A sum of money at simple interest amounts to Rs. 14,160 in 3 years. If the rate of interset is increased to 25 % the same sum amounts to Rs. 14,700 in the same time. The rate of interest is

Options

A. 0.05
B. `5 (1)/(2)` %
C. 0.06
D. 0.07

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Define the variables
Let the principal amount be \( P \) and the rate of interest be \( R \% \).

### Step 2: Write the equations based on the information provided
1. The amount after 3 years at the original rate \( R \% \) is:
\[ A_1 = P + \text{SI} = P + \frac{P \times R \times 3}{100} = 14,160 \]
This can be rearranged to:
\[ P \left(1 + \frac{3R}{100}\right) = 14,160 \quad \text{(Equation 1)} \]

2. When the rate is increased to 25%, the amount after 3 years is:
\[ A_2 = P + \text{SI} = P + \frac{P \times 25 \times 3}{100} = 14,700 \]
This can be rearranged to:
\[ P \left(1 + \frac{75}{100}\right) = 14,700 \quad \text{(Equation 2)} \]

### Step 3: Simplify Equation 2
From Equation 2:
\[ P \left(1 + 0.75\right) = 14,700 \]
\[ P \times 1.75 = 14,700 \]
\[ P = \frac{14,700}{1.75} = 8,400 \]

### Step 4: Substitute \( P \) back into Equation 1
Now substitute \( P \) into Equation 1:
\[ 8400 \left(1 + \frac{3R}{100}\right) = 14,160 \]
\[ 1 + \frac{3R}{100} = \frac{14,160}{8400} \]
\[ 1 + \frac{3R}{100} = 1.68 \]
\[ \frac{3R}{100} = 1.68 - 1 \]
\[ \frac{3R}{100} = 0.68 \]

### Step 5: Solve for \( R \)
Multiply both sides by 100:
\[ 3R = 68 \]
Now divide by 3:
\[ R = \frac{68}{3} \approx 22.67\% \]

### Step 6: Conclusion
The rate of interest \( R \) is approximately \( 22.67\% \).