VMware Clocks and Calendar Questions and Answers Quiz-1

as 259th day of year N and 222th day of year N+1 is same
i.e saturday, so number of day between above 2 day
is must be divided by 7..
no. of days btwn dis 2 day may b (366-
259)+222=329----(1) or (365-259)+222=328------(2)
as 329 is divided by 7 so year N is leap year son Year N-1 is not a leap year....
now number of day btwn 119th of N-1 and 259th of
N is (365-119)+259 =505
now 505/7 gives remainder 1 , so 504 is divided by 7,
so 120th day of N-1 is saturday...so 119th day of N-1 is Friday

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As, we have 36 hour in a day. So, the dial has 18 hours(36/2=18)
to cover daily twice. Now a dial has 360*. So, in one hour
hand moves 360/18=20*.
So, for 12 hour hand moves 12x20=240*.
Now for minutes, hour hand moves 20/90
for 40
minutes it moves 20/90x40=8.88*
So for 12:40 hour hand moves 240+8.88=248.88*.

Now, minute hand moves 90 minutes to complete one hour.
for one minute it moves 360/90=4*
for 40 minutes it moves 4x40=160*

So, the angle between hour hand and minute hand is .
248.88-160=88.88

The ans is 89*

Each day of the week is repeated after 7 days.

So, after 63 days, it will be Monday.

After 61 days, it will be Saturday.

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating
from March 2004 to March 2005. So it has 1 odd day only.

The day on 6th March, 2005 will be 1 day beyond the
day on 6th March, 2004.

Given that, 6th March, 2005 is Monday.

6th March, 2004 is Sunday (1 day before to 6th March, 2005).

x weeks x days = (7x + x) days = 8x days.

100 years contain 5 odd days.

Last day of 1st century is Friday.

200 years contain (5 x 2) 3 odd days.

Last day of 2nd century is Wednesday.

300 years contain (5 x 3) = 15 1 odd day.

Last day of 3rd century is Monday.

400 years contain 0 odd day.

Last day of 4th century is Sunday.

This cycle is repeated.

Last day of a century cannot be Tuesday or Thursday
or Saturday.

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