Lesson plan Grade 7 Health & Consumer Science
Hydration for Larkfield Middle School's Spring Sports Day
A grade 7 health lesson using a fictional track team's fluid log to build proportional reasoning about how activity time changes how much water the body needs.
Objectives
- Students will test whether a set of fluid-log data shows a proportional relationship.
- Students will identify the constant of proportionality as a drinking rate and write its equation.
- Students will predict water needs for a longer session using the proportional model.
Vocabulary
Key words: proportional, constant of proportionality, hydration, rate
Teacher overview
Athletes lose water as they work, and replacing it steadily keeps the body performing. This lesson uses a fictional fluid log kept by a Larkfield Middle School track runner to turn hydration into a proportional-reasoning problem. Students test whether the runner's water intake grows in step with jogging time, find the constant of proportionality, and use it to predict how much water a longer practice would take. A common misconception the lesson addresses is that any table of increasing numbers is proportional; students learn to check that the ratio of cups to minutes stays constant before trusting the model.
Procedure
Hook (5 min): Ask students how they decide when to drink during a game or a run. Collect a few answers. Explain that a Larkfield runner kept a log of how much water she drank during steady jogging, and that the class will use it to build a rule for predicting fluid needs.
Direct instruction (10 min): Present the fictional log as a table: 15 minutes of steady jogging with 1 cup, 30 minutes with 2 cups, 45 minutes with 3 cups, 60 minutes with 4 cups. Model the proportionality test by dividing cups by minutes at each row (1 divided by 15, 2 divided by 30, 3 divided by 45, 4 divided by 60). Show that every ratio equals the same value, so the relationship is proportional. Name that shared value, one fifteenth of a cup per minute, as the constant of proportionality, and rewrite it as the more useful 4 cups per hour. Students record the ratio test and this hourly rate as handout questions 1 and 2.
Guided practice (12 min): Together, write the equation c equals t divided by 15, where c is cups and t is minutes. Use it to predict intake for a 90-minute practice, and confirm the answer two ways: through the equation (90 divided by 15 is 6 cups) and through the hourly rate (1.5 hours times 4 cups is 6 cups). Students record this equation and prediction as handout question 3. Pose the check-for-understanding question: "If a second runner drinks 5 cups in 60 minutes, is that the same rate?" Have students compute and thumbs-up or thumbs-down; if more than a quarter are unsure, re-model the ratio test.
Independent practice (12 min): Students turn to handout question 4, a second runner on a hot day who drinks 3 cups every 30 minutes, find that rate in cups per hour, and compare it with the cool-day rate.
Closure / exit ticket (6 min): Each student completes handout question 5, the exit ticket: "A runner drinks 2 cups in 20 minutes at a steady rate. How many cups would 60 minutes take?" Collect and check that students scaled by the constant rate rather than adding a fixed amount.
Student Handout
Answer each question in order, and show the ratio or the calculation you used.
Cool-day fluid log. A Larkfield runner recorded how much water she drank during steady jogging on a mild afternoon.
| Jogging time | Water |
|---|---|
| 15 minutes | 1 cup |
| 30 minutes | 2 cups |
| 45 minutes | 3 cups |
| 60 minutes | 4 cups |
- Divide cups by minutes for each row of the cool-day log. Are all four ratios equal? State whether water and jogging time form a proportional relationship, and give the reason.
- Write the constant of proportionality for the cool-day log as a drinking rate in cups per hour.
- Write the equation that gives cups (c) from jogging minutes (t) for the cool-day log, then use that equation to predict how much water a 90-minute practice would take.
Hot-day fluid log. On a hot afternoon, a second runner drinks water at a steady rate of 3 cups every 30 minutes.
- Find the constant of proportionality for the hot-day log, written as a drinking rate in cups per hour.
Exit ticket.
- A runner drinks 2 cups in 20 minutes at a steady rate. At that same rate, how many cups would 60 minutes of jogging take?
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
- Answer: Yes, it is proportional. Every row gives the same ratio: 1 over 15, 2 over 30, 3 over 45, and 4 over 60 all equal one fifteenth of a cup per minute.
- Answer: 4 cups per hour. One fifteenth of a cup per minute times 60 minutes per hour equals 4 cups per hour.
- Answer: 6 cups. Using c equals t divided by 15, 90 divided by 15 is 6 cups, which matches 1.5 hours times 4 cups per hour.
- Answer: 6 cups per hour. On the hot day, 3 cups in 30 minutes is one tenth of a cup per minute, or 6 cups per hour, a higher rate than the cooler day.
- Exit ticket answer: 6 cups. The steady rate is 2 cups per 20 minutes, or one tenth of a cup per minute; 60 minutes times one tenth is 6 cups.
Differentiation
Support
Provide the completed ratio for the first row and a filled-in equation frame, so students focus on checking the remaining rows against it. Allow calculators and let students describe the rate in words ("one cup every fifteen minutes") before writing the symbolic equation.
Extension
Ask students to graph the cool-day and hot-day logs on the same axes, identify which line is steeper, and connect the steeper slope to the higher drinking rate. Have them explain what a flat portion of a real log might mean about a rest break.
ELL supports
Pre-teach "proportional," "rate," and "constant" using a picture of a cup filling at a steady pace. Provide a sentence frame for the proportionality test: "The ratio of cups to minutes is ___ in every row, so the relationship is or is not proportional."