Arun has a certain sum deposited in a bank at `5%` per annum. The bank increases the rate of interest from `5%` to `6%`. Arun deposits `₹2000` more in his account. The annual interest received by him now is `₹220` more than before. Find his original deposite.

Answer

To find Arun's original deposit, we will follow these steps:

### Step 1: Define the variables
Let the original deposit be \( P \).

### Step 2: Calculate the initial simple interest
The initial rate of interest is \( 5\% \) per annum. The formula for simple interest (SI) is given by:
\[ \text{SI} = \frac{P \times R \times T}{100} \]
Where:
- \( P \) = Principal amount (original deposit)
- \( R \) = Rate of interest (5%)
- \( T \) = Time (1 year)

So, the initial simple interest is:
\[ \text{SI}_{\text{initial}} = \frac{P \times 5 \times 1}{100} = \frac{5P}{100} \]

### Step 3: Calculate the new simple interest after the changes
After the bank increases the interest rate to \( 6\% \) and Arun deposits an additional \( ₹2000 \), the new principal becomes \( P + 2000 \). The new simple interest is calculated as:
\[ \text{SI}_{\text{new}} = \frac{(P + 2000) \times 6 \times 1}{100} = \frac{6(P + 2000)}{100} \]

### Step 4: Set up the equation based on the information given
According to the problem, the new simple interest is \( ₹220 \) more than the initial simple interest. Therefore, we can write the equation:
\[ \text{SI}_{\text{new}} - \text{SI}_{\text{initial}} = 220 \]
Substituting the expressions we found:
\[ \frac{6(P + 2000)}{100} - \frac{5P}{100} = 220 \]

### Step 5: Simplify the equation
To eliminate the fraction, multiply the entire equation by \( 100 \):
\[ 6(P + 2000) - 5P = 22000 \]
Expanding the left side:
\[ 6P + 12000 - 5P = 22000 \]
Combining like terms:
\[ P + 12000 = 22000 \]

### Step 6: Solve for \( P \)
Subtract \( 12000 \) from both sides:
\[ P = 22000 - 12000 \]
\[ P = 10000 \]

### Conclusion
Arun's original deposit is \( ₹10000 \).

---