Question

For the same principal amount, the compound interest for two years at 5% per annum exceeds the simple interest for three years at 3% per annum by Rs 1125. Then the principal amount in rupees is ?
[This Question was asked as TITA]

• A. Rs. 60000
• B. Rs. 70000
• C. Rs. 100000
• D. Rs. 90000

Answer 1

The Correct Option is D

To determine the principal amount, we need to calculate both compound interest (CI) and simple interest (SI) and use the given information that the difference between them is Rs 1125. For compound interest compounded annually with amount \( A = P(1+\frac{r}{100})^n \), and for simple interest, \( A = P + \frac{P \times r \times t}{100} \). For CI: \[ A_{CI} = P \left(1 + \frac{5}{100}\right)^2 = P \left(1.05\right)^2 = 1.1025P \] This makes the compound interest \( CI = A_{CI} - P = 1.1025P - P = 0.1025P \). For SI over three years: \[ A_{SI} = P + \frac{P \times 3 \times 3}{100} = P + 0.09P = 1.09P \] This gives simple interest as \( SI = A_{SI} - P = 0.09P \). According to the problem \( 0.1025P - 0.09P = 1125 \). This results in: \[ 0.0125P = 1125 \] Solving for \( P \): \[ P = \frac{1125}{0.0125} = 90000 \] Therefore, the principal amount is Rs 90000.

Answer 2

Let the principal be \( P \). 

Compound Interest (CI) for 2 years at 5% per annum:
The formula for CI is:
\( \text{CI} = P \left(1 + \frac{r}{100}\right)^n - P \)
Here, \( r = 5\% \), and \( n = 2 \)
So,
\( \text{CI} = P(1.05)^2 - P = P(1.1025 - 1) = 0.1025P \)

Simple Interest (SI) for 3 years at 3% per annum:
The formula for SI is:
\( \text{SI} = \frac{P \times R \times T}{100} \)
Where \( R = 3\% \), \( T = 3 \) years
So,
\( \text{SI} = P \times \frac{3 \times 3}{100} = 0.09P \)

Given:
Difference between CI and SI = ₹1125
So,
\( \text{CI} - \text{SI} = 0.1025P - 0.09P = 0.0125P \)
Hence,
\( 0.0125P = 1125 \)

Solving for \( P \):
\( \frac{125}{10000}P = 1125 \)
\( P = \frac{1125 \times 10000}{125} = 90000 \)

Therefore, the principal is ₹90000.
Correct option: (D) ₹90000