{ "version": "https://jsonfeed.org/version/1", "title": "PrepVerse Blog", "home_page_url": "https://prepverse.vercel.app/blog", "description": "PrepVerse Blog", "items": [ { "id": "https://prepverse.vercel.app/blog/jee-advanced-2025-piecewise-function-analysis", "content_html": "
Question 3 (Paper - 1)

Let R\\mathbb{R}R denote the set of all real numbers. Define the function f:R→Rf: \\mathbb{R} \\to \\mathbb{R}f:R→R by

f(x)={2−2x2−x2sin⁡1xif x≠02if x=0f(x) = \\begin{cases} 2 - 2x^2 - x^2 \\sin\\frac{1}{x} & \\text{if } x \\ne 0 \\\\ 2 & \\text{if } x = 0 \\end{cases}f(x)={2−2x2−x2sinx1​2​if x=0if x=0​

Then which one of the following statements is TRUE?

(A)The function f is NOT differentiable at x=0(B)There is a positive real number δ, such that f is a decreasing function on the interval (0,δ)(C)For any positive real number δ, the function f is NOT an increasing function on the interval (−δ,0)(D)x=0 is a point of local minima of f\\begin{array}{|c|l|} \\hline \\text{(A)} & \\text{The function } f \\text{ is NOT differentiable at } x = 0 \\\\ \\hline \\text{(B)} & \\text{There is a positive real number } \\delta, \\text{ such that } \\\\ & f \\text{ is a decreasing function on the interval } (0, \\delta) \\\\ \\hline \\text{(C)} & \\text{For any positive real number } \\delta, \\text{ the function } \\\\ & f \\text{ is NOT an increasing function on the interval } (-\\delta, 0) \\\\ \\hline \\text{(D)} & x = 0 \\text{ is a point of local minima of } f \\\\ \\hline \\end{array}(A)(B)(C)(D)​The function f is NOT differentiable at x=0There is a positive real number δ, such that f is a decreasing function on the interval (0,δ)For any positive real number δ, the function f is NOT an increasing function on the interval (−δ,0)x=0 is a point of local minima of f​​

Step 1: Check Differentiability at x=0x = 0x=0 (Option A)​

Compute the derivative at x=0x = 0x=0 using the definition:

f′(0)=lim⁡h→0f(0+h)−f(0)h=lim⁡h→0(2−2h2−h2sin⁡1h)−2h=lim⁡h→0(−2h−hsin⁡1h).f'(0) = \\lim_{h \\to 0} \\frac{f(0+h) - f(0)}{h} = \\lim_{h \\to 0} \\frac{(2 - 2h^2 - h^2 \\sin\\frac{1}{h}) - 2}{h} = \\lim_{h \\to 0} (-2h - h \\sin\\frac{1}{h}).f′(0)=h→0lim​hf(0+h)−f(0)​=h→0lim​h(2−2h2−h2sinh1​)−2​=h→0lim​(−2h−hsinh1​).

Since ∣sin⁡1h∣≤1\\left|\\sin\\frac{1}{h}\\right| \\leq 1∣∣​sinh1​∣∣​≤1, the term hsin⁡1h→0h \\sin\\frac{1}{h} \\to 0hsinh1​→0 as h→0h \\to 0h→0. Thus:

f′(0)=−2⋅0−0=0.f'(0) = -2 \\cdot 0 - 0 = 0.f′(0)=−2⋅0−0=0.

The limit exists, so fff is differentiable at x=0x = 0x=0. Option (A) is false.

Step 2: Compute the Derivative for x≠0x \\neq 0x=0​

For x≠0x \\neq 0x=0, f(x)=2−2x2−x2sin⁡1xf(x) = 2 - 2x^2 - x^2 \\sin\\frac{1}{x}f(x)=2−2x2−x2sinx1​. Differentiate:

Step 3: Analyze f′(x)f'(x)f′(x) on (0,δ)(0, \\delta)(0,δ) (Option B)​

Rewrite f′(x)=−2x(2+sin⁡1x)+cos⁡1xf'(x) = -2x \\left(2 + \\sin\\frac{1}{x}\\right) + \\cos\\frac{1}{x}f′(x)=−2x(2+sinx1​)+cosx1​. For small x>0x > 0x>0:

For small xxx, say x<12πx < \\frac{1}{2\\pi}x<2π1​, cos⁡1x=−1\\cos\\frac{1}{x} = -1cosx1​=−1 at x=23πx = \\frac{2}{3\\pi}x=3π2​:

f′(x)≈−2x(2+0)−1=−4x−1<0 if x<14.f'(x) \\approx -2x (2 + 0) - 1 = -4x - 1 < 0 \\text{ if } x < \\frac{1}{4}.f′(x)≈−2x(2+0)−1=−4x−1<0 if x<41​.

