What was the day on 31st Dec,2011 ?
Answer: B
31st Dec, 2011 = (2010 years + Period from 1.1.2011 to 31.12.2011)
Odd days in 2000 years = 0
10 years has 2 leap years + 8 ordinary years.
Number of odd days in 10 years ( 2 x 2 + 8) = 5 odd days.
31st,Dec => complete year of 2011 (non-leap year) = 1 odd day.
Total number of odd days = (0 + 5 + 1) = 6.
Given day is saturday.
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If 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004?
Answer: A
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).
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What is two weeks from today?
Answer: A
We know that the day repeats every 7 days, 14 days, 21 days,...
So if today is Monday, after 7 days it is again Monday, after 14 days again it is Monday.
Hence, after 2 weeks i.e, 14 days the day repeats and is the same day.
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Today is Monday. After 61 days, it will be :
Answer: D
Only the remainder on division by 7 matters - the odd days.
\[ 61 = 8 \times 7 + 5 \quad\Longrightarrow\quad 5 \text{ odd days} \]
Counting five days on from Monday:
\[ \text{Tue, Wed, Thu, Fri, } \mathbf{Sat} \]
The eight complete weeks return to Monday and contribute nothing.
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On what dates of April, 2001 did Wednesday fall?
Answer: D
We shall find the day on 1st April, 2001.
1st April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)
Odd days in 1600 years = 0
Odd days in 400 years = 0
Jan. Feb. March April
(31 + 28 + 31 + 1) = 91 days 
0 odd days.
Total number of odd days = (0 + 0 + 0) = 0
On 1st April, 2001 it was Sunday.
In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th.
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How much does a clock lose per day, if its hands coincide ever 65 minutes ?
Answer: A
In a correct clock the hands coincide every
\[ \frac{720}{11} = 65\tfrac{5}{11} \text{ minutes} \]
Here they take 65 minutes, slightly less, so the clock is running fast - and the conventional measure of the discrepancy is
\[ \frac{720}{11} - 65 = \frac{5}{11} \text{ minutes per interval} \]
There are \( \tfrac{1440}{65} \) such intervals in a day:
\[ \frac{5}{11} \times \frac{1440}{65} = \frac{7200}{715} = \frac{1440}{143} \text{ minutes} \]
That is about 10.07 minutes - a much smaller discrepancy than the 64-minute case, because 65 is far closer to \( 65\tfrac{5}{11} \).
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The calendar of the year 1897 can be used again in the year?
Answer: C
Given Year is 1897, Which is a non leap year.
Step 1 : Add +11 to the given year (i.e 1897 + 11) = 1908, Which is a leap year.
Step 2 : Add +6 to the given year (i.e 1897 + 6) = 1903
Therfore, The calendar of the year 1897 can be used again in the year 1903
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