Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 7β’4 Lesson 5: Finding One Hundred Percent Given Another Percent Student Outcomes ο§ Students find 100% of a quantity (the whole) when given a quantity that is a percent of the whole by using a variety of methods including finding 1%, equations, mental math using factors of 100, and double number line models. ο§ Students solve word problems involving finding 100% of a given quantity with and without using equations. Classwork Opening Exercise (5 minutes) Students recall factors of 100 and their multiples to complete the table below. The discussion that follows introduces students to a means of calculating whole quantities through the use of a double number line. Opening Exercise What are the whole number factors of πππ? What are the multiples of those factors? How many multiples are there of each factor (up to πππ)? Factors of πππ ο§ Number of Multiples Multiples of the Factors of πππ πππ πππ π ππ ππ, πππ π ππ ππ, ππ, ππ, πππ π ππ ππ, ππ, ππ, ππ, πππ π ππ ππ, ππ, ππ, ππ, ππ, ππ, ππ, ππ, ππ, πππ ππ π π, ππ, ππ, ππ, ππ, ππ, ππ, ππ, ππ, ππ, β¦ , ππ, ππ, ππ, ππ, ππ, πππ ππ π π, π, ππ, ππ, ππ, ππ, ππ, ππ, ππ, ππ, β¦ , ππ, ππ, ππ, ππ, ππ, πππ ππ π π, π, π, π, ππ, ππ, ππ, ππ, ππ, ππ, ππ, β¦ , ππ, ππ, ππ, ππ, ππ, ππ, πππ ππ π π, π, π, π, π, π, β¦ , ππ, ππ, πππ πππ How do you think we can use these whole number factors in calculating percents on a double number line? οΊ The factors represent all ways by which we could break 100% into equal-sized whole number intervals. The multiples listed would be the percents representing each cumulative interval. The number of multiples would be the number of intervals. Lesson 5: Finding One Hundred Percent Given Another Percent This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M4-TE-1.3.0-09.2015 75 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 7β’4 Example 1 (5 minutes): Using a Modified Double Number Line with Percents The use of visual models is a powerful strategy for organizing and solving percent problems. In this example (and others that follow), the double number line is modified so that it is made up of a percent number line and a bar model. This model provides a visual representation of how quantities compare and what percent they correspond with. We use the greatest common factor of the given percent and 100 to determine the number of equal-sized intervals to use. Example 1: Using a Modified Double Number Line with Percents The ππ students who play wind instruments represent ππ% of the students who are in band. How many students are in band? ο§ Which quantity in this problem represents the whole? ο§ Draw the visual model shown with a percent number line and a tape diagram. ο§ Use the number line and tape diagram to find the total number of students in band. οΊ The total number of students in band is the whole, or 100%. MP.2 & MP.4 οΊ 3 100% represents the total number of students in band, and 75% is of 100%. The greatest common 4 factor of 75 and 100 is 25. ππ β ππ% ππ β ππ% π ππ π ( ) β πππ% π π(ππ) β πππ% ππ β πππ% Lesson 5: Finding One Hundred Percent Given Another Percent This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M4-TE-1.3.0-09.2015 76 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 5 7β’4 Exercises 1β3 (10 minutes) Solve Exercises 1β3 using a modified double number line. Exercises 1β3 1. Bobβs Tire Outlet sold a record number of tires last month. One salesman sold πππ tires, which was ππ% of the tires sold in the month. What was the record number of tires sold? The salesmanβs total is being compared to the total number of tires sold by the store, so the total number of tires sold is the whole quantity. The greatest common factor of ππ and πππ is ππ, so I divided the percent line into five equal-sized intervals of ππ%. ππ% is three of the ππ% intervals, so I divided the salesmanβs πππ tires by π and found that ππ tires corresponds with each ππ% interval. πππ% consists of five ππ% intervals, which corresponds to five groups of ππ tires. Since π β ππ = πππ, the record number of tires sold was πππ tires. 