{"id":59717,"date":"2022-06-04T14:24:03","date_gmt":"2022-06-04T08:54:03","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=59717"},"modified":"2022-05-10T14:18:55","modified_gmt":"2022-05-10T08:48:55","slug":"eureka-math-grade-4-module-3-lesson-11","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-4-module-3-lesson-11\/","title":{"rendered":"Eureka Math Grade 4 Module 3 Lesson 11 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 4th Grade Module 3 Lesson 11 Answer Key<\/h2>\n

Eureka Math Grade 4 Module 3 Lesson 11 Problem Set Answer Key<\/h3>\n

Question 1.
\nSolve the following expressions using the standard algorithm, the partial products method, and the area model.
\na. 4 2 5 \u00d7 4
\n\"Eureka
\n4 (400 + 20 + 5)
\n(4 \u00d7 __400___ ) + (4 \u00d7 _20____ ) + (4 \u00d7 __5___ )
\nAnswer:
\n425 X 4 =
\nStandard Algorithm =
\n\u00a01,2<\/span>
\n425
\nX 4
\n1,700
\n<\/span>Partial Products =
\n425
\nX 4
\n<\/span>\u00a0 20
\n80
\n+ 1600
\n1,700<\/span>
\nArea Model =
\n\"Eureka
\n425 X 4 = 1,700,<\/p>\n

Explanation:
\nSolved the following expression 425 X 4 using the standard algorithm\u00a0=
\nwe add same time of multiplying
\n1,2
\n425
\nX 4
\n1,700
\n<\/span>(4 X 5 ones = 20 ones )+ (4 X 2 tens = 8 tens) +
\n(4 X 4 hundreds = 16 hundreds) = 20 + 80 + 1,600 = 1,700,
\npartial products =
\n<\/span>425
\nX 4
\n<\/span>\u00a0 20— 4 X 5
\n80—-4 X 20
\n+ 1600–4 X 400
\n1,700
\n<\/span>and the area model is as shown in the picture above,
\nSo 425 X 4 = 1,700.<\/p>\n

b. 5 3 4 \u00d7 7
\n\"Eureka
\n7 ( __500_ + _30___ + _4___ )
\n( _7_ \u00d7 __500___ ) + ( 7__ \u00d7 __30___ ) + ( 7__ \u00d7 _4___ )
\nAnswer:
\n534\u00a0 X 7 =
\nStandard Algorithm =
\n\u00a02,2<\/span>
\n534
\nX 7
\n3,738
\n<\/span>Partial Products =
\n534
\nX 7
\n<\/span>\u00a0 28
\n210
\n+3500
\n3,738<\/span>
\nArea Model =
\n\"Eureka
\n534\u00a0 X 7 = 3,738,<\/p>\n

Explanation:
\nSolved the following expression 534 X 7 using the standard algorithm\u00a0=
\nwe add same time of multiplying
\n2,2<\/span>
\n534
\nX 7
\n3,738
\n<\/span>(7 X 4 ones = 28 ones )+ (7 X 3 tens = 21 tens) +
\n(7 X 5 hundreds = 35 hundreds) = 28 + 210 + 3,500 = 3,738,
\npartial products =
\n<\/span>534
\nX 7
\n<\/span>\u00a0 28— 7 X 4
\n210—-7 X 30
\n+3500–7 X 500
\n3,738
\n<\/span>and the area model is as shown in the picture above
\nSo 534 X 7 = 3,738.<\/p>\n

c. 2 0 9 \u00d7 8
\n\"Eureka
\n8__ ( _200___ + _9___ )
\n( _8_ \u00d7 _200____ ) + ( _8_ \u00d7 __9___ )
\nAnswer:
\n209\u00a0 X 8 =
\nStandard Algorithm =
\n\u00a07,<\/span>
\n209
\nX 8
\n1,672
\n<\/span>Partial Products =
\n209
\nX 8
\n<\/span>\u00a0 72
\n0000
\n+1600
\n1,672<\/span>
\nArea Model =
\n\"Eureka
\n209\u00a0 X 8 = 1,672,<\/p>\n

