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What was the day of the week on 28th May, 2006?

A.

Thursday

B.

Friday

C.

Saturday

D.

Sunday

Answer with explanation

Answer: Option DExplanation

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

(31[jan] + 28 [Feb]+ 31[Mar] + 30[April] + 28[May] ) = 148 dayss = (21 weeks + 1 day) = 1 odd day.

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

Given day is Sunday

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How many times does the 29th day of the month occur in 400 consecutive years ?

A.

B.

C.

D.

Answer with explanation

Answer: Option AExplanation

Consider a set of every four years (So as to include the leap year with Feb 29). So in 3 years, 29 will occur 11 times, and once it would accur 12 times. So in 4 years, there will be 3*11 + 12 = 45 days in which you will have 29th. So in 400 years, it would be 4500 times.

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If 09/12/2001 happens to be Sunday, then 09/12/1971 would have been at

A.

Wednesday

B.

Tuesday

C.

Saturday

D.

Thursday

Answer with explanation

Answer: Option DExplanation

Every year day increase by 1 and if leap year come then by 2

=> leap year increase on day extra

2001 – 1971 = 30 years

so increase of 30 days

Leap years 1972 , 1976 , 1980 , 1984 , 1988 , 1992 , 1996 ( divisible by 4) , 2000 (divisible by 400)

8 Leap years

Total days increase = 30 + 8 = 38 days

38 = 5*7 + 3

3 days increased from 09/12/1971 to 09/12/1971

so it would be Thursday on 09/12/1971

Read more on Brainly.in – https://brainly.in/question/6133112#readmore

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How many times does the 29th day of the month occur in 400 consecutive years?

A.

5012

B.

4497

C.

4126

D.

1237

Answer with explanation

Answer: Option BExplanation

In 400 consecutive years there are 97 leap years. Hence, in 400 consecutive years February has the 29th day 97 times and the remaining eleven months have the 29th day 400 × 11 or 4400 times.

Therefore, the 29th day of the month occurs (4400 + 97) or 4497 times.

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On 2007, What was the date of last Saturday in May month?

A.

25

B.

21

C.

26

D.

22

Answer with explanation

Answer: Option CExplanation

1 – May – 2007

=(1 + 2 + 7 + 1 + 6)/7 = 17/7 = 3 = Tuesday

= May 1st –> Tuesday + 5 days = Saturday = 5th may

5th may + 7 days = Saturday = 12th may

12th may + 7 days = Saturday = 19th may

19th may + 7 days = Saturday = 26th may

= Answer = 26th may

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John was born on Feb 29^{th} of 2012 which happened to be a Wednesday. If he lives to be 101 years old, how many birthdays would he celebrate on a Wednesday?

A.

5

B.

3

C.

1

D.

4

Answer with explanation

Answer: Option DExplanation

Let us do this iteratively. Feb 29^{th} 2012 = Wednesday => Feb 28^{th} 2012 = Tuesday

Feb 28^{th} 2013 = Thursday (because 2012 is a leap year, there will be 2 odd days)

Feb 28^{th} 2014 = Friday, Feb 28^{th} 2015 = Saturday, Feb 28^{th} 2016 = Sunday, Feb 29^{th} 2016 = Monday

Or, Feb 29^{th} to Feb 29^{th} after 4 years, we have 5 odd days.

So, every subsequent birthday, would come after 5 odd days.

2016 birthday – 5 odd days

2020 birthday – 10 odd days = 3 odd days

2024 birthday – 8 odd days = 1 odd day

2028 birthday – 6 odd days

2032 birthday – 11 odd days = 4 odd days

2036 birthday – 9 odd days = 2 odd days

2040 birthday – 7 odd days = 0 odd days. So, after 28 years he would have a birthday on Wednesday

The next birthday on Wednesday would be on 2068 (further 28 years later), the one after that would be on 2096. His 84th birthday would again be a leap year.

Now, there is a twist again, as 2100 is not a leap year. So, he does not have a birthday in 2100. His next birthday in 2104 would be after 9 odd days since 2096, or 2 odd days since 2096, or on a Thursday.

From now on the same pattern continues. 2108 would be 2 + 5 odd days later = 7 odd days later. Or, 2108 Feb 29^{th} would be a Wednesday.

So, there are 4 occurrences of birthday falling on Wednesday – 2040, 2068 and 2096, 2108.

**Correct Answer:** 4.

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The calendar of the year 1897 can be used again in the year?

A.

1903

B.

1908

C.

1926

D.

1901

Answer with explanation

Answer: Option AExplanation

Given year 1908, when divided by 4 leaves a remainder of 1.

**NOTE:** When remainder is 1, 6 is added to the given year to get the result.

So, 1897 + 6 = 1903

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A.

Friday

B.

Sunday

C.

Thursday

D.

Tuesday

Answer with explanation

Answer: Option DExplanation

Today is Friday.

Lets solve this problem is reverse order

Then we have day after tomorrow = Sunday.

The before day of the day after tomorrow = Saturday

Three days after the before day of the day after tomorrow = Tuesday

Hence Tuesday is the correct answer

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A.

May 8, Thursday

B.

May 10, Thursday

C.

Cannot be determined

D.

June 8, Friday

Answer with explanation

Answer: Option AExplanation

Radha: May 8, Thursday

Geeta: May 10, Thursday

Revathi: June 8, Friday

We know that April 1^{st} = Tuesday

⇒ May 1^{st} = Tuesday + Remainder of 30/7 = Tuesday + 2days = Thursday.

Also, May 1^{st} = Thursday

⇒ May 8^{th} = Thursday

Also, June 8^{th} = Thursday + Remainder of 31/7 = Thursday + 3 days = Sunday

⇒ Revathi is wrong.

