Simple interest for the sum of X 1230 for 2 yr is ? 10 more than the simple interest for X 1130 for the same duration. Find the rate of interest.

Options

A. 0.05
B. 0.06
C. 0.08
D. 0.02

Answer

To solve the problem step by step, we will use the formula for Simple Interest (SI):

\[ \text{SI} = \frac{P \times R \times T}{100} \]

where:
- \( P \) = Principal amount
- \( R \) = Rate of interest per annum
- \( T \) = Time in years

### Step 1: Set up the equations for Simple Interest

1. For the first case with principal \( P_1 = 1230 \) for \( T = 2 \) years:
\[ \text{SI}_1 = \frac{1230 \times R \times 2}{100} \]

2. For the second case with principal \( P_2 = 1130 \) for \( T = 2 \) years:
\[ \text{SI}_2 = \frac{1130 \times R \times 2}{100} \]

### Step 2: Write the equation based on the information given

According to the problem, the simple interest for the first principal is 10 more than that for the second principal:
\[ \text{SI}_1 = \text{SI}_2 + 10 \]

Substituting the expressions for \( \text{SI}_1 \) and \( \text{SI}_2 \):
\[ \frac{1230 \times R \times 2}{100} = \frac{1130 \times R \times 2}{100} + 10 \]

### Step 3: Simplify the equation

1. Multiply through by 100 to eliminate the fraction:
\[ 1230 \times R \times 2 = 1130 \times R \times 2 + 1000 \]

2. Distribute:
\[ 2460R = 2260R + 1000 \]

### Step 4: Solve for \( R \)

1. Rearranging the equation:
\[ 2460R - 2260R = 1000 \]
\[ 200R = 1000 \]

2. Divide both sides by 200:
\[ R = \frac{1000}{200} = 5 \]

### Conclusion

The rate of interest \( R \) is \( 5\% \).

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