A sum of Rs. `2800` is divided into two parts in such a way that the interest on both the parts is equal. If the first part is lent at `9 %` p.a. for `5` years and second part is for `6` years at `10%` p.a. find the two sums.

Options

A. Rs. 1800, Rs.1000
B. Rs. 1600, Rs.1200
C. Rs. 1400, Rs. 1400
D. Rs. 1300, Rs. 1500

Answer

To solve the problem, we need to divide the total sum of Rs. 2800 into two parts such that the interest earned from both parts is equal. Let's denote the first part as \( P_1 \) and the second part as \( P_2 \).

### Step 1: Set up the equations
We know that:
- \( P_1 + P_2 = 2800 \) (Equation 1)

The interest for the first part \( P_1 \) at 9% per annum for 5 years can be calculated using the formula for simple interest:
\[ \text{Interest} = \frac{P \times R \times T}{100} \]
So, the interest on \( P_1 \) is:
\[ I_1 = \frac{P_1 \times 9 \times 5}{100} = \frac{45P_1}{100} = 0.45P_1 \]

For the second part \( P_2 \) at 10% per annum for 6 years, the interest is:
\[ I_2 = \frac{P_2 \times 10 \times 6}{100} = \frac{60P_2}{100} = 0.6P_2 \]

Since the interest on both parts is equal, we can set up the equation:
\[ 0.45P_1 = 0.6P_2 \quad \text{(Equation 2)} \]

### Step 2: Substitute \( P_2 \) in terms of \( P_1 \)
From Equation 1, we can express \( P_2 \) in terms of \( P_1 \):
\[ P_2 = 2800 - P_1 \]

### Step 3: Substitute \( P_2 \) in Equation 2
Now substitute \( P_2 \) in Equation 2:
\[ 0.45P_1 = 0.6(2800 - P_1) \]

### Step 4: Solve for \( P_1 \)
Expanding the equation:
\[ 0.45P_1 = 1680 - 0.6P_1 \]
Now, add \( 0.6P_1 \) to both sides:
\[ 0.45P_1 + 0.6P_1 = 1680 \]
\[ 1.05P_1 = 1680 \]
Now, divide both sides by 1.05:
\[ P_1 = \frac{1680}{1.05} = 1600 \]

### Step 5: Find \( P_2 \)
Now substitute \( P_1 \) back into Equation 1 to find \( P_2 \):
\[ P_2 = 2800 - P_1 = 2800 - 1600 = 1200 \]

### Conclusion
The two sums are:
- First part \( P_1 = Rs. 1600 \)
- Second part \( P_2 = Rs. 1200 \)