Clocks and Calendar Questions and Answers

The year 2004 is a leap year. So, it has 2 odd days.
Feb 2004 is not included because we are calculating from April 2004 to April 2005.
Thus, it has 1 odd day only.
8th April, 2004 will be 1 day before the day on 8th April, 2005.
Given that, 8th March, 2005 is Monday.
So 8th March 2004 is Sunday (1 day before to 8th March, 2005).

Option 1 is the correct answer.

We know that the angle traced by the hour hand in one hour is 30º.

In 19 hours the total angle traced by the hour hand (2 am to 9 pm) is 30 x 19 = 570º.

Option 4 is the correct answer.

Total hours from Monday 10.00 AM to Friday 3.00 PM =101 Hours.
The clock loses 6 minutes every hour. In 101 hours it will lose 101 x 6 =606 minutes. i.e. 10 hours 06 minutes.
To find the actual time, calculate 10 hours 6 min backwards from Friday 3.00 PM which is 4 .54 AM
Hence, the actual time would be 4 .54 AM when clock shows 3.00 PM on Friday.

100 years contain 5 odd days.
Last day of 1st century is Friday.
200 years have (5 x 2) 3 odd days.
Last day of 2nd century is Wednesday.
300 years have (5 x 3) = 15 = 1 odd day.
Last day of 3rd century is Monday.
400 years contain 0 odd day.
Last day of 4th century is Sunday.
This cycle is repeated.
Tuesday or Thursday or Saturday can’t be the last day of century as per Asha’s calculation.

Option 3 is the correct answer

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.

As per the question 9th of the Month=Saturday. (Day preceding Sunday)

Day name repeats after 7 days.

So 9-7=2 which is 2nd of the given month is also Saturday.

Therefore 1st of the given month equates to Saturday Minus one day which will be Friday

 55 min. spaces are covered in 60 min
60 min. spaces are covered
in ((60/55) x 60) min = 720/11 min.
in 64 min. the loss is = (720/11 - 64). = 16/11
In 24 hrs the loss is = (16/11 x 1/64 x 24 x 60) min = 32 x 8/11 = 256/11

Option 1 is the correct answer.

The clock loses 1% time during the first week. A day comprises of 24 hours and a week has 7 days. Therefore, there are 7 * 24 = 168 hours in a week. During the first week if the clock has lost 1% time, then it will show a time which is 1% of 168 hours less than 12 Noon by the ending of the first week = 1.68 hours less. So the clock is gaining 2% during the next week. 2^nd week has 168 hrs and the clock gains 2% time = 2% of 168 hours = 3.36 hours more than the actual time. After losing 1.68 hours during the first week and then gaining 3.36 hours during the next week, the net result will be a -1.68 + 3.36 = 1.68 hour net gain in time. Then the clock will display time, which is 1.68 hours more than 12 Noon two weeks from the time it was set right. 1.68 hrs = 1 hr and 40.8 min = 1 hr + 40 min+ 48 sec. I.e. 1: 40: 48 P.M.

Option 2 is the correct answer.

57 after 12 o’clock 12:21……=1 from 1 to 9 it is 1.01, 1.11, 1.21, 1.31, 1.41, 1.51= 6 similarly 2.02, 2.12… 6*9=54+1=55 after 10 o’clock 10:01, 11:11 Ans: 55+2=57

Option 3 is the correct answer.

If a month has 31 days, and it starts with Sunday, Then Sundays, Mondays, Tuesdays are 5 for that month. Starting of this month begins with Monday, then Mondays, Tuesdays, and Wednesdays are 5 and remaining days are 4 each. So this month start with Monday.

Option 1 is the correct answer.

2/7 in a leap year there are 366 days means 52 weeks and 2days. So already we have 52 Fridays. Now the rest two days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, wed), (Wednesday, Thursdays), (Thursday, Friday), (Friday, sat), (sat, sun) so, the probability of 53 Fridays= 2/7

Option 3 is the correct answer

The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o’clock only). The hands point in the opposite directions 22 times in a day.

