A focused Grade 7 learning session for Percent Increase and Decrease using visual models, worked examples, strategy guidance, and a short mastery check.
Student goal: I can solve problems involving percent increase and decrease.
Percent Increase and Decrease 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 percent increase and decrease 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 Percent Increase and Decrease, 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 percent increase and decrease.
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.
A pair of jeans originally costs \($50\). The store raises the price to \($60\). What is the percent increase?
Answer C: \(20\%\)
Percent increase \(=\frac{\text{change}}{\text{original}}\times100\%=\frac{60-50}{50}\times100\%=20\%\).
A tablet costs \($400\) and is discounted by \(25\%\). What is the new price?
Answer D: \($300\)
Discount amount \(= 25\% \times $400 = 0.25 \times $400 = $100\). New price \(= $400 - $100 = $300\).
A restaurant bill was originally \($80\), and after a \(15\%\) increase, what is the new bill?
Answer D: \($92\)
Increase \(= 15\% \times $80 = 0.15 \times $80 = $12\). New bill \(= $80 + $12 = $92\).
The value of a car decreased from \($24\,000\) to \($18\,000\). What is the percent decrease?
Answer D: \(25\%\)
Percent decrease \(= \frac{24000-18000}{24000}\times100\% = \frac{6000}{24000}\times100\% = 25\%\).
A student's test score improved from \(72\) to \(90\). What is the percent increase in the score?
Answer C: \(25\%\)
Percent increase \(= \frac{90-72}{72}\times100\% = \frac{18}{72}\times100\% = 25\%\).
A store advertises "Save 30% on all winter coats!" If a coat originally costs \($120\), what is the sale price?
Answer D: \($84\)
Savings \(= 30\% \times $120 = $36\). Sale price \(= $120 - $36 = $84\).
The population of a town was \(5\,000\) and increased to \(6\,500\). What is the percent increase?
Answer C: \(30\%\)
Percent increase \(= \frac{6500-5000}{5000}\times100\% = \frac{1500}{5000}\times100\% = 30\%\).
An item originally marked at \($80\) is reduced to \($60\). What is the percent decrease?
Answer D: \(25\%\)
Percent decrease \(= \frac{80-60}{80}\times100\% = \frac{20}{80}\times100\% = 25\%\).
A basketball player's scoring average was \(18\) points per game. After training, it increased to \(27\) points per game. What is the percent increase?
Answer D: \(50\%\)
Percent increase \(= \frac{27-18}{18}\times100\% = \frac{9}{18}\times100\% = 50\%\).
A book costs \($15\). After a \(40\%\) price increase, what is the new cost?
Answer C: \($21\)
Increase \(= 40\% \times $15 = 0.40 \times $15 = $6\). New cost \(= $15 + $6 = $21\).
A membership fee was \($200\) per year. This year it was raised by \(12\%\). How much is the new fee?
Answer C: \($224\)
Increase \(= 12\% \times $200 = 0.12 \times $200 = $24\). New fee \(= $200 + $24 = $224\).
