A person deposited a certain money in bank. Bank offers him 10% per annum compounded annually. If he deposits₹3500 at the end of 1st year and withdraws₹5000 at the end of 2nd year. Finally, at the end of 3rd year the amount] he gets₹ 18700. Then find his initial investment.

Options

A. ₹ 13700
B. ₹ 16000
C. ₹ 17200
D. ₹ 15000

Answer

To solve the problem step by step, let's denote the initial investment as \( P \).

### Step 1: Calculate the amount at the end of the first year
The interest rate is 10% per annum compounded annually.
At the end of the first year, the amount \( A_1 \) can be calculated using the formula:
\[ A_1 = P + \text{Interest} = P + \left(\frac{10}{100} \times P\right) = P + 0.1P = 1.1P \]

**Hint:** Remember that the interest is calculated on the initial principal for the first year.

### Step 2: Add the deposit made at the end of the first year
At the end of the first year, the person deposits ₹3500. Therefore, the new principal for the second year becomes:
\[ P_2 = A_1 + 3500 = 1.1P + 3500 \]

**Hint:** Ensure to add the new deposit to the amount at the end of the first year.

### Step 3: Calculate the amount at the end of the second year
Now, we calculate the amount at the end of the second year \( A_2 \):
\[ A_2 = P_2 + \text{Interest} = (1.1P + 3500) + \left(\frac{10}{100} \times (1.1P + 3500)\right) \]
\[ A_2 = (1.1P + 3500) + 0.1(1.1P + 3500) = (1.1P + 3500) + 0.11P + 350 = 1.21P + 3850 \]

**Hint:** Remember to calculate interest on the new principal which includes the deposit.

### Step 4: Withdraw ₹5000 at the end of the second year
After withdrawing ₹5000 at the end of the second year, the new principal for the third year becomes:
\[ P_3 = A_2 - 5000 = (1.21P + 3850) - 5000 = 1.21P - 1150 \]

**Hint:** Subtract the withdrawal from the amount at the end of the second year.

### Step 5: Calculate the amount at the end of the third year
Now, we calculate the amount at the end of the third year \( A_3 \):
\[ A_3 = P_3 + \text{Interest} = (1.21P - 1150) + \left(\frac{10}{100} \times (1.21P - 1150)\right) \]
\[ A_3 = (1.21P - 1150) + 0.1(1.21P - 1150) = (1.21P - 1150) + 0.121P - 115 = 1.331P - 1265 \]

**Hint:** Again, apply the interest to the new principal after the withdrawal.

### Step 6: Set the equation for the final amount
According to the problem, the amount at the end of the third year is ₹18700. Thus, we can set up the equation:
\[ 1.331P - 1265 = 18700 \]

**Hint:** This equation relates the final amount to the initial investment.

### Step 7: Solve for \( P \)
Now, we will solve for \( P \):
\[ 1.331P = 18700 + 1265 \]
\[ 1.331P = 19965 \]
\[ P = \frac{19965}{1.331} \approx 15000 \]

**Hint:** Ensure to perform the division carefully to find the initial investment.

### Conclusion
The initial investment \( P \) is ₹15000.

**Final Answer:** ₹15000