Aptitude - Calendar - Discussion
Discussion Forum : Calendar - General Questions (Q.No. 1)
1.
It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
Answer: Option
Explanation:
On 31st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
On 31st December 2009, it was Thursday.
Thus, on 1st Jan, 2010 it is Friday.
Discussion:
258 comments Page 22 of 26.
Subhashree said:
1 decade ago
@Sridhar, how we know the centuries table value or months table value?
Abhinav said:
1 decade ago
It is Thursday on 25th of September. What day will it be on the 25th of October in the same year?
Satishmohan said:
1 decade ago
Sridhar my quetion for you how you wrote century table we have to remember that or else is there any calculation but your expalanation so nice I have that doubt only, so please clarify that and before 1700 how you know value ?
Vijay said:
1 decade ago
365 day = 1 year, day364/7days = 54weeks, balance 1 day that will be conce considered as ODD days.
If it is leap year total no of days for = 366, 364/7 = 54 weeks balnce 2 days.
If it is leap year total no of days for = 366, 364/7 = 54 weeks balnce 2 days.
Girija said:
1 decade ago
Sridhar my quetion for you how you wrote century table we have to remember that or else is there any calculation but your expalanation so nice I have that doubt only, so please clarify that and before 1700 how you know value ?
Sekhar said:
1 decade ago
Thanks pavithra
Sridhar said:
1 decade ago
Centuries Table
1700-1799 4
1800-1899 2
1900-1999 0
2000-2099 6
2100-2199 4
2200-2299 2
2300-2399 0
2400-2499 6
2500-2599 4
2600-2699 2
Months Table:
January 0 (in leap year 6)
February 3 (in leap year 2)
March 3
April 6
May 1
June 4
July 6
August 2
September 5
October 0
November 3
December 5
Days Table:
Sunday 0
Monday 1
Tuesday 2
Wednesday 3
Thursday 4
Friday 5
Saturday 6
For example
April 24,1982.
1. Look up the 1900s in the centuries table: 0
2. Note the last two digits of the year: 82
3. Divide the 82 by 4: 82/4 = 20.5 and drop the fractionalpart: 20
4. Look up April in the months table: 6
5. Add all numbers from steps 1-4 to the day of the month (in this case, 24): 0+82+20+6+24=132.
6. Divide the sum from step 5 by 7 and find the remainder: 132/7=18 remainder 6
7. Find the remainder in the days table: 6=Saturday.
1700-1799 4
1800-1899 2
1900-1999 0
2000-2099 6
2100-2199 4
2200-2299 2
2300-2399 0
2400-2499 6
2500-2599 4
2600-2699 2
Months Table:
January 0 (in leap year 6)
February 3 (in leap year 2)
March 3
April 6
May 1
June 4
July 6
August 2
September 5
October 0
November 3
December 5
Days Table:
Sunday 0
Monday 1
Tuesday 2
Wednesday 3
Thursday 4
Friday 5
Saturday 6
For example
April 24,1982.
1. Look up the 1900s in the centuries table: 0
2. Note the last two digits of the year: 82
3. Divide the 82 by 4: 82/4 = 20.5 and drop the fractionalpart: 20
4. Look up April in the months table: 6
5. Add all numbers from steps 1-4 to the day of the month (in this case, 24): 0+82+20+6+24=132.
6. Divide the sum from step 5 by 7 and find the remainder: 132/7=18 remainder 6
7. Find the remainder in the days table: 6=Saturday.
Priya said:
1 decade ago
I would like to know , how many leap years are there in 99 years and 100 years.
i.e How to calculate number of leap years from 'n' number of years ?
I too want to know the same.
i.e How to calculate number of leap years from 'n' number of years ?
I too want to know the same.
Pranay goud said:
1 decade ago
Thanks pavithra and sundar.
Gowthami.s said:
1 decade ago
We want what day is jan 1,2010 for that
2000+9+jan1;
0+(9/4)+1
=(2+7)+1(2=leap year,7=non leap year)
=(2*2+7*1)+1
=4+7+1=12/7=5 odd days
= Friday
2000+9+jan1;
0+(9/4)+1
=(2+7)+1(2=leap year,7=non leap year)
=(2*2+7*1)+1
=4+7+1=12/7=5 odd days
= Friday
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