Prasanna

Engage NY Eureka Math 5th Grade Module 6 Lesson 1 Answer Key

Eureka Math Grade 5 Module 6 Lesson 1 Problem Set Answer Key

Question 1.
Each shape was placed at a point on the number line s. Give the coordinate of each point below.

Answer:

Explanation :
The given number line is divided into 3 equals parts from 0-1,1-2 and so on.
Fraction of each part is \(\frac{1}{3}\).
So, the coordinates of each point located is written in above figure .

Question 2.
Plot the points on the number lines.
a.

Plot A so that its distance from the origin is 2.
Answer :

Explanation :
The given Number line is divided into 3 parts from 0 -3 that means each unit is 1 .
Point A is marked at 2 and is shown in above figure .

b.

Plot R so that its distance from the origin is \(\frac{5}{2}\).
Answer :

Explanation :
The Each part of the above number line is \(\frac{1}{2}\).
R is plotted at \(\frac{5}{2}\). and is shown in above figure .

c.

Plot L so that its distance from the origin is 20.
Answer :

Explanation :
The Each part of the above number line is 5 .
The representation of number line is in decreasing order of 5 . so L is plotted below 35
L is plotted at 20 and is shown in above figure .

d.

Plot a point T so that its distance from the origin is \(\frac{2}{3}\) more than that of S.
Answer:

Explanation :
The given Number line is divided into 6 Equal parts from 4 – 5 that means each unit is \(\frac{1}{6}\).
Point S is marked at  4\(\frac{1}{6}\) and is shown in above figure .
point T so that its distance from the origin is \(\frac{2}{3}\) more than that of S.
Point T = 4\(\frac{1}{6}\) + \(\frac{2}{3}\) = \(\frac{25}{6}\) + \(\frac{4}{6}\)
= \(\frac{29}{6}\) = 4 \(\frac{5}{6}\)
Point T = 4 \(\frac{5}{6}\)  . is marked and is shown in above figure .

Question 3.
Number line g is labeled from 0 to 6. Use number line g below to answer the questions.

a. Plot point A at \(\frac{3}{4}\).
b. Label a point that lies at 4\(\frac{1}{2}\) as B.
c. Label a point, C, whose distance from zero is 5 more than that of A.
The coordinate of C is _______.
d. Plot a point, D, whose distance from zero is 1\(\frac{1}{2}\) less than that of B.
The coordinate of D is ________.
e. The distance of E from zero is 1\(\frac{3}{4}\) more than that of D. Plot point E.
f. What is the coordinate of the point that lies halfway between A and D? _______ Label this point F.

Answer c :
Point A = \(\frac{3}{4}\)
Point C = 5 + \(\frac{3}{4}\)=5 \(\frac{3}{4}\)
Answer d :
Point B = 4\(\frac{1}{2}\)
Point D = 1\(\frac{1}{2}\) less than that of B = 4\(\frac{1}{2}\)  – 1\(\frac{1}{2}\) = 3 .
Answer e :
Point D = 3
Point E = 1\(\frac{3}{4}\) more than that of D = 3 + 1\(\frac{3}{4}\) = 4\(\frac{3}{4}\).
Answer f :
The Coordinate of the point that lies halfway between A and D = Point F = 1.875

Question 4.
Mrs. Fan asked her fifth-grade class to create a number line. Lenox created the number line below:

Parks said Lenox’s number line is wrong because numbers should always increase from left to right. Who is correct? Explain your thinking.
Answer:
Lenox is right .
Explanation :
Any Orientation is acceptable .

Question 5.
A pirate marked the palm tree on his treasure map and buried his treasure 30 feet away. Do you think he will be able to easily find his treasure when he returns? Why or why not? What might he do to make it easier to find?

Answer:
No,
Explanation :
Because point or location will help to find the treasure easily .

Eureka Math Grade 5 Module 6 Lesson 1 Exit Ticket Answer Key

Question 1.
Use number line l to answer the questions.

a. Plot point C so that its distance from the origin is 1.
b. Plot point E \(\frac{4}{5}\) closer to the origin than C. What is its coordinate? _________
c. Plot a point at the midpoint of C and E. Label it H.
Answer:
a .Point C is plotted .
b .Point E = \(\frac{4}{5}\)closer to the origin than C. = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\) .
c. Point H = the midpoint of C and E = 1 + \(\frac{1}{5}\) = \(\frac{6}{5}\)/2 =\(\frac{6}{10}\) = \(\frac{2}{3}\) .

Eureka Math Grade 5 Module 6 Lesson 1 Homework Answer Key

Question 1.
Answer the following questions using number line q below.
a. What is the coordinate, or the distance from the origin, of the ? ___________
b. What is the coordinate of the ? _________
c. What is the coordinate of the ? __________
d. What is the coordinate at the midpoint of the and the ? __________

Answer:
a the coordinate, or the distance from the origin, of the is 3
b the coordinate of the is  8
c the coordinate of the is 14
d the coordinate at the midpoint of the and the is (8 + 14)/2 = 11
Explanation :
The number line is divided into 3 parts from 0-3,3-6 and so on . . . .
So Each unit is 1 .

