Interview questions and answer of Clocks with explanation for fresher

Interview questions and answer of Clocks with explanation for fresher

The dial of a clock is a circle whose circumference is divided into 12 parts, called hour spaces. Each hour space is further divided into 5 parts, called minute spaces. This way, the whole circumference is divided into 12 X 5 = 60 minute spaces.

 The time taken by the hour hand (smaller hand) to  cover a distance of an hour space is equal to the time taken by the minute hand (longer hand) to cover a distance of the whole circumference. Thus, we may conclude that in 60 minutes, the minute hand gains 55 minutes on the hour hand.
 Note : The above statement (underlined) is very much useful in solving the problems in this chapter, so it should be remembered. The above statement wants to say that :

 “ In an hour, the hour-hand moves a distance of 5 minute spaces whereas the minute-hand a distance of 60 minute spaces. Thus the minute-hand remains 60 - 5 = 55 minute spaces ahead of the hour-hand.”

Some other facts :

1. In every hour, both the hands coincide once.
2. When the two hands are at right angle, they are 15 minute spaces apart. This happens twice in every hour.
3. When the hands are in opposite directions, they are 30 minute spaces apart. This happens once in every hour.
4. The hands are in the same straight line when they are coincident or opposite to each other.
5. The hour hand moves around the whole circumference of clock once in 12 hours. So the minute hand is twelve times faster than hour hand.
6. The clock is divided into 60 equal minute divisions.
7. 1 minute division = 360060=60${\displaystyle \frac{{360}^0}{60}=6^0}$ apart
8. The clock has 12 hours numbered from 1 to 12 serially arranged.
9. Each hour number evenly and equally separated by five minute divisions 5×60=300$5\times 6^0={30}^0$  apart.
10. In one minute, the minute hand moves one minute division or 60$6^0$.
11. In one minute, the hour hand moves 120${\displaystyle \frac{1}{2^0}}$
12. In one minute the minute hand gains 5120$5{\displaystyle \frac{1}{2^0}}$ more than hour hand.
13. When the hands are together, they are 00$0^0$ apart. Hence,

Too Fast And Too Slow :
 If a watch indicates 9.20, when the correct time is 9.10, it is said to be 10 minutes too fast. And if  is said to be 10 minutes too fast. And if it indicates 9.00, when the correct time is 9.10, it is said to be 10 minutes too slow.

2 important shortcut techniques:

The angle between the hour hand and minute hand at a given time H:MM is given by θ=∣∣∣30×H−112×MM∣∣∣$\theta =\left|30\times H-{\displaystyle \frac{11}{2}\times MM}\right|$

The time after H hours, hour hand and minute hand are at θ$\theta$ degrees = MM=211×(30×H±θ)$MM={\displaystyle \frac{2}{11}\times \left(30\times H\pm \theta \right)}$

Remember any angle less than 180 degrees comes 2 times in 24 hours.

Practice Problems

1. At what time, in minutes, between 3o’ clock and 4o’clock, both the needles will  coincide each other
At 3o’clock, the minute hand is 15 min. spaces apart from the hour hand. To be coincident, it must gain 15 min. spaces.
55 min. are gained in 60 min.
15 min. are gained in (1555×60)$\left({\displaystyle \frac{15}{55}\times 60}\right)$ =16411$=16{\displaystyle \frac{4}{11}}$ min
The hands are coincident at 16411$16{\displaystyle \frac{4}{11}}$  min. past 3.

Alternate method:

We can also solve this problem using degrees.  At 3'O clock, Hour hand and minute hand are seperated by 90 degrees.  Now to meet the Hour hand minute hand has to gain 90 degrees.  We know that for every minute, minute hand gains 5120$5{\displaystyle {\frac{1}{2}}^0}$.  To gain 90 degree it takes 9005120=90112=90×211=18011⇒16411${\displaystyle \frac{{90}^0}{5{{\displaystyle \frac{1}{2}}}^0}={\displaystyle \frac{90}{{\displaystyle \frac{11}{2}}}=90\times {\displaystyle \frac{2}{11}={\displaystyle \frac{180}{11}\Rightarrow 16{\displaystyle \frac{4}{11}}}}}}$ minutes

Alternate method:

Use formula θ$\theta$ degrees = MM=211×(30×H±θ)$MM={\displaystyle \frac{2}{11}\times \left(30\times H\pm \theta \right)}$

MM=211×(30×3±0)=18011=16411$MM={\displaystyle \frac{2}{11}\times \left(30\times 3\pm 0\right)={\displaystyle \frac{180}{11}=16{\displaystyle \frac{4}{11}}}}$ min

2. At what time between 7 and 8o’clock will the hands of a clock be in the same straight line but, not together ?
When the hands of the clock are in the same straight line but not together, they are 30 minute spaces apart.  At 7o’clock, they are 25 min. spaces apart.
 Minute hand will have to gain only 5 min. spaces
55 min. spaces are gained in 60 min.
5 min. spaces are gained in (555×60)$\left({\displaystyle \frac{5}{55}\times 60}\right)$⇒5511$\Rightarrow 5{\displaystyle \frac{5}{11}}$ min. past 7

Alternate method:
At 7'O clock minute hand and hour hand are 150 degrees apart. To be in the same line minute hand has to gain another 30 degrees.  But we know that miute hand gains 5120$5{\displaystyle {\frac{1}{2}}^0}$ degrees per minute.  So 3005120=30112=30×211=6011⇒5511${\displaystyle \frac{{30}^0}{5{{\displaystyle \frac{1}{2}}}^0}={\displaystyle \frac{30}{{\displaystyle \frac{11}{2}}}=30\times {\displaystyle \frac{2}{11}={\displaystyle \frac{60}{11}\Rightarrow 5{\displaystyle \frac{5}{11}}}}}}$

Alternate method:

Use formula θ$\theta$ degrees = MM=211×(30×H±θ)$MM={\displaystyle \frac{2}{11}\times \left(30\times H\pm \theta \right)}$

MM=211×(30×7±180)=210±18011=39011or3011$MM={\displaystyle \frac{2}{11}\times \left(30\times 7\pm 180\right)={\displaystyle \frac{210\pm 180}{11}={\displaystyle \frac{390}{11}or{\displaystyle \frac{30}{11}}}}}$

We regect 39011${\displaystyle \frac{390}{11}}$ as hour hand and minute hand are at 0 degrees.

3. The minute hand of a clock overtakes the hour hand at intervals of 65 minutes of correct time.  How much does the clock gain or lose per day?
If a clock is running on time, Its hour hand and minute hand meets exactly for every 65511$65{\displaystyle \frac{5}{11}}$ min.
But in this clock both hand are meeting at intervals of 65 min. so this clock is gaining time.
or 65 minutes in the correct clock = 65511$65{\displaystyle \frac{5}{11}}$ minutes  in this clock.
Or for every 65 minutes this clock is gaining 5/11 minutes.
for every minute this clock is gaining 511×165${\displaystyle \frac{5}{11}\times \frac{1}{65}}$
In 24 hours or 1440 min it gains = 511×165×24×60${\displaystyle \frac{5}{11}\times \frac{1}{65}\times 24\times 60}$ = 1010143$10{\displaystyle \frac{10}{143}}$

Alternatively:
The minute hand of a clock overtakes the hour hand at intervals of M minutes of correct time. The clock gains or loses in a day by (72011−M)(60×24M)$\left({\displaystyle \frac{720}{11}}-M\right)\left({\displaystyle \frac{60\times 24}{M}}\right)$ minutes.

4. A watch which gains time uniformly is 5 minutes slow at 8'O clock in the morning on Sunday and is 5 minutes 48 seconds fast at 8 PM the following Sunday.  When was it correct?
This sunday morning at 8:00 AM, the watch is 5 min. Slow, and the next sunday at 8:00PM it becomes 5 min 48 sec fast.  The watch gains 5 + 54860$5{\displaystyle \frac{48}{60}}$ = 545${\displaystyle \frac{54}{5}}$ min in a time of  (7×24)+12$\left(7\times 24\right)+12$ = 180 hours.
To show the correct tine, it has to gain 5 min.
545${\displaystyle \frac{54}{5}}$ min  -------180 hours
5 min -------?
⇒5545×180$\Rightarrow {\displaystyle \frac{5}{{\displaystyle \frac{54}{5}}}\times 180}$
⇒8313hrs=72hrs+1113hrs=3days+11hrs+20min$\Rightarrow 83{\displaystyle \frac{1}{3}\mathrm{h}\mathrm{r}\mathrm{s}=72\mathrm{h}\mathrm{r}\mathrm{s}+11{\displaystyle \frac{1}{3}\mathrm{h}\mathrm{r}\mathrm{s}=3\mathrm{d}\mathrm{a}\mathrm{y}\mathrm{s}+11\mathrm{h}\mathrm{r}\mathrm{s}+20\mathrm{m}\mathrm{i}\mathrm{n}}}$

