Mathematics for FinanceMTP May 18Question 1276 of 512
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A machine worth of 4,90,740\displaystyle 4,90,740 is depreciated at 15%\displaystyle 15\% on its opening value each year. When its value reduces to 2,00,000\displaystyle 2,00,000.

Options

A5\displaystyle 5 years 6\displaystyle 6 months
B6\displaystyle 6 years 7\displaystyle 7 months
C5\displaystyle 5 years 5\displaystyle 5 months
DNone of these
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Correct Answer

✅ Option a — 5\displaystyle 5 years 6\displaystyle 6 months

All Options:

  • A5\displaystyle 5 years 6\displaystyle 6 months
  • B6\displaystyle 6 years 7\displaystyle 7 months
  • C5\displaystyle 5 years 5\displaystyle 5 months
  • DNone of these

Detailed Solution & Explanation

**Derivation of Depreciation Period** Given: - Initial Value (V0\displaystyle V_0) = Rs. 4,90,740\displaystyle \text{Rs. }4,90,740 - Rate of Depreciation (d\displaystyle d) = 15%\displaystyle 15\% per annum - Depreciated Value (Vt\displaystyle V_t) = Rs. 2,00,000\displaystyle \text{Rs. }2,00,000 **Step 2: Set up the depreciation formula** Vt=V0(1−d)tV_t = V_0(1 - d)^t 200000=490740(1−0.15)t200000 = 490740(1 - 0.15)^t 200000490740=(0.85)t\frac{200000}{490740} = (0.85)^t 0.407547=(0.85)t0.407547 = (0.85)^t **Step 3: Solve for t\displaystyle t using logarithms** ln⁡(0.407547)=tln⁡(0.85)\ln(0.407547) = t \ln(0.85) −0.89757=t(−0.16252)-0.89757 = t (-0.16252) t=−0.89757−0.16252≈5.52 yearst = \frac{-0.89757}{-0.16252} \approx 5.52 \text{ years} 5.52 years≈5 years and 6 months5.52 \text{ years} \approx 5 \text{ years and } 6 \text{ months} Hence, **Option A** is the correct answer.

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