Let's set this problem up step by step:
Assume Amal purchases \( x \) pens at 8 rupees each.
Total cost of the pens = \( 8x \) rupees. He hires an employee at a fixed wage \( W \).
He sells 100 pens at 12 rupees each. The revenue from this sale is:
\[ \text{Revenue} = 100 \times 12 = 1200 \, \text{rupees}. \]
Now, there are \( x - 100 \) pens left.
If the remaining pens are sold at 11 rupees each, the revenue from this sale is:
\[ \text{Revenue} = 11(x - 100) \, \text{rupees}. \]
Total revenue from both sales is:
\[ \text{Total Revenue} = 1200 + 11(x - 100). \]
The net profit is the total revenue minus the total cost and wage, which is given as 300 rupees. Hence, the equation becomes:
\[ 1200 + 11x - 1100 - 8x - W = 300 \]
Simplifying the equation:
\[ 3x - W = 200 \quad \text{...(i)} \]
If the remaining pens are sold at 9 rupees each, the revenue from this sale is:
\[ \text{Revenue} = 9(x - 100) \, \text{rupees}. \]
Total revenue from both sales is:
\[ \text{Total Revenue} = 1200 + 9(x - 100). \]
The net loss is the total cost and wage minus the total revenue, which is given as 300 rupees. Hence, the equation becomes:
\[ 8x + W - (1200 + 9x - 900) = 300 \]
Simplifying the equation:
\[ -x + W = 400 \quad \text{...(ii)} \]
We now solve equations (i) and (ii) simultaneously:
Adding both equations (i) and (ii):
\[ (3x - W) + (-x + W) = 200 + 400 \]
Simplifying:
\[ 2x = 600 \]
Solving for \( x \):
\[ x = \frac{600}{2} = 300 \]
Substitute \( x = 300 \) into equation (i):
\[ 3(300) - W = 200 \]
Simplifying:
\[ 900 - W = 200 \]
Solving for \( W \):
\[ W = 900 - 200 = 700 \]
The wage of the employee is 700 rupees.
When $10^{100}$ is divided by 7, the remainder is ?