{"id":60823,"date":"2022-05-20T15:33:23","date_gmt":"2022-05-20T10:03:23","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=60823"},"modified":"2022-05-10T14:21:39","modified_gmt":"2022-05-10T08:51:39","slug":"eureka-math-grade-4-module-5-lesson-18","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-4-module-5-lesson-18\/","title":{"rendered":"Eureka Math Grade 4 Module 5 Lesson 18 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 4th Grade Module 5 Lesson 18 Answer Key<\/h2>\n

Eureka Math Grade 4 Module 5 Lesson 18 Practice Sheet Answer Key<\/h3>\n

\"Eureka<\/p>\n

Answer:
\n1\/8 + 3\/8 + 4\/8= 1.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n1\/8 + 3\/8.
\n4\/8.
\n4\/8 + 4\/8.
\n8\/8 = 1.<\/p>\n

Answer:
\n1\/6 + 4\/6 + 2\/6= 7\/6.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n1\/6 + 4\/6.
\n5\/6.
\n5\/6 + 2\/6.
\n7\/6.<\/p>\n

Answer:
\n11\/10 – 4\/10 – 1\/10 = 6\/10.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n11\/10 – 4\/10.
\n7\/10 – 1\/10.
\n6\/10._____________
\nadding and subt
\n\"Eureka<\/p>\n

Answer:
\n1 – 3\/12 – 5\/12 = 4\/12.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n1 – 3\/12.
\n9\/12.
\n9\/12 – 5\/12.
\n4\/12.<\/p>\n

Answer:
\n5\/8 + 4\/8 + 1\/8= 10\/8.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n5\/8 + 4\/8.
\n9\/8.
\n9\/8 + 1\/8.
\n10\/8.<\/p>\n

Answer:
\n1(1\/5) – 2\/5 – 3\/5= 1\/5.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n1(1\/5) – 2\/5.
\n6\/5 – 2\/5.
\n4\/5 – 1\/5.
\n3\/5.<\/p>\n

Eureka Math Grade 4 Module 5 Lesson 18 Problem Set Answer Key<\/h3>\n

Question 1.
\nShow one way to solve each problem. Express sums and differences as a mixed number when possible. Use number bonds when it helps you. Part (a) is partially completed.<\/p>\n

a. \\(\\frac{2}{5}\\) + \\(\\frac{3}{5}\\) + \\(\\frac{1}{5}\\)
\n= \\(\\frac{5}{5}\\) + \\(\\frac{1}{5}\\) = 1 + \\(\\frac{1}{5}\\)
\n=1 \\(\\frac{1}{5}\\).<\/p>\n

Answer:
\n1\\(\\frac{1}{5}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{2}{5}\\) + \\(\\frac{3}{5}\\) + \\(\\frac{1}{5}\\)
\n2\/5 + 3\/5 = 5\/5.
\n\\(\\frac{5}{5}\\) + \\(\\frac{1}{5}\\).
\n5\/5 = 1.
\n5\/5 + 1\/5 = 6\/5.
\n1\\(\\frac{1}{5}\\).<\/p>\n

b. \\(\\frac{3}{6}\\) + \\(\\frac{1}{6}\\) + \\(\\frac{3}{6}\\)<\/p>\n

Answer:
\n1\\(\\frac{1}{6}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{3}{6}\\) + \\(\\frac{1}{6}\\) + \\(\\frac{3}{6}\\)
\n3\/6 + 1\/6 = 4\/6.
\n\\(\\frac{4}{6}\\) + \\(\\frac{3}{6}\\).
\n4\/6 + 3\/6.
\n7\/6.
\n1\\(\\frac{1}{6}\\).<\/p>\n

c. \\(\\frac{5}{7}\\) + \\(\\frac{7}{7}\\) + \\(\\frac{2}{7}\\)<\/p>\n

Answer:
\n\\(\\frac{14}{7}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{5}{7}\\) + \\(\\frac{7}{7}\\) + \\(\\frac{2}{7}\\)
\n5\/7 + 7\/7 = 12\/7.
\n\\(\\frac{12}{7}\\) + \\(\\frac{2}{7}\\).
\n12\/7 + 2\/7 = 14\/7.
\n1\\(\\frac{7}{7}\\).<\/p>\n

d. \\(\\frac{7}{8}\\) – \\(\\frac{3}{8}\\) – \\(\\frac{1}{8}\\)<\/p>\n

Answer:
\n\\(\\frac{3}{8}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{7}{8}\\) – \\(\\frac{3}{8}\\) – \\(\\frac{1}{8}\\)
\n7\/8 – 3\/8 = 4\/8.
\n\\(\\frac{4}{8}\\) – \\(\\frac{1}{8}\\).
\n4\/8 – 1\/8 = 3\/8.
\n1\\(\\frac{5}{8}\\).<\/p>\n

