lesson 15 homework 3rd grade

Curriculum  /  Math  /  3rd Grade  /  Unit 6: Fractions  /  Lesson 15

Lesson 15 of 24

Criteria for Success

Tips for teachers, anchor tasks.

Problem Set

Target Task

Additional practice.

Explain equivalence by manipulating units and reasoning about their size.

Common Core Standards

Core standards.

The core standards covered in this lesson

Number and Operations—Fractions

3.NF.A.3.A — Understand two fractions as equivalent (equal) if they are the same size, or the same point on a number line.

3.NF.A.3.B — Recognize and generate simple equivalent fractions, e.g., 1/2 = 2/4, 4/6 = 2/3). Explain why the fractions are equivalent, e.g., by using a visual fraction model.

3.NF.A.3.C — Express whole numbers as fractions, and recognize fractions that are equivalent to whole numbers. Example: express 3 in the form 3 = 3/1; recognize that 6/1 = 6. Example: locate 4/4 and 1 at the same point of a number line diagram.

Foundational Standards

The foundational standards covered in this lesson

Measurement and Data

2.MD.A.2 — Measure the length of an object twice, using length units of different lengths for the two measurements; describe how the two measurements relate to the size of the unit chosen.

The essential concepts students need to demonstrate or understand to achieve the lesson objective

• Understand that since equivalent fractions represent the same-sized part of the same-sized whole, the whole that is partitioned into more pieces must have more relevant pieces that constitute its equivalent fraction (MP.7, MP.8). Begin to see this relationship as a multiplicative one (although this explicit understanding is not required until Grade 4).
• Generate simple equivalent fractions in all cases, including those with whole numbers.
• Explain the equivalence of fractions in all cases, including those with whole numbers, using an area model, number line, or other method (MP.3, MP.5).

Suggestions for teachers to help them teach this lesson

This lesson previews the work of Grade 4 of developing an algorithm for finding equivalent fractions, but it also recaps every case of equivalent fractions students have seen thus far in the unit. Therefore, this lesson is optional though highly encouraged since it serves to summarize their work thus far as well as connect to future work on the topic.

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Tasks designed to teach criteria for success of the lesson, and guidance to help draw out student understanding

a.   

Partition the following area model into thirds. Then write a fraction to represent the whole.

Partition the following area model into sixths. Then write a fraction to represent the whole.

Partition the following area model into ninths. Then write a fraction to represent the whole.

b.   What do you notice about the number of parts and the size of each part in each model in Part (a)? What do you wonder?

Guiding Questions

• Partition the following number line into wholes. Label each tick mark with a fraction.

• Partition the following number line into halves. Label each tick mark with a fraction.
• Partition the following number line into fourths. Label each tick mark with a fraction.
• Partition the following number line into eighths. Label each tick mark with a fraction.

Unlock the answer keys for this lesson's problem set and extra practice problems to save time and support student learning.

Discussion of Problem Set

• How did you use the patterns we noticed in the Anchor Tasks to solve #1 without needing to draw a model for every fraction?
• What happened to the size of the equal parts in #2a? What happened to the number of equal parts in #2a? How are those related?
• How did you share the chocolate bars equally? What fraction of a chocolate bar did each friend get? What fraction of all the chocolate bars collectively did each friend get? How do these questions demonstrate the importance of specifying the whole?
• Describe the approach you took to solving #5. Is there more than one correct answer?

A task that represents the peak thinking of the lesson - mastery will indicate whether or not objective was achieved

Two fractions have different numerators and denominators. Is it possible for the two fractions to be located at the same point on the number line? Why or why not?

Student Response

An example response to the Target Task at the level of detail expected of the students.

The Extra Practice Problems can be used as additional practice for homework, during an intervention block, etc. Daily Word Problems and Fluency Activities are aligned to the content of the unit but not necessarily to the lesson objective, therefore feel free to use them anytime during your school day.

Extra Practice Problems

Answer keys for Problem Sets and Extra Practice Problems are available with a Fishtank Plus subscription.

Word Problems and Fluency Activities

Help students strengthen their application and fluency skills with daily word problem practice and content-aligned fluency activities.

Topic A: Understanding Unit Fractions and Building Non-Unit Fractions

Partition a whole into equal parts using area models, identifying fractional units.

3.G.A.2 3.NF.A.1

Partition a whole into equal parts using tape diagrams (i.e., fraction strips), identifying and writing unit fractions in fraction notation.

Partition a whole into equal parts using area models and tape diagrams, identifying and writing non-unit fractions in fraction notation.

