Question

Find the compound interest on ₹5000 at 10% per annum for 2 years.

• A. ₹1000
• B. ₹1025
• C. ₹1050
• D. ₹1100

Hint:

For 2 years at 10%, the effective interest rate is always $21\%$ ($10 + 10 + \frac{10 \times 10}{100}$).
$21\% \text{ of } 5000 = 1050$. This is much faster for multiple-choice questions!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Compound interest is calculated on the initial principal, which also includes all of the accumulated interest of previous periods. It is "interest on interest."
Step 2: Key Formula or Approach:
The formula for Amount (\(A\)) is: \[ A = P \left(1 + \frac{r}{100}\right)^n \]
Where:
P = Principal = 5000
r = Rate = 10
n = Time = 2 years
Compound Interest (CI) = \( A - P \).
Step 3: Detailed Explanation:
1. Using the Formula:
\[ A = 5000 \left(1 + \frac{10}{100}\right)^2 \]
\[ A = 5000 (1.1)^2 \]
\[ A = 5000 \times 1.21 \]
\[ A = 6050 \]
Now, CI = \( 6050 - 5000 = 1050 \).
2. Step-by-Step Method (Manual Calculation):
Interest for the 1st year: \( 10% \text{ of } 5000 = 500 \).
Amount at end of 1st year: \( 5000 + 500 = 5500 \).
Interest for the 2nd year: \( 10% \text{ of } 5500 = 550 \).
Total interest over 2 years: \( 500 + 550 = 1050 \).
3. Successive Percentage Method:
Effective rate for 2 years at 10% = \( x + y + \frac{xy}{100} \)
\( = 10 + 10 + \frac{10 \times 10}{100} = 21% \).
CI = \( 21% \text{ of } 5000 = \frac{21}{100} \times 5000 = 1050 \).
Step 4: Final Answer:
The compound interest is 1050.