So the correct time will be shown on wednesdy at 7:20 PM

5. A clock is set right at 8 AM.  The clock gains 10 minutes in 24 hours. What will be the true time when the clock indicates 1 PM the following day?
Between 8 AM and 1 PM total 29 hours have passed.
This clock shows 24 hr 10 min or 1456${\displaystyle \frac{145}{6}}$ for 24 hours in correct clock.  In 29 hours in this clock = 291456×24${\displaystyle \frac{29}{{\displaystyle \frac{145}{6}}}\times 24}$ hours in actual clock = 1445${\displaystyle \frac{144}{5}}$ = 2845$28{\displaystyle \frac{4}{5}}$ = 28 hrs 48 min.
So actual time is 12: 48 PM

6. How much does a watch lose per day, if its hands coincide every 64 minutes?
Ans: If a clock is running on time, Its hour hand and minute hand meets exactly for every 65511$65{\displaystyle \frac{5}{11}}$ min.
But in this clock both hand are meeting at intervals of 64 min. So this clock is gaining time.
or 64 minutes in the correct clock = 65511$65{\displaystyle \frac{5}{11}}$ minutes in this clock.
Or for every 64 minutes this clock is gaining 1511$1{\displaystyle \frac{5}{11}}$ minutes. or 1611${\displaystyle \frac{16}{11}}$ minutes
For every minute this clock is gaining 1611×164${\displaystyle \frac{16}{11}}\times {\displaystyle \frac{1}{64}}$
In 24 hours or 1440 min it gains = 1611×164×1440${\displaystyle \frac{16}{11}}\times {\displaystyle \frac{1}{64}}\times 1440$ = 32811$32{\displaystyle \frac{8}{11}}$ minutes

Alternatively:
The minute hand of a clock overtakes the hour hand at intervals of M minutes of correct time. The clock gains or loses in a day by (72011−M)(60×24M)$\left({\displaystyle \frac{720}{11}}-M\right)\left({\displaystyle \frac{60\times 24}{M}}\right)$ minutes.
So ⇒(72011−M)(60×24M)=(72011−64)(60×2464)$\Rightarrow \left({\displaystyle \frac{720}{11}}-M\right)\left({\displaystyle \frac{60\times 24}{M}}\right)=\left({\displaystyle \frac{720}{11}}-64\right)\left({\displaystyle \frac{60\times 24}{64}}\right)$ = 1611(60×38)=211(60×3)${\displaystyle \frac{16}{11}}\left({\displaystyle \frac{60\times 3}{8}}\right)={\displaystyle \frac{2}{11}}\left(60\times 3\right)$ = =36011=32811$={\displaystyle \frac{360}{11}}=32{\displaystyle \frac{8}{11}}$

Level - 2

7. A person who left home between 4 p.m. and 5 p.m. returned between 5 p.m. and 6 p.m. and found that the hands of his watch had exactly changed places. When did he go out?
We know that the dial of the clock has 60 equal divisions (Minute divisions).  In one hour the minute hand makes one complete revolution, i.e., it moves through 60 divisions, and the hour hand moves trough 5 divisions,.  Suppose that when the man went out the hour-hand was x divisions ahead after 4'O clock.  Also suppose that when the man came back, the hour hand was y divisions ahead of 5'O clock.

Since the minute-hand and hour-hand exactly interchanged places during the interval that the man remained out, it is clear that when the man went out, the minute-hand was at y and hour-hand was at x, and when the man came back the minute-hand was at x and the hour-hand was at y.