An hours + C hours = B hours … (I)

A, C and B can’t possess values greater than or equal to 24 B minutes + A minutes = C minutes … (ii)

From the above two equations, it is assumed that no value of A satisfies both the equations.

Hence, option 1 is the correct answer.

There are 60 minutes in an hour.

¾ of an hr has (60 * ¾) minutes = 45 minutes.

¾ of an hr has (60 * 45) seconds = 2700 seconds.

Light flashed for every 6 seconds.

In 2700 seconds 2700/6 = 450 times.

Starting of the counter from the first flash, the light will flash 451 times in ¾ of an hr.

Option 2 is the correct answer.

This time period was 1998 which was not a leap year. Now, we calculate the no. of odd days in 1999 up to February 19th
January 1999 gives 3 odd days
19 February 1999 gives 5 odd days
1998, being ordinary year, gives 1 odd day
1997 December 30^th and 31^th have 2 odd days
The totalling of number of odd days = 3 + 5 + 1 + 2 = 11 days = 4 odd days
Hence December 30, 1997 will fall 4 days prior Saturday i.e. on Tuesday.

Option 2 is the correct answer.

Time is 2:20. Position of the hands: Hr hand is at 2 approx. Min hand is at 4

Angle between 2 and 4 is 60 degrees [(360/12) * (4-2)]

In 20 minutes the angle made by the hr hand is 10 degrees, since it turns through ½ degrees in a minute. The angle between the clock hands is 60 degrees - 10 degrees = 50 degrees

Option 4 is the correct answer.

The clock loses 1% time during the first week.

A day comprises of 24 hours and a week has 7 days. Therefore, there are 7 * 24 = 168 hours in a week.

During the first week the clock loses 1% time, 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.

The clock gains 2% during the next week subsequently. The clock gains 2% time in the second week as it has 168 hours = 2% of 168 hours = 3.36 hours more than the actual time.

It had lost 1.68 hrs during the first week and then again gained 3.36 hrs during the next week, the net result will be a -1.68 + 3.36 = 1.68 hour net gain in time.

Hence the clock will display time which is 1.68 hours more than 12 Noon two weeks from the time it was set.

1.68 hrs = 1 hr and 40.8 min = 1 hour + 40 min + 48 sec.

I.e. 1: 40: 48 P.M. Option 2 is the correct answer.

The total angle made by the minute hand during an hour is 360^o. For an hour if it takes 360^o, it will take   = 240^o. Angle between the hr. hand the min. hand will then be somewhere between 240 - 90 = 150^o, as the hour hand is between 3 and 4.

The angle made by the hour hand assuming that it moves from 3 to 4 will be 30^o. So the hour hand will make 30^o during the course of a hr.

the hour hand will therefore, move  (40/60)*30 = 20^o.

Hence the total angle between the hour hand and the minute hand will be 150 - 20 = 130^o.

Option 4 is the correct answer.

After every 400 years, the same day occurs. Before 400 years if 27th February 2003 is Thursday then on 27th February 1603 has to be Thursday.

Option 2 is the correct answer.

At 2 O’ clock, the hour hand is at 2 and the minute hand is at 12, i.e. they are 10 min space apart. The minute hand must gain 10 minutes over the other hand to remain together. So 55 min are gained by it in 60 min. Hence 10 min will be gained in [(60/55) * 10] min = 10 10/11 min.
finally the hands will coincide at 10 10/11 min. past 2.

Option 1 is the correct answer.

Normal year jump +1 gain; Leap year jump +2 gain.

Ans. The year 2004 is a leap year. So, it has 2 odd days. February 2004 is not included in it because we are calculating from March 2004 to March 2005.So it has 1 odd day only. Given that, 6^th March 2005 is Monday.6^th March 2004 is Sunday (1 day before to 6^th March 2005).

