Let's set this problem up step by step:
Let's 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. Revenue from this sale = \( 1200 \) rupees.
Now, there are \( x - 100 \) pens left.
Scenario 1:
If the remaining pens are sold at 11 rupees each:
Revenue = \( 11(x - 100) \) rupees.
Total Revenue = \( 1200 + 11(x - 100) \).
Net Profit = Revenue - Total Cost - Wage = \( 300 \).
\( 1200 + 11x - 1100 - 8x - W = 300 \)
\( 3x - W = 200 \) ...(i)
Scenario 2:
If the remaining pens are sold at 9 rupees each:
Revenue = \( 9(x - 100) \) rupees.
Total Revenue = \( 1200 + 9(x - 100) \).
Net Loss = Total Cost + Wage - Revenue = \( 300 \).
\( 8x + W - (1200 + 9x - 900) = 300 \) \
( -x + W = 400 \) ...(ii)
Solving equations (i) and (ii) simultaneously, we get:
Adding both equations:
\[ 2x = 600 \]
\[ x = 300 \]
Substituting \( x = 300 \) in equation (i):
\[ 3(300) - W = 200 \]
\[ 900 - W = 200 \]
\[ W = 700 \]
So, the wage of the employee is 700 INR.
A furniture trader deals in tables and chairs. He has Rs. 75,000 to invest and a space to store at most 60 items. A table costs him Rs. 1,500 and a chair costs him Rs. 1,000. The trader earns a profit of Rs. 400 and Rs. 250 on a table and chair, respectively. Assuming that he can sell all the items that he can buy, which of the following is/are true for the above problem:
(A) Let the trader buy \( x \) tables and \( y \) chairs. Let \( Z \) denote the total profit. Thus, the mathematical formulation of the given problem is:
\[ Z = 400x + 250y, \]
subject to constraints:
\[ x + y \leq 60, \quad 3x + 2y \leq 150, \quad x \geq 0, \quad y \geq 0. \]
(B) The corner points of the feasible region are (0, 0), (50, 0), (30, 30), and (0, 60).
(C) Maximum profit is Rs. 19,500 when trader purchases 60 chairs only.
(D) Maximum profit is Rs. 20,000 when trader purchases 50 tables only.
Choose the correct answer from the options given below: