Problem:
The rock group Loco Moco was scheduled for a concert at the Blaisdell Concert Hall. Because it was the concert highlight of the year, there was excitement in the air and people were eager to get tickets, which could only be purchased in person. Hours before tickets were to go on sale, people were lined up to buy tickets. In fact, the first person came 12 hours before the ticket booth was open. A new group of ticket buyers joined the line every 30 minutes.
Mission:
If each new group has four persons more than the previous group, how many people were in line after the 20th group joined, assuming the first person is the first group?
What did you notice about the number of people who arrived in each successive group?
How many times did four more people arrive?
If each new group had two persons more, how do you think the number in line after the 20th group joined compares to the number of people in line when four people joined as described in problem 1? Explain your reasoning.
What patterns do you see in the numbers of people in each group in problem 2 and the numbers of people in each group in problem 1?
Why do those patterns make sense?
What will the graph look like for the number of people arriving in each group in problem 2?
How will it compare to the graph for problem 1?
How many people were in line 3 hours before the ticket booth opened?
How did your model for solving problems 1 and 2 help you to solve problem 3? Explain your method.
To solve the problem, let's break it down step by step.
1. How many people were in line after the 20th group joined, assuming the first person is the first group?
To find the total number of people in line, we need to calculate the sum of all the groups that have joined. The first group has one person, and each subsequent group has four more people than the previous group.
We can use the formula for the sum of an arithmetic series to calculate this. The formula is:
Sn = (n/2)[2a + (n-1)d]
Where Sn is the sum of the first n terms, a is the first term, and d is the common difference.
In this case, a = 1 (the number of people in the first group) and d = 4 (the difference between each subsequent group).
Plugging in the values, we get:
S20 = (20/2)[2(1) + (20-1)(4)]
Simplifying,
S20 = 10[2 + 19(4)]
S20 = 10[2 + 76]
S20 = 10(78)
S20 = 780
Therefore, there were 780 people in line after the 20th group joined.
2. What did you notice about the number of people who arrived in each successive group? How many times did four more people arrive?
From the problem statement, we are told that each new group has four more persons than the previous group. This means that the number of people in each group increases by four each time.
To find how many times four more people arrive, we can divide the number of people in the 20th group (780) by four:
780 / 4 = 195
So, four more people arrived 195 times.
3. If each new group had two persons more, how do you think the number in line after the 20th group joined compares to the number of people in line when four people joined as described in problem 1? Explain your reasoning.
If each new group has two persons more than the previous group instead of four, the number of people in line would still increase with each group, but at a slower rate.
Since the difference between groups is smaller (two instead of four), there would be fewer people in line after the 20th group compared to the first problem. This is because the groups in the second problem would have fewer people in each group, resulting in a smaller total.
4. What patterns do you see in the numbers of people in each group in problem 2 and the numbers of people in each group in problem 1? Why do those patterns make sense?
In problem 1, the number of people in each group increases by four each time. This creates an arithmetic sequence where the common difference is four. As a result, the total number of people in line after each group follows an arithmetic series.
In problem 2, the number of people in each group increases by two each time. Similarly, this also creates an arithmetic sequence, but with a smaller common difference. Therefore, the total number of people in line after each group in problem 2 follows a different arithmetic series with a slower growth rate compared to problem 1.
These patterns make sense because they reflect the conditions set in the problem statement. In problem 1, each successive group has four more people, resulting in a faster growth rate. In problem 2, each successive group has two more people, resulting in a slower growth rate.
5. How many people were in line 3 hours before the ticket booth opened? How did your model for solving problems 1 and 2 help you to solve problem 3? Explain your method.
To find the number of people in line 3 hours before the ticket booth opened, we need to consider the number of groups that have joined in that time.
Since each new group arrived every 30 minutes, 3 hours can be divided into 6 groups. Therefore, we need to calculate the sum of the first 6 groups using the same formula as before.
Using the formula:
Sn = (n/2)[2a + (n-1)d]
where n = 6 (number of groups that have joined), a = 1 (number of people in the first group), and d = 4 (difference between each subsequent group), we can calculate:
S6 = (6/2)[2(1) + (6-1)(4)]
S6 = 3[2 + 5(4)]
S6 = 3[2 + 20]
S6 = 3(22)
S6 = 66
Therefore, there were 66 people in line 3 hours before the ticket booth opened.
By applying the same formula and approach used to solve problems 1 and 2, we can find the solution to problem 3. The formula for the sum of an arithmetic series allows us to calculate the total number of people in line based on the number of groups that have joined and the difference between each group.
We used the information provided in the problem statement about the time intervals and the number of people in each group to solve for the number of people in line at a specific time.