Question 2.
Use the number lines to answer the questions.

a Plot T so that its distance from the origin is 10.

b Plot M so that its distance is \(\frac{11}{4}\) from the origin. What is the distance from P to M?

c Plot a point that is 0.15 closer to the origin than Z.

d Plot U so that its distance from the origin is \(\frac{3}{6}\) less than that of W.
Answer a :

Explanation :
The each part of the above number line is 1 unit .
T is plotted at 11 and is shown in above figure .

Answer b :

Explanation :
The each part of the above number line is \(\frac{1}{4}\) unit .
Point P = 1\(\frac{1}{4}\) = \(\frac{5}{4}\)
Point M= \(\frac{11}{4}\)
Distance between P and M =\(\frac{11}{4}\) – \(\frac{5}{4}\) = \(\frac{6}{4}\)

Answer c :

Explanation :
Point Z = 0.95
Point Y = point that is 0.15 closer to the origin than Z. = 0.95- 0.15 = 0.8
Point Y is plotted .

Answer d :

Explanation :
Point W = 9 \(\frac{4}{6}\)
Plot U so that its distance from the origin is \(\frac{3}{6}\) less than that of W
Point U = 9\(\frac{4}{6}\) – \(\frac{3}{6}\) = 9 \(\frac{1}{6}\).
Point U is plotted .

Question 3.
Number line k shows 12 units. Use number line k below to answer the questions.

a. Plot a point at 1. Label it A.
b. Label a point that lies at 3\(\frac{1}{2}\) as B.
c. Label a point, C, whose distance from zero is 8 units farther than that of B.
The coordinate of C is __________.
d. Plot a point, D, whose distance from zero is \(\frac{6}{2}\) less than that of B.
The coordinate of D is __________.
e. What is the coordinate of the point that lies \(\frac{17}{2}\) farther from the origin than D? Label this point E.
f. What is the coordinate of the point that lies halfway between F and D? Label this point G.
Answer:
All the points are plotted and shown in the below figure .

Explanation :
a Point A = 1
b Point B = 3\(\frac{1}{2}\)
c Point C = 8 units farther than that of B = 8 + 3\(\frac{1}{2}\) = 11 \(\frac{1}{2}\)
The coordinate of C is 11 \(\frac{1}{2}\)
d Point D = \(\frac{6}{2}\) less than that of B = 3\(\frac{1}{2}\) – \(\frac{6}{2}\)
= \(\frac{7}{2}\) – \(\frac{6}{2}\)  = \(\frac{1}{2}\)
The coordinate of D is \(\frac{1}{2}\)
e Point E = \(\frac{17}{2}\) farther from the origin than D = \(\frac{1}{2}\)  + \(\frac{17}{2}       \)  = \(\frac{18}{2}\) = 9
f  Point F = 9 \(\frac{1}{2}\)
Point G = halfway between F and D =( 9 \(\frac{1}{2}\) + \(\frac{1}{2}\) )/2 = 10/2 = 5

Question 4.
Mr. Baker’s fifth-grade class buried a time capsule in the field behind the school. They drew a map and marked the location of the capsule with an X so that his class can dig it up in ten years. What could Mr. Baker’s class have done to make the capsule easier to find?

Answer:
The Location of the capsule is marked with x in the map. but to find easier he could add the number of feet’s from the school or a land mark or depth of the time capsule buried etc if given will help to find the time capsule easier even after ten years .

Students of Grade 4 can get a strong foundation on mathematics concepts by referring to the Eureka Math Book. It was developed by highly professional mathematics educators and the solutions prepared by them are in a concise manner for easy grasping. To achieve high scores in Grade 4, students need to solve all questions and exercises included in Eureka’s Math Grade 4 Book.

Engage NY Eureka Math 4th Grade Module 5 Lesson 2 Answer Key

Eureka Math Answer Key for Grade 4 aids teachers to differentiate instruction, building, and reinforcing foundational mathematics skills that alter from the classroom to real life.
With the help of the Eureka Primary School Grade 4 Answer Key, You can think deeply regarding what you are learning, and you will really learn math easily just like that. So teachers and students can find this Eureka Answer Key for Grade 4 more helpful in raising students’ scores and supporting teachers to educate the students.

Eureka Math Grade 4 Module 5 Lesson 2 Problem Set Answer Key

Question 1.
Step 1: Draw and shade a tape diagram of the given fraction.
Step 2: Record the decomposition as a sum of unit fractions.
Step 3: Record the decomposition of the fraction two more ways.
(The first one has been done for you.)
a. \(\frac{5}{8}\)

b. \(\frac{9}{10}\)

Answer:
9/10 = 1/10 + 2/10 + 2/10 + 3/10 + 1/10.

Explanation:
In the above-given question,
given that,
draw and shade a tape diagram.
9/10 = 1/10 + 2/10 + 2/10 + 3/10 + 1/10.
9/10 = 5 /10 + 4 /10 + 1/10.
9/10 = 3/10 + 2/10 + 3/10 + 1/10.

c. \(\frac{3}{2}\)

Answer:
3/2 = 1/2 + 2/2.