Explanation:
\nSolved the following expression 209 X 8 using the standard algorithm\u00a0=
\nwe add same time of multiplying
\n7<\/span>
\n209
\nX 8
\n1,672
\n<\/span>(8 X 9 ones = 72 ones )+ (8 X 0 tens = 0 tens) +
\n(8 X 2 hundreds = 16 hundreds) = 72 + 0 + 1,600 = 1,672,
\npartial products =
\n<\/span>209
\nX 8
\n<\/span>\u00a0 72— 8 X 9
\n000—-8 X 0
\n+1600–8 X 200
\n1,672
\n<\/span>and the area model is as shown in the picture above
\nSo 209 X 8 = 1,672.<\/p>\n

Question 2.
\nSolve using the partial products method. Cayla\u2019s school has 258 students. Janet\u2019s school has 3 times as many students as Cayla\u2019s. How many students are in Janet\u2019s school?
\nAnswer:
\nThere are in Janet’s school are 774 students,<\/p>\n

Explanation:
\nGiven Cayla\u2019s school has 258 students. Janet\u2019s school has 3 times as many students as Cayla\u2019s.
\nSo there are number of students in Janet’s school are
\n3 X 258 students =
\nPartial Products =
\n258
\nX 3
\n<\/span>024
\n150
\n+600
\n774
\n<\/span>So there are in Janet’s school are 774 students.
\n
\n<\/span>Question 3.
\nModel with a tape diagram and solve. 4 times as much as 467
\nAnswer:
\n\"Eureka
\n4 X 467 = 1,868,<\/p>\n

Explanation:
\nModeled with a tape diagram as shown above and 4 times as much as 467 is 4 X 467 =
\n2,2<\/span>
\n467
\nX 4
\n1,868
\n<\/span>So, 4 X 467 = 1,868.<\/p>\n

Solve using the standard algorithm, the area model, the distributive property, or the partial products method.<\/p>\n

Question 4.
\n5,131 \u00d7 7
\nAnswer:
\nStandard algorithm:
\n5,131
\nX\u00a0 \u00a0 7
\n35,917<\/span><\/p>\n

Explanation:
\nGiven expression as 5,131 X 7 solving using standard algorithm method as
\n2<\/span>
\n5,131
\nX\u00a0 \u00a0 7
\n35,917
\n<\/span>First we multiply (7 X 1 one = 7 ones) + (7 X 3 tens = 21 tens) +
\n(7 X 1 hundred = 7 hundreds) + (7 X 5 thousands = 35 thousands),
\nSo 7 X 5,131 = 35,917.<\/p>\n

Question 5.
\n3 times as many as 2,805,<\/p>\n

Answer:
\n3 X 2,805 = 8,415,<\/p>\n

Explanation:
\nGiven to find 3 times as many as 2,805 using the distributive property as shown below
\n3 X 2,000 + 3 X 800 + 3 X 0 + 3 X 5 =
\n6,000 + 2,400 + 0 + 15 = 8,415,
\nSo 3 times as many as 2,805 = 8,415.<\/p>\n

Question 6.
\nA restaurant sells 1,725 pounds of spaghetti and 925 pounds of linguini every month. After 9 months, how many pounds of pasta does the restaurant sell?
\nAnswer:
\n\"Eureka
\n23,850 pounds of pasta the restaurant saled in 9 months,<\/p>\n

Explanation:
\nGiven a restaurant sells 1,725 pounds of spaghetti and 925 pounds of linguini every month. After 9 months, number of pounds of pasta does the restaurant sell is
\n9 X (1,725 +925) pounds = 9 X 2,650 =
\n\u00a0 5,4<\/span>
\n2,650
\nX\u00a0 \u00a0 9
\n23,850
\n<\/span>Solved using the area model as shown above.
\n<\/span><\/p>\n

Eureka Math Grade 4 Module 3 Lesson 11 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nSolve using the standard algorithm, the area model, the distributive property, or the partial products method.
\n2,809 \u00d7 4
\nAnswer:
\n2,809
\nX 4
\n11,236
\n<\/span>2,809 X 4 = 11,236,
\n
\n<\/span>Explanation:
\n<\/span>Solved using the distributive property as shown below
\n(4 X 2,000) + (4 X 800) + (4 X 0) + (4 X 9) =
\n8,000 + 3,200 + 0 + 36 =11,236,
\nSo, 2,809 X 4 = 11,236.<\/p>\n

