Question

Find the compound interest on ₹10,000 at 10% per annum for 2 years, compounded annually.

• A. \( \text{₹}2,000 \)
• B. \( \text{₹}2,100 \)
• C. \( \text{₹}2,200 \)
• D. \( \text{₹}2,310 \)

Hint:

For a 2-year compounding timeline, you can bypass fractional equations entirely by using the net effective percentage formula: \[ \text{Net Effective Rate} = R + R + \frac{R \times R}{100} \] With a given interest rate of \(10%\): \[ \text{Effective Rate} = 10 + 10 + \frac{10 \times 10}{100} = 20 + 1 = 21% \] Simply find \(21%\) of the principal to get your answer directly: \[ 21% \text{ of } 10000 = 21 \times 100 = \text{₹}2,100 \]

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Compound interest is the interest calculated on the initial principal value plus all accumulated interest from previous periods. Unlike simple interest, which remains constant each year, compound interest grows exponentially because the earner makes "interest on interest."

Step 2: Key Formula or Approach:
1. \(\text{Amount } (A) = P \left(1 + \frac{R}{100}\right)^n\)
2. \(\text{Compound Interest } (CI) = A - P\)
where \(P\) is the principal amount, \(R\) is the annual rate of interest, and \(n\) is the time duration in years.

Step 3: Detailed Explanation:
Given the parameters: Principal (\(P\)) = ₹\(10,000\)
Rate (\(R\)) = \(10%\) per annum
Time (\(n\)) = \(2\) years
Substitute these values into the total maturity amount formula: \[ A = 10000 \left(1 + \frac{10}{100}\right)^2 \] \[ A = 10000 \left(1 + \frac{1}{10}\right)^2 \] \[ A = 10000 \left(\frac{11}{10}\right)^2 \] \[ A = 10000 \left(\frac{121}{100}\right) \] Cancel out the two zeros from both the numerator and the denominator: \[ A = 100 \times 121 = \text{₹}12,100 \] Now, calculate the absolute net compound interest accumulated over the 2-year period: \[ CI = A - P \] \[ CI = 12100 - 10000 = \text{₹}2,100 \]

Step 4: Final Answer:
The compound interest is ₹2,100.