Explanation:
In the above-given question,
given that,
draw and shade a tape diagram.
3/2 = 1/2 + 2/2.
3/2 = 1/2 + 1/2 + 1/2.

Question 2.
Step 1: Draw and shade a tape diagram of the given fraction.
Step 2: Record the decomposition of the fraction in three different ways using number sentences.
a. \(\frac{7}{8}\)

Answer:
7/8 = 1/8 + 2/8 + 2/8 + 1/8 + 1/8.

Explanation:
In the above-given question,
given that,
draw and shade a tape diagram.
7/8 = 1/8 + 2/8 + 2/8 + 1/8 + 1/8.
7/8 = 1 /8 + 4 /10 + 2/8.
7/8 = 1/8 + 2/8 + 3/8 + 1/8.

b. \(\frac{5}{3}\)

Answer:
5/3 = 1/3 + 1/3 + 1/3 + 1/3 + 1/3.

Explanation:
In the above-given question,
given that,
draw and shade a tape diagram.
5/3 = 1/3 + 1/3 + 1/3 + 1/3 + 1/3.
5/3 = 1 /3 + 2/3 + 2/3.
5/3 = 1/3 + 2/3 + 1/3 + 1/3.

c. \(\frac{7}{5}\)

Answer:
5/7 = 1/7 + 1/7 + 1/7 + 1/7 + 1/7.

Explanation:
In the above-given question,
given that,
draw and shade a tape diagram.
5/7 = 1/7 + 1/7 + 1/7 + 1/7 + 1/7.
5/7 = 1 /7 + 2/7 + 2/7.
5/7 = 1/7 + 1/7 + 2/7 + 1/7.

d. 1\(\frac{1}{3}\)

Answer:
4/3 = 1/3 + 1/3 + 1/3 + 1/3.

Explanation:
In the above-given question,
given that,
draw and shade a tape diagram.
4/3 = 1/3 + 1/3 + 1/3 + 1/3.
4/3 = 1 /3 + 2/3 + 1/3.
4/3 = 1/3 + 1/3 + 2/3.

Eureka Math Grade 4 Module 5 Lesson 2 Exit Ticket Answer Key

Step 1: Draw and shade a tape diagram of the given fraction.
Step 2: Record the decomposition of the fraction in three different ways using number sentences.
\(\frac{4}{7}\)

Answer:
4/7 = 1/7 + 1/7 + 1/7 + 1/7.

Explanation:
In the above-given question,
given that,
draw and shade a tape diagram.
4/7 = 1/7 + 1/7 + 1/7 + 1/7.
4/7 = 1 /7 + 2/7 + 1/7.
4/7 = 1/7 + 1/7 + 2/7.

Eureka Math Grade 4 Module 5 Lesson 2 Homework Answer Key

Question 1.
Step 1: Draw and shade a tape diagram of the given fraction.
Step 2: Record the decomposition as a sum of unit fractions.
Step 3: Record the decomposition of the fraction two more ways.
(The first one has been done for you.)
a. \(\frac{5}{6}\)

b. \(\frac{6}{8}\)

Answer:
6/8 = 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8.

Explanation:
In the above-given question,
given that,
draw and shade a tape diagram.
6/8 = 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8.
6/8 = 1 /8 + 2/8 + 2/8 + 1/8.
6/8 = 1/8 + 2/8 + 3/8.

c. \(\frac{7}{10}\)

Answer:
7/10 = 1/10 + 2/10 + 2/10 + 1/10 + 2/10 + 2/10.

Explanation:
In the above-given question,
given that,
draw and shade a tape diagram.
7/10 = 1/10 + 2/10 + 2/10 + 1/10 + 2/10 + 2/10.
7/10 = 1 /10 + 2/10 + 4/10 + 3/10.
7/10 = 1/10 + 2/10 + 7/10.

Question 2.
Step 1: Draw and shade a tape diagram of the given fraction.
Step 2: Record the decomposition of the fraction in three different ways using number sentences.
a. \(\frac{10}{12}\)

Answer:
10/12 = 1/12 + 2/12 + 3/12 + 1/12 + 2/12 + 1/12.

Explanation:
In the above-given question,
given that,
draw and shade a tape diagram.
10/12 = 1/12 + 2/12 + 3/12 + 1/12 + 2/12 + 1/12.
10/12 = 1 /12 + 2/12 + 4/12 + 3/12.
10/12 = 1/12 + 2/12 + 7/12.

b. \(\frac{5}{4}\)

Answer:
5/4 = 1/4 + 2/4 + 1/4 + 1/4.

Explanation:
In the above-given question,
given that,
draw and shade a tape diagram.
5/4 = 1/4 + 2/4 + 1/4 + 1/4.
5/4 = 1 /4 + 24 + 1/4 + 1/4.
5/4 = 1/4 + 2/4 + 1/4.

c. \(\frac{6}{5}\)

Answer:
5/6 = 1/6 + 2/6 + 1/6 + 1/6.

Explanation:
In the above-given question,
given that,
draw and shade a tape diagram.
5/6 = 1/6 + 2/6 + 1/6 + 1/6.
5/6 = 1 /6 + 2/6+ 1/6 + 1/6.
5/6 = 1/6 + 2/6 + 2/6.

d. 1\(\frac{1}{4}\)

Answer:
5/4 = 1/4 + 1/4 + 1/4 + 1/4 + 1/4.

