Eureka Math

Engage NY Eureka Math 4th Grade Module 3 Lesson 18 Answer Key

Eureka Math Grade 4 Module 3 Lesson 18 Problem Set Answer Key

Solve using the standard algorithm. Check your quotient and
remainder by using multiplication and addition.

Question 1.
46 ÷ 2
Answer:

Explanation:
Solved 46 ÷ 2 using the standard algorithm.
Checked quotient and remainder by using multiplication as
shown above, quotient = 23, remainder = 0.

Question 2.
96 ÷ 3
Answer:

Explanation:
Solved 96 ÷ 3 using the standard algorithm.
Checked quotient and remainder by using multiplication as
shown above, quotient = 32, remainder = 0.

Question 3.
85 ÷ 5
Answer:

Explanation:
Solved 85 ÷ 5 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient = 17, remainder = 0.

Question 4.
52 ÷ 4
Answer:

Explanation:
Solved 52 ÷ 4 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient = 13, remainder = 0.

Question 5.
53 ÷ 3
Answer:

Explanation:
Solved 53 ÷ 3 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient = 17, remainder = 2.

Question 6.
95 ÷ 4
Answer:

Explanation:
Solved 95 ÷ 4 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient = 23, remainder = 3.

Question 7.
89 ÷ 6
Answer:

Explanation:
Solved 89 ÷ 6 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient = 14, remainder = 5.

Question 8.
96 ÷ 6
Answer:

Explanation:
Solved 96 ÷ 6 using the standard algorithm.
Checked quotient and remainder by using multiplication as
shown above, quotient = 16, remainder = 0.

Question 9.
60 ÷ 3
Answer:

Explanation:
Solved 60 ÷ 3 using the standard algorithm.
Checked quotient and remainder by using multiplication as
shown above, quotient = 20, remainder = 0.

Question 10.
60 ÷ 4
Answer:

Explanation:
Solved 60 ÷ 4 using the standard algorithm.
Checked quotient and remainder by using multiplication as
shown above, quotient =15, remainder = 0.

Question 11.
95 ÷ 8
Answer:

Explanation:
Solved 95 ÷ 8 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient = 11, remainder = 7.

Question 12.
95 ÷ 7
Answer:

Explanation:
Solved 95 ÷ 7 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient =13, remainder = 4.

Eureka Math Grade 4 Module 3 Lesson 18 Exit Ticket Answer Key

Solve using the standard algorithm. Check your quotient and remainder by using multiplication and addition.

Question 1.
93 ÷ 7
Answer:

Explanation:
Solved 93 ÷ 7 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient =13, remainder = 2.

Question 2.
99 ÷ 8
Answer:

Explanation:
Solved 99 ÷ 8 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient =12, remainder = 3.

Eureka Math Grade 4 Module 3 Lesson 18 Homework Answer Key

Solve using the standard algorithm. Check your quotient and remainder by using multiplication and addition.

Question 1.
84 ÷ 2
Answer:

Explanation:
Solved 84 ÷ 2 using the standard algorithm.
Checked quotient and remainder by using multiplication as
shown above, quotient =42, remainder = 0.

Question 2.
84 ÷ 4
Answer:

Explanation:
Solved 84 ÷ 4 using the standard algorithm.
Checked quotient and remainder by using multiplication as
shown above, quotient =21, remainder = 0.

Question 3.
48 ÷ 3
Answer:

Explanation:
Solved 48 ÷ 3 using the standard algorithm.
Checked quotient and remainder by using multiplication as
shown above, quotient =16, remainder = 0.

Question 4.
80 ÷ 5
Answer:

Explanation:
Solved 80 ÷ 5 using the standard algorithm.
Checked quotient and remainder by using multiplication as
shown above, quotient =16, remainder = 0.

Question 5.
79 ÷ 5
Answer:

Explanation:
Solved 79 ÷ 5 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient =15, remainder = 4.

Question 6.
91 ÷ 4
Answer:

Explanation:
Solved 91 ÷ 4 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient =22, remainder = 3.

Question 7.
91 ÷ 6
Answer:

Explanation:
Solved 91 ÷ 6 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient =15, remainder = 1.

Question 8.
91 ÷ 7
Answer:

Explanation:
Solved 91 ÷ 7 using the standard algorithm.
Checked quotient and remainder by using multiplication as
shown above, quotient =13, remainder = 0.

Question 9.
87 ÷ 3
Answer:

Explanation:
Solved 87 ÷ 3 using the standard algorithm.
Checked quotient and remainder by using multiplication as
shown above, quotient = 29, remainder = 0.

Question 10.
87 ÷ 6
Answer:

Explanation:
Solved 87 ÷ 6 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient =14, remainder = 3.

Question 11.
94 ÷ 8
Answer:

Explanation:
Solved 94 ÷ 8 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient =11, remainder = 6.

Question 12.
94 ÷ 6
Answer:

Explanation:
Solved 94 ÷ 6 using the standard algorithm. Checked quotient
and remainder by using multiplication and addition as
shown above, quotient =15, remainder = 4.

Engage NY Eureka Math 4th Grade Module 3 Lesson 17 Answer Key

Eureka Math Grade 4 Module 3 Lesson 17 Problem Set Answer Key

Show the division using disks. Relate your model to long division. Check your quotient and remainder by using multiplication and addition.

Question 1.
5 ÷ 2


quotient = ____2______
remainder = __1________
Answer:

Explanation:
Shown the division 5 ÷ 2 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 2.
50 ÷ 2


quotient = _____25_____
remainder = ____0______
Answer:
50 ÷ 2 =

Explanation:
Shown the division 5 ÷ 2 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 3.
7 ÷ 3


quotient = _____2_____
remainder = ____1______
Answer:
7 ÷ 3 =

Explanation:
Shown the division 7 ÷ 3 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 4.
75 ÷ 3


quotient = ____25____
remainder = ___0____
Answer:
75 ÷ 3 =

Explanation:
Shown the division 75 ÷ 3 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 5.
9 ÷ 4


quotient = ___2_______
remainder = ____1______
Answer:
9 ÷ 4 =

Explanation:
Shown the division 9 ÷ 4 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 6.
92 ÷ 4


quotient = ____23____
remainder = ___0_____
Answer:
92 ÷ 4 =

Explanation:
Shown the division 9 ÷ 4 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Eureka Math Grade 4 Module 3 Lesson 17 Exit Ticket Answer Key

Show the division using disks. Relate your model to long division.
Check your quotient by using multiplication and addition.

Question 1.
5 ÷ 4


quotient = ____1______
remainder = ____1______
Answer:
5 ÷ 4 =

Explanation:
Shown the division 5 ÷ 4 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 2.
56 ÷ 4


quotient = ____14______
remainder = ____0______
Answer:

Explanation:
Shown the division 56 ÷ 4 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Eureka Math Grade 4 Module 3 Lesson 17 Homework Answer Key

Show the division using disks. Relate your model to long division.
Check your quotient by using multiplication and addition.

Question 1.
7 ÷ 2


quotient = ____3_____
remainder = ___1_______
Answer:
7 ÷ 2 =

Explanation:
Shown the division 7 ÷ 2 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 2.
73 ÷ 2


quotient = ____24______
remainder = ___1_______
Answer:
73 ÷ 2 =

Explanation:
Shown the division 73 ÷ 2 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 3.
6 ÷ 4


quotient = ____1______
remainder = ____2______
Answer:

Explanation:
Shown the division 6 ÷ 4 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 4.
62 ÷ 4


quotient = _____15_____
remainder = ____2______
Answer:

Explanation:
Shown the division 62 ÷ 4 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 5.
8 ÷ 3


quotient = ____2______
remainder = ____2______
Answer:

Explanation:
Shown the division 8 ÷ 3 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 6.
84 ÷ 3


quotient = ____28______
remainder = ____3______
Answer:
84 ÷ 3 =

Explanation:
Shown the division 84 ÷ 3 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Engage NY Eureka Math 4th Grade Module 3 Lesson 16 Answer Key

Eureka Math Grade 4 Module 3 Lesson 16 Problem Set Answer Key

Show the division using disks. Relate your work on the
place value chart to long division. Check your quotient and
remainder by using multiplication and addition.

Question 1.
7 ÷ 2

quotient = ____3______
remainder = ___1_______
Answer:
7 ÷ 2 =

Explanation:
Shown the division 7 ÷ 2 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 2.
27 ÷ 2

quotient = ____13______
remainder = ____1______
Answer:
27 ÷ 2 =

Explanation:
Shown the division 27 ÷ 2 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 3.
8 ÷ 3

quotient = ___2______
remainder = ____2______
Answer:
8 ÷ 3  =

Explanation:
Shown the division 8 ÷ 3 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 4.
38 ÷ 3

quotient = ____12______
remainder = ____2______
Answer:
38 ÷ 3 =

Explanation:
Shown the division 38 ÷ 3 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 5.
6 ÷ 4

quotient = ____1___
remainder = ___2____
Answer:
6 ÷ 4 =

Explanation:
Shown the division 6 ÷ 4 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 6.
86 ÷ 4

quotient = ____21______
remainder = ___2_______
Answer:
86 ÷ 4 =

Explanation:
Shown the division 86 ÷ 4 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Eureka Math Grade 4 Module 3 Lesson 16 Exit Ticket Answer Key

Show the division using disks. Relate your work on the place value chart to long division. Check your quotient and remainder by using multiplication and addition.

Question 1.
5 ÷ 3

quotient = ____1______
remainder = ___2_______
Answer:
5 ÷ 3 =

Explanation:
Shown the division 5 ÷ 3 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 2.
65 ÷ 3

quotient = ____21______
remainder = ____2______
Answer:
65 ÷ 3 =

Explanation:
Shown the division 65 ÷ 3 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Eureka Math Grade 4 Module 3 Lesson 16 Homework Answer Key

Show the division using disks. Relate your work on the
place value chart to long division. Check your quotient and
remainder by using multiplication and addition.

Question 1.
7 ÷ 3

quotient = ____2______
remainder = ___1_______
Answer:
7 ÷ 3 =

Explanation:
Shown the division 7 ÷ 3 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 2.
67 ÷ 3

quotient = ____22______
remainder = ____1______
Answer:
67 ÷ 3=

Explanation:
Shown the division 67 ÷ 3 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 3.
5 ÷ 2

quotient = ____2______
remainder = ___1_______
Answer:
5 ÷ 2 =

Explanation:
Shown the division 5 ÷ 2 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 4.
85 ÷ 2

quotient = ___42_______
remainder = ___1_______
Answer:
85 ÷ 2 =

Explanation:
Shown the division 85 ÷ 2 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 5.
5 ÷ 4

quotient = ____1______
remainder = ____1______
Answer:
5 ÷ 4 =

Explanation:
Shown the division 5 ÷ 4 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Question 6.
85 ÷ 4

quotient = ___21_______
remainder = ____1______
Answer:
85 ÷ 4 =

Explanation:
Shown the division 85 ÷ 4 using disks. Related my work on the
place value chart to long division. Checking my quotient and
remainder by using multiplication and addition as shown above.

Eureka Math Grade 4 Module 3 Lesson 16 Template Set Answer Key

_______________________________________________________
tens place value chart

Explanation:
In math, every digit in a number has a place value.
Place value can be defined as the value represented by
a digit in a number on the basis of its position in the number.
In above example 42, 2 is at ones place and 4 is at tens place
as shown above.

Engage NY Eureka Math 4th Grade Module 3 Lesson 15 Answer Key

Eureka Math Grade 4 Module 3 Lesson 15 Problem Set Answer Key

Question 1.

Show division using an array.	Show division using an area model.
18 ÷ 6 Quotient = ____3_____ Remainder = __0_____	Can you show 18 ÷ 6 with one rectangle? __Yes____

Answer:

Explanation:
Shown division using an array as shown above,
18 ÷ 6 = Quotient = 3 and Remainder = 0,
Shown division using an area model above,
Yes, shown 18 ÷ 6 with one rectangle above.

Question 2.

Show division using an array.	Show division using an area model.
19 ÷ 6 Quotient = ____3_____ Remainder = ___1____	Can you show 19 ÷ 6 with one rectangle? ___Yes___ Explain how you showed the remainder:

Answer:

Explanation:
Shown division using an array as shown above,
19 ÷ 6 = Quotient = 3 and Remainder = 1,
Show division using an area model above,
Yes, shown 19 ÷ 6 with one rectangle above and showed
remainder as one square as shown above as.

Question 3.
29 ÷ 3
a.
Answer:

Explanation:
Shown division using an array as shown above,
29 ÷ 3 = Quotient = 9 and Remainder = 2.

b.
Answer:

Explanation:
Shown division 29 ÷ 3 using an area model above.

Question 4.
22 ÷ 5
a.
Answer:

Explanation:
Shown division using an array as shown above,
22 ÷ 5 = Quotient = 4 and Remainder = 2.

b.
Answer:

Explanation:
Shown division 22 ÷ 5 using an area model above.

Question 5.
43 ÷ 4
a.
Answer:

Explanation:
Shown division using an array as shown above,
43 ÷ 4 = Quotient = 10 and Remainder = 3.

b.
Answer:

Explanation:
Shown division 43 ÷ 4 using an area model above.

Question 6.
59 ÷ 7
a.
Answer:

Explanation:
Shown division using an array as shown above,
59 ÷ 7 = Quotient = 8 and Remainder = 3.

b.
Answer:

Explanation:
Shown division 59 ÷ 7 using an area model above.

Eureka Math Grade 4 Module 3 Lesson 15 Exit Ticket Answer Key

Solve using an array and area model.

Question 1.
27 ÷ 5
a.
Answer:

Explanation:
Shown division using an array as shown above,
27 ÷ 5 = Quotient = 5 and Remainder = 2.

b.
Answer:

Explanation:
Shown division 27 ÷ 5 using an area model above.

Question 2.
32 ÷ 6
a.
Answer:

Explanation:
Shown division using an array as shown above,
32 ÷ 6 = Quotient = 5 and Remainder = 2.

b.
Answer:

Explanation:
Shown division 32 ÷ 6 using an area model above.

Eureka Math Grade 4 Module 3 Lesson 15 Homework Answer Key

Question 1.

Show division using an array.	Show division using an area model.
24 ÷ 4 Quotient = ____6_____ Remainder = ___0____	Can you show 24 ÷ 4 with one rectangle? ___yes___

Answer:

Explanation:
Shown division using an array as shown above,
24 ÷ 4 = Quotient = 6 and Remainder = 0,
Shown division using an area model above,
Yes, shown 24 ÷ 4 with one rectangle above.

Question 2.

Show division using an array.	Show division using an area model.
25 ÷ 4 Quotient = _________ Remainder = _______	Can you show 25 ÷ 4 with one rectangle? ______ Explain how you showed the remainder:

Answer:

Shown division using an array as shown above,
25 ÷ 4 = Quotient = 6 and Remainder = 1,
Shown division using an area model above,
Yes, shown 25 ÷ 4 with one rectangle above and showed
remainder as one square as shown above as.

Question 3.
44 ÷ 7
a.
Answer:

Explanation:
Shown division using an array as shown above,
44 ÷ 7 = Quotient = 6 and Remainder = 2.

b.
Answer:

Explanation:
Shown division 44 ÷ 7 using an area model above.

Question 4.
34 ÷ 6
a.
Answer:

Explanation:
Shown division using an array as shown above,
34 ÷ 6 = Quotient = 5 and Remainder = 4.

b.
Answer:

Explanation:
Shown division 34 ÷ 6 using an area model above.

Question 5.
37 ÷ 6
a.
Answer:

Explanation:
Shown division using an array as shown above,
37 ÷ 6 = Quotient = 6 and Remainder = 1.

b.
Answer:

Explanation:
Shown division 34 ÷ 6 using an area model above.

Question 6.
46 ÷ 8
a.
Answer:

Explanation:
Shown division using an array as shown above,
46 ÷ 8 = Quotient = 5 and Remainder = 6.

b.
Answer:

Explanation:
Shown division 46 ÷ 8 using an area model above.

Engage NY Eureka Math 4th Grade Module 3 Lesson 14 Answer Key

Eureka Math Grade 4 Module 3 Lesson 14 Problem Set Answer Key

Use the RDW process to solve the following problems.

Question 1.
There are 19 identical socks. How many pairs of socks are there?
Will there be any socks without a match? If so, how many?
Answer:
9 pair of socks are there, 1 sock without a match,

Explanation:
Given there are 19 identical socks, number of pairs of
socks are there are 19 ÷ 2 = 9 quotient (2 X 9 = 18) and
remainder 1, means there are 9 pair of socks are there,
1 sock without a match,
Solved the problems using RDW process as shown above.

Question 2.
If it takes 8 inches of ribbon to make a bow, how many bows can
be made from 3 feet of ribbon (1 foot = 12 inches)?
Will any ribbon be left over? If so, how much?
Answer:
4 bows can be made from 3 feet of ribbon,
Left over ribbon will be 4 inches,

Explanation
Given If it takes 8 inches of ribbon to make a bow,
So how many bows can be made from 3 feet of ribbon
(1 foot = 12 inches) are 3 X 12 = 36 inches ÷ 8 inches =
4 quotient (8 X 4 = 32) and 4 inches remainder means|
4 bows can be made from 3 feet of ribbon,
Left over ribbon will be 4 inches,
Solved the problems using RDW process as shown above.

Question 3.
The library has 27 chairs and 5 tables. If the same number of chairs is
placed at each table, how many chairs can be placed at each table?
Will there be any extra chairs? If so, how many?
Answer:
5 chairs can be placed at each table and there will be 2 extra chairs,

Explanation:
Given the library has 27 chairs and 5 tables. If the same
number of chairs is placed at each table, how many chairs
can be placed at each table are 27 ÷ 5 = 5 quotient (5 X 5 = 25)
and 2 remainder means 5 chairs can be placed at
each table and there will be 2 extra chairs,
Solved the problems using RDW process as shown above.

Question 4.
The baker has 42 kilograms of flour. She uses 8 kilograms each day.
After how many days will she need to buy more flour?
Answer:
She needs to buy more flour after 5 days,

Explanation:
Given the baker has 42 kilograms of flour. She uses
8 kilograms each day.
After number of days will she need to buy more
flour is 42 ÷ 8 = 5 quotient (8 X 5 = 40) and
2 remainder it means that she needs to buy
more flour after 5 days,
Solved the problems using RDW process as shown above.

Question 5.
Caleb has 76 apples. He wants to bake as many pies as he can.
If it takes 8 apples to make each pie, how many apples will he use?
How many apples will not be used?
Answer:
Caleb uses 72 apples, 4 apples will not be used,

Explanation:
Given Caleb has 76 apples. He wants to bake as
many pies as he can. If it takes 8 apples to make each pie,
number of apples will he uses are
76 ÷ 8 = 9 quotient (8 X 9 = 72), 4 remainder it means
number of apples used are
8 X 9 = 72 apples and apples not used are 4,
Solved the problems using RDW process as shown above.

