Eureka Math Grade 4 Module 3 Lesson 10 Answer Key (2024)

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 × 42
Answer:
42
X 3
126

3 X 42 = 126,

Explanation:
In standard algorithm we add same time of multiplying as
42
X 3
126 ,
here 3 X 2 ones = 6 ones then 3 X 4 tens = 12 tens
we write 2 at tens place and write 1 at hundred place,
So 3 X 42 = 126 as shown above.

b. 6 × 42
Answer:
1
42
X 6
252

6 X 42 = 252,

Explanation:
In standard algorithm we add same time of multiplying as
1
42
X 6
252 ,
here 6 X 2 ones = 12 ones we write 2 at ones place and
1 at tens places then 6 x 4 tens = 24 tens, 24 tens +1 ten =
25 tens we write 5 at tens place and write 2 at hundred place,
So 6 X 42 = 252 as shown above.

c. 6 × 431
Answer:
1
431
X 6
2,586

6 X 42 = 2,586,

Explanation:
In standard algorithm we add same time of multiplying as
1
431
X 6
2,586 ,
here 6 X 1 ones = 6 ones then 6 X 3 tens = 18 tens,
we write 8 at tens place and write 1 at hundreds place,
6 X 4 hundreds = 24 hundreds, 24 hundreds + 1 hundred =
25 hundreds we write 25 hundreds as 5 at hundreds place and
2 at thousands place, So 6 X 431 = 2,586 as shown above.

d. 3 × 431
Answer:
431
X 3
1,293

3 X 431 = 1,293,

Explanation:
In standard algorithm we add same time of multiplying as
431
X 3
1,293 ,
here 3 X 1 ones = 3 ones then 3 X 3 tens = 9 tens,
we write 9 at tens place and 3 X 4 hundreds = 12 hundreds,
we write 12 hundreds as 2 at hundreds place and 1 at
thousands place, So 3 X 431 = 1,293 as shown above.

e. 3 × 6,212
Answer:
6,212
X 3
18,636
3 X 6,212 = 18,636,

Explanation:
In standard algorithm we add same time of multiplying as
6,212
X 3
18,636 ,
here 3 X 2 ones = 6 ones then 3 X 1 ten = 3 tens,
we write 3 at tens place, 3 X 2 hundreds = 6 hundreds,
3 X 6 thousands = 18 thousands we write 18 thousands as
8 at thousands place and 1 at ten thousands place,
So 3 X 6,212 = 18,636 as shown above.

f. 3 × 3,106
Answer:
1
3,106
X 3
9,318
3 X 3,106 = 9,318,

Explanation:
In standard algorithm we add same time of multiplying as
1
3,106
X 3
9,318 ,
here 3 X 6 ones = 18 ones we write 8 at ones place and
1 at tens places then 3 X 0 tens = 0 tens, 0 tens + 1 ten = 1 ten
at ten place, 3 X 1 hundred = 3 hundreds
we write 3 at hundreds place and 3 X 3 thousands =
9 thousands we write 9 at thousands place,
So 3 X 3,106 = 9,318 as shown above.

g. 4 × 4,309
Answer:
1,3
4,309
X 4
17,236
4 X 4,309 = 17,236,

Explanation:
In standard algorithm we add same time of multiplying as
1,3
4,309
X 4
17,236 ,here 4 X 9 ones = 36 ones we write 6 at ones place and
3 at tens places then 4 X 0 tens = 0 tens, 0 tens + 3 tens = 3 tens
at ten place, 4 X 3 hundreds = 12 hundreds
we write 12 hundreds as 2 at hundreds place and 1 at
thousands place, 4 X 4 thousands = 16 thousands,
16 thousands + 1 thousand = 17 thousands we write 7 at
thousands place and 1 at ten thousands place,
So 4 X 4,309 = 17,236 as shown above.

h. 4 × 8,618
Answer:
2,3
8,618
X 4
34,472
4 X 8,618 = 34,472,

Explanation:
In standard algorithm we add same time of multiplying as
2,3
8,618
X 4
34,472 ,here 4 X 8 ones = 32 ones we write 2 at ones place and
3 at tens places then 4 X 1 ten = 4 tens, 4 tens + 3 tens = 7 tens
at ten place, 4 X 6 hundreds = 24 hundreds
we write 24 hundreds as 4 at hundreds place and 2 at
thousands place, 4 X 8 thousands = 32 thousands,
32 thousands + 2 thousands = 34 thousands we write 4 at
thousands place and 3 at ten thousands place,
So 4 X 8,618 = 34,472 as shown above.

