For complete understanding of the concept of clocks, go
through

**Concept of Clocks**

*Aptitude Clocks - Solved Problem 1*

What is the angle made
by the minute angle in 16 minutes?

a) 8

^{0}b)32^{0}c)96^{0 }d) 48^{0 }e)90^{0}
Answer: C

Explanation:

The angle made by the
minute hand in 1 minute is 6

^{0.}
In 16 minutes, the angle
made by the minute hand = 16 x 6

^{0 }=96^{0}

*Aptitude Clocks - Solved Problem 2*

2. What is the angle
made by the hour hand in 30 minutes?

a) 5

^{0}b)10^{0}c)180^{0}d) 15^{0 }e)28^{0}**Answer : D**

**Explanation:**

The angle made by the hour hand in 1 minute is
$\frac{1}{2}$

^{0}.
In 16 minutes,
the angle made by 30 minutes is = 30 x (½)

^{0 }
=15

^{0}

*Aptitude Problem on Clocks - Solved Problem 3*

3. What is the
difference between angles made by minute hand and hour hand in 24 minutes?

a) 185

^{0}b)120^{0}c)180^{0}d) 132^{0 }e)208^{0}
Answer: D

Explanation : The angle made by the minute hand in 24 minutes = 24 x 6

^{0}=144^{0}
The angle made by the hour
hand in 24 minutes =24 x (½)

^{0 }=12^{0}
The difference is = 144

^{0}-12^{0}=132^{0}

*Aptitude Problems on Clocks - Solved Problem 4*

4. How often the hands of clock at right angle everyday ?

a)11 times b)22
times c)44 times d)55 times e)66
times

Answer : C

Explanation:

In onehour there are two positions in which the minute hand and
hour hand are at right angle.

Each of these
positions is repeated 22 times in every 12 hours.

Therefore in a day ( 24 hours ),
the two hands are perpendicular to each other

ð 22 + 22 =44 times.

*Aptitude Clocks - Solved Problem 5*5. How many times the hands of a clock are in a straight line every day?

a)
11 times b)22 times
c)33 times d)44 times c)55 times

Answer : D

Explanation: Any relative
position of the two hands of a clock is repeated 11 times in every 12 hours.

In every 12 hours,
two hands coincide ( when the two hands coincide, they are always in the same
straight line ) 11 times and two hands are opposite to each other (but
in the same straight line )11 times .

∴ In 12 hours , two hands are in straight line
11 +11 =22 times.

∴ In every 24 hours , two hands are in straight line
44 times.

*Aptitude Questions on Clocks - Solved Problem 6*__6. A clock strikes 5 takes 16 seconds. In order to strike 10 at the same rate, the time taken is__

a) 24 seconds b)30
seconds c)36 seconds
d)32 seconds e)40 seconds

Answer : C

Explanation :

There are 4 intervals when
the clock strikes 5.

Time taken for 4 intervals = 16 seconds

∴ Time taken for
1 interval = 4 seconds

In order to strike
10, there are 9 intervals, for which the time taken is 9 x 4 =36 seconds

*Aptitude questions on Clocks - Solved Problem 7*7. What is the angle between the minute hand and the hour hand at 20 minutes past 4 O' clock ?

a)
5

^{0}b)10^{0}c)180^{0}d) 15^{0}e)20^{0}
Answer : B

Explanation : To
find the angle between hour hand and minutes at H hours M minutes
,

θ = |(30 X H) - $\frac{11}{2}$ M |

= |(30 X 4) -($\frac{11}{2}$ )x 20 |

= |120 -
110 | = 10

^{0}

*Aptitude questions on Clocks - Solved Problem 8*
8.At what time between 4 and 5’O clock are the two hands of the
clock coincide ?

Explanation : At 4'O clock
, the hour hand is at 4 and the minute hand is at 12. Hence minute hand is 20
minutes spaces behind the hour hand.

The two hands will
coincide each other, when the minute hand gains 20 minutes spaces over the hour
hand.

The minute hand gains 55
minute spaces in 60 minutes .

To gain 1 minute space ,
it will take $\frac{60}{55}$ minutes.

To gain 20 minute spaces,
it will take 20 x $\frac{60}{55}$ = 20 x $\frac{12}{11}$ =21 $\frac{9}{11}$ minutes

∴ The two hands coincide at 21
$\frac{9}{11}$ minutes past 4.

