{"id":59706,"date":"2022-06-04T14:24:00","date_gmt":"2022-06-04T08:54:00","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=59706"},"modified":"2022-05-10T14:18:55","modified_gmt":"2022-05-10T08:48:55","slug":"eureka-math-grade-4-module-3-lesson-10","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-4-module-3-lesson-10\/","title":{"rendered":"Eureka Math Grade 4 Module 3 Lesson 10 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 4th Grade Module 3 Lesson 10 Answer Key<\/h2>\n

Eureka Math Grade 4 Module 3 Lesson 10 Problem Set Answer Key<\/h3>\n

Question 1.
\nSolve using the standard algorithm.
\na. 3 \u00d7 42
\nAnswer:
\n42
\nX 3
\n126<\/span>
\n3 X 42 = 126,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n42
\nX 3
\n126 ,<\/span>here 3 X 2 ones = 6 ones then 3 X 4 tens = 12 tens
\nwe write 2 at tens place and write 1 at hundred place,
\nSo 3 X 42 = 126 as shown above.<\/p>\n

b. 6 \u00d7 42
\nAnswer:
\n1
\n<\/u>42
\nX 6
\n252<\/span>
\n6 X 42 = 252,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n1<\/span>
\n42
\nX 6
\n252 ,<\/span>here 6 X 2 ones = 12 ones we write 2 at ones place and
\n1 at tens places then 6 x 4 tens = 24 tens, 24 tens +1 ten =
\n25 tens we write 5 at tens place and write 2 at hundred place,
\nSo 6 X 42 = 252 as shown above.<\/p>\n

c. 6 \u00d7 431
\nAnswer:
\n1
\n<\/u>431
\nX 6
\n2,586<\/span>
\n6 X 42 = 2,586,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n1<\/span>
\n431
\nX 6
\n2,586 ,<\/span>here 6 X 1 ones = 6 ones then 6 X 3 tens = 18 tens,
\nwe write 8 at tens place and write 1 at hundreds place,
\n6 X 4 hundreds = 24 hundreds, 24 hundreds + 1 hundred =
\n25 hundreds we write 25 hundreds as 5 at hundreds place and
\n2 at thousands place, So 6 X 431 = 2,586 as shown above.<\/p>\n

d. 3 \u00d7 431
\nAnswer:
\n431
\nX 3
\n1,293<\/span>
\n3 X 431 = 1,293,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n431
\nX 3
\n1,293 ,<\/span>here 3 X 1 ones = 3 ones then 3 X 3 tens = 9 tens,
\nwe write 9 at tens place and 3 X 4 hundreds = 12 hundreds,
\nwe write 12 hundreds as 2 at hundreds place and 1 at
\nthousands place, So 3 X 431 = 1,293 as shown above.<\/p>\n

e. 3 \u00d7 6,212
\nAnswer:
\n6,212
\nX 3
\n18,636
\n<\/span>3 X 6,212 = 18,636,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n6,212
\nX 3
\n18,636 ,<\/span>here 3 X 2 ones = 6 ones then 3 X 1 ten = 3 tens,
\nwe write 3 at tens place, 3 X 2 hundreds = 6 hundreds,
\n3 X 6 thousands = 18 thousands we write 18 thousands as
\n8 at thousands place and 1 at ten thousands place,
\nSo 3 X 6,212 = 18,636 as shown above.
\n
\n<\/span>f. 3 \u00d7 3,106
\nAnswer:
\n1<\/span>
\n3,106
\nX 3
\n9,318
\n<\/span>3 X 3,106 = 9,318,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n1<\/span>
\n3,106
\nX 3\u00a0 \u00a0<\/span>
\n9,318<\/span> ,
\nhere 3 X 6 ones = 18 ones we write 8 at ones place and
\n1 at tens places then 3 X 0 tens = 0 tens, 0 tens + 1 ten = 1 ten
\nat ten place, 3 X 1 hundred = 3 hundreds
\nwe write 3 at hundreds place and 3 X 3 thousands =
\n9 thousands we write 9 at thousands place,
\nSo 3 X 3,106 = 9,318 as shown above.
\n<\/span><\/p>\n

