Mr. Bajaj needs `₹ 30,000` after `2` years. What least money (in multiple of `₹ 5`) must be deposit every month in a recurring deposit account to get required money at the end of `2` years, the rate of interest being `8%` p.a. ?

Answer

To solve the problem, we need to determine the monthly deposit Mr. Bajaj must make in a recurring deposit account to accumulate ₹30,000 after 2 years at an interest rate of 8% per annum.

### Step-by-step Solution:

1. **Identify the Required Information:**
- Maturity Amount (A) = ₹30,000
- Time (T) = 2 years = 24 months (since 2 years × 12 months/year)
- Rate of Interest (R) = 8% per annum

2. **Use the Formula for Simple Interest in Recurring Deposits:**
The formula for the simple interest earned on a recurring deposit is:
\[ SI = P \times \frac{n(n + 1)}{2} \times \frac{R}{100 \times 12} \]
where:
- \( P \) = monthly deposit (installment)
- \( n \) = number of months
- \( R \) = rate of interest

3. **Substitute the Known Values into the Formula:**
Here, \( n = 24 \) and \( R = 8 \):
\[ SI = P \times \frac{24(24 + 1)}{2} \times \frac{8}{100 \times 12} \]
Simplifying this:
\[ SI = P \times \frac{24 \times 25}{2} \times \frac{8}{1200} \]
\[ SI = P \times 300 \times \frac{8}{1200} \]
\[ SI = P \times \frac{2400}{1200} = P \times 2 \]

4. **Determine the Total Amount at Maturity:**
The maturity amount (A) is given by:
\[ A = P \times n + SI \]
Substituting the values we have:
\[ 30000 = P \times 24 + 2P \]
\[ 30000 = 24P + 2P = 26P \]

5. **Solve for P:**
\[ P = \frac{30000}{26} \approx 1153.84 \]

6. **Round to the Nearest Multiple of ₹5:**
Since the deposit must be in multiples of ₹5, we round ₹1153.84 to the nearest multiple of ₹5. The nearest multiple is ₹1155.

### Final Answer:
The least money Mr. Bajaj must deposit every month is **₹1155**.