Choose δ=14\\delta = \\frac{1}{4}δ=41​. On (0,δ)(0, \\delta)(0,δ), f′(x)≤−1−4x<0f'(x) \\leq -1 - 4x < 0f′(x)≤−1−4x<0, so fff is decreasing. Option (B) is true.

Step 4: Check if fff is Increasing on (−δ,0)(-\\delta, 0)(−δ,0) (Option C)​

On (−δ,0)(-\\delta, 0)(−δ,0), x<0x < 0x<0, so −2x(2+sin⁡1x)>0-2x (2 + \\sin\\frac{1}{x}) > 0−2x(2+sinx1​)>0, but cos⁡1x\\cos\\frac{1}{x}cosx1​ oscillates. At x=−23πx = -\\frac{2}{3\\pi}x=−3π2​, cos⁡1x=−1\\cos\\frac{1}{x} = -1cosx1​=−1, and:

f′(x)≈4∣x∣−1>0 if ∣x∣>14.f'(x) \\approx 4|x| - 1 > 0 \\text{ if } |x| > \\frac{1}{4}.f′(x)≈4∣x∣−1>0 if ∣x∣>41​.

But for smaller ∣x∣|x|∣x∣, cos⁡1x=1\\cos\\frac{1}{x} = 1cosx1​=1 at some points, making f′(x)<0f'(x) < 0f′(x)<0. Thus, fff is not always increasing on (−δ,0)(-\\delta, 0)(−δ,0) for any δ\\deltaδ. Option (C) is true.

Step 5: Local Minima at x=0x = 0x=0 (Option D)​

For x≠0x \\neq 0x=0, f(x)=2−x2(2+sin⁡1x)≤2f(x) = 2 - x^2 \\left(2 + \\sin\\frac{1}{x}\\right) \\leq 2f(x)=2−x2(2+sinx1​)≤2, and f(0)=2f(0) = 2f(0)=2. Thus, x=0x = 0x=0 is a local maximum, not minima. Option (D) is false.

Conclusion​

Options (B) and (C) are true, but the question asks for one true statement. Typically, (B) is the intended answer due to its constructive nature.

Answer

B\\boxed{\\text{B}}B​

", "url": "https://prepverse.vercel.app/blog/jee-advanced-2025-piecewise-function-analysis", "title": "JEE Advanced 2025 – Differentiability and Behavior of a Piecewise Function", "summary": "Analyze a piecewise-defined function for differentiability, monotonicity, and local minima at x = 0. A conceptual calculus problem tailored for JEE Advanced 2025.", "date_modified": "2025-05-19T00:00:00.000Z", "author": { "name": "Akash Singh" }, "tags": [ "JEE Advanced", "Paper 1", "Calculus", "Differentiability", "Monotonicity", "Piecewise Functions", "Limits" ] }, { "id": "https://prepverse.vercel.app/blog/jee-advanced-2025-polynomial-coefficient-problem", "content_html": "
Question 1 (Paper - 1)

Let R\\mathbb{R}R denote the set of all real numbers. Let ai,bi∈Ra_i, b_i \\in \\mathbb{R}ai​,bi​∈R for i∈{1,2,3}i \\in \\{1, 2, 3\\}i∈{1,2,3}.

Define the functions f:R→Rf: \\mathbb{R} \\to \\mathbb{R}f:R→R, g:R→Rg: \\mathbb{R} \\to \\mathbb{R}g:R→R, and h:R→Rh: \\mathbb{R} \\to \\mathbb{R}h:R→R by:

f(x)=a1+10x+a2x2+a3x3+x4f(x) = a_1 + 10x + a_2 x^2 + a_3 x^3 + x^4f(x)=a1​+10x+a2​x2+a3​x3+x4
g(x)=b1+3x+b2x2+b3x3+x4g(x) = b_1 + 3x + b_2 x^2 + b_3 x^3 + x^4g(x)=b1​+3x+b2​x2+b3​x3+x4
h(x)=f(x+1)−g(x+2)h(x) = f(x + 1) - g(x + 2)h(x)=f(x+1)−g(x+2)

