Question:

A loan of Rs 1000 is fully repaid by two installments of Rs 530 and Rs 594, paid at the end of the first and second year, respectively. If the interest is compounded annually, then the rate of interest, in percentage, is:

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Whenever loan repayments occur in installments, convert each installment into its present value and sum them. If the interest is compounded annually, the discount factor for the \(n\)-th year is \((1+r)^n\).
Updated On: Dec 4, 2025
  • \(6%\)
  • \(7%\)
  • \(8%\)
  • \(9%\)
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The Correct Option is C

Solution and Explanation

Step 1: Use the Present Value principle. The loan amount equals the sum of the present values of all future installments. \[ 1000 = \frac{530}{1+r} + \frac{594}{(1+r)^2}. \] Let \[ x = 1+r. \] Then the equation becomes \[ 1000 = \frac{530}{x} + \frac{594}{x^2}. \]
Step 2: Clear denominators. Multiply through by \(x^2\): \[ 1000x^2 = 530x + 594. \] Rearrange: \[ 1000x^2 - 530x - 594 = 0. \] Divide by 2 for simplicity: \[ 500x^2 - 265x - 297 = 0. \]
Step 3: Solve the quadratic using the formula. Here, \[ a=500,\; b=-265,\; c=-297. \] \[ x = \frac{265 \pm \sqrt{(-265)^2 - 4(500)(-297)}}{1000}. \] Compute the discriminant: \[ (-265)^2 = 70225,\qquad 4 \cdot 500 \cdot 297 = 594000, \] \[ 70225 + 594000 = 664225 = 815^2. \] Thus, \[ x = \frac{265 \pm 815}{1000}. \] The valid root: \[ x = \frac{1080}{1000} = 1.08. \] The negative root is discarded because \(x = 1+r>0\).
Step 4: Find the rate of interest. \[ 1 + r = 1.08 \quad\Rightarrow\quad r = 0.08 = 8%. \] Therefore, the interest rate is \(8%\).
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