2. Nick currently has π, πππ points in his fantasy baseball league, which is ππ% more points than Adam. How many points does Adam have? Nickβs points are being compared to Adamβs points, so Adamβs points are the whole quantity. Nick has ππ% more points than Adam, so Nick really has πππ% of Adamβs points. The greatest common factor of πππ and πππ is ππ, so I divided the πππ% on the percent line into six equal-sized intervals. I divided Nickβs π, πππ points by π and found that π, πππ points corresponds to each ππ% interval. Five intervals of ππ% make πππ%, and five intervals of π, πππ points totals π, πππ points. Adam has π, πππ points in the fantasy baseball league. 3. Kurt has driven πππ miles of his road trip but has ππ% of the trip left to go. How many more miles does Kurt have to drive to get to his destination? With ππ% of his trip left to go, Kurt has only driven ππ% of the way to his destination. The greatest common factor of ππ and πππ is ππ, so I divided the percent line into ten equal-sized intervals. ππ% is three of the ππ% intervals, so I divided πππ miles by π and found that ππ miles corresponds to each ππ% interval. Ten intervals of ππ% make πππ%, and ten intervals of ππ miles totals πππ miles. Kurt has already driven πππ miles, and πππ β πππ = πππ, so Kurt has πππ miles left to get to his destination. Lesson 5: Finding One Hundred Percent Given Another Percent This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M4-TE-1.3.0-09.2015 77 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 7β’4 Example 2 (10 minutes): Mental Math Using Factors of πππ Students use mental math and factors of 100 to determine the whole quantity when given a quantity that is a percent of that whole. Example 2: Mental Math Using Factors of πππ Answer each part below using only mental math, and describe your method. a. If ππ is π% of a number, what is that number? How did you find your answer? ππ is π% of π, πππ. I found my answer by multiplying ππ β πππ because ππ corresponds with each π% in πππ%, and π% β πππ = πππ%, so ππ β πππ = π, πππ. b. If ππ is ππ% of a number, what is that number? How did you find your answer? ππ is ππ% of πππ. ππ is a factor of πππ, and there are ten ππ% intervals in πππ%. The quantity ππ corresponds to ππ%, so there are ππ β ππ in the whole quantity, and ππ β ππ = πππ. c. If ππ is π% of a number, what is that number? How did you find your answer? ππ is π% of πππ. π is a factor of πππ, and there are twenty π% intervals in πππ%. The quantity ππ corresponds to π%, so there are twenty intervals of ππ in the whole quantity. ππ β ππ ππ β π β ππ Factored ππ for easier mental math ππ β ππ πππ d. If ππ is ππ% of a number, what is that number? How did you find your answer? ππ is ππ% of πππ. ππ is not a factor of πππ, but ππ and πππ have a common factor of π. If ππ% is ππ, then because π = ππ ππ , π% is ππ = . There are twenty π% intervals in πππ%, so there are twenty intervals of π π ππ in the whole. ππ β ππ ππ β π β ππ Factored ππ for easier mental math ππ β ππ πππ e. If ππ is ππ% of a number, what is that number? How did you find your answer? ππ is ππ% of πππ. ππ is a factor of πππ, and there are four intervals of ππ% in πππ%. The quantity ππ corresponds with ππ%, so there are ππ β π in the whole quantity. ππ β π ππ β π β π Factored π for easier mental math ππ β π πππ Lesson 5: Finding One Hundred Percent Given Another Percent This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M4-TE-1.3.0-09.2015 78 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 7β’4 Exercises 4β5 (8 minutes) Solve Exercises 4 and 5 using mental math and factors of 100. Describe your method with each exercise. Exercises 4β5 Derrick had a π. πππ batting average at the end of his last baseball season, which means that he got a hit ππ% of the times he was up to bat. If Derrick had ππ hits last season, how many times did he bat? 4. The decimal π. πππ is ππ%, which means that Derrick had a hit ππ% of the times that he batted. His number of hits is being compared to the total number of times he was up to bat. The ππ hits corresponds with ππ%, and since ππ is a factor of πππ, πππ = ππ β π. I used mental math to multiply the following: ππ β π (ππ β π) β π Used the distributive property for easier mental math πππ β ππ πππ Derrick was up to bat πππ times last season. Nelson used ππ% of his savings account for his class trip in May. If he used $πππ from his savings account while on his class trip, how much money was in his savings account before the trip? 5. ππ% of Nelsonβs account was spent on the trip, which was $πππ. The amount that he