g. 4 \u00d7 4,309
\nAnswer:
\n1,3<\/span>
\n4,309
\nX 4
\n17,236
\n<\/span>4 X 4,309 = 17,236,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n1,3<\/span>
\n4,309
\nX 4\u00a0 \u00a0<\/span>
\n17,236<\/span> ,here 4 X 9 ones = 36 ones we write 6 at ones place and
\n3 at tens places then 4 X 0 tens = 0 tens, 0 tens + 3 tens = 3 tens
\nat ten place, 4 X 3 hundreds = 12 hundreds
\nwe write 12 hundreds as 2 at hundreds place and 1 at
\nthousands place, 4 X 4 thousands = 16 thousands,
\n16 thousands + 1 thousand = 17 thousands we write 7 at
\nthousands place and 1 at ten thousands place,
\nSo 4 X 4,309 = 17,236 as shown above.
\n<\/span><\/p>\n

h. 4 \u00d7 8,618
\nAnswer:
\n2,3<\/span>
\n8,618
\nX 4
\n34,472
\n<\/span>4 X 8,618 = 34,472,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n2,3<\/span>
\n8,618
\nX 4\u00a0 \u00a0<\/span>
\n34,472<\/span> ,here 4 X 8 ones = 32 ones we write 2 at ones place and
\n3 at tens places then 4 X 1 ten = 4 tens, 4 tens + 3 tens = 7 tens
\nat ten place, 4 X 6 hundreds = 24 hundreds
\nwe write 24 hundreds as 4 at hundreds place and 2 at
\nthousands place, 4 X 8 thousands = 32 thousands,
\n32 thousands + 2 thousands = 34 thousands we write 4 at
\nthousands place and 3 at ten thousands place,
\nSo 4 X 8,618 = 34,472 as shown above.
\n<\/span><\/p>\n

Question 2.
\nThere are 365 days in a common year. How many days are in 3 common years?
\nAnswer:
\nThere are 1,095 days in 3 common years,<\/p>\n

Explanation:
\nGiven there are 365 days in a common year.
\nSo number of days in 3 common years are 3 X 365 days =
\n1<\/span>
\n365
\nX 3
\n1,095
\n<\/span>here 3 X 5 ones = 15 ones we write 1 at ones place and
\n5 at tens places then 3 X 6 tens = 18 tens,
\nwe write 8 at ten place and 1 at hundreds place and
\n3 X 3 hundreds = 9 hundreds, 9 hundreds + 1 hundered =
\n10 hundereds we write 10 hundreds as 0 at hundreds place and
\n1 at thousands place, So 3 X 365 = 1,095.
\ntherefore, there are 1,095 days in 3 common years.<\/p>\n

Question 3.
\nThe length of one side of a square city block is 462 meters. What is the perimeter of the block?
\nAnswer:
\nThe perimeter of the block is 1,848 meters,<\/p>\n

Explanation:
\nGiven the length of one side of a square city block is 462 meters,
\nSo the perimeter of the block is as we know perimeter of the
\nsquare is 4 X length, So 4 X 462 meters =
\n2<\/span>
\n462
\nX 4
\n1,848
\n<\/span>here first we write 4 X 2 ones = 8 ones, 4 X 6 tens = 24 tens we
\nwrite 4 at tens place and 2 at hundreds place and
\n4 X 4 hundreds = 16 hundreds, 16 hundreds + 2 hundreds =
\n18 hundreds we write 8 at hundreds place and 1 at thousands
\nplace so 4 X 462 = 1,848. Therefore the perimeter of
\nthe square block is 1,848 meters.<\/p>\n

Question 4.
\nJake ran 2 miles. Jesse ran 4 times as far. There are 5,280 feet in a mile. How many feet did Jesse run?
\nAnswer:
\nJesse ran 42,240 feet,<\/p>\n

Explanation:
\nGiven Jake ran 2 miles. Jesse ran 4 times as far.
\nThere are 5,280 feet in a mile. So Jesse ran 4 X 2 = 8 miles,
\n8 miles X 5,280 feet =
\n\u00a0 2,6<\/span>
\n5,280
\nX 8
\n42,240
\n<\/span>here 8 X 0 ones = 0 ones we write 0 at ones place
\nthen 8 X 8 tens = 64 tens, 4 at ten place, 6 at hundreds
\nplace, 8 X 2 hundreds = 16 hundreds, 16 hundreds + 6 hundreds =
\n22 hundreds we write 22 hundreds as 2 at hundreds place and
\n2 at thousands place, 8 X 5 thousands = 40 thousands,
\n40 thousands + 2 thousands = 42 thousands we write 2 at
\nthousands place and 4 at ten thousands place, So 8 X 5,280 =
\n42,240, therefore Jesse ran 42,240 feet.<\/p>\n

