Quantitative Aptitude
CALENDAR MCQs
Total Questions : 273
| Page 28 of 28 pages
Answer: Option A. -> Wednesday
9th May, 2001 = (2000 years + Period from 1.1.2001 to 9.5.2001)
Odd days in 1600 years = 0
Odd days in 400 years = 0
Jan. Feb. March April May
(31 + 28 + 31 + 30 + 9) = 129 days
129 days = (18 weeks + 3 day) = 3 odd day
Total number of odd days = (0 + 0 + 3 ) = 3 odd day
Given day is Wednesday
9th May, 2001 = (2000 years + Period from 1.1.2001 to 9.5.2001)
Odd days in 1600 years = 0
Odd days in 400 years = 0
Jan. Feb. March April May
(31 + 28 + 31 + 30 + 9) = 129 days
129 days = (18 weeks + 3 day) = 3 odd day
Total number of odd days = (0 + 0 + 3 ) = 3 odd day
Given day is Wednesday
Answer: Option D. -> Wednesday
Here from 4th December 1999 to 3rd January 2000,
there is a difference of 30 days
Number of odd days = $$\frac{{30}}{7}$$ = 2 odd days
Then no. of odd days = 2 ⇒ tuesday and wednesday (from monday)
Hence on 3rd january it is wednesday.
Here from 4th December 1999 to 3rd January 2000,
there is a difference of 30 days
Number of odd days = $$\frac{{30}}{7}$$ = 2 odd days
Then no. of odd days = 2 ⇒ tuesday and wednesday (from monday)
Hence on 3rd january it is wednesday.
Answer: Option A. -> 1828
Given year 1856, when divided by 4 leaves a remainder of 0.
NOTE: When remainder is 0, 28 is subtracted to the given year to get the result.
So, 1856 - 28 = 1828
Given year 1856, when divided by 4 leaves a remainder of 0.
NOTE: When remainder is 0, 28 is subtracted to the given year to get the result.
So, 1856 - 28 = 1828
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