Option 1 is correct answer.

Let’s assume the race started at 5 hours X minutes and ended at 8 hours Y minutes. Angles made by both the hour’s hands are 150+X/2 and 240+Y/2. Angles made by the minutes hands are 6X and 6Y.

We know that the difference between the angles made by the hands of the clock is same, so we equate that: 6X – (150 + X/2) = (240 + Y/2) – 6Y => 12X – 300 – X = 480 + Y – 12 Y => 11X + 11Y = 480+300 => X+Y = (780/11) => X+Y ≈ ≈ 71

We know that the sum of the angles made by the hands of the clock is also same, so we equate that as well: 6X+150+X/2 = 240+Y/2+6Y => 13X + 300 = 480 + 13Y => 13(X-Y) = 180 => X-Y = (180/13) => X-Y = ≈ ≈ 14

Solving we two equations, we get X = 42.5 mins = 42 mins 30 secs and Y = 28.5 mins = 28 mins 30 secs

Hence, duration of the race is = 8:28-5:42 = 2 hours 46 mins

Option 2 is the correct answer

In the amount of time taken by Aron to run one side of the square, Donnie runs all four sides and reaches where he was initially standing. Hence, Aron and Donnie can only meet at Donnie’s initial position. When Aron travels 3s, Braun travels 6s and reaches the point where Donnie is initially standing. Similarly, Cathy travels 9s and reaches Donnie’s starting point. After the first meeting, all three meet after Aron completes an entire round around the square.

So, total distance covered by Aron =3s+4s+4s = 11s.

Option 2 is the correct answer.

At any time, A hours and B minutes, the angle made by the hours hand is 30 A + B / 2. 30A+B/2. Similarly, the angle made by the minutes hand is 6B.

For a person to not be able to find the correct time, there should be another time     A₁:B₁ such that 30 A₁ + B₁ / 2 = 6B and 6B₁ = 30 A + B / 2 6B1=30A+B/2

There is no integral solution existing when 0 < B < 60

Option 4 is the correct answer.

We are given that If Arun's birthday on May 25.The day on Arun's birthday date is Monday.

His sister's birthday is on July 13.We have to find the day on his sister's birthday date. We know that 7 days in a week Now add 7 to 25 then the date is I June and day on I July is Monday. Because total days in May=31Now again add 7 to I July then date is 8th June and the day is Monday Now again add 7 days then the date is 15th June and the day is Monday Now Monday is on dates22th June, 29th June, 6th July, 13th July

Total days in June month=30

Hence, the day on his sister's birthday is Monday. Option 1 is the correct answer.

Second hand will move 360 degree in one minutes. Hence time is =3960/360=11 minutes so hr hand will travel 11*30/60=5.5 degree.

Option 2 is the correct answer.

The number of hours from 8:00 p.m. on Thursday to 8:00 a.m. on Monday = 84 hours.

In 84 hours, the clock gained 12 minutes.

The clock has to gain 5 minutes to show the correct time.

Therefore, 5/12 *84 = 35 hours.

35 hours from 8:00 p.m. on Thursday is 7:00 a.m. on Saturday.

Option 1 is the correct answer.

730/7 = 2 odd days. Hence Saturday will be the day of the week.

Option 3 is the correct answer.

A year is usually 365 days.

That is 52 weeks + 1 day

Indeed, 52 * 7 = 364.

A year has 364 days or 52 weeks exact and if this year starts on Monday then it would end on a Sunday. A normal year starts on a Monday and will end on a Monday. Therefore a year mostly ends on the same week days as it had started except when it is a leap year, then the year has 366 days.

That is 52 weeks + 2 days

If a year starts on Monday and ends on Tuesday indicates that it is a leap year. Therefore in this case February has 29 days.

The answer is: 15 days in January. There are 29 days in February and 15 days are in March.

Total: 15+29+15 = 59 days.  Option 3 is the correct answer.