- Home
- Aptitude Tests
- Calender

### Aptitude Topics

- Height and Distance
- Simple Interest
- Profit and Loss
- Percentage
- Calender
- Average
- Volume and Surface Area
- Numbers
- Problems on H.C.F and L.C.M
- Simplification
- Surds and Indices
- Chain Rule
- Boat and Streams
- Logarithm
- Stocks and Shares
- True Discount
- Odd Man Out and Series
- Time and Distance
- Time and Work
- Compound Interest
- Partnership
- Problems on Ages
- Clock
- Area
- Permutation and Combination
- Problems on Numbers
- Decimal Fraction
- Square Root and Cube Root
- Ratio and Proportion
- Pipes and Cistern
- Alligation or Mixture
- Races and Games
- Probability
- Banker's Discount

### Calender

Answer: Option C

Explanation:

On 31^{st} December, 2005 it was Saturday.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.

On 31^{st} December 2009, it was Thursday.

Thus, on 1^{st} Jan, 2010 it is Friday.

Answer: Option D

Explanation:

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)

Odd days in 1600 years = 0

Odd days in 400 years = 0

5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 6 odd days

Jan. Feb. March April May (31 + 28 + 31 + 30 + 28 ) = 148 days

148 days = (21 weeks + 1 day) 1 odd day.

Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day.

Given day is Sunday.

Answer: Option C

Explanation:

17^{th} June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)

Odd days in 1600 years = 0

Odd days in 300 years = (5 x 3) 1

97 years has 24 leap years + 73 ordinary years.

Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan. Feb. March April May June (31 + 28 + 31 + 30 + 31 + 17) = 168 days

168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.

Given day is Wednesday.

Answer: Option A

Explanation:

15^{th} August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)

Odd days in 1600 years = 0

Odd days in 400 years = 0

9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days 4 odd days.

Jan. Feb. March April May June July Aug. (31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days

227 days = (32 weeks + 3 days) 3 odd days.

Total number of odd days = (0 + 0 + 4 + 3) = 7 0 odd days.

Given day is Sunday.

Answer: Option B

Explanation:

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.

Answer: Option A

Explanation:

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 6^{th} March, 2005 will be 1 day beyond the day on 6^{th} March, 2004.

Given that, 6^{th} March, 2005 is Monday.

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

Answer: Option D

Explanation:

^{st}April, 2001.

1^{st} 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 1^{st} April, 2001 it was Sunday.

In April, 2001 Wednesday falls on 4^{th}, 11^{th}, 18^{th} and 25^{th}.

Answer: Option B

Explanation:

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

Answer: Option C

Explanation:

100 years contain 5 odd days.

Last day of 1^{st} century is Friday.

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

Last day of 2^{nd} century is Wednesday.

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

Last day of 3^{rd} century is Monday.

400 years contain 0 odd day.

Last day of 4^{th} century is Sunday.

This cycle is repeated.

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

Answer: Option C

Explanation:

The year 2004 is a leap year. It has 2 odd days.

The day on 8^{th} Feb, 2004 is 2 days before the day on 8^{th} Feb, 2005.

Hence, this day is Sunday.

Answer: Option D

Explanation:

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.

Year : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 Odd day : 1 2 1 1 1 2 1 1 1 2 1

Sum = 14 odd days 0 odd days.

Calendar for the year 2018 will be the same as for the year 2007.

Answer: Option A

Explanation:

The century divisible by 400 is a leap year.

The year 700 is not a leap year.

Answer: Option D

Explanation:

The year 2006 is an ordinary year. So, it has 1 odd day.

So, the day on 8^{th} Dec, 2007 will be 1 day beyond the day on 8^{th} Dec, 2006.

But, 8^{th} Dec, 2007 is Saturday.

8^{th} Dec, 2006 is Friday.

Answer: Option C

Explanation:

The year 2008 is a leap year. So, it has 2 odd days.

1^{st} day of the year 2008 is Tuesday (Given)

So, 1^{st} day of the year 2009 is 2 days beyond Tuesday.

Hence, it will be Thursday.

Answer: Option B

Explanation:

The year 2007 is an ordinary year. So, it has 1 odd day.

1^{st} day of the year 2007 was Monday.

1^{st} day of the year 2008 will be 1 day beyond Monday.

Hence, it will be Tuesday.