,
apart and to be in the straight line both the hands have to be 30 minutes apart.
Now as per properties of the clock,
minute hand gains 55 minutes space in 60 minutes.
Hence it will gain 5 minutes space in minutes or minutes.
Therefore, the hands are in the same straight line, but not together at minutes past 7.
Illustration 2
At what time between 5.30 and 6 will the hands of a clock be at right angle ?
(a) minutes past 5 (b) minutes past 5
(c) 40 minutes past 5 (d) 45 minutes past 5
Solution
At 5 both the hands are 25 minutes apart and to be at right angle both the hands have to be 15 minutes apart
as shown in fig. (2) and fig. (3). Since we have to take the position of clock between 5.30 and 6 therefore, the
positions of hands of clock as per fig. (3) is our answer. Now it is clear from fig. (1) and fig. (3) that minute hand
will have to travel (25 + 15) = 40 minutes space in order to form a right angle witf-the hour hand.
55 minutes space is gained in 60 minutes.
Therefore, 40 minutes space will be gained in minutes
or minutes.
Therefore, the hands are at right angle at minutes past 5.
So, the correct option is (b).
Illustration 3
At what time between 4 and 5 will the hands of a watch point in opposite directions ?
(a) 45 minutes past 4 (b) 40 minutes past 4
(c) minutes past 4 (d) minutes past 4
Solution
At 4 o’clock both the hands are 20 minutes apart and for having in the opposite direction they have to be 30 minutes apart.
From figure (1) and (2) it is clear, that minute hand has to travel (20 + 30) minutes space in order to be in opposite direction to each other.
Now 55 minutes space is gained in 60 minutes.
Therefore, 50 minutes space will be gained in minutes
or minutes.
Hence, the hands of the clock will be in opposite direction at minutes past 4. Therefore, (d) is the answer.
Illustration 4
A watch, which gains uniformly, is 2 minutes slow at noon on monday, and in 4 minutes, 48 seconds fast at 2 PM. on the
following monday. What time it was correct ?
(a) 2 p.m. on tuesday (b) 2 p.m. on Wednesday
(c) 3 p.m. on thursda (d) 1 p.m. on friday
Solution
Time from monday noon (12 p.m.) to 2 p.m. the following monday
= 7 days 2 hours =170 hours
Now, the watch gains minutes from monday (12 p.m.) to 2 p.m. on the following monday.
or in other words, the watch gains minutes in 170 hours,
Therefore, it will gain 2 minutes in hours = 50 hours
= 2 days 2 hours.
Therefore, the watch is correct after 2 days 2 hours from monday noon or at 2 p.m. on Wednesday. So the correct option is (b).
Illustration 5
A clock is set right at 5 a.m. The clock loses 16 minutes in 24 hours. What will be the right time when the clock indicates 10 p.m. on the 4th day?
(a) 11.15 p.m. (b) 11 p.m. (c) 12 p.m. (d) 12.30 p.m.
Solution
Time from 5 a.m. of a particular day to 10 p.m. on the 4th day is 89 hours. Now, the clock loses 16 minutes in
24 hours or in other words we can say that 23 hours 44 minutes of this clock is equal to 24 hours of the correct clock.
or hours of this clock = 24 hours of the correct clock.
89 hours of this clock = hrs. of correct clock.
= 90 hours of the correct clock
or 89 hours of this clock = 90 hours of the correct clock.
Therefore, it is clear that in 89 hours this clock loses 1 hour and hence the correct time is 11 p.m. When this clock shows 10 p.m.
Illustration 6
How many times do the hands of a clock coincide in a day ?
(a) 24 (b) 22
(c) 21 (d) 20
Solution
From the properties of the clock we know that hands of a clock coincide once in every hour but between 11 o’clock and
1 o’clock they coincide only once. Therefore, the hands of a clock coincide 11 times in every 12 hours.
Hence they will coincide (11 × 2) 22 times in 24 hours. So our answer is (b).
Illustration 7
How many times are the hands of a clock at right angles in a day ?
(a) 24 (b) 48
(c) 22 (d) 44
Solution
We know that hands of a clock are at right angle twice in every hour. But two positions of the hands of clock
i.e. at 3 o’clock and 9 o’clock are identical. So, they are at right angles 22 times in 12 hours and therefore,
in 24 hours or in a day they are at right angle 44 times.
