50+ Calendar Problems with Solutions Pdf - 1

Ans: C

There are 3 intervals when the clock strikes 4

Time taken in 3 intervals = 9 seconds

∴ Time taken for 1 interval = 3 seconds

In order to strike 12, there are 11 intervals,
for which the time taken is 11 x 3 seconds = 33 seconds.

Ans: B

400 consecutive years contain 97 leap years.

∴ In 400 consecutive years February has 29 days 97 times and the remaining 11 months have 29th day 400 x 11 = 4400 times

∴ 29th day of the month occurs 4400 + 97 = 4497 times.

Ans: C

Number of days after 3^rd November, 1994 will be

Nov. Dec. Jan. Feb. March

27 + 31 + 31 + 28 + 20 = 137 days ⇒ 19 weeks + 4 days

∴ Number of odd days = 4.

∴ The day on 3rd November, 1994 is (7 – 4) days beyond the day on 20th March, 1995. So, the required day is Thursday.

Ans: B

1600 years contain 0 odd day.

300 years contain 1 odd day.

94 years = (23 leap years + 71 ordinary years)

= (46 + 71) odd days

= 117 odd days, i.e., 5odd days.

Days from 1st January 1995 to 28th February 1995

= (31 + 28) days = 59 days

= ( 8weeks + 3 days) = 3 odd days

Total number of odd days = (0 + 1 + 5 + 3) = 9 odd days

i.e., 2 odd days.

So, the required day is Tuesday.

Ans: D

1992 was a leap year.

Hence it had 2 odd days.

So, the first day of the year 1993 must be two days after Wednesday.

So, it was Friday.