We know that the speed of the hour hand and minute hand are in the ratio 1 : 12.
From the above diagram, In the time hour hand moves x divisions, hour hand moves 25 + y divisions.  (calculate from 4'O clock)
⇒112=x25+y$\Rightarrow {\displaystyle \frac{1}{12}={\displaystyle \frac{x}{25+y}}}$ ----------- (1)
Also in the time hour hand moves y divisions, minute hand moves 10 + x divisions (calculate from 5'O clock)
⇒112=y20+x$\Rightarrow {\displaystyle \frac{1}{12}={\displaystyle \frac{y}{20+x}}}$ ----------- (2)

From equation (1) we get 25 + y = 12x and from equation (2) we get 20 + x = 12y or x = 12 y - 20
Substituting x value in equation (1)
⇒$\Rightarrow$25 + y = 12 (12y - 20)
⇒$\Rightarrow$25 + y = 144y - 240
⇒$\Rightarrow$143 y = 265
⇒$\Rightarrow$ y = 1122143$1{\displaystyle \frac{122}{143}}$

So the person went out y divisions after 5.  So 25 + 1122143$1{\displaystyle \frac{122}{143}}$

So the time he went out = 4 hours 26122143$26{\displaystyle \frac{122}{143}}$

Alternate method:

From the above diagram it is clear that Hour hand and Minute hand together covered 60 minute spaces.
We know that the speeds of hour hand and minute hand are in the ratio 1 : 12.  So our of these 60 minute spaces hour hand would have covered 60×113=4813$60\times {\displaystyle \frac{1}{13}=4{\displaystyle \frac{8}{13}}}$ minute spaces.

i.e., Minute hand is 4813$4{\displaystyle \frac{8}{13}}$ minute spaces ahead of hour hand when the man went out.

At 4'O clock Minute hand is 20 minute spaces behind hour hand..  When the man went out it was 4813$4{\displaystyle \frac{8}{13}}$ spaces ahead of hour hand.  So it has gained 20 + 4813$4{\displaystyle \frac{8}{13}}$ = 20 + 6013${\displaystyle \frac{60}{13}}$ = 32013${\displaystyle \frac{320}{13}}$

But we know that minute hand gains 55 minute spaces over the hour hand in 60 minutes.
It gains 1 minute space in 6055${\displaystyle \frac{60}{55}}$ minutes.
To gain 32013${\displaystyle \frac{320}{13}}$ minute spaces it takes 32013×6055=26122143${\displaystyle \frac{320}{13}\times {\displaystyle \frac{60}{55}=26{\displaystyle \frac{122}{143}}}}$ minutes.

So the man went out at 4 hours 26122143$26{\displaystyle \frac{122}{143}}$ minutes.

8. A person who left home between 2 p.m. and 3 p.m. returned between 4 p.m. and 5 p.m. and found that the hands of his watch had exactly changed places. How much time did he out?

In this questions, Hour hand and minute hand together covered 120 minute spaces together.  (2:mm to 4:mm)
Of these minute hand would have covered 12/13 part.
So total time he went out given by ⇒120×1213=1101013$\Rightarrow 120\times {\displaystyle \frac{12}{13}=110\frac{10}{13}}$ minutes.

9. Between 5 and 6 a lady looked at her watch and mistaking hour hand for the minute hand, she thought that she was 57 minutes earlier than the correct time.  When was the correct time?

Let hour hand is x minute spaces ahead of 5. As we know hour hand speed is 12 times of hour hand, Minute hand moved 12x minute spaces.
So correct time = 5: 12x or (300 + 12 x) minutes
But she mistook this time and assumed 4 : (25 + x) or (265 + x) minutes
⇒$\Rightarrow$ (300 + 12x) - (265 + x) = 57 minutes
⇒$\Rightarrow$ x = 12 min

So correct time = 5: 24 minutes.


10. When asked about the time, Amit replied; "If you add one quarter of the time from midnight till now to half the time from now till the next midnight, you get the time".  what is the time now?
Let the time be "t" hours.  From mid night till this time "t" hours passed.  From now to next midnight there are (24-t) hours.

