Practice Calendar problems - verbal reasoning Online Quiz (set-1) For All Competitive Exams

Q-1)   How many Monday's are there in a particular month of a particular year if the month ends on Wednesday?

(a)

(b)

(c)

(d)

Explanation:

There are months of 30, 31 and 28 days and last day of the month are Wednesday.

So, using 28 and 30 days, there are 4 Mondays.

Using 31 days, there are 5 Mondays

So, it cannot be specified.


Q-2)   How many days are there in x weeks x days?

(a)

(b)

(c)

(d)


Q-3)   Which of the following is not a leap year?

(a)

(b)

(c)

(d)


Q-4)   Which of the following is a leap year?

(a)

(b)

(c)

(d)

(e)

Explanation:

The century year which is completely divisible by 400, is a leap year.

Thus, the year 2800 is a leap year.


Q-5)   The calendar for the year 2007 will be the same for the year:

(a)

(b)

(c)

(d)

Explanation:

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

Sum = 14 odd days = 0 odd day.

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


Q-6)   On which day of the week does 28th August 2009 fall?

(a)

(b)

(c)

(d)

Explanation:

28th August 2009 means,

2008 complete years + First 7 months of the year 2009 + 28 days of August

Number of odd days in 2000 yrs = 0

Number of odd days from 2001 yrs to 2008 yrs

 

Year Number of odd days
2001 1
2002 1
2003 1
2004 2
2005 1
2006 1
2007 1
2008 2

 

2001 2002 2003 2004 2005 2006 2007 2008  1 1 1 2 1 1 1 2

=1+1+1+ 2+ 1+1+1+2 = 10

= 7 x1+ 3 = 3 odd days

Number of odd days in 2009,

Month Odd days
January 3
February

0

(ordinary year) 

March 3
April 2
May 3
June 2
July 3
August 0

January February March April May July  August 3 0 3 2 3 2 3 0

= 3 + 0 + 3 + 2 + 3 + 2 + 3 + 0

= 16 = 7 x 2 + 2 = 2 odd days

Total number of odd days till 28th August 2009

= 0 + 3 + 2 = 5

So, the required day is Friday.


Q-7)   The last day of a century cannot be

(a)

(b)

(c)

(d)


Q-8)   If January 1 is a Friday, then what is the first day of the month of March in a leap year?

(a)

(b)

(c)

(d)

Explanation:

Total number of days from January 1 to March 1

= 30 + 29 + 1 = 60 days

(February in leap years = 29 days)

60 ÷ 7 = 8 weeks and 4 odd days

So, the fourth day from Friday = Tuesday


Q-9)   The day before the day before yesterday is three days after Saturday. What day is it today?

(a)

(b)

(c)

(d)

Explanation:

Three days after Saturday is Tuesday and Tuesday is the day before a day before yesterday

So, yesterday is Thursday and today is Friday.


Q-10)   If Tuesday falls 3 days after today, then what day of the week was it on 4 days before yesterday?

(a)

(b)

(c)

(d)

Explanation:

Three days after today = Tuesday

Today = Tuesday - 3 = Saturday

Yesterday = Saturday - 1 = Friday

Therefore, 4 days before yesterday = Friday - 4 = Monday


Q-11)   If day before yesterday was Saturday, then what day of the week will it be after tomorrow?

(a)

(b)

(c)

(d)

Explanation: r: (d)

Day before yesterday = Saturday

Today = Saturday + 2 = Monday

Therefore, Tomorrow = Monday + 1 = Tuesday

Therefore, Day after tomorrow = Tuesday + 1 = Wednesday


Q-12)   If 26 January 2011 was Wednesday, then what day of the week was it on 26th January 2012?

(a)

(b)

(c)

(d)

Explanation:

26th January 2011 to 26th January 2012 will be considered as an ordinary year because 26th January in 2012 (a leap year) comes before 29th February.

Hence, the period of this one year will have only 1 odd day.

Since 26th January 2011 = Wednesday

∴ 26th January 2012 = Wednesday + 1 odd day

= Thursday


Q-13)   If 1st day of a year which is not a leap year is Friday, then find the last day of that year,

(a)

(b)

(c)

(d)

Explanation:

As we know that, first and last day of an ordinary year is the same.

Since, 1st day = Friday

⇒ Last day = Friday


Q-14)   If it was Saturday on December 17, 1899, then what will be the day on December 22, 1901?

(a)

(b)

(c)

(d)

Explanation:

Since, December 17, 1899 - Saturday

December 17, 1900 - Sunday

December 18, 1901 - Tuesday

∴ December 22, 1901 - Saturday


Q-15)   If Republic day was celebrated in 1996 on Friday, on which day in 2000 Independence day was celebrated?

(a)

(b)

(c)

(d)

Explanation:

Number of days in 1996 (366-26) = 340

Number of days in 1997 = 365

Number of days in 1998 = 365

Number of days in 1999 = 365 

Number of days from January 2000 to July 2000 = 31 + 29 + 31 + 30 + 31 + 30 + 31

= 213 

Number of days from 1st to 15th August, 2000 = 15

∴ Total days = 340 + 365 + 365 + 365 + 213 + 15 = 1663

∴ 1663 ÷ 7 = remainder 4

∴ 1663 days = (237 X 7 + 4) days = 237 weeks + 4 days

∴  Number of odd days = 4

∴  Day on 15th August, 2000 = Friday + 4 Odd days = Tuesday


Q-16)   If 1st January 2001 was Monday, then what day of the week was it on 31st December 2001?

(a)

(b)

(c)

(d)

Explanation:

The year 2001 was an ordinary year and in an ordinary year 1st day = Last day  

(remember) 1st January = 31st December   

As, given that, 1st January = Monday 

Hence, 31st December = Monday


Q-17)   The first day of a leap year is Wednesday, then what day of the week was it on 31st December in that year?

(a)

(b)

(c)

(d)

Explanation:

In a leap year, Last day = 1st day + 1 odd day (remember)

As given, 1st day = Wednesday  

Last day = Wednesday + 1 odd day = Thursday


Q-18)   If 1st January 2007 was Monday, then what day of the weeklies on 1st January 2008?

(a)

(b)

(c)

(d)

Explanation:

2007 is an ordinary year and in an ordinary year 1st January = 31st December

As, 1st January = Monday  

∴  31st December = Monday

∴  1st January 2008 = Monday + 1odd day = Tuesday


Q-19)   If 15th August 2011 was Tuesday, then what day of the week was it on 17th September 2011?

(a)

(b)

(c)

(d)

Explanation:

Since Total days from 15th August, 2011 to 17 September, 2011 = 33

33 ÷ 7 => 7)33(4 = remainder 5 odd days

∴ Required day = Tuesday + 5 odd days

= Sunday


Q-20)   What was the day of the week on 28th May 2006?

(a)

(b)

(c)

(d)

Explanation:

Odd days in 1600 yrs = 0

Odd days in 400 yrs = 0 5 yrs = (4 ordinary year + 1 leap year)

= (4 x 1+ 1 x 2) = 6 odd days

Month Odd days
January 3
February

0

(ordinary year) 

March 3
April 2
May 0 i.e,(28 ÷ 7)
Total 8

Total odd days = 8 + 6 = 14 = 0 odd day

∴ Required day Sunday