__Problem on Clocks - TCS Placement Model Questions__For a complete understanding of CLOCKS, Concept of Clocks

For more solved problems and examples, go through Problems on Clocks

__T__

__CS Placement Papers - New Pattern Question 1__

Between 9 a.m
and 9 p.m of a particular day for how many times are the minute and hour hands
together?

a)10
b)11
c)12
d)13

**Answer: B**

**Explanation:**

**The hands of a clock together (coincide each other) once in an hour.**

From 9 a.m to 9 p.m, there are 12 hours. But between 11 a.m and 1 p.m,
the two hands coincide only once at exactly 12'o clock.

Hence between 9 am and 9 pm, the two hands will be
together 11 times.

__TCS Placement Papers - New Pattern Question 2__

At what time between 6 O’clock and 7 O’ clock are
the hands of the clock coincide?

a) 6:13
b
)6:36
c) 6:32 $\frac{8}{11}$
$
d) 6:35 $\frac{8}{11}$

**Answer: C**

**Explanation:**

**Here the two hands coincide => the angle between the two hands is 0**

^{0}

To find the angle between two hands

^{ }
Î¸ =|30X
H –$\frac{11}{2}$ x M)|

0

^{0}=| 30X 6 - $\frac{11}{2}$ x M|
Ã° $\frac{11}{2}$ M = 180

Ã° M= $\frac{360}{11}$

Ã° M= 32 $\frac{8}{11}$ minutes

Therefore, the two hands coincide at 6 : 32 $\frac{8}{11}$

__TCS Placement Papers - New Pattern Question 3__

How many palindromes are there in a clock from noon
to midnight ( For Example 6.06 is a palindrome)?

a)42 b)51 c)53 d)57

a)42 b)51 c)53 d)57

**Answer: D**

**Explanation:**

From 12.00 to 1.00, there is only one palindrome
12.21

From 1.00 to 2.00 => 1.01, 1.11, 1.21, 1.31,
1.41, 1.51 => 6 palindromes

From 2.00 to 3.00 => 2.02, 2.12, 2.22, 2.32, 2.42,
2.52 => 6 palindromes

Similarly from 3 to 4, 4 to 5, 5 to 6, 6 to 7, 7 to
8 , 8 to 9 , 9 to 10 => In each hour there are 6 palindromes

From 10.00 to 11.00 =>10.01 is only palindrome
=> 1 palindrome

From 11.00 to 12.00 => 11.11 is only
palindrome => 1 palindrome

Therefore Number of palindromes = 1 + (9 x 6)+ 1 +
1 =57

__TCS Placement Papers - New Pattern Question 4__

The length of the minute hand of a clock is 5.4 cm.
What is the area covered by this in 10 minutes

a)14 cm

a)14 cm

^{2}b)15.27cm^{2}c) 16.12 cm^{2}d)17.73^{2}**Answer: B**

**Explanation:**

Solution: If the minute
hand moves 60 minutes, it makes a circle. Here it is covered only 10
minutes. So it makes a sector.

Length of the minute hand = Radius r = 5.4 cm

The angle made by minute hand in 10 minutes = 10 x
6

^{0}=60^{0}
Area covered by minutes hand in 10 minutes = Î¸/360 x

**Ï€**, x r^{2}
=

^{ }$\frac{60}{360}$ x $\frac{22}{7}$ x 5.4^{2}
= 15.27cm

^{2}

__TCS Placement Papers - New Pattern Question 5.__

The
famous church in the city of Kumbakonnam has a big clock tower and is said to
be over 300 years old. Every Monday 10.00 A M the clock is set by John, doing
service in the church. The Clock loses 6 minutes every hour. What will be the
actual time when the faulty clock shows 3 P.M on Friday?

a. 2:12 AM Saturday b.3.16 PM Friday

a. 2:12 AM Saturday b.3.16 PM Friday

c.
4.54 AM Sunday
d. 3 AM Saturday

**Answer: A**

**Explanation:**

Total number of hours from Monday 10 am
to Friday at 3 pm in the faulty clock

= (24 hours x 4 days) + 5 hours = 96 hours + 5 hours = 101 hours.

The clock loses 6 minutes in an hour.

54 minutes of the faulty clock = 60 minutes of the correct clock

101 hours of the faulty clock =$\frac{60}{54}$ x 101 = 112.2 Hours.= 96 Hrs + 16.2 Hrs

Friday 10 am + 16 hrs = Saturday 2am

0.2 x 60 min = 12 min

Therefore, the actual time is Saturday at 2.12 min AM

54 minutes of the faulty clock = 60 minutes of the correct clock

101 hours of the faulty clock =$\frac{60}{54}$ x 101 = 112.2 Hours.= 96 Hrs + 16.2 Hrs

Friday 10 am + 16 hrs = Saturday 2am

0.2 x 60 min = 12 min

Therefore, the actual time is Saturday at 2.12 min AM

__TCS Placement Papers - New Pattern Question 6__

One-quarter of the time till now from midnight and half of the time remaining from
now up to midnight adds to the present time. What is the present time?

a)8.30 a.m
b)9.36
a.m
c)10.14
a.m d)11.32 a.m

**Answer: B**

**Explanation:**

Let the present time = t hrs

One-quarter of the time till now from midnight=$\frac{t}{4}$

Half of the time remaining from now up to midnight =(24-t)/2

Hence
=>
$\frac{t}{4}$ + $\frac{24-t}{2}$=t

Ã° 5t= 48

=> 9.6 hours

The time was 9 hours 36 minutes = 9.36 a.m

__TCS Placement Papers - New Pattern Question 7__

In a clock the
longhand is of 8cm and the shorthand is of 7cm. if the clock runs for 4 days
find out the total distance covered by both the hands

a) 1824 Ï€ cm
b)1648Ï€
cm
c)1724Ï€ cm d)2028Ï€ cm

**Answer : B**

**Explanation:**

__Short Hand ( Hour hand)__

The hour hand makes a full rotation in 12 hours.