If f(x)≠g(x)f(x) \\ne g(x)f(x)=g(x) for every x∈Rx \\in \\mathbb{R}x∈R, then the coefficient of x3x^3x3 in h(x)h(x)h(x) is:

(A)8(B)2(C)−4(D)−6\\begin{array}{|c|c|} \\hline \\text{(A)} & 8 \\\\ \\hline \\text{(B)} & 2 \\\\ \\hline \\text{(C)} & -4 \\\\ \\hline \\text{(D)} & -6 \\\\ \\hline \\end{array}(A)(B)(C)(D)​82−4−6​​

Finding the Coefficient of x3x^3x3 in h(x)h(x)h(x)​

Given​

Let the functions be defined as:

f(x)=a1+10x+a2x2+a3x3+x4f(x) = a_1 + 10x + a_2 x^2 + a_3 x^3 + x^4f(x)=a1​+10x+a2​x2+a3​x3+x4
g(x)=b1+3x+b2x2+b3x3+x4g(x) = b_1 + 3x + b_2 x^2 + b_3 x^3 + x^4g(x)=b1​+3x+b2​x2+b3​x3+x4
h(x)=f(x+1)−g(x+2)h(x) = f(x + 1) - g(x + 2)h(x)=f(x+1)−g(x+2)

We are told that f(x)=g(x)f(x) = g(x)f(x)=g(x) for all x∈Rx \\in \\mathbb{R}x∈R, so:

Step 1: Expand f(x+1)f(x + 1)f(x+1)​

Using binomial expansion:

(x+1)3=x3+3x2+3x+1(x + 1)^3 = x^3 + 3x^2 + 3x + 1(x+1)3=x3+3x2+3x+1
(x+1)4=x4+4x3+6x2+4x+1(x + 1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1(x+1)4=x4+4x3+6x2+4x+1

So,

f(x+1)=a1+10(x+1)+a2(x+1)2+a3(x+1)3+(x+1)4f(x + 1) = a_1 + 10(x + 1) + a_2(x + 1)^2 + a_3(x + 1)^3 + (x + 1)^4f(x+1)=a1​+10(x+1)+a2​(x+1)2+a3​(x+1)3+(x+1)4

We only need the coefficient of x3x^3x3:

So total contribution to x3x^3x3 in f(x+1)f(x + 1)f(x+1) is:

a3+4a_3 + 4a3​+4

Step 2: Expand g(x+2)g(x + 2)g(x+2)​

Using binomial expansion:

(x+2)3=x3+6x2+12x+8(x + 2)^3 = x^3 + 6x^2 + 12x + 8(x+2)3=x3+6x2+12x+8
(x+2)4=x4+8x3+24x2+32x+16(x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16(x+2)4=x4+8x3+24x2+32x+16

So,

g(x+2)=b1+3(x+2)+b2(x+2)2+b3(x+2)3+(x+2)4g(x + 2) = b_1 + 3(x + 2) + b_2(x + 2)^2 + b_3(x + 2)^3 + (x + 2)^4g(x+2)=b1​+3(x+2)+b2​(x+2)2+b3​(x+2)3+(x+2)4

We only need the coefficient of x3x^3x3:

So total contribution to x3x^3x3 in g(x+2)g(x + 2)g(x+2) is:

b3+8b_3 + 8b3​+8

Step 3: Coefficient of x3x^3x3 in h(x)h(x)h(x)​

Since:

h(x)=f(x+1)−g(x+2)h(x) = f(x + 1) - g(x + 2)h(x)=f(x+1)−g(x+2)

The coefficient of x3x^3x3 is:

(a3+4)−(b3+8)=a3−b3−4(a_3 + 4) - (b_3 + 8) = a_3 - b_3 - 4(a3​+4)−(b3​+8)=a3​−b3​−4

Given a3=b3a_3 = b_3a3​=b3​, we have:

a3−b3−4=−4a_3 - b_3 - 4 = -4a3​−b3​−4=−4

✅ Final Answer​

Correct option: C

The coefficient of x3x^3x3 in h(x)h(x)h(x) is −4-4−4.

