A man invested Rs.P in three different schemes - A, B and C in the ratio 2:1:3 respectively. He invested in scheme A at the rate of 10% p.a. at Si for 2 years, in scheme B at the rate of 5% p.a at C.I compounded annually for 2 years and in scheme C at the rate of 6% p.a at CI compounded half yearly for 1 year and received total interest of Rs. 6852. Find the value of P.

Options

A. Rs.60000
B. Rs.72000
C. Rs.48000
D. Rs.120000

Answer

To solve the problem step by step, we will follow the given information and apply the formulas for simple interest and compound interest.

### Step 1: Determine the amounts invested in each scheme
The man invested in three schemes A, B, and C in the ratio 2:1:3. Let the total investment be Rs. P. Therefore, the amounts invested in each scheme are:
- Scheme A: \( \frac{2}{6}P = \frac{P}{3} \)
- Scheme B: \( \frac{1}{6}P = \frac{P}{6} \)
- Scheme C: \( \frac{3}{6}P = \frac{P}{2} \)

### Step 2: Calculate the interest from Scheme A
Scheme A has a principal of \( \frac{P}{3} \), a rate of 10% per annum, and the time period is 2 years. The formula for simple interest (SI) is:
\[ SI = \frac{P \times R \times T}{100} \]
Substituting the values:
\[ SI_A = \frac{\left(\frac{P}{3}\right) \times 10 \times 2}{100} = \frac{20P}{300} = \frac{P}{15} \]

### Step 3: Calculate the interest from Scheme B
Scheme B has a principal of \( \frac{P}{6} \), a rate of 5% per annum, and the time period is 2 years. The formula for compound interest (CI) is:
\[ A = P \left(1 + \frac{R}{100}\right)^T \]
Substituting the values:
\[ A_B = \frac{P}{6} \left(1 + \frac{5}{100}\right)^2 = \frac{P}{6} \left(1.05\right)^2 = \frac{P}{6} \times 1.1025 = \frac{1.1025P}{6} \]
The interest from Scheme B is:
\[ SI_B = A_B - \text{Principal} = \frac{1.1025P}{6} - \frac{P}{6} = \frac{1.1025P - P}{6} = \frac{0.1025P}{6} \]

### Step 4: Calculate the interest from Scheme C
Scheme C has a principal of \( \frac{P}{2} \), a rate of 6% per annum compounded half-yearly for 1 year. The rate for half-yearly compounding is \( \frac{6}{2} = 3\% \) and the time becomes \( 2 \) half-years. Using the compound interest formula:
\[ A_C = \frac{P}{2} \left(1 + \frac{3}{100}\right)^2 = \frac{P}{2} \left(1.03\right)^2 = \frac{P}{2} \times 1.0609 = \frac{1.0609P}{2} \]
The interest from Scheme C is:
\[ SI_C = A_C - \text{Principal} = \frac{1.0609P}{2} - \frac{P}{2} = \frac{1.0609P - P}{2} = \frac{0.0609P}{2} \]

### Step 5: Sum up the total interest
The total interest received from all three schemes is given as Rs. 6852. Therefore, we can write:
\[ SI_A + SI_B + SI_C = 6852 \]
Substituting the values we calculated:
\[ \frac{P}{15} + \frac{0.1025P}{6} + \frac{0.0609P}{2} = 6852 \]

### Step 6: Find a common denominator and solve for P
The common denominator for 15, 6, and 2 is 30. Rewriting each term:
\[ \frac{2P}{30} + \frac{0.1025P \times 5}{30} + \frac{0.0609P \times 15}{30} = 6852 \]
Calculating the coefficients:
\[ \frac{2P + 0.5125P + 0.9135P}{30} = 6852 \]
\[ \frac{3.426P}{30} = 6852 \]
Multiplying both sides by 30:
\[ 3.426P = 205560 \]
Finally, solving for P:
\[ P = \frac{205560}{3.426} \approx 60000 \]

### Conclusion
The value of P is Rs. 60,000.