Question:medium

There are 25 rooms in a hotel. Each room can accommodate at most three people. For each room, the single occupancy charge is Rs. 2000 per day, the double occupancy charge is Rs. 3000 per day, and the triple occupancy charge is Rs. 3500 per day. If there are 55 people staying in the hotel today, what is the maximum possible revenue from room occupancy charges today?

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Formulate constraints as linear equations and maximize revenue using the given conditions.
Updated On: Jun 15, 2026
  • Rs. 72500
  • Rs. 77500
  • Rs. 87500
  • Rs. 82500
  • Rs. 92500
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
To maximize revenue, we need to distribute 55 people into 25 rooms such that the total price is maximized. We compare the revenue generated per person: Single: \(2000/1 = 2000\) per person. Double: \(3000/2 = 1500\) per person. Triple: \(3500/3 \approx 1167\) per person. However, since we have a room limit (25), maximizing revenue means spreading people across as many rooms as possible at the highest room rate.
Step 2: Key Formula or Approach:
Revenue = \(2000 \times S + 3000 \times D + 3500 \times T\). Constraints: \(S + D + T = 25\) and \(1S + 2D + 3T = 55\).
Step 3: Detailed Explanation:
Using all 25 rooms generally yields higher revenue than packing people into fewer rooms. Let's try to maximize the higher-priced room types (Double and Triple) while using all 25 rooms. If we use \(T\) triple rooms and \(D\) double rooms: \(D + T = 25 \implies D = 25 - T\) Substitute into the capacity equation: \(2(25 - T) + 3T = 55\) \(50 - 2T + 3T = 55 \implies T = 5\). So, \(T = 5\) and \(D = 25 - 5 = 20\). Revenue = \(20 \times 3000 + 5 \times 3500 = 60000 + 17500 = 77500\). If we used single rooms, we would run out of rooms much faster (e.g., 25 singles only hold 25 people). Spreading 55 people over 25 rooms forces a mix of doubles and triples.
Step 4: Final Answer:
The maximum possible revenue is Rs. 77500.
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