Eureka Math Grade 4 Module 3 Lesson 10 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nSolve using the standard algorithm.
\na. 2,348 \u00d7 6
\nAnswer:
\n2,2,4<\/span>
\n2,348
\nX 6
\n14,088
\n<\/span>6 X 2,348 = 14,088,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n2,2,4<\/span>
\n2,348
\nX 6\u00a0 \u00a0<\/span>
\n14,088<\/span> ,here 6 X 8 ones = 48 ones we write 8 at ones place and
\n4 at tens places then 6 X 4 tens = 24 tens, 24 tens + 4 tens = 28 tens,
\n8 at tens place and 2 at hundreds place, 6 X 3 hundreds =
\n18 hundreds, 18 hundreds + 2 hundreds = 20 hundreds,
\nwe write 20 hundreds as 0 at hundreds place and 2 at
\nthousands place, 6 X 2 thousands = 12 thousands,
\n12 thousands + 2 thousands = 14 thousands we write 4 at
\nthousands place and 1 at ten thousands place,
\nSo 6 X 2,348 = 14,088 as shown above.
\n<\/span><\/p>\n

b. 1,679 \u00d7 7
\nAnswer:
\n4,5,6<\/span>
\n1,679
\nX 7
\n11,753
\n<\/span>1,679 X 7 = 11,753,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n4,5,6<\/span>
\n1,679
\nX 7\u00a0 \u00a0<\/span>
\n11,753<\/span> ,here 7 X 9 ones = 63 ones we write 3 at ones place and
\n6 at tens places then 7 X 7 tens = 49 tens, 49 tens + 6 tens = 55 tens,
\n5 at tens place and 5 at hundreds place, 7 X 6 hundreds =
\n42 hundreds, 42 hundreds + 5 hundreds = 47 hundreds,
\nwe write 47 hundreds as 7 at hundreds place and 4 at
\nthousands place, 7 X 1 thousands = 7 thousands,
\n7 thousands + 4 thousands = 11 thousands we write 1 at
\nthousands place and 1 at ten thousands place,
\nSo 7 X 1,679 = 11,753 as shown above.
\n<\/span><\/p>\n

Question 2.
\nA farmer planted 4 rows of sunflowers. There were 1,205 plants in each row. How many sunflowers did he plant?
\nAnswer:
\nNumber of sunflowers planted are 4,820,<\/p>\n

Explanation:
\nGiven a farmer planted 4 rows of sunflowers.
\nThere were 1,205 plants in each row.
\nSo number of sunflowers did he planted are 4 X 1,205 =
\n\u00a0 \u00a0 \u00a02<\/span>
\n1,205
\nX 4
\n4,820
\n<\/span>here 5 X 4 ones = 20 ones we write 0 at ones place and
\n2 at tens places then 4 X 0 tens = 0 tens, 0 tens + 2 tens = 2 tens,
\n2 at tens place, 4 X 2 hundreds = 8 hundreds,
\nwe write 8 hundreds at hundreds place and
\n4 X 1 thousands = 4 thousands, 4 at thousands place,
\nSo 4 X 1,205 = 4,820, therefore number of sunflowers planted are 4,820.
\n<\/span><\/p>\n

Eureka Math Grade 4 Module 3 Lesson 10 Homework Answer Key<\/h3>\n

Question 1.
\nSolve using the standard algorithm.
\na. 3 \u00d7 41
\nAnswer:
\n41
\nX 3
\n123<\/span>
\n3 X 41 = 123,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n41
\nX 3
\n123 ,<\/span>here 3 X 1 one = 3 ones then 3 X 4 tens = 12 tens
\nwe write 2 at tens place and write 1 at hundred place,
\nSo 3 X 41 = 123 as shown above.<\/p>\n

b. 9 \u00d7 41
\nAnswer:
\n<\/u>41
\nX 9
\n369<\/span>
\n9 X 41 = 369,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n41
\nX 9
\n369 ,<\/span>here 9 X 1 one = 9 ones we write 9 at ones place
\nthen 9 x 4 tens = 36 tens, we write 6 at tens place and
\nwrite 3 at hundred place,
\nSo 9 X 41 = 369 as shown above.<\/p>\n

