Mr. Kumar has a recurring deposit account in a bank for 4 years at 10% p.a. rate of interest. If he gets Rs 21.560 as interest at the time of maturity, find :
the monthly instalment paid by Mr. Kumar.

Answer

To find the monthly installment paid by Mr. Kumar in his recurring deposit account, we can follow these steps:

### Step 1: Identify the given values
- Interest earned (I) = Rs 21,560
- Rate of interest (r) = 10% per annum
- Duration (n) = 4 years

### Step 2: Convert the duration into months
Since the duration is given in years, we convert it to months:
\[ n = 4 \text{ years} \times 12 \text{ months/year} = 48 \text{ months} \]

### Step 3: Use the formula for interest in a recurring deposit
The formula for the interest earned in a recurring deposit is:
\[ I = \frac{P \times r \times n \times (n + 1)}{2400} \]
Where:
- \(I\) = Interest earned
- \(P\) = Monthly installment
- \(r\) = Rate of interest per annum
- \(n\) = Total number of months

### Step 4: Substitute the known values into the formula
Substituting the values we have:
\[ 21560 = \frac{P \times 10 \times 48 \times (48 + 1)}{2400} \]

### Step 5: Simplify the equation
Calculate \(48 + 1 = 49\):
\[ 21560 = \frac{P \times 10 \times 48 \times 49}{2400} \]
Now, simplify the right side:
\[ 21560 = \frac{P \times 10 \times 48 \times 49}{2400} \]
\[ = \frac{P \times 480 \times 49}{2400} \]
\[ = \frac{P \times 23520}{2400} \]

### Step 6: Cross-multiply to solve for P
Cross-multiplying gives:
\[ 21560 \times 2400 = P \times 23520 \]
Calculating \(21560 \times 2400\):
\[ = 51744000 \]
So we have:
\[ 51744000 = P \times 23520 \]

### Step 7: Solve for P
Now, divide both sides by 23520:
\[ P = \frac{51744000}{23520} \]
Calculating this gives:
\[ P = 2200 \]

### Conclusion
The monthly installment paid by Mr. Kumar is Rs 2200.