Step 1: Calculate total revenue. The theatre has 200 seats, and the total occupancy was 80%, so:
Seats occupied per show = 200 × 0.8 = 160
Revenue from the first two shows (at Rs. 250 per ticket):
Revenue from 2 shows = 160 × 250 × 2 = 80,000 Rs.
Revenue from the late-night show (at Rs. 200 per ticket):
Revenue from 1 show = 160 × 200 = 32,000 Rs.
Total revenue:
80,000 + 32,000 = 1,12,000 Rs.
Step 2: Calculate total cost. Fixed cost for the day:
10,000 Rs.
Cost per show:
5000 × 3 = 15,000 Rs.
Total cost:
10,000 + 15,000 = 25,000 Rs.
Step 3: Calculate profit. Profit:
Profit = Total Revenue − Total Cost
Profit = 1,12,000 − 25,000 = 87,000 Rs.
Answer: Rs. 1,16,000
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:
A | B | C | D | Average |
---|---|---|---|---|
3 | 4 | 4 | ? | 4 |
3 | ? | 5 | ? | 4 |
? | 3 | 3 | ? | 4 |
? | ? | ? | ? | 4.25 |
4 | 4 | 4 | 4.25 |