c. 7 \u00d7 143
\nAnswer:
\n3,2
\n<\/u>143
\nX 7
\n1,001<\/span>
\n7 X 143 = 1,001,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n3,2<\/span>
\n143
\nX 7
\n1,001 ,<\/span>here 7 X 3 ones = 21 ones we write 1 at ones place and
\n2 at tens places then 7 x 4 tens = 28 tens, 28 tens +2 tens =
\n30 tens we write 0 at tens place and write 3 at hundreds place,
\nNow 7 X 1 hundred = 7 hundreds, 7 hundreds + 3 hundreds =
\n10 hundreds we write 0 at hundreds place and 1 at
\nthousands places, So 7 X 143 = 1,001 as shown above.<\/p>\n

d. 7 \u00d7 286
\nAnswer:
\n6,4
\n<\/u>286
\nX 7
\n2,002<\/span>
\n7 X 286 = 2,002,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n6,4<\/span>
\n286
\nX 7
\n2,002 ,<\/span>here 7 X 6 ones = 42 ones we write 2 at ones place and
\n4 at tens places then 7 x 8 tens = 56 tens, 56 tens +4 tens =
\n60 tens we write 0 at tens place and write 6 at hundreds place,
\nNow 7 X 2 hundreds = 14 hundreds, 14 hundreds + 6 hundreds =
\n20 hundreds we write 0 at hundreds place and 2 at
\nthousands places, So 7 X 286 = 2,002 as shown above.<\/p>\n

e. 4 \u00d7 2,048
\nAnswer:
\n1,3<\/span>
\n2,048
\nX 4
\n8,192
\n<\/span>4 X 2,048 = 8,192,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n1,3
\n<\/span>2,048
\nX 4
\n8,192\u00a0,<\/span>here 4 X 8 ones = 32 ones, we write 2 at ones place,
\n3 at tens place then 4 X 4 tens = 16 tens, 16 tens + 3 tens =
\n19 tens we write 9 at tens place, 1 at hundreds place,
\n4 X 0 hundreds = 0 hundreds, 0 hundreds +1 hundred =
\n1 hundred, 4 X 2 thousands = 8 thousands we write 8 at
\nthousands place, So 4 X 2,048 = 8,192 as shown above.<\/u><\/p>\n

f. 4 \u00d7 4,096
\nAnswer:
\n3,2<\/span>
\n4,096
\nX 4
\n16,384
\n<\/span>4 X 4,096 = 16,384,
\n<\/span><\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n3,2
\n<\/span>4,096
\nX 4
\n16,384\u00a0,<\/span>here 4 X 6 ones = 24 ones, we write 4 at ones place,
\n2 at tens place then 4 X 9 tens = 36 tens, 36 tens + 2 tens =
\n38 tens we write 8 at tens place, 3 at hundreds place,
\n4 X 0 hundreds = 0 hundreds, 0 hundreds +3 hundreds =
\n3 hundreds at hundreds place, 4 X 4 thousands =
\n16 thousands we write 6 at thousands place and
\n1 at ten thousands place,
\nSo 4 X 4,096 = 16,384 as shown above.<\/u><\/p>\n

g. 8 \u00d7 4,096
\nAnswer:
\n7,4<\/span>
\n4,096
\nX 8
\n32,768
\n<\/span>8 X 4,096 = 32,768,
\n<\/span><\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n7,4
\n<\/span>4,096
\nX 8
\n32,768 ,<\/span>here 8 X 6 ones = 48 ones, we write 8 at ones place,
\n4 at tens place then 8 X 9 tens = 72 tens, 72 tens + 4 tens =
\n76 tens we write 6 at tens place, 7 at hundreds place,
\n8 X 0 hundreds = 0 hundreds, 0 hundreds +7 hundreds =
\n7 hundreds at hundreds place, 8 X 4 thousands =
\n32 thousands we write 2 at thousands place and
\n3 at ten thousands place,
\nSo 8 X 4,096 = 32,768 as shown above.<\/u><\/p>\n

h. 4 \u00d7 8,192
\nAnswer:
\n3<\/span>
\n8,192
\nX\u00a0 \u00a04
\n32,768
\n<\/span>4 X 8,192 = 32,768,<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n3
\n<\/span>8,192
\nX 4
\n32,768 ,<\/span>here 4 X 2 ones = 8 ones, we write 8 at ones place
\nthen 4 X 9 tens = 36 tens we write 6 at tens place,
\n3 at hundreds place, 4 X 1 hundred = 4 hundreds,
\n4 hundreds +3 hundreds = 7 hundreds at hundreds place,
\n4 X 8 thousands = 32 thousands we write 2 at thousands place and
\n3 at ten thousands place, So 4 X 8,192 = 32,768 as shown above.<\/u><\/p>\n

