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If Feb 12th,1986 falls on Wednesday then Jan 1st,1987 falls on which day?

A.

B.

Wednesday

C.

Thursday

D.

Saturday

Answer with explanation

Answer: Option CExplanation

The number of days more than the complete weeks are called odd days in a given period.

1 ordinary year ≡ 365 days ≡ (52 weeks + 1 day)

Hence the number of odd days in 1 ordinary year= 1.

1 leap year ≡ 366 days ≡ (52 weeks + 2 days)

Hence the number of odd days in 1 leap year= 2.

100 years ≡ (76 ordinary years + 24 leap years )

= (76 x 1 + 24 x 2) odd days = 124 odd days.

=> (17 weeks + 5 days)

≡ 5 odd days.

Hence the number of odd days in 100 years = 5.

Number of odd days—->Day of the week

==>0—->Sunday

==>1—->Monday

==>2—->Tuesday

==>3—->Wednesday

==>4—->Thursday

==>5—->Friday

==>6—->Saturday

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

because 100 century consists 5 odd days, means Friday is the last day.

200 century consists 3 odd days, means Wednesday is the last day.

300 century consists 1 odd day, means Monday is the last day.

400 century doesn’t consist any odd day.

the cycle continues, hence proof.

Solution: First, we count the number of odd days for the leftover days in the given period.

Here,given period is 12.2.1986 to 1.1.1987

Feb Mar Apr May June July Aug Sept Oct Nov Dec Jan

16 31 30 31 30 31 31 30 31 30 31 1 (left days)

2 + 3 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 1(odd days) = 1 odd day

So,given day Wednesday + 1 = Thursday is the required result.

Workspace

The calendar of the year 2024 can be used again in the year?

A.

B.

C.

D.

Answer with explanation

Answer: Option BExplanation

**Repetition of leap year ===>** Add +28 to the Given Year.

**Repetition of non leap year**

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2: Add +6 to the Given Year.

**Solution :**

Given Year is 2024, Which is a leap year.

So, Add +28 to the given year (i.e 2024 + 28) = 2052

Therfore, The calendar of the year 2024 can be used again in the year 2052.

Workspace

The calendar for the year 1993 will be same for the year

A.

B.

C.

D.

Answer with explanation

Answer: Option AExplanation

**Repetition of leap year ===>** Add +28 to the Given Year.

**Repetition of non leap year**

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2: Add +6 to the Given Year.

**Solution :**

Given Year is 1993, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 1993 + 11) = 2004, Which is a leap year.

Step 2 : Add +6 to the given year (i.e 1993 + 6) = 1999

Therfore, The calendar for the year 1993 will be same for the year 1999

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If Mar 18th,1994 falls on Friday then Feb 25th,1995 falls on which day?

A.

B.

C.

D.

Saturday

Answer with explanation

Answer: Option DExplanation

The number of days more than the complete weeks are called odd days in a given period.

1 ordinary year ≡ 365 days ≡ (52 weeks + 1 day)

Hence the number of odd days in 1 ordinary year= 1.

1 leap year ≡ 366 days ≡ (52 weeks + 2 days)

Hence the number of odd days in 1 leap year= 2.

100 years ≡ (76 ordinary years + 24 leap years )

= (76 x 1 + 24 x 2) odd days = 124 odd days.

=> (17 weeks + 5 days)

≡ 5 odd days.

Hence the number of odd days in 100 years = 5.

Number of odd days—->Day of the week

==>0—->Sunday

==>1—->Monday

==>2—->Tuesday

==>3—->Wednesday

==>4—->Thursday

==>5—->Friday

==>6—->Saturday

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

because 100 century consists 5 odd days, means Friday is the last day.

200 century consists 3 odd days, means Wednesday is the last day.

300 century consists 1 odd day, means Monday is the last day.

400 century doesn’t consist any odd day.

the cycle continues, hence proof.