Explanation:
In the above-given question,
given that,
draw and shade a tape diagram.
5/4 = 1/4 + 1/4 + 1/4 + 1/4 + 1/4.
5/4 = 1 /4 + 2/4 + 1/4 + 1/4.
5/4 = 1/4 + 2/4 + 2/4.

Students of Grade 4 can get a strong foundation on mathematics concepts by referring to the Eureka Math Book. It was developed by highly professional mathematics educators and the solutions prepared by them are in a concise manner for easy grasping. To achieve high scores in Grade 4, students need to solve all questions and exercises included in Eureka’s Math Grade 4 Book.

Engage NY Eureka Math 4th Grade Module 6 Lesson 15 Answer Key

Eureka Math Answer Key for Grade 4 aids teachers to differentiate instruction, building, and reinforcing foundational mathematics skills that alter from the classroom to real life.
With the help of the Eureka Primary School Grade 4 Answer Key, You can think deeply regarding what you are learning, and you will really learn math easily just like that. So teachers and students can find this Eureka Answer Key for Grade 4 more helpful in raising students’ scores.

Eureka Math Grade 4 Module 6 Lesson 15 Problem Set Answer Key

Question 1.
100 pennies = $___.______ 100₵ = \(\frac{}{100}\) dollar
Answer:

Question 2.
1 penny = $___.______ 1₵ = \(\frac{}{100}\) dollar
Answer:

Question 3.
6 pennies = $___.______ 6₵ = \(\frac{}{100}\) dollar
Answer:

Question 4.
10 pennies = $___.______ 10₵ = \(\frac{}{100}\) dollar
Answer:

Question 5.
26 pennies = $___.______ 26₵ = \(\frac{}{100}\) dollar
Answer:

Question 6.
10 dimes = $___.______ 100₵ = \(\frac{}{10}\) dollar
Answer:

Question 7.
1 dime = $___.______ 10₵ = \(\frac{}{10}\) dollar
Answer:

Question 8.
3 dimes = $___.______ 30₵ = \(\frac{}{10}\) dollar
Answer:

Question 9.
5 dimes = $___.______ 50₵ = \(\frac{}{10}\) dollar
Answer:

Question 10.
6 dimes = $___.______ 60₵ = \(\frac{}{10}\) dollar
Answer:

Question 11.
4 quarters = $___.______ 100₵ = \(\frac{}{100}\) dollar
Answer:

Question 12.
1 quarter = $___.______ 25₵ = \(\frac{}{100}\) dollar
Answer:

Question 13.
2 quarters = $___.______ 50₵ = \(\frac{}{100}\) dollar
Answer:

Question 14.
3 quarters = $___.______ 75₵ = \(\frac{}{100}\) dollar
Answer:

Solve. Give the total amount of money in fraction and decimal form.
Question 15.
3 dimes and 8 pennies
Answer:

Question 16.
8 dimes and 23 pennies
Answer:

Question 17.
3 quarters 3 dimes and 5 pennies
Answer:

Question 18.
236 cents is what fraction of a dollar?
Answer:

Solve. Express the answer as a decimal.
Question 19.
2 dollars 17 pennies + 4 dollars 2 quarters
Answer:

Question 20.
3 dollars 8 dimes + 1 dollar 2 quarters 5 pennies
Answer:

Question 21.
9 dollars 9 dimes + 4 dollars 3 quarters 16 pennies
Answer:

Eureka Math Grade 4 Module 6 Lesson 15 Exit Ticket Answer Key

Solve. Give the total amount of money in fraction and decimal form.
Question 1.
2 quarters and 3 dimes
Answer:

Question 2.
1 quarter 7 dimes and 23 pennies
Answer:

Solve. Express the answer as a decimal.
Question 3.
2 dollars 1 quarter 14 pennies + 3 dollars 2 quarters 3 dimes
Answer:

Eureka Math Grade 4 Module 6 Lesson 15 Homework Answer Key

Question 1.
100 pennies = $___.______ 100₵ = \(\frac{}{100}\) dollar
Answer:

Question 2.
100 pennies = $___.______ 100₵ = \(\frac{}{100}\) dollar
Answer:

Question 3.
3 pennies = $___.______ 3₵ = \(\frac{}{100}\) dollar
Answer:

Question 4.
20 pennies = $___.______ 20₵ = \(\frac{}{100}\) dollar
Answer:

Question 5.
37 pennies = $___.______ 37₵ = \(\frac{}{100}\) dollar
Answer:

Question 6.
10 dimes = $___.______ 100₵ = \(\frac{}{10}\) dollar
Answer:

Question 7.
2 dimes = $___.______ 20₵ = \(\frac{}{10}\) dollar
Answer:

Question 8.
4 dimes = $___.______ 40₵ = \(\frac{}{10}\) dollar
Answer:

Question 9.
6 dimes = $___.______ 60₵ = \(\frac{}{10}\) dollar
Answer:

Question 10.
9 dimes = $___.______ 90₵ = \(\frac{}{10}\) dollar
Answer:

