What is the principal amount which earns Rs. 210 as compound interest for the second year at 5% per annum?
Rs. 4000
This question asks us to find the initial principal amount that generates a specific amount of compound interest specifically in the second year at a given annual interest rate. Let's break down how compound interest works and how to use the information provided to find the principal.
Compound interest means that the interest earned in the first year is added to the principal, and then in the second year, interest is calculated on this new, larger amount. The interest earned in the second year is the interest on the original principal plus the interest on the interest earned in the first year.
Let the principal amount be \( P \).
The annual rate of interest is \( R = 5\% \).
Step 1: Calculate Interest for the First Year (I1)
The interest for the first year is simple interest on the principal \( P \) at \( 5\% \) for 1 year.
\( I_1 = P \times \frac{R}{100} \times 1 \)
\( I_1 = P \times \frac{5}{100} \times 1 = 0.05P \)
Step 2: Calculate the Amount at the end of the First Year
This is the principal plus the interest earned in the first year.
Amount after 1 year \( = P + I_1 = P + 0.05P = 1.05P \)
Step 3: Calculate Interest for the Second Year (I2)
The interest for the second year is the simple interest on the amount at the end of the first year (\( 1.05P \)) at \( 5\% \) for 1 year.
\( I_2 = (1.05P) \times \frac{R}{100} \times 1 \)
\( I_2 = (1.05P) \times \frac{5}{100} \times 1 = 1.05P \times 0.05 \)
\( I_2 = 0.0525P \)
Step 4: Use the Given Information to Find P
We are given that the compound interest for the second year is Rs. 210.
So, \( I_2 = 210 \).
We have the equation: \( 0.0525P = 210 \)
Step 5: Solve for P
To find \( P \), divide 210 by 0.0525.
\( P = \frac{210}{0.0525} \)
To make the division easier, we can multiply the numerator and denominator by 10000 to remove the decimal:
\( P = \frac{210 \times 10000}{0.0525 \times 10000} = \frac{2100000}{525} \)
Now, simplify the fraction. We can divide both numerator and denominator by common factors. For example, divide by 25:
\( 2100000 \div 25 = 84000 \)
\( 525 \div 25 = 21 \)
So, \( P = \frac{84000}{21} \)
Now, divide 84000 by 21:
\( P = \frac{84}{21} \times 1000 = 4 \times 1000 = 4000 \)
The principal amount is Rs. 4000.
Let's verify this:
| Year | Starting Amount | Interest (5%) | Ending Amount |
|---|---|---|---|
| 1 | Rs. 4000 | \( 4000 \times \frac{5}{100} = \) Rs. 200 | \( 4000 + 200 = \) Rs. 4200 |
| 2 | Rs. 4200 | \( 4200 \times \frac{5}{100} = \) Rs. 210 | \( 4200 + 210 = \) Rs. 4410 |
The interest earned in the second year is indeed Rs. 210, which matches the information given in the question. Therefore, the principal amount is Rs. 4000.
| Concept | Explanation | Formula (P=Principal, R=Rate, n=Time) |
|---|---|---|
| Simple Interest (SI) | Interest calculated only on the principal amount. | \( SI = \frac{P \times R \times n}{100} \) |
| Compound Interest (CI) | Interest calculated on the principal and accumulated interest from previous periods. | \( \text{Amount (A)} = P \left(1 + \frac{R}{100}\right)^n \) \( CI = A - P \) |
| CI for a Specific Year (k) | Interest earned only during the k-th year. Calculated as Amount after k years - Amount after (k-1) years. | \( CI_{\text{year k}} = P \left(1 + \frac{R}{100}\right)^k - P \left(1 + \frac{R}{100}\right)^{k-1} \) |
It's important to understand the difference between simple and compound interest. Simple interest is a fixed amount calculated yearly on the original principal. Compound interest grows faster because the interest earned is reinvested, earning interest itself. In this problem, the interest for the second year is higher than the interest for the first year (Rs. 210 vs Rs. 200, if P=4000) because it includes interest on the first year's interest. This compounding effect is key to financial growth over time.
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