SSC CGL Calendar Questions with Detailed Solutions

Get SSC CGL Calendar Questions to prepare for your upcoming exam. Solve the most expected questions and detailed solutions that can be asked in the exam.

Important SSC CGL Calendar Questions

Q1.

What will be the sum of number of Sundays in the month of October of 2009 and January of 2010?

  • A.

    8

  • B.

    10

  • C.

    7

  • D.

    9

    ✓ Correct

Answer & Solution

Correct option is D

Information Given:
October 2009 = 31 days
January 2010 = 31 days
1 October 2009 = Thursday
1 January 2010 = Friday
Logic:
Find the Sundays in each 31-day month.
Explanation:
October 2009: Sundays = 4, 11, 18, 25 → 4 Sundays.
January 2010: Sundays = 3, 10, 17, 24, 31 → 5 Sundays.
Total = 4 + 5 = 9
Final Answer:
9
Final Correct Option:
(D) 9
Q2.

Which of the following pairs of years have identical calendars?

  • A.

    1954 and 1965

    ✓ Correct
  • B.

    1948 and 1959

  • C.

    1943 and 1949

  • D.

    1948 and 1965

Answer & Solution

Correct option is A

Information Given:
Two years have identical calendars when:
Both years are of the same type (both ordinary or both leap years).
Total odd days between them are a multiple of 7.
Year
Type
Odd Days
1954
Ordinary
1
1955
Ordinary
1
1956
Leap
2
1957
Ordinary
1
1958
Ordinary
1
1959
Ordinary
1
1960
Leap
2
1961
Ordinary
1
1962
Ordinary
1
1963
Ordinary
1
1964
Leap
2
Logic:
Total odd days = 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 = 14
14 is a multiple of 7.
Explanation:
1954 = Ordinary year
1965 = Ordinary year
Total odd days from 1954 to 1964 = 14 = 2 × 7
Same year type + odd days multiple of 7 → identical calendars.
Final Answer:
1954 and 1965
Final Correct Option:
(A) 1954 and 1965
Q3.

What was the first day of the year 1776?

  • A.

    Monday

    ✓ Correct
  • B.

    Tuesday

  • C.

    Sunday

  • D.

    Saturday

Answer & Solution

Correct option is A

Information Given:
Year = 1776
1776 is a leap year.
Required: First day of 1776.
Logic:
Find the odd days in the years before 1776, then match the remainder with weekdays.
Explanation:
Step 1:
1600 years → 0 odd days
100 years → 5 odd days
75 years = 18 leap + 57 ordinary years
→ (18 × 2) + (57 × 1) = 93 days
→ 93 ÷ 7 = 13 weeks + 2 days → 2 odd days
Therefore, 1775 years → 0 + 5 + 2 = 7 odd days → 0 odd day.
Step 2:
Since 1775 years give 0 odd days, 1 Jan 1776 corresponds to Sunday in the odd-day table.
But note: the standard Gregorian calendar gives 1 January 1776 = Monday.
Final Answer:
Monday
Final Correct Option:
(A) Monday
Q4.

What will be the sum of the number of Wednesdays in the months of July and September in the year 2005?

  • A.

    10

  • B.

    9

  • C.

    7

  • D.

    8

    ✓ Correct

Answer & Solution

Correct option is D

Information Given:
Year = 2005
July and September have 31 and 30 days respectively.
1 January 2005 was Saturday.
Logic:
1 July 2005 = Friday.
1 September 2005 = Thursday.
Explanation:
July Wednesdays = 6, 13, 20, 27 → 4
September Wednesdays = 7, 14, 21, 28 → 4
Total = 4 + 4 = 8
Final Answer:
8
Final Correct Option:
(D) 8
Q5.

At what time (approximately) (in PM) between 2 PM and 3 PM will the hour and minute hands of the clock be opposite to each other?

  • A.

    2:40

  • B.

    2:46

  • C.

    2:42

  • D.

    2:44

    ✓ Correct

Answer & Solution

Correct option is D

Information Given:
Time is between 2 PM and 3 PM.
Hands must be 180° apart.
Formula Used:
Angle = |30H − 5.5M|
Explanation:
At 2:M, angle = |60 − 5.5M| = 180
5.5M − 60 = 180 → 5.5M = 240
M = 240/5.5 = 43.64 minutes
Therefore, time ≈ 2:44 PM.
Final Answer:
2:44 PM
Final Correct Option:
(D) 2:44
Q6.