spent is being compared to the total amount of savings, so the total savings represents the whole. The greatest common factor of ππ and πππ is π. ππ% is seven intervals of π%, so I divided $πππ by π to find that $ππ corresponds to π%. πππ% = π% β ππ, so the whole quantity is $ππ β ππ = $πππ. Nelsonβs savings account had $πππ in it before his class trip. Closing (2 minutes) ο§ What does the modified double number line method and the factors of 100 method have in common? οΊ ο§ Describe a situation where you would prefer using the modified double number line. οΊ ο§ Both methods involve breaking 100% into equal-sized intervals using the greatest common factor of 100 and the percent corresponding to the part. Answers will vary. Describe a situation where you would prefer using the factors of 100. οΊ Answers will vary. Lesson Summary To find πππ% of the whole, you can use a variety of methods, including factors of πππ (π, π, π, π, ππ, ππ, ππ, ππ, and πππ) and double number lines. Both methods will require breaking πππ% into equal-sized intervals. Use the greatest common factor of πππ and the percent corresponding to the part. Exit Ticket (5 minutes) Lesson 5: Finding One Hundred Percent Given Another Percent This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M4-TE-1.3.0-09.2015 79 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM Name 7β’4 Date Lesson 5: Finding One Hundred Percent Given Another Percent Exit Ticket 1. A tank that is 40% full contains 648 gallons of water. Use a double number line to find the maximum capacity of the water tank. 2. Loretta picks apples for her grandfather to make apple cider. She brings him her cart with 420 apples. Her grandfather smiles at her and says, βThank you, Loretta. That is 35% of the apples that we need.β Use mental math to find how many apples Lorettaβs grandfather needs. Describe your method. Lesson 5: Finding One Hundred Percent Given Another Percent This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M4-TE-1.3.0-09.2015 80 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 7β’4 Exit Ticket Sample Solutions 1. A tank that is ππ% full contains πππ gallons of water. Use a double number line to find the maximum capacity of the water tank. I divided the percent line into intervals of ππ% making five intervals of ππ% in πππ%. I know that I have to divide ππ π to get ππ, so I divided πππ π to get πππ that corresponds with ππ%. Since there are five ππ% intervals in πππ%, there are five πππ gallon intervals in the whole quantity, and πππ β π = π, πππ. The capacity of the tank is π, πππ gallons. 2. Loretta picks apples for her grandfather to make apple cider. She brings him her cart with πππ apples. Her grandfather smiles at her and says βThank you, Loretta. That is ππ% of the apples that we need.β Use mental math to find how many apples Lorettaβs grandfather needs. Describe your method. πππ is ππ% of π, πππ. ππ is not a factor of πππ, but ππ and πππ have a common factor of π. There are seven intervals of π% in ππ%, so I divided πππ apples into seven intervals; πππ π = ππ. There are ππ intervals of π% in πππ%, so I multiplied as follows: ππ β ππ ππ β π β ππ πππ β ππ π, πππ Lorettaβs grandfather needs a total of π, πππ apples to make apple cider. Problem Set Sample Solutions Use a double number line to answer Problems 1β5. 1. Tanner collected πππ cans and bottles while fundraising for his baseball team. This was ππ% of what Reggie collected. How many cans and bottles did Reggie collect? The greatest common factor of ππ and πππ is ππ. π π π (ππ%) = ππ%, and (πππ) = πππ, so πππ corresponds with ππ%. π There are five intervals of ππ% in πππ%, and π(πππ) = πππ, so Reggie collected πππ cans and bottles. Lesson 5: Finding One Hundred Percent Given Another Percent This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M4-TE-1.3.0-09.2015 81 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 2. 7β’4 Emilio paid $πππ. ππ in taxes to the school district that he lives in this year. This yearβs taxes were a ππ% increase from last year. What did Emilio pay in school taxes last year? The greatest common factor of πππ and πππ is π. There are ππ intervals of π% in πππ%, and πππ.