Mathematics for FinanceMTP Jun 23 Series IIQuestion 1384 of 512
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A certain sum of money amounts to 5,000\displaystyle 5,000 in 5\displaystyle 5 years at 10%\displaystyle 10\% p.a. In how many years will it amount to 6,000\displaystyle 6,000 at same rate of S.I. p.a.

Options

A10\displaystyle 10 years
B8\displaystyle 8 years
C6\displaystyle 6 years
DNone of these
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Correct Answer

✅ Option b — 8\displaystyle 8 years

All Options:

  • A10\displaystyle 10 years
  • B8\displaystyle 8 years
  • C6\displaystyle 6 years
  • DNone of these

Detailed Solution & Explanation

**Derivation of Required Period** Given: - Amount in 5\displaystyle 5 years (A1\displaystyle A_1) = Rs. 5,000\displaystyle \text{Rs. }5,000 - Rate of simple interest (R\displaystyle R) = 10%\displaystyle 10\% per annum - Target Amount (A2\displaystyle A_2) = Rs. 6,000\displaystyle \text{Rs. }6,000 **Step 1: Calculate the Principal (P\displaystyle P) from the first scenario** A1=P(1+R×t1100)A_1 = P\left(1 + \frac{R \times t_1}{100}\right) 5000=P(1+10×5100)5000 = P\left(1 + \frac{10 \times 5}{100}\right) 5000=P(1+0.50)=1.50P5000 = P(1 + 0.50) = 1.50 P P=50001.50=Rs. 100003≈Rs. 3,333.33P = \frac{5000}{1.50} = \text{Rs. }\frac{10000}{3} \approx \text{Rs. }3,333.33 **Step 2: Find the time (t2\displaystyle t_2) to reach Rs. 6,000** A2=P(1+R×t2100)A_2 = P\left(1 + \frac{R \times t_2}{100}\right) 6000=100003(1+10×t2100)6000 = \frac{10000}{3}\left(1 + \frac{10 \times t_2}{100}\right) 1.80=1+0.10t21.80 = 1 + 0.10 t_2 0.10t2=0.80  ⟹  t2=8 years0.10 t_2 = 0.80 \implies t_2 = 8 \text{ years} Hence, **Option B** is the correct answer.

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