eureka math grade 5 lesson 2 homework 5.3

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Eureka Math Student Materials: Grades K–5

Learn, Practice, Succeed

Learn, Practice, and Succeed from   Eureka Math™   offer teachers multiple ways to differentiate instruction, provide extra practice, and assess student learning. These versatile companions to   A Story of Units®   (Grades K–5) guide teachers in response to intervention (RTI), provide extra practice, and inform instruction.

Also available for Grades 6–8 . 

Learn, Practice, Succeed can be purchased all together or bundled in any configuration. Contact your account solutions manager for more information and pricing.  

The Learn book serves as a student’s in-class companion where they show their thinking, share what they know, and watch their knowledge build every day!

Application Problems:  Problem solving in a real-world context is a daily part of   Eureka Math , building student confidence and perseverance as students apply their knowledge in new and varied ways.

Problem Sets :  A carefully sequenced Problem Set provides an in-class opportunity for independent work, with multiple entry points for differentiation.

Exit Tickets:   These exercises check student understanding, providing the teacher with immediate, valuable evidence of the efficacy of that day’s instruction and informing next steps.

Templates:   Learn   includes templates for the pictures, reusable models, and data sets that students need for   Eureka Math   activities.

With   Practice , students build competence in newly acquired skills and reinforce previously learned skills in preparation for tomorrow’s lesson.   Together,   Learn   and   Practice   provide all the print materials a student uses for their core instruction.

Eureka Math  contains multiple daily opportunities to build fluency in   mathematics . Each is designed with the same notion—growing every student’s ability to use mathematics   with ease . Fluency experiences are generally fast-paced and energetic, celebrating improvement and focusing on recognizing patterns and connections within the material.

Eureka Math   fluency activities provide differentiated practice through a variety of formats—some are conducted orally, some use manipulatives, others use a personal whiteboard, or a handout and paper-and-pencil format.

Sprints:  Sprint fluency activities in  Eureka Math Practice  build speed and accuracy with already acquired skills. Used when students are nearing optimum proficiency, Sprints leverage tempo to build a low-stakes adrenaline boost that increases memory and recall. Their intentional design makes Sprints inherently differentiated – the problems build from simple to complex, with the first quadrant of problems being the simplest, and each subsequent quadrant adding complexity.

Eureka Math Succeed   enables students to work individually toward mastery.  Teachers and tutors can use  Succeed   books from prior grade levels as curriculum-consistent tools for filling gaps in foundational knowledge. Students will thrive and progress more quickly, as familiar models facilitate connections to their current, grade-level content.

Additional Problem Sets:  Ideal for Homework or extra practice, these additional problem sets align lesson-by-lesson with what is happening in the classroom. These problems are sequenced from simple-to-complex to naturally scaffold student practice. They align with   Eureka Math   and use the curriculum’s mathematical models and language, ensuring that students feel the connections and relevance to their daily instruction, whether they are working on foundational skills or getting extra practice on the current topic.

Homework Helpers:   Each problem set is accompanied by a Homework Helper, a set of worked examples that illustrate how similar problems are solved. The examples, viewed side by side with the homework, support students as they reinforce the day’s learning. Homework Helpers are also a great way to keep parents informed about math class.

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Eureka Math Grade 5 Module 3 Lesson 13 Answer Key

Engage ny eureka math 5th grade module 3 lesson 13 answer key, eureka math grade 5 module 3 lesson 13 problem set answer key.

Question 1. Are the following expressions greater than or less than 1? Circle the correct answer. a. \(\frac{1}{2}\) + \(\frac{2}{7}\)    greater than 1    less than 1

b. \(\frac{5}{8}\) + \(\frac{3}{5}\)   greater than 1     less than 1

c. 1\(\frac{1}{4}\) – \(\frac{1}{3}\)   greater than 1     less than 1

b. \(\frac{5}{8}\) + \(\frac{3}{5}\) lcm of 8 and 5 is 40. \(\frac{25}{40}\) + \(\frac{24}{40}\)   = \(\frac{49}{40}\) = 1\(\frac{9}{40}\)

c. 1\(\frac{1}{4}\) – \(\frac{1}{3}\)   = \(\frac{5}{4}\) – \(\frac{1}{3}\) lcm of 4 and 3 is 12. \(\frac{15}{12}\) – \(\frac{4}{12}\) = \(\frac{11}{12}\)

d. 3\(\frac{5}{8}\) – 2\(\frac{5}{9}\)  = \(\frac{29}{8}\) – \(\frac{23}{9}\) lcm of 8 and 9 is 72 . \(\frac{261}{72}\) – \(\frac{184}{72}\)  = \(\frac{77}{72}\) = 1\(\frac{5}{72}\) .