Question 6.
Forty-five people are going to the beach. Seven people can ride in each van.
How many vans will be required to get everyone to the beach?
Answer:
7 vans are required to get everyone to the beach,

Explanation:
Given Forty-five people are going to the beach.
Seven people can ride in each van.
Number of vans will be required to get everyone to the
beach are 45 ÷ 7 = 6 quotient (7 X 6 = 42), 3 remainder
it means 6 vans full and 3 people remaining therefore
7 vans are required to get everyone to the beach,
Solved the problems using RDW process as shown above.

Eureka Math Grade 4 Module 3 Lesson 14 Exit Ticket Answer Key

Use the RDW process to solve the following problem.

Fifty-three students are going on a field trip. The students are divided into
groups of 6 students. How many groups of 6 students will there be?
If the remaining students form a smaller group, and one chaperone
is assigned to every group, how many total chaperones are needed?
Answer:
8 groups of 6 students will be there and 1 chaperones are needed,

Explanation:
Given Fifty-three students are going on a field trip.
The students are divided into groups of 6 students.
Number of groups of 6 students will there be are
53 ÷ 6 = 8 quotient (6 X 8 = 48), 5 remainder it means
that 8 groups of 6 students will be there
and If the remaining students form a smaller group,
and one chaperone is assigned to every group,
number of chaperones are needed is 1 as
5 students form one smaller group,
Solved the problems using RDW process as shown above.

Eureka Math Grade 4 Module 3 Lesson 14 Homework Answer Key

Use the RDW process to solve the following problems.

Question 1.
Linda makes booklets using 2 sheets of paper.
She has 17 sheets of paper.
How many of these booklets can she make?
Will she have any extra paper?How many sheets?
Answer:
Linda can make 8 booklets, Yes, she will have
extra paper, 1 sheet,

Explanation:
Given Linda makes booklets using 2 sheets of paper.
She has 17 sheets of paper. Number of these booklets
can she make are 17 ÷ 2 = 8 quotient (2 X 8 = 16),
1 remainder means Linda can make 8 booklets,
Yes, she will have extra paper, 1 sheet,
Solved the problems using RDW process as shown above.

Question 2.
Linda uses thread to sew the booklets together.
She cuts 6 inches of thread for each booklet.
How many booklets can she stitch with 50 inches of thread?
Will she have any unused thread after stitching up
the booklets? If so, how much?
Answer:
Linda can stitch 8 booklets with 50 inches of thread,
Yes, she will have any unused thread after stitching up
the booklets, 2 inches of the thread,

Explanation:
Given Linda uses thread to sew the booklets together.
She cuts 6 inches of thread for each booklet.
Number of booklets can she stitch with 50 inches of thread is
50 inches ÷ 6 inches = 8 quotient (6 X 8 = 48),
2 remainder means Linda can stitch 8 booklets
with 50 inches of thread,
Yes, she will have any unused thread after stitching up
the booklets, 2 inches of the thread,
Solved the problems using RDW process as shown above.

Question 3.
Ms. Rochelle wants to put her 29 students into groups of 6.
How many groups of 6 can she make?
If she puts any remaining students in a smaller group,
how many students will be in that group?
Answer:
Ms.Rochelle can make 4 groups of 6 students,
In smaller group there will be 5 students,
Explanation:
Given Ms. Rochelle wants to put her 29 students
into groups of 6. Number of groups she can make
are 29 ÷ 6 = 4 quotient (6 X 4 = 24) and 5 remainder means
Ms.Rochelle can make 4 groups of 6 students,
In smaller group there will be 5 students,
Solved the problems using RDW process as shown above.

Question 4.
A trainer gives his horse, Caballo, 7 gallons of water
every day from a 57-gallon container.
How many days will Caballo receive his full portion
of water from the container?
On which number day will the trainer need to
refill the container of water?
Answer:
Number of days Caballo receive his full portion of
water from the container is 8 days,
On day 9th the trainer need to refill the conatiner of water,

Explanation:
Given a trainer gives his horse, Caballo, 7 gallons of water
every day from a 57-gallon container.
Number of days Caballo receive his full portion of
water from the container is 57 ÷ 7 = 8 quotient (7 X 8 = 56)
and 1 remainder means 8 days and on day 9th the
trainer need to refill the conatiner of water,
Solved the problems using RDW process as shown above.

Question 5.
Meliza has 43 toy soldiers. She lines them up in
rows of 5 to fight imaginary zombies.
How many of these rows can she make?
After making as many rows of 5 as she can,
she puts the remaining soldiers in the last row.
How many soldiers are in that row?
Answer:
Meliza makes 8 number of rows of 5 fight imaginary zombies,
In the last row number of soldiers are 3,

Explanation:
Given Meliza has 43 toy soldiers. She lines them up in
rows of 5 to fight imaginary zombies.
So number of rows she can make are 43 ÷ 5 =
8 quotient (5 X 8 = 40) and 3 remainder means
Meliza makes 8 number of rows of 5 fight imaginary zombies,
In the last row number of soldiers are 3,
Solved the problems using RDW process as shown above.

Question 6.
Seventy-eight students are separated into groups of 8
for a field trip.
How many groups are there? The remaining students form
a smaller group of how many students?
Answer:
There are 9 groups of 8 students for a field trip and in
smaller group there are 6 students,

Explanation:
Given Seventy-eight students are separated into groups of 8
for a field trip. So number of groups are 78 ÷ 8 =
9 quotient (8 X 9 = 72) and 6 remainder means
there are 9 groups of 8 students for a field trip and in
smaller group there are 6 students,
Solved the problems using RDW process as shown above.

Engage NY Eureka Math 4th Grade Module 3 Lesson 13 Answer Key

Eureka Math Grade 4 Module 3 Lesson 13 Sprint Answer Key

Mental Multiplication

Question 1.
1 × 4 =
Answer:
1 X 4 = 4,

Explanation:
Given expression as 1 X 4, multiplying 1 with 4
we get 4, So 1 X 4 = 4.

Question 2.
10 × 4 =
Answer:
10 X 4 = 40,

Explanation:
Given expression as 10 X 4, multiplying 10 with 4
we get 40, So 10 X 4 = 40.

Question 3.
11 × 4 =
Answer:
11 X 4 = 44,

Explanation:
Given expression as 11 X 4, multiplying 11 with 4
we get 44, So 11 X 4 = 44.

Question 4.
1 × 2 =
Answer:
1 X 2 = 2,

Explanation:
Explanation:
Given expression as 1 X 2, multiplying 1 with 2
we get 2, So 1 X 2 = 2.

Question 5.
20 × 2 =
Answer:
20 X 2 = 100,

Explanation:
Given expression as 20 X 2, multiplying 20 with 2
we  get 40, So 20 X 2 = 40.

Question 6.
21 × 2 =
Answer:
21 X 2 = 42,

Explanation:
Given expression as 21 X 2, multiplying 21 with 2
we  get 42, So 21 X 2 = 42.

Question 7.
2 × 3 =
Answer:
2 X 3 = 6,

Explanation:
Given expression as 2 X 3, multiplying 2 with 3
we get 6, So 2 X 3 = 6.

Question 8.
30 × 3 =
Answer:
30 X 3 = 90,

Explanation:
Given expression as 30 X 3, multiplying 30 with 3
we get 90, So 30 X 3 = 90.

Question 9.
32 × 3 =
Answer:
32 X 3 = 96,

Explanation:
Given expression as 32 X 3, multiplying 32 with 3
we get 96, So 32 X 3 = 96.

Question 10.
3 × 5=
Answer:
3 X 5 = 15,

Explanation:
Given expression as 3 X 5, multiplying 3 with 5
we get 15, So 3 X 5 = 15.

Question 11.
20 × 5 =
Answer:
20 X 5 = 100,

Explanation:
Given expression as 20 X 5, multiplying 20 with 5
we get 100, So 20 X 5 = 100.

Question 12.
23 × 5 =
Answer:
23 X 5 = 115,

Explanation:
Given expression as 23 X 5, multiplying 23 with 5
we get 115, So 23 X 5 = 115.

Question 13.
3 × 3 =
Answer:
3 X 3 = 9,

Explanation:
Given expression as 3 X 3, multiplying 3 with 3
we get 9, So 3 X 3 = 9.

Question 14.
40 × 3 =
Answer:
40 X 3 = 120,

Explanation:
Given expression as 40 X 3, multiplying 40 with 3
we get 120, So 40 X 3 = 120.

Question 15.
43 × 3 =
Answer:
43 X 3 = 129,

Explanation:
Given expression as 43 X 3, multiplying 43 with 3
we get 129, So 43 X 3 = 129.

Question 16.
4 × 2 =
Answer:
4 X 2 = 8,

Explanation:
Given expression as 4 X 2, multiplying 4 with 2
we get 8, So 4 X 2 = 8.

Question 17.
70 × 2 =
Answer:
70 X 2 = 140,

Explanation:
Given expression as 70 X 2, multiplying 70 with 2
we  get 140, So 70 X 2 = 140.

Question 18.
74 × 2 =
Answer:
74 X 2 = 148,

Explanation:
Given expression as 74 X 2, multiplying 74 with 2
we get 148, So 74 X 2 = 148.

Question 19.
2 × 3 =
Answer:
2 X 3 = 6,

Explanation:
Given expression as 2 X 3, multiplying 2 with 3
we get 6, So 2 X 3 = 6.

Question 20.
60 × 3 =
Answer:
60 X 3 = 180,

Explanation:
Given expression as 60 X 3, multiplying 60 with 3
we get 180, So 60 X 3 = 180.

Question 21.
62 × 3 =
Answer:
62 X 3 = 186,

Explanation:
Given expression as 62 X 3 , multiplying 62 with 3
we get 186, So 62 X 3 = 186.

Question 22.
63 × 3 =
Answer:
63 X 3 = 189,

Explanation:
Given expression as 63 X 3, multiplying 63 with 3
we get 189, So 63 X 3 = 189.

Question 23.
21 × 3 =
Answer:
21 X 3 = 63,

Explanation:
Given expression as 21 X 3, multiplying 21 with 3
we get 63, So 21 X 3 = 63.

Question 24.
121 × 3 =
Answer:
121 X 3 = 363,

Explanation:
Given expression as 121 X 3, multiplying 121 with 3
we get 363, So 121 X 3 = 363.

Question 25.
42 × 2 =
Answer:
42 X 2 = 84,

Explanation:
Given expression as 42 X 2, multiplying 42 with 2
we get 84, So 42 X 2 = 84.

Question 26.
142 × 2 =
Answer:
142 X 2 = 284,

Explanation:
Given expression as 142 X 2, multiplying 142 with 2
we get 284, So 142 X 2 = 284.

Question 27.
242 × 2 =
Answer:
242 X 2 = 484,

Explanation:
Given expression as 242 X 2, multiplying 242 with 2
we get 484, So 242 X 2 = 484.

Question 28.
342 × 2 =
Answer:
342 X 2 = 684,

Explanation:
Given expression as 342 X 2, multiplying 342 with 2
we get 684, So 342 X 2 = 684.

Question 29.
442 × 2 =
Answer:
442 X 2 = 884,

Explanation:
Given expression as 442 X 2, multiplying 442 with 2
we get 884, So 442 X 2 = 884.

Question 30.
3 × 3 =
Answer:
3 X 3 = 9,

Explanation:
Given expression as 3 X 3, multiplying 3 with 3
we get 9, So 3 X 3 = 9.

Question 31.
13 × 3 =
Answer:
13 X 3 = 39,

Explanation:
Given expression as 13 X 3, multiplying 13 with 3
we get 39, So 13 X 3 = 39.

Question 32.
213 × 3 =
Answer:
213 X 3 =639,

Explanation:
Given expression as 213 X 3, multiplying 213 with 3
we get 639, So 213 X 3 = 639.

Question 33.
1,213 × 3 =
Answer:
1,213 X 3 = 3,639,

Explanation:
Given expression as 1,213 X 3, multiplying 1,213 with 3
we get 1,213, So 1,213 X 3 = 3,639.

Question 34.
2,113 × 3 =
Answer:
2,113 X 3 = 6,339,

Explanation:
Given expression as 2,113 X 3, multiplying 2,113 with 3
we get 6,339, So 2,113 X 3 = 6,339.

Question 35.
2,131 × 3 =
Answer:
2,131 X 3 = 6,393,
Explanation:
Given expression as 2,131 X 3, multiplying 2,131 with 3
we get 6,393, So 2,131 X 3 = 6,393.

Question 36.
2,311 × 3 =
Answer:
2,311 X 3 = 6,933,

Explanation:
Given expression as 2,311 X 3, multiplying 2,311 with 3
we get 6,933, So 2,311 X 3 = 6,933.

Question 37.
24 × 4 =
Answer:
24 X 4 = 96,

Explanation:
Given expression as 24 X 4, multiplying 24 with 4
we get 96, So 24 X 4 = 96.

Question 38.
35 × 5 =
Answer:
35 X 5 = 175,

Explanation:
Given expression as 35 X 5, multiplying 35 with 5
we get 175, So 35 X 5 = 175.

Question 39.
54 × 3 =
Answer:
54 X 3 = 162,

Explanation:
Given expression as 54 X 3, multiplying 54 with 3
we get 162, So 54 X 3 = 162.

Question 40.
63 × 6 =
Answer:
63 X 6 = 378,

Explanation:
Given expression as 63 X 6, multiplying 63 with 6
we get 378, So 63 X 6 = 378.

Question 41.
125 × 4 =
Answer:
125 X 4 = 500,

Explanation:
Given expression as 125 X 4, multiplying 125 with 4
we get 500, So 125 X 4 = 500.

Question 42.
214 × 3 =
Answer:
214 X 3 = 642,

Explanation:
Given expression as 214 X 3, multiplying 214 with 3
we get 642, So 214 X 3 = 642.

Question 43.
5,213 × 2 =
Answer:
5,213 X 2 = 10,426,

Explanation:
Given expression as 5,213 X 2, multiplying 5,213 with 2
we get 10,426, So 5,213 X 2 = 10,426.

Question 44.
2,135 × 4 =
Answer:
2,135 X 4 = 8,540,

Explanation:
Given expression as 2,135 X 4, multiplying 2,135 with 4
we get 8,540, So 2,135 X 4 = 8,540.

Mental Multiplication

Question 1.
1 × 6 =
Answer:
1 X 6 = 6,

Explanation:
Given expression as 1 X 6, multiplying 1 with 6
we get 6, So 1 X 6 = 6.

Question 2.
10 × 6 =
Answer:
10 X 6 = 60,

Explanation:
Given expression as 10 X 6, multiplying 10 with 6
we get 60, So 10 X 6 = 60.

Question 3.
11 × 6 =
Answer:
11 X 6 = 66,

Explanation:
Given expression as 11 X 6, multiplying 11 with 6
we get 66, So 11 X 6 = 66.

Question 4.
1 × 2 =
Answer:
1 X 2 = 2,

Explanation:
Given expression as 1 X 2, multiplying 1 with 2
we get 2, So 1 X 2 = 2.

Question 5.
30 × 2 =
Answer:
30 X 2 = 60,

Explanation:
Given expression as 30 X 2, multiplying 30 with 2
we get 60, So 30 X 2 = 60.

Question 6.
31 × 2 =
Answer:
31 X 2 = 62,

Explanation:
Given expression as 31 X 2, multiplying 31 with 2
we get 62, So 31 X 2 = 62.

Question 7.
3 × 3 =
Answer:
3 X 3 = 9,

Explanation:
Given expression as 3 X 3, multiplying 3 with 3
we get 9, So 3 X 3 = 9.

Question 8.
20 × 3 =
Answer:
20 X 3 = 60,

Explanation:
Given expression as 20 X 3, multiplying 20 with 3
we get 60, So 20 X 3 = 60.

Question 9.
23 × 3 =
Answer:
23 X 3 = 69,

Explanation:
Given expression as 23 X 3, multiplying 23 with 3
we get 69, So 23 X 3 = 69.

Question 10.
5 × 5 =
Answer:
5 X 5 = 25,

Explanation:
Given expression as 5 X 5, multiplying 5 with 5
we get 25, So 5 X 5 = 25.

Question 11.
20 × 5 =
Answer:
20 X 5 = 100,

Explanation:
Given expression as 20 X 5, multiplying 20 with 5
we get 100, So 20 X 5 = 100.

Question 12.
25 × 5 =
Answer:
25 X 5 = 125,

Explanation:
Given expression as 25 X 5, multiplying 25 with 5
we get 125, So 25 X 5 = 125.

Question 13.
4 × 4 =
Answer:
4 X 4 = 16,

Explanation:
Given expression as 4 X 4, multiplying 4 with 4
we get 16, So 4 X 4 = 16.

Question 14.
30 × 4 =
Answer:
30 X 4 = 120,

Explanation:
Given expression as 30 X 4, multiplying 30 with 4
we get 120, So 30 X 4 = 120.

Question 15.
34 × 4 =
Answer:
34 X 4 = 136,
Explanation:
Given expression as 34 X 4, multiplying 34 with 4
we get 136, So 34 X 4 = 136.

Question 16.
4 × 2 =
Answer:
4 X 2 = 8,

Explanation:
Given expression as 4 X 2, multiplying 4 with 2
we get 8, So 4 X 2 = 8.

Question 17.
90 × 2 =
Answer:
90 X 2 = 180,

Explanation:
Given expression as 90 X 2, multiplying 90 with 2
we get 180, So 90 X 2 = 180.

Question 18.
94 × 2 =
Answer:
94 X 2 = 188,

Explanation:
Given expression as 94 X 2, multiplying 94 with 2
we get 188, So 94 X 2 = 188.

Question 19.
2 × 3 =
Answer:
2 X 3 = 6,

Explanation:
Given expression as 2 X 3, multiplying 2 with 3
we get 6, So 2 X 3 = 6.

Question 20.
40 × 3 =
Answer:
40 X 3 = 120,

Explanation:
Given expression as 40 X 3, multiplying 40 with 3
we get 120, So 40 X 3 = 120.

Question 21.
42 × 3 =
Answer:
42 X 3 = 126,

Explanation:
Given expression as 42 X 3, multiplying 42 with 3
we get 126, So 42 X 3 = 126.

Question 22.
43 × 3 =
Answer:
43 X 3 = 129,

Explanation:
Given expression as 43 X 3, multiplying 43 with 3
we get 129, So 43 X 3 = 129.

Question 23.
21 × 4 =
Answer:
21 X 4 = 84,

Explanation:
Given expression as 21 X 4, multiplying 21 with 4
we get 84, So 21 X 4 = 84.

Question 24.
121 × 4 =
Answer:
121 X 4 = 484,

Explanation:
Given expression as 121 X 4, multiplying 121 with 4
we get 484, So 121 X 4 = 484.