So, the answer is (d).
13.2 Calender
This is another important chapter in Mental Ability. Questions based on calender are of set pattern and hence follow
set rules. We are giving below the definition/meaning of certain words which are used in solving questions on calender.
Odd days : We know that a week contains 7 days counting from monday to Sunday. So, any number of days, which are
more than complete number of a week in a given period are called odd days. For example a period of 10 days contains 3
odd days, 11 days contains 4 odd days, 12 days contains 5 odd days. But period of 14 days contains zero odd day.
Therefore, in finding number of odd days in a given period of time, one has to divide that period by 7. If it is completely
divisible by 7, it contains zero odd day and if it is not divisible by 7 then remaining number of days are the odd days.
Leap year : Every year which is divisible by 4 is called a leap year. But every century which is divisible by 4 is not a leap year.
Every fourth century is a leap year. For example 400, 800, 1200, 1600 ........... are all leap years but centuries like 100, 300, 500,
600...........are not leap years.
An ordinary year has 365 days i.e. (52 weeks + 1 day)
A leap year has 366 days i.e. (52 weeks + 2 days)
When we divide 365 (an ordinary year) by 7, we get remainder 1, it means that it has 1 odd day. Like wise 366 days
(leap year) has 2 odd days.
A century has 100 years and every fourth year is a leap year. We can break a century in the leap year as follows :
4,8,12,16,20............96.
Now number of terms contained by the above series.
96 = a + (n – 1) d
[Arithmetic progression where a = first term, d = common difference]
96 = 4 + (n – 1) 4
92 = (n – l) 4
(n – 1) =
n = 24
Therefore, a century has 76 ordinary years
100 years = 24 leap year + 76 ordinary year
= (24 x 366 + 76 x 365) days
= 36524 days = weeks
= 5217 weeks + 5 days
= 5 odd days.
So, 100 years contain 5 odd days
200 years contain 10 odd days or 3 odd days
300 years contain 15 odd days or 1 odd day
400 year contain 0 odd day.
Like wise years 800, 1200, 1600, 2000........contain zero odd day.
Counting of day in respect of odd day :
Illustration 8
Find the day of the week on :
(a) 27th December, 1985
(b) 15th August, 1947
(c) 12th January, 1979
Solution
(a) 27th December 1985 has (1984 years, 11 months and 27 days).
Now 1600 years have 0 odd days.
300 years have 1 odd day
84 years contains = (21 leap years and 63 ordinary years)
= (21 × 366 + 63 × 365) days
= (7686 + 22995) = 30681 days = 4383 weeks.
i.e. 84 years contains 0 odd days.
11 months and 27 days =361 days
51 weeks + 4 days = 4 odd days.
= 4 + 1 = 5
Therefore,
27 December 1985 has (6 + 1) = 7 odd days.
Now counting Sunday as 0 ouv day, Tuesday as 2 odd day and so on.
Friday will have 5 odd day. Therefore, 27th December 1985 will be Friday.
(b) 15th August 1947
(1946 years, 7 months and 15 days)
Now 1600 years have 0 odd day
300 years have 1 odd day
1900 years have 1 odd day.
46 year have (11 leap years and 35 ordinary years)
= (11 × 366 + 35 × 365) days = 16801 days
= (2400 weeks + 1 day) = 1 odd day.
7 months and 15 days = 227 days
= (32 weeks + 3 days) = 3 odd days
(1946 years + 7 months + 15 days)
have (1 + 1 + 3) = 5 odd days
Which is Friday.
(c) 12th January, 1979 = (1978 years + 12 days)
Now, 1600 years have 0 odd days
300 years have 1 odd day
78 years have (19 leap years + 59 ordinary years)
= (19 × 366 + 59 × 365) days = 28489 days
= 4069 weeks + 6 days = 6 odd days
12 days = 5 odd days
Total number of odd days = (1 + 6 + 5)
= 12
= 5 odd days
So, the day on 12th January 1979 was Friday.
Illustration 9
On what days of July 1776 did Sunday fall ?
Solution
First of all find the day on 1st July 1976.