Question 2. Are the following expressions greater than or less than \(\frac{1}{2}\) ? Circle the correct answer. a. \(\frac{1}{4}\) + \(\frac{2}{3}\)      greater than \(\frac{1}{2}\)          less than \(\frac{1}{2}\)

b.\(\frac{3}{7}\) – \(\frac{1}{8}\)        greater than \(\frac{1}{2}\)          less than \(\frac{1}{2}\)

c. 1\(\frac{1}{7}\) – \(\frac{7}{8}\)       greater than \(\frac{1}{2}\)        less than \(\frac{1}{2}\)

b.\(\frac{3}{7}\) – \(\frac{1}{8}\) lcm of 7 and 8 is 56. \(\frac{24}{56}\) – \(\frac{7}{56}\) =  \(\frac{17}{56}\) less than \(\frac{1}{2}\)

c. 1\(\frac{1}{7}\) – \(\frac{7}{8}\) = \(\frac{8}{7}\) – \(\frac{7}{8}\) lcm of 7 and 8 is 56. \(\frac{64}{56}\) – \(\frac{49}{56}\) = \(\frac{15}{56}\) less than \(\frac{1}{2}\)

d. \(\frac{3}{7}\) + \(\frac{2}{6}\) lcm of 7 and 6 is 42. \(\frac{18}{42}\) + \(\frac{14}{42}\)  = \(\frac{32}{42}\) = \(\frac{16}{21}\) greater than \(\frac{1}{2}\) .

Question 3. Use > , < , or = to make the following statements true. a. 5\(\frac{2}{3}\) + 3\(\frac{3}{4}\) _______ 8\(\frac{2}{3}\) b. 4\(\frac{5}{8}\) – 3\(\frac{2}{5}\) _______ 1\(\frac{5}{8}\) + \(\frac{2}{5}\) c. 5\(\frac{1}{2}\) + 1\(\frac{3}{7}\) _______ 6 + \(\frac{13}{14}\) d. 15\(\frac{4}{7}\) – 11\(\frac{2}{5}\) _______ 4\(\frac{4}{7}\) + \(\frac{2}{5}\) Answer: a. 5\(\frac{2}{3}\) + 3\(\frac{3}{4}\) = 8\(\frac{2}{3}\) b. 4\(\frac{5}{8}\) – 3\(\frac{2}{5}\) < 1\(\frac{5}{8}\) + \(\frac{2}{5}\) c. 5\(\frac{1}{2}\) + 1\(\frac{3}{7}\) = 6 + \(\frac{13}{14}\) d. 15\(\frac{4}{7}\) – 11\(\frac{2}{5}\) > 4\(\frac{4}{7}\) + \(\frac{2}{5}\) Explanation : a. 5\(\frac{2}{3}\) + 3\(\frac{3}{4}\) = \(\frac{17}{3}\) + \(\frac{12}{4}\) lcm of 3 and 4 is 12. \(\frac{68}{12}\) + \(\frac{36}{12}\) = \(\frac{104}{12}\) = \(\frac{26}{3}\) =8\(\frac{2}{3} \)

b. 4\(\frac{5}{8}\) – 3\(\frac{2}{5}\) = \(\frac{37}{8}\) – \(\frac{17}{5}\) lcm of 8 and 5 is 40 . \(\frac{185}{40}\) – \(\frac{136}{40}\) = \(\frac{49}{40}\) =1\(\frac{9}{40}\) 1\(\frac{5}{8}\) + \(\frac{2}{5}\) = \(\frac{13}{8}\) + \(\frac{2}{5}\) lcm of 8 and 5 is 40 . \(\frac{65}{40}\) + \(\frac{16}{5}\) = \(\frac{81}{40}\) = 2\(\frac{1}{40}\)