Question 2.
\nThe monthly school newspaper is 9 pages long. Mrs. Smith needs to print 675 copies. What will be the total number of pages printed?
\nAnswer:
\nTotal number of pages printed are 6,075,<\/p>\n

Explanation:
\nGiven the monthly school newspaper is 9 pages long.
\nand Mrs. Smith needs to print 675 copies,
\nSo the total number of pages printed are 9 X 6,075,
\nSolving using Partial Products as shown below
\n6,075
\nX\u00a0 9
\n<\/span>45—9\u00a0 X 5
\n630–9 X 7
\n000– 9 X 0
\n5400–9 X 6
\n6,075<\/span>
\ntherefore, total number of pages printed are 6,075.<\/p>\n

Eureka Math Grade 4 Module 3 Lesson 11 Homework Answer Key<\/h3>\n

Question 1.
\nSolve the following expressions using the standard algorithm, the partial products method, and the area model.
\na. 3 0 2 \u00d7 8
\n\"Eureka
\n8 (300 + 2)
\n(8 \u00d7 _300____ ) + (8 \u00d7 __2___ )
\nAnswer:
\n302 X 8 =
\nStandard Algorithm =
\n\u00a01<\/span>
\n302
\nX 8
\n2,416
\n<\/span>Partial Products =
\n302
\nX 8
\n<\/span>\u00a0 16
\n00
\n+2400
\n2,416<\/span>
\nArea Model =
\n\"Eureka
\n302 X 8 = 2,416,<\/p>\n

Explanation:
\nSolved the following expression 302 X 8 using the standard algorithm\u00a0=
\nwe add same time of multiplying
\n\u00a01
\n<\/span>302
\nX 8
\n2,416
\n<\/span>(8 X 2 ones = 16 ones )+ (8 X 0 tens = 0 tens) +
\n(8 X 3 hundreds = 24 hundreds) = 16 + 0 + 2,400 = 2,416,
\npartial products =
\n<\/span>302
\nX 8
\n<\/span> 16
\n00
\n+2400
\n2,416
\n<\/span>and the area model is as shown in the picture above,
\nSo 302 X 8 = 2,416.<\/p>\n

b. 2 1 6 \u00d7 5
\n\"Eureka
\n5 ( _200___ + _10___ + _6___ )
\n( 5__ \u00d7 __200___ ) + ( _5_ \u00d7 __10___ ) + ( 5__ \u00d7 _6___ )
\nAnswer:
\n216 X 5 =
\nStandard Algorithm =
\n\u00a03<\/span>
\n216
\nX 5
\n1,080
\n<\/span>Partial Products =
\n216
\nX 5
\n<\/span>\u00a0 30
\n50
\n+1000
\n1,080<\/span>
\nArea Model =
\n\"Eureka
\n216 X 5 = 1,080,<\/p>\n

Explanation:
\nSolved the following expression 302 X 8 using the standard algorithm\u00a0=
\nwe add same time of multiplying
\n\u00a03
\n<\/span>216
\nX 5
\n1,080
\n<\/span>(5 X 6 ones = 30 ones )+ (5 X 1 tens = 5 tens) +
\n(5 X 2 hundreds = 10 hundreds) = 30 + 50 + 1,000 = 1,080,
\npartial products =
\n<\/span>216
\nX 5
\n<\/span>\u00a0 30
\n50
\n+1000
\n1,080
\n<\/span>and the area model is as shown in the picture above,
\nSo 216 X 5 = 1,080.<\/p>\n

c. 5 9 3 \u00d7 9
\n\"Eureka
\n_9_ ( _500___ + _90___ + _3___ )
\n( _9_ \u00d7 __500___ ) + ( _9_ \u00d7 __90___ ) + ( _9_ \u00d7 __3__ )
\nAnswer:
\n593 X 9 =
\nStandard Algorithm =
\n\u00a08,2<\/span>
\n593
\nX 9
\n5,337
\n<\/span>Partial Products =
\n593
\nX 9
\n<\/span>\u00a0 27
\n810
\n+4500
\n5,337<\/span>
\nArea Model =
\n\"Eureka
\n593 X 9 = 5,337,<\/p>\n