Question 2.
There are 365 days in a common year. How many days are in 3 common years?
Answer:
There are 1,095 days in 3 common years,

Explanation:
Given there are 365 days in a common year.
So number of days in 3 common years are 3 X 365 days =
1
365
X 3
1,095
here 3 X 5 ones = 15 ones we write 1 at ones place and
5 at tens places then 3 X 6 tens = 18 tens,
we write 8 at ten place and 1 at hundreds place and
3 X 3 hundreds = 9 hundreds, 9 hundreds + 1 hundered =
10 hundereds we write 10 hundreds as 0 at hundreds place and
1 at thousands place, So 3 X 365 = 1,095.
therefore, there are 1,095 days in 3 common years.

Question 3.
The length of one side of a square city block is 462 meters. What is the perimeter of the block?
Answer:
The perimeter of the block is 1,848 meters,

Explanation:
Given the length of one side of a square city block is 462 meters,
So the perimeter of the block is as we know perimeter of the
square is 4 X length, So 4 X 462 meters =
2
462
X 4
1,848
here first we write 4 X 2 ones = 8 ones, 4 X 6 tens = 24 tens we
write 4 at tens place and 2 at hundreds place and
4 X 4 hundreds = 16 hundreds, 16 hundreds + 2 hundreds =
18 hundreds we write 8 at hundreds place and 1 at thousands
place so 4 X 462 = 1,848. Therefore the perimeter of
the square block is 1,848 meters.

Question 4.
Jake ran 2 miles. Jesse ran 4 times as far. There are 5,280 feet in a mile. How many feet did Jesse run?
Answer:
Jesse ran 42,240 feet,

Explanation:
Given Jake ran 2 miles. Jesse ran 4 times as far.
There are 5,280 feet in a mile. So Jesse ran 4 X 2 = 8 miles,
8 miles X 5,280 feet =
2,6
5,280
X 8
42,240
here 8 X 0 ones = 0 ones we write 0 at ones place
then 8 X 8 tens = 64 tens, 4 at ten place, 6 at hundreds
place, 8 X 2 hundreds = 16 hundreds, 16 hundreds + 6 hundreds =
22 hundreds we write 22 hundreds as 2 at hundreds place and
2 at thousands place, 8 X 5 thousands = 40 thousands,
40 thousands + 2 thousands = 42 thousands we write 2 at
thousands place and 4 at ten thousands place, So 8 X 5,280 =
42,240, therefore Jesse ran 42,240 feet.

Eureka Math Grade 4 Module 3 Lesson 10 Exit Ticket Answer Key

Question 1.
Solve using the standard algorithm.
a. 2,348 × 6
Answer:
2,2,4
2,348
X 6
14,088
6 X 2,348 = 14,088,

Explanation:
In standard algorithm we add same time of multiplying as
2,2,4
2,348
X 6
14,088 ,here 6 X 8 ones = 48 ones we write 8 at ones place and
4 at tens places then 6 X 4 tens = 24 tens, 24 tens + 4 tens = 28 tens,
8 at tens place and 2 at hundreds place, 6 X 3 hundreds =
18 hundreds, 18 hundreds + 2 hundreds = 20 hundreds,
we write 20 hundreds as 0 at hundreds place and 2 at
thousands place, 6 X 2 thousands = 12 thousands,
12 thousands + 2 thousands = 14 thousands we write 4 at
thousands place and 1 at ten thousands place,
So 6 X 2,348 = 14,088 as shown above.

b. 1,679 × 7
Answer:
4,5,6
1,679
X 7
11,753
1,679 X 7 = 11,753,

Explanation:
In standard algorithm we add same time of multiplying as
4,5,6
1,679
X 7
11,753 ,here 7 X 9 ones = 63 ones we write 3 at ones place and
6 at tens places then 7 X 7 tens = 49 tens, 49 tens + 6 tens = 55 tens,
5 at tens place and 5 at hundreds place, 7 X 6 hundreds =
42 hundreds, 42 hundreds + 5 hundreds = 47 hundreds,
we write 47 hundreds as 7 at hundreds place and 4 at
thousands place, 7 X 1 thousands = 7 thousands,
7 thousands + 4 thousands = 11 thousands we write 1 at
thousands place and 1 at ten thousands place,
So 7 X 1,679 = 11,753 as shown above.

