SSC CGL Reasoning Cube & Cuboid: Formulas and Important Questions

Calendar is one of the most scoring topics in the SSC CGL Reasoning section because it follows fixed rules and requires minimal memorisation. Questions are commonly asked on leap years, odd days, day calculations, and determining the day for a given date. 

Once you understand the basic concepts and shortcuts, you can solve these questions within seconds. Here, we have covered all the important calendar concepts, formulas, tricks, and solved examples required for SSC CGL preparation. 

Calendar Concepts for SSC CGL

Understanding calendar-based reasoning questions is crucial for competitive exams like SSC CGL. It explains key principles and problem-solving techniques, from identifying leap years to calculating specific dates, providing a solid foundation for tackling various calendar puzzles effectively.

Leap Year Rules

A leap year contains 366 days, with February having 29 days. The rules for identifying a leap year differ for century and non-century years. 

Century Leap Years

A century year (ending in 00) is a leap year only if it is divisible by 400 (example 1600, 2000)

• If a century year is completely divisible by 400, it is a leap year.

• If it is not completely divisible by 400, it is not a leap year.

Examples:

• 1200 / 400 = 3 (divisible) → Leap Year

• 1600 / 400 = 4 (divisible) → Leap Year

• 2000 / 400 = 5 (divisible) → Leap Year

• 1900 / 400 = 4.75 (not divisible) → Not a Leap Year

Calculating February Occurrences in 400 Years

The occurrence of February 28th and 29th differs in a 400-year cycle. 

• If the question asks how many times February 29th occurs in 400 years, the answer is 97 times, corresponding to the 97 leap years in a 400-year cycle.

• However, February 28th occurs in every single year, regardless of whether it's a leap year or a normal year.

• Therefore, in 400 years, February 28th will occur 400 times.

Identifying Non-Century Leap Years

For a non-century year (a year not divisible by 100), the rule for identifying a leap year is simpler.

• A non-century year is a leap year if it is divisible by 4.

• Trick: To check divisibility by 4, only look at the last two digits of the year. If the number formed by the last two digits is divisible by 4, the entire year is a leap year.

• Example: For the year 1972, the last two digits '72' are divisible by 4 (72 / 4 = 18). Thus, 1972 is a leap year.

Day of First and Last Day of a Year

The relationship between the first and last day of a year depends on whether it's a normal year or a leap year.

Normal Year Principle

In a Normal Year, the first day (January 1st) and the last day (December 31st) are the same.

• Example: If January 1st, 2002, is a Monday, then December 31st, 2002, will also be a Monday.

Application: Same Date in the Next Year

If a specific date falls on a certain day in a Normal Year, then the same date in the next year will be one day ahead. This happens because a normal year has 365 days, which is 52 weeks and 1 odd day.

• Example:

• August 15th, 1997 = Friday (1997 is a Normal Year)

• August 15th, 1998 = Saturday (Friday + 1 day)

Application: Same Date Two Years Apart

If a specific date in a Normal Year is given, and you need to find the day two years later, it will generally be two days ahead, assuming no intervening leap year between the two identical dates.

• Example:

• February 18th, 1997 = Monday

• February 18th, 1999 = Wednesday (Monday + 2 days, as 1998 is a Normal Year, adding one day per year).

Determining Days in Months

A common question involves the number of months containing 28 days.

All 12 months in any given year (normal or leap) contain at least 28 days. The question is often posed to test understanding of month lengths. 

Comparative Analysis of Month Lengths: 

• 7 months have 31 days (January, March, May, July, August, October, December).

• 4 months have 30 days (April, June, September, November).

• 1 month (February) has either 28 days (in a normal year) or 29 days (in a leap year).

Therefore, all 12 months include 28 days.

Calculating Days After/Before a Given Date

This involves the concept of Odd Days, which are the remainders after dividing the total number of days by 7.

Example 1: Days After a Specific Day

• Problem: If today is Monday, what day will it be 100 days from now?

• Solution:

• Divide the number of days (100) by 7: 100 / 7 = 14 with a remainder of 2. These are the odd days.

• Add this remainder to the current day: Monday + 2 days = Wednesday.

Example 2: Days Before a Specific Day

• Problem: If yesterday was Thursday, what day was it 73 days ago from today?

• Solution:

• Step 1: Find Today's Day. If yesterday was Thursday, then today is Friday.

• Step 2: Calculate Odd Days for 73 days. Divide 73 by 7: 73 / 7 = 10 with a remainder of 3.

• Step 3: Subtract Odd Days from Today's Day. Since we are going "73 days before," we subtract the odd days: Friday - 3 days = Tuesday (Friday - 1 = Thursday, Thursday - 1 = Wednesday, Wednesday - 1 = Tuesday).

Date Calculation with Year and Month Changes

This is a frequently encountered and most important type of question as it involves changes in date, month, and year.

• Calculate the number of days between the two dates within the initial year context: This involves finding remaining days in the starting month, adding days for full intervening months, and adding days up to the target date in the final month.

• Calculate the number of years between the start and end years. 

• Identify the number of Leap Years within that year range. Ensure that February 29th of the leap year falls within the specified date range.

• Sum all three values: total days from step 1, years difference from step 2, and leap years from step 3.

• Divide the total sum by 7 to find the odd days (remainder). 

• Add the odd days to the starting day to find the final day.

Example 1: Year Change Only

• Problem: If April 10th, 2019, was Wednesday, what day was April 10th, 2023?

• Solution:

• Years difference: 2023 - 2019 = 4 years.

• Leap Years in between: 2020 is a leap year (only 1).

• Total shift: 4 (years) + 1 (leap year) = 5 days.

• Final Day: Wednesday + 5 days = Monday.

Example 2: Date, Month, and Year Change

• Problem: If March 20th, 2001, was Tuesday, what day was April 23rd, 2021?

• Solution:

• Step 1: Days between dates (March 20th to April 23rd in the same year context):

• Days remaining in March (March has 31 days): 31 - 20 = 11 days.

• Days in April up to 23rd: 23 days.

• Total days = 11 + 23 = 34 days.

• Step 2: Years difference: 2021 - 2001 = 20 years.

• Step 3: Leap Years between 2001 and 2021: These are 2004, 2008, 2012, 2016, 2020. There are 5 leap years.

• Step 4: Total sum of days/shifts: 34 (days) + 20 (years) + 5 (leap years) = 59.

• Step 5: Calculate Odd Days: 59 / 7 = 8 with a remainder of 3.

• Step 6: Final Day: Tuesday + 3 days = Wednesday, Thursday, Friday.

Finding a Birthday from Clues (Date Range)

When multiple individuals provide date ranges for an event like a birthday, the actual date is the common date that falls within all the stated ranges.

Example 1:

• Mother's Statement: After the 26th and before the 29th (Possible dates: 27, 28).

• Father's Statement: After the 26th and before the 28th (Possible dates: 27).

• Common Date: The only date common to both ranges is 27.

• Therefore, the birthday is on the 27th.

Example 2:

• Hari's Memory: After the 13th and before the 16th (Possible dates: 14, 15).

• Sister's Memory: After the 14th and before the 18th (Possible dates: 15, 16, 17).

• Common Date: The only date common to both ranges is 15.

• Therefore, the birthday is on the 15th of June.