A person deposited Rs. 6000 in a bank for 2 years. At the end of the first year, he withdrew Rs. 500. How much does he get from the bank at the end of the second year interest paid at a rate of `8(1)/(3)%` per annum compounded annually ?

Options

A. 6500
B. 7000
C. 6725
D. 6025

Explanation

To solve the problem step by step, we will calculate the amount the person receives from the bank at the end of the second year after making a withdrawal at the end of the first year.

### Step 1: Identify the principal amount and the interest rate.
- Principal (P) = Rs. 6000
- Rate of interest (R) = 8⅓% = 25/3% = 25/300 = 1/12 (as a decimal)

### Step 2: Calculate the amount at the end of the first year.
Using the formula for compound interest:
\[ A = P \left(1 + \frac{R}{100}\right)^n \]
Where:
- \( A \) = Amount after time \( n \)
- \( P \) = Principal
- \( R \) = Rate of interest
- \( n \) = Number of years

For the first year:
- \( n = 1 \)

Substituting the values:
\[ A = 6000 \left(1 + \frac{25}{300}\right)^1 \]
\[ A = 6000 \left(1 + \frac{1}{12}\right) \]
\[ A = 6000 \left(\frac{12 + 1}{12}\right) \]
\[ A = 6000 \left(\frac{13}{12}\right) \]
\[ A = 6500 \]

### Step 3: Withdraw Rs. 500 at the end of the first year.
After the withdrawal:
\[ Amount after withdrawal = 6500 - 500 = 6000 \]

### Step 4: Calculate the amount at the end of the second year.
Now, we will calculate the amount for the second year using the new principal (after withdrawal):
- New Principal (P) = Rs. 6000
- Rate of interest (R) = 8⅓% = 25/3% = 25/300 = 1/12 (as a decimal)

For the second year:
- \( n = 1 \)

Substituting the values again:
\[ A = 6000 \left(1 + \frac{25}{300}\right)^1 \]
\[ A = 6000 \left(1 + \frac{1}{12}\right) \]
\[ A = 6000 \left(\frac{12 + 1}{12}\right) \]
\[ A = 6000 \left(\frac{13}{12}\right) \]
\[ A = 6500 \]

### Final Result
At the end of the second year, the total amount the person receives from the bank is Rs. 6500.

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