1. The bus for Chennai leaves
every 30 minutes from a bus depot. The inquiry clerk told a passenger that the
bus for Chennai left 10 minutes ago and the next bus will leave at 10: 30 am,
what was the time when the inquiry clerk told this?

**[SSC (10 + 2) 2012]**
a) 10: 10 am b)10 : 20 am c)9 :50 am d)10 : 00 am

**Answer: A**

**Explanation :**

Next bus is at 10:30 a.m. And the bus for Chennai leaves every 30 minutes.

So last bus has left 30 minutes before 10:
30 a.m. => 10:00 a.m

When the passenger enquired, the clerk
answer the bus left 10 minutes ago.

\
Enquiry time = 10:00 a.m + 10 min

= 10: 10 a.m

2.
In Ravi’s clock shop, two
clocks were brought for repairs. One clock has the cuckoo coming out every 16
min, while the other one has the cuckoo coming out every 18 min. Both cuckoos
come out at 12: 00 noon. When will they both come out together again?

__[SSC CGL 2011]__
a)
2: 32 PM b)2: 24 PM c)2: 04 PM d)2: 06 PM

**Answer: B**

**Explanation :**

The cuckoos of both
the clocks will come out together after intervals which are common multiples of
16 min and 18 min.

16 = 2

^{4}18 = 2 x 3^{2}
LCM of 16 and 18 = 2

^{4}x 3^{2}= 144 minutes
144 minutes = 2 hours + 24 minutes

Both the cuckoos come out
together at 12:00 p.m

Again they will
immediately come out at => 12: 00 PM + 2 hours 24 minutes

=
2: 24 PM

3.
At what time are the hands
of the clock together between 6 and 7?

__[SSC CGL 2011]__
a)32 $\frac{8}{11}$
minutes past 6 b)34
$\frac{8}{11}$ minutes past 6

c)30$\frac{8}{11}$
minutes past 6 d)32 $\frac{8}{11}$ minutes
past 6

**Answer : A**

**Explanation :**

At
6 O’ clock, the two hands are 30-minute spaces apart.

To be together, the minute hand has to gain 30-minute spaces over the hour hand.

The minute hand gains 55-minute spaces in 60
minutes.

To gain 1-minute space, the minute hand
takes $\frac{60}{55}$ minutes.

To gain 30 minute spaces, it takes 30 x $\frac{60}{55}$ = 30 x $\frac{12}{11}$

= 32 $\frac{8}{11}$ minutes

At
6: 32 $\frac{8}{11}$, the two hands of a clock are together between 6 and 7

4.
A clock gains five minutes
every hour. What will be the angle traversed by the second hand in 1 min?

__[SSC CGL 2011]__
a)360 b)360.5

^{0}c)390 d)380**Answer : C**

**Explanation :**

In one second, the second-hand makes an angle of 6

^{0}.
The clock gains 5 minutes every hour. Then
the minute hand of the clock moves 65 minutes in one hour.

So seconds hand also moves 65 seconds.

Therefore, the angle covered by
the second hand = 65 x 6

^{0}= 390^{0}
5.
The chairman of the
Selection Committee arrived at the interview- room for conducting an interview
at 10 minutes to 12: 30 h. He was earlier by 20 minutes than the other members
of the board, who arrived late by 30 min. At what time was the interview
scheduled?

__[SSC Previous Question 2005]__
a)12: 10 b)12: 20 c)12 : 30
d) 12: 40

**Answer : A**

**Explanation :**

The chairman of the
selection committee arrived at 10 minutes to 12 :30 => 12 : 20

He was earlier by
20 minutes than the other members.

So other members arrived at =>
(12 : 20 + 20 minutes)= 12 : 40

Other members were
late by 30 minutes

So scheduled time of interview = (12 : 40 – 30
minutes) = 12 :10

6.
By looking in a mirror, it
appears that it is 6: 30 in the clock. What is real-time?

__[SSC Past Qustion 2005]__
a)6
: 30 b)5 : 30
c)6 : 40 d)5 : 00

**Answer : B**

**Explanation :**

The
mirror reflection shows the time 6 :30

The actual time = (12 hours - 6 : 30 )
or (11 : 60 – 30 minutes)

= 5 :
30

7.
Reaching the place of
meeting 20 minutes before 8: 50 hours, Satish found himself 30 minutes earlier
than the man who came 40 minutes late. What was the scheduled time of the
meetings?

__[SSC (10 + 2) 2002]__
a) 08: 20 b)08
: 10 c) 08: 05
d) 8: 00

**Answer : A**

**Explanation :**

Satish reached the
place of meeting 20 minutes before 8: 50

So he reached at =>
(8: 50 – 20 minutes) = 08: 30

Satish was 30 minutes earlier than the man who came 40
minutes late.

