Dharam invested Rs.10000 in two schemes for two years and both schemes offer R% S.I. If difference between S.I. earned on both schemes is Rs.480 and ratio of interest earned from both schemes is 3 : 2. Then, find the value of R.

Options

A. A)0.15
B. B)0.1
C. C)0.2
D. D)0.12

Answer

To solve the problem step by step, we will follow the information given in the question and apply the concepts of simple interest.

### Step 1: Understanding the Problem
Dharam invested Rs. 10,000 in two schemes for 2 years, both offering R% simple interest. The difference in the interest earned from both schemes is Rs. 480, and the ratio of interest earned is 3:2.

### Step 2: Setting Up the Ratios
Let’s denote the interest earned from Scheme A as \( I_A \) and from Scheme B as \( I_B \). According to the problem:
- The ratio of interest earned is given as:
\[ \frac{I_A}{I_B} = \frac{3}{2} \]
This implies:
\[ I_A = 3x \quad \text{and} \quad I_B = 2x \]
for some value \( x \).

### Step 3: Finding the Total Interest
The difference between the interests earned from both schemes is given as Rs. 480:
\[ I_A - I_B = 480 \]
Substituting the values of \( I_A \) and \( I_B \):
\[ 3x - 2x = 480 \]
This simplifies to:
\[ x = 480 \]

### Step 4: Calculating Individual Interests
Now we can find \( I_A \) and \( I_B \):
\[ I_A = 3x = 3 \times 480 = 1440 \]
\[ I_B = 2x = 2 \times 480 = 960 \]

### Step 5: Finding the Principal Amounts
Since the ratio of the principal amounts is the same as the ratio of the interests (3:2), let’s denote the principal amounts for Scheme A and Scheme B as \( P_A \) and \( P_B \) respectively:
\[ \frac{P_A}{P_B} = \frac{3}{2} \]
Let \( P_A = 3k \) and \( P_B = 2k \). The total principal is:
\[ P_A + P_B = 3k + 2k = 5k = 10000 \]
Thus, we can solve for \( k \):
\[ k = \frac{10000}{5} = 2000 \]
Now substituting back to find \( P_A \) and \( P_B \):
\[ P_A = 3k = 3 \times 2000 = 6000 \]
\[ P_B = 2k = 2 \times 2000 = 4000 \]

### Step 6: Using the Simple Interest Formula
The formula for simple interest is:
\[ I = \frac{P \times R \times T}{100} \]
For Scheme A:
\[ I_A = \frac{6000 \times R \times 2}{100} = 120R \]
For Scheme B:
\[ I_B = \frac{4000 \times R \times 2}{100} = 80R \]

### Step 7: Setting Up the Equation
From our earlier calculations, we know:
\[ I_A = 1440 \quad \text{and} \quad I_B = 960 \]
Thus, we can set up the equations:
\[ 120R = 1440 \quad \text{and} \quad 80R = 960 \]

### Step 8: Solving for R
From \( 120R = 1440 \):
\[ R = \frac{1440}{120} = 12 \]
From \( 80R = 960 \):
\[ R = \frac{960}{80} = 12 \]
Both methods confirm that \( R = 12\% \).

### Step 9: Converting to Decimal
To express \( R \) in decimal form:
\[ R = 12\% = \frac{12}{100} = 0.12 \]

### Final Answer
The value of \( R \) is \( 0.12 \).

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