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A Learning Portal from Recruitment India

On 8th Feb 2005 it was Tuesday. What was the day of the week on 8th Feb 2004?

A.

Tuesday

B.

Sunday

C.

Monday

D.

Saturday

Answer with explanation

Answer: Option BExplanation

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.

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On July 2, 1985, it was Wednesday. The day of the week on July 2, 1984 was

A.

Wednesday

B.

Tuesday

C.

Monday

D.

Thursday

Answer with explanation

Answer: Option BExplanation

Let us calculate the number of odd days between these two dates.

July month of 1984 has 29 days left. So odd days are 1.

Now August 1984 to June 1985 ===>> 3 + 2 + 3 + 2 + 3 + 2 + 0 + 3 + 2 + 3 + 2.

Now July 1985 till 2nd July contains 2 odd days.

Total 29 odd days. or 1 odd day.

So July 2, 1984, is 1 day before Wednesday. It is Tuesday.

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On 8th march,2005,wednesday falls what day of the week was it on 8th march,2004?

A.

Monday

B.

Tuesday

C.

Wednesday

D.

Friday

Answer with explanation

Answer: Option BExplanation

As a year contains 365 days with factors of 7(1week) gives 1(day) as a reminder which adjusts one day extra to the next year.

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How many leap years do 300 years have?

A.

74

B.

75

C.

76

D.

72

Answer with explanation

Answer: Option DExplanation

Given year is divided by 4, and the quotient gives the number of leap years. But, as 100,200 and 300 are not leap years => 75 – 3= 72 leap years.

OR

A 300-year period can have either 72 or 73 leap years. Each 400-year cycle of the Gregorian calendar has 97 leap years, so the average number of leap years for a 300-year period is 72.75.

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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.

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The calendar for 1990 is the same as for

A.

1992

B.

1995

C.

1996

D.

2001

Answer with explanation

Answer: Option DExplanation

For a year to have the same calendar with 1990 ,total odd days from 1990 should be 0.

Take the year 1992 from the given choices.

Total odd days in the period 1990-1991= 2 normal years

≡ 2 x 1 = 2 odd days

Take the year 1995 from the given choices**.**

Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year

≡ 4 x 1 + 1 x 2 = 6 odd days

Take the year 1996 from the given choices.

Number of odd days in the period 1990-1995= 5 normal years + 1 leap year

≡ 5 x 1 + 1 x 2 = 7 odd days ≡ 0 odd days

As we can reduce multiples of 7 from odd days which will not change anything

Though number of odd days in the period 1990-1995 is 0, there is a catch here.

1990 is not a leap year whereas 1996 is a leap year.

Hence calendar for 1990 and 1996 will never be the same.

Take the year 2001 from the given choices.

Number of odd days in the period 1990-2000= 8 normal years + 3 leap years

≡ 8 x 1 + 3 x 2 = 14 odd days ≡ 0 odd days

Also, both 1990 and 2001 are normal years.

Hence 1990 will have the same calendar as that of 2001

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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?

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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