Mr. Richard has a recurring deposite account in a bank for 3 years at 7.5% p.a. simple interest. If he gets Rs 8325 as interest at the time of maturity, find the monthly deposite.

Answer

To find the monthly deposit in Mr. Richard's recurring deposit account, we can follow these steps:

### Step 1: Understand the given information
- Total interest earned (I) = Rs. 8325
- Time period (T) = 3 years
- Rate of interest (R) = 7.5% per annum
- Total number of months (N) = 3 years × 12 months/year = 36 months

### Step 2: Use the formula for interest in a recurring deposit
The formula for the interest earned in a recurring deposit account is given by:
\[ I = \frac{P \times R \times N \times (N + 1)}{2400} \]
Where:
- \(I\) = Total interest
- \(P\) = Monthly deposit
- \(R\) = Rate of interest per annum
- \(N\) = Total number of months

### Step 3: Substitute the known values into the formula
Substituting the known values into the formula:
\[ 8325 = \frac{P \times 7.5 \times 36 \times (36 + 1)}{2400} \]

### Step 4: Simplify the equation
Calculating \(N + 1\):
\[ N + 1 = 36 + 1 = 37 \]
Now substituting this back into the equation:
\[ 8325 = \frac{P \times 7.5 \times 36 \times 37}{2400} \]

### Step 5: Calculate the constants
Calculating \(7.5 \times 36 \times 37\):
\[ 7.5 \times 36 = 270 \]
\[ 270 \times 37 = 9990 \]
So, we have:
\[ 8325 = \frac{P \times 9990}{2400} \]

### Step 6: Cross-multiply to solve for P
Cross-multiplying gives:
\[ 8325 \times 2400 = P \times 9990 \]
Calculating \(8325 \times 2400\):
\[ 8325 \times 2400 = 19980000 \]
So we have:
\[ 19980000 = P \times 9990 \]

### Step 7: Solve for P
Now, divide both sides by 9990:
\[ P = \frac{19980000}{9990} \]
Calculating this gives:
\[ P \approx 2000 \]

### Conclusion
The monthly deposit \(P\) is Rs. 2000.