If the compound interest on a certain sum of 2 years at 4% per annum is ₹1530. What would be the simple interest on the same sum for the same period and at the same rate?
The question provides information about the compound interest earned on a sum of money over a specific period at a given rate and asks for the simple interest on the same sum for the same period and rate. To solve this, we first need to determine the principal sum on which the interest is calculated.
Find the Simple Interest (SI) on the same principal, for the same period, and at the same rate.
The formula for the amount (A) with compound interest is:
\(A = P \left(1 + \frac{R}{100}\right)^T\)
The compound interest (CI) is the difference between the amount and the principal:
\(CI = A - P = P \left(1 + \frac{R}{100}\right)^T - P\)
\(CI = P \left[ \left(1 + \frac{R}{100}\right)^T - 1 \right]\)
Substitute the given values into the formula:
\(1530 = P \left[ \left(1 + \frac{4}{100}\right)^2 - 1 \right]\)
\(1530 = P \left[ \left(1 + 0.04\right)^2 - 1 \right]\)
\(1530 = P \left[ \left(1.04\right)^2 - 1 \right]\)
Calculate \((1.04)^2\):
\(1.04 \times 1.04 = 1.0816\)
Now substitute this back into the equation:
\(1530 = P \left[ 1.0816 - 1 \right]\)
\(1530 = P \times 0.0816\)
To find P, divide 1530 by 0.0816:
\(P = \frac{1530}{0.0816}\)
Let's perform the division:
\(P = 18750\)
So, the principal sum is ₹18750.
Now that we have the principal (P), rate (R), and time (T), we can calculate the simple interest using the formula:
\(SI = \frac{P \times R \times T}{100}\)
Substitute the values: P = ₹18750, R = 4%, T = 2 years.
\(SI = \frac{18750 \times 4 \times 2}{100}\)
\(SI = \frac{18750 \times 8}{100}\)
\(SI = \frac{150000}{100}\)
\(SI = 1500\)
The simple interest on the sum for the same period and at the same rate is ₹1500.
The simple interest on the sum of ₹18750 for 2 years at 4% per annum is ₹1500.
| Aspect | Compound Interest (CI) | Simple Interest (SI) |
|---|---|---|
| Calculation Basis | Calculated on the principal amount plus accumulated interest from previous periods. | Calculated only on the original principal amount. |
| Interest Growth | Grows faster over time (for T > 1 year) as interest earns interest. | Grows linearly over time; the amount of interest is the same each period. |
| Formula (Amount) | \(A = P(1 + \frac{R}{100})^T\) | \(A = P + SI = P + \frac{PRT}{100}\) |
| Formula (Interest) | \(CI = P[(1 + \frac{R}{100})^T - 1]\) | \(SI = \frac{PRT}{100}\) |
For a principal amount P, rate R, and time T:
In this problem, the difference between CI and SI is \(1530 - 1500 = 30\). Using the difference formula for 2 years: Difference \(= 18750 \times (\frac{4}{100})^2 = 18750 \times (0.04)^2 = 18750 \times 0.0016 = 30\). This confirms our calculations.
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