π ππ = ππ. π, so ππ. π corresponds with π%. There are ππ intervals of π% in πππ%, and ππ(ππ. π) = πππ, so Emilio paid $πππ in school taxes last year. 3. A snowmobile manufacturer claims that its newest model is ππ% lighter than last yearβs model. If this yearβs model weighs πππ π₯π., how much did last yearβs model weigh? ππ% lighter than last yearβs model means ππ% less than πππ% of last yearβs modelβs weight, which is ππ%. The greatest common factor of ππ and πππ is π. There are ππ intervals of π% in ππ%, and πππ ππ = ππ, so ππ corresponds with π%. There are ππ intervals of π% in πππ%, and ππ(ππ) = πππ, so last yearβs model weighed πππ pounds. 4. Student enrollment at a local school is concerning the community because the number of students has dropped to πππ, which is a ππ% decrease from the previous year. What was the student enrollment the previous year? A ππ% decrease implies that this yearβs enrollment is ππ% of last yearβs enrollment. The greatest common factor of ππ and πππ is ππ. There are π intervals of ππ% in ππ%, and πππ π = πππ, so πππ corresponds to ππ%. There are π intervals of ππ% in πππ%, and π(πππ) = πππ, so the student enrollment from the previous year was πππ students. Lesson 5: Finding One Hundred Percent Given Another Percent This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M4-TE-1.3.0-09.2015 82 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 5. 7β’4 The color of paint used to paint a race car includes a mixture of yellow and green paint. Scotty wants to lighten the color by increasing the amount of yellow paint ππ%. If a new mixture contains π. π liters of yellow paint, how many liters of yellow paint did he use in the previous mixture? The greatest common factor of πππ and πππ is ππ. There are ππ intervals of ππ% in πππ%, and π.π ππ = π. π, so π. π corresponds to ππ%. There are ππ intervals of ππ% in πππ%, and ππ(π. π) = π, so the previous mixture included π liters of yellow paint. Use factors of πππ and mental math to answer Problems 6β10. Describe the method you used. 6. Alexis and Tasha challenged each other to a typing test. Alexis typed ππ words in one minute, which was πππ% of what Tasha typed. How many words did Tasha type in one minute? The greatest common factor of πππ and πππ is ππ, and there are π intervals of ππ% in πππ%, so I divided ππ into π equal-sized intervals to find that π corresponds to ππ%. There are five intervals of ππ% in πππ%, so there are five intervals of π words in the whole quantity. π β π = ππ, so Tasha typed ππ words in one minute. 7. Yoshi is π% taller today than she was one year ago. Her current height is πππ ππ¦. How tall was she one year ago? π% taller means that Yoshiβs height is πππ% of her height one year ago. The greatest common factor of πππ and πππ is π, and there are ππ intervals of π% in πππ%, so I divided πππ into ππ equal-sized intervals to find that π ππ¦ corresponds to π%. There are ππ intervals of π% in πππ%, so there are ππ intervals of π ππ¦ in the whole quantity. ππ β π ππ¦ = πππ ππ¦, so Yoshi was πππ ππ¦ tall one year ago. 8. Toya can run one lap of the track in π π¦π’π§. π π¬ππ., which is ππ% of her younger sister Nikiβs time. What is Nikiβs time for one lap of the track? π π¦π’π§. π π¬ππ = ππ π¬ππ. The greatest common factor of ππ and πππ is ππ, and there are nine intervals of ππ in ππ, so I divided ππ π¬ππ. by π to find that π π¬ππ. corresponds to ππ%. There are ππ intervals of ππ% in πππ%, so ππ intervals of π π¬ππ. represents the whole quantity, which is ππ π¬ππ. ππ π¬ππ. = π π¦π’π§. ππ π¬ππ. Niki can run one lap of the track in π π¦π’π§. ππ π¬ππ. 9. An animal shelter houses only cats and dogs, and there are ππ% more cats than dogs. If there are ππ cats, how many dogs are there, and how many animals are there total? ππ% more cats than dogs means that the number of cats is πππ% the number of dogs. The greatest common factor of πππ and πππ is ππ. There are π intervals of ππ% in πππ%, so I divided the number of cats into π intervals to find that π corresponds to ππ%. There are four intervals of ππ% in πππ%, so there are four intervals of π in the whole quantity. π β π = ππ. There are ππ dogs in the animal shelter. The number of animals combined is ππ + ππ = ππ, so there are ππ animals in the animal shelter. 