CAT 2025

Slot 2

QA

Question & Solution

1. Concept Used

• Topic: Compound Interest — Present Value of Installments
• Formula: $$\text{Loan Amount} = \frac{I_1}{(1+r)} + \frac{I_2}{(1+r)^2}$$

where $$I_1$$ and $$I_2$$ are installments paid at the end of year 1 and year 2 respectively, and $$r$$ is the annual rate of interest (in decimal).

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2. Calculation

We are given that a loan of Rs 1000 is repaid by two installments: Rs 530 at the end of year 1 and Rs 594 at the end of year 2. Using the present value equation:

$$\frac{530}{1+r} + \frac{594}{(1+r)^2} = 1000$$

Let $$x = \frac{1}{1+r}$$. Substituting:

$$530x + 594x^2 = 1000$$

Rearranging into standard quadratic form:

$$594x^2 + 530x - 1000 = 0$$

Now applying the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a = 594,\ b = 530,\ c = -1000$$:

$$\Delta = b^2 - 4ac = (530)^2 + 4 \times 594 \times 1000$$

$$\Delta = 280900 + 2376000 = 2656900$$

$$\sqrt{\Delta} = \sqrt{2656900} = 1630$$

Taking the positive root (since $$x > 0$$):

$$x = \frac{-530 + 1630}{2 \times 594} = \frac{1100}{1188} = \frac{275}{297}$$

Since $$x = \frac{1}{1+r}$$, we get:

$$1 + r = \frac{297}{275}$$

$$r = \frac{297 - 275}{275} = \frac{22}{275} = \frac{2}{25} = 0.08$$

$$r = 8%$$

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3. Solution

Answer = Option D (8%) ✅

The annual rate of interest is 8%.