Reasoning Clock And CalenderPage 4
31. | | The year next to 1990 will have the same calendar as that of the year 1990? | | (a)1995 | | (b)1997 | | (c)1996 | | (d)1992 |
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Answer is: CThe year 1990 has 365 days, i.e. 1 odd day, year 1991 has 365 days, 1 odd day, year 1992 has 366 days, i.e. 2 odd days. Likewise year 1993, 1994, 1995 have 1 odd day each. The sum of odd days.
So, Calculated from years 1990 – 95 = (1+1+2+1+1+1) = 7 days = 0 odd day
Hence, the year 1996 will have the same calendar as that of the year 1990.
32. | | If 27 March, 1995 was a Monday, then what day of week was 1 November, 1994? | | (a)Sunday | | (b)Monday | | (c)Tuesday | | (d)Wednesday |
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Answer is: CHere, 27th March 1995 was Monday. Now for calculating total number of odd days. First, we calculate total number of days till 1 November 1994.
Number of days in March 1995 = 27
Number of days in February 1995 = 28
Number of days in January 1995 = 31
Number of days in December 1994 = 31
Number of days in November 1994 = 29/146
Number of odd days = 146/7 = 20(6/7) days
So, 6 odd days
On November 1, 1994 = Monday – 6 = Tuesday
33. | | If the Republic day of India in 1980, falls on Saturday, X was born on March 3, 1980 and Y's birthday fell on? | | (a)Thursday | | (b)Friday | | (c)Wednesday | | (d)Saturday |
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Answer is: ARepublic day in 1980 i.e. January 26, 1980 = Saturday
Number of odd days till March 3, 1980
January February March = 37 odd days = 2 odd days
So, March 3 is Monday and Y is 4 days order to X.
So, his birthday must be four days ahead of X i.e. on Thursday.
34. | | If it was Saturday on December 17, 1899. What will be the day on December 22, 1901? | | (a)Friday | | (b)Saturday | | (c)Sunday | | (d)Monday |
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Answer is: BTotal number of odd days from December 17, 1899 to December 22, 1901.
14 + 365 + 356 = 735
Or, 735/7 = 0 odd days
It was Saturday on December 17, 1899.
So, It will be Saturday on December 22, 1901.
35. | | At what time between 5 o’clock and 6 o’clock, are the hands of a clock coincide? | | (a)22 min past 5 | | (b)30 min past 5 | | (c)22(8/11) min past 5 | | (d)27(3/11) min past 5 |
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Answer is: DWe know that minute hand is 25 min spaces apart from hour hand. In order to coincide, it has to gain 25 min spaces. Now, 55 min are gained by minute hand in 60 min. 25 min will gained in = 60/55 x 25 = 27(3/11) min
So, the hands will coincide at 27(3/11) min past 5.
36. | | At what time between 9 o’clock and 10 o’clock, will be the hands of a watch be together? | | (a)45 min past 9 | | (b)50 min past 9 | | (c)49(1/11) min past 9 | | (d)48(2/11) min past 9 |
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Answer is: CBoth the hands are 15 min space part at 9 O’ clock. To be together between 9 and 10, minute hand has to gain 45 min.
Now, minute hand gains 55 min in 60 min.
it will gains 45 min in (60/55) x 45 = 49(1/11) min.
Therefore, the hands are together at 49(1/11) min past 9.
37. | | At what time between 4 o’clock and 5 o’clock, will the hands of watch point in opposite directions? | | (a)45 min past 4 | | (b)40 min past 4 | | (c)50(4/11) min past 4 | | (d)54(6/11) min past 4 |
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Answer is: DAt 4,o,clock and hands are 20 min apart from each other and for having in the opposite direction they needs to be 30 min apart.
At point in opposite direction = (300 x 2)/11 = 54(6/11)
38. | | Find at what time between 8 o’clock and 9 o’clock will the hands of a clock be in the same straight line but not together? | | (a)10(10/11) min past 8 | | (b)50(10/11) min past 8 | | (c)10(1/11) min past 8 | | (d)10 min past 8 |
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Answer is: AThe positions of the hands of the clock and it are clear that they are 20 min apart. To be in the straight line, they have to be 30 min apart. So, the minute hand will have to move 10 min space in order to be 30 min apart from hour hand. 55 min are gained in 60 min.
10 min will be gained in = (60/55) x10 = 120/11 min
Therefore, the hands will be at right angle but not together at 10(10/11) min past 8.
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