Question 25.
24 × 2 =
Answer:
24 X 2 = 48,

Explanation:
Given expression as 24 X 2, multiplying 24 with 2
we get 48, So 24 X 2 = 48.

Question 26.
124 × 2 =
Answer:
124 X 2 = 248,

Explanation:
Given expression as 124 X 2, multiplying 124 with 2
we get 248, So 124 X 2 = 248.

Question 27.
224 × 2 =
Answer:
224 X 2 = 448,

Explanation:
Given expression as 224 X 2, multiplying 224 with 2
we get 448, So 224 X 2 = 448.

Question 28.
324 × 2 =
Answer:
324 X 2 = 648,

Explanation:
Given expression as 324 X 2, multiplying 324 with 2
we get 648, So 324 X 2 = 648.

Question 29.
424 × 2 =
Answer:
424 X 2 = 848,

Explanation:
Given expression as 424 X 2, multiplying 424 with 2
we get 848, So 424 X 2 = 848.

Question 30.
3 × 2 =
Answer:
3 X 2 = 6,

Explanation:
Given expression as 3 X 2, multiplying 3 with 2
we get 6, So 3 X 2 = 6.

Question 31.
13 × 2 =
Answer:
13 X 2 =

Explanation:
Given expression as 13 X 2, multiplying 13 with 2
we get 26, So 13 X 2 = 26.

Question 32.
213 × 2 =
Answer:
213 X 2 = 426,

Explanation:
Given expression as 213 X 2, multiplying 213 with 2
we get 426, So 213 X 2 = 426.

Question 33.
1,213 × 2 =
Answer:
1,213 X 2 = 2,426,

Explanation:
Given expression as 1,213 X 2, multiplying 1,213 with 2
we get 2,426, So 1,213 X 2 = 2,426.

Question 34.
2,113 × 2 =
Answer:
2,113 X 2 = 4,226,

Explanation:
Given expression as 2,113 X 2, multiplying 2,113 with 2
we get 4,226, So 2,113 X 2 = 4,226.

Question 35.
2,131 × 2 =
Answer:
2,131 X 2 = 4,262,

Explanation:
Given expression as 2,131 X 2, multiplying 4,262 with 2
we get 4,262, So 2,131 X 2 = 4,262.

Question 36.
2,311 × 2 =
Answer:
2,311 X 2 = 4,262,

Explanation:
Given expression as 2,311 X 2, multiplying 2,311 with 2
we get 4,262, So 2,311 X 2 = 4,262.

Question 37.
23 × 4 =
Answer:
23 X 4 = 92,

Explanation:
Given expression as 23 X 4, multiplying 23 with 4
we get 92, So 23 X 4 = 92.

Question 38.
53 × 5 =
Answer:
53 X 5 = 265,

Explanation:
Given expression as 53 X 5, multiplying 53 with 5
we get 265, So 53 X 5 = 265.

Question 39.
45 × 3 =
Answer:
45 X 3 = 135,

Explanation:
Given expression as 45 X 3, multiplying 45 with 3
we get 135, So 45 X 3 = 135.

Question 40.
36 × 6 =
Answer:
36 X 6 = 216,

Explanation:
Given expression as 36 X 6, multiplying 36 with 6
we get 216, So 36 X 6  = 216.

Question 41.
215 × 3 =
Answer:
215 X 3 = 645,

Explanation:
Given expression as 215 X 3, multiplying 215 with 3
we get 645, So 215 X 3 = 645.

Question 42.
125 × 4 =
Answer:
125 X 4 = 500,

Explanation:
Given expression as 125 X 4, multiplying 125 with 4
we get 500, So 125 X 4 = 500.

Question 43.
5,312 × 2 =
Answer:
5,312 X 2 = 10,624,

Explanation:
Given expression as 5,312 X 2, multiplying 5,312 with 2
we get 10,624, So 5,312 X 2 = 10,624.

Explanation:
Given expression as 5,312 X 2, multiplying 5,312 with 2
we get 10,624, So 5,312 X 2 = 10,624.

Question 44.
1,235 × 4 =
Answer:
1,235 X 4 = 4,940,

Explanation:
Given expression as 1,235 X 4, multiplying 1,234 with 4
we get 4,940 So 1,235 X 4 = 4,940.

Eureka Math Grade 4 Module 3 Lesson 13 Problem Set Answer Key

Solve using the RDW process.

Question 1.
Over the summer, Kate earned $180 each week for 7 weeks.
Of that money, she spent $375 on a new computer and
$137 on new clothes. How much money did she have left?
Answer:
Kate is left with money of $748,

Explanation:
Given over the summer, Kate earned $180 each week for 7 weeks.
So Kate earned $180 X 7 =
   5
$180
X   7
$1,260
Of that money, she spent $375 on a new computer and
$137 on new clothes. So Kate has spent $375 + $137 = $512,
Now money did Kate have left with is $1,260 – $512 = $748,
Solved using the RDW process as shown above.

Question 2.
Sylvia weighed 8 pounds when she was born.
By her first birthday, her weight had tripled.
By her second birthday, she had gained 12 more pounds.
At that time, Sylvia’s father weighed 5 times as much as she did.
What was Sylvia and her dad’s combined weight?
Answer:
Sylvia and her dad’s combined weight is 216 pounds,

Explanation:
Given Sylvia weighed 8 pounds when she was born.
By her first birthday, her weight had tripled, So Sylvia
weighed on her first birthday is 8 pounds X 3 = 24 pounds,
By her second birthday, she had gained 12 more pounds means
24 pounds + 12 pounds = 36 pounds,
At that time, Sylvia’s father weighed 5 times as much as she did,
So 36 pounds X 5 =
3
36
X5
180 pounds
therefore, Sylvia and her dad’s combined weight is
180 ponds + 36 pounds = 216 pounds.
Solved using the RDW process as shown above.

Question 3.
Three boxes weighing 128 pounds each and one
box weighing 254 pounds were loaded onto the back
of an empty truck. A crate of apples was then loaded onto
the same truck. If the total weight loaded onto the truck
was 2,000 pounds, how much did the crate of apples weigh?

Answer:
The crate of apples weigh 1,362 pounds,

Explanation:
Given Three boxes weighing 128 pounds each so weight
of 3 boxes are 128 pounds X 3 =
2
128
X 3
384 pounds and one box weighing 254 pounds were
loaded onto the back of an empty truck. So now weight is
384 pounds + 254 pounds =
384 pounds
+254 pounds
638 pounds,
A crate of apples was then loaded onto the same truck.
If the total weight loaded onto the truck was 2,000 pounds,
therefore crate of apples weigh is
2,000 pounds
– 638 pounds
1,362 pounds.
Solved using the RDW process as shown above.

Question 4.
In one month, Charlie read 814 pages. In the same month,
his mom read 4 times as many pages as Charlie,
and that was 143 pages more than Charlie’s dad read.
What was the total number of pages read by Charlie and his parents?
Answer:
Total number of pages Charlie and his parents read are 7,183,

Explanation:
Given In one month, Charlie read 814 pages. In the same month,
his mom read 4 times as many pages as Charlie,
means Charlie’s mom read 814 X 4 =
1
814
X 4
3,256 pages, Charlie’s dad read 3,256 pages – 143 pages =
3,256
– 143
3,113 pages
Now total number of pages Charlie and his parents read are
814 + 3,256 + 3,113 =
814
3,256
+3,113
7,183 pages,Solved using the RDW process as shown above.

Eureka Math Grade 4 Module 3 Lesson 13 Exit Ticket Answer Key

Solve using the RDW process.

Question 1.
Michael earns $9 per hour. He works 28 hours each week.
How much does he earn in 6 weeks?
Answer:
Michael earns in 6 weeks is $1,512,

Explanation:
Given Michael earns $9 per hour. He works 28 hours each week,
means Michael earns in each week is 28 hours X $9 = $252,
Now in 6 weeks Michael earns $252 X 6 = $1,512.
Solved using the RDW process as shown above.

Question 2.
David earns $8 per hour. He works 40 hours each week.
How much does he earn in 6 weeks?
Answer:
David earns in 6 weeks is $1,920,

Explanation:
Given David earns $8 per hour. He works 40 hours each week,
means David earns in each week is 40 hours X $8 = $320,
Now in 6 weeks David earns $320 X 6 = $1,920.
Solved using the RDW process as shown above.

Question 3.
After 6 weeks, who earned more money?
How much more money?
Answer:
David earned more $408,

Explanation:
We got Michael earns in 6 weeks is $1,512 and
David earns in 6 weeks is $1,920 so more amount is earned
by David by $1,920 – $1,512 = $408 as shown above.

Eureka Math Grade 4 Module 3 Lesson 13 Homework Answer Key

Solve using the RDW process.

Question 1.
A pair of jeans costs $89. A jean jacket costs twice as much.
What is the total cost of a jean jacket and 4 pairs of jeans?
Answer:
The total cost of a Jean jacket and 4 pairs of jeans is $534,

Explanation:
Given a pair of jeans costs $89. A jean jacket costs twice as much,
So jean jcket costs 2 X $89 = $178,
The total cost of a jean jacket and 4 pairs of jeans is
$178 + 4 x $89 = $178 + $356 = $534.

Question 2.
Sarah bought a shirt on sale for $35. The original price
of the shirt was 3 times that amount. Sarah also bought a
pair of shoes on sale for $28. The original price of the
shoes was 5 times that amount. Together, how much money
did the shirt and shoes cost before they went on sale?
Answer:
Together,the shirt and shoes cost before they
went on sale is $245,

Explanation:
Given Sarah bought a shirt on sale for $35. The original price
of the shirt was 3 times that amount means the original price of
the shirt is 3 X $35 =
 1
$35
X 3
$105
Sarah also bought a pair of shoes on sale for $28.
The original price of the shoes was 5 times that amount means
5 X $28 =
4
$28
X 5
$140,
Together,the shirt and shoes cost before they went on sale is
$105 + $140 = $245 as shown above.

Question 3.
All 3,000 seats in a theater are being replaced.
So far, 5 sections of 136 seats and a sixth section
containing 348 seats have been replaced. How many more
seats do they still need to replace?
Answer:
Total more 1,972 seats needs to be still replaced,

Explanation:
Given all 3,000 seats in a theater are being replaced,
So far, 5 sections of 136 seats and a sixth section
containing 348 seats have been replaced means
5 X 136 seats + 348 seats are replaced = 680 + 348 = 1,028 seats
are replaced, So more seats needed to be still replaced are
3,000 seats – 1,028 seats = 1,972 seats as shown above.

Question 4.
Computer Depot sold 762 reams of paper. Paper Palace
sold 3 times as much paper as Computer Depot and
143 reams more than Office Supply Central.
How many reams of paper were sold by all three stores combined?
Answer:
Total 5,191 reams of paper were sold by all three stores combined,

Explanation:
Given Computer Depot sold 762 reams of paper. Paper Palace
sold 3 times as much paper as Computer Depot means
3 X 762 reams =
1
762
X 3
2,286 reams sold by Paper Palace,
Now 2,286 reams – 143 reams = 2,143 reams
by Office Supply Central, therefore all three stores combinedly
sold 762 + 2,286 + 2,143 = 5,191 reams of paper as shown above.

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 × 4

4 (400 + 20 + 5)
(4 × __400___ ) + (4 × _20____ ) + (4 × __5___ )
Answer:
425 X 4 =
Standard Algorithm =
 1,2
425
X 4
1,700
Partial Products =
425
X 4
  20
80
+ 1600
1,700
Area Model =

425 X 4 = 1,700,

Explanation:
Solved the following expression 425 X 4 using
the standard algorithm =
we add same time of multiplying
1,2
425
X 4
1,700
(4 X 5 ones = 20 ones )+ (4 X 2 tens = 8 tens) +
(4 X 4 hundreds = 16 hundreds) = 20 + 80 + 1,600 = 1,700,
partial products =
425
X 4
  20— 4 X 5
80—-4 X 20
+ 1600–4 X 400
1,700
and the area model is as shown in the picture above,
So 425 X 4 = 1,700.

b. 5 3 4 × 7

7 ( __500_ + _30___ + _4___ )
( _7_ × __500___ ) + ( 7__ × __30___ ) + ( 7__ × _4___ )
Answer:
534  X 7 =
Standard Algorithm =
 2,2
534
X 7
3,738
Partial Products =
534
X 7
  28
210
+3500
3,738
Area Model =

534  X 7 = 3,738,

Explanation:
Solved the following expression 534 X 7 using
the standard algorithm =
we add same time of multiplying
2,2
534
X 7
3,738
(7 X 4 ones = 28 ones )+ (7 X 3 tens = 21 tens) +
(7 X 5 hundreds = 35 hundreds) = 28 + 210 + 3,500 = 3,738,
partial products =
534
X 7
  28— 7 X 4
210—-7 X 30
+3500–7 X 500
3,738
and the area model is as shown in the picture above
So 534 X 7 = 3,738.

c. 2 0 9 × 8

8__ ( _200___ + _9___ )
( _8_ × _200____ ) + ( _8_ × __9___ )
Answer:
209  X 8 =
Standard Algorithm =
 7,
209
X 8
1,672
Partial Products =
209
X 8
  72
0000
+1600
1,672
Area Model =

209  X 8 = 1,672,

Explanation:
Solved the following expression 209 X 8 using
the standard algorithm =
we add same time of multiplying
7
209
X 8
1,672
(8 X 9 ones = 72 ones )+ (8 X 0 tens = 0 tens) +
(8 X 2 hundreds = 16 hundreds) = 72 + 0 + 1,600 = 1,672,
partial products =
209
X 8
  72— 8 X 9
000—-8 X 0
+1600–8 X 200
1,672
and the area model is as shown in the picture above
So 209 X 8 = 1,672.

Question 2.
Solve using the partial products method.
Cayla’s school has 258 students. Janet’s school has 3 times as many students as Cayla’s. How many students are in Janet’s school?
Answer:
There are in Janet’s school are 774 students,

Explanation:
Given Cayla’s school has 258 students. Janet’s school has
3 times as many students as Cayla’s.
So there are number of students in Janet’s school are
3 X 258 students =
Partial Products =
258
X 3
024
150
+600
774
So there are in Janet’s school are 774 students.

Question 3.
Model with a tape diagram and solve.
4 times as much as 467
Answer:

4 X 467 = 1,868,

Explanation:
Modeled with a tape diagram as shown above and
4 times as much as 467 is 4 X 467 =
2,2
467
X 4
1,868
So, 4 X 467 = 1,868.

Solve using the standard algorithm, the area model, the distributive property, or the partial products method.

Question 4.
5,131 × 7
Answer:
Standard algorithm:
5,131
X    7
35,917

Explanation:
Given expression as 5,131 X 7 solving using standard algorithm
method as
2
5,131
X    7
35,917
First we multiply (7 X 1 one = 7 ones) + (7 X 3 tens = 21 tens) +
(7 X 1 hundred = 7 hundreds) + (7 X 5 thousands = 35 thousands),
So 7 X 5,131 = 35,917.

Question 5.
3 times as many as 2,805,

Answer:
3 X 2,805 = 8,415,

Explanation:
Given to find 3 times as many as 2,805 using the
distributive property as shown below
3 X 2,000 + 3 X 800 + 3 X 0 + 3 X 5 =
6,000 + 2,400 + 0 + 15 = 8,415,
So 3 times as many as 2,805 = 8,415.

Question 6.
A 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?
Answer:

23,850 pounds of pasta the restaurant saled in 9 months,

Explanation:
Given 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
9 X (1,725 +925) pounds = 9 X 2,650 =
  5,4
2,650
X    9
23,850
Solved using the area model as shown above.

Eureka Math Grade 4 Module 3 Lesson 11 Exit Ticket Answer Key

Question 1.
Solve using the standard algorithm, the area model, the distributive property, or the partial products method.
2,809 × 4
Answer:
2,809
X 4
11,236
2,809 X 4 = 11,236,

Explanation:
Solved using the distributive property as shown below
(4 X 2,000) + (4 X 800) + (4 X 0) + (4 X 9) =
8,000 + 3,200 + 0 + 36 =11,236,
So, 2,809 X 4 = 11,236.

Question 2.
The monthly school newspaper is 9 pages long.
Mrs. Smith needs to print 675 copies.
What will be the total number of pages printed?
Answer:
Total number of pages printed are 6,075,

Explanation:
Given the monthly school newspaper is 9 pages long.
and Mrs. Smith needs to print 675 copies,
So the total number of pages printed are 9 X 6,075,
Solving using Partial Products as shown below
6,075
X  9
45—9  X 5
630–9 X 7
000– 9 X 0
5400–9 X 6
6,075
therefore, total number of pages printed are 6,075.

Eureka Math Grade 4 Module 3 Lesson 11 Homework Answer Key

Question 1.
Solve the following expressions using the standard algorithm, the partial products method, and the area model.
a. 3 0 2 × 8

8 (300 + 2)
(8 × _300____ ) + (8 × __2___ )
Answer:
302 X 8 =
Standard Algorithm =
 1
302
X 8
2,416
Partial Products =
302
X 8
  16
00
+2400
2,416
Area Model =

302 X 8 = 2,416,

Explanation:
Solved the following expression 302 X 8 using
the standard algorithm =
we add same time of multiplying
 1
302
X 8
2,416
(8 X 2 ones = 16 ones )+ (8 X 0 tens = 0 tens) +
(8 X 3 hundreds = 24 hundreds) = 16 + 0 + 2,400 = 2,416,
partial products =
302
X 8
16
00
+2400
2,416
and the area model is as shown in the picture above,
So 302 X 8 = 2,416.

b. 2 1 6 × 5

5 ( _200___ + _10___ + _6___ )
( 5__ × __200___ ) + ( _5_ × __10___ ) + ( 5__ × _6___ )
Answer:
216 X 5 =
Standard Algorithm =
 3
216
X 5
1,080
Partial Products =
216
X 5
  30
50
+1000
1,080
Area Model =

216 X 5 = 1,080,

Explanation:
Solved the following expression 302 X 8 using
the standard algorithm =
we add same time of multiplying
 3
216
X 5
1,080
(5 X 6 ones = 30 ones )+ (5 X 1 tens = 5 tens) +
(5 X 2 hundreds = 10 hundreds) = 30 + 50 + 1,000 = 1,080,
partial products =
216
X 5
  30
50
+1000
1,080
and the area model is as shown in the picture above,
So 216 X 5 = 1,080.

c. 5 9 3 × 9

_9_ ( _500___ + _90___ + _3___ )
( _9_ × __500___ ) + ( _9_ × __90___ ) + ( _9_ × __3__ )
Answer:
593 X 9 =
Standard Algorithm =
 8,2
593
X 9
5,337
Partial Products =
593
X 9
  27
810
+4500
5,337
Area Model =

593 X 9 = 5,337,

Explanation:
Solved the following expression 593 X 9 using
the standard algorithm =
we add same time of multiplying
8,2
593
X 9
5,337
(9 X 3 ones = 27 ones )+ (9 X 9 tens = 81 tens) +
(9 X 5 hundreds = 45 hundreds)= 27 + 810 + 4,500 = 5,337,
partial products =
593
X 9
  27
810
+4500
5,337
and the area model is as shown in the picture above,
So 593 X 9 = 5,337.