Identify fractions of a whole that is not partitioned into equal parts.

Draw the whole when given the unit fraction.

Identify a shaded fractional part in different ways, depending on the designation of the whole.

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Topic B: Fractions on a Number Line

Partition a number line from 0 to 1 into fractional units.

Place any fraction on a number line with endpoints 0 and 1.

Place any fraction on a number line with endpoints 0 and another whole number greater than 1.

Place any fraction on a number line with endpoints greater than 0.

3.NF.A.2 3.NF.A.3.C

Place various fractions on a number line where the given interval is not a whole.

3.NF.A.2 3.NF.A.3.D

Topic C: Equivalent Fractions

Understand two fractions as equivalent if they are the same point on a number line referring to the same whole. Use this understanding to generate simple equivalent fractions.

3.NF.A.3.A 3.NF.A.3.B

Understand two fractions as equivalent if they are the same sized pieces of the same sized wholes, though not necessarily the same shape. Use this understanding to generate simple equivalent fractions.

Express whole numbers as fractions, and recognize fractions that are equivalent to whole numbers.

3.NF.A.3.A 3.NF.A.3.B 3.NF.A.3.C

Topic D: Comparing Fractions

Compare unit fractions (a unique case of fractions with the same numerators) by reasoning about the size of their units. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols >, =, or <.

Compare fractions with the same numerators by reasoning about the size of their units. Record the results of comparisons with the symbols >, =, or <.

Compare fractions with the same denominators by reasoning about their number of units. Record the results of comparisons with the symbols >, =, or <.

Compare and order fractions using various methods.

Understand fractions as numbers.

Topic E: Line Plots

Measure lengths to the nearest half inch.

Measure lengths to the nearest quarter inch. 

Generate measurement data and represent it in a line plot.

Create line plots (dot plots). 

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Unit 1: Intro to multiplication

Unit 2: 1-digit multiplication, unit 3: addition, subtraction, and estimation, unit 4: intro to division, unit 5: understand fractions, unit 6: equivalent fractions and comparing fractions, unit 7: more with multiplication and division, unit 8: arithmetic patterns and problem solving, unit 9: quadrilaterals, unit 10: area, unit 11: perimeter, unit 12: time, unit 13: measurement, unit 14: represent and interpret data.

Eureka Math Grade 3 Module 3 Lesson 15 Answer Key

Eureka Grade 3 ensures students with the learning targets and success criteria. The Diverse opportunities will develop problem-solving skills among the primary school kids practicing Bigideas Math Grade 3 Solutions. Eureka Textbook Grade 3 answers help Students Master the subject on Consistent practice.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 15 Answer Key

Grade 3 Eureka Math Answer Key is provided by subject experts as per the Common Core curriculum. Grade 3 Eureka Math Solutions provided educates the primary school Kids on all Math concepts of the chapter efficiently. Eureka Math Grade 3 Textbook Answers makes it easy for kids to Grasp the Concepts as well as solve any problem from chapter Tests, Practice Tests, Performance tests, cumulative practice, etc.

Eureka Math Grade 3 Module 3 Lesson 15 Pattern Sheet Answer Key

Answer: 9 x 1 = 9, 9 x 2 = 18, 9 x 3 = 27, 9 x 4 = 36, 9 x 5 = 45, 9 x 6 = 54, 9 x 7 = 63, 9 x 8 = 72, 9 x 9 = 81, 9 x 10 = 90, 9 x 5 = 45, 9 x 6 = 54, 9 x 5 = 45, 9 x 7 = 63, 9 x 5 = 45, 9 x 8 = 72, 9 x 5 = 45, 9 x 9 = 81, 9 x 5 = 45, 9 x 10 = 90, 9 x 6 = 54, 9 x 5 = 45, 9 x 6 = 54, 9 x 7 = 63, 9 x 6 = 54, 9 x 8 = 72, 9 x 6 = 54, 9 x 9 = 81, 9 x 6 = 54, 9 x 7 = 63, 9 x 6 = 54, 9 x 7 = 63, 9 x 8 = 72, 9 x 7 = 63, 9 x 9 = 81, 9 x 7 = 63, 9 x 8 = 72, 9 x 6 = 54, 9 x 8 = 72, 9 x 7 = 63, 9 x 8 = 72, 9 x 9 = 81, 9 x 9 = 81, 9 x 6 = 54, 9 x 9 = 81, 9 x 7 = 63, 9 x 9 = 81, 9 x 8 = 72, 9 x 9 = 81, 9 x 8 = 72, 9 x 6 = 54, 9 x 9 = 81, 9 x 7 = 63, 9 x 9 = 81, 9 x 6 = 54, 9 x 8 = 72, 9 x 9 = 81, 9 x 7 = 63, 9 x 6 = 54, 9 x 8 = 72.

