Mr Shamin's salary increases every year by 10% in June . If there is not other increase or reduction in the salary and his salary in June 2011 was Rs. 22,385 , what was his salary in June 2009 ?

Options

A. Rs. 18,650
B. Rs. 18000
C. Rs. 19,250
D. Rs. 18,500

Explanation

To find Mr. Shamin's salary in June 2009 given his salary in June 2011 was Rs. 22,385 and that it increases by 10% every year, we can follow these steps:

### Step 1: Understand the Salary Increase
Mr. Shamin's salary increases by 10% each year. This means that if his salary in one year is \( S \), then his salary in the next year will be:
\[ S_{\text{next}} = S + 0.10 \times S = 1.10 \times S \]

### Step 2: Calculate the Salary for Previous Years
Let \( S_{2009} \) be the salary in June 2009. Then:
- Salary in June 2010:
\[ S_{2010} = 1.10 \times S_{2009} \]
- Salary in June 2011:
\[ S_{2011} = 1.10 \times S_{2010} = 1.10 \times (1.10 \times S_{2009}) = (1.10)^2 \times S_{2009} = 1.21 \times S_{2009} \]

### Step 3: Set Up the Equation
We know that \( S_{2011} = 22,385 \). Therefore, we can set up the equation:
\[ 1.21 \times S_{2009} = 22,385 \]

### Step 4: Solve for \( S_{2009} \)
To find \( S_{2009} \), we divide both sides of the equation by 1.21:
\[ S_{2009} = \frac{22,385}{1.21} \]

### Step 5: Perform the Calculation
Now, we will calculate \( S_{2009} \):
\[ S_{2009} = \frac{22,385}{1.21} \approx 18,500 \]

### Conclusion
Thus, Mr. Shamin's salary in June 2009 was approximately Rs. 18,500.

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