Now
t4+24−t2=t⇒t+48−2t4=t${\displaystyle \frac{t}{4}+\frac{24-t}{2}=t\Rightarrow \frac{t+48-2t}{4}=t}$
⇒48−t=4t$\Rightarrow 48-t=4t$
⇒5t=48⇒t=485$\Rightarrow 5t=48\Rightarrow t={\displaystyle \frac{48}{5}}$hours or 935$9{\displaystyle \frac{3}{5}}$ hours

11. The inhabitants of planet Rahu measure time in hours and minutes which are different from the hours and minutes of our earth.  Their day consists of 36 hours with each hour having 120 minutes.  The dials of their clocks show 36 hours.  What is the angle between the hour hand and the minutes hand of a Rahuian clock when it shows a time of 9:48? [Rahuians measure angles in degrees the way we do on earth.  But for them, the angle around a point is 720 degrees instead of 360 degrees. ]
In rahuian degrees the minutes hand travels a full circle in 1 hour.  i.e., 120 minutes
i.e., 720 degrees in 120 min or 6 degrees per min
and the hours hand travels 720/36= 20 degrees per hour
and 20/120 = 1/6 degrees/minute.
At 9:48 the hours:
Angle covered by hour hand = 9 x 20 + 48 x 1/6 = 188 deg
Angle covered by min hand = 48 x 6 = 288 deg
Angle between hour hand and minute hands = 288 - 188 = 100 degrees

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Calendars

        Calendars

 “Today is 15 August 1995”. And you are asked to find the day of the week on 15 August 2001.

 If you don’t know the method, it will prove a tough job for you. This type of question is sometimes asked in competitive exams. The process of finding it lies in obtaining the number of odd days. So, we should be familiar with odd days.

The number of days more than the complete number of weeks in a given period, are called odd days. For example :
 (1) In an ordinary year (of 365 days) there are 52 weeks and one odd day.
 (2) In a leap year (of 366 days) there are 52 weeks and two odd days.

What is the Leap and Ordinary year ?
Every year which is exactly divisible by 4 such as 1988, 1992, 1996 et. is called a leap year.
Also every 4th century is a leap year. The other centuries, although divisible by 4, are not leap years. Thus, for a century to be a leap year, it should be exactly divisible by 400. For example :
 (1) 400, 800, 1200, etc are leap years since they are exactly divisible by 400.
 (2) 700, 600, 500 etc are not leap years since they are not exactly divisible by 400.

How to find number of odd days :An ordinary year has 365 days. If we divide 365 by 7, we get, 52 as quotient and 1 as remainder. Thus, we may say that an ordinary year of 365 days has 52 weeks and 1 day. Since, the remainder day is left odd-out we call it odd day.

Therefore, an ordinary year has 1 odd day.
A leap year has 366 days, i.e. 52 weeks and 2 days.
Therefore, a leap year has 2 odd days.

 A century, ie, 100 years has :
 76 ordinary years and 24 leap years.
 = [(76 X 1 day] + [(24 X 2 days]
 =  124 days
if we devide 124 with 7 there are 17 full weeks and 5 odd days remains.
Therefore, 100 years contain 5 odd days.
 Now, (i) 200 years contain 10 odd days, ie, 3 odd days.
 (ii) 300 years contain 8 odd days, ie, 1 odd day.
 (iii) 400 years contain 6 + 1 = 21, ie, no odd day.
 (Nore: 400th year is a leap year therefore, one additional day is added, So number of odd days in year 301 to 400 = 6 instead of 5)

Similarly, 800, 1200 etc contain no odd day.

Practice Problems

1. January 1, 1992 was a Wednesday. What day of the week will it be on January 1, 1993
Ans: 1992 being a leap year, it has 2 odd days. So, the first day of the year 1993 will be two days beyond Wednesday. ie it will be Friday

2. On January 12, 1980, it was Saturday. The day of the week on January 12, 1979  was :
Ans: The year 1979 being an ordinary year, it has 1 odd day. So, the day on 12th January 1980 is one day beyond the day on 12th January, 1979. But, January 12, 1980 being Saturday.  January 12, 1979 was Friday

3. On July 2, 1985, it was Wednesday. The day of the week on July 2, 1984 was : 
Ans: Let us calculate the number of odd days between these two dates.  July month of 1984 has 29 days left. So odd days are 1. Now august 1984 to June 1985 we have 3 + 2 + 3 + 2 + 3 + 2 + 0 + 3 + 2 + 3 + 2.  Now July, 1985 till 2nd July contains 2 odd days. Total 29 odd days. or 1 odd day.  So July 2, 1984 is 1 day before Wednesday. It is Tuesday.
(2nd July 1985 is in fact Tuesday, and 2nd July 1984 is Monday)

4. Monday falls on 4th April, 1988. What was the day on 3rd November, 1987 ?
Ans: Counting the number of days after 3rd November, 1987 we have :
           Nov   Dec    Jan      Feb    March     April
days    27 + 31   +  31    + 29   +    3    +     4
= 153 days containing 6 odd days
i.e., (7-6) = 1 day beyond the day on 4th April, 1988. So, the day was Tuesday.

