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An airplane can hold 325 passengers, 30 in first class and the rest in coach. If a first-class ticket costs \(700 and a coach ticket costs \)250, then what is the minimum revenue that the airplane will gross on a flight in which exactly 3 seats remain empty?

Short Answer

Expert verified
The minimum gross revenue on a flight with exactly 3 empty seats is $92,650.

Step by step solution

01

Determine the number of coach seats and first-class seats with 3 empty seats

If 30 seats are for first-class passengers and the rest are coach, then there are 325 - 30 = 295 coach seats on the airplane. We are given that exactly 3 seats remain empty, so we'll assume these empty seats are in first-class to minimize revenue. Therefore, there are 30 - 3 = 27 first-class seats occupied.
02

Calculate the revenue from the occupied first-class seats

Each first-class ticket costs \(700. So, the revenue from the first-class tickets is 27 * \)700 = $18,900.
03

Calculate the revenue from the occupied coach seats

Assuming all coach seats are occupied, there are 295 occupied coach seats. Each coach ticket costs \(250, so the total revenue from coach tickets is 295 * \)250 = $73,750.
04

Calculate the total minimum revenue

To find the total minimum revenue, add the revenue from occupied first-class and coach seats: \(18,900 + \)73,750 = $92,650. The minimum gross revenue on a flight with exactly 3 empty seats is $92,650.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mathematical Reasoning
Mathematical reasoning helps us make sense of problems by logically evaluating the information provided. It is the process of identifying relationships and drawing conclusions based on facts. In our exercise, we first needed to determine the distribution of seats and understand what the problem was asking. The problem indicated a total of 325 seats on the airplane, with a separate categorization for first-class and coach passengers. We understood that three seats were empty, and we aimed to minimize the revenue, starting by supposing the empty seats were the highest-priced ones (first-class).

Breaking the problem into smaller parts allowed us to focus on specific details, like realizing there are more coach seats compared to first-class. This recognition directed our strategy to assume the emptier category would be first-class, leaving more affordable seats filled to the full capacity.

Using mathematical reasoning made it clear how to organize our approach and prioritize calculations, ensuring each step logically followed the previous one.
Problem-Solving Strategies
Problem-solving strategies are essential tools for overcoming mathematical challenges. They involve steps such as understanding the problem, devising a plan, executing that plan, and evaluating the solution. In this exercise, the initial step was understanding that minimizing revenue meant considering which seats remained empty.

Steps employed include:
  • Identifying the total seats and separating them into first class and coach.
  • Communicating the goal of minimizing the revenue.
  • Choosing a tactic to achieve this, like keeping the most expensive seats unoccupied.
This strategy called for assuming empty first-class seats, which, although hypothetical, logically directed the lowest revenue possible given the ticket prices.

This logical sequencing and clear strategy allowed for simple arithmetic calculations at the final stage. It highlighted how choosing the correct lines of reasoning leads to efficiently solving the problem.
Arithmetic Calculations
Arithmetic calculations are the backbone of solving numerical problems. This involves basic operations like addition, subtraction, multiplication, and division. For the airplane revenue problem, we encountered multiplication and addition, two primary arithmetic calculations.

Firstly, we determined the number of occupied first-class seats as 27, and multiplied by the price per ticket to find the total revenue from these seats: \[ 27 \times 700 = 18,900 \]Next, we calculated revenue from coach tickets, with 295 seats fully occupied:\[ 295 \times 250 = 73,750 \]Finally, these partial revenues were added to determine total minimum gross revenue:\[ 18,900 + 73,750 = 92,650 \]Every numerical step was crucial as they combined to form the final solution. By organizing and executing arithmetic calculations, we ensured an accurate and concise conclusion to our problem-solving journey.

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