Eureka Math Grade 3 Module 3 Lesson 15 Problem Set Answer Key

Write an equation, and use a letter to represent the unknown for Problems 1 – 6.

Question 1. Mrs. Parson gave each of her grandchildren $9. She gave a total of $36. How many grandchildren does Mrs. Parson have?

Answer: The number of grandchildren = 4.

Explanation: In the above-given question, given that, Mrs. Parson gave each of her grandchildren $9. she gave a total of $36. 9 / c = 36. where c indicates children. c = 36 / 9. c = 4.

Question 2. Shiva pours 27 liters of water equally into 9 containers. How many liters of water are in each container?

Answer: The number of liters of water in each container = 3.

Explanation: In the above-given question, given that, shiva pours 27 liters of water equally into 9 containers. 27 / w = 9. w = 27 / 9. where w indicates water. w = 3.

Question 3. Derek cuts 7 pieces of wire. Each piece is 9 meters long. What is the total length of the 7 pieces?

Answer: The total length of the 7 pieces = 63.

Explanation: In the above-given question, given that, Derek cuts 7 pieces of wire. Each piece is 9 meters long. 7 x l = 9. where l indicates length. l = 7 x 9. l = 63.

Question 4. Aunt Deena and Uncle Chris share the cost of a limousine ride with their 7 friends. The ride cost a total of $63. If everyone shares the cost equally, how much does each person pay?

Answer: The cost each person pays = 9.

Explanation: In the above-given question, given that, Aunt Deena and Uncle Chris share the cost of a limousine ride with their 7 friends. The ride cost a total of $63. 7 / r = 63. where r indicates ride. r = 63 / 7. r = 9.

Question 5. Cara bought 9 packs of beads. There are 10 beads in each pack. She always uses 30 beads to make each necklace. How many necklaces can she make if she uses all the beads?

Answer: The number of necklaces = 3.

Explanation: In the above-given question, given that, Cara bought 9 packs of beads. There are 10 beads in each pack. she always uses 30 beads to make each necklace. 9 / n = 90. where c indicates children. n = 90 / 9. n = 30.

Question 6. There are 8 erasers in a set. Damon buys 9 sets. After giving some erasers away, Damon has 35 erasers left. How many erasers did he give away?

Answer: The number of erasers = 37.

Explanation: In the above-given question, given that, There are 8 erasers in a set. Damon buys 9 sets. After giving some erasers away. Damon has 35 erasers left. where e indicates eraser. e = 72 / 9. e = 8.

Eureka Math Grade 3 Module 3 Lesson 15 Exit Ticket Answer Key

Use a letter to represent the unknown.

Question 1. Mrs. Aquino pours 36 liters of water equally into 9 containers. How much water is in each container?

Answer: The water in each container = 4.

Explanation: In the above-given question, given that, Mrs. Aquino pours 36 liters of water equally into 9 containers. where c indicates container. c = 36 / 9. c = 4.

Question 2. Marlon buys 9 packs of hot dogs. There are 6 hot dogs in each pack. After the barbeque, 35 hot dogs are left over. How many hot dogs were eaten?

Answer: The number of hot dogs were eaten = 19.

Explanation: In the above-given question, given that, Marlon buys 9 packs of hot dogs. There are 6 hot dogs in each pack. 9 x 6 = 54. where n indicates a number of hot dogs. n = 54 – 35. n = 19. Eureka Math Grade 3 Module 3 Lesson 15 Homework Answer Key

Question 1. The store clerk equally divides 36 apples among 9 baskets. Draw a tape diagram, and label the number of apples in each basket as a. Write an equation, and solve for a.

Answer: The number of apples in each basket = 4.

Explanation: In the above-given question, given that, The store clerk equally divides 36 apples among 9 baskets. 9 / n = 36. where n indicates apples. n = 36 / 9. n = 4.

Question 2. Elijah gives each of his friends a pack of 9 almonds. He gives away a total of 45 almonds. How many packs of almonds did he give away? Model using a letter to represent the unknown, and then solve.

Answer: The number of packs of almonds = 5.

Explanation: In the above-given question, given that, Elijah gives each of his friends a pack of 9 almonds. He gives away a total of 45 almonds. 9 / p = 45. where p indicates pack. p = 45 / 9. p = 5.

