MCQ: Clock & Calendar - 3(Hard) - SSC CGL MCQ

The last day of century cannot be either Tuesday, Thursday or Saturday.

26 January, 1996 to 15 August, 2000 days were = 1663
1663 having odd day = 4
∴ In the year 2000, Independence day will on Tuesday.

To find the day on which Mohan's birthday will fall, follow these steps:

1. Determine Mohan’s birthday:

   • Since Mohan is younger by 11 days, his birthday is on July 31 (August 11 - 11 days).

2. Determine the day of the week for July 31:

   • This year’s Independence Day (August 15) falls on a Monday. So, we need to determine the day of the week 15 days before August 15.

3. Calculate the day of the week for July 31:

   • If August 15 is a Monday, then August 1 is 14 days before August 15, which is also a Monday.
   • Therefore, July 31, which is one day before August 1, will be one day before Monday, which is a Sunday.

So, Mohan's birthday (July 31) will fall on a Sunday this year.

The correct answer is:

3. Sunday

Total number of odd days
30 September, 1997-98 = 1
30 September, 1998-99 = 1
30 September, 1999-2000 = 2
30 September, 2000-02 = 1
30 September, 2001-02 = 1
30 September, 2002-03 = 1/7
∴ Tuesday + 7 = Tuesday
So, the next Tuesday will come on the Mrs Susheela's wedding anniversary in 30^th September, 2003

18^th February, 1997 is Tuesday. So, 17^th February, 1998 will also be Tuesday.
Again 16^th February, 1999 will also be Tuesday.
Hence, 18^th February, 1999 will be Thursday.

Time from 5 am of a particular day to 10 pm on the 4th day is 89 h. Now, the clock loses 16 min in 24 h or on other words, we can say that 23 h 44 min of this clock is equal to 24 h of the correct clock. or (23 + 44/60)
⇒ 356 h of this clock = 24 h of the correct clock
∴ 89 h of this clock
= (24 x 15/356 x 89) h of correct clock
= 90 h of the correct clock

The year 2004 is a leap year. It has 2 odd days. The day on 8th February, 2004 is 2 days before the day on 8th February, 2005.
Hence, this day is Sunday

26^th January (Republic day) 2008
→ Friday
Clary, it will be on Wednesday in 2012.

2^nd July, 1984 means (1983 years 6 months and 2 days) 1900 years have 1 odd day 83 years have 20 leap years and 63 ordinary years
= (40 + 63) odd days
= 103
= 5 odd days
6 months and 2 days
Jan 31
Feb 29
Mar 31
Apr 30
May 31
June 30
July 02
= 184 days = 2 odd days
Total number of days = (1 + 5 + 2)
= 8 odd days
= 1 odd day
Hence, it was Monday on 2nd July,1984.

The year 1990 has 356 days, i.e., 1 odd day, year 1991 has 356 days, v 1 odd day, year 1992 has 366 days, i.e., 2 odd days. Likewise year 1993, 1994, 1995 have 1 odd days, so calculated from years 1990-95 = (1 + 1 + 2 + 1 + 1 + 1)
= 7 odd days
= 0 odd day
Hence, the year 1996 will have the same calendar as that of the year 1990.

Total number of odd days from December 17, 1899 to December 22, 1901
14 + 356 + 356 = 735
or, 735/7 = 105
= 0 odd days
it was Saturday on December 17, 1899.
So, it will be Saturday on December 22, 1901.

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 direction at 6 'o clock only). So, in a day, the hands point in the opposite direction 22 times.

In a period of 100 years there are 23 or 24 leap years (as for century year it might be or might not be a leap year, as 1900 was not a leap year)
Number of days in a period of 100 years is 365 x 100 + 23 or 365 x 100 + 24.
If century year is not a leap year then number of days = 365 x 100 + 23, and number of weeks is 5217 and 4 odd and for leap year it will be 5217 weeks and 5 odd days, hence number of Sundays is either 5217 or 5218.

Number of odd days between 2013 and 2015 is 2, so we have to find the year which will have 2 odd days between 1977 and required year.
Consider options one by one -
(a) Number of odd days between 1977 and 1981 is 1 + 1 + 1 + 2 = 5 odd days
(b) Number of odd days between 1977 and 1985 is 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 = 10 odd days or 3 odd days
(c) Number of odd days between 1977 and 1990 is 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 = 16 odd days or 2 odd days, hence calendar of 1990 is the answer.

Let us take two cases
Case (i) : When year start with Sunday then next 4 years will always have 52 Sundays hence total number of Sundays are 53 + 3 x 52 = 209 Sundays
Case (ii) : When year start with Sunday and then we have 53 Sundays that means year is a leap year then next 4 years will always have 52 Sundays hence total number of Sundays are 53 + 3 x 52 = 209 Sundays.