MBA Quantitative Aptitude Questions & Answers – Calendar Set-I
‘Clocks and Calendar’ is an important topic of Quantitative Section for your CAT and other MBA entrances. But, in this MCQ sheet, we will practice the questions related to Calendar only.
1. The Calendar for the year 2291 will be the same for the year.
a) 2297
b) 2304
c) 2301
d) 2298
2. If 13th January, 1821 was Tuesday, what will be the day on 14th Feb, 1894?
a) Monday
b) Tuesday
c) Wednesday
d) Friday
3. If 22nd February 2024 was Saturday, what will be the day on 19th November 2054?
a) Saturday
b) Friday
c) Thursday
d) Wednesday
4. The starting and ending days of 10th century were:-
a) Monday & Friday
b) Tuesday & Sunday
c) Thursday & Monday
d) Saturday & Wednesday
5. How many years have 29 days in February from 1901 to 2000?
a) 26
b) 25
c) 23
d) 24
6. The last day of a century cannot be
a) Monday
b) Sunday
c) Tuesday
d) Saturday
7. On 10th Feb, 2005 it was Thursday. What was the day of the week on 10th Feb, 2004?
a) Tuesday
b) Wednesday
c) Sunday
d) None of these
8. On what dates of April, 2001 did Tuesday fall?
a) 1st, 8th, 15th, 22nd, 29th
b) 2nd, 9th, 16th, 23rd, 30th
c) 3rd, 10th, 17th, 24th
d) 3th, 10th, 17th, 24th
9. What was the day of the week on 26th May, 2006?
a) Thursday
b) Friday
c) Saturday
d) Sunday
10. On what date of February 2007 did Friday fall?
a) 1st February
b) 2nd February
c) 3rd February
d) 4th February
Answers:
Ques 1 |
Ques 2 |
Ques 3 |
Ques 4 |
Ques 5 |
Ques 6 |
Ques 7 |
Ques 8 |
Ques 9 |
Ques 10 |
---|---|---|---|---|---|---|---|---|---|
c |
b |
a |
d |
d |
c |
a |
a |
b |
b |
Explanation:
In this section we will explain the rationale for choosing the answer pertaining to every question. After practicing these MCQ(s), you would be able to understand the concepts of Cloze Test.
Explanation (1):
Year |
Odd Day |
Total |
---|---|---|
2291 |
1 |
1 |
2292 |
2 |
3 |
2293 |
1 |
4 |
2294 |
1 |
5 |
2295 |
1 |
6 |
2296 |
2 |
8 |
2297 |
1 |
9 |
2298 |
1 |
10 |
2299 |
1 |
11 |
2300 |
2 |
13 |
2301 |
1 |
14 |
Explanation (2):
13th Jan 1821 – 31st Dec 1821 = 2 odd days
1822 – 1893 = 3 odd days
1st Jan 1894 – 14th Feb 1894 = 3 odd days
Total odd days = 8 = 1 odd day
= Tuesday (as reference is Tuesday)
Explanation (3):
22nd Feb 2024 – 31st Dec. 2024 = 6 odd days
2025 – 2053 = 1 odd day
1st Jan 2054 – 19th Nov. 2054 = 1 odd day
Total odd days = 8 = 1 odd day
= Saturday (as reference is Saturday).
Explanation (4):
10th century is of form 4n + 2
Century No. |
Starting day |
Ending day |
---|---|---|
4n + 1 |
Monday |
Friday |
4n + 1 |
Saturday |
Wednesday |
4n + 3 |
Thursday |
Monday |
4n |
Tuesday |
Sunday |
Explanation (5):
100th year is not a leap year. So 24 February’s has 29 days.
Explanation (6):
100 years contain 5 odd days.
Last day of 1st century is Friday.
200 years contain (5 x 2) 3 odd days.
Last day of 2nd century is Wednesday.
300 years contain (5 x 3) = 15 1 odd day.
Last day of 3rd century is Monday.
400 years contain 0 odd day.
Last day of 4th century is Sunday.
This cycle is repeated.
Last day of a century cannot be Tuesday or Thursday or Saturday.
Explanation (7):
The year 2004 is a leap year. It has 2 odd days.
The day on 10th Feb, 2004 is 2 days before the day on 10th Feb, 2005.
Hence, this day is Tuesday.
Explanation (8):
We shall find the day on 1st April, 2001.
1st April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)
Odd days in 1600 years = 0
Odd days in 400 years = 0
Jan. Feb. March April
(31 + 28 + 31 + 1) = 91 days 0 odd days.
Total number of odd days = (0 + 0 + 0) = 0
On 1st April, 2001 it was Sunday.
In April, 2001 Wednesday falls on 3th, 10th, 17th and 24th.
Explanation (9):
28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 6 odd days
Jan. Feb. March April May
(31 + 28 + 31 + 30 + 28) = 148 days
148 days = (21 weeks + 1 day) 1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day.
Given day is Friday.
Explanation (10):
For this find the day of 1.2.2007
1600 + 400 years has 0 odd days
From 2001 to 2006 there are 1 leap year + 5 ordinary years
So number of odd days = 1*2 + 5*1 = 2 + 5 = 7 = 1 week = 0 odd day
Now from 1.1.2007 to 1.2.2007 number of days = 32 = 4 weeks + 4 odd days = 4 odd days
So, total number of odd days = 4, so 1.2.2007 will be Thursday
Now Friday will be on 2.2.2007.
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