Question 2.
\nRobert\u2019s family brings six gallons of water for the players on the football team. If one gallon of water contains 128 fluid ounces, how many fluid ounces are in six gallons?
\nAnswer:
\nThere are 768 fluid ounces in six gallons,<\/p>\n

Explanation:
\nGiven Robert\u2019s family brings six gallons of water for the players on the football team. If one gallon of water contains 128 fluid ounces, Number of fluid ounces are in six gallons are 6 X 128 fluid ounces =
\n1,4<\/span>
\n128
\nX 6
\n768 ,<\/span> here 6 X 8 ones = 48 ones we write 8 at ones place,
\n4 at tens place then 6 x 2 tens = 12 tens, 12 tens + 4 tens =
\n16 tens we write 6 at tens place and write 1 at hundreds place,
\n6 X 1 hundred = 6 hundreds, 6 hundreds + 1 hundred =
\n7 hundreds, So 6 X 128 = 768,
\ntherefore there are 768 fluid ounces in six gallons.<\/p>\n

Question 3.
\nIt takes 687 Earth days for the planet Mars to revolve around the sun once. How many Earth days does it take Mars to revolve around the sun four times?
\nAnswer:
\nIt will take 2,748 Earth daysfor Mars to revolve around the sun four times,<\/p>\n

Explanation:
\nGiven It takes 687 Earth days for the planet Mars to revolve around the sun once. So Earth days does it take Mars to revolve around the sun four times is
\n4 X 687 Earth days =
\n3,2<\/span>
\n687
\nX 4<\/span>
\n2,748
\n<\/span>here 4 X 7 ones = 28 ones, we write 8 at ones place,
\n2 at tens place then 4 X 8 tens = 32 tens, 32 tens + 2 tens =
\n34 tens we write 4 at tens place, 3 at hundreds place,
\n4 X 6 hundreds = 24 hundreds, 24 hundreds +3 hundreds =
\n27 hundreds, 7 at hundreds place, 2 at thousands place
\nSo 4 X 687 = 2,748 as shown above, therefore, it will take
\n2,748 Earth daysfor Mars to revolve around the sun four times.
\n<\/span><\/p>\n

Question 4.
\nTammy buys a 4-gigabyte memory card for her camera. Dijonea buys a memory card with twice as much storage as Tammy\u2019s. One gigabyte is 1,024 megabytes. How many megabytes of storage does Dijonea have on her memory card?
\nAnswer:
\n8,192 megabytes of storage does Dijonea have on her memory card,<\/p>\n

Explanation:
\nGiven Tammy buys a 4-gigabyte memory card for her camera.
\nDijonea buys a memory card with twice as much storage as Tammy\u2019s.
\nOne gigabyte is 1,024 megabytes. Dijonea has storage of 2 X 4 – gigabyte = 8 gigabyte memory card now Dijonea has number of megabytes of storage is 8 X 1,024 megabytes =
\n1,3<\/span>
\n1,024
\nX 8
\n8,192
\n<\/span>here 8 X 4 ones = 32 ones, we write 2 at ones place,
\n3 at tens place then 8 X 2 tens = 16 tens, 16 tens + 3 tens =
\n19 tens we write 9 at tens place, 1 at hundreds place,
\n8 X 0 hundreds = 0 hundreds, 0 hundreds + 1 hundred =
\n1 hundred, 1 at hundreds place, 8 X 1 thousand =
\n8 thousand, 8 at thousands place, So 8 X 1,024 =
\n8,192 as shown above
\ntherefore, 8,192 megabytes of storage does Dijonea have on her memory card.
\n<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 4th Grade Module 3 Lesson 10 Answer Key Eureka Math Grade 4 Module 3 Lesson 10 Problem Set Answer Key Question 1. Solve using the standard algorithm. a. 3 \u00d7 42 Answer: 42 X 3 126 3 X 42 = 126, Explanation: In standard algorithm we add same time of multiplying … 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\/59706"}],"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=59706"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/59706\/revisions"}],"predecessor-version":[{"id":137374,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/59706\/revisions\/137374"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=59706"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=59706"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=59706"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}