Problem 79 - Algebra and Trigonometry

Question

A hotel has 200 rooms. Those with kitchen facilities rent for \(\$ 100\) per night and those without kitchen facilities rent for \(\$ 80\) per night. On a night when the hotel was completely occupied, revenues were \(\$ 17,000 .\) How many of each type of room does the hotel have?

Step-by-Step Solution

Answer

The hotel has 150 rooms with kitchen facilities and 50 rooms without kitchen facilities.

Step 1: Establish the two unknowns

Let \(x\) be the number of rooms with kitchen facilities and \(y\) be the number of rooms without kitchen facilities.

Step 2: Set up the first equation

This equation represents the total number of rooms in the hotel. So, \(x + y = 200\). This is because the hotel has 200 rooms in total.

Step 3: Set up the second equation

This equation represents the total revenue for the hotel. So, \(100x + 80y = 17000\). This is because rooms with kitchen facilities rent for $100 per night and those without rent for $80 per night. When the hotel was completely occupied, revenues were $17,000.

Step 4: Solve the system of equations

You can solve this system of equations using various methods such as substitution or elimination. Here, the substitution method will be used. Solve the first equation \(x + y = 200\) for \(x\) so \(x = 200 - y\). Then, substitute this into second equation, we get \(100(200 - y) + 80y = 17000\). Solving this, we find \(y = 50\). Substituting \(y = 50\) into the first equation, we get \(x = 150\). So, there are 150 rooms with kitchen facilities and 50 rooms without kitchen facilities.