Clocks and Calendar Quiz-1

8: 30. Angle between 6 and 8 hour spaces is 30 × 2 = 60°.

For an additional 30 min, angle = 30 = 15°. Hence, required angle = 60 + 15 = 75°

 

From 10 am on the 1st day to 4 am on the 5th day is a total of 114 hours.

When a clock loses 15 min, then 23 hours 45 min of this clock are the same as 24 hours of correct clock

i.e. 95/4 hours of this clock = 24 hours of correct clock.

95/4 hours of this clock =((24×4/95)×114) hours of correct clock

= 115.2 hours of correct clock which is equal to 115 hours 11 min.

The correct time is 5:11 am.

 

In a correct clock, the minute hand gains 55 min spaces over the hour hand in 60 min.

To be together again, the minute hand must gain 60 min over the hour hand.

We know that 60 min are gained in (60/55)×60 = 65(5/11)min

But they are together after 63 min. gain in 63 min = 65 – 63 = 2 = 65(5/11) - 63=2(5/11) = (27/11)min.

Gain in 24 hrs = = 27×60×24/11×63 = 56(8/77)min

At 3:00 the hour needle would be on 3 and the minute hand would be on 12

Which meant there is a 15-minute space between them and to collide the minute needle should advance 15 mins. Above the hour needle.

Therefore 55 minutes will be advanced in 60/55 * 15 = 16 4/11 minutes

Hence the hands will coincide at  16 4/11 past 3

 

The needles of a watch coincide with each other 11 times in every 12 hours (Since in between 11:00 and 1: 00 the needles coincide with each other only 1 time that is at 12:00) Therefore in total, the needles coincide each other in total 22 times per day.

 

The hands collide or would be in the opposite position (22+22)= 44 times daily

 

Time covered between 7:00 AM and 4:15 PM= 9 hours 15 mins. Or 9 15/60 hours or 37/4 hours

Now mentioned that the clock goes ahead by 5 secs every 3 mins

Or 3 min 5 sec of the incorrect clock= 3 minutes of the correct one.

Or 37/720 hours of this watch=  1/20 hr of the correct one.

And  37/4 hrs of this clock= 720/37 * 1/20 * 37/4= 9 hrs  of the correct clock

Hence the correct time will be 9 hours past 7: AM= 4:00 PM

 

Odd days in 100 years = 5

Therefore, we can say that the last day of the first century is Friday.

Odd days in 200 years = (2 x 5) Ξ 3 odd days

Therefore, the last day of the second century is Wednesday

Odd days in 300 years = (5 x 3) = 15  Ξ 1 odd days

Therefore, we can say that the last day of the third century is Monday.

Odd days in 400 years = 0

Therefore, we can say that the last day of the 3rd centaury is Sunday.

The above cycle is repeated.

We can say that a century’s last day cannot be Tuesday, Thursday, or Saturday.

2006 is an ordinary year. Hence, the number of odd days = 1.

Also, the day of 7th December 2007 will be one day after the day on 7th December 2006.

Day on 7th December 2007= Friday.

Hence, 8th December 2006 = Thursday.

First we need to calculate the day on April 1st 2002.

= (2001 + time from 1.1.2002 to 1.4.2002)

Odd days in 2000 years = 0

Odd days in 2001 years = 1

Number of days for Jan, Feb, March, and April = 31 + 28 + 31 + 1 = 91 days = 0 odd days

Total odd days = 1+0 = 1

Day on April 1, 2002 = Monday

Therefore, in the month of April 2002, Tuesday will be on dates 2, 9, 16, and 23.