Question 2.
Solve 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?
Answer:
Number of people visited the museum on
Saturday are 1,900,
Partial Products =
475
X 4
  20
280
+1600
1,900

Explanation:
Given to solve 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.
475
X 4
20
280
+1600
1,900
Number of people visited the museum on Saturday are 1,900.

Question 3.
Model with a tape diagram and solve.
6 times as much as 384
Answer:

6 X 384 = 2,304,

Explanation:
Modeled with a tape diagram as shown above and
6 times as much as 384 is 4 X 467 =
5,2
384
X 6
2,304
So, 4 X 467 = 2,304.

Solve using the standard algorithm, the area model,
the distributive property, or the partial products method.

Question 4.
6,253 × 3
Answer:
6,253 X 3 = 18,759,

Explanation:
Given expression 6,253 X 3 using the distributive property method
we solve
6000 X 3 + 200 X 3 + 50 X 3 + 3 X 3 =
18,000 + 600 + 150 + 9 = 18,759,
theerfore 6,253 X 3 = 18,759.

Question 5.
7 times as many as 3,073
Answer:
7 times as many as 3,073 is 21,511,

Explanation:
We solve 7 times as many as 3,073 the partial products method as
3,073
X   7
  21—- 7 x 3
490—–7 X 70
0000—- 7 X 0
21000—-7 X 3,000
21,511
So, 7 times as many as 3,073 is 21,511.

Question 6.
A 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?
Answer:
After 6 months 18,744 pounds of rice the cafeteria make,

Explanation:
Given 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,
Now number of pounds of rice does the cafeteria make after
6 months is 6 X 3,124 using the standard algorithm we solve as
1,2
3,124
X    6
18,744
First we multiply 6 X 4 ones = 24 ones, 6 X 2 tens = 12 tens,
6 X 1 hundred = 6 hundreds, 6 X 3 thousands = 18 thousands,
24 + 120 + 600 + 1,8000 = 18,744,
therefore, after 6 months 18,744 pounds of rice the cafeteria make.

Engage NY Eureka Math 4th Grade Module 3 Lesson 12 Answer Key

Eureka Math Grade 4 Module 3 Lesson 12 Problem Set Answer Key

Use the RDW process to solve the following problems.

Question 1.
The table shows the cost of party favors. Each party guest receives a bag with 1 balloon, 1 lollipop, and 1 bracelet. What is the total cost for 9 guests?

Answer:
The total cost for 9 guests is 657¢,

Explanation:
Given each party guest receives a bag with 1 balloon,
1 lollipop, and 1 bracelet, So each guest costs
26¢ + 14¢ + 33 ¢ = 73¢, Now the total cost for 9 guests is
9 X 73¢ = 657¢ as shown in the picture above.

Question 2.
The Turner family uses 548 liters of water per day.
The Hill family uses 3 times as much water per day.
How much water does the Hill family use per week?
Answer:
The Hill family uses 11,508 liters of water per week,

Explanation:
Given the Turner family uses 548 liters of water per day.
The Hill family uses 3 times as much water per day,
So per day Hill family uses 3 X 548 liters = 1,644 liters,
So number of liters of water does the Hill family use per week is
7 X 1,644 liters = 11,508 liters as shown in the picture above.

Question 3.
Jayden has 347 marbles. Elvis has 4 times as many as Jayden.
Presley has 799 fewer than Elvis.
How many marbles does Presley have?
Answer:
Number of marbles Presley have are 589,

Explanation:
Given Jayden has 347 marbles. Elvis has 4 times as many as Jayden.
So Elvis have 4 X 347 marbles = 1,388 marbles,
Now Presley has 799 fewer than Elvis means number
of marbles Presley have are 1,388 – 799 = 589 marbles
as shown in the picture above.

Question 4.
a. Write an equation that would allow someone to find the value of R.

Answer:
R + $239 = $1,167 X 3,
R = ($1,167 X 3) – $239,

Explanation:
Wrote an equation that would allow someone to
find the value of R as R + $239 = $1,167 X 3,
R = ($1,167 X 3) – $239.

b. Write your own word problem to correspond
to the tape diagram, and then solve
Answer:
Joy brought 3 books each of cost $1,167,
all the 3 books cost $239 more than her existing books,
What is the cost of existing books?
Joy’s existing books cost $3,262,

Explanation:
Wrote my own word problem to correspond
to the tape diagram, and then solved as below,
Joy brought 3 books each of cost $1,167,
all the 3 books cost $239 more than her existing books,
What is the cost of existing books?
Solution :
       2,2
$1,167
X     3
$3,501
Now $3,501 – $239 = $3,262,
So Joy’s existing books cost $3,262.

Eureka Math Grade 4 Module 3 Lesson 12 Exit Ticket Answer Key

Use the RDW process to solve the following problem.

Jennifer has 256 beads. Stella has 3 times as many
beads as Jennifer. Tiah has 104 more beads than Stella.
How many beads does Tiah have?
Answer:
Tiah have 872 beads,

Explanation:
Given Jennifer has 256 beads. Stella has 3 times as many
beads as Jennifer. So stella have 3 X 256 beads = 768 beads,
Now Tiah has 104 more beads than Stella,
So number of beads does Tiah have are 768 beads + 104 beads =
872 beads as shown above.

Eureka Math Grade 4 Module 3 Lesson 12 Homework Answer Key

Use the RDW process to solve the following problems.

Question 1.
The table shows the number of stickers of various
types in Chrissy’s new sticker book. Chrissy’s
six friends each own the same sticker book.
How many stickers do Chrissy and her six friends have altogether?

Answer:
644 stickers does Chrissy and her six friends have altogether,

Explanation:
Given the number of stickers of various
types in Chrissy’s new sticker book.
Chrissy’s new sticker book has flowers + smiley faces + hearts =
32 + 21 + 39 = 92 stickers,
given six friends each own the same sticker book.
So number of  stickers does Chrissy and her six friends
have altogether are 92 stickers X 7 =
1
92
x 7
644 stickers as shown above.

Question 2.
The small copier makes 437 copies each day.
The large copier makes 4 times as many copies each day.
How many copies does the large copier make each week?
Answer:
12,236 copies does the the large copier make each week,

Explanation:
Given the small copier makes 437 copies each day.
The large copier makes 4 times as many copies each day.
So each day the large copier makes are
4 X 437 copies =
  1,2
437
X 4
1,748 copies, Now number of copies does the large
copier make each week are 1,748 copies X 7 =
5,3,5
1,748
X   7
12,236 copies as shown above.

Question 3.
Jared sold 194 Boy Scout chocolate bars. Matthew sold
three times as many as Jared. Gary sold 297 fewer than
Matthew. How many bars did Gary sell?
Answer:
Gary sells 285 Boy Scout chocolate bars,

Explanation:
Given Jared sold 194 Boy Scout chocolate bars.
Matthew sold three times as many as Jared.
So Matthew sold 194 X 3 =
   2,1
194
X 3
582, Now Gary sold 297 fewer than Matthew.
So number of bars did Gary sell are 582 – 297 = 285,
therefore Gary sells 285 Boy Scout chocolate bars as shown above.

Question 4.
a. Write an equation that would allow someone to find the value of M.

Answer:
M = (723 meters X 3) +  973 meters,

Explanation:
Wrote an equation that would allow someone to
find the value of M as M = (723 meters X 3) +  973 meters.

b. Write your own word problem to correspond to
the tape diagram, and then solve.
Answer:
Peter has 2 rope pieces of one piece of length 973 meters and
other 3 times 723 meters of length altogether howm much
length of rope in meters do Peter have?
Peter have altogether rope of length 3,142 meters,

Explanation:
Wrote my own word problem to correspond
to the tape diagram, and then solved as below,
Peter has 2 rope pieces of one piece of length 973 meters and
other 3 times 723 meters of length altogether howm much
length of rope in meters do Peter have?
Solution :
One piece of length is 723 meters X 3 =
723
X 3
2,169 meters,
Now adding this to other piece of length 973 meters ,
2,169 + 973 = 3,142 meters,
therefore Peter have altogether rope of length 3,142 meters.

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.

Engage NY Eureka Math 4th Grade Module 3 Lesson 8 Answer Key

Eureka Math Grade 4 Module 3 Lesson 8 Problem Set Answer Key

Question 1.
Represent the following expressions with disks,
regrouping as necessary, writing a matching expression, and
recording the partial products vertically as shown below.
a. 1 × 213

1 × _2__ hundreds + 1 × _1__ ten + 1 × _3__ ones

Answer:

Explanation:
Represented the following expression 1 X 213 with disks,
regrouped as necessary. Wrote a matching expression 1 X 213 = 213
and recorded the partial products vertically as shown above.

b. 2 × 213

Answer:

Explanation:
Represented the following expression 2 X 213 with disks,
regrouped as necessary. Wrote a matching expression 2 X 213 = 426
and recorded the partial products vertically as shown above.

c. 3 × 214

Answer:

Explanation:
Represented the following expression 3 X 214 with disks,
regrouped as necessary. Wrote a matching expression 3 X 214 = 642
and recorded the partial products vertically as shown above.

d. 3 × 1,254

Answer:

Explanation:
Represented the following expression 3 X 1,254 with disks,
regrouped as necessary. Wrote a matching expression 3 X 1,254 = 3,762
and recorded the partial products vertically as shown above.

Question 2.
Represent the following expressions with disks,
using either method shown during class, regrouping as necessary.
To the right, record the partial products vertically.
a. 3 × 212
Answer:

Explanation:
Represented the following expression 3 X 212 with disks,
regrouped as necessary and recorded the partial products
vertically as shown above.

b. 2 × 4,036
Answer:

Explanation:
Represented the following expression 2 X 4,036 = 8,072 with disks,
regrouped as necessary and recorded the partial products
vertically as shown above.

c. 3 × 2,546
Answer:

Explanation:
Represented the following expression 3 X 2,546 = 7,638 with disks,
regrouped as necessary and recorded the partial products
vertically as shown above.

d. 3 × 1,407
Answer:
Explanation:
Represented the following expression 3 X 1,407 = 4,221 with disks,
regrouped as necessary and recorded the partial products
vertically as shown above.

Question 3.
Every day at the bagel factory, Cyndi makes 5 different
kinds of bagels. If she makes 144 of each kind,
what is the total number of bagels that she makes?
Answer:
Total number of bagels Cyndi makes are 720 every day,

Explanation:
Given every day at the bagel factory, Cyndi makes 5 different
kinds of bagels. If she makes 144 of each kind,
The total number of bagels Cyndi makes are
5 X 144 = 720 every day.

Eureka Math Grade 4 Module 3 Lesson 8 Exit Ticket Answer Key

Represent the following expressions with disks, regrouping as necessary. To the right, record the partial products vertically.

Question 1.
4 × 513
Answer:

Explanation:
Represented the following expression 4 X 513 = 2,052 with disks,
regrouped as necessary and recorded the partial products
vertically as shown above.

Question 2.
3 × 1,054
Answer:

Explanation:
Represented the following expression 3 X 1,054 = 3,162 with disks,
regrouped as necessary and recorded the partial products
vertically as shown above.

Eureka Math Grade 4 Module 3 Lesson 8 Homework Answer Key

Question 1.
Represent the following expressions with disks, regrouping as necessary, writing a matching expression, and recording the partial products vertically as shown below.
a. 2 × 424

2 × ___ ________ + 2 × ___ _____ + 2 × ___ ones

Answer:

Explanation:
Represented the following expression 2 X 424 with disks,
regrouped as necessary. Wrote a matching expression 2 X 424 = 848
and recorded the partial products vertically as shown above.

b. 3 × 424

Answer:

Explanation:
Represented the following expression 3 X 424 with disks,
regrouped as necessary. Wrote a matching expression 3 X 424 = 1,272
and recorded the partial products vertically as shown above.

c. 4 × 1,424
Answer:

Explanation:
Represented the following expression 4 X 1,424 with disks,
regrouped as necessary. Wrote a matching expression 4 X 1,424 = 5,696
and recorded the partial products vertically as shown above.

Question 2.
Represent the following expressions with disks, using either
method shown in class, regrouping as necessary.
To the right, record the partial products vertically.
a. 2 × 617
Answer:

Explanation:
Represented the following expression 2 X 617 with disks,
regrouped as necessary. Wrote a matching expression 2 X 617 = 1,234
and recorded the partial products vertically as shown above.

b. 5 × 642
Answer:

Explanation:
Represented the following expression 5 X 642 with disks,
regrouped as necessary. Wrote a matching expression 5 X 642 = 3,210
and recorded the partial products vertically as shown above.

c. 3 × 3,034
Answer:

Explanation:
Represented the following expression 3 X 3,034 with disks,
regrouped as necessary. Wrote a matching expression 3 X 3,034 = 9,102
and recorded the partial products vertically as shown above.

Question 3.
Every day, Penelope jogs three laps around the playground
to keep in shape. The playground is rectangular with a
width of 163 m and a length of 320 m.
a. Find the total amount of meters in one lap.
Answer:
The total amount of meters in one lap is 52,160 square meters,

Explanation:
Given the playground is rectangular with a
width of 163 m and a length of 320 m, So the
total area of the playground is 163 m X 320 m =
52,160 square meters.Therefore the total amount of meters
in one lap is 52,160 square meters penelope jogs.

b. Determine how many meters Penelope jogs in three laps.
Answer:
Penelope jogs 1,56,480 square meters in three laps,

Explanation:
Given every day, Penelope jogs three laps around the
playground to keep in shape. The playground is rectangular
with a width of 163 m and a length of 320 m, so total area
of the rectangular playground is 163 m X 320 m =
52,160 square meters, Now for 3 laps it is 3 X 52,160 sq mt =
1,56,480 square meters.

Engage NY Eureka Math 4th Grade Module 3 Lesson 9 Answer Key

Eureka Math Grade 4 Module 3 Lesson 9 Problem Set Answer Key

Question 1.
Solve using each method.

Answer:

Explanation:
Given expression as 4 X 34 =
Partial Products
34
X 4
  16
+120
136 , Here we first write multiplication of 4 X 4 ones, then
4 X 3 tens then add as shown above 16 + 120 = 136,
In standard algorithm we add same time of multiplying as
1
34
X 4
136 ,here 4 X 4 ones= 16 ones we write 6 at ones place and take 1
at tens place then 4 x 3 at tens place = 12 tens
add 1 ten to 12 tens we get 13 tens so 4 X 34 = 136.

Answer:

Explanation:
Given expression as 3 X 224 =
Partial Products
224
X 3
  12
60
+600
672 , Here we first write multiplication of 3 X 4 ones, then
3 X 2 tens and 3 X 2 hundreds then add as shown above
12 + 60 + 600 = 672,
In standard algorithm we add same time of multiplying as
1
224
X 3
672 ,here 3 X 4 ones = 12 ones we write 2 at ones place and
take 1 at tens place then 3 x 2 tens = 6 tens add 1 ten to 6 tens
we get 7 tens and 3 X 2 hundreds = 6 hundreds,  so 4 X 244 = 672.

Question 2.
Solve. Use the standard algorithm.
a.

Answer:

   7     5     3      

Explanation:
In standard algorithm we add same time of multiplying as
1
251
X 3
753 ,here 3 X 1 ones = 3 ones then 3 x 5 tens = 15 tens
we write 5 at tens place and take 1 to hundred place and
3 X 2 hundreds = 6 hundreds, now adding 1 hundred to
6 hundreds we get 7 hundreds , so 3 X 251 = 753.

b.

Answer:

  8      1      0

Explanation:
In standard algorithm we add same time of multiplying as
1, 3
135
X 6
810 ,here 6 X 5 ones = 30 ones, we write 0 at ones place and
take 3 to tens places then 6 x 3 tens = 18 tens + 3 tens = 21 tens,
we write 1 at tens place and take 2 to hundred place and
6 X 1 hundred = 6 hundreds, now adding 2 hundreds to
6 hundreds we get 8 hundreds , so 6 X 135 = 810.

c.

Answer:

2,   7   3   6

Explanation:
In standard algorithm we add same time of multiplying as
 3
304
X 9
2,736 ,here 9 X 4 ones = 36 ones, we write 6 at ones place and
take 3 to tens places then 9 x 0 ten = 0 ten + 3 tens = 3 tens,
9 X 3 hundreds = 27 hundreds, now we write 7 at hundreds place
and 2 at thousands place as shown above, So 9 X 304 = 2,736.    

d.

Answer:

1,  6    2   0

Explanation:
In standard algorithm we add same time of multiplying as
2
405
X 4
1,620 ,here 4 X 5 ones = 20 ones, we write 0 at ones place and
take 2 to tens places then 4 x 0 ten = 0 ten + 2 tens = 2 tens,
4 X 4 hundreds = 16 hundreds, now we write 6 at hundreds place
and 1 at thousand place as shown above, So 4 X 405 = 1,620.

e.

Answer:

1,  5    8   0

Explanation:
In standard algorithm we add same time of multiplying as
3
316
X 5
1,580 ,here 5 X 6 ones = 30 ones, we write 0 at ones place and
take 3 to tens places then 5 x 1 ten = 5 tens + 3 tens = 8 tens,
5 X 3 hundreds = 15 hundreds, now we write 5 at hundreds place
and 1 at thousand place as shown above, So 5 X 316 = 1,580.

f.

Answer:

2,  3   5    2

Explanation:
In standard algorithm we add same time of multiplying as
5,1
392
X 6
2,352 ,here 6 X 2 ones = 12 ones, we write 2 at ones place and
take 1 to tens places then 6 x 9 tens = 54 tens + 1 tens = 55 tens,
we write 5 at tens place and take another 5 to hundreds place,
6 X 3 hundreds = 18 hundreds,18 hundreds + 5 hundreds =
23 hundreds now we write 3 at hundreds place
and 2 at thousands place as shown above, So 6 X 392 = 2,352.

Question 3.
The product of 7 and 86 is ________.
Answer:
The product of 7 and 86 is 602,

Explanation:
Give to find the product of 7 and 86 is
4
86
X 7
602
first we multiply 7 with 6 ones = 42 ones we write 2 at ones
place and take 4 to tens place now 7 X 8 = 56 tens adding
4 tens we get 60 tens, So 7 X 86 = 602.

Question 4.
9 times as many as 457 is _________.
Answer:
9 times as many as 457 is 4,113,

Explanation:
Given to find 9 times as many as 457 is
6
457
X 9
4,113 ,here 9 X 7 ones = 63 ones, we write 3 at ones place and
take 6 to tens places then 9 x 5 tens = 45 tens + 6 tens = 51 tens,
we write 1 at tens place and take 5 to hundreds place,
9 X 4 hundreds = 36 hundreds, 36 hundreds + 5 hundreds =
41 hundreds now we write 1 at hundred place
and 4 at thousands place as shown above, So 9 X 457 = 4,113.

