Clocks and Calendars
- A watch which gains uniformly, is 5 min slow at 7 o'clock in the morning on Sunday and it is 5 min 48 s fast at 7 pm on following Sunday. Which day was it correct ?
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Time from 7 am on Sunday to 7 pm on following Sunday
= 7 days 12 h = 180 h
∴ Watch gains (5 + 54/5) min or 54/5 in 180 h.
Now, 54/5 min are gained in 180 h.
∴ 5 min gained in (180 x 5/54 x 5) h
= 83 h 20 min.Correct Option: D
Time from 7 am on Sunday to 7 pm on following Sunday
= 7 days 12 h = 180 h
∴ Watch gains (5 + 54/5) min or 54/5 in 180 h.
Now, 54/5 min are gained in 180 h.
∴ 5 min gained in (180 x 5/54 x 5) h = 83 h 20 min.
= 3 days 11 h and 20 min
∴ Watch is correct after 3 days 11 h and 20 min after 7 am of Sunday.
∴ It will be correct at 20 min past 6 pm on Wednesday.
- On the 6 March, 2005, Monday falls. What was the day of the week on 7th March 2004 ?
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6th March 2005 = Monday.
Then, 6th March 2004 = Monday - 1 day = Sunday
[ ∵ 2004 is a leap year but it does not cross 29th February of 2004, so only 1 is taken as odd day. ]Correct Option: B
6th March 2005 = Monday.
Then, 6th March 2004 = Monday - 1 day = Sunday
[ ∵ 2004 is a leap year but it does not cross 29th February of 2004, so only 1 is taken as odd day. ]
6th March 2004 = Sunday
7th march 2004 = Monday.
- The year next to 1991 having the same calender as that of 1990 is ?
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We go on counting the odd day from 1991 onwards till the sum is divisible by 7 .
Correct Option: C
We go on counting the odd day from 1991 onwards till the sum is divisible by 7 . The number of such days are 14 up to the year 2001. So, the calendar for 1991 will be repeated in the year 2002.
- What was the day of the week on 17th July, 1776 ?
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Period upto 17th July, 1776 = 1775 yr + Period from 1st January 1776 to 17th July 1776
Counting of odd days
In 1600 yr = 0
In 100 yr = 5
75 yr = 18 leap years + 57 ordinary years = (18 x 2 + 57 x 1) odd days
= 93 odd days
= 13 weeks + 2 days
= 2 odd days
∴ 1775 yr have(0 + 5 + 2) odd days
= 7 odd days = 1 week + 0 odd day
= 0 odd day
Number of days between 1.1.1776 to 17.7.1776
January + February + March + April + May + June + July
31 + 29 + 31 + 30 + 31 + 30 + 17 = 199 days = 28 weeks + 3 odd daysCorrect Option: A
Period upto 17th July, 1776 = 1775 yr + Period from 1st January 1776 to 17th July 1776
Counting of odd days
In 1600 yr = 0
In 100 yr = 5
75 yr = 18 leap years + 57 ordinary years = (18 x 2 + 57 x 1) odd days
= 93 odd days
= 13 weeks + 2 days
= 2 odd days
∴ 1775 yr have(0 + 5 + 2) odd days
= 7 odd days = 1 week + 0 odd day
= 0 odd day
Number of days between 1.1.1776 to 17.7.1776
January + February + March + April + May + June + July
31 + 29 + 31 + 30 + 31 + 30 + 17 = 199 days = 28 weeks + 3 odd days
∴ Total number of odd days = (3 + 0) = 3
Hence, the required day is Wednesday.
- What was the days of the week on 17th August , 2010 ?
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Period up to 17th August, 2010 = (2009 yr + Period from 1.1.2010 to 17.8.2010)
Counting of odd days
Odd day in 1600 yr = 0
Odd days in 400 yr = 0
9 yr = (2 leap year + 7 ordinary years) = (2 x 2 + 7 x 1 ) = 1 Week + 4 days = 4 odd days
Number of days between 1.1.2010 to 17.8.2010
January + February + March + April + May + June + July + August.
= (31 + 28 + 31 + 30 + 31 + 30 + 31 + 17) days
= 229 days = 32 weeks + 5 odd daysCorrect Option: C
Period up to 17th August, 2010 = (2009 yr + Period from 1.1.2010 to 17.8.2010)
Counting of odd days
Odd day in 1600 yr = 0
Odd days in 400 yr = 0
9 yr = (2 leap year + 7 ordinary years) = (2 x 2 + 7 x 1 ) = 1 Week + 4 days = 4 odd days
Number of days between 1.1.2010 to 17.8.2010
January + February + March + April + May + June + July + August.
= (31 + 28 + 31 + 30 + 31 + 30 + 31 + 17) days
= 229 days = 32 weeks + 5 odd days
Total number of odd days
= (0 + 0 + 4 + 5) days = 9 days
= 1 week + 2 odd days
Hence, the required day is Tuesday.