10. Angie scored ππ points on a test but only received a ππ% grade on the test. How many points were possible on the test? The greatest common factor of ππ and πππ is π. There are 13 intervals of π% in ππ%, so I divided ππ points into ππ intervals and found that π points corresponds to π%. There are ππ intervals of π% in πππ%, so I multiplied π points times ππ, which is πππ points. There were πππ points possible on Angieβs test. Lesson 5: Finding One Hundred Percent Given Another Percent This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M4-TE-1.3.0-09.2015 83 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 7β’4 For Problems 11β17, find the answer using any appropriate method. 11. Robbie owns ππ% more movies than Rebecca, and Rebecca owns ππ% more movies than Joshua. If Rebecca owns πππ movies, how many movies do Robbie and Joshua each have? Robbie owns πππ movies, and Joshua owns πππ movies. π π 12. ππ% of the seventh-grade students have math class in the morning. ππ % of those students also have science class in the morning. If ππ seventh-grade students have math class in the morning but not science class, find how many seventh-grade students there are. There are πππ seventh-grade students. 13. The school bookstore ordered three-ring notebooks. They put ππ% of the order in the warehouse and sold ππ% of the rest in the first week of school. There are ππ notebooks left in the store to sell. How many three-ring notebooks did they originally order? The store originally ordered πππ three-ring notebooks. 14. In the first game of the year, the modified basketball team made ππ. π% of their foul shot free throws. Matthew made all π of his free throws, which made up ππ% of the teamβs free throws. How many free throws did the team miss altogether? The team attempted ππ free throws, made ππ of them, and missed π. 15. Aidenβs mom calculated that in the previous month, their family had used ππ% of their monthly income for gasoline, and ππ% of that gasoline was consumed by the familyβs SUV. If the familyβs SUV used $πππ. ππ worth of gasoline last month, how much money was left after gasoline expenses? The amount of money spent on gasoline was $πππ; the monthly income was $π, πππ. ππ. The amount left over after gasoline expenses was $πππ. ππ. 16. Rectangle A is a scale drawing of Rectangle B and has ππ% of its area. If Rectangle A has side lengths of π ππ¦ and π ππ¦, what are the side lengths of Rectangle B? A ππ«πππ¨ = π₯ππ§π ππ‘ × π°π’πππ‘ π cm ππ«πππ¨ = (π ππ¦)(π ππ¦) B ππ«πππ¨ = ππ ππ¦π π cm The area of Rectangle A is ππ% of the area of Rectangle B. ππ% × π = πππ% ππ × π = ππ So, the area of Rectangle B is ππ ππ¦π. The value of the ratio of area π¨ to area π© is the square of the scale factor of the side lengths π¨: π©. The value of the ratio of area π¨: π© is ππ π π π π π = , and = ( ) , so the scale factor of the side lengths π¨: π© is . ππ π π π π So, using the scale factor: π (π₯ππ§π ππ‘π© ) = π ππ¦; π₯ππ§π ππ‘π© = ππ ππ¦ π π (π°π’πππ‘π© ) = π ππ¦; π°π’πππ‘π© = π ππ¦ π The dimensions of Rectangle B are π ππ¦ and ππ ππ¦. Lesson 5: Finding One Hundred Percent Given Another Percent This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M4-TE-1.3.0-09.2015 84 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 7β’4 17. Ted is a supervisor and spends ππ% of his typical work day in meetings and ππ% of that meeting time in his daily team meeting. If he starts each day at 7:30 a.m., and his daily team meeting is from 8:00 a.m. to 8:20 a.m., when does Tedβs typical work day end? ππ minutes is Ted spends π π π π of an hour since ππ ππ π = . π hour in his daily team meeting, so π π π π ππ% in πππ%, and π ( ) = , so Ted spends π π π π π π corresponds to ππ% of his meeting time. There are π intervals of hours in meetings. of an hour corresponds to ππ% of Tedβs work day. π π There are π intervals of ππ% in πππ%, and π ( ) = Since π π ππ ππ ππ π , so Ted spends hours working. hours = π hours. π π π π hour = ππ minutes, Ted works a total of π hours ππ minutes. If he starts at 7:30 a.m., he works π π π π π π π π π hours ππ minutes until 12:00 p.m., and since π β π = π , Ted works another π hours after 12:00 p.m. π π hour = ππ minutes, and π π hour = ππ minutes, so Ted works π hours ππ minutes after 12:00 p.m., which is 3:50 p.m. Therefore, Tedβs typical work day ends at 3:50 p.m. Lesson 5: Finding One Hundred Percent Given Another Percent This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M4-TE-1.3.0-09.2015 85 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
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