Question 5.
Jashawn wants to make 5 airplane propellers. He needs
18 centimeters of wood for each propeller. How many
centimeters of wood will he use?

Answer:
Jashawn needs 90 centimeters of wood to make
5 airplane propellers,

Explanation:
Given Jashawn wants to make 5 airplane propellers.
He needs 18 centimeters of wood for each propeller.
So number of centimeters of wood will he use is 5 X 18 cms =
4
18
X 5
90
here first we multiply 5 X 8 ones = 40 ones, we write 0
at ones place and take 4 to tens place then 5 X 1 ten =
5 tens+ 4 tens = 9 tens, So 5 X 18 = 90 centimeters,
therefore, Jashawn needs 90 centimeters of wood to make
5 airplane propellers.

Question 6.
One game system costs $238. How much will 4 game
systems cost?
Answer:
4 game systems will cost $952,

Explanation:
Given one game system costs $238. So for 4 game
systems it will cost 4 X $238 =
 1, 3
$238
X  4
952
here first we multiply 4 X 8 ones = 32 ones,
we write 2 at ones place and take 3 to tens places
then 4 x 3 tens = 12 tens + 3 tens = 15 tens,
we write 5 at tens place and take 1 to hundred place,
4 X 2 hundreds = 8 hundreds, now 8 hundreds + 1 hundred =
9 hundreds, So 5 X $238 = 952, therefore 4 game systems will cost $952.

Question 7.
A small bag of chips weighs 48 grams. A large bag of
chips weighs three times as much as the small bag.
How much will 7 large bags of chips weigh?

Answer:
7 large bags of chips weigh 1,008 grams,

Explanation:
Given a small bag of chips weighs 48 grams.
A large bag of chips weighs three times as much as
the small bag means 3 X 48 grams=
2
48
X 3
144
here first we multiply 3 X 8 ones = 24 ones, we write 4
at ones place and take 2 to tens place then 3 X 4 ten =
12 tens+ 2 tens = 14 tens, we write 14 tens as 4 at tens place
and 1 at hundreds place, So 3 X 48 grams = 144 grams,
Now 7 large bags of chips weigh 7 X 144 grams =
 3,2
144
X 7
1,008
here first we multiply 7 X 4 ones = 28 ones, we write 8
at ones place and take 2 to tens place then 7 X 4 ten =
28 tens+ 2 tens = 30 tens, we write 30 tens as 0 at tens place
and other 3 at hundreds place,now 7 X 1 hundred = 7 hundred,
7 hundred + 3 hundreds = 10 hundreds we write 0 at
hundreds place at 1 at thousands place, So 7 X 144 = 1,008 grams,
therefore, 7 large bags of chips weigh 1,008 grams.

Eureka Math Grade 4 Module 3 Lesson 9 Exit Ticket Answer Key

Question 1.
Solve using the standard algorithm.
a.

Answer:

5,  4     7   2
Explanation:
In standard algorithm we add same time of multiplying as
7
608
X 9
5,472 ,here 9 X 8 ones = 72 ones, we write 2 at ones place and
take 7 to tens places then 9 x 0 tens = 0 tens + 7 tens = 7 tens,
we write 7 at tens place and 9 X 6 hundreds = 54 hundreds,
now we write 4 at hundreds place and 5 at thousands place as shown above, So 9 X 608 = 5,472.

b.

Answer:

4, 0 1 8
Explanation:
In standard algorithm we add same time of multiplying as
2
574
X 7
4,018 ,here 7 X 4 ones = 28 ones, we write 8 at ones place and
take 2 to tens places then 7 x 7 tens = 49 tens + 2 tens = 51 tens,
we write 1 at tens place and 5 at hundreds place and
7 X 5 hundreds = 35 hundreds, 35 hundreds + 5 hundreds =
40 hundreds,now we write 0 at hundreds place and 4
at thousands place as shown above, So 7 X 574 = 4,018.

Question 2.
Morgan is 23 years old. Her grandfather is 4 times as old.
How old is her grandfather?
Answer:
Morgan’s grandfather is 92 years old,

Explanation:
Given Morgan is 23 years old and her grandfather is 4 times
as old. So Morgan’s grandfather’s is 4 X 23 years =
1
23
X 4
92, here 4 X 3 ones = 12 ones, we write 2 at ones place and
take 1 to tens places then 4 x 2 tens = 8 tens, 8 tens + 1 ten =
9 tens, So Morgan’s grandfather is 92 years old.

Eureka Math Grade 4 Module 3 Lesson 9 Homework Answer Key

Question 1.
Solve using each method.

Answer:

Explanation:
Given expression as 2 X 46 =
Partial Products
46
X 2
  12
+80
92 , Here we first write multiplication of 2 X 6 ones, then
2 X 4 tens then add as shown above 12 + 80 = 92,
In standard algorithm we add same time of multiplying as
1
46
X 2
92 ,here 2 X 6 ones= 12 ones we write 2 at ones place and take 1
at tens place then 2 x 4 at tens place = 8 tens
add 1 ten to 8 tens we get 9 tens so 2 X 46 = 92.

Answer:

Explanation:
Given expression as 4 X 315 =
Partial Products
315
X 4
 20
40
+1200
1,260 , Here we first write multiplication of 4 X 5 ones, then
4 X 1 ten and 4 X 3 hundreds then add as shown above
20 + 40 + 1,200 = 1,260,
In standard algorithm we add same time of multiplying as
2
315
X 4
1,260 ,here 4 X 5 ones = 20 ones we write 0 at ones place and
take 2 at tens place then 4 x 1 tens = 4 tens add 2 tens to 4 tens
we get 6 tens and 4 X 3 hundreds = 12 hundreds, we write 2 at
hundreds and 1 at thousands places,  so 4 X 315 = 1,260.

Question 2.
Solve using the standard algorithm.
a.

Answer:

  9         2       8  

Explanation:
In standard algorithm we add same time of multiplying as
1
232
X 4
928 ,here 4 X 2 ones = 8 ones then 4 x 3 tens = 12 tens
we write 2 at tens place and take 1 to hundred place and
4 X 2 hundreds = 8 hundreds, now adding 1 hundred to
8 hundreds we get 9 hundreds , so 4 X 232 = 928.

b.

Answer:

     8      5       2
Explanation:
In standard algorithm we add same time of multiplying as
2, 1
142
X 6
852 ,here 6 X 2 ones = 12 ones, we write 2 at ones place and
take 1 to tens places then 6 x 4 tens = 24 tens + 1 tens = 25 tens,
we write 5 at tens place and take 2 to hundred place and
6 X 1 hundred = 6 hundreds, now adding 2 hundreds to
6 hundreds we get 8 hundreds , so 6 X 142 = 852.

c.

Answer:

 2,      1      9    8
Explanation:
In standard algorithm we add same time of multiplying as
 2
314
X 7
2,198 ,here 7 X 4 ones = 28 ones, we write 8 at ones place and
take 2 to tens places then 7 x 1 ten = 7 tens + 2 tens = 9 tens,
7 X 3 hundreds = 21 hundreds, now we write 1 at hundreds place
and 2 at thousands place as shown above, So 7 X 314 = 2,198.    

d.

Answer:

1,   3     2     0

Explanation:
In standard algorithm we add same time of multiplying as
1
440
X 3
1,320 ,here 3 X 0 ones = 0 ones, we write 0 at ones place,
3 x 4 tens = 12 tens we write 2 at tens place and take 1
at hundreds place, 3 X 4 hundreds = 12 hundreds,
12 hundreds + 1 hundred = 13 hundreds ,now we write 3
at hundreds place and 1 at thousand place as shown above,
So 3 X 440 = 1,320.

e.

Answer:

4,    0     5     6
Explanation:
In standard algorithm we add same time of multiplying as
5
507
X 8
4,056 ,here 8 X 7 ones = 56 ones, we write 6 at ones place and
take 5 to tens places then 8 x 0 tens = 0 tens + 5 tens = 5 tens,
we write 5 at tens place, 8 X 5 hundreds = 40 hundreds,now we write
0 at hundreds place and 4 at thousands place as shown above,
So 8 X 507 = 4,056.

f.

Answer:

3,    4      5      6
Explanation:
In standard algorithm we add same time of multiplying as
7, 3
384
X 9
3,456 ,here 9 X 4 ones = 36 ones, we write 6 at ones place and
take 3 to tens places then 9 x 8 tens = 72 tens + 3 tens = 75 tens,
we write 5 at tens places 7 at hundreds place and
9 X 3 hundreds = 27 hundreds, 27 hundreds + 7 hundreds =
34 hundreds now we write 4 at hundreds place and
3 at thousands place as shown above,
So 9 X 384 = 3,456.

Question 3.
What is the product of 8 and 54?
Answer:
The product of 8 and 54 is 432,

Explanation:
Give to find the product of 8 and 54 is
3
54
X 8
432
first we multiply 8 with 4 ones = 32 ones we write 2 at ones
place and take 3 to tens place now 8 X 5 = 40 tens adding
3 tens we get 43 tens, So 8 X 54 = 432.

Question 4.
Isabel earned 350 points while she was playing Blasting Robot.
Isabel’s mom earned 3 times as many points as Isabel.
How many points did Isabel’s mom earn?
Answer:
Isabel’s mom earn 1050 points,

Explanation:
Given Isabel earned 350 points while she was
playing Blasting Robot. Isabel’s mom earned 3 times
as many points as Isabel. So  points did Isabel’s mom earn
is 3 X 350 =
1
350
X 3
1050,
first we multiply 3 with 0 ones = 0 ones ,
3 X 5 = 15 tens we write 15 tens as 5 at tens place and
1 at hundreds place and 3 X 3 = 9 hundreds + 1 hundred =
10 hundreds we write 10 hundreds as 0 at hundreds place and
1 at thousands place, So 3 X 350 = 1050.
Therefore, Isabel’s mom earn 1050 points.

Question 5.
To get enough money to go on a field trip, every student
in a club has to raise $53 by selling chocolate bars.
There are 9 students in the club. How much money does
the club need to raise to go on the field trip?
Answer:
The club needs to raise $477 to go on the field trip,

Explanation:
Given to get enough money to go on a field trip, every student
in a club has to raise $53 by selling chocolate bars.
There are 9 students in the club. So money does
the club need to raise to go on the field trip is 9 X $53 =
2
$53
X 9
$477
first we multiply 9 with 3 ones = 27 ones , we write 7 at
ones place and 2 at tens place, 9 X 5 tens= 45 tens,
45 tens + 2 tens = 47 tens we write 47 tens as
7 at tens place and 4 at hundreds place so 9 X $53 = $477,
therefore,the club needs to raise $477 to go on the field trip.

Question 6.
Mr. Meyers wants to order 4 tablets for his classroom.
Each tablet costs $329. How much will all four tablets cost?
Answer:
Total cost for all four tablets are $1,316,

Explanation:
Given Mr. Meyers wants to order 4 tablets for his classroom.
Each tablet costs $329. So total cost for four tablets are
4 X $329 =
1,3
$329
X 4
$ 1,316
first we multiply 4 with 9 ones = 36 ones , we write 6 at
ones place and 3 at tens place, 4 X 2 tens= 8 tens,
8 tens + 3 tens = 11 tens we write 11 tens as
1 at tens place and 1 at hundreds place, Now 4 X 3 hundreds =
12 hundreds, 12 hundreds + 1 hundred = 13 hundreds, we write
3 at hundreds place and 1 at thousands place, So 4 X $329 = $1,316.

Question 7.
Amaya read 64 pages last week. Amaya’s older brother,
Rogelio, read twice as many pages in the same amount of time.
Their big sister, Elianna, is in high school and read 4 times as
many pages as Rogelio did. How many pages did
Elianna read last week?
Answer:
Elianna read 512 pages last week,

Explanation:
Given Amaya read 64 pages last week. Amaya’s older brother,
Rogelio, read twice as many pages in the same amount of time.
So Rogelio reads 2 X 64 pages =
64
X 2
128
2 X 4 ones = 8  and 2 X 6 tens= 120, So 2 X 64 = 8 + 120 = 128,
So Rogelio reads 128 pages,
Their big sister, Elianna, is in high school and read 4 times as
many pages as Rogelio did. So number of pages did
Elianna read last week are 4 X 128 pages=
1, 3
128
x 4
512
here we first we multiply 4 with 8 ones = 32 ones ,
we write 2 at ones place and 3 at tens place, 4 X 2 tens= 8 tens,
8 tens + 3 tens = 11 tens we write 11 tens as
1 at tens place and 1 at hundreds place and 4 X 1 hundred =
4 hundreds, 4 hundreds +1 hundred = 5 hundreds, So
4 X 128 pages = 512 pages,
therefore, Elianna read 512 pages last week.

Engage NY Eureka Math 4th Grade Module 3 Lesson 7 Answer Key

Eureka Math Grade 4 Module 3 Lesson 7 Sprint Answer Key

Multiply Multiples of 10, 100, and 1,000

Question 1.
3 × 2 =
Answer:
3 X 2 = 6,

Explanation:
Given 3 X 2 = multiplying 3 with 2 we get 6,
3 X 2 = 6.

Question 2.
30 × 2 =
Answer:
30 X 2 =
3 X 10 X 2 = 60,

Explanation:
Given 30 X 2 = writing 30 as 3 X 10 and multiplying with 2
as  3 X 10 X 2 we get 60, So 30 X 2 = 60.

Question 3.
300 × 2 =
Answer:
300 X 2 =
3 X 100 X 2 = 600,

Explanation:
Given 300 X 2 = writing 300 as 3 X 100 and multiplying 2 we get 600,
3 X 100 X 2 = 600, so 300 X 2 = 600.

Question 4.
3,000 × 2 =
Answer:
3,000 X 2 =
3 X 1,000 X 2 = 6,000,

Explanation:
Given 3,000 X 2 = writing 3,000 as 3 X 1,000 and multiplying with 2
we get 6,000 as 3 X 1,000 X 2 = 6,000, So 3,000 X 2 = 6,000.

Question 5.
2 × 3,000 =
Answer:
2 X 3,000 =
2 X 3 X 1,000 = 6,000,

Explanation:
Given 2 X 3,000 = writing 3,000 as 3 X 1,000 and multiplying with
2 we get 6,000,2 X 3 X 1,000 = 6,000, So 2 X 3,000 = 6,000.

Question 6.
2 × 4 =
Answer:
2 X 4 = 8,

Explanation:
Given 2 X 4 = multiplying 2 with 4 we get 8,
2 X 4 = 8.

Question 7.
2 × 40 =
Answer:
2 X 40 =

2 X 4 X 10 = 80,

Explanation:
Given 2 X 40 = writing 40 as 4 X 10 and multiplying with 2 we get 80,
as 2 X 4 x 10 = 80, So 2 X 40 = 80.

Question 8.
2 × 400 =
Answer:
2 X 400 =
2 X 4 X 100 = 800,

Explanation:
Given 2 X 400 = writing 400 as 4 X 100 and multiplying with 2
we get 800 as 2 X 4 X 100 = 800, So 2 X 400 = 800.

Question 9.
2 × 4,000 =
Answer:
2 X 4,000 =
2 X 4 X 1,000 = 8,000,

Explanation:
Given 2 X 4,000 = writing 4,000 as 4 X 1,000 and multiplying with 2,
we get 8,000, as 2 X 4 X 1,000 = 8,000, So 2 X 4,000 = 8,000.

Question 10.
3 × 3 =
Answer:
3 X 3 = 9,

Explanation:
Given 3 X 3 = multiplying 3 with 3 we get 9,
3 X 3 = 9.

Question 11.
30 × 3 =
Answer:
30 X 3 =
3 X 10 X 3 = 90,

Explanation:
Given 30 X 3 = writing 30 as 3 X 10 and multiplying with 3 we get 90
as 3 X 10 X 3 = 90, So 30 X 3 = 90.

Question 12.
300 × 3 =
Answer:
300 X 3 =
3 X 100 X 3 = 900,

Explanation:
Given 300 X 3 = writing 300 as 3 X 100 and multiplying with 3
we get 900 as 3 X 100 X 3 = 900, So 300 X 3 = 900.

Question 13.
3,000 × 3 =
Answer:
3,000 X 3 =
3 X 1,000 X 3 = 9,000,

Explanation:
Given 3,000 X 3 = writing 3,000 as 3 X 1,000 and multiplying with 3 we get 9,000 as 3 X 1,000 X 3 = 9,000, So 3,000 X 3 = 9,000.

Question 14.
4,000 × 3 =
Answer:
4,000 X 3 =
4 X 1,000 X 3 = 12,000,

Explanation:
Given 4,000 X 3 = writing 4,000 as 4 X 1,000 and multiplying
with 3 we get 12,000 so 4 X 1,000 X 3 = 12,000, So 4,000 X 3 = 12,000.

Question 15.
400 × 3 =
Answer:
400 X 3 =
4 X 100 X 3 = 1,200,

Explanation:
Given 400 X 3 = writing 400 as 4 X 100 and multiplying
with 3 we get 1,200 as 4 X 100 X 3 = 1,200, So 400 X 3 = 1,200.

Question 16.
40 × 3 =
Answer:
40 X 3 =
4 X 10 X 3 = 120,

Explanation:
Given 40 X 3 = writing 40 as 4 X 10 multiplying with 3 we get 120
as 4 X 10 X 3 = 120, So 40 X 3 = 120.

Question 17.
5 × 3 =
Answer:
5 X 3 = 15,

Explanation:
Given 5 X 3 = multiplying 5 with 3 we get 15,
5 X 3 = 15.

Question 18.
500 × 3 =
Answer:
500 X 3 =
5 X 100 X 3 = 1,500,

Explanation:
Given 500 X 3 = writing 500 as 5 X 100 and multiplying with 3 we get 1,500 as 5 x 100 X 3 = 1,500, So 500 X 3 = 1,500.

Question 19.
7 × 2 =
Answer:
7 X 2 = 14,

Explanation:
Given 7 X 2 = multiplying 7 with 2 we get 14,
7 X 2 = 14.

Question 20.
70 × 2 =
Answer:
70 X 2 =
7 X 10 X 2 = 140,

Explanation:
Given 70 X 2 = writing 70 as 7 X 10 and multiplying with 2
we get 140 as 7 X 10 X 2 = 140, So 70 X 2 = 140.

Question 21.
4 × 4 =
Answer:
4 X 4 = 16,

Explanation:
Given 4 X 4 = multiplying 4 with 4 we get 16,
4 X 4 = 16.

Question 22.
4,000 × 4 =
Answer:
4,000 X 4 =
4 X 1,000 X 4 = 1,600,

Explanation:
Given 4,000 X 4 = writing 4,000 as 4 X 1,000 and multiplying
with 4 we get 1,600 as 4 X 1,000 X 4 = 1,600, So 4,000 X 4 = 1,600.

