A sum of money amounts to Rs 5,200 in 5 years and to Rs 5,680 in 7 years at simple interset. The rate of interst per annum is

Options

A. 3%
B. 4%
C. 5%
D. 6%

Answer

To solve the problem of finding the rate of interest per annum given the amounts at different times, we can follow these steps:

### Step 1: Identify the given information
- Amount after 5 years (A1) = Rs 5,200
- Amount after 7 years (A2) = Rs 5,680
- Time period for A1 (T1) = 5 years
- Time period for A2 (T2) = 7 years

### Step 2: Calculate the simple interest for the two time periods
The simple interest (SI) can be calculated using the formula:
\[ SI = A - P \]
Where \( A \) is the amount and \( P \) is the principal.

Let \( P \) be the principal amount.

From the first amount:
\[ SI_1 = A_1 - P = 5200 - P \]
For 5 years:
\[ SI_1 = \frac{R \times P \times T_1}{100} \]
\[ SI_1 = \frac{R \times P \times 5}{100} \]

From the second amount:
\[ SI_2 = A_2 - P = 5680 - P \]
For 7 years:
\[ SI_2 = \frac{R \times P \times T_2}{100} \]
\[ SI_2 = \frac{R \times P \times 7}{100} \]

### Step 3: Set up the equations
From the two expressions for simple interest, we can write:
1. \( 5200 - P = \frac{R \times P \times 5}{100} \) (Equation 1)
2. \( 5680 - P = \frac{R \times P \times 7}{100} \) (Equation 2)

### Step 4: Subtract the two equations
Subtract Equation 1 from Equation 2:
\[ (5680 - P) - (5200 - P) = \frac{R \times P \times 7}{100} - \frac{R \times P \times 5}{100} \]
\[ 5680 - 5200 = \frac{R \times P \times (7 - 5)}{100} \]
\[ 480 = \frac{R \times P \times 2}{100} \]

### Step 5: Solve for \( R \times P \)
Rearranging gives:
\[ R \times P = \frac{480 \times 100}{2} \]
\[ R \times P = 24000 \]

### Step 6: Substitute back to find \( R \)
Now we can substitute \( P \) back into one of the original equations. We can use Equation 1:
\[ 5200 - P = \frac{R \times P \times 5}{100} \]
Substituting \( R \times P = 24000 \):
\[ 5200 - P = \frac{24000 \times 5}{100} \]
\[ 5200 - P = 1200 \]
\[ P = 5200 - 1200 \]
\[ P = 4000 \]

### Step 7: Calculate the rate \( R \)
Now substituting \( P \) back into \( R \times P = 24000 \):
\[ R \times 4000 = 24000 \]
\[ R = \frac{24000}{4000} \]
\[ R = 6 \]

### Conclusion
The rate of interest per annum is **6%**.

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