Mohan has a recurring deposite account in a bank for 2 years at 6% p.a. simple interest. If he gets Rs 1200 as interest at the time of maturity, find :
(i) the monthly instalment

Answer

To find the monthly installment that Mohan deposits in his recurring deposit account, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Total interest earned (I) = Rs 1200
- Time period (T) = 2 years = 24 months
- Rate of interest (R) = 6% per annum

2. **Use the Formula for Simple Interest:**
The formula for simple interest is:
\[ I = \frac{P \cdot n \cdot (n + 1)}{2 \cdot 12} \cdot \frac{R}{100} \]
where:
- \(I\) = total interest
- \(P\) = monthly installment
- \(n\) = total number of months (24 in this case)
- \(R\) = rate of interest

3. **Substituting the Known Values:**
Substitute \(I = 1200\), \(n = 24\), and \(R = 6\) into the formula:
\[ 1200 = \frac{P \cdot 24 \cdot (24 + 1)}{2 \cdot 12} \cdot \frac{6}{100} \]

4. **Simplifying the Equation:**
- Calculate \(24 + 1 = 25\).
- The equation becomes:
\[ 1200 = \frac{P \cdot 24 \cdot 25}{2 \cdot 12} \cdot \frac{6}{100} \]
- Simplifying further:
\[ 1200 = \frac{P \cdot 24 \cdot 25}{24} \cdot \frac{6}{100} \]
- The \(24\) cancels out:
\[ 1200 = P \cdot 25 \cdot \frac{6}{100} \]

5. **Rearranging the Equation:**
- Multiply both sides by \(100\) to eliminate the fraction:
\[ 1200 \cdot 100 = P \cdot 25 \cdot 6 \]
- This simplifies to:
\[ 120000 = P \cdot 150 \]

6. **Solving for \(P\):**
- Divide both sides by \(150\):
\[ P = \frac{120000}{150} \]
- Calculating the value:
\[ P = 800 \]

7. **Conclusion:**
The monthly installment that Mohan deposits is **Rs 800**.