Question 23.
7 × 5 =
Answer:
7 X 5 = 35,

Explanation:
Given 7 X 5 = multiplying 7 with 5 we get 35,
7 X 5 = 35.

Question 24.
700 × 5 =
Answer:
700 X 5 =
7 x 100 X 5 = 3,500,

Explanation:
Given 700 X 5 = writing 700 as 7 X 100 and multiplying
with 5 we get 3,500 as 7 X 100 X 5 = 700, So 700 X 5 = 3,500.

Question 25.
8 × 3 =
Answer:
8 X 3 = 24,

Explanation:
Given 8 X 3 = multiplying 8 with 3 we get 24,
8 X 3 = 24.

Question 26.
80 × 3 =
Answer:
80 X 3 =
8 X 10 X 3 = 240,

Explanation:
Given 80 X 3 = writing 80 as 8 X 10 and multiplying
with 3 we get 240 as 8 X 10 X 3 = 240, So 80 X 3 = 240.

Question 27.
9 × 4 =
Answer:
9 X 4 = 36,

Explanation:
Given 9 X 4 = multiplying 9 with 4 we get 36,
9 X 4 = 36.

Question 28.
9,000 × 4 =
Answer:
9,000 x 4 =
9 X 1,000 X 4 = 36,000,

Explanation:
Given 9,000 X 4 = writing 9,000 as 9 X 1,000 and multiplying
with 4 we get 36,000 as 9 X 1,000 X 4 = 36,000,
So 9,000 X 4 = 36,000.

Question 29.
7 × 6 =
Answer:
7 X 6 = 42,

Explanation:
Given 7 X 6 = multiplying 7 with 6 we get 42,
7 X 6 = 42.

Question 30.
7 × 600 =
Answer:
7 X 600 =
7 X 6 X 100 = 4,200,

Explanation:
Given 7 X 600 = writing 600 as 6 X 100 and multiplying
with 7 we get 4,200 as 7 X 6 X 100 = 4,200 or 7 X 600 = 4,200.

Question 31.
8 × 9 =
Answer:
8 X 9 = 72,

Explanation:
Given 8 X 9 = multiplying 8 with 9 we get 72,
8 X 9 = 72.

Question 32.
8 × 90 =
Answer:
8 X 90 =
8 X 9 X 10 = 720,

Explanation:
Given 8 X 90 = writing 90 as 9 X 10 and multiplying
with 8 we get 720 as 8 x 9 X 10 = 720, So 8 X 90 = 720.

Question 33.
6 × 9 =
Answer:
6 X 9 = 54,

Explanation:
Given 6 X 9 = multiplying 6 with 9 we get 54,
6 X 9 = 54.

Question 34.
6 × 9,000 =
Answer:
6 X 9,000 =
6 X 9 X 1,000 = 54,000,

Explanation:
Given 6 X 9,000 = writing 9,000 as 9 X 1,000 and multiplying
with 6 we get 54,000 as 6 x 9 X 1,000 = 54,000, So 6 X 9,000 = 54,000.

Question 35.
900 × 9 =
Answer:
900 X 9 =
9 X 100 X 9 = 8,100,

Explanation:
Given 900 X 9 = writing 900 as 9 X 100 and multiplying
with 9 we get 8,100 as 9 X 100 X 9 = 8,100, So 900 X 9 = 8,100.

Question 36.
8,000 × 8 =
Answer:
8,000 X 8 =
8 X 1,000 X 8 = 64,000,

Explanation:
Given 8,000 X 8 = writing 8,000 as 8 X 1,000 and multiplying
with 8 we get 64,000 as 8 X 1,000 X 8 = 64,000, So 8,000 X 8 = 64,000.

Question 37.
7 × 70 =
Answer:
7 X 70 =
7 X 7 X 10 = 490,

Explanation:
Given 7 X 70 = writing 70 as 7 X 10 and multiplying
with 7 we get 490 as 7 X 7 X 10 = 490, So 7 X 70 = 490.

Question 38.
6 × 600 =
Answer:
6 X 600 =
6 X 6 X 100 = 3,600,

Explanation:
Given 6 X 600 = writing 600 as 6 X 100 and multiplying
with 6 we get 3,600 as 6 x 6 X 100 = 3,600, So 6 X 600 = 3,600.

Question 39.
800 × 7 =
Answer:
800 X 7 =
8 X 100 X 7 = 5,600,

Explanation:
Given 800 X 7 = writing 800 as 8 x 100 and multiplying
with 7 we get 5,600 as 8 X 100 X 7 = 5,600, So 800 X 7 = 5,600.

Question 40.
7 × 9,000 =
Answer:
7 X 9,000 =
7 X 9 X 1,000 = 63,000,

Explanation:
Given 7 X 9,000 = writing 9,000 as 9 X 1,000 and multiplying
with 7 we get 63,000 as 7 X 9 X 1,000 = 63,000, So 7 X 9,000 = 63,000.

Question 41.
200 × 5 =
Answer:
200 X 5 =
2 X 100 X 5 = 1,000,

Explanation:
Given 200 X 5 = writing 200 as 2 X 100 and multiplying
with 5 we get 1,000 as 2 X 100 X 5 = 1,000, So 200 X 5 = 1,000.

Question 42.
5 × 60 =
Answer:
5 X 60 =
5 X 6 X 10 =300,

Explanation:
Given 5 X 60 = writing 60 as 6 X 10 and multiplying
with 5 we get 300 as 5 X 6 X 10 = 300, So 5 X 60 = 300.

Question 43.
4,000 × 5 =
Answer:
4,000 X 5 =
4 X 1,000 X 5 = 20,000,

Explanation:
Given 4,000 X 5 = writing 4,000 as 4 X 1,000 and multiplying
with 5 we get 20,000 as 4 x 1,000 X 5 = 20,000, So 4,000 X 5 = 20,000.

Question 44.
800 × 5 =
Answer:
800 X 5 = 4,000,
8 X 100 X 5 = 4,000,

Explanation:
Given 800 X 5 = writing 800 as 8 X 100 and multiplying
with 5 we get 4,000 as 8 X 100 X 5 = 4,000, So 800 X 5 = 4,000.

Multiply Multiples of 10, 100, and 1,000

Question 1.
4 × 2 =
Answer:
4 X 2 = 8,

Explanation:
Given 4 X 2 = multiplying 4 with 2 we get 8,
4 X 2 = 8.

Question 2.
40 × 2 =
Answer:
40 X 2 =
4 X 10 X 2 = 80,

Explanation:
Given 40 X 2 = writing 40 as 4 X 10 and multiplying
with 2 we get 80 as 4 X 10 X 2 = 80, So 40 X 2 = 80.

Question 3.
400 × 2 =
Answer:
400 X 2 =
4 X 100 X 2 = 800,

Explanation:
Given 400 X 2 = writing 400 as 4 X 100 and multiplying
with 2 we get 800 as 4 X 100 X 2 = 800, So 400 X 2 = 800.

Question 4.
4,000 × 2 =
Answer:
4,000 X 2 =
4 X 1,000 X 2 = 8,000,

Explanation:
Given 4,000 X 2 = writing 4,000 as 4 X 1,000 and multiplying with 2,
we get 8,000, as 4 X 1,000 X 2 = 8,000, So 4,000 X 2 = 8,000.

Question 5.
2 × 4,000 =
Answer:
2 X 4,000 =
2 X 4 X 1,000 = 8,000,

Explanation:
Given 2 X 4,000 = writing 4,000 as 4 X 1,000 and multiplying with 2,
we get 8,000, as 2 X 4 X 1,000 = 8,000, So 2 X 4,000 = 8,000.

Question 6.
3 × 3 =
Answer:
3 X 3 = 9,

Explanation:
Given 3 X 3 = multiplying 3 with 3 we get 9,
3 X 3 = 9.

Question 7.
3 × 30 =
Answer:
3 X 30 =
3 X 3 X 10 = 90,

Explanation:
Given 3 X 30 = writing 30 as 3 X 10 and multiplying with 3 we get 90
as 3 X 3 X 10 = 90, So 3 X 30 = 90.

Question 8.
3 × 300 =
Answer:
3 X 300 =
3 X 3 X 100 = 900,

Explanation:
Given 3 X 300 = writing 300 as 3 X 100 and multiplying with 3
we get 900 as 3 X 3 X 100 = 900, So 3 X 300 = 900.

Question 9.
3 × 3,000 =
Answer:
3 X 3,000 =
3 X 3 X 1,000 = 9,000,

Explanation:
Given 3 X 3,000 = writing 3,000 as 3 X 1,000 and multiplying with 3 we get 9,000 as 3 X 3 X 1,000 = 9,000, So 3 X 3,000 = 9,000.

Question 10.
2 × 3 =
Answer:
2 X 3 = 6,

Explanation:
Given 2 X 3 = multiplying 2 with 3 we get 6,
2 X 3 = 6.

Question 11.
20 × 3 =
Answer:
20 X 3 =
2 x 10 X 3 = 60,

Explanation:
Given 20 X 3 = writing 20 as 2 X 10 and multiplying with 3 we get 60
as 2 X 10 X 3 = 60, So 20 X 3 = 60.

Question 12.
200 × 3 =
Answer:
200 X 3 =
2 X 100 X 3 = 600,

Explanation:
Given 200 X 3 = writing 200 as 2 X 100 and multiplying
with 3 we get 600 as 2 X 100 X 3 = 600, So 200 X 3 = 600.

Question 13.
2,000 × 3 =
Answer:
2,000 X 3 =
2 X 1,000 X 3 = 6,000,

Explanation:
Given 2,000 X 3 = writing 2,000 as 2 X 1,000 and
multiplying with 3 we get 6,000 as 2 X 1,000 X 3 = 6,000,
So 2,000 X 3 = 6,000.

Question 14.
3,000 × 4 =
Answer:
3,000 X 4 =
3 X 1,000 X 4 = 12,000,

Explanation:
Given 3,000 X 4 = writing 3,000 as 3 X 1,000 and
multiplying with 4 we get 12,000 as 3 X 1,000 X 4 = 12,000,
So 3,000 X 4 = 12,000.

Question 15.
300 × 4 =
Answer:
300 X 4 =
3 X 100 X 4 = 1,200,

Explanation:
Given 300 X 4 = writing 300 as 3 X 100 and multiplying
with 4 we get 1,200 as 3 X 100 X 4 = 1,200, So 300 X 4 = 1,200.

Question 16.
30 × 4 =
Answer:
30 X 4 =
3 X 10 X 4 = 120,

Explanation:
Given 30 X 4 = writing 30 as 3 X 10 and multiplying
with 4 we get 120 as 3 X 10 X 4 = 120, So 30 X 4 = 120.

Question 17.
3 × 5 =
Answer:
3 X 5 = 15,

Explanation:
Given 3 X 5 = multiplying 3 with 5 we get 15
as 3 X 5 = 15, So 3 X 5 = 15.

Question 18.
30 × 5 =
Answer:
30 X 5 =
3 X 10 X 5 = 150,

Explanation:
Given 30 X 5 = writing 30 as 3 X 10 and multiplying
with 5 we get 150 as 3 X 10 X 5 = 150, So 30 X 5 = 150.

Question 19.
6 × 2 =
Answer:
6 X 2 = 12,

Explanation:
Given 6 X 2 = multiplying 6 with 2 we get 12
as 6 X 2 = 12, So 6 X 2 = 12.

Question 20.
60 × 2 =
Answer:
60 X 2 =
6 X 10 X 2 = 120,

Explanation:
Given 60 X 2 = writing 60 as 6 X 10 and multiplying
with 2 we get 120 as 6 X 10 X 2 = 120, So 60 X 2 = 120.

Question 21.
4 × 4 =
Answer:
4 X 4 = 16,

Explanation:
Given 4 X 4 = multiplying 4 with 4 we get 16
as 4 X 4 = 16, So 4 X 4 = 16.

Question 22.
400 × 4 =
Answer:
400 X 4 =
4 X 100 X 4 = 1,600,

Explanation:
Given 400 X 4 = writing 400 as 4 X 100 and multiplying
with 4 we get 1,600 as 4 X 100 X 4 = 1,600, So 400 X 4 = 1,600.

Question 23.
9 × 5 =
Answer:
9 X 5 = 45,

Explanation:
Given 9 X 5 = multiplying 9 with 5 we get 45 as
9 X 5 = 45.

Question 24.
900 × 5 =
9 X 100 X 5 = 4,500,

Explanation:
Given 900 X 5 = writing 900 as 9 X 100 and multiplying
with 5 we get 4,500 as 9 X 100 X 5 = 4,500, So 900 X 5 = 4,500.

Question 25.
8 × 4 =
Answer:
8 X 4 = 32,

Explanation:
Given 8 X 4 = multiplying 8 with 4 we get 32 as
8 X 4 = 32.

Question 26.
80 × 4 =
Answer:
80 X 4 =
8 X 10 X 4 = 320,

Explanation:
Given 80 X 4 = writing 80 as 8 X 10 and multiplying
with 4 we get 320 as 8 X 10 X 4 = 320, So 80 X 4 = 320.

Question 27.
9 × 3 =
Answer:
9 X 3 = 27,

Explanation:
Given 9 X 3 = multiplying 9 with 3 we get 27 as
9 X 3 = 27.

Question 28.
9,000 × 3 =
Answer:
9,000 X 3 =
9 X 1,000 X 3 = 27,000,

Explanation:
Given 9,000 X 3 = writing 9,000 as 9 X 1,000 and
multiplying with 3 we get 27,000 as 9 X 1,000 X 3 = 27,000,
So 9,000 X 3 = 27,000.

Question 29.
6 × 7 =
Answer:
6 X 7 = 42,

Explanation:
Given 6 X 7 = multiplying 6 with 7 we get 42 as
6 X 7 = 42.

Question 30.
6 × 700 =
Answer:
6 X 700 =
6 X 7 X 100 = 4,200,

Explanation:
Given 6 X 700 = writing 700 as 7 X 100 and multiplying
with 6 we get 4,200 as 6 X 7 X 100 = 4,200, So 6 X 700 = 4,200.

Question 31.
8 × 7 =
Answer:
8 X 7 = 56,

Explanation:
Given 8 X 7 = multiplying 8 with 7 we get 56 as
8 X 7 = 56.

Question 32.
8 × 70 =
Answer:
8 X 70 =
8 X 7 X 10 = 560,

Explanation:
Given 8 X 70 = writing 70 as 7 X 10 and multiplying
with 8 we get 560 as 8 x 7 X 10 = 560, So 8 X 70 = 560.

Question 33.
9 × 6 =
Answer:
9 X 6 = 54,

Explanation:
Given 9 X 6 = multiplying 9 with 6 we get 54 as
9 X 6 = 54.

Question 34.
9 × 6,000 =
Answer:
9 X 6,000 =
9 X 6 X 1,000 = 54,000,

Explanation:
Given 9 X 6,000 = writing 6,000 as 6 X 1,000 and multiplying with 9,
we get 54,000, as 9 X 6 X 1,000 = 54,000, So 9 X 6,000 = 54,000.

Question 35.
800 × 8 =
Answer:
800 X 8 =
8 X 100 X 8 = 6,400,

Explanation:
Given 800 X 8 = writing 800 as 8 X 100 and multiplying
with 8 we get 6,400 as 8 X 100 X 8 = 6,400, So 800 X 8 = 6,400.

Question 36.
9,000 × 9 =
Answer:
9,000 x 9 =
9 X 1,000 X 9 = 81,000,

Explanation:
Given 9,000 X 9 = writing 9,000 as 9 X 1,000 and multiplying
with 9 we get 81,000 as 9 X 1,000 X 9 = 81,000,
So 9,000 X 9 = 81,000.

Question 37.
7 × 700 =
Answer:
7 X 700 =
7 X 7 X 100 = 4,900,

Explanation:
Given 7 X 700 = writing 700 as 7 X 100 and multiplying
with 7 we get 4,900 as 7 X 7 X 100 = 4,900, So 7 X 700 = 4,900.

Question 38.
6 × 60 =
Answer:
6 X 60 =
6 X 6 X 10 = 360,

Explanation:
Given 6 X 60 = writing 60 as 6 X 10 and multiplying
with 6 we get 360 as 6 x 6 X 10 = 360, So 6 X 60 = 360.

Question 39.
700 × 8 =
Answer:
700 X 8 =
7 X 100 X 8 = 5,600,

Explanation:
Given 700 X 8 = writing 700 as 7 X 100 and multiplying
with 8 we get 5,600 as 7 X 100 X 8 = 5,600, So 700 X 8 = 5,600.

Question 40.
9 × 7,000 =
Answer:
9 X 7000 =
9 X 7 X 1,000 = 63,000,

Explanation:
Given 9 X 7,000 = writing 7,000 as 7 X 1,000 and multiplying with 9,
we get 63,000, as 9 X 7 X 1,000 = 63,000, So 9 X 7,000 = 63,000.

Question 41.
20 × 5 =
Answer:
20 X 5 =
2 X 10 X 5 = 100,

Explanation:
Given 20 X 5 = writing 20 as 2 X 10 and multiplying
with 5 we get 100 as 2 X 10 X 5 = 100, So 20 X 5 = 100.

Question 42.
5 × 600 =
Answer:
5 X 600 =
5 X 6 X 100 = 3,000,

Explanation:
Given 5 X 600 = writing 600 as 6 X 100 and multiplying
with 5 we get 3,000 as 5 X 6 X 100 = 3,000, So 5 X 600 = 3,000.

Question 43.
400 × 5 =
Answer:
400 X 5 =
4 X 100 X 5 = 2,000,

Explanation:
Given 400 X 5 = writing 400 as 4 X 100 and multiplying
with 5 we get 2,000 as 4 X 100 X 5 = 2,000, So 400 X 5 = 2,000.

Question 44.
8,000 × 5 =
Answer:
8,000 x 5 =
8 X 1,000 X 5 = 40,000,

Explanation:
Given 8,000 X 5 = writing 8,000 as 8 X 1,000 and multiplying
with 5 we get 40,000 as 8 X 1,000 X 5 = 40,000,
So 8,000 X 5 = 40,000.

Eureka Math Grade 4 Module 3 Lesson 7 Problem Set Answer Key

Question 1.
Represent the following expressions with disks, regrouping as necessary, writing a matching expression, and recording the partial products vertically as shown below.
a. 1 × 43


Answer:

43
X1
    3—-1 X 3 ones
+40—-1 X 4 tens
43

Explanation:
Represented the following expression 1 X43 with disks,
regrouped as necessary, wrote a matching expression, and
recorded the partial products vertically as shown above 1 X 43 = 43.

b. 2 × 43

Answer:

Explanation:
Represented the following expression 2 X 43 with disks,
regrouped as necessary, wrote a matching expression, and
recorded the partial products vertically as shown above 2 X 43 = 86.

c. 3 × 43

Answer:

Explanation:
Represented the following expression 3 X 43 with disks,
regrouped as necessary, wrote a matching expression, and
recorded the partial products vertically as shown above 3 X 43 = 129.

d. 4 × 43

Answer:

Explanation:
Represented the following expression 4 X 43 with disks,
regrouped as necessary, wrote a matching expression, and
recorded the partial products vertically as shown above 4 X 43 = 172.

