Ashish deposits a certain sum of money every month in a Recurring Deposit Account for a period of `12` months. If the bank pays interest at the rate of `11%` p.a. and Ashish gets `₹ 12,715` as the maturity value of this account, what sum of money did he pay every month ?

Answer

To find the monthly installment Ashish paid into his Recurring Deposit Account, we can follow these steps:

### Step 1: Define Variables
Let the monthly installment be \( y \).
The number of months \( n = 12 \).
The rate of interest \( r = 11\% \) per annum.

### Step 2: Use the Formula for Simple Interest (SI)
The formula for Simple Interest (SI) for a Recurring Deposit is given by:
\[ SI = P \times \frac{n(n + 1)}{2} \times \frac{r}{100 \times 12} \]
Where:
- \( P \) is the monthly installment (which is \( y \)),
- \( n \) is the number of months (which is \( 12 \)),
- \( r \) is the rate of interest (which is \( 11\% \)).

### Step 3: Substitute Values into the Formula
Substituting the values into the formula:
\[ SI = y \times \frac{12(12 + 1)}{2} \times \frac{11}{100 \times 12} \]
This simplifies to:
\[ SI = y \times \frac{12 \times 13}{2} \times \frac{11}{100 \times 12} \]
The \( 12 \) in the numerator and denominator cancels out:
\[ SI = y \times \frac{13 \times 11}{200} \]
\[ SI = y \times \frac{143}{200} \]

### Step 4: Calculate Maturity Amount
The maturity amount \( A \) is given by:
\[ A = \text{Total Principal} + SI \]
The total principal for 12 months is \( 12y \):
\[ A = 12y + SI \]
Substituting the expression for SI:
\[ A = 12y + y \times \frac{143}{200} \]
\[ A = 12y + \frac{143y}{200} \]

### Step 5: Combine the Terms
To combine the terms, we need a common denominator:
\[ A = \frac{2400y}{200} + \frac{143y}{200} \]
\[ A = \frac{2400y + 143y}{200} \]
\[ A = \frac{2543y}{200} \]

### Step 6: Set the Maturity Amount Equal to the Given Value
We know from the problem that the maturity amount \( A = 12,715 \):
\[ \frac{2543y}{200} = 12715 \]

### Step 7: Solve for \( y \)
To find \( y \), multiply both sides by \( 200 \):
\[ 2543y = 12715 \times 200 \]
\[ 2543y = 2543000 \]
Now, divide both sides by \( 2543 \):
\[ y = \frac{2543000}{2543} \]
Calculating this gives:
\[ y = 1000 \]

### Conclusion
Thus, the monthly installment Ashish paid is \( \text{₹} 1000 \).

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