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                  Problem on Clocks - TCS Placement Model Questions For a complete understanding of  CLOCKS,  Concept of Clocks   For m...

Model questions for TCS Placement Papers - Latest Pattern Clocks

                 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


TCS 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 00 
To find the angle between two hands   
                                                         θ =|30X H –$\frac{11}{2}$ x M)|
                                                         00 =| 30X 6 -  $\frac{11}{2}$  x M|
ð        $\frac{11}{2}$ M = 180
ð  M= $\frac{360}{11}$
ð  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
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 cm2                        b)15.27cm2                    c) 16.12 cm2                  d)17.732
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 60 =600
Area covered by minutes hand in 10 minutes =   θ/360  x π, x r2
                                                                 = $\frac{60}{360}$ x $\frac{22}{7}$ x 5.42
                                                                  = 15.27cm2

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
 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


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
 ð  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 π cm                     b)1648π cm                     c)1724π cm       d)2028π cm
      Answer : B
Explanation:   
Short Hand ( Hour hand)
The hour hand makes a full rotation in 12 hours.
One full rotation in 12 hours =>2πr=14π cm distance is covered in 12 hours.
In one day(24 hours ),  it covers the distance 2 x 14π  = 28 π
In  4 days, the hour hand covers 4 x 28 π=112π cm
Long Hand ( Minute hand)
One full rotation in 1 hour =>2πr=16π cm distance is covered in one hour.
In one day ( 24 hours ), the minute hand covers  24 x  16 π=384π cm.
In 4 days, it covers the distance 4 x 384π=1536π cm.
Total Distance covered by both hands
For the total, we have 112π+1536π=1648π 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)15th position         b)20th position               c)21st position                    d)27th 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
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
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
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 = 3600  => 1 hour = $\frac{360^{0}}{18}$  = 200
In 12 hours, the angle made by hour hand = 12 x 20 = 2400
Given 1 hour = 90 minutes => 90 minutes =200 => 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=248.880

Taking minute hand into consideration:
90 min = 3600  =>  1 min= $\frac{360^{0}}{90^{0}}$ =40
In  40 minutes= 40*4= 1600

The difference between the angle made by the hour hand and minute hand
              = 248.88 – 160 = 88.880= 89 (Approximately)