eureka math grade 5 lesson 6 homework 5.3

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5th grade (Eureka Math/EngageNY)

Unit 1: module 1: place value and decimal fractions, unit 2: module 2: multi-digit whole number and decimal fraction operations, unit 3: module 3: addition and subtractions of fractions, unit 4: module 4: multiplication and division of fractions and decimal fractions, unit 5: module 5: addition and multiplication with volume and area, unit 6: module 6: problem solving with the coordinate plane.

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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.

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Eureka Math Lesson 6 Homework 5.3 Answer Key

Most students start learning the Math subject with Eureka Math. Every student will pass the exercise and take the exam. In fact, not a few students have difficulty in answering practice questions. Well, they will need some answer keys that can help them to answer the exams and also practice questions.

Unfortunately, not all answer keys for Eureka’s practices and exams  are available on the internet. Certainly, the students will look for an alternative way to get the answer keys for the lesson that they are facing.

In this chance, we will show you the answer keys for  Eureka Math Lesson 6 Homework 5.3 that you can learn to answer the Eureka Math practice. So, let’s see the answer keys below!

Here’s a List of Answer Keys for Eureka Math Lesson 6 Homework 5.3!

We got  the Eureka Math Lesson Answer Keys from ewa5thgrade.pbworks,com that you can access here . The answer key for Eureka Math Lesson 6 Homework 5.3 comes in PDF format, so you can download it or print it easily. For more information, this site actually provides a number of practice and homeworks answer keys for Eureka Math.

Here you go!

• For the following problems, draw a picture using the rectangular fraction and write the answer. Simplify your answer, if possible.

a. 1 ¼  – 1/3  =

b. 1 1/5 – 1/3 =

c. 1 3/8 – ½ =

d. 1 2/5 – ½ =

e. 1 2/7 – 1/3 =

f. 1 2/3 – 2/3 =

a. 1 ¼  – 1/3  = 5/4 – 1/3 = 15/12 – 4/12 = 11/12

b. 1 1/5 – 1/3 = 6/5 – 1/3 = 18/15 – 5/15 = 13/15

c. 1 3/8 – ½ = 11/8 – ½ = 22/6 – 8/16 = 14/ 16

14/16 : 2/2 = 7/8

d. 1 2/5 – ½ = 7/5 – ½ = 14/ 10 – 7/10 = 7/10

e. 1 2/7 – 1/3 = 9/7 – 1/3 = 27/ 21 – 7/21 = 20/21

f. 1 2/3 – 3/5 =  5/3 – 3/5 = 25/ 15 – 9/15 = 16/15

• Jean-Luc jogged around the lake in 1 ¼ hour. William jogged the same distance in 5/6 hour. How much longer did Jean-Luc take than William in hours?

1 ¼ – 5/6 = 5/4 – 5/6 = 15/12 – 10/12 = 5/12

30/24 – 20/ 24 = 10/ 24 : 2/2 = 5/12

He jogged 5/12 of an hour more.

• Is it true that 1 2/5 – ¾ = ¼ + 2/5?. Prove the answer!

7/5 – ¾ = 28/ 20 – 15/ 20 = 13/20

¼ + 2/5 = 5/20 + 8/20= 13/ 20

Yes, it is true.

Well, that’s a list of answer keys for Eureka Math Lesson 6 Homework 5.3.

What Is Eureka Math?

In education, Eureka Math can be mentioned as a  holistic curriculum that is intended for Pre-Kindergarten through Grade 12. The Eureka Math curriculum sequences mathematical progressions in the modules created by the experts. Of course, this curriculum actually makes the math subject enjoyable and fun to learn.

The Eureka Math curriculum provides  a community of support, in-depth professional development and also learning materials. This curriculum is always committed to support the learners throughout an implementation. In this case, the professional development created by the Eureka curriculum is actually designed to equip the educators with any tools that they have to support both new and continuing learners of the Eureka Math curriculum.

Additionally, the Eureka Math also provides the Teacher Resources Pack that includes a selection of instructional tools and materials. Here are they:

• Pacing and Preparation Guides
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• Standards Checklists
• Materials Lists

For more information, the Eureka Math also provides a Digital Suite that offers the learners with two exceptional online resources. The Digital Suite here includes the Navigator that is an interactive digital version of the Prekindergarten through Grade 12 Curriculum. It also provides the Teach Eureka Video Series which is a collection of on-demand PD videos which accompanies the curriculum.

How Worthy Is the Eureka Math Answer Keys for Learners?

It cannot be denied if the Eureka Math answer keys are very important for learners. The answer keys actually facilitate them to learn outside the classroom. In other words, the answer keys for some exams and practice is such a great solution for identifying the way of math was ever taught,

Moreover, the Eureka Math answer keys really help the learners to understand the mathematical concepts easier. Well, this way will enhance the problem-solving skills for the math subject.

Furthermore, the Eureka Math answer keys are totally intended to establish fluency in Math subjects in a fun way. The learners will then find the Homework Help with the help of Eureka Math answer keys that they very need to use the quick resources available.

Aside from that, the Eureka Math answer keys will assist the learners to be proficient in Math subject. To obtain the answer keys, it’s very easy to do, due to a number of sources that will provide the answer keys for Eureka Math in PDF. Well, the answer keys can be downloaded and printed, so the learners can use the answer keys to prepare their exams offline when they want.

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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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