Mathematics for FinanceMTP Dec 22 Series IIQuestion 1376 of 512
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8,829\displaystyle 8,829 is invested into three different sectors in such a way that their amounts at 4%\displaystyle 4\% p.a. S.I. after 5\displaystyle 5 years; 6\displaystyle 6 and 8\displaystyle 8 years are equal. Find each part of the sum.

Options

A3,069;2,970;2,790\displaystyle 3,069; 2,970; 2,790
B3,089;2,970;2,790\displaystyle 3,089; 2,970; 2,790
C3,690;2,970;2,790\displaystyle 3,690; 2,970; 2,790
D3,069;2,960;2,760\displaystyle 3,069; 2,960; 2,760
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Correct Answer

✅ Option a — 3,069;2,970;2,790\displaystyle 3,069; 2,970; 2,790

All Options:

  • A3,069;2,970;2,790\displaystyle 3,069; 2,970; 2,790
  • B3,089;2,970;2,790\displaystyle 3,089; 2,970; 2,790
  • C3,690;2,970;2,790\displaystyle 3,690; 2,970; 2,790
  • D3,069;2,960;2,760\displaystyle 3,069; 2,960; 2,760

Detailed Solution & Explanation

**Derivation of Sector Investments** Given: - Total sum to invest = Rs. 8,829\displaystyle \text{Rs. }8,829 - Rate of Interest (R\displaystyle R) = 4%\displaystyle 4\% per annum simple interest - Time periods for the three parts: t1=5\displaystyle t_1 = 5 years, t2=6\displaystyle t_2 = 6 years, t3=8\displaystyle t_3 = 8 years - Amounts after these periods are equal. **Step 1: Set up the equal amounts relation** Let the three parts be x\displaystyle x, y\displaystyle y, and z\displaystyle z. A1=x(1+0.04×5)=1.20xA_1 = x(1 + 0.04 \times 5) = 1.20 x A2=y(1+0.04×6)=1.24yA_2 = y(1 + 0.04 \times 6) = 1.24 y A3=z(1+0.04×8)=1.32zA_3 = z(1 + 0.04 \times 8) = 1.32 z Since the amounts are equal: 1.20x=1.24y=1.32z1.20 x = 1.24 y = 1.32 z Dividing by 0.04\displaystyle 0.04: 30x=31y=33z30 x = 31 y = 33 z **Step 2: Find the ratio of the parts** x:y:z=130:131:133x : y : z = \frac{1}{30} : \frac{1}{31} : \frac{1}{33} Multiplying by 30×31×33=30690\displaystyle 30 \times 31 \times 33 = 30690: x:y:z=1023:990:930x : y : z = 1023 : 990 : 930 Sum of the ratio parts = 1023+990+930=2943\displaystyle 1023 + 990 + 930 = 2943. **Step 3: Calculate the value of each part** x=8829×10232943=3×1023=Rs. 3,069x = 8829 \times \frac{1023}{2943} = 3 \times 1023 = \text{Rs. }3,069 y=8829×9902943=3×990=Rs. 2,970y = 8829 \times \frac{990}{2943} = 3 \times 990 = \text{Rs. }2,970 z=8829×9302943=3×930=Rs. 2,790z = 8829 \times \frac{930}{2943} = 3 \times 930 = \text{Rs. }2,790 Hence, **Option A** is the correct answer.

More Questions from Mathematics for Finance

A sum of money invested in compounded interest doubles itself in four years. In how many years it becomes 32 times of itself as the same rate of compound interest?

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A certain sum of money was put at S.I. for 2.5\displaystyle 2.5 years at a certain rate of S.I. p.a. Had it been put at 4%\displaystyle 4\% higher rate, it would have fetched 500\displaystyle 500 more. Find the sum of money.

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If Rs. 1,000\displaystyle \text{Rs. }1,000 be invested at interest rate of 5%\displaystyle 5\% and the interest be added to the principal every 10\displaystyle 10 years, then the number in years in which it will amount to Rs. 2,000\displaystyle \text{Rs. }2,000 is:

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