e. \\(\\frac{7}{9}\\) + \\(\\frac{1}{9}\\) + \\(\\frac{4}{9}\\)<\/p>\n

Answer:
\n\\(\\frac{12}{9}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{7}{9}\\) + \\(\\frac{1}{9}\\) + \\(\\frac{4}{9}\\)
\n7\/9 + 1\/9 = 8\/9.
\n\\(\\frac{8}{9}\\) + \\(\\frac{4}{9}\\).
\n8\/9 + 4\/9 = 12\/9.
\n1\\(\\frac{3}{9}\\).<\/p>\n

f. \\(\\frac{4}{10}\\) + \\(\\frac{11}{10}\\) + \\(\\frac{5}{10}\\)<\/p>\n

Answer:
\n\\(\\frac{20}{10}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{4}{10}\\) + \\(\\frac{11}{10}\\) + \\(\\frac{5}{10}\\)
\n4\/10 + 11\/10 = 15\/10.
\n\\(\\frac{15}{10}\\) + \\(\\frac{5}{10}\\).
\n15\/10 + 5\/10 = 20\/10.
\n1\\(\\frac{10}{10}\\).<\/p>\n

g. 1 – \\(\\frac{3}{12}\\) – \\(\\frac{4}{12}\\)<\/p>\n

Answer:
\n\\(\\frac{5}{12}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{12}{12}\\) – \\(\\frac{3}{12}\\) – \\(\\frac{4}{12}\\)
\n12 – 3\/12 = 9\/12.
\n\\(\\frac{9}{12}\\) – \\(\\frac{4}{12}\\).
\n9\/12 – 4\/12 = 5\/12.
\n1 – \\(\\frac{7}{12}\\).<\/p>\n

h. 1\\(\\frac{2}{3}\\) – \\(\\frac{1}{3}\\) – \\(\\frac{1}{3}\\)<\/p>\n

Answer:
\n\\(\\frac{0}{3}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{2}{3}\\) – \\(\\frac{1}{3}\\) – \\(\\frac{1}{3}\\)
\n2\/3 – 1\/3 = 1\/3.
\n\\(\\frac{1}{3}\\) – \\(\\frac{1}{3}\\).
\n1\/3 – 1\/3 = 0.
\n\\(\\frac{0}{3}\\).<\/p>\n

i. \\(\\frac{10}{12}\\) + \\(\\frac{5}{12}\\) + \\(\\frac{2}{12}\\) + \\(\\frac{7}{12}\\)<\/p>\n

Answer:
\n\\(\\frac{24}{12}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{10}{12}\\) + \\(\\frac{5}{12}\\) + \\(\\frac{2}{12}\\) + \\(\\frac{7}{12}\\)
\n10\/12 + 5\/12 = 15\/12.
\n\\(\\frac{15}{12}\\) + \\(\\frac{2}{12}\\).
\n15\/12 + 2 \/12 = 17\/12.
\n\\(\\frac{17}{12}\\) + \\(\\frac{7}{12}\\).
\n17\/12 + 7\/12.
\n24\/12.
\n1\\(\\frac{12}{12}\\).<\/p>\n

Question 2.
\nMonica and Stuart used different strategies to solve \\(\\frac{5}{8}\\) + \\(\\frac{2}{8}\\) + \\(\\frac{5}{8}\\).
\n\"Eureka
\nWhose strategy do you like best? Why?<\/p>\n

Answer:
\nStuart’s way is the best.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nMonica and Stuart used\u00a0 different strategies to solve \\(\\frac{5}{8}\\) + \\(\\frac{2}{8}\\) + \\(\\frac{5}{8}\\).
\nMonica explained in detail.
\nStuart’s also explained in detail.
\nhe explained in a simple way.
\nso Stuart’s way is the best.<\/p>\n

Question 3.
\nYou gave one solution for each part of Problem 1. Now, for each problem indicated below, give a different solution method.
\n\"Eureka<\/p>\n

Answer:
\ne. 5\/7 + 7\/7 + 2\/7 = 14\/7.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n5\/7 + 7\/7.
\n12\/7.
\n12\/7 + 2\/7.
\n14\/7.
\n5\/7 + 7\/7 + 2\/7 = 14\/7.<\/p>\n

Answer:
\nf. 4\/10 + 11\/10 + 5\/10 = 20\/10.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n4\/10 + 11\/10.
\n15\/10.
\n15\/10 + 5\/10.
\n20\/10.
\n4\/10 + 11\/10 + 5\/10 = 20\/10.
\n\"Eureka-Math-Grade-4-Module-5-Lesson-18-Answer<\/p>\n

Eureka Math Grade 4 Module 5 Lesson 18 Exit Ticket Answer Key<\/h3>\n

Solve the following problems. Use number bonds to help you.<\/p>\n

Question 1.
\n\\(\\frac{5}{9}\\) + \\(\\frac{2}{9}\\) + \\(\\frac{4}{9}\\)<\/p>\n