*Aptitude Problems on Clocks - Solved Problem 9*

9. Find the time between 8 and 9’O Clock when
the two hands of a clock are in the same straight line.

a) 8.41 $\frac{9}{11}$ b)8.43 $\frac{7}{11}$ c)8.47 $\frac{3}{11}$ d)8.44 $\frac{3}{11}$ e)None the above

Answer: B

Explanation :

Two hands of a clock are
in the same straight line in two cases:

__Case 1:__**.**

__When the two hands are in exactly opposite direction__
This equals 180

^{o}=$\frac{180^{0}}{6}$ =30 minute spaces apart.
At 8’O Clock, the minute
hand is (8×5)=40 minute spaces behind the hour hand. Therefore, the minute hand
will have to gain (40-30)=10 minute spaces over the hour hand.

55 minutes spaces are
gained in 60 minutes

To Gain of 40 minute spaces => 10 x $\frac{60}{55}$ =$\frac{120}{11}$ =
10 $\frac{10}{11}$ minutes.

The minute hand will be is exactly opposite
direction to the hour hand at 10 $\frac{10}{11}$ minutes past 8’O Clock.

__Case 2:__

__When the two hands coincide i.e, 0 minute spaces apart.__
In this case, the minute hand will have to gain (8×5)=40 minute spaces over the
hour hand.

To gain 40 minute spaces=>$\frac{60}{55}$ ×40=$\frac{480}{11}$ = 43 $\frac{7}{11}$ times

The
two hands will hands will coincide at 43 $\frac{7}{11}$ minutes past 8 ‘O Clock.

*Aptitude questions on Clocks - Solved Problem 10*

10. At what time between 5 and 6’O clock will the
two hands of a clock be at right angle?

a) 5.43 $\frac{7}{11}$ b)5.10 $\frac{10}{11}$ c)5.12
$\frac{8}{11}$ d)5.13 $\frac{8}{11}$ e) 5.09 $\frac{7}{11}$

Answer : Both A and B

Explanation: At 5’O Clock the minute hand is (5×5)=25 minute spaces behind the
hour hand.

The two hands will be at a
right angle when either (i) the minute hand is 15 minute-spaces behind the hour
hand or (ii) the minute hands are 15 minute-spaces ahead of the hour hand.

__Case1:__

__When the minute hand 15 minute-spaces behind the hour hand.__
For the two hands to be in
this position, the minute hand must gain (25-15) =10 minute-spaces over the
hour hand.

55 minute spaces are
gained in 60 minutes.

So10 minute-spaces are gained in => $\frac{60}{55}$ ×10=$\frac{120}{11}$ =10 $\frac{10}{11}$ minutes.

Therefore , the two hands will be at right
angle at 10 $\frac{10}{11}$ minute past 5’O clock.

__Case 2:__

__when the minute hand 15 minute-spaces ahead of the hour hand.__
To be in this position, the minute hand must gain (25+15)=40 minute spaces.

So 40 minute-spaces are gained in=>
$\frac{60}{55}$ ×40 =$\frac{480}{11}$ =43 $\frac{7}{11}$ times.

Therefore , the two hands will be at right
angle at 43 $\frac{7}{11}$ minutes past 5’O Clock.

*Aptitude questions on Clocks - Solved Problem 11*
11. At what time between 4 and 5 are the hands 2
minutes spaces apart?

a) 4 19 $\frac{7}{11}$ and 4.22

b)4.21 $\frac{7}{11}$ and 4.24

c) 4. 19 $\frac{7}{11}$ and 4.24

d) 4.18 $\frac{9}{11}$ and 4.24 $\frac{2}{11}$

e) None of the above

Answer : C

Explanation: 4’O Clock, the two hands are 20 minute spaces apart.

__Case 1:__

__When the minute hand is 2 minute spaces behind the hour hand.__
In this case, the
minute hand will have to gain (20-2) i.e 18 minute spaces. Now, we know that 18
minute spaces will be gained in 18 × $\frac{12}{11}$ = $\frac{216}{11}$ =19 $\frac{7}{11}$ minute.

Therefore , the two hands
will be 2 minutes apart at 19 $\frac{7}{11}$ minutes past 4.

__Case 2:__

**.**

__when the minute hand is 2 minute spaces ahead of the hour hand__
In this case,
the minute hand will have to gain (20+2) i.e., 22 minute spaces.