First,we count the number of odd days for the left over days in the given period.Here,given period is 18.3.1994 to 25.2.1995

Month | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | Jan | Feb |
---|---|---|---|---|---|---|---|---|---|---|---|---|

Days | 13 | 30 | 31 | 30 | 31 | 31 | 30 | 31 | 30 | 31 | 31 | 25 |

Odd Days | 6 | 2 | 3 | 2 | 3 | 3 | 2 | 3 | 2 | 3 | 3 | 4 |

Therefore, No. of Odd Days = 6 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 3 + 4 = 36 = 1 odd day

So, given day Friday + 1 = Saturday is the required result.

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If the day after tomorrow is Monday, what day was tomorrow’s day before yesterday?

If the day, after tomorrow is Sunday, what day was tomorrow’s day before yesterday ?…

** **

A.

Saturday

B.

Friday

C.

Thursday

D.

Wednesday

Answer with explanation

Answer: Option BExplanation

As day after tomorrow is Monday, today will be Saturday and yesterday will be Friday…

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Today is Monday. After 61 days it will be

A.

Wednesday

B.

Saturday

C.

Monday

D.

Friday

Answer with explanation

Answer: Option BExplanation

We know, after every 7 days the same day is repeated. So, we must divide 61 by 7 and have to take

the remainder.

the remainder.

Here, after 8 complete weeks again we will get Monday and 5 more days. So, Monday+5=

Saturday

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If 2 January 2019 is Wednesday. Which day of the week will be on 22 January 2019?

A.

Saturday

B.

Friday

C.

Tuesday

D.

Wednesday

Answer with explanation

Answer: Option CExplanation

Since within month ( only date different month and year same), we must count the total day after 2, that is from 3 January to 22 January and then divide by 7. Take remainder as ODD /Extra day. Then add an extra day with the given day. That is, 22-2=20/7, and it will go 2 times and the remainder will be 6. Hence given day +6= Wednesday+6= Tuesday

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If 31 March 2008 is Wednesday. Which day of the week will be on 12 July 2008?

A.

Saturday

B.

Friday

C.

Monday

D.

Wednesday

Answer with explanation

Answer: Option CExplanation

Always mind its day will be count after the given date(31 March).So after 31 March, we get complete April, May, June, and July up to 12 days.

The extra day of April =2

The extra day of May=3

The extra day of June =2

The extra day of July =5 (July month has only 12 days so in 12÷7, remainder 5)

Total extra days=12 ( Which is greater than 7, so again must divide by 7 to get a total extra day)

Therefore, 12÷7 provide remainder 5. Hence 5 will be the final extra day of whole period)

Now given day+5=Wednesday+5=Monday

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If 1 January 2024 is a Friday. Which day of the week will be on 5 April 2024?

A.

Saturday

B.

Friday

C.

Monday

D.

Tuesday

Answer with explanation

Answer: Option DExplanation

Always check that the given year is Leap Year or normal Year. If Leap Year and February encluded then the extra/ odd day of that Feb. month will be 1.If Normal Year and February encluded then the extra/ odd day of that Feb. month will be 0. ( As per table no-II).

**Here, **the day will be count after the given date(1 January). So after 1 January, we get 30 days for January month and complete Feb., Mar., and April up to 5 days.

The extra day of Jan =2 (If 30÷7, the remainder will be 2)

The extra day of Feb.=1(**2024 is Leap Year **and so** Feb’s** extra day is 1) ☆ **SeeTable-II**

The extra day of Mar =3

The extra day of April =5 (April month has only 5 days so the extra day is 5, less than 7)

Total extra days=11 ( Which is greater than 7, so again must divide by 7 to get a total extra day)

Therefore, 11÷7 provide the remainder 4. Hence 4will be the final extra day of whole period)

Now given day+4=Friday+4=Tuesday.

Workspace

If the day before yesterday was Friday. What will be the third day after the day after tomorrow?

A.

**Thursday**

B.

** Friday**

C.

** d.Sunday **

D.

