Can someone check my answer on the essentials of a partnership and give me marks out of 3?

BrunoAI answer

Good answer! I'd give about 1.5 out of 2 (an ICAI examiner may vary). What's right: - Correct Section 4 definition and the three essentials. - Facts applied, with a clear conclusion: no partnership. What cost marks: 1. You don't say why there is no profit sharing. Add: 5% of the contract amount is fixed commission (remuneration), not a share of profits. 2. Add: Roop Singh bears no risk or loss (Manak bears all costs) and cannot bind Manak. 3. Skip the struck-out Section 6 line, or use it to support point 1. Also write "Roop Singh" in full. See note 5 for past-year law answers.

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BrunoAI answer

Your answer is correct: Ram is liable only for his share of Rs. 1,500. I'd give about 4.5 to 5 out of 6. What you did well: - Section 4 definition and the three essentials (agreement, sharing profits of a business, mutual agency) are written clearly. - The facts are applied neatly and the conclusion is right. Where you lost marks: 1. You did not cite Section 6 (mode of determining the existence of partnership). Its Explanation 1 says that sharing the gross returns of jointly owned property does not, by itself, make the owners partners. This is the key provision in the ICAI answer. 2. Add one line that Ram and Raheem are only co-owners of the flat, not partners, so there is no joint liability. 3. Minor point: keep "existence" and "Raheem" spelled correctly and the conclusion tidy.

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VVeer

1. Right side me log rule lagao: \(\frac{1}{2}(\log a + \log b)\) ban jayega \(\log(\sqrt{ab})\). 2. Dono taraf se \(\log \) hata do, toh bachega \(\frac{a+b}{3} = \sqrt{ab}\). 3. Ab dono side square karo, toh \((a+b)^2 = 9ab\) aa jayega. 4. Formula kholo: \(a^2 + b^2 + 2ab = 9ab\), yani \(a^2 + b^2 = 7ab\). 5. Bas ab poore equation ko \(ab\) se divide kar do, seedha \(\frac{a}{b} + \frac{b}{a} = 7\) mil jayega

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AAkshuuu

With pratice bro

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IIshaan· 3 found helpful

You should go for May 27 or September 27

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VVeer

• Set equal to \(k\). Let \(3^a = 5^b = 75^c = k\), which gives \(3 = k^{1/a}\), \(5 = k^{1/b}\), and \(75 = k^{1/c}\). • Factor \(75\). Since \(75 = 3 \times 5^2\), substitute the \(k\) values to get \(k^{1/c} = k^{1/a} \cdot (k^{1/b})^2\). • Equate exponents and simplify. Add the powers on the right to get \(\frac{1}{c} = \frac{1}{a} + \frac{2}{b}\). Cross-multiplying gives \(ab = c(2a + b)\), so \(ab - c(2a + b) = 0\).

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