c. 5\(\frac{1}{2}\) + 1\(\frac{3}{7}\) = \(\frac{11}{2}\) + \(\frac{10}{7}\) lcm of 2 and 7 is 14. \(\frac{77}{14}\) + \(\frac{20}{14}\) = \(\frac{97}{14}\) = 6 \(\frac{13}{14}\) 6 + \(\frac{13}{14}\) =\(\frac{84}{14}\) + \(\frac{13}{14}\) = \(\frac{97}{14}\) = 6 \(\frac{13}{14}\)

d. 15\(\frac{4}{7}\) – 11\(\frac{2}{5}\) =\(\frac{109}{7}\) – \(\frac{57}{5}\) lcm of 7 and 5 is 35 . \(\frac{545}{35}\) – \(\frac{399}{35}\) = \(\frac{944}{35}\) = 26\(\frac{34}{35}\) . 4\(\frac{4}{7}\) + \(\frac{2}{5}\) = \(\frac{32}{7}\) + \(\frac{2}{5}\) lcm of 7 and 5 is 35 . \(\frac{160}{35}\) + \(\frac{14}{35}\) = \(\frac{174}{35}\)= 4\(\frac{34}{35}\)

Question 4. Is it true that 4\(\frac{3}{5}\) – 3\(\frac{2}{3}\) = 1 + \(\frac{3}{5}\) + \(\frac{2}{3}\)? Prove your answer. Answer: No it is wrong 4\(\frac{3}{5}\) – 3\(\frac{2}{3}\) < 1 + \(\frac{3}{5}\) + \(\frac{2}{3}\) Explanation : 4\(\frac{3}{5}\) – 3\(\frac{2}{3}\) = \(\frac{23}{5}\) – \(\frac{11}{3}\) lcm of 5 and 3 is 15 . \(\frac{69}{15}\) – \(\frac{55}{15}\) = \(\frac{14}{15}\)

1 + \(\frac{3}{5}\) + \(\frac{2}{3}\) lcm of 5 and 3 is 15 . \(\frac{15}{15}\) + \(\frac{9}{15}\) + \(\frac{10}{15}\) = latex]\frac{34}{15}[/latex] = 2latex]\frac{4}{15}[/latex] .

Question 5. Jackson needs to be 1\(\frac{3}{4}\) inches taller in order to ride the roller coaster. Since he can’t wait, he puts on a pair of boots that add 1\(\frac{1}{6}\) inches to his height and slips an insole inside to add another \(\frac{1}{8}\) inch to his height. Will this make Jackson appear tall enough to ride the roller coaster? Answer: Fraction of Height required to ride a roller coaster for Jackson = 1\(\frac{3}{4}\) inches. Fraction of his height = 1\(\frac{1}{6}\) inches = \(\frac{7}{6}\) Fraction of his boots length = \(\frac{1}{8}\) inches Total fraction of his height with boots = \(\frac{7}{6}\) + \(\frac{1}{8}\) = \(\frac{28}{24}\) + \(\frac{3}{24}\) = \(\frac{31}{24}\) = 1\(\frac{7}{24}\) . 1\(\frac{3}{4}\) = multiply by 6 both numerator and denominator = 1\(\frac{18}{24}\) therefore, 1\(\frac{18}{24}\) > is greater than 1\(\frac{7}{24}\)  So, he is not taller enough to ride roller coaster . So, he cant ride the roller coaster .

Question 6. A baker needs 5 lb of butter for a recipe. She found 2 portions that each weigh 1\(\frac{1}{6}\) lb and a portion that weighs 2\(\frac{2}{7}\) lb. Does she have enough butter for her recipe? Answer: Fraction of butter required for a recipe = 5 lb Fraction of 2 portions that weigh = 2 × \(\frac{7}{6}\) lb = \(\frac{7}{3}\) Fraction of portions that weighs = 2\(\frac{2}{7}\) lb. = \(\frac{16}{7}\) lb. Fraction of butter of portions = \(\frac{7}{3}\) + \(\frac{16}{7}\) = \(\frac{49}{21}\) + \(\frac{48}{21}\) = \(\frac{97}{21}\) = 4\(\frac{13}{21}\) Therefore, she doesnot have enough butter for the recipe =  4\(\frac{13}{21}\)