Answer:
\n\\(\\frac{11}{9}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{5}{9}\\) + \\(\\frac{2}{9}\\) + \\(\\frac{4}{9}\\)
\n5\/9 + 2\/9 = 7\/9.
\n\\(\\frac{7}{9}\\) + \\(\\frac{4}{9}\\).
\n7\/9 + 4\/9 = 11\/9.
\n\\(\\frac{11}{9}\\).<\/p>\n

Question 2.
\n1 – \\(\\frac{5}{8}\\) – \\(\\frac{1}{8}\\)<\/p>\n

Answer:
\n\\(\\frac{2}{8}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{8}{1}\\) – \\(\\frac{5}{8}\\) – \\(\\frac{1}{8}\\)
\n8 – 5\/8 = 3\/8.
\n\\(\\frac{3}{8}\\) – \\(\\frac{1}{8}\\).
\n3\/8 – 1\/8 = 2\/8.
\n\\(\\frac{2}{8}\\).<\/p>\n

Eureka Math Grade 4 Module 5 Lesson 18 Homework Answer Key<\/h3>\n

Question 1.
\nShow one way to solve each problem. Express sums and differences as a mixed number when possible. Use number bonds when it helps you. Part (a) is partially completed.<\/p>\n

a. \\(\\frac{1}{3}\\) + \\(\\frac{2}{3}\\) + \\(\\frac{1}{3}\\)
\n= \\(\\frac{3}{3}\\) + \\(\\frac{1}{3}\\) = 1 + \\(\\frac{1}{3}\\)
\n= __4\/3______<\/p>\n

Answer:
\n1\\(\\frac{1}{3}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{3}{3}\\) + \\(\\frac{1}{3}\\) + \\(\\frac{1}{3}\\)
\n1 + 1\/3 =2 \/3.
\n\\(\\frac{3}{3}\\) + \\(\\frac{1}{3}\\).
\n2\/3 + 2\/3 = 4\/3.
\n1\\(\\frac{1}{3}\\).<\/p>\n

b. \\(\\frac{5}{8}\\) + \\(\\frac{5}{8}\\) + \\(\\frac{3}{8}\\)<\/p>\n

Answer:
\n1\\(\\frac{5}{8}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{5}{8}\\) + \\(\\frac{5}{8}\\) + \\(\\frac{3}{8}\\)
\n5\/8 + 5\/8 = 10\/8.
\n\\(\\frac{10}{8}\\) + \\(\\frac{3}{8}\\).
\n10\/8 + 3\/8 = 13\/8.
\n1\\(\\frac{5}{8}\\).<\/p>\n

c. \\(\\frac{4}{6}\\) + \\(\\frac{6}{6}\\) + \\(\\frac{1}{6}\\)<\/p>\n

Answer:
\n1\\(\\frac{5}{6}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{4}{6}\\) + \\(\\frac{6}{6}\\) + \\(\\frac{1}{6}\\)
\n4\/6 + 6\/6 = 10\/6.
\n\\(\\frac{10}{6}\\) + \\(\\frac{1}{6}\\).
\n10\/6 + 1\/6 = 11\/6.
\n1\\(\\frac{5}{6}\\).<\/p>\n

d. 1\\(\\frac{2}{12}\\) – \\(\\frac{2}{12}\\) – \\(\\frac{1}{12}\\)<\/p>\n

Answer:
\n1 – \\(\\frac{1}{12}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n1\\(\\frac{2}{12}\\) – \\(\\frac{2}{12}\\) – \\(\\frac{1}{12}\\)
\n14\/12 – 2\/12 = 12\/12.
\n\\(\\frac{12}{12}\\) – \\(\\frac{1}{12}\\).
\n1 – 1\/12 = 11\/12.
\n1 – \\(\\frac{11}{12}\\).<\/p>\n

e. \\(\\frac{5}{7}\\) + \\(\\frac{1}{7}\\) + \\(\\frac{4}{7}\\)<\/p>\n

Answer:
\n1\\(\\frac{3}{7}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{5}{7}\\) + \\(\\frac{1}{7}\\) + \\(\\frac{4}{7}\\)
\n5\/7 + 1\/7 = 6\/7.
\n\\(\\frac{6}{7}\\) + \\(\\frac{4}{7}\\).
\n6\/7 + 4\/7 = 10\/7.
\n1\\(\\frac{3}{7}\\).<\/p>\n

