Quantitative Aptitude
CALENDAR MCQs
Total Questions : 273
| Page 27 of 28 pages
Answer: Option A. -> 700
The century divisible by 400 is a leap year.
∴ The year 700 is not a leap year.
The century divisible by 400 is a leap year.
∴ The year 700 is not a leap year.
Answer: Option D. -> Friday
The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 8th Dec, 2007 will be 1 day beyond the day on 8th Dec, 2006.
But, 8th Dec, 2007 is Saturday.
∴ 8th Dec, 2006 is Friday.
The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 8th Dec, 2007 will be 1 day beyond the day on 8th Dec, 2006.
But, 8th Dec, 2007 is Saturday.
∴ 8th Dec, 2006 is Friday.
Answer: Option B. -> Tuesday
The year 2007 is an ordinary year. So, it has 1 odd day.
1st day of the year 2007 was Monday.
1st day of the year 2008 will be 1 day beyond Monday.
Hence, it will be Tuesday.
The year 2007 is an ordinary year. So, it has 1 odd day.
1st day of the year 2007 was Monday.
1st day of the year 2008 will be 1 day beyond Monday.
Hence, it will be Tuesday.
Answer: Option C. -> Thursday
The year 2008 is a leap year. So, it has 2 odd days.
1st day of the year 2008 is Tuesday (Given)
So, 1st day of the year 2009 is 2 days beyond Tuesday.
Hence, it will be Thursday.
The year 2008 is a leap year. So, it has 2 odd days.
1st day of the year 2008 is Tuesday (Given)
So, 1st day of the year 2009 is 2 days beyond Tuesday.
Hence, it will be Thursday.
Answer: Option C. -> Thursday
Formula : (Date + Month code + No.of years + No.of leap year + Century code)/7
$$\eqalign{
& = \frac{{26 + 1 + 50 + 12 + 0}}{7} \cr
& = \frac{{89}}{7} \cr
& = 5 \cr
& = {\text{Thursday}} \cr} $$
Formula : (Date + Month code + No.of years + No.of leap year + Century code)/7
$$\eqalign{
& = \frac{{26 + 1 + 50 + 12 + 0}}{7} \cr
& = \frac{{89}}{7} \cr
& = 5 \cr
& = {\text{Thursday}} \cr} $$
Answer: Option A. -> Sunday
59 days = 8 weeks 3 days = 3 odd days
Hence if today is Thursday, After 59 days, it will be
= (Thursday + 3 odd days)
= Sunday
59 days = 8 weeks 3 days = 3 odd days
Hence if today is Thursday, After 59 days, it will be
= (Thursday + 3 odd days)
= Sunday
Answer: Option C. -> 24.12.91
Given that 1.12.91 is the first Sunday
Hence we can assume that 3.12.91 is the first Tuesday
If we add 7 days to 3.12.91, we will get second Tuesday
If we add 14 days to 3.12.91, we will get third Tuesday
If we add 21 days to 3.12.91, we will get fourth Tuesday
⇒ Fourth Tuesday = (3.12.91 + 21 days) = 24.12.91
Given that 1.12.91 is the first Sunday
Hence we can assume that 3.12.91 is the first Tuesday
If we add 7 days to 3.12.91, we will get second Tuesday
If we add 14 days to 3.12.91, we will get third Tuesday
If we add 21 days to 3.12.91, we will get fourth Tuesday
⇒ Fourth Tuesday = (3.12.91 + 21 days) = 24.12.91
Answer: Option B. -> Tuesday
Given that January 1, 2007 was Monday.
Odd days in 2007 = 1 (we have taken the complete year 2007 because we need to find out the odd days from 01-Jan-2007 to 31-Dec-2007, that is the whole year 2007)
Hence January 1, 2008 = (Monday + 1 Odd day) = Tuesday
Given that January 1, 2007 was Monday.
Odd days in 2007 = 1 (we have taken the complete year 2007 because we need to find out the odd days from 01-Jan-2007 to 31-Dec-2007, that is the whole year 2007)
Hence January 1, 2008 = (Monday + 1 Odd day) = Tuesday
Answer: Option C. -> Tuesday
30 days = 4 x 7 + 2 = 2 odd days
The day is 2 days before Thursday i.e Tuesday
30 days = 4 x 7 + 2 = 2 odd days
The day is 2 days before Thursday i.e Tuesday
Answer: Option D. -> 6011 hrs
From March 6th to November 11th,
there are 250 days and
Thus 250 x 24 hours
So from Mar 6th 9AM to Nov 11th 8PM, there will be
250 x 24 + 11
hours = 6011 hrs.
From March 6th to November 11th,
there are 250 days and
Thus 250 x 24 hours
So from Mar 6th 9AM to Nov 11th 8PM, there will be
250 x 24 + 11
hours = 6011 hrs.
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