Question 3. Denice buys 7 movies. Each movie costs $9. What is the total cost of 7 movies? Use a letter to represent the unknown. Solve.

Answer: The total cost of 7 movies = 63.

Explanation: In the above-given question, given that, Denice buys 7 movies. Each movie costs $9. 7 x  m= 9. where m indicates total movies. m = 7 x 9. m = 63.

Question 4. Mr. Doyle shares 1 roll of bulletin board paper equally with 8 teachers. The total length of the roll is 72 meters. How much bulletin board paper does each teacher get?

Answer: The bulletin board does each teacher get = 9 meters.

Explanation: In the above-given question, given that, Mr. Doyle shares 1 roll of bulletin board paper equally with 8 teachers. The total length of the roll is 72 meters. 72 / 8 = 9.

Question 5. There are 9 pens in a pack. Ms. Ochoa buys 9 packs. After giving her students some pens, she has 27 pens left. How many pens did she give away?

Answer: The number of pens she give = 54.

Explanation: In the above-given question, given that, There are 9 pens in a pack. Ms. Ochoa buys 9 packs. After giving her students some pens, she has 27 pens left. 9 x 9 = 81. 81 – 27 = 54.

Question 6. Allen buys 9 packs of trading cards. There are 10 cards in each pack. He can trade 30 cards for a comic book. How many comic books can he get if he trades all of his cards?

Answer: The number of comic books can he get if he trades all of his cards = 60.

Explanation: In the above-given question, given that, Allen buys 9 packs of trading cards. There are 10 cards in each pack. He can trade 30 cards for a comic book. 9 x 10 = 90. 90 – 30 = 60.

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Eureka Math Grade 5 Module 5 Lesson 15 Answer Key

Engage ny eureka math 5th grade module 5 lesson 15 answer key, eureka math grade 5 module 5 lesson 15 problem set answer key.

Question 1. The length of a flowerbed is 4 times as long as its width. If the width is “\(\frac{3}{8}\) meter, what is the area? Answer:

The width of the flower bed = 3/8 meters

The length of the flower bed is 4 times as long as its width

Which means, 3/8 x 4 = 12/ 8 = 3/2

Now, Area = length x width

A= 3/2 x 3/8

Therefore, the area of the flowerbed is 9/16 m 2 .

The measurement of basil plot on each side = 5/9 yard

A= 5/8 x 5/8 = 25/64

Therefore, the area of the basil plot = 25/64 yd 2

The measurement of basil plot in each side = 5/9 yard

=  15/8 or 1 7/8 feet

Now, the length of one side

2 feet + 2 feet + 1 7/8 feet

Perimetre = 4 x sides

= 4 x 5 7/8

= 20 + 3 4/8

Therefore, the perimetre of the fence = 23 1/2 feet.

The total area that the fence encloses

= length x width

= 5 7/8 x 5 7/8

=  ( 5 x 5 ) + ( 5 x 7/8 ) +( 5 x 7/8 ) + ( 7/8 x 7/8 )

= 25 + 4 3/8 + 4 3/8 + 49/64

= 33 + 48/64 + 49/64

= 33 + 97/64

Therefore, the area that fence encloses = 34 33/64 square feet.

Question 3. Janet bought 5 yards of fabric 2\(\frac{1}{4}\)-feet wide to make curtains. She used \(\frac{1}{3}\) of the fabric to make a long set of curtains and the rest to make 4 short sets. a. Find the area of the fabric she used for the long set of curtains. b. Find the area of the fabric she used for each of the short sets. Answer:

Area = length x width

= 5 x 2 1/4

= 11 1/4 square feet

Therefore, the area of long set is 11 1/4 square feet

15 feet – 5 feet = 10 feet

1/4 of 10 is 2 1/2 feet

Now, the area of each set of shorter curtain

A = 2 1/2 x 2 1/2

( 2 x 2 ) + ( 2 x 1/2 ) + ( 2 x 1/2 ) + ( 1/2 x 1/2 )

= 4 + 1/2 + 1 + 1/8

= 5 + 1/2 +1/8

= 5 5/8 square feet

Therefore, the area of each set of shorter curtain is 5 5/8 square feet.