Question 11.
3 quarters = $___.______ 75₵ = \(\frac{}{100}\) dollar
Answer:

Question 12.
2 quarters = $___.______ 50₵ = \(\frac{}{100}\) dollar
Answer:

Question 13.
4 quarters = $___.______ 100₵ = \(\frac{}{100}\) dollar
Answer:

Question 14.
1 quarter = $___.______ 25₵ = \(\frac{}{100}\) dollar
Answer:

Solve. Give the total amount of money in fraction and decimal form.
Question 15.
5 dimes and 8 pennies
Answer:

Question 16.
3 quarters and 13 pennies
Answer:

Question 17.
3 quarters 7 dimes and 16 pennies
Answer:

Question 18.
187 cents is what fraction of a dollar?
Answer:

Solve. Express the answer in decimal form.
Question 19.
1 dollar 2 dimes 13 pennies + 2 dollars 3 quarters
Answer:

Question 20.
2 dollars 6 dimes + 2 dollars 2 quarters 16 pennies
Answer:

Question 21.
8 dollars 8 dimes + 7 dollars 1 quarter 8 dimes
Answer:

Engage NY Eureka Math 4th Grade Module 3 Lesson 19 Answer Key

Eureka Math Grade 4 Module 3 Lesson 19 Sprint Answer Key

Mental Division

Question 1.
20 ÷ 2 =
Answer:
20 ÷ 2 = 10,

Explanation :
Given expression 20 ÷ 2  when 20 is divided by 2
we get 10, So quotient is 10 and remainder is 0 and
(2 X 10 = 20).

Question 2.
4 ÷ 2 =
Answer:
4 ÷ 2 = 2,

Explanation :
Given expression 4 ÷ 2  when 4 is divided by 2
we get 2, So quotient is 2 and remainder is 0 and
(2 X 2 = 4).

Question 3.
24 ÷ 2 =
Answer:
24 ÷ 2 = 12,

Explanation :
Given expression 24 ÷ 2  when 24 is divided by 2
we get 12, So quotient is 12 and remainder is 0 and
(2 X 12 = 24).

Question 4.
30 ÷ 3 =
Answer:
30 ÷ 3 = 10,

Explanation :
Given expression 30 ÷ 3  when 30 is divided by 3
we get 10, So quotient is 10 and remainder is 0 and
(3 X 10 = 30).

Question 5.
6 ÷ 3 =
Answer:
6 ÷ 3 = 2,

Explanation :
Given expression 6 ÷ 3  when 6 is divided by 3
we get 2, So quotient is 2 and remainder is 0 and
(3 X 2 = 6).

Question 6.
36 ÷ 3 =
Answer:
36 ÷ 3 = 12,

Explanation :
Given expression 36 ÷ 3  when 6 is divided by 3
we get 12, So quotient is 12 and remainder is 0 and
(3 X 12 = 36).

Question 7.
40 ÷ 4 =
Answer:
40 ÷ 4 = 10,

Explanation :
Given expression 40 ÷ 4  when 40 is divided by 4
we get 10, So quotient is 10 and remainder is 0 and
(4 X 10 = 40).

Question 8.
8 ÷ 4 =
Answer:
8 ÷ 4 = 2,

Explanation :
Given expression 40 ÷ 4  when 40 is divided by 4
we get 10, So quotient is 10 and remainder is 0 and
(4 X 10 = 40).

Question 9.
48 ÷ 4 =
Answer:
48 ÷ 4 = 12,

Explanation :
Given expression 48 ÷ 4  when 48 is divided by 4
we get 12, So quotient is 12 and remainder is 0 and
(4 X 12 = 46).

Question 10.
2 ÷ 2 =
Answer:
2 ÷ 2 = 1,

Explanation :
Given expression 2 ÷ 2  when 2 is divided by 2
we get 1, So quotient is 1 and remainder is 0 and
(2 X 1 = 2).

Question 11.
40 ÷ 2 =
Answer:
40 ÷ 2 = 20,

Explanation :
Given expression 40 ÷ 2  when 40 is divided by 2
we get 20, So quotient is 20 and remainder is 0 and
(2 X 20 = 40).

Question 12.
42 ÷ 2 =
Answer:
42 ÷ 2 = 21,

Explanation :
Given expression 42 ÷ 2  when 42 is divided by 2
we get 21, So quotient is 21 and remainder is 0 and
(2 X 21 = 42).

Question 13.
3 ÷ 3 =
Answer:
3 ÷ 3 = 1,

Explanation :
Given expression 3 ÷ 3 when 1 is divided by 3
we get 3, So quotient is 1 and remainder is 0 and
(3 X 1 = 3).

Question 14.
60 ÷ 3 =
Answer:
60 ÷ 3 =20,

Explanation :
Given expression 60 ÷ 3 when 60 is divided by 3
we get 20, So quotient is 20 and remainder is 0 and
(3 X 20 = 60).

Question 15.
63 ÷ 3 =
Answer:
63 ÷ 3 = 21,

Explanation :
Given expression 63 ÷ 3 when 63 is divided by 3
we get 21, So quotient is 21 and remainder is 0 and
(3 X 21 = 63).

