A sum is invested at compound interest payable annually. The interest in two successive years was Rs. 500 and Rs. 540. The sum is-

Options

A. Rs.3750
B. Rs.5000
C. Rs.5600
D. Rs.6250

Answer

To solve the problem, we need to find the principal amount (sum) invested based on the compound interest earned in two successive years. Let's break it down step by step.

### Step 1: Understand the Given Information
We know that:
- The interest for the first year (I1) = Rs. 500
- The interest for the second year (I2) = Rs. 540

### Step 2: Relate the Interests to the Principal and Rate of Interest
The interest for the first year can be expressed as:
\[ I1 = P \times r \]
Where:
- \( P \) = Principal amount
- \( r \) = Rate of interest (in decimal)

For the second year, the interest is calculated on the amount after the first year, which includes the principal and the interest earned in the first year:
\[ I2 = (P + I1) \times r \]
Substituting \( I1 \):
\[ I2 = (P + 500) \times r \]

### Step 3: Set Up the Equations
From the above, we can set up two equations:
1. \( 500 = P \times r \) (Equation 1)
2. \( 540 = (P + 500) \times r \) (Equation 2)

### Step 4: Solve for the Rate of Interest
From Equation 1, we can express \( r \) in terms of \( P \):
\[ r = \frac{500}{P} \]

Now, substitute this expression for \( r \) into Equation 2:
\[ 540 = (P + 500) \times \frac{500}{P} \]

### Step 5: Simplify the Equation
Multiply both sides by \( P \) to eliminate the fraction:
\[ 540P = (P + 500) \times 500 \]
\[ 540P = 500P + 250000 \]

### Step 6: Rearrange and Solve for \( P \)
Now, rearranging the equation gives:
\[ 540P - 500P = 250000 \]
\[ 40P = 250000 \]
\[ P = \frac{250000}{40} \]
\[ P = 6250 \]

### Conclusion
The sum invested is Rs. 6250.

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