Lesson 27-29 NYS COMMON CORE MATHEMATICS CURRICULUM 6β’1 T Lesson 27: Solving Percent Problems Student Outcomes ο§ Students find the percent of a quantity. Given a part and the percent, students solve problems involving finding the whole. Classwork Example (10 minutes) Example Solve the following three problems. Write the words PERCENT, WHOLE, or PART under each problem to show which piece you were solving for. ππ% of πππ = ππ×π πππ×π = πππ ππ% of πππ ππ×π πππ πππ×π πππ = = πππ ππ out of πππ = πππ ππ÷π πππ πππ÷π PART WHOLE = ππ % ππ πππ PERCENT How did your solving method differ with each problem? Solutions will vary. A possible answer may include: When solving for the part, I need to find the missing number in the numerator. When solving for the whole, I solve for the denominator. When I solve for the percent, I need to find the numerator when the denominator is πππ. ο§ What are you trying to find in each example? οΊ ο§ How are the problems different from each other? οΊ ο§ Part, whole, percent Answers will vary. How are the problems alike? οΊ Answers will vary. Take time to discuss the clues in each problem including the placement of the word βof.β The word βofβ lets students know which piece of information is the whole amount compared to the part. In the first example, 60% of 300 tells us that we are looking for part of 300. Therefore, 300 is the whole. In the second example where 60% of 500 is 300, 300 is the part, and 500 is the whole. In the third example, 60 out of 300 tells us that now, 60 is the part, and 300 is the whole. Structure the conversation around the part-whole relationship. Lesson 27 Solving Percent Problems This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M1-TE-1.3.0-07.2015 215 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27-29 NYS COMMON CORE MATHEMATICS CURRICULUM 6β’1 T ο§ In the first question, what is 60% of 300? οΊ Students should understand that 60 100 is the same ratio as unknown number 300 to determine an answer of 180. ο§ In this case, is 180 the part or the whole? οΊ ο§ 180 is the part. It is part of 300. In the second question, we are given 60% of some value equals 300 β οΊ ο§ 300 ? . What is that value? 500 500 is the whole, and 300 is the part. In the third question, we are asked, 60 out of 300 equals what percent β οΊ ο§ = In this case, is 500 the part or the whole? What about 300? Is that a part or the whole? οΊ ο§ 60 100 ? 60 = 300 100 . What percent is that? The percent is 20%. In this case, is 300 the part or the whole? οΊ 300 is the whole. Exercise (20 minutes) At this time, students break out into pairs or small groups to solve the problem. Exercise Use models, such as ππ × ππ grids, ratio tables, tape diagrams, or double number line diagrams, to solve the following situation. Priya is doing her back-to-school shopping. Calculate all of the missing values in the table below, rounding to the nearest penny, and calculate the total amount Priya will spend on her outfit after she receives the indicated discounts. Shirt (ππ% discount) Pants (ππ% discount) Shoes (ππ% discount) Necklace (ππ% discount) Sweater (ππ% discount) Original Price $ππ $ππ $ππ $ππ $ππ Amount of Discount $ππ $ππ $π $π $π Lesson 27 Solving Percent Problems This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M1-TE-1.3.0-07.2015 216 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 27-29 6β’1 T What is the total cost of Priyaβs outfit? Shirt ππ% = ππ π ππ = = The discount is $ππ. The cost of the shirt is $ππ because $ππ β $ππ = $ππ. πππ π ππ Pants ππ% = ππ ππ = The original price is $ππ. The price of the pants is $ππ because $ππ β $ππ = $ππ. πππ ππ Shoes ππ% = ππ π π = = The original price is $ππ. The cost of the shoes is $ππ because $ππ β $π = $ππ. πππ ππ ππ Necklace ππ% = Sweater ππ% = π π = The discount is $π. The cost of the necklace is $ππ because $ππ β $π = $ππ. ππ ππ ππ π π = = The original price is $ππ. The cost of the sweater is $ππ because $ππ β $π = $ππ. πππ π ππ The total outfit would cost the following: $ππ + $ππ + $ππ + $ππ + $ππ = $πππ. Closing (10 minutes) Give students time to share samples of how they solved the problem and describe the methods they chose to use when solving. Lesson Summary