1st July 1776 = (1775 years + 6 months + 1 days)
Now, 1600 years have 0 odd days
100 years have 5 odd day
75 years have 18 leap years and 57 ordinary years
Which have 2 odd days
1775 years have (0 + 5 + 2) = 7 = 0 odd day
Now, 6 months + 1 day
= Jan + Feb + March + April + May + June + 1
= (31 + 29* + 31 + 30 + 31 + 301 + 1
= 183 days
= 1 odd day
1st July 1776 will be Monday and hence first Sunday for the month of July will fall on 7th. Therefore,
other Sundays will fall on 14th, 21st and 28th.
Illustration 10
What was the day on 26th January 1950, when first Republic Day of India was celebrated ?
(a) Monday (b) Tuesday (c) Thursday (d) Friday
Solution
26th January 1950 means
(1949 years and 26 days)
1600 years have 0 odd days
300 years have 1 odd day
49 years have (12 leap years and 37 ordinary years)
(12 x 366 + 37 x 365)
days – (4392 + 13505) days
(17897) days = 2556 weeks + 5 days
So, 49 years have 5 odd days
and 26 days have 5 odd days
Total number of odd days = 0 + 1 + 5 + 5
= 11 days
= 4 odd days.
Hence the day on 26th January 1950 was Thursday.
Illustration 11
What is the number of odd days in a leap year?
(a) 1 (b)2
(c) 3 (d) 4
Solution
A leap year has 366 days
Now if we divide 366 by 7 it gives 2 as remainder.
Hence number of odd days in 366 days is 2.
Illustration 12
Prove that the calender for 1990 will serve for 2001 also.
Solution
There are two ways of proving it. One find the day on 1st January 1990 and 1st January 2001. Both the days should be
identical. Second find the number of odd days between 31st Dec. 1989 and 31st Dec. 2000. The sum of odd
days should be zero. Odd days are calculated as below :
Sum of odd days = 14 i.e. 0 odd days.
EXERCISE-I
1. 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 ?
(A) min. past 8 (B) min. past 8
(C) min. past 8 (D) 10 min. past 8
2. At what time between 3 and 4 o’clock will the hands of a clock coincide ?
(A) 15 min. past 3 (B) min. past 3
(C) min. past 3 (D) min. past 3
3. A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours. What will be the right time when the clock indicates 1 p.m. on the following day ?
(A) 11.40 p.m. (B) 12.48 p.m.
(C) 12 p.m. (D) 10 p.m.
4. At what time between 4 and 5 o’clock will the hands of a clock be at right angle ?
(A) 30 min. past 4 (B) min. past 4
(C) min. past 4 (D) 33 min. past 4
5. At what time between 5 and 6 are the hands of a clock coincident ?
(A) 22 minutes past 5 (B) 30 minutes past 5
(C) minutes past 5 (D) minutes past 5
6. At what time between 9 and 10 will the hands of a watch be together ?
(A) 45 minutes past 9 (B) 50 minute - past 9
(C) minutes past 9 (D) minutes past 9
7. How many times do the hands of a clock point towards each other in a day ?
(A) 24 (B) 20
(C) 12 (D) 22
EXERCISE-II
1. What is the day on 1st January 1901 ?
(A) Monday (B) Wednesday
(C) Sunday (D) Tuesday
2. What is the day on 31st October 1984 ?
(A) Friday (B) Sunday
(C) Wednesday (D) Monday.
3. What is the day on 14th March, 1993 ?
(A) Friday (B) Thursday
(C) Sunday (D) Saturday.
