An amusement park, that you are excited to try is $28 per visit. You have the option to purchase a monthly membership for $90 and then pay $15 for each visit. (4 points)

Question

1. You have plans to attend 3 times this summer. Which option is best?

2. You think you will visit probably about 8 times this summer. Which option is best?

3. You have budgeted $150 for visiting the park this summer. Which option is best?

4. How many visits would be approximately the break even point where both options would cost about the same?

A. Pay per visit 
B. Buy a membership 
C. 2 
D. 15 
E. 7 
F. 11

GPT 3.5 · Answer

To determine which option is best, we need to compare the total cost of each option for the given scenarios.

1. If you plan to attend 3 times this summer:
- Option 1: Pay per visit
Total cost = $28 x 3 = $84
- Option 2: Buy a membership
Total cost = $90 (membership fee) + $15 x 3 (visit cost) = $135
In this case, Option 1 (Pay per visit) is the best option.

2. If you think you will visit about 8 times this summer:
- Option 1: Pay per visit
Total cost = $28 x 8 = $224
- Option 2: Buy a membership
Total cost = $90 (membership fee) + $15 x 8 (visit cost) = $210
In this case, Option 2 (Buy a membership) is the best option.

3. If you have budgeted $150 for visiting the park this summer:
- Option 1: Pay per visit
You can calculate approximately how many visits you can have with $150:
Number of visits = $150 / $28 ≈ 5.36 (rounded down to 5 visits)
Total cost = $28 x 5 = $140
- Option 2: Buy a membership
Total cost = $90 (membership fee) + $15 x 5 (visit cost) = $165
In this case, Option 1 (Pay per visit) is the best option.

4. To find the break-even point where both options cost about the same:
Let the number of visits be N.
Option 1 (Pay per visit) cost = $28N
Option 2 (Buy a membership) cost = $90 + $15N
Set the costs equal to each other and solve for N:
$28N = $90 + $15N
$28N - $15N = $90
$13N = $90
N ≈ 6.92
Since you can't have a fraction of a visit, the break-even point is approximately 7 visits.
Therefore, the answer to question 4 is E (7).