Percent problems include the part, whole, and percent. When one of these values is missing, we can use tables, diagrams, and models to solve for the missing number. Lesson 27 Solving Percent Problems This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M1-TE-1.3.0-07.2015 217 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 27-29 6β’1 T Lesson 28: Solving Percent Problems Student Outcomes ο§ Given a part and the percent, students find the percent of a quantity and solve problems involving finding the whole. Classwork Example (5 minutes) Read the questions from the example one by one. Example If an item is discounted ππ%, the sale price is what percent of the original price? πππ β ππ = ππ ππ% If the original price of the item is $πππ, what is the dollar amount of the discount? ππ% = πππ × ππ π = πππ ππ π πππ = = $ππ ππ ππ $ππ discount How much is the sale price? ππ% = πππ × ππ π = πππ ππ π ππππ = = $πππ, or πππ β ππ = $πππ ππ ππ $πππ sale price ο§ What are some different ways that we can solve this question? οΊ Answers will vary. Some students may draw diagrams that they can share with the class. Others may have found the value by finding equivalent fractions or by multiplying a quantity by the percent written as a fraction. Lesson 27 Solving Percent Problems This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M1-TE-1.3.0-07.2015 218 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27-29 NYS COMMON CORE MATHEMATICS CURRICULUM 6β’1 T Be sure to discuss different models that could be used. Exercise (20 minutes) Have students work in pairs or small groups to solve the problems. Students are given the sale price and the percent that was saved. They need to come up with the original price. Students should create models in order to prove that their answers are correct. Exercise The following items were bought on sale. Complete the missing information in the table. Item Original Price Sale Price Amount of Discount Percent Saved Percent Paid Television $ππππ $πππ $πππ ππ% ππ% Sneakers $ππ $ππ $ππ ππ% ππ% Video Games $ππ $ππ $π ππ% ππ% MP3 Player $ππ $ππ. ππ $ππ. ππ ππ% ππ% Book $ππ. ππ $ππ. ππ $π. ππ ππ% ππ% Snack Bar $π. ππ $π. ππ $π. ππ ππ% ππ% Closing (10 minutes) ο§ Have students showcase some of the models used to solve the problems. One possible way to showcase the work, if time allows, would be to hang the work on the walls and have students do a gallery walk to view the diagrams. Ask students how they could check their work. οΊ The answers may vary according to which values are given and which values are missing. Students may mention that the discount and the sale price should add to be the original amount. The percents should add to 100%. They could solve the problem using the answer to see if they can work back to a given amount. Lesson Summary Percent problems include the part, whole, and percent. When one of these values is missing, we can use tables, diagrams, and models to solve for the missing number. Lesson 27 Solving Percent Problems This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M1-TE-1.3.0-07.2015 219 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27-29 NYS COMMON CORE MATHEMATICS CURRICULUM 6β’1 T Lesson 29: Solving Percent Problems Student Outcomes ο§ Students find the percent of a quantity. ο§ Given a part and the percent, students solve problems involving finding the whole. Classwork Exploratory Challenges (25 minutes): Group/Partner Students explore what it means to have 10%. Students recognize the equivalence between 10%, 10 100 , and 1 10 and use this relationship to quickly calculate 10% of different quantities. Being able to calculate 10% of a quantity can be an efficient tool or strategy when calculating other percents. Exploratory Challenge 1 Claim: To find ππ% of a number, all you need to do is move the decimal to the left once. Use at least one model to solve each problem (e.g., tape diagram, table, double number line diagram, ππ × ππ grid). a. Make a prediction. Do you think the claim is true or false? Explain why. Answers will vary. One could think the claim is true because ππ% as a fraction is happens when one divides by ππ or multiplies by π π . The same thing ππ . A student may think