A teacher is choosing between two options for a class field trip to an orchard.

Question

At orchard A, admission costs $9 per person, and 3 chaperones are required.
At orchard B, the cost is cost is $10 per person, but only 1 chaperone is required.
At each orchard, the same price applies to both chaperones and students.
To help her compare the cost of her two options, the teacher first writes the equation 9(n+3)=10(n+1), and the she writes the inequality 9(n+3)<10(n+1)

Answer 1

Let's analyze the teacher's equations and inequalities step by step to understand them better.

1. Understanding the Variables:

   • Let \( n \) be the number of students attending the trip.
   • At Orchard A, each person (students and chaperones) costs $9, and she needs 3 chaperones, leading to \( n + 3 \) total people.
   • At Orchard B, each person costs $10, and only 1 chaperone is needed, leading to \( n + 1 \) total people.

2. Setting Up the Equation: The teacher's equation \( 9(n + 3) = 10(n + 1) \) is used to find the point at which the total costs of both orchards are equal.

   • Calculating the left side: \( 9(n + 3) = 9n + 27 \)
   • Calculating the right side: \( 10(n + 1) = 10n + 10 \)

   So, the equation becomes: \[ 9n + 27 = 10n + 10 \] Rearranging gives: \[ 27 - 10 = 10n - 9n \implies 17 = n \] This means that when there are 17 students, the costs of both orchards are the same.

3. Setting Up the Inequality: The teacher's inequality \( 9(n + 3) < 10(n + 1) \) is used to find out when Orchard A is cheaper than Orchard B. Using the same expressions as before, the inequality becomes: \[ 9n + 27 < 10n + 10 \] Rearranging gives: \[ 27 - 10 < 10n - 9n \implies 17 < n \] This means Orchard A will be cheaper when the number of students \( n \) is greater than 17.

4. Conclusions:

   • If there are exactly 17 students, both orchards cost the same.
   • If there are fewer than 17 students, Orchard B is cheaper (the inequality \( 9(n + 3) > 10(n + 1) \)).
   • If there are more than 17 students, Orchard A becomes the cheaper option (the inequality \( 9(n + 3) < 10(n + 1) \)).

So, the teacher can use this information to make a decision based on the number of students attending.