Mr. Anand deposited a total amount of Rs. 65000 in three different schemes A, B and C with rates of interset 12 % p.a. 16 % p.a. and 18 % p.a. respectively and earned a total interset of Rs. 10180 in one year. If the amount invested in scheme A was 72 % of the amount invested in scheme C , what was the amount invested in scheme B ?

Options

A. Rs. 25000
B. Rs. 22000
C. Rs. 18000
D. Rs. 16000

Answer

To solve the problem step by step, we will use the information given about the investments in schemes A, B, and C.

### Step 1: Define the Variables
Let:
- Amount invested in scheme A = \( A \)
- Amount invested in scheme B = \( B \)
- Amount invested in scheme C = \( C \)

From the problem, we know:
1. \( A + B + C = 65000 \) (total investment)
2. The interest rates are:
- Scheme A: 12% per annum
- Scheme B: 16% per annum
- Scheme C: 18% per annum
3. The total interest earned in one year is \( 10180 \).
4. \( A = 0.72C \) (amount invested in scheme A is 72% of the amount invested in scheme C).

### Step 2: Express A in terms of C
From the relationship \( A = 0.72C \), we can express \( A \) as:
\[ A = \frac{72}{100}C = \frac{18}{25}C \]

### Step 3: Substitute A in the Total Investment Equation
Now, substituting \( A \) in the total investment equation:
\[ \frac{18}{25}C + B + C = 65000 \]

Combining terms:
\[ B + \frac{18}{25}C + \frac{25}{25}C = 65000 \]
\[ B + \frac{43}{25}C = 65000 \]

### Step 4: Express B in terms of C
Rearranging the equation gives:
\[ B = 65000 - \frac{43}{25}C \]

### Step 5: Set Up the Interest Equation
Now, we can set up the interest equation using the interest rates:
\[ \text{Interest from A} + \text{Interest from B} + \text{Interest from C} = 10180 \]
This translates to:
\[ 0.12A + 0.16B + 0.18C = 10180 \]

Substituting \( A = \frac{18}{25}C \):
\[ 0.12\left(\frac{18}{25}C\right) + 0.16B + 0.18C = 10180 \]
\[ \frac{2.16}{25}C + 0.16B + 0.18C = 10180 \]

### Step 6: Combine Like Terms
Combining the terms involving \( C \):
\[ \left(\frac{2.16}{25} + 0.18\right)C + 0.16B = 10180 \]
Converting \( 0.18 \) to a fraction:
\[ 0.18 = \frac{18}{100} = \frac{9}{50} = \frac{45}{250} \]
So,
\[ \frac{2.16}{25} = \frac{2.16 \times 10}{250} = \frac{21.6}{250} \]
Thus,
\[ \left(\frac{21.6 + 45}{250}\right)C + 0.16B = 10180 \]
\[ \frac{66.6}{250}C + 0.16B = 10180 \]

### Step 7: Substitute B from Earlier
Now, substitute \( B = 65000 - \frac{43}{25}C \) into the interest equation:
\[ \frac{66.6}{250}C + 0.16\left(65000 - \frac{43}{25}C\right) = 10180 \]

### Step 8: Solve for C
Expanding the equation:
\[ \frac{66.6}{250}C + 10400 - \frac{6.88}{25}C = 10180 \]
Combining terms and solving for \( C \):
\[ \left(\frac{66.6 - 68.8}{250}\right)C + 10400 = 10180 \]
\[ -\frac{2.2}{250}C = -220 \]
\[ C = \frac{220 \times 250}{2.2} = 25000 \]

### Step 9: Find A and B
Now that we have \( C \):
\[ A = 0.72C = 0.72 \times 25000 = 18000 \]
\[ B = 65000 - A - C = 65000 - 18000 - 25000 = 22000 \]

### Final Answer
The amount invested in scheme B is **Rs. 22000**.