Question 16.
4 ÷ 4 =
Answer:
4 ÷ 4 = 1,

Explanation :
Given expression 4 ÷ 4 when 4 is divided by 4
we get 1, So quotient is 1 and remainder is 0 and
(4 X 1 = 4).

Question 17.
80 ÷ 4 =
Answer:
80 ÷ 4 = 20,

Explanation :
Given expression 80 ÷ 4 when 80 is divided by 4
we get 20, So quotient is 20 and remainder is 0 and
(4 X 20 = 80).

Question 18.
84 ÷ 4 =
Answer:
84 ÷ 4 = 21,

Explanation :
Given expression 84 ÷ 4 when 84 is divided by 4
we get 21, So quotient is 21 and remainder is 0 and
(4 X 21 = 84).

Question 19.
40 ÷ 5 =
Answer:
40 ÷ 5 = 8,

Explanation :
Given expression 40 ÷ 5 when 40 is divided by 5
we get 8, So quotient is 8 and remainder is 0 and
(5 X 8 = 40).

Question 20.
50 ÷ 5 =
Answer:
50 ÷ 5 = 10,

Explanation :
Given expression 50 ÷ 5 when 50 is divided by 5
we get 10, So quotient is 10 and remainder is 0 and
(5 X 10 = 50).

Question 21.
60 ÷ 5 =
Answer:
60 ÷ 5 = 12,

Explanation :
Given expression 60 ÷ 5 when 60 is divided by 5
we get 12, So quotient is 12 and remainder is 0 and
(5 X 12 = 60).

Question 22.
70 ÷ 5 =
Answer:
70 ÷ 5 = 14,

Explanation :
Given expression 70 ÷ 5 when 70 is divided by 5
we get 14, So quotient is 14 and remainder is 0 and
(5 X 14 = 70).

Question 23.
68 ÷ 2 =
Answer:
68 ÷ 2 = 34,

Explanation :
Given expression 68 ÷ 2 when 68 is divided by 2
we get 34, So quotient is 34 and remainder is 0 and
(2 X 34 = 68).

Question 24.
96 ÷ 3 =
Answer:
96 ÷ 3 = 32,

Explanation :
Given expression 96 ÷ 3 when 96 is divided by 3
we get 32, So quotient is 32 and remainder is 0 and
(3 X 32 = 96).

Question 25.
86 ÷ 2 =
Answer:
86 ÷ 2 = 43,

Explanation :
Given expression 86 ÷ 2 when 86 is divided by 2
we get 43, So quotient is 43 and remainder is 0 and
(2 X 43 = 86).

Question 26.
93 ÷ 3 =
Answer:
93 ÷ 3 = 31,

Explanation :
Given expression 93 ÷ 3 when 93 is divided by 3
we get 31, So quotient is 31 and remainder is 0 and
(3 X 31 = 93).

Question 27.
88 ÷ 4 =
Answer:
88 ÷ 4 = 22,

Explanation :
Given expression 88 ÷ 4 when 88 is divided by 4
we get 22, So quotient is 22 and remainder is 0 and
(4 X 22 = 88).

Question 28.
99 ÷ 3 =
Answer:
99 ÷ 3 = 33,

Explanation :
Given expression 99 ÷ 3 when 99 is divided by 3
we get 33, So quotient is 33 and remainder is 0 and
(3 X 33 = 99).

Question 29.
66 ÷ 3 =
Answer:
66 ÷ 3 = 22,

Explanation :
Given expression 66 ÷ 3 when 66 is divided by 3
we get 22, So quotient is 22 and remainder is 0 and
(22 X 3 = 66).

Question 30.
66 ÷ 2 =
Answer:
66 ÷ 2 = 33,

Explanation :
Given expression 66 ÷ 2 when 66 is divided by 2
we get 33, So quotient is 33 and remainder is 0 and
(33 X 2 = 66).

Question 31.
40 ÷ 4 =
Answer:
40 ÷ 4 = 10,

Explanation :
Given expression 40 ÷ 4 when 40 is divided by 4
we get 10, So quotient is 10 and remainder is 0 and
(4 X 10 = 40).

Question 32.
80 ÷ 4 =
Answer:
80 ÷ 4 = 20,

Explanation :
Given expression 80 ÷ 4 when 80 is divided by 4
we get 20, So quotient is 20 and remainder is 0 and
(4 X 20 = 80).

Question 33.
60 ÷ 4 =
Answer:
60 ÷ 4 = 15,

Explanation :
Given expression 60 ÷ 4 when 60 is divided by 4
we get 15, So quotient is 15 and remainder is 0 and
(4 X 15 = 60).

Question 34.
68 ÷ 4 =
Answer:
68 ÷ 4 = 17,

Explanation :
Given expression 68 ÷ 4 when 68 is divided by 4
we get 17, So quotient is 17 and remainder is 0 and
(4 X 17 = 68).

Question 35.
20 ÷ 2 =
Answer:
20 ÷ 2 = 10,

Explanation :
Given expression 20 ÷ 2 when 20 is divided by 2
we get 10, So quotient is 10 and remainder is 0 and
(2 X 10 = 20).