the claim is false because it ππ depends on what whole amount represents the number from which the percentage is taken. Determine ππ% of πππ. b. πππ × ππ π πππ = = ππ ππ ππ ππ% of e. π ππ = =π ππ ππ Find ππ% of π. π× πππ π π π. π π ππ × = π. π ππ f. Find ππ% of ππ. ππ × Determine ππ% of ππ. d. c. is ππ. g. π π π π = = ππ ππ π ππ% of ππ is π. π × ππ = ππ ππ ππ ππ ππ ππ ππ ππ ππ ππ ππ ππ × ππ = πππ Lesson 27 Solving Percent Problems This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M1-TE-1.3.0-07.2015 220 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27-29 NYS COMMON CORE MATHEMATICS CURRICULUM 6β’1 T h. Gary read ππ pages of a πππ pages book. What percent did he read? ππ ÷ ππ π ππ = = = ππ% πππ ÷ ππ ππ πππ i. Micah read ππ pages of his book. If this is ππ% of the book, how many pages are in the book? ππ π × ππ ππ = = πππ ππ × ππ πππ There are πππ pages in the book. j. Using the solutions to the problems above, what conclusions can you make about the claim? The claim is true. When I find ππ% of a number, I am really finding π ππ of the amount or dividing by ππ, which is the same as what occurred when I moved the decimal point in the number one place to the left. ο§ Using the solutions to the problems above, what conclusions can you make about the claim? οΊ Answers will vary. However, students are required to share what is mathematically happening when the decimal is moved over once to help make connections to why it works. Students may relate back to using place value and regrouping with the concept of decimals. Students read a claim that two separate discounts give the same results as the sum of the two discounts taken off the original price at the same time. Students need to conclude that they are not the same because the second discount is being taken off a new amount not the original price. Exploratory Challenge 2 Claim: If an item is already on sale, and then there is another discount taken off the sale price, this is the same as taking the sum of the two discounts off the original price. Use at least one model to solve each problem (e.g., tape diagram, table, double number line diagram, ππ × ππ grid). a. Make a prediction. Do you think the claim is true or false? _____________ Explain. The answer is false. They will be different because when two discounts are taken off, the second discount is taken off a new amount. b. Sam purchased π games for $πππ after a discount of ππ%. What was the original price? ππ ππ ππ ππ Sale price: $πππ ππ ππ ππ ππ ππ ππ Discount: $ππ $πππ is the original price. c. If Sam had used a ππ% off coupon and opened a frequent shopper discount membership to save ππ%, would the games still have a total of $πππ? ππ% = ππ π = πππ ππ Lesson 27 $πππ × π ππ = Solving Percent Problems This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M1-TE-1.3.0-07.2015 $πππ ππ = $ππ saved. The price after the coupon is $πππ. 221 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 27-29 NYS COMMON CORE MATHEMATICS CURRICULUM 6β’1 T ππ% = ππ π = πππ ππ $πππ × π $πππ = = $ππ saved. The price after the coupon and ππ ππ discount membership is $πππ. No, the games would now total $πππ. d. Do you agree with the claim? support your claim. NO Explain why or why not. Create a new example to help When two discounts are taken off, the shopper pays more than if both were added together and taken off. Example: $πππ original price ππ%: πππ × Two ππ% off discounts: π ππ = πππ ππ = ππ saved $πππ β $ππ = $ππ sale price πππ × ππ × π ππ π ππ = = πππ ππ ππ ππ = ππ =π $πππ β $ππ β $π = $ππ sale price Closing (15 minutes) Give students time to share samples of how they solved the problem. Take time to point out similarities in the different models. Ask students to reflect on which models they like to use most and why. Lesson Summary Percent problems have three parts: whole, part, percent. Percent problems can be solved using models such as ratio tables, tape diagrams, double number line diagrams, and ππ × ππ grids. . Lesson 27 Solving Percent Problems This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G6-M1-TE-1.3.0-07.2015 222 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
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