Question 2.
A farmer planted 4 rows of sunflowers. There were 1,205 plants in each row. How many sunflowers did he plant?
Answer:
Number of sunflowers planted are 4,820,

Explanation:
Given a farmer planted 4 rows of sunflowers.
There were 1,205 plants in each row.
So number of sunflowers did he planted are 4 X 1,205 =
2
1,205
X 4
4,820
here 5 X 4 ones = 20 ones we write 0 at ones place and
2 at tens places then 4 X 0 tens = 0 tens, 0 tens + 2 tens = 2 tens,
2 at tens place, 4 X 2 hundreds = 8 hundreds,
we write 8 hundreds at hundreds place and
4 X 1 thousands = 4 thousands, 4 at thousands place,
So 4 X 1,205 = 4,820, therefore number of sunflowers planted are 4,820.

Eureka Math Grade 4 Module 3 Lesson 10 Homework Answer Key

Question 1.
Solve using the standard algorithm.
a. 3 × 41
Answer:
41
X 3
123

3 X 41 = 123,

Explanation:
In standard algorithm we add same time of multiplying as
41
X 3
123 ,
here 3 X 1 one = 3 ones then 3 X 4 tens = 12 tens
we write 2 at tens place and write 1 at hundred place,
So 3 X 41 = 123 as shown above.

b. 9 × 41
Answer:
41
X 9
369

9 X 41 = 369,

Explanation:
In standard algorithm we add same time of multiplying as
41
X 9
369 ,
here 9 X 1 one = 9 ones we write 9 at ones place
then 9 x 4 tens = 36 tens, we write 6 at tens place and
write 3 at hundred place,
So 9 X 41 = 369 as shown above.

c. 7 × 143
Answer:
3,2
143
X 7
1,001

7 X 143 = 1,001,

Explanation:
In standard algorithm we add same time of multiplying as
3,2
143
X 7
1,001 ,
here 7 X 3 ones = 21 ones we write 1 at ones place and
2 at tens places then 7 x 4 tens = 28 tens, 28 tens +2 tens =
30 tens we write 0 at tens place and write 3 at hundreds place,
Now 7 X 1 hundred = 7 hundreds, 7 hundreds + 3 hundreds =
10 hundreds we write 0 at hundreds place and 1 at
thousands places, So 7 X 143 = 1,001 as shown above.

d. 7 × 286
Answer:
6,4
286
X 7
2,002

7 X 286 = 2,002,

Explanation:
In standard algorithm we add same time of multiplying as
6,4
286
X 7
2,002 ,
here 7 X 6 ones = 42 ones we write 2 at ones place and
4 at tens places then 7 x 8 tens = 56 tens, 56 tens +4 tens =
60 tens we write 0 at tens place and write 6 at hundreds place,
Now 7 X 2 hundreds = 14 hundreds, 14 hundreds + 6 hundreds =
20 hundreds we write 0 at hundreds place and 2 at
thousands places, So 7 X 286 = 2,002 as shown above.

e. 4 × 2,048
Answer:
1,3
2,048
X 4
8,192
4 X 2,048 = 8,192,

Explanation:
In standard algorithm we add same time of multiplying as
1,3
2,048
X 4
8,192,
here 4 X 8 ones = 32 ones, we write 2 at ones place,
3 at tens place then 4 X 4 tens = 16 tens, 16 tens + 3 tens =
19 tens we write 9 at tens place, 1 at hundreds place,
4 X 0 hundreds = 0 hundreds, 0 hundreds +1 hundred =
1 hundred, 4 X 2 thousands = 8 thousands we write 8 at
thousands place, So 4 X 2,048 = 8,192 as shown above.

