A focused Grade 7 learning session for Simple Interest: Earning and Paying Interest using visual models, worked examples, strategy guidance, and a short mastery check.
Student goal: I can solve problems involving simple interest: earning and paying interest.
Simple Interest: Earning and Paying Interest 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 simple interest: earning and paying interest 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 Simple Interest: Earning and Paying Interest, 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 simple interest: earning and paying interest.
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.
Using the simple interest formula \(I=Prt\), how much interest does \($800\) earn at an annual rate of \(5\%\) over \(3\) years?
Answer D: \($120\)
\(I=Prt=800\times0.05\times3=$120\).
Maya deposits \($1200\) into a savings account with a simple interest rate of \(4\%\) per year. How much interest will she earn after \(2\) years?
Answer D: \($96\)
STEP 1: Convert percentage to decimal: \(4\% = 4 \div 100 = 0.04\). STEP 2: \(I=1200\times0.04\times2=$96\).
A loan of \($5000\) has a simple interest rate of \(6\%\) per year. What will be the total amount owed after \(4\) years?
Answer C: \($6200\)
First find interest: \(I=5000\times0.06\times4=$1200\). Then \(A=P+I=5000+1200=$6200\).
Jordan invests \($2000\) at a simple interest rate of \(3.5\%\) per year for \(5\) years. How much interest will he earn?
Answer C: \($350\)
\(I=2000\times0.035\times5=$350\). Convert \(3.5\%\) to \(0.035\).
A savings account earns \($75\) in interest from a principal of \($500\) at an annual rate of \(5\%\). How many years has the money been in the account?
Answer D: \(3\) years
Using \(I=Prt\): \(75=500\times0.05\times t \Rightarrow 75=25t \Rightarrow t=3\).
An investment of \($3000\) earns \($270\) in simple interest over \(2\) years. What is the annual interest rate?
Answer D: \(4.5\%\)
\(270=3000\times r\times2 \Rightarrow 270=6000r \Rightarrow r=0.045=4.5\%\).
A car loan of \($15000\) is taken at a simple interest rate of \(7\%\) per year. How much total interest will be paid over \(6\) years?
Answer D: \($6300\)
\(I=15000\times0.07\times6=$6300\).
Sofia opens a savings account with \($400\) at a simple interest rate of \(2.5\%\) per year. What will be the total amount in her account after \(4\) years?
Answer D: \($440\)
\(I=400\times0.025\times4=$40\). Then \(A=400+40=$440\).
A principal amount earns \($200\) in interest at \(8\%\) annual rate over \(5\) years. What was the original principal?
Answer D: \($500\)
\(200=P\times0.08\times5 \Rightarrow 200=0.4P \Rightarrow P=$500\).
Which scenario produces more interest: \($2000\) at \(4\%\) for \(2\) years, or \($1000\) at \(6\%\) for \(2\) years?
Answer D: \($2000\) at \(4\%\)
First scenario: \(I=2000\times0.04\times2=$160\). Second: \(I=1000\times0.06\times2=$120\). First produces \($160\), which is \($40\) more than the second.
A student borrows \($2500\) for a bike at \(9\%\) simple interest per year. If the loan must be repaid in \(18\) months, how much interest will the student pay?
Answer D: \($337.50\)
STEP 1: Convert \(18\) months to years: \(t = 18 \div 12 = 1.5\) years. STEP 2: \(I=2500\times0.09\times1.5=$337.50\).