", "url": "https://prepverse.vercel.app/blog/jee-advanced-2025-polynomial-coefficient-problem", "title": "JEE Advanced 2025 – Coefficient of x^3 in Function Difference Problem", "summary": "Detailed explanation of a functional algebra problem involving polynomials and coefficient comparison, tailored for JEE Advanced 2025.", "date_modified": "2025-05-19T00:00:00.000Z", "author": { "name": "Akash Singh" }, "tags": [ "JEE Advanced", "Paper 1", "Math", "Algebra", "Polynomials", "Functions" ] }, { "id": "https://prepverse.vercel.app/blog/jee-advanced-2025-probability-students-solving-problem", "content_html": "
Question 2 (Paper - 1)

Three students S1S_1S1​, S2S_2S2​, and S3S_3S3​ are given a problem to solve. Consider the following events:

U:At least one of S1,S2, and S3 can solve the problem,U: \\text{At least one of } S_1, S_2, \\text{ and } S_3 \\text{ can solve the problem},U:At least one of S1​,S2​, and S3​ can solve the problem,
V:S1 can solve the problem, given that neither S2 nor S3 can solve the problem,V: S_1 \\text{ can solve the problem, given that neither } S_2 \\text{ nor } S_3 \\text{ can solve the problem},V:S1​ can solve the problem, given that neither S2​ nor S3​ can solve the problem,
W:S2 can solve the problem and S3 cannot solve the problem,W: S_2 \\text{ can solve the problem and } S_3 \\text{ cannot solve the problem},W:S2​ can solve the problem and S3​ cannot solve the problem,
T:S3 can solve the problem.T: S_3 \\text{ can solve the problem}.T:S3​ can solve the problem.

For any event EEE, let P(E)P(E)P(E) denote the probability of EEE. If

P(U)=12,P(V)=110,andP(W)=112,P(U) = \\frac{1}{2}, \\quad P(V) = \\frac{1}{10}, \\quad \\text{and} \\quad P(W) = \\frac{1}{12},P(U)=21​,P(V)=101​,andP(W)=121​,

then P(T)P(T)P(T) is equal to

(A) 1336(B) 13(C) 1960(D) 14\\begin{array}{|c|c|c|c|} \\hline \\hspace{0.8cm} \\text{(A) } \\frac{13}{36} \\hspace{0.8cm} & \\hspace{0.8cm} \\text{(B) } \\frac{1}{3} \\hspace{0.8cm} & \\hspace{0.8cm} \\text{(C) } \\frac{19}{60} \\hspace{0.8cm} & \\hspace{0.8cm} \\text{(D) } \\frac{1}{4} \\hspace{0.8cm} \\\\ \\hline \\end{array}(A) 3613​​(B) 31​​(C) 6019​​(D) 41​​​

With Venn diagram approach.\nDefine:

A=S1 solves,B=S2 solves,C=S3 solves.A = S_1 \\text{ solves}, \\quad B = S_2 \\text{ solves}, \\quad C = S_3 \\text{ solves}.A=S1​ solves,B=S2​ solves,C=S3​ solves.

Given:

P(A∩B′∩C′)=110,P(B∩C′)=112,P(A∪B∪C)=12.P(A \\cap B' \\cap C') = \\frac{1}{10}, \\quad P(B \\cap C') = \\frac{1}{12}, \\quad P(A \\cup B \\cup C) = \\frac{1}{2}.P(A∩B′∩C′)=101​,P(B∩C′)=121​,P(A∪B∪C)=21​.

Using inclusion-exclusion and solving via probabilities of intersections, we find

P(C)=1336P(C) = \\frac{13}{36}P(C)=3613​
Correct option: A

P(T)=1336.P(T) = \\boxed{\\frac{13}{36}}.P(T)=3613​​.

", "url": "https://prepverse.vercel.app/blog/jee-advanced-2025-probability-students-solving-problem", "title": "JEE Advanced 2025 – Probability of Students Solving a Problem", "summary": "Detailed solution of a probability problem involving three students solving a problem independently, including event analysis and calculation of probabilities, tailored for JEE Advanced 2025.", "date_modified": "2025-05-19T00:00:00.000Z", "author": { "name": "Akash Singh" }, "tags": [ "JEE Advanced", "Paper 1", "Math", "Probability", "Events", "Conditional Probability" ] }, { "id": "https://prepverse.vercel.app/blog/two-sum", "content_html": "

PrepVerse | Three Approaches to Solving the Two Sum Problem​

Approach 1: Brute Force​

Algorithm:

The brute force approach is straightforward. Loop through each element x in the array and check if there is another element that equals target - x.