** Sunday**

Answer with explanation

Answer: Option BExplanation

First, write down DBY then Y, TD, T, DAT, 1, 2, 3 serially and then count from Thursday to 3. 3 is Friday. Hence the third day after the day after tomorrow** **is Friday

Workspace

On what dates of March 2005 did Friday fall ?

A.

**4,11,18&25 **

B.

**5,12,19&26**

C.

**6,13,20&27 **

D.

**7,14,21&28**

Answer with explanation

Answer: Option AExplanation

First of all, we have to find the day on 01.03.2005

01.03.2005 = (2004 years + Period from 01.01.2005 to 01.03.2005)

Number of odd days in 2000 years = 0 —-(1)

Number of odd days in 4 years = 1 leap year + 3 ordinary year

= 1X2 + 3X1 = 5 days —-(2)

From 01.01.2005 to 01.03.2005, we have,

Jan-31days + Feb-28days + Mar-1day = 60days

= 8weeks + 4days = 4 odd days —-(3)

Adding (1),(2)&(3), we get,total no.of odd days = 0+5+4

= 9 days = 1 week + 2days = 2 odd days.

2 odd days corresponding to Tuesday.So 01.03.2005 was Tuesday. Therefore, Friday lies on 04.03.2005.

Hence, the dates of March,2015 on which Fridays fell are 4,11,18&25

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What was the day of the week on 16th July,1776?

A.

** Monday **

B.

**Tuesday**

C.

**Wednesday **

D.

**Thursday**

Answer with explanation

Answer: Option BExplanation

16th July,1776 = (1775 years + Period from 01.01.1776 to 16.07.1776)

1775 = 1600 + 100 + 75

No.of odd days in 1600 years = 0 —-(1)

No.of odd days in 100 years = 5 —-(2)

No.of odd days in 75 years = 18 leap years + 57 ordinary years = 18X2+57X1=93days=13weeks+2days=2 odd days —-(3)

From 01.01.1776 to 16.07.1776, we have,

Jan-31days+Feb-29days+Mar-31days+Apr-30days+May-31days+Jun-30days+July-16days = 198days = 28weeks + 2 days = 2 odd days —-(4)

Adding (1),(2),(3) & (4), we get,

Total no.of odd days = 0+5+2+2 = 9days = 1 week + 2days = 2 odd days. 2 odd days corresponding to Tuesday

Hence, the day of the week on 16th July,1776 is Tuesday

Workspace

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

A.

**2014 **

B.

** 2016**

C.

** 2017 **

D.

**2018**

Answer with explanation

Answer: Option DExplanation

For example,if our answer is 2018, we must have the same day on 01.01.07 & 01.01.18 and number of odd days from 01.01.07 to 31.12.17 must be zero. Let us check the same thing for each of the given options.

Option (a)2015: No.of odd days from 01.01.2007 to 31.12.2014 = 2 leap years + 6 ordinary years = 2X2+6X1 = 10days = 1week + 3days = 3 odd days[not correct]

Option (b)2016: No.of odd days from 01.01.2007 to 31.12.2015 = 2 leap years + 7 ordinary years = 2X2+7X1 = 11 days = 1 week + 4 days = 4 odd days [not correct]

Option (c)2017: No.of odd days from 01.01.2007 to 31.12.2016 = 3 leap years + 7 ordinary years = 3X2+7X1 = 13 days = 2 weeks + 1 day = 1 odd day [not correct]

Option (d)2018: No.of odd days from 01.01.2007 to 31.12.2017 = 3 leap years + 8 ordinary years = 3X2+8X1 = 14 days = 2 weeks + 0 day = 0 odd day [correct]

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How many days are there in “x” weeks “x” days?

A.

**7x ^{2} **

B.

**7x**

C.

**8x ^{2} **

D.

**8x**

Answer with explanation

Answer: Option DExplanation

Number of days in “x” weeks = 7x

Number of days in “x” days = x

Number of days in “x”weeks”x”days = 7x+x = 8x

Hence, the number of days in “x”weeks”x”days = 8x days

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