Also, May 10^{th} = May 8 + 2 days = Thursday + 2 days = Saturday

⇒ Geeta is also wrong.

Hence, only Radha is correct.

Hence, correct date is May 8, Thursday.

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Today is Thursday. The day after 59 days will be?

A.

Wednesday

B.

Sunday

C.

Monday

D.

Tuesday

Answer with explanation

Answer: Option BExplanation

59 days = 8 weeks 3 days

Every day of the week is repeated after 7 days hence after 56 days it

would be Thursday again .And after 59 day it would be Sunday

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Prove that any date in March of a year is the same day of the week corresponding date in November that year.

A.

Not same day

B.

Same day

C.

Previous day

D.

Next day

Answer with explanation

Answer: Option BExplanation

We will show that the number of odd days between last day of February and last day of October is zero. .

March April May June July Aug. Sept. Oct.

31 + 30 + 31 + 30 + 31 + 31 + 30 + 31

= 241 days = 35 weeks = 0 odd day. ,Number of odd days during this period = 0.

Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows

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The calendar for the year 2007 will be the same for the year:

A.

2018

B.

2016

C.

2014

D.

2012

Answer with explanation

Answer: Option AExplanation

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.

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How many leap years do 300 years have?

A.

72

B.

73

C.

74

D.

75

Answer with explanation

Answer: Option AExplanation

Given year is divided by 4, and the quotient gives the number of leap years.

Here, 300%4 = 75.

But, as 100,200 and 300 are not leap years => 75 – 3= 72 leap years.

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Today is 5^{th} August. The day of the week is Wednesday. This is a leap year. What will be the day of the week on this date after 3 years?

A.

Tuesday

B.

Saturday

C.

Thursday

D.

Sunday

Answer with explanation

Answer: Option BExplanation

This is a leap year.

So none of the next 3 years will be leap years.

Each ordinary year has one odd day, so there are 3 odd days in next 3 years.

So the day of the week will be 3 odd days beyond Wednesday i.e. it will be Saturday

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What will be the day on week 4th April 2001?

A.

Friday

B.

Sunday

C.

Monday

D.

Wednesday

Answer with explanation

Answer: Option DExplanation

April 4th 2001 =(2000 years+Period of 1.1.2001 to 4.4.2001)

Odd days in 2000 years = 0 days

From January to April 4

Total no.of.days: 31+28+31+4=94 days =(13 weeks +3 days)=3 odd days

So the day will be Wednesday

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What will be the day of the week 15^{th} August 2010?

A.

Monday

B.

Friday

C.

Sunday

D.

Tuesday

Answer with explanation

Answer: Option CExplanation

15th 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 = 4 odd days.

Jan. Feb. Mar. Apr. May. Jun. Jul. Aug.

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 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

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Today is Monday. After 61 days, it will be:

A.

Monday

B.

wednesday

C.

Friday

D.

Saturday

Answer with explanation

Answer: Option DExplanation

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.

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If 1^{st} Jan 2005 is Saturday then what will be 1^{st} Jan 2008?

A.

Monday

B.

Tuesday

C.

Saturday

D.

Sunday

Answer with explanation

Answer: Option BExplanation

On 31st December it was Friday.Number of Odd days between 2006 and 2008 = (1+1)=2days

Then December 31st 2007 will be Monday then 1st Jan 2008 will be Tuesday.

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On what dates of April 2009 did Thursday fall

A.

4,11,18,25

B.

1,8,15,22,29

C.

2,9,16,23,30

D.

3,10,17,24

Answer with explanation

Answer: Option CExplanation

We first find the day of the week on 1^{st} April 2009.

1^{st} April 2009 means, 2008 years 3 months and 1 day

= 2000 years + 8 years + January + February + March + 1^{st} April

= 0 odd days + 2 leap years + 6 ordinary years + 31 days + 28 days +31 days + 1 day

= 0+ 4 odd days + 6 odd days + 3 odd days + 0 odd days +3 odd days + 1 odd day

=17 odd days = 3 odd days

From the date in the above problem, when number of days is 3, the day of the week becomes Wednesday.

Thus, first Thursday falls on 2^{nd} April.

In that month, Thursday falls on 2nd, 9th, 16^{th}, 23^{rd}, and 30^{th}.

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What was the day of the week on 26-January-1950?

A.

Tuesday

B.

Wednesday

C.

Thursday

D.

Sunday

Answer with explanation

Answer: Option CExplanation

Formula:-(Date + Month code + No.of years + No.of leap year + Century code)/7

= (26 + 1 + 50 + 12 + 0)/7 = 89/7 = 12

= Thursday

Day Codes:-

Sat = 0

Sun = 1

Mon = 2

Tue = 3

Wed = 4

Thu = 5

Fri = 6

Month Codes:-

Jan = 1

Feb = 4

Mar = 4

Apr = 0

May = 2

Jun = 5

Jul = 0

Aug = 3

Sep = 6

Oct = 1

Nov = 4

Dec = 6

Century Codes:-

1500 – 1599 —> 0

1600 – 1699 —> 6

1700 – 1799 —> 4

1800 – 1899 —> 2

1900 – 1999 —> 0

2000 – 2099 —> 6

2100 – 2199 —> 4

2200 – 2299 —> 2

2300 – 2399 —> 0

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What will be the day of the week 15th August, 2010?

A.

Monday

B.

Wednesday

C.

Saturday

D.

Sunday

Answer with explanation

Answer: Option DExplanation

15th 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 = 4 odd days.

Jan. Feb. Mar. Apr. May. Jun. Jul. 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

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The last day of a century cannot be

A.

Wednesday

B.

Tuesday

C.

Friday

D.

Monday

Answer with explanation

Answer: Option BExplanation

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.

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