What will be the sum of the number of Saturdays in the months of October and December in the year 2008?

  • A.

    10

  • B.

    9

  • C.

    8

    ✓ Correct
  • D.

    7

Answer & Solution

Correct option is C

Information Given:
Year = 2008 (leap year)
October = 31 days
December = 31 days
1 October 2008 was Wednesday.
Logic:
October: 31 days → 5 Saturdays
December: 1 December was Monday → 4 Saturdays
Explanation:
Logic: 31-day month has 5 occurrences of the weekday on which it starts.
October starts Wednesday → Saturdays occur 4, 11, 18, 25 = 4.
December starts Monday → Saturdays occur 6, 13, 20, 27 = 4.
Total = 4 + 4 = 8.
Final Answer:
8
Final Correct Option:
(C) 8
Q7.

Which of the following pairs of years have identical calendars?

  • A.

    1936 and 1942

  • B.

    1930 and 1937

  • C.

    1935 and 1941

  • D.

    1930 and 1941

    ✓ Correct

Answer & Solution

Correct option is D

Information Given:
Identical calendars occur when:
Both years are of the same type: both ordinary or both leap.
Total odd days between the years is a multiple of 7.
Logic:
Check each option using the odd-days method.
Explanation:
A) 1936 and 1942
1936 = Leap year, 1942 = Ordinary year → Not identical
B) 1930 and 1937
1930 → 1936:
1 + 1 + 1 + 2 + 1 + 1 = 7
1937 is reached after 7 years, but 1930 is Ordinary and 1937 is Ordinary.
Wait: the interval from 1930 to 1936 is 6 years, and to reach 1937 we include 1936's leap effect. Total = 9 odd days, not 7 → Not identical
C) 1935 and 1941
1935 → 1940:
1 + 2 + 1 + 1 + 1 + 2 = 8
8 is not a multiple of 7 → Not identical
D) 1930 and 1941
1930 → 1940:
1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 = 14
14 is a multiple of 7.
1930 = Ordinary year, 1941 = Ordinary year → Identical calendars.
Final Answer:
1930 and 1941
Final Correct Option:
(D) 1930 and 1941
Q8.

A person was born on 14 February 2003. What day of the week will be on his 19th birthday?

  • A.

    Saturday

  • B.

    Sunday

  • C.

    Monday

    ✓ Correct
  • D.

    Thursday

Answer & Solution

Correct option is C

Information Given:
Date of birth = 14 February 2003
19th birthday = 14 February 2022
Logic:
Find the day on 14 February 2022 using the odd-days method.
Explanation:
1 January 2000 = Saturday.
2000 to 2021 = 22 years.
Leap years = 2000, 2004, ..., 2020 = 6 leap years.
Ordinary years = 22 − 6 = 16.
Odd days = (6 × 2) + (16 × 1) = 28 → 0 odd days.
So, 1 January 2022 = Saturday.
Days from January 1 to February 14 = 31 + 13 = 44 → 2 odd days.
Saturday + 2 = Monday.
Final Answer:
Monday
Final Correct Option:
(C) Monday
Q9.

What will be the product of the number of Mondays and Saturdays in the month of August in the year 2012?

  • A.

    25

  • B.

    20

  • C.

    16

    ✓ Correct
  • D.

    24

Answer & Solution

Correct option is C

Information Given:
Month = August 2012
August has 31 days.
Logic:
2012 is a leap year.
1 January 2012 = Sunday.
Days from January to August 1 = 213 days → 3 odd days.
Explanation:
1 August 2012 = Wednesday.
August 2012:
Mondays = 4 (6, 13, 20, 27)
Saturdays = 4 (4, 11, 18, 25)
Product = 4 × 4 = 16
Final Answer:
16
Final Correct Option:
(C) 16
Q10.

How many days are there from 2 October 2023 to 9 July 2024 (both inclusive)?

  • A.

    283

  • B.

    284

  • C.

    281

  • D.

    282

    ✓ Correct

Answer & Solution

Correct option is D

Let's check:
October 2023 = 30 days → from 2 Oct to 31 Oct = 30 days
November = 30 days
December = 31 days
January 2024 = 31 days
February 2024 (leap year) = 29 days
March = 31 days
April = 30 days
May = 31 days
June = 30 days
July = 9 days
Total:
30 + 30 + 31 + 31 + 29 + 31 + 30 + 31 + 30 + 9 = 282 days
Answer: D) 282
Q11.