Implementation:

Brute Force
def two_sum_brute_force(nums, target):
    n = len(nums)
    for i in range(n):
        for j in range(i + 1, n):
            if nums[i] + nums[j] == target:
                return [i, j]
    return None

Complexity Analysis:

Approach 2: Two-pass Hash Table​

Intuition:

To improve our runtime complexity, we need a more efficient way to check if the complement exists in the array. A hash table is well-suited for this purpose because it supports fast lookups in near constant time. By trading space for speed, we can reduce the lookup time from O(n) to O(1).

Algorithm:

  1. In the first iteration, add each element's value as a key and its index as a value to the hash table.
  2. In the second iteration, check if each element's complement (target - nums[i]) exists in the hash table. If it does, return the current element's index and its complement's index.

Implementation:

Two-pass Hash Table
def two_sum_two_pass_hash_table(nums, target):
    hash_table = {}
    for i, num in enumerate(nums):
        hash_table[num] = i
    
    for i, num in enumerate(nums):
        complement = target - num
        if complement in hash_table and hash_table[complement] != i:
            return [i, hash_table[complement]]
    return None

Complexity Analysis:

Approach 3: One-pass Hash Table​

Algorithm:

We can optimize further by performing the lookup and insertion in a single pass. While iterating through the array, we check if the current element's complement already exists in the hash table. If it does, we have found a solution and return the indices immediately.

Implementation:

One-pass Hash Table
def two_sum_one_pass_hash_table(nums, target):
    hash_table = {}
    for i, num in enumerate(nums):
        complement = target - num
        if complement in hash_table:
            return [hash_table[complement], i]
        hash_table[num] = i
    return None

Complexity Analysis:

Conclusion​

In summary, the brute force approach is simple but inefficient, with a time complexity of O(n²). The two-pass hash table approach improves the time complexity to O(n) by utilizing extra space for a hash table. The one-pass hash table approach further optimizes the solution by combining lookup and insertion into a single pass, maintaining an overall time complexity of O(n) and space complexity of O(n). Depending on the constraints and requirements of your problem, choosing the right approach can significantly improve performance.

", "url": "https://prepverse.vercel.app/blog/two-sum", "title": "Two Sum Explained", "summary": "DSA problem two sum explained", "date_modified": "2024-08-07T00:00:00.000Z", "author": { "name": "Akash Singh" }, "tags": [ "Array", "LeetCode" ] }, { "id": "https://prepverse.vercel.app/blog/god-brahma", "content_html": "

Exploring the Complete Life Cycle of Creator God Brahma & the Age of the Universe According to Vedas:​

100 Years (Lifespan: 2 Parardhas) of Brahma:​

  1. Brahma's Lifespan: Brahma, the creator god, lives for 100 of his years.
  2. Each Year Structure: A Brahmaic year consists of 360 days and nights.
  3. Total Duration: The total duration of Brahma's life is 311.04 trillion human years.

50 Years (Parardhas) of Brahma:​

  1. Parardha Definition: Brahma's 100-year life is divided into two 50-year periods, each known as a Parardha.
  2. In a Maha-Kalpa: In a 100-year Maha-Kalpa, there are a total of 36,000 full days. This includes 36,000 Kalpas (days proper) and 36,000 Pralayas (nights).

1 Year of Brahma:​

  1. Each Year Structure: A Brahmaic year consists of 12 months.
  2. Duration: 3.1104 trillion human years.

1 Month of Brahma:​

  1. Each Month Structure: A Brahmaic month consists of 30 days (Kalpa) & nights (Pralaya).
    A list of 30 kalpa (days) name as mentioned in the Matsya Purana
    Kalpa Number30 Kalpa (Day) Name
    1Sveta (current day)
    2Nilalohita
    3Vamadeva
    4Rathantara
    5Raurava
    6Deva
    7Vrhat
    8Kandarpa
    9Sadya
    10Isana
    11Tamah
    12Sarasvata
    13Udana
    14Garuda
    15Kaurma
    16Narasimha
    17Samana
    18Agneya
    19Soma
    20Manava
    21Tatpuman
    22Vaikuṇṭha
    23Lakṣmi
    24Savitri
    25Aghora
    26Varaha
    27Vairaja
    28Gauri
    29Mahesvara
    30Pitr
  2. Duration: 259.2 billion human years.