Question 2.
Represent the following expressions with disks,
regrouping as necessary. To the right, record the partial
products vertically.
a. 2 × 36

Answer:

Explanation:
Represented the following expression 2 X 36 with disks,
regrouped as necessary. To the right, record the partial
products vertically as 2 X 36 = 72 as shown above.

b. 3 × 61

Answer:

Explanation:
Represented the following expression 3 X 61 with disks,
regrouped as necessary. To the right, recorded the partial
products vertically as 3 X 61 = 183 as shown above.

c. 4 × 84

Answer:

Explanation:
Represented the following expression 4 X 84 with disks,
regrouped as necessary. To the right, recorded the partial
products vertically as 4 X 84 = 336 as shown above.

Eureka Math Grade 4 Module 3 Lesson 7 Exit Ticket Answer Key

Represent the following expressions with disks, regrouping as necessary. To the right, record the partial products vertically.

Question 1.
6 × 41

Answer:

Explanation:
Represented the following expression 6 X 41 with disks,
regrouped as necessary. To the right, recorded the partial
products vertically as 6 X 41 = 246 as shown above.

Question 2.
7 × 31

Answer:

Explanation:
Represented the following expression 7 X 31 with disks,
regrouped as necessary. To the right, recorded the partial
products vertically as 7 X 31 = 217 as shown above.

Eureka Math Grade 4 Module 3 Lesson 7 Homework Answer Key

Question 1.
Represent the following expressions with disks, regrouping as necessary, writing a matching expression, and recording the partial products vertically.
a. 3 × 24

Answer:

Explanation:
Represented the following expression 3 X 24 with disks,
regrouped as necessary. To the right, recorded the partial
products vertically as 3 X 24 = 72 as shown above.

b. 3 × 42

Answer:

Explanation:
Represented the following expression 3 X 42 with disks,
regrouped as necessary. To the right, recorded the partial
products vertically as 3 X 42 = 126 as shown above.

c. 4 × 34

Answer:

Explanation:
Represented the following expression 4 X 34 with disks,
regrouped as necessary. To the right, recorded the partial
products vertically as 4 X 34 = 136 as shown above.

Question 2.
Represent the following expressions with disks, regrouping as necessary. To the right, record the partial products vertically.
a. 4 × 27

Answer:

Explanation:
Represented the following expression 4 X 27 with disks,
regrouped as necessary. To the right, recorded the partial
products vertically as 4 X 27 = 108 as shown above.

b. 5 × 42

Answer:

Explanation:
Represented the following expression 5 X 42 with disks,
regrouped as necessary. To the right, recorded the partial
products vertically as 5 X 42 = 210 as shown above.

Question 3.
Cindy says she found a shortcut for doing multiplication problems.
When she multiplies 3 × 24, she says, “3 × 4 is 12 ones,
or 1 ten and 2 ones. Then, there’s just 2 tens left in 24,
so add it up, and you get 3 tens and 2 ones.”
Do you think Cindy’s shortcut works?
Explain your thinking in words, and justify your
response using a model or partial products.
Answer:
No, Cindy’s shortcut will not work,

Explanation:
Given Cindy says she found a shortcut for doing multiplication problems.
When she multiplies 3 × 24, she says, “3 × 4 is 12 ones,
or 1 ten and 2 ones. Then, there’s just 2 tens left in 24,
so add it up, and you get 3 tens and 2 ones.”
No I don’t think Cindy’s shortcut works,
because 3 X 24 = 72 not 32,
Explained my thinking in words, and justified
using a model or partial products as show above.

Eureka Math Grade 4 Module 3 Lesson 7 Template Answer Key

_10,000____________
ten thousand value chart
Answer:

Explanation:
A place value chart is diagram that helps us to find and
compare the place value of the digits in numbers through
millions. The place value of a digit in the place value chart
increases by ten times as we shift to the left and decreases
by ten times as we shift to the right.
For example for ten thousand place value chart where we write
10,000 as 10 in thousands place, 0 in hundreds place, 0 in tens place
and 0 in ones place as shown above.

Engage NY Eureka Math 4th Grade Module 3 Lesson 6 Answer Key

Eureka Math Grade 4 Module 3 Lesson 6 Problem Set Answer Key

Represent the following problem by drawing disks in the place value chart.

Question 1.
To solve 20 × 40, think
(2 tens × 4) × 10 = __800___
20 × (4 × 10) = ___800_____
20 × 40 = _____800__

Answer:

20 X 40 = 800,
(2 tens × 4) × 10 = 8 hundreds,
20 × (4 × 10) = 800,
20 × 40 = 800,

Explanation:
Drawn place value disks to represent the value of the
following expression as shown above as 20 X 40 = 800,
(2 tens × 4) × 10 =8 hundreds, 20 × (4 × 10) = 800 and
20 × 40 = 800.

Question 2.
Draw an area model to represent 20 × 40.
2 tens × 4 tens = _____ __8 hundreds__________
Answer:

Explanation:
Drawn an area model to represent 20 × 40,
as shown above 2 tens X 4 tens = 8 hundreds.

Question 3.
Draw an area model to represent 30 × 40.
3 tens × 4 tens = _____ twelve hundreds_____________
30 × 40 = __12,00____
Answer:

Explanation:
Drawn an area model to represent 30 × 40,
as shown above 3 tens X 4 tens = 12 hundreds.

Question 4.
Draw an area model to represent 20 × 50.
2 tens × 5 tens = _____ ___1 thousand__________
20 × 50 = __1,000_____
Answer:

Explanation:
Drawn an area model to represent 20 × 50,
as shown above 2 tens X 5 tens = 1 thousand.

Rewrite each equation in unit form and solve.

Question 5.
20 × 20 = __400______
2 tens × 2 tens = ___4__ hundreds
Answer:
20 X 20 = 400,
2 tens × 2 tens = 4 hundreds,

Explanation:
Given equation is 20 X 20, Rewrote in unit form as
2 tens X 2 tens = 4 hundreds, So 20 X 20 = 40.

Question 6.
60 × 20 = ___1,200____
6 tens × 2 __tens___ = __12__ hundreds
Answer:
60 X 20 = 1,200,
6 tens × 2 tens = 12 hundreds,

Explanation:
Given equation is 60 X 20 = 1,200,  Rewrote in unit form as
6 tens X 2 tens = 12 hundreds, So 60 X 20 = 1,200.

Question 7.
70 × 20 = __1,400_____
__7___ tens × __2___ tens = 1,400__
Answer:
70 X 20 = 1,400,
7 tens X 2 tens = 1,400,

Explanation:
Given equation is 70 X 20 = 1,400, Rewrote in unit form as
7 tens X 2 tens = 14 hundreds, So 70 X 20 = 1,400.

Question 8.
70 × 30 = ___2,100____
__7__ __tens_____ × __3__ _tens______ = __21__ hundreds
Answer:
70 X 30 = 2,100,
7 tens X 3 tens = 21 hundreds,

Explanation:
Given equation is 70 X 30 = 2,100, Rewrote in unit form as
7 tens X 3 tens = 21 hundreds, So 70 X 30 = 2,100.

Question 9.
If there are 40 seats per row, how many seats are in 90 rows?
Answer:
There are 3,600 seats in 90 rows,

Explanation:
Given there are 40 seats per row, So number of seats in 90 rows are
40 seats X 90 = 3,600 seats, therefore there are 3,600 seats in 90 rows.

Question 10.
One ticket to the symphony costs $50. How much
money is collected if 80 tickets are sold?
Answer:
$4,000 amount money is collected if 80 tickets are sold,

Explanation:
Given one ticket to the symphony costs $50. So
money collected if 80 tickets are sold are $50 X 80 = $4,000.
Therefore, $4,000 amount money is collected if 80 tickets are sold.

Eureka Math Grade 4 Module 3 Lesson 6 Exit Ticket Answer Key

Represent the following problem by drawing disks in the place value chart.
Question 1.
To solve 20 × 30, think
(2 tens × 3) × 10 = ____600____
20 × (3 × 10) = ____600____
20 × 30 = ___600____

Answer:

20 X 30 = 600,
(2 tens × 3) × 10 =6 hundreds,
20 × (3 × 10) = 600,
20 × 30 = 600,

Explanation:
Drawn place value disks to represent the value of the
following expression as shown above as 20 X 30 = 600,
(2 tens × 3) × 10 =3 hundreds, 20 × (3 × 10) = 600 and
20 × 30 = 600.

Question 2.
Draw an area model to represent 20 × 30.
2 tens × 3 tens = _____ _6 hundreds___________
Answer:

Explanation:
Drawn an area model to represent 20 × 30,
as shown above 2 tens X 3 tens = 6 hundreds.

Question 3.
Every night, Eloise reads 40 pages. How many total pages
does she read at night during the 30 days of November?
Answer:
1,200 pages she read at night during the 30 days
of November,

Explanation:
Given every night, Eloise reads 40 pages, Total number of
pages she read at night during the 30 days of November are
40 pages X 30 = 1,200 pages, So Eloise 1,200 pages she read at
night during the 30 days of November.

Eureka Math Grade 4 Module 3 Lesson 6 Homework Answer Key

Represent the following problem by drawing disks in the place value chart.

Question 1.
To solve 30 × 60, think
(3 tens × 6) × 10 = ___1,800_____
30 × (6 × 10) = __1,800______
30 × 60 = __1,800_____

Answer:

30 X 60 = 1,800,
(3 tens × 6) × 10 =18 hundreds,
30 × (6 × 10) = 1,800,
30 × 60 = 1,800,

Explanation:
Drawn place value disks to represent the value of the
following expression as shown above as 30 X 60 = 1,800,
(3 tens × 6) × 10 =18 hundreds, 30 × (6 × 10) = 1,800 and
30 × 60 = 1,800.

Question 2.
Draw an area model to represent 30 × 60.
3 tens × 6 tens = _____18 hundreds_____________
Answer:

Explanation:
Drawn an area model to represent 30 × 60,
as shown above 3 tens X 6 tens = 18 hundreds.

Question 3.
Draw an area model to represent 20 × 20.
2 tens × 2 tens = _____ 4 hundreds_____________
20 × 20 = __400____
Answer:

Explanation:
Drawn an area model to represent 20 × 20,
as shown above 2 tens X 2 tens = 4 hundreds.

Question 4.
Draw an area model to represent 40 × 60.
4 tens × 6 tens = __24___ _hundreds____________
40 × 60 = ___240____
Answer:

Explanation:
Drawn an area model to represent 40 × 60,
as shown above 4 tens X 6 tens = 24 hundreds.

Rewrite each equation in unit form and solve.

Question 5.
50 × 20 = ___1,000_____
5 tens × 2 tens = _10____ hundreds
Answer:
50 X 20 = 1,000,
5 tens X 2 tens = 10 hundreds,

Explanation:
Given equation is 50 X 20 = 1,000, Rewrote in unit form as
5 tens X 2 tens = 10 hundreds, So 50 X 20 = 1,000.

Question 6.
30 × 50 = _1,500____
3 tens × 5 _tens___ = _15___ hundreds
Answer:
30 X 50 = 1,500,
3 tes X 5 tens = 15 hundreds,

Explanation:
Given equation is 30 X 50 = 1,500, Rewrote in unit form as
3 tens X 5 tens = 15 hundreds, So 30 X 50 = 1,500.

Question 7.
60 × 20 = __1,200______
__6___ tens × __2___ tens = 12 __hundreds_______
Answer:
60 X 20 = 1,200,
6 tens X 2 tens = 12 hundreds,

Explanation:
Given equation is 60 X 20 = 1,200, Rewrote in unit form as
6 tens X 2 tens = 12 hundreds, So 60 X 20 = 1,200.

Question 8.
40 × 70 = __2,800______
_4___ ___tens____ × _7___ __tens_____ = __28___ hundreds
Answer:
40 X 70 = 2,800,
4 tens X 7 tens = 28 hundreds,

Explanation:
Given equation is 40 X 70 = 2,800, Rewrote in unit form as
4 tens X 7 tens = 28 hundreds, So 40 X 70 = 2,800.

Question 9.
There are 60 seconds in a minute and 60 minutes in an hour.
How many seconds are in one hour?
Answer:
There are 3,600 seconds in one hour,

Explanation:
There are 60 seconds in a minute and 60 minutes in an hour.
Number of seconds are there in one hour is 60 seconds X 60 =
3,600 seconds,therefore, there are 3,600 seconds in one hour.

Question 10.
To print a comic book, 50 pieces of paper are needed.
How many pieces of paper are needed to print 40 comic books?
Answer:
Number of pieces of paper needed to print 40 comic books are
2,000 pieces,

Explanation:
Given to print a comic book, 50 pieces of paper are needed.
So number of pieces of paper are needed to print 40 comic books
are 50 pieces X 40 = 2,000 pieces.

Engage NY Eureka Math 4th Grade Module 3 Lesson 5 Answer Key

Eureka Math Grade 4 Module 3 Lesson 5 Problem Set Answer Key

Draw place value disks to represent the value of the following expressions.

Question 1.
2 × 3 = __6____
2 times __3___ ones is __6___ ones.


Answer:

2 X 3 = 6,
2 times 3 ones is 6 ones,
3
X 2
6__

Explanation:
Drawn place value disks to represent the value of the
following expression as shown above as 2 X 3 = 6.

Question 2.
2 × 30 = ___60___
2 times __3___ tens is __6 tens_______ .


Answer:

2 X 30 = 60,
2 times 3 tens is 6 tens,
30
X 2
60_

Explanation:
Drawn place value disks to represent the value of the
following expression as shown above as 2 X 30 = 60.

Question 3.
2 × 300 = ___600___
2 times ___hundreds ___ is _____6 hundreds________.


Answer:

2 X 300 = 600,
2 times 3 hundreds is 6 hundreds,
300
X 2
600_

Explanation:
Drawn place value disks to represent the value of the
following expression as shown above as 2 X 300 = 600.

Question 4.
2 × 3,000 = __6,000____
__2__ times ___3 thousands___ is ____6 thousands____ .


Answer:

2 X 3,000 = 6,000,
2 times 3 thousands is 6 thousands,
3,000
X   2
6,000_

Explanation:
Drawn place value disks to represent the value of the
following expression as shown above as 2 X 3,000 = 6,000.

Question 5.
Find the product

Answer:

Explanation:
Given a. 20 X 7 = 140, multiplying 20 with 7 we get 140,

b. 3 X 60 = 180, multiplying 3 with 60 we get 180,

c. 3 X 400 = 1,200, multiplying 3 with 400 we get 1,200,

d. 2 X 800 = 1,600, multiplying 2 with 800 we get 1,600,

e. 7 X 30 = 210, multiplying 7 with 30 we get 210,

f. 60 X 6 = 360, multiplying 60 with 6 we get 360,

g. 400 X 4 = 1,600, multiplying 400 with 4 we get 1,600,

h. 4 X 8,000 = 32,000, multiplying 4 with 8,000 we get 32,000,

i. 5 X 30 = 150, multiplying 5 with 30 we get 150,

j. 5 X 60 = 300, multiplying 5 with 60 we get 300,

k. 5 X 400 = 2,000, multiplying 5 with 400 we get 2,000,

l. 8,000 X 5 = 40,000, multiplying 8,000 with 5 we get 40,000.

Question 6.
Brianna buys 3 packs of balloons for a party.
Each pack has 60 balloons. How many balloons does Brianna have?
Answer:
Brianna have 180 balloons in all,

Explanation:
Given Brianna buys 3 packs of balloons for a party,
Each pack has 60 balloons. So number of balloons
does Brianna have are 3 X 60 balloons = 180 balloons.

Question 7.
Jordan has twenty times as many baseball cards as his brother.
His brother has 9 cards. How many cards does Jordan have?
Answer:
Jordan have 180 cards in total,

Explanation:
Given Jordan has twenty times as many baseball cards
as his brother. His brother has 9 cards. So number of
cards does Jordan have are 20 X 9 cards = 180 cards.

Question 8.
The aquarium has 30 times as many fish in one tank
as Jacob has. The aquarium has 90 fish.
How many fish does Jacob have?
Answer:
Jacob has 3 fishes in all,

Explanation:
Given the aquarium has 30 times as many fish in one tank
as Jacob has. The aquarium has 90 fish.
Number of fishes does Jacob have are 90 fish ÷ 30 = 3 fishes,
(3 X 30 = 90), Therefore, Jacob has 3 fishes in all.

Eureka Math Grade 4 Module 3 Lesson 5 Exit Ticket Answer Key

Draw place value disks to represent the value of the following expressions.

Question 1.
4 × 200 = ___800___
4 times ____2 hundreds_________ is _______800_________.


Answer:

4 X 200 = 800,
4 times 2 hundreds is 800,
200
X 4
800_

Explanation:
Drawn place value disks to represent the value of the
following expression as shown above as 4 X 200 = 800.

Question 2.
4 × 2,000 = _8,000_____
___4___ times ____2 thousands_______ is __8 thousands_______.


Answer:

4 X 2,000 = 8,000,
4 times 2 thousands is 8 thousands,
2,000
X   4
8,000

Explanation:
Drawn place value disks to represent the value of the
following expression as shown above as 4 X 2,000 = 8,000.

Question 3.
Find the product.

Answer:

Explanation:
Given a. 30 X 3 = 90, multiplying 30 with 3 we get 90,

b. 8 X 20 = 160, multiplying 8 with 20 we get 160,

c. 6 X 400 = 2,400, multiplying 6 with 400 we get 2,400,

d. 2 X 900 = 1,800, multiplying 2 with 900 we get 1,800,

e. 8 X 80 = 640, multiplying 8 with 80 we get 640,

f. 30 X 4 = 120, multiplying 30 with 4 we get 120,

g. 500 X 6 = 3,000, multiplying 500 with 6 we get 3,000,

h. 8 X 5,000 = 40,000, multiplying 8 with 5,000 we get 40,000.

Question 4.
Bonnie worked for 7 hours each day for 30 days.
How many hours did she work altogether?
Answer:
Bonnie has worked for 210 hours altogether in 30 days,

Explanation:
Given Bonnie worked for 7 hours each day for 30 days.
So number of hours did she work altogether is 7 hours X 30 =
210 hours, Therefore, Bonnie has worked for 210 hours
altogether in 30 days.

Eureka Math Grade 4 Module 3 Lesson 5 Homework Answer Key

Question 1.
5 × 2 = __10____
5 times __2__ ones is __10___ ones.


