The Algebra Teacher\'s Guide to Reteaching Essential Concepts and Skills

Teaching Notes 2.17: Solving Percent Problems

Three types of percent problems are addressed in most algebra curriculums: finding a percent of a

number, finding what percent a number is of another, and finding a number when a percent of it

is known. Students often confuse the methods used to solve each type of problem.

1. Explain that percent problems can be solved using a proportion.


2. Review the information and examples on theworksheet with your students. Make sure that
   your students understand the proportion and are able to write it correctly. Note that stu-
   dents may simplify a ratio prior to solving the proportion. This will not affect the answer and
   usually makes the computation easier. Depending on the abilities of your students, you may
   wish to review the steps for solving proportions.


3. Encourage your students to solve percent problems using another method if they feel the
   method is easier. For example, problems such as 30 percent of 80 can easily be found by
   multiplying 80 times 0.3. This method is often used in the lower grades.


4. Encourage your students to double-check their work by placing their answer back in the
   problem to see that it is reasonable. For example, 30 percent of 80=240. Because 30 percent
   is a part of 80, the answer cannot be larger than 80. In this case, the decimal point was
   omitted. 30 percent of 80=24.

EXTRA HELP:

If the percent is less than 100 percent, the ‘‘part’’ is less than the ‘‘whole.’’ If the percent is greater

than 100 percent, the ‘‘part’’ is greater than the ‘‘whole.’’ 100 percent means the part and the whole

are equal.

ANSWER KEY:

(1)16.5 (2) 52 (3) 400 (4)48% (5)60% (6)14.4

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(Challenge)Answers may vary. Carla knew that, mathematically, ‘‘of’’ means ‘‘times.’’ Because

every percent can be written as a fraction, she could set up a multiplication problem that was

easy to solve. She also knew that 66

2

3

percent can be written as a fraction equal to

2

3

, making

this problem particularly easy to solve.

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