So
Satish was 10 minutes late.

\ Scheduled time of the
meeting = (8: 30 - 10 minutes)

= 8: 20

8. A clock goes fast by 1 minute during the 1

^{st}hour, by 2 minutes at the end of the 2^{nd}hour, by 4 minutes at the end of 3^{rd}hour, by 8 minutes by the end of 4^{th}hour, and so on. At the end of which hour, will it be fast by just over 60 min?__[SSC (10 + 2) 2002]__
a)Fifth b)Sixth c)Seventh
d) Eighth

Answer:

**C****Explanation :**

A clock goes fast
by 1 minute during the 1

^{st}hour, by 2 minutes at the end of the 2^{nd}hour, by 4 minutes at the end of 3^{rd}hour, by 8 minutes by the end of 4^{th}hour and so on.
Every hour, the clock gains the double the time it gained in the previous hour.

At the end of the first
hour => 1 minute

At the end of
second hour => 2 minutes

At the end of the third
hour => 4 minutes

At the end of
fourth hour => 8 minutes

At the end of the fifth
hour => 16 minutes

At the end of the sixth
hour => 32 minutes

At the end of
seventh hour => 64 minutes

Therefore,
at the end of 7

^{th}hour, the clock will be fast by just over 60 min.
9. A
clock goes slow from mid-night by minutes. At the end of the 1

^{st}hour, by 10 min at the end of the 2^{nd}hour by 10 min, by 15 min at the end of the 3^{rd}hour and so on. What will be the time by this clock after 6 h?__[SSC (10 + 2) 2002]__
a)6 : 00 a.m b)5 : 30 a.m c)6
: 30 a.m d) 5 : 15 a.m

**Answer : B**

**Explanation:**

The clock loses
5 minutes in 1

^{st}hour, l10 minutes at the end of the second the hour and loses 15 minutes at the end of the third hour and so on.
The clock loses
5 minutes every hour.

In 6 hours,
the clock loses by ( 6 x 5 ) = 30
minutes

The clock was correct at midnight, and after 6 hours, the clock loses 30 minutes

So after 6 hours , the correct time is 6 a.m

The clock shows the time after 6 hours =>
6 a.m – 30 minutes = 5 : 30 a.m

10.
Reaching the place of meeting on Tuesday 15 min before 08: 30 h. I found myself half an hour earlier
than the man who was 40 min late. What was the scheduled time of the meeting?

__[SSC (10 + 2) 2000]__
a) 08: 00 h b)08
: 05 h c) 08: 15 h d) 08: 45 h

**Answer : B**

**Explanation :**

I reached 15 minutes before 8:30 = 8:30
– 15 minutes = 8 : 15

I was 30 minutes earlier than the man
who was 40 minutes late

I was late by=> 40 min –
30 min = 10 minutes

Therefore, scheduled time = 08: 15 – 10
minutes

= 08: 05 h

11. Kamala would like to complete all her homework before 10.00 pm to watch and important TV program. She has 40 min assignment in each of her five prepared subjects. What is the latest time at which she can start and still complete her homework in the time of the program?

11. Kamala would like to complete all her homework before 10.00 pm to watch and important TV program. She has 40 min assignment in each of her five prepared subjects. What is the latest time at which she can start and still complete her homework in the time of the program?

__[SSC (10 + 2) 2000]__
a) 6: 40 pm b) 6: 30 pm c) 7: 10 pm d) 7: 20 pm

**Answer: A**

TV Programme is at 10:00 p.m

She has to complete 5 assignments each of 40 minutes.

Time is taken to complete the assignments = 5
x 40 minutes

= 3 h 20 min

So
the latest time at which she can start is = (10: 00 - 03: 20) = 06: 40 pm

12.
After 9 O’clock at what time between 9 pm and 10 pm will the hour and minute hands of a clock point in
opposite directions?

__[SSC CGL 2000]__
a)15 min past 9 b)16 min past 9 c)16 $\frac{4}{11}$ min past 9
d) 17 $\frac{1}{11}$ min past 9

**Answer : C**

**Explanation:**

At
9 O’ clock, the two hands are 45 minutes spaces apart.

So
the minute hand has to gain 15-minute spaces over the hour hand

At
9 O’clock, the min hand is 9 x 5 = 45 min spaces behind the hour hand.

To gain 1-minute space, the minute hand
takes $\frac{60}{55}$ minutes

To gain 15
minutes spaces, it will take => 15 x $\frac{60}{55}$ = 16 $\frac{4}{11}$

Therefore,
the two hands of a clock point in opposite directions after 9 O’clock at 16
$\frac{4}{11}$ min past 9.