5. Today is 1st August. The day of the week is Monday. This is a leap year. The day of the week on this day after 3 years will be :
Ans: This being a leap year none of the next 3 years is a leap year. So, the day of the week will be 3 days beyond Monday ie, it will be Thursday.

6. January 16, 1997 was a Thursday. What day of the week will it be on January 4, 2000 ?
Ans: First we look for the leap years during this period.
1997, 1998, 1999 are not leap years.
1998 and 1999 together have net 2 odd days.
No, of days remaining in 1997 = 365 - 16 = 349 days = 49 weeks 6 odd days.
Total no. of odd days = 2 + 6 + 4 = 12 days = 7 days (1 week) + 5 odd days
Hence, January 4, 2000 will be 5 days beyond Thursday ie it will be on Tuesday.

7. February 20, 1999 was Saturday. What day of the week was on December 30, 1997 ?
Ans: The year during this interval was 1998 and it was not a leap year. Now, we calculate the no. of odd days in 1999 up to February 19 :
January 1999 gives 3 odd days
19 February 1999 gives 5 odd days
1998, being ordinary year, gives 1 odd day
In 1997, December 30 and 31 give  2 odd days
Total no. of odd days = 3 + 5 + 1 + 2 = 11 days = 4 odd days
Therefore, December 30, 1997 will fall 4 days before Saturday ie on Tuesday.

8. March 5, 1999 was on Friday, what day of the week will be on March 5, 2000 ?
Ans: Year 2000 is a leap year
 No. of remaining days in 1999 = 365 - [31 days in January + 28 days in February + 5 days in March] = 301 days = 43 weeks ie 0 odd day. No. of days passed in 2000 = January (31 days) gives 3 odd days.
 February (29 days, being a leap year) gives 1 odd day March (5 days) gives 5 odd days
 Total no. of odd days = 0 + 3 + 1 + 5 = 9 days ie 2 odd days.
 Therefore, March 5, 2000 will be two days beyond Friday, ie on Sunday.

9. On which week day August 15, 1947 falls?
Ans: We know that odd days upto 1600 years are zero.
For the years 1601 to 1700 there exist 5 odd days, 1701 to 1800 there exist 5 odd days, 1801 to 1900 there are another 5 odd days.  So upto 1900 there are 15 odd days or 1 odd day

Now from 1901 to 1946 there are 11 leap and 36 non leap years.  So number of odd days for these 46 years will be 11 X 2 + 35 X 1 = 57 .  After deviding this with 7 we get 1 odd day.

Now we entered into year 1947.  January contains 3 odd days, february 2, march 3, april 2, may 3, june 2, july 3, august 15 = 3 +0+3+2+3+2+3+15 = 31 = 3 odd days

So total odd days = 1 + 1+3 = 5
If odd days are 0 then it is sunday, 1 monday,........... so It is friday

Short Cut:

Remember this shortcut technique: Month Code: 033 614 625 035 Century Code: 6420
15 - 08 - 1947

Century code explanation:  We must consider every 400 years as a set, and of these if the given years fall in between first 100 years then CC = 6, 101 to 200 then CC = 4... So on.  For 1947 we should take 1601 to 2000 as a 400 year set. Of these 1947 fall in the last century
Total odd days = 1 + 2+ 0+ 11 + 5 = 19 = 5
5 odd days means friday. 

10. When do we use same calender of the year 1968?
Ans: To solve this problem we need to calculate the odd days for consecutive years upto the odd days become 0.  This is a tedius job.  

Shortcut:

If the given year is a leap year then add 28
If the given year is an year next to a leap year then add 6
For other years add 11.

Here 1968 is a leap year so add 28 to it.  So we can use the same calender for year 1996