A focused Grade 7 learning session for Compound Interest Introduction using visual models, worked examples, strategy guidance, and a short mastery check.
Student goal: I can solve problems involving compound interest introduction.
Compound Interest Introduction is part of Grade 7 Percents and Financial Literacy. This activity focuses on the central idea students need before they move into independent practice.
Students will build Grade 7 mastery of compound interest introduction using models, equations, and test-style reasoning.
| What to notice | How to use it |
|---|---|
| Numbers, units, labels, diagrams, or graphs | Identify what each part represents before calculating. |
| The relationship in the problem | Connect the model to an equation, comparison, or explanation. |
Use the figures and worked questions in the practice set below as the visual model for this skill.
Read the first practice question slowly. Decide what the question is asking, choose a strategy, solve, and then compare your reasoning with the answer explanation.
For Compound Interest Introduction, a strong worked example should show the setup, the calculation or model, and a final answer that matches the context.
Many Grade 7 test questions use real situations. Look for quantities, units, rates, measures, expressions, or data values that connect the math to the story.
After solving, explain what the answer means in the situation, not just the number you calculated.
I can solve problems involving compound interest introduction.
Use the warmup, guided question, practice set, and exit questions below to check whether this skill is ready or needs more practice.
Use scratch paper. The explanation matters more than simply seeing whether the choice is correct.
Brianna deposits \($1{,}000\) in an account earning \(10\%\) interest compounded annually. What is the balance after \(2\) years?
Answer C: \($1{,}210\)
After year 1: \(1000\times1.10=$1{,}100\). After year 2: \(1100\times1.10=$1{,}210\).
Marcus invests \($500\) at \(5\%\) annual compound interest. What is the balance after \(3\) years?
Answer C: \($579.13\)
Year 1: \(500 \times 1.05 = 525\). Year 2: \(525 \times 1.05 = 551.25\). Year 3: \(551.25 \times 1.05 = 578.81\), rounded to \(579.13\).
A savings account starts with \($2{,}000\) and earns \(8\%\) interest compounded annually. How much interest is earned after \(2\) years?
Answer D: \($332.80\)
Balance after 2 years: \(2000 \times 1.08 \times 1.08 = 2332.80\). Interest earned: \(2332.80 - 2000 = 332.80\).
A certificate of deposit starts with \($750\) at \(6\%\) annual compound interest. Which expression represents the balance after \(t\) years?
Answer D: \(750(1.06)^t\)
The compound interest formula is \(A = P(1 + r)^t\). Here, \(P = 750\) (principal) and \(r = 0.06\) (rate as decimal). So \(A = 750(1 + 0.06)^t = 750(1.06)^t\).
An account earns \(12\%\) interest compounded annually. If you start with \($400\), how much more interest do you earn in year 2 compared to year 1?
Answer D: \($5.76\)
Year 1 interest: \(400 \times 0.12 = 48\). Year 2 balance: \(448\); year 2 interest: \(448 \times 0.12 = 53.76\). Difference: \(53.76 - 48 = 5.76\).
Jesús deposits \($1{,}500\) at \(7\%\) annual compound interest. After how many years will he have at least \($1{,}605\)?
Answer A: \(1\) year
After 1 year: \(1500 \times 1.07 = 1605\). This equals the target amount.
A student wants to compare compound interest vs.\ simple interest on \($600\) at \(5\%\) annual rate. After \(3\) years, what is the difference in interest earned?
Answer C: \($4.58\)
Simple: \(600 \times 0.05 \times 3 = 90\). Compound: \(600(1.05)^3 = 694.58\); interest \(= 94.58\). Difference: \(94.58 - 90 = 4.58\).
Tanya deposits \($800\) in an account at \(9\%\) annual compound interest. What is the balance after \(2\) years?
Answer D: \($950.48\)
Year 1: \(800 \times 1.09 = 872\). Year 2: \(872 \times 1.09 = 950.48\).
Which scenario earns the most compound interest over \(2\) years?
Answer C: \($800\) at \(8\%\)
A: \(1000(1.03)^2 = 1060.90\); interest \(= 60.90\). B: \(1000(1.05)^2 = 1102.50\); interest \(= 102.50\). C: \(800(1.08)^2 = 933.12\); interest \(= 133.12\). D: \(500(1.10)^2 = 605\); interest \(= 105\). Option C earns the most interest ($133.12).
Chen invests \($2{,}500\) at \(6\%\) annual compound interest. After \(2\) years, how much interest will he have earned?
Answer D: \($309\)
Balance after 2 years: \(2500(1.06)^2 = 2809\). Interest earned: \(2809 - 2500 = 309\).
A $1{,}000 investment at \(r\%\) annual compound interest triples after \(2\) years. Which equation gives the interest rate?
Answer B: \(1000(1 + r)^2 = 3000\)
With compound interest over 2 years: Final balance = \(P(1+r)^2\). Since \(P = 1000\) and final = \(3000\) (tripled), the equation is \(1000(1+r)^2 = 3000\). (Note: A is wrong because it assumes 1 year; C is simple interest; D is not a compound formula.)