Explanation:
\nSolved the following expression 593 X 9 using
\nthe standard algorithm\u00a0=
\nwe add same time of multiplying
\n8,2
\n<\/span>593
\nX 9
\n5,337
\n<\/span>(9 X 3 ones = 27 ones )+ (9 X 9 tens = 81 tens) +
\n(9 X 5 hundreds = 45 hundreds)= 27 + 810 + 4,500 = 5,337,
\npartial products =
\n<\/span>593
\nX 9
\n<\/span>\u00a0 27
\n810
\n+4500
\n5,337
\n<\/span>and the area model is as shown in the picture above,
\nSo 593 X 9 = 5,337.<\/p>\n

Question 2.
\nSolve using the partial products method. On Monday, 475 people visited the museum. On Saturday, there were 4 times as many visitors as there were on Monday. How many people visited the museum on Saturday?
\nAnswer:
\nNumber of people visited the museum on Saturday are 1,900,
\nPartial Products =
\n475
\nX 4
\n<\/span>\u00a0 20
\n280
\n+1600
\n1,900<\/span><\/p>\n

Explanation:
\nGiven to solve using the partial products method. On Monday, 475 people visited the museum.
\nOn Saturday, there were 4 times as many visitors as there were on Monday.
\n475
\nX 4<\/span>
\n20
\n280
\n+1600
\n1,900
\n<\/span>Number of people visited the museum on Saturday are 1,900.<\/p>\n

Question 3.
\nModel with a tape diagram and solve. 6 times as much as 384
\nAnswer:
\n\"Eureka
\n6 X 384 = 2,304,<\/p>\n

Explanation:
\nModeled with a tape diagram as shown above and 6 times as much as 384 is 4 X 467 =
\n5,2<\/span>
\n384
\nX 6
\n2,304
\n<\/span>So, 4 X 467 = 2,304.<\/p>\n

Solve using the standard algorithm, the area model, the distributive property, or the partial products method.<\/p>\n

Question 4.
\n6,253 \u00d7 3
\nAnswer:
\n6,253 X 3 = 18,759,<\/p>\n

Explanation:
\nGiven expression 6,253 X 3 using the distributive property method
\nwe solve
\n6000 X 3 + 200 X 3 + 50 X 3 + 3 X 3 =
\n18,000 + 600 + 150 + 9 = 18,759,
\ntheerfore 6,253 X 3 = 18,759.<\/p>\n

Question 5.
\n7 times as many as 3,073
\nAnswer:
\n7 times as many as 3,073 is 21,511,<\/p>\n

Explanation:
\nWe solve 7 times as many as 3,073 the partial products method as
\n3,073
\nX\u00a0 \u00a07
\n<\/span>\u00a0 21—- 7 x 3
\n490—–7 X 70
\n0000—- 7 X 0
\n21000—-7 X 3,000
\n21,511
\n<\/span>So, 7 times as many as 3,073 is 21,511.
\n<\/span><\/p>\n

Question 6.
\nA cafeteria makes 2,516 pounds of white rice and 608 pounds of brown rice every month. After 6 months, how many pounds of rice does the cafeteria make?
\nAnswer:
\nAfter 6 months 18,744 pounds of rice the cafeteria make,<\/p>\n

Explanation:
\nGiven a cafeteria makes 2,516 pounds of white rice and 608 pounds of brown rice every month. Total number of pounds of rice every month is 2,516 + 608 = 3,124 pounds,
\nNow number of pounds of rice does the cafeteria make after 6 months is 6 X 3,124 using the standard algorithm we solve as
\n1,2<\/span>
\n3,124
\nX\u00a0 \u00a0 6
\n18,744<\/span>
\nFirst we multiply 6 X 4 ones = 24 ones, 6 X 2 tens = 12 tens,
\n6 X 1 hundred = 6 hundreds, 6 X 3 thousands = 18 thousands,
\n24 + 120 + 600 + 1,8000 = 18,744,
\ntherefore, after 6 months 18,744 pounds of rice the cafeteria make.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 4th Grade Module 3 Lesson 11 Answer Key Eureka Math Grade 4 Module 3 Lesson 11 Problem Set Answer Key Question 1. Solve the following expressions using the standard algorithm, the partial products method, and the area model. a. 4 2 5 \u00d7 4 4 (400 + 20 + 5) (4 … 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\/59717"}],"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=59717"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/59717\/revisions"}],"predecessor-version":[{"id":137373,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/59717\/revisions\/137373"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=59717"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=59717"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=59717"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}