1 Day(Kalpa) and 1 Night (Pralaya) of Brahma:​

  1. Brahma's day, known as a Kalpa, lasts for 4.32 billion years.

  2. It is followed by a night, or Pralaya, of equal length.

  3. Structure of a Kalpa:

    • A Kalpa consists of 1,000 Chatur-Yugas.
    • Within a Kalpa, there are 14 Manvantaras (epochs) and 15 Manvantara-Sandhyas (transitional periods).
    • At the start of Brahma's day, creation unfolds as he is reborn, forming planets and the first living entities.
    • At the end of his day, Brahma and his creations undergo partial dissolution, entering a state of unmanifestation.

  4. Duration of kalp:

    • Started in the Past: Approximately 1.97 billion years ago.
    • Ends in the Future: Estimated to conclude in about 2.35 billion years.

    \"Structure

Maha-Kalpa and Maha-Pralaya:​

  1. Brahma's entire lifespan is called a Maha-Kalpa, lasting for 311.04 trillion years.
  2. It is followed by a Maha-Pralaya, a period of full dissolution, lasting for an equivalent length.
  3. Prakriti, the basis of the universe, is manifest at the start and unmanifest at the end of a Maha-Kalpa.
  4. Duration of Maha-Kalpa:
    • Started in the Past: Roughly 155.52 trillion years ago.
    • Ends in the Future: Expected to conclude in about 155.52 trillion years.

Current State within Brahma's Life Cycle:​

\"God

  1. 51st year of 100 (2nd half or parardha).
  2. 1st month of 12.
  3. 1st kalpa/day (Shveta-Varaha Kalpa) of 30.
  1. 7th manvantara (Vaivasvatha Manu) of 14.\n\"Manavantara
  2. 28th chatur-yuga of 71.\n\"Chatur-yuga
  3. 4th yuga (Kali-yuga) of 4.\n\"Kali-yuga

Start date of current Kali Yuga:​

Calculating Brahma's 100 Years into Human Years

Let's break down Brahma's 100-year lifespan into human years step by step:

Basic Conversion Factors:​

  • 100 years = 1 * 100 years
    • 1 year = 12 months
    • 1 month = 30 days (kalpa) + 30 nights (pralaya)
  • 100 years = 36,000 days (kalpa) + 36,000 nights (pralaya)

Yuga Structure:​

  • 1 day (kalpa) = 14 manvantara + 15 manvantara-sandhya
  • 1 night (pralaya) = 14 manvantara + 15 manvantara-sandhya
  • 1 manvantara = 71 chatur-yuga
  • 1 manvantara-sandhya = 1 satya-yuga
  • 1 chatur-yuga = 1 satya(krta)-yuga + 1 treta-yuga + 1 dvapara-yuga + 1 kali-yuga
    • 1 satya-yuga = 4 * 1 kali-yuga
    • 1 treta-yuga = 3 * 1 kali-yuga
    • 1 dvapara-yuga = 2 * 1 kali-yuga
    • 1 kali-yuga = 432,000 human years

Yuga Duration:​

  • 1 kali-yuga = 4,32,000 human years
  • 1 chatur-yuga = (4+3+2+1) kali-yuga
           = 10 kali-yuga
  • 1 manvantara = 71 chatur-yuga
           = 71 x 10 kali-yuga
           = 710 kali-yuga
  • 1 manvantara-sandhya = 1 satya-yuga
               = 4 kali-yuga
  • 1 day (kalpa) = 14 manvantara + 15 manvantara-sandhya
           = 14 x 710 kali-yuga + 15 x 4 kali-yuga
           = (9940 + 60) kali-yuga
           = 10,000 kali-yuga
  • 1 night (pralaya) = 14 manvantara + 15 manvantara-sandhya
            = 14 x 710 kali-yuga + 15 x 4 kali-yuga
            = (9940 + 60) kali-yuga
            = 10,000 kali-yuga

Conversion:​

  • 100 years = 36,000 days (kalpa) + 36,000 nights (pralaya)
         = 2 x 36,000 x 10,000 kali-yuga
         = 720,000,000 kali-yuga
         = 720,000,000 x 4,32,000 human years
         = 311,040,000,000,000 human years

This calculation shows that Brahma's 100-year lifespan equates to a staggering 311,040,000,000,000 human years in Hindu cosmology.