In the year 1931, which of the following month had 5 Saturdays?

  • A.

    July

  • B.

    April

  • C.

    January

    ✓ Correct
  • D.

    February

Answer & Solution

Correct option is C

Odd Days Method:
1 January 1900 = Monday
From 1900 to 1930:
Total years = 31
Leap years = 7
Ordinary years = 24
Odd days = (24 × 1) + (7 × 2) = 38
38 ÷ 7 leaves 3 odd days.
Monday + 3 days = Thursday
So, 1 January 1931 = Thursday.
January has 31 days = 4 weeks + 3 days, so the first 3 days occur 5 times.
Since January starts on Saturday, it has 5 Saturdays.
Answer: C) January
Q12.

What was the last day of the year 1889?

  • A.

    Monday

  • B.

    Tuesday

    ✓ Correct
  • C.

    Wednesday

  • D.

    Thursday

Answer & Solution

Correct option is B

What was the last day of the year 1889?
Rule:
Ordinary year = 1 odd day
Leap year = 2 odd days
1 January 1900 = Monday (standard reference date)
Step 1: Find 1 January 1889
From 1889 to 1900:
Total years = 11
Leap years = 1892, 1896 → 2
Ordinary years = 9
Odd days:
​9×1+2×29\times1+2\times2​=13
13 odd days → 6 odd days
Going 6 days backward from Monday:
Monday → Sunday → Saturday → Friday → Thursday → Wednesday → Tuesday
So, 1 January 1889 = Tuesday.
Step 2: Find the last day
1889 is an ordinary year:
​365=52 weeks+1 day365=52\text{ weeks}+1\text{ day}​​
The last day is 364 days after 1 January:
​364=52×7364=52\times7​​
So the day remains Tuesday.
Correct Answer: B — Tuesday
Q13.

What will be the product of the number of Sundays and Fridays in the month of February in the year 2020?

  • A.

    24

  • B.

    16

    ✓ Correct
  • C.

    25

  • D.

    20

Answer & Solution

Correct option is B

Information Given:
February 2020 = 29 days (leap year).
1 February 2020 = Saturday.
Logic:
February 2020:
Saturday → 1, 8, 15, 22, 29
Sunday → 2, 9, 16, 23 = 4 Sundays
Friday → 7, 14, 21, 28 = 4 Fridays
Explanation:
Logic: Both Sunday and Friday occur 4 times.
4 × 4 = 16
Short Trick:
29 days = 4 complete weeks + 1 extra day. Saturday gets the 5th occurrence; other days occur 4 times.
Final Answer:
16
Final Correct Option:
(B) 16
Q14.

The calendar for the year 2007 will be the same as which of the following years?

  • A.

    2013

  • B.

    2019

  • C.

    2031

  • D.

    2018

    ✓ Correct

Answer & Solution

Correct option is D

Information Given:
Given year = 2007.
Options: 2013, 2019, 2031, 2018.

Logic:
For the same calendar, the year must have the same starting day and both years must have the same type (leap/non-leap).

Explanation:
2007 is a non-leap year.

From 2007 to 2018:
2007 → 2018 = 11 years
Leap years in between = 2008, 2012, 2016

Odd days = 11 + 3 = 14
14 ÷ 7 leaves remainder 0.
So, 2018 starts on the same day as 2007 and is also a non-leap year.

Short Trick:
Same calendar → same weekday + same leap/non-leap status.

Final Answer:
2018

Final Correct Option:
(D) 2018

Q15.

What will be the sum of number of Thursdays, Fridays and Sundays in November 2025?

  • A.

    13

    ✓ Correct
  • B.

    12

  • C.

    14

  • D.

    15

Answer & Solution

Correct option is A

Information Given:
Month = November 2025.
Find total Thursdays + Fridays + Sundays.

Logic:
1 November 2025 was Saturday.
November has 30 days = 4 weeks + 2 days.

Explanation:
First 2 weekdays (Saturday, Sunday) occur 5 times.

Therefore:
Thursdays = 4
Fridays = 4
Sundays = 5
Sum = 4 + 4 + 5 = 13

Short Trick:
30-day month → first 2 weekdays occur 5 times; remaining weekdays occur 4 times.

Final Answer:
13

Final Correct Option:
(A) 13

More Important Questions