David invested certain amount in three different schemes A, B and C with the rate of interest 10% p.a., 12% p.a and 15% p.a. respectively. If the the total interest accrued in one year was 3200 and the amount invested in Scheme C was 150 % of the amount invested in Scheme A and 240% of the amount invested in Scheme B, what was the amount invested in Scheme B?

Options

A. 5000
B. 6500
C. 8000
D. cannot be determined

Answer

To solve the problem step by step, we will define the amounts invested in schemes A, B, and C, and then set up equations based on the information given.

### Step 1: Define the variables
Let:
- Amount invested in Scheme A = \( x \)
- Amount invested in Scheme B = \( y \)
- Amount invested in Scheme C = \( z \)

### Step 2: Set up relationships based on the problem
According to the problem:
1. The amount invested in Scheme C is 150% of the amount invested in Scheme A:
\[ z = 1.5x \]
2. The amount invested in Scheme C is also 240% of the amount invested in Scheme B:
\[ z = 2.4y \]

### Step 3: Set up the equation for total interest
The total interest accrued in one year from all three schemes is given as 3200. The interest from each scheme can be calculated as follows:
- Interest from Scheme A (10% of \( x \)):
\[ \text{Interest from A} = 0.1x \]
- Interest from Scheme B (12% of \( y \)):
\[ \text{Interest from B} = 0.12y \]
- Interest from Scheme C (15% of \( z \)):
\[ \text{Interest from C} = 0.15z \]

Thus, the total interest equation can be written as:
\[ 0.1x + 0.12y + 0.15z = 3200 \]

### Step 4: Substitute \( z \) in the interest equation
From the relationships established earlier, we can substitute \( z \) in terms of \( x \) or \( y \). Let's use \( z = 1.5x \):
\[ 0.1x + 0.12y + 0.15(1.5x) = 3200 \]
This simplifies to:
\[ 0.1x + 0.12y + 0.225x = 3200 \]
Combining like terms:
\[ 0.325x + 0.12y = 3200 \]

### Step 5: Substitute \( z \) in terms of \( y \)
Now, we can also express \( z \) in terms of \( y \) using \( z = 2.4y \):
\[ 0.1x + 0.12y + 0.15(2.4y) = 3200 \]
This simplifies to:
\[ 0.1x + 0.12y + 0.36y = 3200 \]
Combining like terms:
\[ 0.1x + 0.48y = 3200 \]

### Step 6: Solve the equations
Now we have two equations:
1. \( 0.325x + 0.12y = 3200 \) (Equation 1)
2. \( 0.1x + 0.48y = 3200 \) (Equation 2)

We can solve these equations simultaneously. Let's express \( x \) in terms of \( y \) from Equation 1:
\[ 0.325x = 3200 - 0.12y \implies x = \frac{3200 - 0.12y}{0.325} \]

Substituting this expression for \( x \) into Equation 2:
\[ 0.1\left(\frac{3200 - 0.12y}{0.325}\right) + 0.48y = 3200 \]

### Step 7: Solve for \( y \)
Now, we can solve for \( y \). This will require some algebraic manipulation. After solving, we find the value of \( y \).

### Final Calculation
After performing the calculations, we find:
\[ y = 5000 \]

### Conclusion
Thus, the amount invested in Scheme B is **5000**.

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