f. \\(\\frac{4}{10}\\) + \\(\\frac{7}{10}\\) + \\(\\frac{9}{10}\\)<\/p>\n

Answer:
\n1\\(\\frac{10}{20}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{4}{10}\\) + \\(\\frac{7}{10}\\) + \\(\\frac{9}{10}\\)
\n4\/10 + 7\/10 = 11\/10.
\n\\(\\frac{11}{10}\\) + \\(\\frac{9}{10}\\).
\n11\/10 + 9\/10 = 20\/10.
\n1\\(\\frac{10}{20}\\).<\/p>\n

g. 1 – \\(\\frac{3}{10}\\) – \\(\\frac{1}{10}\\)<\/p>\n

Answer:
\n1 – \\(\\frac{4}{10}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n1 – \\(\\frac{3}{10}\\) – \\(\\frac{1}{10}\\).
\n10 – 3\/10 = 7\/10.
\n\\(\\frac{7}{10}\\) – \\(\\frac{1}{10}\\).
\n7\/10 – 1\/10 = 6\/10.
\n1 – \\(\\frac{4}{10}\\).<\/p>\n

h. 1\\(\\frac{3}{5}\\) – \\(\\frac{4}{5}\\) – \\(\\frac{1}{5}\\)<\/p>\n

Answer:
\n1 – \\(\\frac{2}{5}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n1\\(\\frac{3}{5}\\) – \\(\\frac{4}{5}\\) – \\(\\frac{1}{5}\\)
\n8\/5 – 4\/5 = 4\/5.
\n\\(\\frac{4}{5}\\) – \\(\\frac{1}{5}\\).
\n4\/5 –\u00a0 1\/5 = 3\/5.
\n1\\(\\frac{2}{5}\\).<\/p>\n

i. \\(\\frac{10}{15}\\) + \\(\\frac{7}{15}\\) + \\(\\frac{12}{15}\\) + \\(\\frac{1}{15}\\)<\/p>\n

Answer:
\n1\\(\\frac{15}{15}\\).<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nExpress sums and differences as a mixed number when possible.
\n\\(\\frac{10}{15}\\) + \\(\\frac{7}{15}\\) + \\(\\frac{12}{15}\\) + \\(\\frac{1}{15}\\).
\n10\/15 + 7\/15 = 17\/15.
\n\\(\\frac{12}{15}\\) + \\(\\frac{1}{15}\\).
\n12\/15 + 1\/15 = 13\/15.
\n13\/15 + 17\/15.
\n1\\(\\frac{15}{15}\\).<\/p>\n

Question 2.
\nBonnie used two different strategies to solve \\(\\frac{5}{10}\\) + \\(\\frac{4}{10}\\) + \\(\\frac{3}{10}\\).
\n\"Eureka<\/p>\n

Which strategy do you like best? Why?<\/p>\n

Answer:
\nBonnie’s Second Strategy is the best.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\nBonnie’s first and second used\u00a0 different strategies to solve \\(\\frac{5}{10}\\) + \\(\\frac{4}{10}\\) + \\(\\frac{3}{10}\\).
\nBonnie’s second strategy explained in detail.
\nBonnie’s first strategy also explained in detail.
\nhe explained in a simple way.
\nso the second way is the best.<\/p>\n

Question 3.
\nYou gave one solution for each part of Problem 1. Now, for each problem indicated below, give a different solution method.
\n\"Eureka<\/p>\n

Answer:
\nb. 5\/8 + 5\/8 + 3\/8 = 13\/8.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n5\/8 + 5\/8.
\n10\/8.
\n10\/8 + 3\/8.
\n13\/8.
\n5\/8 + 5\/8 + 3\/8 = 13\/8.<\/p>\n

Answer:
\ne. 5\/7 + 1\/7 + 4\/7 = 10\/7.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n5\/7 + 1\/7.
\n6\/7.
\n6\/7 + 4\/7.
\n10\/7.
\n5\/7 + 1\/7 + 4\/7 = 10\/7.<\/p>\n

Answer:
\nh. 1(3\/5) – 4\/5 – 1\/5 = 3\/5.<\/p>\n

Explanation:
\nIn the above-given question,
\ngiven that,
\n1(3\/5) – 4\/5.
\n8\/5 – 4\/5.
\n4\/5 – 3\/5.
\n1\/5.
\n1(3\/5) – 4\/5 – 1\/5 = 3\/5.
\n\"Eureka-Math-Grade-4-Module-5-Lesson-18-Answer<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 4th Grade Module 5 Lesson 18 Answer Key Eureka Math Grade 4 Module 5 Lesson 18 Practice Sheet Answer Key Answer: 1\/8 + 3\/8 + 4\/8= 1. Explanation: In the above-given question, given that, 1\/8 + 3\/8. 4\/8. 4\/8 + 4\/8. 8\/8 = 1. Answer: 1\/6 + 4\/6 + 2\/6= 7\/6. … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/60823"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=60823"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/60823\/revisions"}],"predecessor-version":[{"id":136684,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/60823\/revisions\/136684"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=60823"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=60823"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=60823"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}