Lesson Plan: Understanding Compound Interest
- The Chairman
- 12 hours ago
- 2 min read
High School Financial Literacy
Lesson Overview
This lesson introduces students to compound interest, a critical financial concept for understanding savings, investments, and debt. Students will learn the compound interest formula, explore real-life examples, and practice solving problems step by step.
Grade Level:
9th – 12th Grade
Subject:
Financial Literacy / Personal Finance
Duration:
90 minutes (adaptable to 45 minutes)
🎯 Learning Objectives
By the end of this lesson, students will be able to:
Understand the concept and importance of compound interest.
Use the compound interest formula to calculate savings or loan balances.
Apply step-by-step problem-solving techniques to real-world examples.
Recognize the long-term effects of compounding on financial decision-making.
📜 Florida Sunshine State Standards Alignment
SS.912.FL.1.1 – Explain how personal financial decisions are influenced by incentives, institutions, and individual choice.
SS.912.FL.1.2 – Analyze the relationship between career choices, income, and financial planning.
SS.912.FL.3.2 – Explain how individuals incur costs and realize benefits when consuming, saving, and investing.
SS.912.FL.4.1 – Calculate and compare simple and compound interest.
🔑 Key Formula for Compound Interest
Where:
A = Final amount after interest
P = Principal (starting amount)
r = Annual interest rate (decimal)
n = Number of compounding periods per year
t = Time in years
📝 Steps to Solve Compound Interest Problems
Step 1 — Write Down the Formula
A=P×(1+nr)n×t
Step 2 — Identify the Variables
Determine P, r, n, and t from the problem.
Step 3 — Plug in the Variables
Substitute the given numbers into the formula.
Step 4 — Solve Using Order of Operations
Remember PEMDAS: Parentheses → Exponents → Multiplication/Division → Addition/Subtraction.
📚 Example 1 — Savings Growth
Problem:You deposit $5,000 into a savings account earning 5% annual interest, compounded quarterly (n = 4). How much will you have after 6 years?
Solution:Step 1: Formula → A=P(1+r/n)ntA = P(1 + r/n)^{nt}A=P(1+r/n)ntStep 2: Variables → P=5000P = 5000P=5000, r=0.05r = 0.05r=0.05, n=4n = 4n=4, t=6t = 6t=6Step 3: Plug in values →
A=5000×(1+0.054)4×6A = 5000 \times \left(1 + \frac{0.05}{4}\right)^{4 \times 6}A=5000×(1+40.05)4×6A=5000×(1+0.0125)24A = 5000 \times \left(1 + 0.0125\right)^{24}A=5000×(1+0.0125)24A=5000×(1.0125)24A = 5000 \times (1.0125)^{24}A=5000×(1.0125)24
Step 4: Solve exponent →
A=5000×1.34935A = 5000 \times 1.34935A=5000×1.34935A=$6,746.75A = \mathbf{\$6,746.75}A=$6,746.75
Answer: After 6 years, you’ll have $6,746.75.
📚 Example 2 — Loan Balance
Problem:You borrow $10,000 at 6% annual interest, compounded monthly (n = 12). What will you owe after 5 years if no payments are made?
Solution:Step 1: Formula → A=P(1+r/n)ntA = P(1 + r/n)^{nt}A=P(1+r/n)ntStep 2: Variables → P=10,000P = 10,000P=10,000, r=0.06r = 0.06r=0.06, n=12n = 12n=12, t=5t = 5t=5Step 3: Plug in values →
A=10,000×(1+0.0612)12×5A = 10,000 \times \left(1 + \frac{0.06}{12}\right)^{12 \times 5}A=10,000×(1+120.06)12×5A=10,000×(1+0.005)60A = 10,000 \times \left(1 + 0.005\right)^{60}A=10,000×(1+0.005)60A=10,000×(1.005)60A = 10,000 \times (1.005)^{60}A=10,000×(1.005)60
Step 4: Solve exponent →
A=10,000×1.34885A = 10,000 \times 1.34885A=10,000×1.34885A=$13,488.50A = \mathbf{\$13,488.50}A=$13,488.50
Answer: After 5 years, you’ll owe $13,488.50.
💡 Class Activity
Split students into pairs.
Each pair gets a compound interest problem with different P, r, n, and t values.
Students solve using the 4-step process and present their solution to the class.
📌 Exit Ticket
Write a short response:
“How does compound interest help build wealth, and how can it also increase debt?”