Eureka Math Grade 5 Module 3 Lesson 13 Exit Ticket Answer Key

Question 1. Circle the correct answer. a. \(\frac{1}{2}\) +\(\frac{5}{12}\)        greater than 1          less than 1

b. 2\(\frac{7}{8}\) – 1\(\frac{7}{9}\)      greater than 1         less than 1

c. 1\(\frac{1}{12}\) – \(\frac{7}{10}\)     greater than \(\frac{1}{2}\)      less than \(\frac{1}{2}\)

Question 2. Use > , < , or = to make the following statement true. 4\(\frac{4}{5}\) + 3\(\frac{2}{3}\) < 8\(\frac{1}{2}\) Answer: 4\(\frac{4}{5}\) + 3\(\frac{2}{3}\) < 8\(\frac{1}{2}\) Explanation : 4\(\frac{4}{5}\) + 3\(\frac{2}{3}\) = \(\frac{24}{5}\) + \(\frac{11}{3}\) lcm of 5 and 3 is 15 . \(\frac{72}{15}\) + \(\frac{55}{15}\) = \(\frac{127}{15}\) = 8 \(\frac{7}{15}\)

Eureka Math Grade 5 Module 3 Lesson 13 Homework Answer Key

Question 1. Are the following expressions greater than or less than 1? Circle the correct answer. a. \(\frac{1}{2}\) + \(\frac{4}{9}\)        greater than 1          less than 1

b. \(\frac{5}{8}\) + \(\frac{3}{5}\)        greater than 1          less than 1

c. 1\(\frac{1}{5}\) – \(\frac{1}{3}\)       greater than 1           less than 1

Question 2. Are the following expressions greater than or less than \(\frac{1}{2}\)? Circle the correct answer. a. \(\frac{1}{5}\) + \(\frac{1}{4}\)        greater than \(\frac{1}{2}\)       less than \(\frac{1}{2}\)

b. \(\frac{6}{7}\) – \(\frac{1}{6}\)         greater than \(\frac{1}{2}\)        less than \(\frac{1}{2}\)

c. 1\(\frac{1}{7}\) – \(\frac{5}{6}\)        greater than \(\frac{1}{2}\)       less than \(\frac{1}{2}\)

Question 3. Use > , < , or = to make the following statements true. a. 5\(\frac{4}{5}\) + 2\(\frac{2}{3}\) _______ 8\(\frac{3}{4}\) b. 3\(\frac{4}{7}\) – 2\(\frac{3}{5}\) _______ 1\(\frac{4}{7}\) + \(\frac{3}{5}\) c. 4\(\frac{1}{2}\) + 1\(\frac{4}{9}\) _______ 5 + \(\frac{13}{18}\) d. 10\(\frac{3}{8}\) – 7\(\frac{3}{5}\) _______ 3\(\frac{3}{8}\) + \(\frac{3}{5}\) Answer: a. 5\(\frac{4}{5}\) + 2\(\frac{2}{3}\) < 8\(\frac{3}{4}\) b. 3\(\frac{4}{7}\) – 2\(\frac{3}{5}\) < 1\(\frac{4}{7}\) + \(\frac{3}{5}\) c. 4\(\frac{1}{2}\) + 1\(\frac{4}{9}\) < 5 + \(\frac{13}{18}\) d. 10\(\frac{3}{8}\) – 7\(\frac{3}{5}\) > 3\(\frac{3}{8}\) + \(\frac{3}{5}\) Explanation : a. 5\(\frac{4}{5}\) + 2\(\frac{2}{3}\) = \(\frac{29}{5}\) + \(\frac{8}{3}\) lcm of 5 and 3 is 15 . \(\frac{57}{15}\) + \(\frac{40}{15}\) = \(\frac{97}{15}\) = 6\(\frac{7}{15}\) .