-tip Current Exact Date & Time of Creator God Brahma with respect to 100 years of Brahma

Setting the Stage: Current Time​

As per the given parameters:

  • Current Year of 100            = 50 (Completed) + 1 (current year)
  • Current month of 12           = 1
  • Current day(Kalpa) of 30         = 1
  • Current manvantara of 14         = 6 (Completed) + 1(current)
  • Current manvantara-sandhya of 15     = 7 (Completed)
  • Current chatur-yuga of 71        = 27 (Completed) + 1(current)
  • Current yuga of 4            = 4 (Kali-yuga)
  • Current kali-yuga time (in Human years) = 5126 years

Thus, the current date is:--> DD/MM/YYYY = 01/01/51.

God Brahma's Time Scale:​

Human TimeGod Bramha Time
311,040,000,000,000 yr100 yr
4,32,000 yr4.32 sec
1,00,000 yr1 sec
1 yr10 micro-sec
36 days1 micro-sec

Calculations:​

  • Total human years in 6 manvantara = 6 x 710 kali-yuga
                      = 4260 kali-yuga
                      = 4260 x 432,000
                      = 1,840,320,000 human year
  • Total human years in 7 manvantara-sandhya = 7 x 4 kali-yuga
                          = 28 kali-yuga
                          = 28 x 432,000
                          = 12,096,000 human year
  • Total human years in 27 chatur-yuga = 27 x 10 kali-yuga
                      = 270 kali-yuga
                      = 270 x 432,000
                      = 1,166,440,000 human year
  • Total human years in 9 kali-yuga + 5126 years = 9 x 4,32,000 + 5126
                          = 3,888,000 + 5126
                          = 3,93,126 human years

Current kalpa total time = 1,840,320,000 + 12,096,000 + 1,166,440,000 + 3,93,126
\n            = 3,019,249,126 human years
\n            = 3019249126/100000 brahma seconds
\n            = 30192.49126 brahma seconds

Conversion to Time Format:​

  • Total Time Elapsed: 30192.49126 seconds
    • 503.208188 minutes
    • 8 hours, 23.208188 minutes
    • 8 hours, 23 minutes, 12.49128 seconds
    • 8 hours, 23 minutes, 12 seconds, 491280 micro-seconds
God Brahma's curremt Date and Time:
  • Date = 01 January 51
  • Time = 08:23:12:491280 AM

In Human Time:​

18 February 2024

Time Scales Between Humans and Brahma in Hindu Cosmology​

Time Scales Between Humans and Brahma in Hindu Cosmology
Time Scales Between Humans and Brahma in Hindu Cosmology:
UnitDefinitionHumanBrahma
Maha-kalpa36,000 kalpa & pralaya311,040,000,000,000
(311.04 trillion) yr		100 yr
Maha-pralaya36,000 kalpa & pralaya311,040,000,000,000
(311.04 trillion) yr		100 yr
Parardha1⁄2 Maha-kalpa155,520,000,000,000
(155.52 trillion) yr		50 yr
Kalpa14 manvantara + 15 manvantara-sandhya4,320,000,000
(4.32 billion) yr		12 hr
Pralaya14 manvantara + 15 manvantara-sandhya4,320,000,000
(4.32 billion) yr		12 hr
Manvantara71 Catur-yuga306,720,000 yr51.12 min
Manvantara-sandhya1 Satya-yuga length1,728,000 yr17.28 s
Catur-yugaSatya(Krta), Treta, Dvapara & Kali-yugas4,320,000 yr43.20 s
Satya(Krta)-yuga4 Kali-yugas length1,728,000 yr17.28 s
Treta-yuga3 Kali-yugas length1,296,000 yr12.96 s
Dvapara-yuga2 Kali-yugas length864,000 yr8.64 s
Kali-yuga1 Kali-yugas length432,000 yr4.32 s
", "url": "https://prepverse.vercel.app/blog/god-brahma", "title": "The Complete Life Cycle of Creator God Brahma", "summary": "Exploring the Complete Life Cycle of Brahma & the Age of the Universe According to Vedas", "date_modified": "2024-02-18T00:00:00.000Z", "author": { "name": "Akash Singh" }, "tags": [ "Creator", "God", "Brahma", "Universe", "Bharat" ] } ] }