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 :
(ii) the amount of maturity.

Answer

To find the amount of maturity for Mohan's recurring deposit account, we will follow these steps:

### Step 1: Identify the given values
- **Interest (I)** = Rs 1200
- **Time (n)** = 2 years = 24 months
- **Rate (r)** = 6% per annum

### Step 2: Use the formula for Simple Interest
The formula for simple interest is given by:
\[ I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \]
Where:
- \(I\) = Interest
- \(P\) = Monthly deposit
- \(n\) = Total number of months
- \(r\) = Rate of interest

### Step 3: Substitute the known values into the formula
We know:
- \(I = 1200\)
- \(n = 24\)
- \(r = 6\)

Substituting these values into the formula:
\[ 1200 = P \times \frac{24(24 + 1)}{2 \times 12} \times \frac{6}{100} \]

### Step 4: Simplify the equation
Calculating \(n(n + 1)\):
\[ 24(24 + 1) = 24 \times 25 = 600 \]
Now substituting back into the equation:
\[ 1200 = P \times \frac{600}{24} \times \frac{6}{100} \]
Calculating \(\frac{600}{24}\):
\[ \frac{600}{24} = 25 \]
So now we have:
\[ 1200 = P \times 25 \times \frac{6}{100} \]
Calculating \(\frac{6}{100}\):
\[ \frac{6}{100} = 0.06 \]
Thus, the equation becomes:
\[ 1200 = P \times 25 \times 0.06 \]
Calculating \(25 \times 0.06\):
\[ 25 \times 0.06 = 1.5 \]
So we have:
\[ 1200 = P \times 1.5 \]

### Step 5: Solve for \(P\)
Now, divide both sides by 1.5:
\[ P = \frac{1200}{1.5} = 800 \]
Thus, the monthly deposit \(P\) is Rs 800.

### Step 6: Calculate the maturity amount
The maturity amount (M) can be calculated using the formula:
\[ M = P \times n + I \]
Substituting the known values:
\[ M = 800 \times 24 + 1200 \]
Calculating \(800 \times 24\):
\[ 800 \times 24 = 19200 \]
Now adding the interest:
\[ M = 19200 + 1200 = 20400 \]

### Final Answer
The amount of maturity is Rs 20,400.

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