b. 3\(\frac{4}{7}\) – 2\(\frac{3}{5}\) = \(\frac{25}{7}\) – \(\frac{13}{5}\) . lcm of 7 and 5 is 35 . \(\frac{125}{35}\) – \(\frac{91}{35}\) = \(\frac{34}{35}\) 1\(\frac{4}{7}\) + \(\frac{3}{5}\) = \(\frac{11}{7}\) + \(\frac{3}{5}\) lcm of 5 and 7 is 35 . \(\frac{55}{35}\) + \(\frac{21}{35}\) = \(\frac{76}{35}\) = 2 \(\frac{6}{35}\)

c. 4\(\frac{1}{2}\) + 1\(\frac{4}{9}\) = \(\frac{9}{2}\) + \(\frac{13}{9}\) lcm of 2 and 9 is 18 . 4\(\frac{1}{2}\) + 1\(\frac{4}{9}\) 5 + \(\frac{13}{18}\) = \(\frac{90}{18}\) + \(\frac{13}{18}\) = \(\frac{103}{18}\) = 5\(\frac{13}{18}\)

d. 10\(\frac{3}{8}\) – 7\(\frac{3}{5}\) = \(\frac{83}{8}\) – \(\frac{38}{5}\) lcm of 8 and 5 is 40 . \(\frac{415}{40}\) – \(\frac{304}{40}\) = \(\frac{311}{40}\) = 7\(\frac{31}{40}\) 3\(\frac{3}{8}\) + \(\frac{3}{5}\) = \(\frac{27}{8}\) + \(\frac{3}{5}\) lcm of 8 and 5 is 40 . \(\frac{135}{40}\) + \(\frac{24}{40}\) = \(\frac{159}{40}\)= 3\(\frac{39}{40}\)

Question 4. Is it true that 5\(\frac{2}{3}\) – 3\(\frac{3}{4}\) = 1 + \(\frac{2}{3}\) + \(\frac{3}{4}\)? Prove your answer. Answer: It is not true . 5\(\frac{2}{3}\) – 3\(\frac{3}{4}\) < 1 + \(\frac{2}{3}\) + \(\frac{3}{4}\) Explanation : 5\(\frac{2}{3}\) – 3\(\frac{3}{4}\) = \(\frac{17}{3}\) – \(\frac{15}{4}\) lcm of 3 and 4 is 12. \(\frac{68}{12}\) – \(\frac{45}{12}\) = \(\frac{23}{12}\) = 1\(\frac{11}{12}\)

1 + \(\frac{2}{3}\) + \(\frac{3}{4}\) lcm of 3 and 4 is 12 . \(\frac{12}{12}\) + \(\frac{8}{12}\) + \(\frac{9}{12}\) = \(\frac{29}{12}\) = 2\(\frac{5}{12}\) .

Question 5. A tree limb hangs 5\(\frac{1}{4}\) feet from a telephone wire. The city trims back the branch before it grows within 2 \(\frac{1}{2}\) feet of the wire. Will the city allow the tree to grow 2\(\frac{3}{4}\) more feet? Answer: Fraction of height at which telephone wire is hung = 5\(\frac{1}{4}\) =\(\frac{21}{4}\)  feet Fraction of height city allow the tree to grow = 2\(\frac{3}{4}\) = \(\frac{11}{4}\) feet . Fraction of height city trims back the branch before it grows = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) feet Fraction of height of telephone wire can be hang = \(\frac{21}{4}\)  – \(\frac{11}{4}\)  = \(\frac{10}{4}\) = \(\frac{5}{2}\) both are equal that means the tree will be trim back .

Question 6. Mr. Kreider wants to paint two doors and several shutters. It takes 2\(\frac{1}{8}\) gallons of paint to coat each door and 1\(\frac{3}{5}\) gallons of paint to coat all of his shutters. If Mr. Kreider buys three 2-gallon cans of paint, does he have enough to complete the job? Answer: Fraction of cost of paint to coat each door = 2\(\frac{1}{8}\) gallons = \(\frac{17}{8}\) Fraction of cost of paint to coat all his shutters = 1\(\frac{3}{5}\) gallons = \(\frac{8}{5}\) Fraction of cost to paint 2 doors and shutters = 2 × \(\frac{17}{8}\) + \(\frac{8}{5}\) = \(\frac{17}{4}\) + \(\frac{8}{5}\) = \(\frac{85}{20}\) + \(\frac{32}{20}\) = \(\frac{117}{20}\) = 5latex]\frac{17}{20}[/latex] Total paint = three 2-gallon cans of paint = 3 × 2 = 6 gallons. Therefore Kreider doesn’t have sufficient amount of paint .

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