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Lesson plan Grade 7 Math

Ketterwick's Standpipe Loan: A Grade 7 Simple Interest Lesson

A 50-minute grade 7 math lesson where students compute simple interest on a fictional town's loan for a new standpipe and share the cost per household.

Objectives

  1. Students will calculate simple interest for one year and for the full term of a loan.
  2. Students will divide a total repayment among households to find an annual and monthly cost.
  3. Students will compare the total interest owed at two different rates.

Vocabulary

Standpipe
A standpipe is a tall, cylindrical water storage tank that sits directly on the ground instead of standing on legs like a water tower. It stores water and helps maintain steady pressure in the pipes nearby.

Also: principal, simple interest, term, percent

Teacher overview

A standpipe is a tall steel storage tank that sits on the ground, and the height of the water it holds supplies the pressure for a small system. The town of Ketterwick, a fictional community used throughout this lesson, has one built in the 1930s whose interior coating has failed, so the council plans to borrow $480,000 over ten years to replace it. Students compute the simple interest on that loan and then divide the yearly payment among the households that will pay for it. The error to watch for is treating simple interest as a one-time charge: students who multiply the principal by the rate and stop have found one year of interest, not the interest for the whole term.

Procedure

Hook (5 min): Describe two ways a town can pay for a large repair, either saving up first or borrowing now and repaying over several years. Explain that the fictional town of Ketterwick has chosen to borrow, and that borrowing adds a cost called interest.

Direct instruction (12 min): Define principal, rate, and term. Model a loan that is not on the handout: a fictional district borrows $200,000 for 5 years at 6 percent simple interest. Show that one year of interest is 0.06 times 200,000, or $12,000, and that five years of interest is 5 times $12,000, or $60,000, so the district repays $260,000. Write the pattern as interest equals principal times rate times years, and say plainly that the rate applies again in every year of the term.

Guided practice (13 min): Distribute the handout. Complete items 1 and 2 together so students see the difference between one year of interest and the interest for the full ten years. Have pairs try item 3 and compare answers before continuing.

Independent practice (12 min): Students work items 4 through 7 on their own. Point out that item 5 divides among households and item 7 changes only the rate, so the principal and term stay the same.

Closure and exit ticket (8 min): Each student answers on an index card: "A fictional village borrows $60,000 for 8 years at 5 percent simple interest. How much interest does it owe in total?" Collect and sort the cards. If more than a quarter of the class reports $3,000, reteach the step that multiplies by the number of years.

Ketterwick's Standpipe Loan: A Grade 7 Simple Interest Lesson — Grade 7 Math
Name: ____________________________ Date: ______________

Student Handout

Use the fictional town of Ketterwick for every question. Ketterwick will borrow $480,000 for 10 years at 4 percent simple interest to replace its standpipe, and the town has 1,600 households.

  1. How much simple interest does Ketterwick owe for one year?
  2. How much simple interest does the town owe over the full 10 years?
  3. What total amount does Ketterwick repay, counting the money borrowed and all of the interest?
  4. If the town repays that total in 10 equal yearly payments, how much is each yearly payment?
  5. Divided evenly among 1,600 households, how much does one yearly payment cost each household?
  6. How much is that per household each month?
  7. If Ketterwick could borrow the same $480,000 for 10 years at 3.5 percent instead, how much less total interest would it owe?

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: $19,200. One year of interest is the principal times the rate: 0.04 times $480,000 equals $19,200.
  2. Answer: $192,000. Ten years of interest is 10 times $19,200, which equals $192,000.
  3. Answer: $672,000. Add the principal to the interest: $480,000 plus $192,000 equals $672,000.
  4. Answer: $67,200. Divide the total repayment by the term: $672,000 divided by 10 equals $67,200 each year.
  5. Answer: $42. Divide the yearly payment among the households: $67,200 divided by 1,600 equals $42 per household per year.
  6. Answer: $3.50. Divide the yearly household cost by 12 months: $42 divided by 12 equals $3.50 per month.
  7. Answer: $24,000. At 3.5 percent the yearly interest is 0.035 times $480,000, or $16,800, so ten years cost $168,000. The saving is $192,000 minus $168,000, which is $24,000.

Exit ticket answer: $24,000. One year is 0.05 times $60,000, or $3,000, and eight years is 8 times $3,000.

Differentiation

Support

Give these students a completed example for the fictional district from the direct instruction, with each step labeled principal, rate, one year of interest, and total interest. Let them use that page as a template and allow calculators throughout. Accept item 6 rounded to the nearest cent without further explanation.

Extension

Ask students to find what yearly household cost the fictional town of Ketterwick would face if it borrowed the same amount over 20 years at 4 percent instead of 10, and to write two sentences on why a longer term lowers the yearly payment while raising the total interest.

ELL supports

Pre-teach "borrow," "interest," "principal," and "term" using a labeled timeline that shows money going out in year zero and payments returning in each of ten years. Provide the frame "The town pays ___ dollars of interest each year for ___ years" for items 1 and 2.