Question 4. Some wire is used to make 3 rectangles: A, B, and C. Rectangle bs dimensions are \(\frac{3}{5}\) cm larger than Rectangle A’s dimensions, and Rectangle C’s dimensions are \(\frac{3}{5}\) cm larger than Rectangle B’s dimensions. Rectangle A is 2 cm by 3\(\frac{1}{5}\) cm. a. What is the total area of all three rectangles? b. If a 40-cm coil of wire was used to form the rectangles, how much wire is left? Answer:

The area of rectangle A =

A = 2 X 3 1/5

= 6 2/5 square centimetres

Area of of rectangle B =

A = 2 3/5 cm x 3 4/5 cm

= 6 + 8/5 + 9/5 + 12/25

= 9 + 21/5 + 12/25

= 9 22/25 square centimetres

Area of rectangle C =

A =  3 1/5 cm x 4 2/5 cm

= 12 + 6/5 + 4/5 + 2/25

= 14 2/25 square centimetres

Total area of three rectangles =

= 6 2/5 + 9 22/25 + 14 2/25

= 29 + 2/5 + 24/25

= 29 +10/25 +24/25

= 30 9/25 square centimetres

Therefore, the area of three rectangles = 30 9/25 square centimetres.

Perimeter of rectangle A =

2 + 2 + 3 1/5 + 3 1/5

= 6 2/5  centimetres

Perimeter of rectangle B =

2 3/5 + 2 3/5 + 3 4/5 + 3 4/5

= 12 4/5 centimetres

Perimeter of rectangle C =

3 1/5 + 3 1/5 + 4 2/5 + 4 2/5

= 15 1/5 centimetres

Total perimeter =

1 2/5 cm + 12 4/5 cm + 15 1/5 cm

= 38 2/5 cm

Now, according to the given condition :

40 cm – 38 2/5 cm

Therefore, the leftover wire length = 1 3/5 cm.

Eureka Math Grade 5 Module 5 Lesson 15 Exit Ticket Answer Key

Wheat grass is grown in planters that are 3\(\frac{1}{2}\) inch by 1\(\frac{3}{4}\) inch. If there is a 6 × 6 array of these planters with no space between them, what is the area covered by the planters? Answer:

The area of planter =

3 1/2 x 1 3/4

( 3 x  1 ) + ( 3 x 3/4 ) + ( 1 x 1/2 ) + ( 1/2 x 3/4 )

= 3 + 9/4 + 1/2 + 3/8

= 24 + 18/8 + 4/8 + 3/8

=49/8 = 6 1/8

According to given condition

= 36 x  6 1/8

= 200 + 1/2

Therefore, the area covered by the planters = 200 1/2 square inches.

Eureka Math Grade 5 Module 5 Lesson 15 Homework Answer Key

Question 1. The width of a picnic table is 3 times its length. If the length is \(\frac{5}{6}\)-yd long, what is the area of the picnic table in square feet? Answer:

1 yard = 3 feet

The area of the picnic table =

5/6 x 3= 15/6 = 5/2 = 2 1/2 feet

Area = 2 1/2 x 7 1/2

A = 5/2 x 15 1/2

= 18 3/4  square feet

Therefore, the area of the table = 2 1/2 square yards.

The area of window A =

= 6 1/4 x  5 3/4

= 30 x  1 1/4 + 4 2/4 + 3/16

= 35 + 3/4 + 3/16

= 35 + 15/16

= 35 15/16 square feet

The area of window B =

= 12 1/2 square feet

The area of D  =

4 x 8 = 32 square feet

The total area of the wall =

52 1/2 x 33

= 1716 + 33/2

= 1732 1/2 square feet

Now, the total area of A,B,C A and D

= 35 15/16 +12 1/2 + 32 + 9 1/2

= 22 + 35 15/16 +32

Now, the area of the painted wall =

1732 1/2 sq. ft – 89 15/16

= 1642 9/16 square feet

Therefor, the area of the wall painted = 1642 9/16 square feet.

The area of the smallest rectangle =

4 1/2 x 7 3/4

= 24 + 3 + 3 1/2 + 3/8

= 34 7/8 square inches

4 1/2 + 2 1/4

= 4 2/4 + 2 1/4

7 3/4 + 2 1/4 = 10

36 3/4 x 10

= 60 + 7 2/4

= 67 1/2 square inches

6 3/4 + 2 1/4 =9

9 x 12 1/4 =

= 108 + 2 1/4 = 110 1/4 square inches

d. 9 + 2 1/4 = 11 1/4

12 1/4 + 2 1/4 = 14 2/4

11 1/4 x 14 2/4

= 154 + 3 2/4 + 5 1/2 + 1/8

= 163 1/8 square inches.

So, the total area of the figure = 110 + 34 + 67 + 163

= 374 + 1 3/4

= 375 3/4 square inches

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