4. Prove that the Calender for 1992 and for 1997 are identical ?
5. On what dates of August 1980 did Monday fall ?
(A) 4th, 11th, 18th and 25th
(B) 3rd, 10th, 17th, 24th
(C) 6th, 13th, 20th and 27th
(D) 9th, 16th, 23rd & 30th
6. On what dates of December, 1984 did Sunday fall ?
(A) 6th, 13th, 20th & 27th
(B) 7th, 14th, 21st, & 28th
(C) 2nd, 9th, 16th, 23rd & 30th
(D) 1st, 8th, 15th & 22nd
7. Today is Wednesday, What will be the day after 94 days ?
(A) Monday (B) Tuesday
(C) Wednesday (D) Sunday
8. Today is Thursday. The day after 59 days will be:
(A) Sunday (B) Monday
(C) Tuesday (D) Wednesday
9. Today is Friday. The day after 63 days will be :
(A) Friday (B) Thursday
(C) Saturday (D) Monday
10. What was the day of the week on 2nd July 1984 ?
(A) Wednesday (B) Tuesday
(C) Monday (D) Thursday
11. The year next to 1990 will have the same calendar as that of the year 1990 :
(A) 1995 (B)1997
(C) 1996 (D) 1992
12. The year next to 1996 will have the same calendar as that of the year 1996 :
(A) 2001 (B)1996
(C)1997 (D)1999
EXERCISE-III
1. In a chess tournament each of six players will play every other player exactly once.
How many matches will beplayed during the tournament ?
(A) 12 (B) 15 (C) 30 (D) 36
2. A man has a certain number of small boxes to pack into parcels. If he packs 3, 4, 5 or 6 in a parcel, he is left
with one over; if he packs 7 in a parcel, none is left over. What is the number of boxes, he may have to pack?
(A) 106 (B) 301 (C) 309 (D) 400
3. A, B, C and D play a game of cards, A says to B, ‘ If I give you 8 cards, you will have as many as C has and I shall
have 3 less than what C has. Also, if I take 6 cards from C, I shall have twice as many as D has.” If B and D together have 50 cards, how many cards has A got ?
(A) 40 (B) 37 (C) 27 (D) 23
4. In a group of cows and hens, the number of legs are 14 more than twice the number of heads. The number of cows is ?
(A) 5 (B) 7 (C) 10 (D) 12
5. A worker may claim Rs. 15 for each km which he travels by taxi and Rs. 5 for each km which he drives his own car.
If in one week he claimed Rs. 500 for travelling 80 km how many kms did he travel by taxi?
(A) 10 (B) 20 (C) 30 (D) 40
6. Reena is twice as old as Sunita. Three years ago, she was three times as old as Sunita. How old is Reena now ?
(A) 6 years (B) 7 years (C) 8 years (D) 12 years
7. The age of a father is twice that of the elder son. Ten years hence the age of the father will be three times that of the
younger son. If the difference of ages of the two sons is 15 years, the age of the father is ?
(A) 50 years (B) 55 years (C) 60 years (D) 70 years
8. A shepherd had 17 sheep. All but nine died. How many was he left with ?
(A) Nil (B) 8 (C) 9 (D) 17
9. A bird shooter was asked how many birds he had in the bag. He replied that there were all sparrows but six,
all pigeons but six, and all docks but six. How many birds had he in all ?
(A) 9 (B) 18 (C) 27 (D) 36
10. What is the smallest number of ducks that could swim in this formation - two ducks in front of a duck,
two ducks behind a duck and a duck between two ducks ?
(A) 3 (B) 5 (C) 7 (D) 9
11. A group of 1200 persons consisting of captains and soldiers in travelling in a train. For every 15 soldiers there is one captain.
The number of captains in the group is ?
(A) 85 (B) 80 (C) 75 (D) 70
12. Aruna cut a cake into two halves and cuts one half into smaller pieces of equal size. Each of the small pieces is twenty
grams in weight. If she has seven pieces of the cake in all with her, how heavy was the original cake ?
(A) 120 grams (B) 140 grams (C) 240 grams
(D) 280 grams (E) None of these
13. First bunch of bananas has 1/4 again as many bananas as a second bunch. If the second bunch has 3
bananas less than the first bunch, then the number of bananas in the first bunch are ?
(A) 9 (B) 10 (C) 12 (D) 15
14. At the end of a business conference the ten peoples present all shake hands with each other once.
How many handshakes will there be altogether ?
(A) 20 (B) 45 (C) 55 (D) 90
15. A student got twice as many sums wrong as he got right. If he attempted 48 sums in all,
how many did he solve correctly ?
(A) 12 (B) 16 (C) 24 (D) 18
16. The number of boys ina class is three times the number of girls. Which one of the following
numbers cannot represent the total number of children in the class ?
(A) 48 (B) 44 (C) 42 (D) 40
17. A motorist knows four different routes from Bristol to Birmingham. From Birmingham to Sheffield he knows
three different routes and from Sheffield to Carlisle he knows two different routes. How many routes does he know from Bristol to Carlisle ?
(A) 4 (B) 8 (C) 12 (D) 24