This section directly relates to and provides a basis for discounts, mark-ups, commissions, taxes and so forth, discussed in the next section. In older textbooks, the percent of change was given by a two part formula. One pertained to increases, and the other to decreases.
Percent of increase was defined as:
Percent of decrease was defined as:
The purpose of this dual approach was to avoid negative numbers in the numerator, since that topic was not taught in lower grades. The current approach is to define it one way which allows some flexibility:
In this way, teachers can elect to teach it as “new minus old divided by old”. A negative number represents a decrease, and a positive number represents an increase. I have done this both ways, and prefer expressing everything as either x percent increase or x percent decrease where x is always positive. In a later lesson, when students are more comfortable with the concept, I will introduce the option of using negative percents.
Warm-Ups. The Warm-Up has a question that asks for the percent increase over an original amount. Other questions review solving percent problems by equation or proportion.
1. Show equation or proportion and solve. 18 is 90% of what number?
Answer: 18= 0.9x; x = 20; or 18/x = 90/100; 18/x = 9/10; 9x = 180; x = 20
2. Mr. Thomas made $1,500 per week. He received a raise that was 10% of his salary. How much was his increase? Answer: 10% of 1500 = 0.1 × 1500 = $150.
3. John Jones caught 15 fish his first time out and 18 fish the second time. The increase in the amount caught represents what percent of his first catch? Write an equation or use a proportion to solve. Answer: Equivalent to “3 is what percent of 15?” 3 = 15x; x =0.2… or 20%; OR 3/15 = x/100; 15x = 300; x = 20, so 20%
4. John scored 95% on his math test. He answered 38 questions correctly. How many questions were on the test? Write an equation or proportion and solve.
Answer: 0.95x = 38; x = 40; OR: 95/100 = 38/x; 95x =3800; x = 40
5. Solve. −3(𝑥−4)−3=−27 Answer: -3x+12-3 = -27; -3x + 9 = -27; -3x=-36; x=12
Problem 1 requires students to either translate directly into an equation or to use a proportion by identifying part and whole—which most will do via the “is over of” mnemonic.
Problem 2 is a sneak peak at the next lesson which is about mark-ups (and discounts). It also serves as a starting point for this lesson.
Problem 4 is similar to Problem 1 in that it is asking for the “whole”. Students have seen this type of problem before so they should be able to recognize it. For those who don’t, a prompt would be to ask whether it’s asking for percent, part, or whole. If this doesn’t do it, they should restate the problem using the same structure as Problem 1.
Problem 5 is a multistep equation. Students will still be making mistakes with -27 – 9. A prompt I’ve used repeatedly is “I lost $27 and lost $9 more. How much have I lost in all?”
Finding a Percent of Increase. In Problem 2 of the Warm-Ups, we’re told that Mr. Thomas received a 10% raise, and asks for what the increase is if his base salary was $1,500 per week.
“Suppose we are given that his new salary is $1,650; we know his old salary was $1,500. But now we are not told what the percent raise was. We can figure it out using the Percent of Change method.”
I then write the formula for Percent of Change on the board:
“When we find the percent of increase, we find the change between an original amount and a new amount. Let’s pretend we don’t know it’s 10%. What is the increase in his raise?” I have them figure it out: $150.
“We plug that value in to the formula. The amount it changed—increased in this case—is $150. What do we put in the denominator?”
Now we have 150/1500 = 0.1. “How do I write that as a percent?” There may be some students who don’t know this, so just to make sure I go over this. “We know that 0.1 is the same as 0.10. And to change a percent, we multiply by 100.”
I write the following steps on the board (using the numbers in the example we just did) and have the copy it in their notebooks:
Percent of Increase
Step 1: Find the amount of change between the two amounts: 1650-1500=150.
Step 2: Divide by the original amount. 150÷1500 or 150/1500; 0.1 = 10%
Examples:
I have them write on mini-whiteboards or notebooks for several examples.
“If I write “20 to 25” what is the original amount?” This may seem obvious, but some students find this wording confusing. “Now what is the change in value? By how much did it increase?” Answer: 5/20 = 0.25 = 25% increase
I advise them to write “increase” after the percent to make it clear what direction the change has gone.
15 to 25 Answer: 10/15 = 2/3 = 0.67 = 67% increase
4 to 8 Answer: 4/4 = 1 = 100% increase. This might raise some eyebrows, but they know from their previous work in sixth grade and recent reviews that percents can be greater than 100. Also, they know that 1 expressed as a percent is 1 × 100 = 100.
I read the next one aloud: “A store owner buys bicycles for $100 per bike and sells them for $150. What is the percent increase in price?” I write down 100 to 150. Answer: 50/100 = 0.5 = 50% increase
Finding a Percent of Decrease. Similarly we can find percents of decrease. We use the same formula.
“Let’s say the registered vehicles in Smalltown for two years is: 2018: 350, 2021: 329
“We know the amount has gone down, so it is a decrease.”
I have them write the steps down in their notebooks and prompt them as I’m writing.
Percent of Decrease
STEP 1: We take the positive amount of change: 350-329 = 21
(I ask “How do we find the change?)
STEP 2: Divide by the original amount, or base.
(I ask “What was the original amount?” 21/350= 6% decrease.
Examples:
20 to 12 Answer: (20-12)/20 = 8/20 = 0.4 or 40% decrease
16 to 10 Answer: 6/16 = 6/16 = 3/8 = 0.375 or 37.5% decrease (I instruct them to take it to three decimal places)
25 to 4 Answer: 21/25 = 0.84 = 84% decrease
Finding Percent of Error. Students may know from their science class that measurement is not precise; there are errors. “If three people measure something, they are likely to get three different answers, each off by just a little bit. The measure of that error is called Percent of Error”.
I write on the board, and they copy in their notebooks:
“Percent error is a measure of the difference between an estimate, prediction or measurement and the actual value.”
“Let’s say we estimate the length of a board as 16 feet but its actual length is 21. So we find the positive amount of the difference between the estimated and actual length. (actual length – estimated length)/actual length.” I have them write the calculation on their mini-whiteboards. Answer: 5/21 = 0.238 = 23.8%.
I work through the next example with them, and then give them independent work.
Examples:
Estimated weight: 8 pounds, actual weight 6.4 pounds: 1.6/6.4 = ¼ = 25% error
Measured length: 2.5 cm; actual length 2.54 cm; Difference: 0.04; percent error: 0.04/2.54 = 0.016 (rounded) = 1.6% error
Before assigning homework, I give them the following problem. “A man gets a raise from $1,000 per week to $1,250 per week. What is the percent increase of his raise?”
Students should get 25% increase.
“Now let’s say that because of cutbacks, the company returns his salary to $1,000 per week. What do you think the percent decrease in his salary will be?” In my experience, everyone says it will be 25%. I now have them figure it out.
“What is the amount of decrease?” $250.
“And what is the original amount?” It is a different “original amount” than when we began; it is now $1,250.
The percent decrease is 250/1250 = 0.2 or 20% decrease.
Some students will be surprised. I will ask why they think it is different. What has changed? After a little back and forth they will see that the base has changed; for the increase, the base was $1,000. For the decrease, it was $1,250.
“Intuition is fine sometimes, but it isn’t always right. Trust the mathematics!”
Homework. Homework is a mixture of percent of increase, decrease and error; some straight numbers, and others word problems.