f. 4 × 4,096
Answer:
3,2
4,096
X 4
16,384
4 X 4,096 = 16,384,

Explanation:
In standard algorithm we add same time of multiplying as
3,2
4,096
X 4
16,384,
here 4 X 6 ones = 24 ones, we write 4 at ones place,
2 at tens place then 4 X 9 tens = 36 tens, 36 tens + 2 tens =
38 tens we write 8 at tens place, 3 at hundreds place,
4 X 0 hundreds = 0 hundreds, 0 hundreds +3 hundreds =
3 hundreds at hundreds place, 4 X 4 thousands =
16 thousands we write 6 at thousands place and
1 at ten thousands place,
So 4 X 4,096 = 16,384 as shown above.

g. 8 × 4,096
Answer:
7,4
4,096
X 8
32,768
8 X 4,096 = 32,768,

Explanation:
In standard algorithm we add same time of multiplying as
7,4
4,096
X 8
32,768 ,
here 8 X 6 ones = 48 ones, we write 8 at ones place,
4 at tens place then 8 X 9 tens = 72 tens, 72 tens + 4 tens =
76 tens we write 6 at tens place, 7 at hundreds place,
8 X 0 hundreds = 0 hundreds, 0 hundreds +7 hundreds =
7 hundreds at hundreds place, 8 X 4 thousands =
32 thousands we write 2 at thousands place and
3 at ten thousands place,
So 8 X 4,096 = 32,768 as shown above.

h. 4 × 8,192
Answer:
3
8,192
X 4
32,768
4 X 8,192 = 32,768,

Explanation:
In standard algorithm we add same time of multiplying as
3
8,192
X 4
32,768 ,
here 4 X 2 ones = 8 ones, we write 8 at ones place
then 4 X 9 tens = 36 tens we write 6 at tens place,
3 at hundreds place, 4 X 1 hundred = 4 hundreds,
4 hundreds +3 hundreds = 7 hundreds at hundreds place,
4 X 8 thousands = 32 thousands we write 2 at thousands place and
3 at ten thousands place, So 4 X 8,192 = 32,768 as shown above.

Question 2.
Robert’s 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?
Answer:
There are 768 fluid ounces in six gallons,

Explanation:
Given Robert’s 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 =
1,4
128
X 6
768 ,
here 6 X 8 ones = 48 ones we write 8 at ones place,
4 at tens place then 6 x 2 tens = 12 tens, 12 tens + 4 tens =
16 tens we write 6 at tens place and write 1 at hundreds place,
6 X 1 hundred = 6 hundreds, 6 hundreds + 1 hundred =
7 hundreds, So 6 X 128 = 768,
therefore there are 768 fluid ounces in six gallons.

Question 3.
It 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?
Answer:
It will take 2,748 Earth daysfor Mars to revolve around the sun four times,

Explanation:
Given 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
4 X 687 Earth days =
3,2
687
X 4
2,748
here 4 X 7 ones = 28 ones, we write 8 at ones place,
2 at tens place then 4 X 8 tens = 32 tens, 32 tens + 2 tens =
34 tens we write 4 at tens place, 3 at hundreds place,
4 X 6 hundreds = 24 hundreds, 24 hundreds +3 hundreds =
27 hundreds, 7 at hundreds place, 2 at thousands place
So 4 X 687 = 2,748 as shown above, therefore, it will take
2,748 Earth daysfor Mars to revolve around the sun four times.

Question 4.
Tammy buys a 4-gigabyte memory card for her camera. Dijonea buys a memory card with twice as much storage as Tammy’s. One gigabyte is 1,024 megabytes. How many megabytes of storage does Dijonea have on her memory card?
Answer:
8,192 megabytes of storage does Dijonea have on her memory card,

Explanation:
Given Tammy buys a 4-gigabyte memory card for her camera.
Dijonea buys a memory card with twice as much storage as Tammy’s.
One 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 =
1,3
1,024
X 8
8,192
here 8 X 4 ones = 32 ones, we write 2 at ones place,
3 at tens place then 8 X 2 tens = 16 tens, 16 tens + 3 tens =
19 tens we write 9 at tens place, 1 at hundreds place,
8 X 0 hundreds = 0 hundreds, 0 hundreds + 1 hundred =
1 hundred, 1 at hundreds place, 8 X 1 thousand =
8 thousand, 8 at thousands place, So 8 X 1,024 =
8,192 as shown above
therefore, 8,192 megabytes of storage does Dijonea have on her memory card.

Eureka Math Grade 4 Module 3 Lesson 10 Answer Key (2024)

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