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Alignments to Content Standards: 8.F.B.4

The SLV High School graduation started at 1:00PM. After some speeches, the principal started reading off the names of the students, alphabetically by last name. When he finishes, the graduation will end.

1. Use the bulletin shown below to estimate when the graduation will end.

2. Estimate how long the speeches took. How do you know?
3. Write an equation that the parents could use to find the approximate time the principal will call their child’s name given the child’s position in the list in the graduation program.
4. Aptos High School and Santa Cruz High School started their graduations at the same time. The graphs shown below show the time of day as a function of the number of names the principal has read at each school. Write down as many differences between the two graduations as you can based on differences in the two graphs. Give your reasons for each.

## IM Commentary

Students will need to make the assumption that the principal reads the names at a constant rate to solve this problem. While not a full-blown modeling problem, this task does address some aspects of modeling as described in Standard for Mathematical Practice 4. Also, students often think that time must always be the independent variable, and so may need some help understanding that one chooses the independent and dependent variable based on the way one wants to view a situation.

This task was submitted by Dan Meyer for the first Illustrative Mathematics task writing contest 2011/12/18.

## Solution

1. The principal had read 40 names by 1:28 and 65 names by 1:33. That yields a differential of 25 names in 5 minutes or 5 names per minute. There are 126 names remaining to be called.

$$\frac{126 \text { names}}{\displaystyle 5\frac{\text{names}}{\text{minute}}}=25.2\text{ minute}$$

The graduation will last approximately 25 minutes after 1:33. If it takes exactly 25 more minutes, the graduation will end at 1:58. It is reasonable to say that the graduation ceremony will end close to 2:00 p.m.

2. Assuming the principal read 5 names per minute, the initial 40 names took

$$40 \text{ names}\cdot \frac{1 \text { minute}}{5 \text{ names}}$$

or 8 minutes to read. We know the 40th name was read at 1:28, which means the principal began reading names at 1:20 or so. Since the graduation itself started at 1:00, the speeches took about 20 minutes.

3. The principal began reading names at 20 minutes past 1:00. Let $n$ be the child’s position in the list in the graduation bulletin and $t$ be the number of minutes past 1:00 the child’s name will be called. Assuming each name requires 1/5 minutes, then one equation would be:

$$t=\frac{n}{5}+20$$

We can emphasize the fact that $t$, the number of minutes after 1:00 is a function of $n$, the position in the graduation list by writing the answer using function notation:

$$t(n)=\frac{n}{5}+20$$
4. Aptos started calling names later because its $y$-intercept is greater; in other words, more time has passed before Aptos started calling names. The principal at Aptos calls out names more slowly because the slope, which corresponds to the number of "minutes per name," is greater. In other words, it takes the principal a greater number of minutes per name to read the list.

#### Carolyn Viss says:

I believe the function rule should be t(n) = (n-1)/5. The first graduate doesn't have to wait 20 seconds before their name is called. It is called right at 1:00. If I am 2nd in line, I hear my name at 20 seconds after 1:00.