Clocks and Calendars
- In every 30 minutes the time of a watch increases by 3 minutes. After setting the correct time at 5 am what time will the watch show after 6 hours?
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We know that ,
In every 30 minutes the time of watch increased by 3 minutes = 12 × 3 = 36 minutes
So the time after 6 hours = 5 am + 6 hours + 30 minutes = 11 : 36 am.Correct Option: C
We know that ,
In every 30 minutes the time of watch increased by 3 minutes = 12 × 3 = 36 minutes
So the time after 6 hours = 5 am + 6 hours + 30 minutes = 11 : 36 am.
Therefore , required time will be 11 : 36 am.
- If the day after tomorrow is Sunday, what day was tomorrow’s day before yesterday?
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Given that , The day after tomorrow is Sunday.
Therefore today is Friday.
Hence, the day on tomorrow’s day before yesterday is given by:
= Friday – 1 day = ThursdayCorrect Option: B
Given that , The day after tomorrow is Sunday.
Therefore today is Friday.
Hence, the day on tomorrow’s day before yesterday is given by:
= Friday – 1 day = Thursday
Therefore , required day will be Thursday .
- The first Republic Day of India was celebrated on January 26, 1950. What was the day of the week on that date?
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According to question , we can say that
Total number of odd days = 1600 years have 0 odd day + 300 years have 1 odd day + 49 years (12 leap years + 37 ordinary years) have 5 odd days + 26 days of January have 5 odd days = 0 + 1 + 5 + 5 = 4 odd days.Correct Option: D
According to question , we can say that
Total number of odd days = 1600 years have 0 odd day + 300 years have 1 odd day + 49 years (12 leap years + 37 ordinary years) have 5 odd days + 26 days of January have 5 odd days = 0 + 1 + 5 + 5 = 4 odd days.
So, the day was Thursday.
- March 5, 1999 was on Friday. What day of the week was on March 5, 2000?
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As per the given above question , we know that
Year 2000 was a leap year.
Number of days remaining in 1999 = 365 - [31 days of January + 28 days of February + 5 days March] = 301 days = 43 weeks, i.e., 0 odd day.
Number of days passed in 2000 = January 31 days have 3 odd days.
February 29 days (being leap year) have 1 odd day
March 5 days have 5 odd days.
∴ Total number of odd days = 0 + 3 + 1 + 5 = 9 days = 1 week + 2 odd daysCorrect Option: B
As per the given above question , we know that
Year 2000 was a leap year.
Number of days remaining in 1999 = 365 - [31 days of January + 28 days of February + 5 days March] = 301 days = 43 weeks, i.e., 0 odd day.
Number of days passed in 2000 = January 31 days have 3 odd days.
February 29 days (being leap year) have 1 odd day
March 5 days have 5 odd days.
∴ Total number of odd days = 0 + 3 + 1 + 5 = 9 days = 1 week + 2 odd days
Therefore, March 5, 2000 would be two days beyond Friday, i.e., on Sunday.
- A clock takes 9 seconds to strike 4 times. In order to strike 12 times at the same rate, the time taken is:
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As we can say that ,
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 × 3 seconds = 33 seconds .Correct Option: D
As we can say that ,
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 × 3 seconds = 33 seconds.
Therefore , Required time taken is 33 seconds .