Now the 22 minute spaces will be gained in

22× $\frac{12}{11}$ = 24 minutes

The hands will be 2 minute
spaces apart at 24 minute past 4

^{ }

__Aptitude Questions on Clocks - Solved Problem 12__

When do the two hands of
a clock of just after 3 pm make 30 º angles between them?

a)3:15:00 b)3:10:54
c)3:01:59 d) 3:20:21 e)3:18:00

**Answer:(B)**

**Explanation:**To find the angle between hour hand and minutes at H hours M minutes,

θ =|30H – $\frac{11}{2}$ M|

Here the angle given
is 30=> - 30 = 30 x 3 – $\frac{11}{2}$ M ( We take + and – angles , because minute
hand is 30

^{0}behind the hour hand once and 30^{0}ahead of the hour hand another time)
$\frac{11}{2}$ M = 60 => M= $\frac{120}{11}$ = 10 $\frac{10}{11}$ minute

= 10 min 54s

(We have the answer in the given choices,
we need not compute the another time with +30

^{0})
∴ Required time = 3:10:54

__Aptitude problems on Clocks - Solved Problem 13__
A clock strikes ones at
1 O’clock, twice at 2 O’clock and so on. What is the total number of striking
in a day

a)
12 b) 156 c) 78 d)
24 e) 48

**Answer:(B)**

**Explanation:**The clock strikes once at 1 O’clock, twice at 2 O’clock, thrice at 3 O’clock and so on.

So in 12 hours, the total number of
striking = 1 + 2 + 3 + 4 + ---- + 12

(Sum of the first n natural numbers= $\frac{n(n+1)}{2}$
)

=
12 x $\frac{(12+1)}{2}$
(12+1)/2 =78

In 12 hours, the total striking are 78.

∴ In a day (24
hours) , total number of striking = 2 x 78 = 156

__Aptitude questions on Clocks - Solved Problem 14__
Three cuckoo clocks are
such that the cuckoos chime after every 9 minutes, 15 minutes and 35 minutes respectively.
If the 3 clocks chime simultaneously at 3 p.m, what time will they chime
together again?

a)8 :15 p.m b)9:15 p.m c)10.30 p.m d) 11.00 p.m e)00.45 p .m

**Answer:B**

**Explanation:**

Three cuckoo clocks are such that the
cuckoos chime after every 9 minutes, 15 minutes and 35 minutes respectively.

All the three clock chime
simultaneously at 3 p.m

They will again chime together =>
LCM of 9 minutes, 15 minutes and 35 minutes

ð 315 minutes or 5
hours 15 minutes

After 5 hours 15 minutes , all the
three clock will chime together

The time at they will chime again together = 3
p.m + 5 hours 15 minutes

= 8:15 p.m

__Aptitude problems on Clocks - Solved Problem 15__
The minute hand of a
clock overtakes the hour hand at intervals of 65 minutes of correct time. How
much in a day does the clock gain or lose?

a)
11 $\frac{11}{143}$ minutes b)10 $\frac{10}{143}$ minutes c)12 $\frac{11}{143}$ minutes

d)
9 $\frac{11}{143}$ minutes e)
13 $\frac{11}{143}$ minutes

**Answer: B**

**Explanation:**

In a correct
clock, the minute hand gains 55 minute spaces over the hour hand in 60 minutes.

To coincide each
other again, the minute hand must gain 60 minutes over the hour hand.

Now, 55 min. are
gained in 60 min.

∴ 60 min. are gained in $\frac{60}{55}$ x 60 min. = 65 $\frac{5}{11}$ min.

But in the given question, the minute hand overtakes the hour hand at
regular intervals of 65 minutes ..

In the given
clock, the two hands meet at 65 minutes instead of 65 $\frac{5}{11}$ min.

Gain in every
65 minutes. =65 $\frac{5}{11}$ -65 min. = $\frac{5}{11}$
min.

In 65 minutes ,
the clock gains $\frac{5}{11}$ minutes.

The clock gains in a day (24 hours) = $\frac{5}{11}$ X
$\frac{1}{65}$ X 24 X 60 = $\frac{1440}{143}$ minutes = 10 $\frac{11}{143}$

__Aptitude difficult problems Clocks - Solved Problem 16__

__.__
There are two clocks A
and B. The hands of the clock A moves normally as clockwise while in clock B
(due to reverse connection) they move anticlockwise. Initially the two hands of
both clocks are at 12. If after some time, the angle between the directions of
two hour hands is 90º for the first time , then at the same instant the angle
between the directions of minute hand will be

a)0

^{0}b) 60º c) 120º d)180º e) 90^{0}**Answer:**

**A**

**Explanation:**The angle between the two hour hands after some time is 90

^{0}.

