The true essence of percent is that it means “parts out of 100”. Thus the fraction, p/100 (p out of 100), which shows its representation, has been taught to the student from early on. Then p % of a whole quantity, w, can be described as q, the part of the quantity. This is the basis for the our main formula (1): q = (p /100) x w = (p x w) / 100
This formula (1) is all the students need to perform all the basic calculations involving percent. Other forms of this formula will be studied but they can all be brought back to our standard percentage formula. This formula allows us to solve for the quantity. By moving letters and numbers around we can solve for the percent, p, and the whole amount, w.
Students often have difficulty working with formulas therefore addressing how to use formulas themselves is a priority to be delved into before the teaching of percent problems. Beginning in seventh grade exploration of the use of formulas is more widely addressed. (For example the use of area, perimeter and volume formulas.) By eighth grade, the use of formulas needs to be a comfortable task for the students in mathematical solving. Using formulas is an objective that is tested on by the Connecticut Mastery Test in September of the eighth grade year. Problem solving and using formulas serve as a precursor for Algebra.
Percents can also involve fractional percent, which can complicate the significance for using them. Not every problem can be solved as easily using the same method. Although they are accurately correct, it can then lose its simplicity. Address these difficulties by calculator use, rounding or teaching other methods for solving. There are enough routine examples to provide for the student in order for them to gain confidence with solving percents.
Regarding the use of the calculator: Consider our first equation:
15% of 60 = q. By grade 8 students should to be comfortable solving this equation both with and without a calculator. However, in many circumstance the ability levels within our classrooms are diverse. It is up to the good judgment of the teacher to know who does and does not have the skills necessary to do these types of problems without use of the calculator. This is where differentiation needs to be taking place within our classroom space. The teacher is trying to maximize student development. “… when an entire class moves forward to study new skills and concepts without any individual adjustments in time or support, some students are doomed to fail.”3 Developing understanding of what needs to be done to calculate should take precedence over the actual ability to calculate. We should be trying to increase students’ success and personal growth. However, for those students who do have the ability to multiply and divide accurately, using these types of problems can reinforce those skills as well. Teachers need to monitor both computational and conceptual understanding in the classroom because both are important to student success in Mathematics. Each student needs to be challenged at an appropriate level. In the eighth grade mathematics classroom, each teacher must carefully decide when student success of understanding a concept is handicapped by a lack of good computational skills and act appropriately.
The first objective in working with percent would be to understand the relationship between the p (%) and the q, (quantity) of the whole, w. Extensive discussion of this percent, part, and whole is important to student understanding of using percent. Students need to know which of the three they will be solving for. Given a great deal of practice, students can see this relation. When beginning to develop this relation, begin with sets of problems such as our first equation:
15% of 60 = q (q
being the part of the quantity, as taught to the students)
q = (p /100) x w = (p x w) / 100
q = (15/ 100) x 60 = (15 x 60 /100) = 900/100 = 9
Next, consider the following problems which students must see are asking essentially the same question: Find 15% of 60. 15% of 60 is what number? What is 15% of 60?
The solving of these three examples above uses the same equation (1). The answer 9 is the same. The terminology used however is different. Use all of these ways to express 15% of 60
because students must understand they will see different ways
.
Then they can take that familiarity and transfer it to word problems using the same numbers or involving estimation. Consider asking:
The teacher announced to the class that only 15% of the 60 students in the eighth grade had made honors that quarter. How many of the students made honors? Or
The teacher announced to the class that approximately 15% of the 58 students in the eighth grade had made honors that quarter. About how many of the students made honors? Or
If nine out of 58 students made honors is it better to say 15% or 16% of the 58 students made honors that quarter?
You can assign these problems independently, sequentially, or simultaneously. These three problems can elicit a variety of responses leading to a variety of considerations. For example, students will interpret the issue of rounding the quantity in varying ways. The students can write their thinking in their journals about the similarities and differences between the problems. A question for their reflection might be to decide which problem is more realistic to this type of situation.
Another consideration for the teacher is being cognizant that students may bring up differing ways to solve instead of using the formula. For example, changing the percent to a decimal or fraction and multiplying by the whole, or using a percent proportion (%/100 = part/ whole). When solving percent problems these may be methods with which students are comfortable. When this occurs, it is a good time to explore formula (1) and address how the different methods they have been taught is related to the formula. This helps students see how this formula works. The students, in groups, can figure out how to justify the relationships of the formula to other methods of solving. “Justification is central to mathematics, and even young children cannot learn mathematics with understanding without engaging in justification. …as they share their ideas and are asked to convince others that a procedure they have used to solve problems is valid, they have to use arguments that are convincing to other people.”4
An added theme with finding part or quantity, q, which is important for investigation, is regarding the complementary part or quantity, q?. When we have a percent of a whole, we also have the complementary part, the part left over. (If 10% made honors, 90% did not.) If the whole is w and the part is q, then the complementary part is q?.
(2) q = w - q = w - p/100 * w = (1 - p/100) * w = (100 - p)/ 100 * w
This equation (2) gives alternative ways to compute the q?. First use formula (1) to find q; then, subtract to find q?. Or compute the p?, by 100 - p, and then use formula (1). Some problems involving complementary part and complementary percent:
Troy and Kevin shot 60 baskets each at the gym. Troy made 75% of his shots. How many shots did Troy miss? Or
There is a 25% discount on the $80 basketball jersey that I want to buy. How much will the jersey cost?
The percent decrease is 25% and the amount of decrease is $20. This is the savings. The complementary percent is the percent left after the decrease. The jersey will actually cost 75% of $80. Students are ready for you to introduce the new terms, percent decrease and increase.
Let us also consider percent increase: 6% of 50 = q; 6% of 50 + 50 = q + w, or, 100% of 50 = w; 106% of 50 = w + q. Either way the answer is the same.
A large exchange regarding the relations of these problems should be discussed. This can lead to the lesson of computing percent increase or adding tax onto a purchase with word problems. Begin with a quick review discussion of 100%.
Try
,
100% of the 710 students at Troup have a lunch period
,
for discussion
.
Reminding students that 100% would mean everyone. Then:
The price of a $60 baseball glove is increased by 10%. Find the new price.
Students should be prepared to use formula (3) to solve for 10%. Remind them that the original price of the glove, $60, is 100% of the cost. Suppressing the $ relate:
(3) q? = w + q = w + p/100 * w = (1 + p/100) * w = (100 + p)/ 100 * w
60 + 6 = 66 is the same as: 100% + 10% = 110% 110/100 * 60 = 66 this represents the original price (100%) and the additional percent increase (10%). (100/100 + 10/100)*60 = (100 + 10)60. Set up a way to display plenty of practice using a table if you choose. See table 1 at end of unit.
