A man received Rs.3456 when he invests P at 12% p.a. at Si for 3 years. If he invested Rs. (P + 1400) at15% p.a. at Cl compounding annually for 2 years then find the interest received by him.

Options

A. Rs.3515.5
B. Rs.3960.5
C. Rs.3725.5
D. Rs.3547.5

Answer

To solve the problem step by step, we will break it down into two parts: calculating the principal amount (P) using simple interest, and then calculating the compound interest for the second investment.

### Step 1: Calculate the Principal Amount (P)

Given:
- Total amount received after 3 years (A) = Rs. 3456
- Rate of interest (R) = 12% per annum
- Time (T) = 3 years

Using the formula for Simple Interest (SI):
\[ SI = \frac{P \times R \times T}{100} \]
The total amount (A) is given by:
\[ A = P + SI \]
Substituting the SI formula into the equation:
\[ A = P + \frac{P \times R \times T}{100} \]
This can be rewritten as:
\[ A = P \left(1 + \frac{R \times T}{100}\right) \]
Substituting the known values:
\[ 3456 = P \left(1 + \frac{12 \times 3}{100}\right) \]
Calculating the term in the parentheses:
\[ 1 + \frac{36}{100} = 1.36 \]
Now, substituting back:
\[ 3456 = P \times 1.36 \]
To find P, divide both sides by 1.36:
\[ P = \frac{3456}{1.36} \approx 2544 \]

### Step 2: Calculate the Compound Interest (CI)

Now, the new principal amount for the second investment is:
\[ P + 1400 = 2544 + 1400 = 3944 \]
Given:
- Rate of interest (R) = 15% per annum
- Time (T) = 2 years

Using the formula for Compound Interest (CI):
\[ CI = P \left(1 + \frac{R}{100}\right)^T - P \]
Substituting the values:
\[ CI = 3944 \left(1 + \frac{15}{100}\right)^2 - 3944 \]
Calculating the term in the parentheses:
\[ 1 + \frac{15}{100} = 1.15 \]
Now, squaring it:
\[ (1.15)^2 = 1.3225 \]
Now substituting back into the CI formula:
\[ CI = 3944 \times 1.3225 - 3944 \]
Calculating the first term:
\[ 3944 \times 1.3225 \approx 5218.78 \]
Now, subtracting P:
\[ CI = 5218.78 - 3944 \approx 1274.78 \]

### Final Answer:
The interest received by him from the second investment is approximately Rs. 1274.78.