Answer:

5 X 2 = 10,
5 times 2 ones is 10 ones,
2
X 5
10__

Explanation:
Drawn place value disks to represent the value of the
following expression as shown above as 5 X 2 = 10.

Question 2.
5 × 20 = _100_____
5 times ____2______ tens is _____10 tens________________.


Answer:

5 X 20 = 100,
5 times 2 tens is 10 tens,
20
X 5
100_

Explanation:
Drawn place value disks to represent the value of the
following expression as shown above as 5 X 20 = 100.

Question 3.
5 × 200 = __1,000____
5 times ____2 hundreds______ is ____10 hundreds_____.


Answer:

5 X 200 = 1,000,
5 times 2 hundreds is 10 hundreds,
200
X 5
1,000_

Explanation:
Drawn place value disks to represent the value of the
following expression as shown above as 5 X 200 = 1,000.

Question 4.
5 × 2,000 = _10,000_____
__5__ times _______2 thousands____ is ___10 thousands________ .


Answer:

5 X 2,000 = 10,000,
5 times 2 thousands is 10 thousands,
2,000
X    5
10,000_

Explanation:
Drawn place value disks to represent the value of the
following expression as shown above as 5 X 2,000 = 10,000.

Question 5.
Find the product.

Answer:

Explanation:
Given a. 20 X 9 = 180, multiplying 20 with 9 we get 180,

b. 6 X 70 = 420, multiplying 6 with 70 we get 420,

c. 7 X 700 = 4,900, multiplying 7 with 700 we get 4,900,

d. 3 X 900 = 2,700, multiplying 3 with 900 we get 2,700,

e. 9 X 90 = 810, multiplying 9 with 90 we get 810,

f. 40 X 7 = 280, multiplying 40 with 7 we get 280,

g. 600 X 6 = 3,600, multiplying 600 with 6 we get 3,600,

h. 8 X 6,000 = 48,000, multiplying 8 with 6,000 we get 48,000.

i. 5 X 70 = 350, multiplying 5 with 70 we get 350,

j. 5 X 80 = 400, multiplying 5 with 80 we get 400,

k. 5 X 200 = 1,000, multiplying 5 with 200 we get 1,000,

l. 6,000 X 5 = 30,000, multiplying 6,000 with 5 we get 30,000.

Question 6.
At the school cafeteria, each student who orders
lunch gets 6 chicken nuggets. The cafeteria staff prepares
enough for 300 kids. How many chicken nuggets does the
cafeteria staff prepare altogether?
Answer:
1,800 chicken nuggets the cafeteria staff prepare altogether,

Explanation:
Given at the school cafeteria, each student who orders
lunch gets 6 chicken nuggets. The cafeteria staff prepares
enough for 300 kids. Number of chicken nuggets does the
cafeteria staff prepare altogether is 6 chicken nuggets X 300 = 1,800,
therefore, 1,800 chicken nuggets the cafeteria staff prepare altogether.

Engage NY Eureka Math 4th Grade Module 3 Lesson 4 Answer Key

Eureka Math Grade 4 Module 3 Lesson 4 Problem Set Answer Key

Draw place value disks and arrows as shown to represent each product.

Question 1.
5 × 100 = ___500___
5 × 10 × 10 = ___500___
5 ones × 100 = ___500____

Answer:
5 x 100 = 500,

5 X 10 X 10 = 500,

5 ones × 100 = 500,

Explanation:
Drawn and  placed value disks and arrow as shown above
to represent the products as 5 x 100 = 500, 5 X 10 X 10 = 500
and 5 ones X 100 = 500.

Question 2.
5 × 1,000 = ____5,000______
5 × 10 × 10 × 10 = __5,000_____
5 ones × 1,000 = ____ _5,000____

Answer:
5 X 1,000 = 5,000,

5 X 10 X 10 X 10 = 5,000,

5 ones × 1,000 = 5,000,

Explanation:
Drawn and  placed value disks and arrow as shown above
to represent the product as 5 x 1,000 = 5,000,
5 X 10 X 10 X 10 = 5,000 and 5 ones X 1,000 = 5,000.

Question 3.
Fill in the blanks in the following equations.
a. 6 × 10 = ____60____
Answer:
6 x 10 = 60,

Explanation:
Given 6 X 10 multiplying 6 with 10 we get 60,
so 6 X 10 = 60.

b. __100____ × 6 = 600
Answer:
100 X 6 = 600,

Explanation:
Given ____ × 6 = 600, so we get unknown factor as
600 ÷ 6 = 100, so 100 X 6 = 600.

c. 6,000 = ______ × 1,000
Answer:
6,000 = 6 X 1,000,

Explanation:
Given 6,000 = ______ × 1,000, so we get unknown factor as
6,000 ÷ 1,000 = 6, so 6,000 = 6 X 1,000.

d. 10 × 4 = ______
Answer:
10 X 4 = 40,

Explanation:
Given 10 X 4 = ____,multiplying 10 with 4 we get 40,
so 10 X 4 = 40.

e. 4 × ______ = 400
Answer:
4 X 100 = 400,

Explanation:
Given 4 x ____  = 400, so we get unknown factor as
400 ÷ 4 = 100, so 4 X 100 = 400.

f. ______ × 4 = 4,000
Answer:
1,000 X 4 = 4,000,

Explanation:
Given ______ × 1,000 = 4,000 so we get unknown factor as
4,000 ÷ 4 = 1,000, so 1,000 X 4 = 4,000.

g. 1,000 × 9 = ______
Answer:
1,000 X 9 = 9,000,

Explanation:
Given 1,000 X 9 =___ ,multiplying 1,000 with 9 we get 9,000,
so 1,000 X 9 = 9,000.

h. ______ = 10 × 9
Answer:
90 = 10 X 9,
Explanation:
Given _____ = 10 X 9 multiplying 10 with 9 we get 90,
so 90 = 10 X 9.

i. 900 = ______ × 100
Answer:
900 = 9 X 100,

Explanation:
Given 900 = ______ × 100 so we get unknown factor as
900 ÷ 100 = 9, so 900 = 9 X 100.

Draw place value disks and arrows to represent each product.

Question 4.
12 × 10 = ____120______
(1 ten 2 ones) × 10 = ______120_________

Answer:
12 x 10 = 120,
(1 ten 2 ones) × 10 = 120,

Explanation:
Drawn and  placed value disks and arrow as shown above
to represent the products as 12 x 10 = 120 and (1 ten 2 ones) = 120.

Question 5.
18 × 100 = ___18,00_______
18 × 10 × 10 = __18,00________
(1 ten 8 ones) × 100 = ____18,00________

Answer:
18 × 100 =18,00,
18 × 10 × 10 = 18,00,
(1 ten 8 ones) × 100 = 18,00,

Explanation:
Drawn and  placed value disks and arrow as shown above
to represent the products as 18 x 100 = 18,00 ,
18 X 10 X 10 = 18,00 and (1 ten 8 ones) × 100 = 18,00.

Question 6.
25 × 1,000 = ____25,000______
25 × 10 × 10 × 10 = ____25,000______
(2 tens 5 ones) × 1,000 = ______25,000__________

Answer:
25 × 1,000 = 25,000,
25 × 10 × 10 × 10 = 25,000,
(2 tens 5 ones) × 1,000 = 25,000,

Explanation:
Drawn and  placed value disks and arrow as shown above
to represent the products as 25 × 1,000 = 25,000,
25 × 10 × 10 × 10 = 25,000 and (2 tens 5 ones) × 1,000 = 25,000.

Decompose each multiple of 10, 100, or 1,000 before multiplying.

Question 7.
3 × 40 = 3 × 4 × __10___
= 12 × __10____
= ___120_______
Answer:

3 X 40 = 3 X 4 X 10
= 12 X 10
= 120,

Explanation:
Given 3 X 40 decomposing into 10 before multiplying
first we write 40 as 4 X 10,
then we multiply with 3, as 3 X 4 X 10,
Now we multiply 3 and 4 then 10
we get 12 X 10 = 120,
So 3 X 40 = 120.

Question 8.
3 × 200 = 3 × ___2__ × __100____
= ___6___ × ___100___
= ____600____
Answer:
3 X 200 = 3 X 2 X 100
= 6 X 100
= 600,

Explanation:
Given 3 X 200 decomposing into 100 before multiplying
first we write 200 as 2 X 100,
then we multiply with 3 as 3 X 2 X 100,
Now we multiply 3 and 2 then 100
we get 6 X 100 = 6,00,
So 3 X 200 = 600.

Question 9.
4 × 4,000 = __4___ × __4___× __1,000_______
= ___16___ × __1,000_______
= ____16,000_____
Answer:
4 × 4,000 = 4 X 4 X 1,000,
=16 X 1,000
= 16,000,

Explanation:
Given 4 X 4,000 decomposing into 1,000 before multiplying
first we write 4,000 as 4 X 1,000,
then we multiply with 4 as 4 X 4 X 1,000,
Now we multiply 4 and 4 then 1,000
we get 16 X 1,000 = 16,000,
So 4 X 4,000 = 16,000.

Question 10.
5 × 4,000 = __5___ × __4___ × ___1,000______
= ___20___ × __1,000______
= __20,000______
Answer:
5 X 4,000 = 5 X 4 X 1,000,
= 20 X 1,000,
= 20,000,

Explanation:
Given 5 X 4,000 decomposing into 1,000 before multiplying
first we write 4,000 as 4 X 1,000,
then we multiply with 5 as 5 X 4 X 1,000,
Now we multiply 5 and 4 then 1,000
we get 20 X 1,000 = 20,000,
So 5 X 4,000 = 20,000.

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

Fill in the blanks in the following equations.

a. 5 × 10 = ___50_____
Answer:
5 X 10 = 50,

Explanation:
Given 5 X 10 multiplying 5 with 10 we get 50,
so 5 X 10 = 50.

b. ______ × 5 = 500
Answer:
100 X 5 = 500,

Explanation:
Given  ____  X 5 = 500 so we get unknown factor as
500 ÷ 5 = 100, so 100 X 5 = 500.

c. 5,000 = ______ × 1000
Answer:
5,000 = 5 x 1,000,

Explanation:
Given 5,000 = _______ X 1,000, now we get unknown factor as
5,000 ÷ 1,000 = 5, so 5,000 = 5 X 1,000.

d. 10 × 2 = ______
Answer:
10 X 2 = 20,

Explanation:
Given 10 X 2 multiplying 10 with 2 we get 20,
so 10 X 2 = 20.

e. ______ × 20 = 2,000
Answer:
100 X 20 = 2,000,

Explanation:
Given_______ X 20 = 2,000, now we get unknown factor as
2,000 ÷ 20 = 100, so 100 X 20 = 2,000.

f. 2,000 = 10 × ______
Answer:
2,000 = 10 X 200,

Explanation:
Given 2,000 = 10 X _________, now we get unknown factor as
2,000 ÷ 10 = 200, so 2,000 = 10 X 200.

g. 100 × 18 = ______
Answer:
100 X 18 = 1,800,

Explanation:
Given 100 X 18 multiplying 100 with 18 we get 1,800
so 100 X 18 = 1,800.

h. ______ = 10 × 32
Answer:
320 = 10 X 32,

Explanation:
Given_______  = 10 X 32, multiplying 10 with 32 we get 320,
so 320 = 10 X 32.

i. 4,800 = ______ × 100
Answer:
4,800 = 48 X 100,

Explanation:
Given 4,800 = _______ X 100, now we get unknown factor as
4,800 ÷ 100 = 48, so 4,800 = 48 X 100.

j. 60 × 4 = ______
Answer:
60 × 4 = 240,

Explanation:
Given 60 X 4, multiplying 60 with 4 we get 240,
so 60 X 4 = 240.

k. 5 × 600 = ______
Answer:
5 X 600 = 3,000,

Explanation:
Given 5 X 600 = _________,multiplying 5 with 600 we get 3,000,
so 5 X 600 = 3,000.

l. 8,000 × 5 = ______,
Answer:
8,000 X 5 = 40,000,

Explanation:
Given 8,000 X 5 = _______, multiplying 8,000 with 5 we get 40,000,
so 8,000 X 5 = 40,000.

Eureka Math Grade 4 Module 3 Lesson 4 Homework Answer Key

Question 1.
7 × 100 = ___700_______
7 × 10 × 10 = __700________
7 ones × 100 = ____700_______

Answer:
7 X 100 = 700,

7 × 10 × 10 = 700,

7 ones × 100 = 700,

Explanation:
Drawn and  placed value disks and arrow as shown above
to represent the products as 7 x 100 = 700, 7 X 10 X 10 = 700
and 7 ones X 100 = 700.

Question 2.
7 × 1,000 = ____7,000______
7 × 10 × 10 × 10 = __7,000________
7 ones × 1,000 = ____7,000_____________

Answer:
7 × 1,000 = 7,000,

7 × 10 × 10 × 10 = 7,000,

7 ones × 1,000 = 7,000,

Explanation:
Drawn and  placed value disks and arrow as shown above
to represent the products as 7 x 1,000 = 7,000, 7 X 10 X 10 X 10 = 7,000
and 7 ones X 1,000 = 7,000.

Question 3.
Fill in the blanks in the following equations.
a. 8 × 10 = ________
Answer:
8 X 10 = 80,

Explanation:
Given 8 X 10 = ________, multiplying 8 with 10 we get 80,
so 8 X 10 = 80.

b. ______ × 8 = 800
Answer:
100 X 8 = 800,

Explanation:
Given ______ X 8 = 800, now we get unknown factor as
800 ÷ 8 = 100, so 100 X 8 = 800.

c. 8,000 = ______ × 1,000
Answer:
8,000 = 8 X 1,000,

Explanation:
Given 8,000 = 8 X 1,000, now we get unknown factor as
8,000 ÷ 1,000 = 8, so 8,000 = 8 X 1,000.

d. 10 × 3 = ______
Answer:
10 X 3 = 30,

Explanation:
Given 10 X 3 = ________, multiplying 10 with 3 we get 30,
so 10 X 3 = 30.

e. 3 × ______ = 3,000
Answer:
3 X 1,000 = 3,000,

Explanation:
Given 3 X ______ = 3,000, now we get unknown factor as
3,000 ÷ 3 = 1,000, so 3 X 1,000 = 3,000.

f. ______ × 3 = 300
Answer:
100 X 3= 300,

Explanation:
Given ______ X 3 = 300, now we get unknown factor as
300 ÷ 3 = 100, so 100 X 3 = 300.

g. 1,000 × 4 = ______
Answer:
1,000 X 4 = 4,000,

Explanation:
Given 1,000 X 4 = ________, multiplying 1,000 with 4 we get 4,000,
so 1,000 X 4 = 4,000.

h. ______ = 10 × 4
Answer:
40 = 10 X 4,

Explanation:
Given ______ = 10 X 4, multiplying 10 with 4 we get 40,
so 40 = 10 X 4.

i. 400 = ______ × 100
Answer:
400 = 4 X 100,

Explanation:
Given 400 = 4 X 100, now we get unknown factor as
400 ÷ 100 = 4, so 400 = 4 X 100.

Draw place value disks and arrows to represent each product.

Question 4.
15 × 10 = ___150_______
(1 ten 5 ones) × 10 = _____150________

Answer:
15 X 10 = 150,
(1 ten 5 ones) × 10 = 150,

Explanation:
Drawn and  placed value disks and arrow as shown above
to represent the products as 15 x 10 = 150 and (1 ten 5 ones) = 150.

Question 5.
17 × 100 = ___1,700_______
17 × 10 × 10 = ___1,700_______
(1 ten 7 ones) × 100 = _______1,700___________

Answer:
17 × 100 = 1,700,
17 × 10 × 10 = 17,00,
(1 ten 7 ones) × 100 = 17,00,

Explanation:
Drawn and  placed value disks and arrow as shown above
to represent the products as 17 x 100 = 17,00 ,
17 X 10 X 10 = 17,00 and (1 ten 7 ones) × 100 = 17,00.

Question 6.
36 × 1,000 = __________
36 × 10 × 10 × 10 = __________
(3 tens 6 ones) × 1,000 = ____________

Answer:
36 × 1,000 = 36,000,
36 × 10 × 10 × 10 = 36,000,
(3 tens 6 ones) × 1,000 = 36,000,

Explanation:
Drawn and  placed value disks and arrow as shown above
to represent the products as 36 × 1,000 = 36,000,
36 × 10 × 10 × 10 = 36,000 and (3 tens 6 ones) × 1,000 = 36,000.

Decompose each multiple of 10, 100, or 1000 before multiplying.

Question 7.
2 × 80 = 2 × 8 × __16___
= 16 × _10_____
= ___160_______
Answer:
2 X 80 = 2 X 8 X 10,
= 16 X 10,
= 160,

Explanation:
Given 2 X 80 decomposing into 10 before multiplying
first we write 80 as 8 X 10,
then we multiply with 2, as 2 X 8 X 10,
Now we multiply 2 and 8 then 10
we get 16 X 10 = 160,
So 2 X 80 = 160.

Question 8.
2 × 400 = 2 × _4___ × __100____
= ___8___ × __100____
= ___800_____
Answer:
2 X 400 = 2 X 4 X 100
= 8 X 100
= 800,

Explanation:
Given 3 X 200 decomposing into 100 before multiplying
first we write 200 as 2 X 100,
then we multiply with 3 as 3 X 2 X 100,
Now we multiply 3 and 2 then 100
we get 6 X 100 = 6,00,
So 3 X 200 = 600.

Question 9.
5 × 5,000 = _____ × _____ × _________
= ______ × _________
= _________
Answer:
5 × 5,000 = 5 X 5 X 1,000,
=25 X 1,000
= 25,000,

Explanation:
Given 5 X 5,000 decomposing into 1,000 before multiplying
first we write 5,000 as 5 X 1,000,
then we multiply with 5 as 5 X 5 X 1,000,
Now we multiply 5 and 5 then 1,000
we get 25 X 1,000 = 25,000,
So 5 X 5,000 = 25,000.

Question 10.
7 × 6,000 = _____ × _____ × _________
= ________ × ________
= ________
Answer:
7 X 6,000 = 7 X 6 X 1,000,
= 42 X 1,000,
= 42,000,

Explanation:
Given 7 X 6,000 decomposing into 1,000 before multiplying
first we write 6,000 as 6 X 1,000,
then we multiply with 7 as 7 X 6 X 1,000,
Now we multiply 7 and 6 then 1,000
we get 42 X 1,000 = 42,000,
So 7 X 6,000 = 42,000.

Eureka Math Grade 4 Module 3 Lesson 4 Template Answer Key

______1,000___________
thousands place value chart
Answer:

Explanation:
A place value chart is diagram that helps us to find and
compare the place value of the digits in numbers through
millions. The place value of a digit in the place value chart
increases by ten times as we shift to the left and decreases
by ten times as we shift to the right.
For example for one thousand place value chart where we write
1,000 as 1 in thousands place, 0 in hundreds place, 0 in tens place
and 0 in ones place as shown above.