One full rotation in 12 hours =>2Ï€r=14Ï€ cm distance is
covered in 12 hours.

In one day(24 hours ), it covers the
distance 2 x 14Ï€ = 28 Ï€

In 4 days, the hour hand covers 4 x 28 Ï€=112Ï€ cm

*Long Hand ( Minute hand)*

One full rotation in 1 hour =>2Ï€r=16Ï€ cm distance is
covered in one hour.

In one day ( 24 hours ), the minute hand covers 24 x 16 Ï€=384Ï€ cm.

In 4 days, it covers the distance 4 x 384Ï€=1536Ï€ cm.

__Total Distance covered by both hands__

For the total, we have 112Ï€+1536Ï€=1648Ï€ cm.

__TCS Placement Papers - New Pattern Question 8__

A circle has 29
points arranged in a clockwise manner from o to 28. A bug moves clockwise manner from 0 to 28. The bug moves clockwise on the circle according to the following rule. If it is at a point i on the circle, it moves clockwise in 1
sec by (1 + r) places, where r is the remainder (possibly 0) when i is divided
by 11. If it starts in 23rd position, at what position will it be after
2012 sec.

a)15

^{th}position b)20^{th}position c)21^{st}position d)27^{th}position**Answer : B**

**Explanation:**

**After 1st second, it moves 1 + $\frac{23}{11}$ = 1 + 1 = 2, So 25th position**

After 2nd second, it moves 1 + $\frac{25}{11}$ = 1 + 3 = 4, So 29th position = 0

After 3rd second, it moves 1 + $\frac{0}{11}$ = 1 + 0 = 1, So 1st position

After 4th second, it moves 1 + 1 = 3rd position

after 5th, 1 + $\frac{3}{11}$ = 4 So 7th

After 6th, 1 + $\frac{7}{11}$ = 8 so 15th

After 7th, 1 + $\frac{15}{11}$ = 5 so 20th

After 8th, 1 + $\frac{20}{11}$ = 10th, So 30th = 1st

So it is on 1st after every 3 + 5n seconds. So it is in 1st position after 2008 seconds (3 + 5 x 401)

Therefore, 20th after 2012 position

__TCS Placement Papers - New Pattern Question 9__

A clock loses
1% time during the first week and then gains 2% time during the next week. if
the clock was set right at 12 noon on Sunday. What will be the time exactly
that the clock will show 14 days from the time it was set right?

a)1.30
pm b)1.42.28
pm c) 1.40.48
pm d)1.52.49
pm

**Answer: C**

**Explanation:**

The clock loses 1% time during the first week.

In a week, there are 7 * 24 = 168 hours

In a week, there are 7 * 24 = 168 hours

The clock loses 1% time during the first week, then it will show a time
which is 1% of 168 hours less than 12 Noon at the end of the first week = 1.68
hours less.

Subsequently, the clock gains 2% during the next week. The second week has 168 hours and the clock gains 2% time = 2% of 168 hours = 3.36 hours more than the actual time.

Total gain or loss in 2 weeks = -1.68 + 3.36 = 1.68 hours gain

So the clock will show a time which is 1.68 hours more than 12 Noon after 14 days

Subsequently, the clock gains 2% during the next week. The second week has 168 hours and the clock gains 2% time = 2% of 168 hours = 3.36 hours more than the actual time.

Total gain or loss in 2 weeks = -1.68 + 3.36 = 1.68 hours gain

So the clock will show a time which is 1.68 hours more than 12 Noon after 14 days

1.68 hours = 1 hour and 40.8 minutes = 1 hour 40
minutes and 48 seconds.

i.e. 12 noon + 1 hour 40 minutes and 48 seconds =1 :
40: 48 P.M.

__TCS Placement Papers - New Pattern Question 10__

On the planet
Oz, there are 8 days in a week- Sunday to Saturday and another day called Oz
day. There are 36 hours in a day and each hour has 90 minutes while each minute
has 60 sec. As on earth, the hour hand covers the dial twice every day. Find
the approximate angle between the hands of a clock on Oz when the time is 12:40
am.

a) 50 b)70 c)89 d)131

a) 50 b)70 c)89 d)131

**Answer: C**

**Explanation:**

**There are 36 hours a day and each hour has 90 minutes and each minute has 60 seconds.**

Given time is 12.40

__Taking Hour hand into consideration:__

The hour hand covers the dial 2 times every day.

So there are 18 hours in dial.

18 hours = 360

^{0}^{ }=> 1 hour = $\frac{360^{0}}{18}$ = 20^{0}
In 12 hours, the angle made by hour hand = 12 x 20
= 240

^{0}
Given 1 hour = 90 minutes => 90 minutes =20

^{0}=> 1 minute = $\frac{20^{0}}{90}$
=> 1 minute = ($\frac{2}{9}$)

^{0}
so 40 Min= 40*($\frac{2}{9}$)=8.88(approximately)

Total angel covered by the hour hand is=240+8.88=248.88

^{0}__Taking minute hand into consideration:__

90 min = 360

^{0}^{ }=> 1 min= $\frac{360^{0}}{90^{0}}$ =4^{0}
In 40 minutes= 40*4= 160

^{0}
The difference between the angle made by the hour hand
and minute hand

= 248.88 – 160 = 88.88

^{0}= 89^{0 }(Approximately)