Mathematics for FinanceMTP Jun 23 Series IQuestion 1378 of 512
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A trust fund has invested 27000\displaystyle 27000 money in two schemes 'A' and 'B' offering compound interest at the rate of 8%\displaystyle 8\% and 9%\displaystyle 9\% per annum respectively. If the total amount of interest accrued through these two schemes together in two years was 4818.30\displaystyle 4818.30. What was the amount invested in schemes 'A'?

Options

A12,000\displaystyle 12,000
B12,500\displaystyle 12,500
C13,000\displaystyle 13,000
D12,500\displaystyle 12,500
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Correct Answer

✅ Option a — 12,000\displaystyle 12,000

All Options:

  • A12,000\displaystyle 12,000
  • B12,500\displaystyle 12,500
  • C13,000\displaystyle 13,000
  • D12,500\displaystyle 12,500

Detailed Solution & Explanation

**Derivation of Investment in Scheme A** Given: - Total investment = Rs. 27,000\displaystyle \text{Rs. }27,000 - Scheme A rate (rA\displaystyle r_A) = 8%\displaystyle 8\% p.a. compounded annually - Scheme B rate (rB\displaystyle r_B) = 9%\displaystyle 9\% p.a. compounded annually - Time (t\displaystyle t) = 2\displaystyle 2 years - Total Compound Interest accrued = Rs. 4,818.30\displaystyle \text{Rs. }4,818.30 **Step 1: Set up the compound interest coefficients** - For Scheme A: CIA=PA[(1.08)2−1]=0.1664PA\displaystyle CI_A = P_A [(1.08)^2 - 1] = 0.1664 P_A - For Scheme B: CIB=PB[(1.09)2−1]=0.1881PB\displaystyle CI_B = P_B [(1.09)^2 - 1] = 0.1881 P_B **Step 2: Formulate the equations** Let x\displaystyle x be the amount invested in Scheme A. Then investment in Scheme B is (27000−x)\displaystyle (27000 - x). 0.1664x+0.1881(27000−x)=4818.300.1664 x + 0.1881(27000 - x) = 4818.30 0.1664x+5078.70−0.1881x=4818.300.1664 x + 5078.70 - 0.1881 x = 4818.30 −0.0217x=4818.30−5078.70-0.0217 x = 4818.30 - 5078.70 −0.0217x=−260.40-0.0217 x = -260.40 **Step 3: Solve for x\displaystyle x** x=260.400.0217=Rs. 12,000x = \frac{260.40}{0.0217} = \text{Rs. }12,000 Hence, **Option A** is the correct answer.

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