As the both started at 12 ‘o clock,
the hour hand of clock A makes 45

^{0}and the hour hand of clock B makes 45^{0}from their initial position.
The hour hand of
clock A moved 45º from its initial position. Therefore, number of minutes the
hour hand makes is 45 x 2 = 90 minutes

In 90 minutes, the minute hand makes
angle of 90 x 6

^{0}= 540^{0}
As 360

^{0}is a complete revolution = > 540^{0}= (360^{0}+ 180º) from its initial position.
As both hands move 180

^{0 }in either direction, the minute hands of two clocks coincide, i.e. the angle between them is zero degree.

__Aptitude questions on Clocks - Solved Problem 17__
A watch was set correct
at 12’O clock. It loses 10 minutes per
hour. What will be the angle between the two hands of the clock after 1 hour?

a)
75º b)
85º c) 90º d) 105º e)120

^{0}**Answer:(B)**

**Explanation:**

**The clock loses 10 minutes per hour**

S the clock shows only 50 minutes for every 60
minutes.

After 1 hour, the
watch show the time 12:50

The angle between
the two hands of the clock after 1 hour i.e 60 minutes

(i.e.
50 minutes according to the clock) = |30H
– $\frac{11}{2}$ M|

=|30 x 12 – $\frac{11}{2}$ x 50|

= |360º – 275º| = 85º

__Aptitude problems on Clocks - Solved Problem 18__

A clock is set right at
7:10 am on Thursday, which gains 12 minutes in a day. Find the true time when
this clocks the shows the 3:50 p.m on the following Sunday?

a)2:50
pm b)
3:10 pm c) 3:30 pm
d)4:30 pm e) 4.10 p.m

**Answer: B**

**Explanation:**Total number of hours from Thursday 7.10 a.m. to the 3 : 50 p.m . on Sunday.

=>24 x 3 + 8 hours 40 minutes = 80
hours 40 minutes. (80 $\frac{40}{60}$ = $\frac{242}{3}$ hours)

The clock gains
12 minutes in every 24 hours.

24 hours 12 min.
of this clock =24 hours of correct clock,

i.e. $\frac{121}{5}$
hours of this clock = 24 hours of correct clock

∴ 80 hours 40 minutes of this clock =24 X $\frac{5}{121}$
X $\frac{242}{3}$ hours of correct clock

=
80 hours of correct clock

Therefore the correct time is 7:10 a.m + 80 hours = 3:10 p.m.

__Aptitude difficult questions on Clocks - Solved Problem 19__
A clock is set right at 10
a.m on Tuesday . The clock gains 10 min in 24 hours. What will be the correct time
when the clock shows the time 8 p.m on the following Thursday?

a)
8.36 p.m. b) 8.40 p.m. c)7.36
p.m. d)
7.52 p.m e)8.12 p.m

**Answer: A**

**Explanation:**

Total number of
hours from Tuesday at 10 a.m. to the following Thursday at 8 p.m.

ð 24 x 2 + 10 = 58 hours

24 hours 10 minutes of this clock = 24 hours of a correct clock.

$\frac{145}{6}
Hours of the incorrect clock = 24
hours of correct clock.

58 hours of the incorrect clock = 24 X $\frac{6}{145}$
X 58 hours of correct clock

= 57 $\frac{3}{5}$ $\frac{3}{5}$ hours of correct clock.

= 57 hours 36 minutes

Thus, the correct time on the following
Wednesday will be 8.36 p.m.

__Aptitude Clocks - Solved Problem 20__

A clock was correct at 2
p.m, but then it began to lose 30 minutes each hour. It now shows 6 pm, but it
stopped 3 hours ago. What is the correct time now?

a)8.30 pm. B)12
midnight
c)11 p.m. d)1.30 a.m. e)1 a.m

**Answer: E**

**Explanation :**The clock loses 30 minutes per hour. And the clock set correct at 12 noon.

30 minutes of this clock = 60
minutes of the correct clock

From 2 p.m to 6 p.m , total number of
hours = 4 hours

4 hours of this clock => 4
x $\frac{60}{30}$ = 8 hours

The correct time when the clock
show 6 p.m = 6 p.m + 4 = 10 p.m

The clock stopped 3 hours ago , So
present time is 10 p.m + 3 hours = 1 a.m