Question 36.
40 ÷ 2 =
Answer:
40 ÷ 2 = 20,

Explanation :
Given expression 40 ÷ 2 when 40 is divided by 2
we get 20, So quotient is 20 and remainder is 0 and
(2 X 20 = 40).

Question 37.
30 ÷ 2 =
Answer:
30 ÷ 2 = 15,

Explanation :
Given expression 30 ÷ 2 when 30 is divided by 2
we get 15, So quotient is 15 and remainder is 0 and
(2 X 15 = 30).

Question 38.
36 ÷ 2 =
Answer:
36 ÷ 2 = 18,

Explanation :
Given expression 36 ÷ 2 when 36 is divided by 2
we get 18, So quotient is 18 and remainder is 0 and
(2 X 18 = 36).

Question 39.
30 ÷ 3 =
Answer:
30 ÷ 3 = 10,

Explanation :
Given expression 30 ÷ 3 when 30 is divided by 3
we get 10, So quotient is 10 and remainder is 0 and
(3 X 10 = 30).

Question 40.
39 ÷ 3 =
Answer:
39 ÷ 3 = 13,

Explanation :
Given expression 39 ÷ 3 when 39 is divided by 3
we get 13, So quotient is 13 and remainder is 0 and
(3 X 13 = 39).

Question 41.
45 ÷ 3 =
Answer:
45 ÷ 3 = 15,

Explanation :
Given expression 45 ÷ 3 when 45 is divided by 3
we get 15, So quotient is 15 and remainder is 0 and
(3 X 15 = 45).

Question 42.
60 ÷ 3 =
Answer:
60 ÷ 3 = 20,

Explanation :
Given expression 60 ÷ 3 when 60 is divided by 3
we get 20, So quotient is 20 and remainder is 0 and
(3 X 20 = 60).

Question 43.
57 ÷ 3 =
Answer:
57 ÷ 3 = 19,

Explanation :
Given expression 57 ÷ 3 when 57 is divided by 3
we get 19, So quotient is 19 and remainder is 0 and
(3 X 19 = 57).

Question 44.
51 ÷ 3 =
Answer:
51 ÷ 3 = 17,

Explanation :
Given expression 51 ÷ 3 when 51 is divided by 3
we get 17, So quotient is 17 and remainder is 0 and
(3 X 17 = 51).

Mental Division

Question 1.
30 ÷ 3 =
Answer:
30 ÷ 3 = 10,

Explanation :
Given expression 30 ÷ 3 when 30 is divided by 3
we get 10, So quotient is 10 and remainder is 0 and
(3 X 10 = 30).

Question 2.
9 ÷ 3 =
Answer:
9 ÷ 3 = 3,

Explanation :
Given expression 9 ÷ 3 when 9 is divided by 3
we get 3, So quotient is 3 and remainder is 0 and
(3 X 3 = 9).

Question 3.
39 ÷ 3 =
Answer:
39 ÷ 3 = 13,

Explanation :
Given expression 39 ÷ 3 when 39 is divided by 3
we get 13, So quotient is 13 and remainder is 0 and
(3 X 13 = 39).

Question 4.
20 ÷ 2 =
Answer:
20 ÷ 2 = 10,

Explanation :
Given expression 20 ÷ 2 when 20 is divided by 2
we get 10, So quotient is 10 and remainder is 0 and
(2 X 10 = 20).

Question 5.
6 ÷ 2 =
Answer:
6 ÷ 2 = 3,

Explanation :
Given expression 6 ÷ 2 when 6 is divided by 2
we get 3, So quotient is 3 and remainder is 0 and
(2 X 3 = 6).

Question 6.
26 ÷ 2 =
Answer:
26 ÷ 2 = 13,

Explanation :
Given expression 26 ÷ 2 when 26 is divided by 2
we get 13, So quotient is 13 and remainder is 0 and
(2 X 13 = 26).

Question 7.
80 ÷ 4 =
Answer:
80 ÷ 4 = 20,

Explanation :
Given expression 80 ÷ 4 when 80 is divided by 4
we get 20, So quotient is 20 and remainder is 0 and
(4 X 20 = 80).

Question 8.
4 ÷ 4 =
Answer:
4 ÷ 4 = 1,

Explanation :
Given expression 4 ÷ 4 when 4 is divided by 4
we get 1, So quotient is 4 and remainder is 0 and
(4 X 1 = 4).

Question 9.
84 ÷ 4 =
Answer:
84 ÷ 4 = 21,

Explanation :
Given expression 84 ÷ 4 when 84 is divided by 4
we get 21, So quotient is 21 and remainder is 0 and
(4 X 21 = 84).

Question 10.
2 ÷ 2 =
Answer:
2 ÷ 2 = 1,

Explanation :
Given expression 2 ÷ 2 when 2 is divided by 2
we get 1, So quotient is 1 and remainder is 0 and
(2 X 1 = 2).

Question 11.
60 ÷ 2 =
Answer:
60 ÷ 2 = 30,

Explanation :
Given expression 60 ÷ 2 when 60 is divided by 2
we get 30, So quotient is 30 and remainder is 0 and
(2 X 30 = 60).

