CET Reasoning Chapter 12: Calendar (कैलेंडर) | CET 12th 2026

Reasoning Chapter 12: Calendar (कैलेंडर) | CET 12th 2026

We've covered this topic in our YouTube class. In this post you'll find the class recording, downloadable PDFs, and practice questions.

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Calendar (कैलेंडर)

1. Fundamental Core Theorems & Priority Rules

All calendar reasoning calculations depend on the modulus 7 remainder operation known as Odd Days (विषम दिन)[cite: 13]:

  1. Ordinary Year vs Leap Year Base: An ordinary year contains \(365 \text{ days} = 52 \text{ weeks} + 1 \text{ odd day}\)[cite: 13]. A leap year contains \(366 \text{ days} = 52 \text{ weeks} + 2 \text{ odd days}\)[cite: 13].
  2. Divisibility Identification Rule: Standard years must be divisible by \(4\)[cite: 13]. Century years (e.g., \(1900, 2000, 2100\)) must be strictly divisible by \(400\) to be a leap year (\(2000\) was a leap year, but \(1900\) and \(2100\) are ordinary years)[cite: 13].
  3. Century Odd-Day Invariant (5-3-1-0 Rule): $$\begin{aligned} 100 \text{ Years} &= 5 \text{ Odd Days (Friday / शुक्रवार)} \\ 200 \text{ Years} &= 3 \text{ Odd Days (Wednesday / बुधवार)} \\ 300 \text{ Years} &= 1 \text{ Odd Day (Monday / सोमवार)} \\ 400 \text{ Years} &= 0 \text{ Odd Days (Sunday / रविवार)} \end{aligned}$$[cite: 13] Consequently, the last day of a century can never be Tuesday, Thursday, or Saturday (T-T-S)[cite: 13].
  4. Century & Month Codes:
    • Century Codes: \(1600\text{s}=0\), \(1700\text{s}=5\), \(1800\text{s}=3\), \(1900\text{s}=1\), \(2000\text{s}=0\)[cite: 13].
    • Month Codes (144-025-036-146): Jan (\(1\)), Feb (\(4\)), Mar (\(4\)), Apr (\(0\)), May (\(2\)), Jun (\(5\)), Jul (\(0\)), Aug (\(3\)), Sep (\(6\)), Oct (\(1\)), Nov (\(4\)), Dec (\(6\))[cite: 13]. In a leap year, Jan becomes \(0\) and Feb becomes \(3\)[cite: 13].
    • Day Codes: Sunday (\(0\)), Monday (\(1\)), Tuesday (\(2\)), Wednesday (\(3\)), Thursday (\(4\)), Friday (\(5\)), Saturday (\(6\))[cite: 13].
  5. Calendar Repetition Framework:
    • Leap Year \(\rightarrow\) repeats after \(+28\) years[cite: 13].
    • Year immediately succeeding a leap year \(\rightarrow\) repeats after \(+6\) years[cite: 13].
    • Ordinary Year \(\rightarrow\) repeats after \(+11\) years (or \(+6\) years if odd-day accumulated sum mod \(7 = 0\))[cite: 13].

2. Master Date Evaluation Formula

To determine the day of the week for any Gregorian date[cite: 13]:

$$\text{Day Index} = \left(\text{Date} + \text{Month Code} + Y_{\text{last2}} + \left\lfloor\frac{Y_{\text{last2}}}{4}\right\rfloor + \text{Century Code} + 5\right) \pmod 7$$[cite: 13]

3. Worked Examples

  • India's Independence Day (15 August 1947): $$\text{Date} = 15, \quad \text{Aug Code} = 3, \quad Y = 47, \quad \left\lfloor\frac{47}{4}\right\rfloor = 11, \quad \text{Century Code (1900s)} = 1$$[cite: 13] $$\text{Sum} = 15 + 3 + 47 + 11 + 1 + 5 = 82$$[cite: 13] $$82 \pmod 7 = 5 \implies \mathbf{Friday\ (शुक्रवार)}$$[cite: 13]
  • Century Repetition Boundary Exception (1897): The general \(+6\) rule fails across the year \(1900\) because \(1900\) is an ordinary non-leap century year[cite: 13]. Calculating year-by-year odd days: \(1897(1) + 1898(1) + 1899(1) + 1900(1) + 1901(1) + 1902(1) + 1903(1) = 7\)[cite: 13]. But \(1904\) is a leap year and cannot duplicate the ordinary calendar of \(1897\)[cite: 13]. Continuing the accumulation: \(1904(2) + 1905(1) + 1906(1) + 1907(1) + 1908(2) = 14 \pmod 7 = 0\), which gives the repeating year 1909[cite: 13].

Practice Quiz (10 Questions)

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