Mathematics of FinanceMCQPYQ Dec 23Question 1254 of 512
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A machine costing 1,00,000\displaystyle 1,00,000 has useful life of 10\displaystyle 10 years. If the rate of depreciation is 12%\displaystyle 12\% what is scrap value of the machine at the end of life? Given (0.88)10=0.27850\displaystyle (0.88)^{10} = 0.27850

Options

A25,850\displaystyle 25,850
B26,850\displaystyle 26,850
C27,850\displaystyle 27,850
D28,850\displaystyle 28,850
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Correct Answer

Option c27,850\displaystyle 27,850

All Options:

  • A25,850\displaystyle 25,850
  • B26,850\displaystyle 26,850
  • C27,850\displaystyle 27,850
  • D28,850\displaystyle 28,850

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Detailed Solution & Explanation

**Derivation of Scrap Value** Given: - Initial Cost of Machine (V0\displaystyle V_0) = Rs. 1,00,000\displaystyle \text{Rs. }1,00,000 - Useful Life (t\displaystyle t) = 10\displaystyle 10 years - Rate of Depreciation (d\displaystyle d) = 12%\displaystyle 12\% per annum - Given: (0.88)10=0.27850\displaystyle (0.88)^{10} = 0.27850 **Step 1: Calculate Scrap Value (Vt\displaystyle V_t)** Vt=V0(1d)tV_t = V_0(1 - d)^t V10=100000(10.12)10V_{10} = 100000(1 - 0.12)^{10} V10=100000(0.88)10V_{10} = 100000(0.88)^{10} Using the given value: V10=100000×0.27850=Rs. 27,850V_{10} = 100000 \times 0.27850 = \text{Rs. }27,850 Hence, **Option C** is the correct answer.

About This Chapter: Mathematics of Finance

Paper

Paper 3: Quantitative Aptitude

Weightage

12-16 Marks

Key Topics

Simple & Compound Interest, Annuity, Perpetuity

The most important mathematical chapter in the entire syllabus. It covers Simple Interest (SI), Compound Interest (CI), Nominal vs Effective rates, Present and Future Value, Annuities (Ordinary and Due), Sinking Funds, and Perpetuities. The concepts learned here are applied heavily in CA Intermediate and Final.

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