Mathematics for FinancePYQ Jan 21Question 1219 of 512
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A man invested one-third of his capital at 7%\displaystyle 7\%, one fourth at 8%\displaystyle 8\% and the remainder at 10%\displaystyle 10\%. If the annual income is 561\displaystyle 561. The capital is -

Options

A4,400\displaystyle 4,400
B5,500\displaystyle 5,500
C6,600\displaystyle 6,600
D5,800\displaystyle 5,800
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Correct Answer

✅ Option c — 6,600\displaystyle 6,600

All Options:

  • A4,400\displaystyle 4,400
  • B5,500\displaystyle 5,500
  • C6,600\displaystyle 6,600
  • D5,800\displaystyle 5,800

Detailed Solution & Explanation

**Derivation of Capital Amount** Given: - Let the total capital be C\displaystyle C. - One-third of capital (C/3\displaystyle C/3) is invested at r1=7%\displaystyle r_1 = 7\% simple interest. - One-fourth of capital (C/4\displaystyle C/4) is invested at r2=8%\displaystyle r_2 = 8\% simple interest. - Remaining capital is invested at r3=10%\displaystyle r_3 = 10\% simple interest. - Total annual income (interest earned in 1 year) = Rs. 561\displaystyle \text{Rs. }561 **Step 1: Express the remaining capital** Remaining Capital=C−C3−C4=C(1−712)=5C12\text{Remaining Capital} = C - \frac{C}{3} - \frac{C}{4} = C \left(1 - \frac{7}{12}\right) = \frac{5C}{12} **Step 2: Set up the total annual interest equation** Total Interest=Interest1+Interest2+Interest3\text{Total Interest} = \text{Interest}_1 + \text{Interest}_2 + \text{Interest}_3 561=C3×7×1100+C4×8×1100+5C12×10×1100561 = \frac{\frac{C}{3} \times 7 \times 1}{100} + \frac{\frac{C}{4} \times 8 \times 1}{100} + \frac{\frac{5C}{12} \times 10 \times 1}{100} 561=C100(73+2+5012)561 = \frac{C}{100} \left( \frac{7}{3} + 2 + \frac{50}{12} \right) Simplify the sum inside the parenthesis: 73+2+256=14+12+256=516=8.5\frac{7}{3} + 2 + \frac{25}{6} = \frac{14 + 12 + 25}{6} = \frac{51}{6} = 8.5 561=C100×8.5561 = \frac{C}{100} \times 8.5 561=0.085C561 = 0.085 C **Step 3: Solve for C\displaystyle C** C=5610.085=Rs. 6,600C = \frac{561}{0.085} = \text{Rs. }6,600 Hence, **Option C** 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.

Rs.1,25,000\displaystyle 1,25,000 is borrowed at compound interest at the rate of 2%\displaystyle 2\% for the 1st year, 3%\displaystyle 3\% for the second year and 4%\displaystyle 4\% for the 3rd year. Find the amount to be paid after 3 years.

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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