Question 12.
62 ÷ 2 =
Answer:
62 ÷ 2 = 31,

Explanation :
Given expression 62 ÷ 2 when 62 is divided by 2
we get 31, So quotient is 31 and remainder is 0 and
(2 X 31 = 62).

Question 13.
3 ÷ 3 =
Answer:
3 ÷ 3 = 1,

Explanation :
Given expression 3 ÷ 3 when 3 is divided by 3
we get 1, So quotient is 1 and remainder is 0 and
(3 X 1 = 3).

Question 14.
90 ÷ 3 =
Answer:
90 ÷ 3 = 30,

Explanation :
Given expression 90 ÷ 3 when 90 is divided by 3
we get 30, So quotient is 30 and remainder is 0 and
(3 X 30 = 90).

Question 15.
93 ÷ 3 =
Answer:
93 ÷ 3 = 31,

Explanation :
Given expression 93 ÷ 3 when 93 is divided by 3
we get 31, So quotient is 31 and remainder is 0 and
(3 X 31 = 93).

Question 16.
8 ÷ 4 =
Answer:
8 ÷ 4 = 2,

Explanation :
Given expression 8 ÷ 4 when 8 is divided by 4
we get 2, So quotient is 2 and remainder is 0 and
(4 X 2 = 8).

Question 17.
40 ÷ 4 =
Answer:
40 ÷ 4 = 10,

Explanation :
Given expression 40 ÷ 4 when 40 is divided by 4
we get 10, So quotient is 10 and remainder is 0 and
(4 X 10 = 40).

Question 18.
48 ÷ 4 =
Answer:
48 ÷ 4 = 12,

Explanation :
Given expression 48 ÷ 4 when 48 is divided by 4
we get 12, So quotient is 12 and remainder is 0 and
(4 X 12 = 48).

Question 19.
50 ÷ 5 =
Answer:
50 ÷ 5 = 10,

Explanation :
Given expression 50 ÷ 5 when 50 is divided by 5
we get 10, So quotient is 10 and remainder is 0 and
(5 X 10 = 50).

Question 20.
60 ÷ 5 =
Answer:
60 ÷ 5 = 12,

Explanation :
Given expression 60 ÷ 5 when 60 is divided by 5
we get 12, So quotient is  and remainder is 0 and
(5 X 12 = 60).

Question 21.
70 ÷ 5 =
Answer:
70 ÷ 5 = 14,

Explanation :
Given expression 70 ÷ 5 when 70 is divided by 5
we get 14, So quotient is 14 and remainder is 0 and
(5 X 14 = 70).

Question 22.
80 ÷ 5 =
Answer:
80 ÷ 5 = 16,

Explanation :
Given expression 80 ÷ 5 when 80 is divided by 5
we get 16, So quotient is 16 and remainder is 0 and
(5 X 16 = 80).

Question 23.
86 ÷ 2 =
Answer:
86 ÷ 2 = 43,

Explanation :
Given expression 86 ÷ 2 when 86 is divided by 2
we get 43, So quotient is 43 and remainder is 0 and
(2 X 43 = 86).

Question 24.
69 ÷ 3 =
Answer:
69 ÷ 3 = 23,

Explanation :
Given expression 69 ÷ 3 when 69 is divided by 3
we get 23, So quotient is 23 and remainder is 0 and
(3 X 23 = 69).

Question 25.
68 ÷ 2 =
Answer:
68 ÷ 2 = 34,

Explanation :
Given expression 68 ÷ 2 when 68 is divided by 2
we get 34, So quotient is 34 and remainder is 0 and
(2 X 34 = 68).

Question 26.
96 ÷ 3 =
Answer:
96 ÷ 3 = 32,

Explanation :
Given expression 96 ÷ 3 when 96 is divided by 3
we get 32, So quotient is 32 and remainder is 0 and
(3 X 32 = 96).

Question 27.
66 ÷ 3 =
Answer:
66 ÷ 3 = 22,

Explanation :
Given expression 66 ÷ 3 when 66 is divided by 3
we get 22, So quotient is 22 and remainder is 0 and
(3 X 22 = 66).

Question 28.
99 ÷ 3 =
Answer:
99 ÷ 3 = 33,

Explanation :
Given expression 99 ÷ 3 when 99 is divided by 3
we get 33, So quotient is 33 and remainder is 0 and
(3 X 33 = 99).

Question 29.
88 ÷ 4 =
Answer:
88 ÷ 4 = 22,

Explanation :
Given expression 88 ÷ 4 when 88 is divided by 4
we get 22, So quotient is 22 and remainder is 0 and
(4 X 22 = 88).

Question 30.
88 ÷ 2 =
Answer:
88 ÷ 2 = 44,

Explanation :
Given expression 88 ÷ 2 when 88 is divided by 2
we get 44, So quotient is 44 and remainder is 0 and
(2 X 44 = 88).

Question 31.
40 ÷ 4 =
Answer:
40 ÷ 4 = 10,

Explanation :
Given expression 40 ÷ 10 when 40 is divided by 4
we get 10, So quotient is 10 and remainder is 0 and
(4 X 10 = 40).

Question 32.
80 ÷ 4 =
Answer:
80 ÷ 4 = 20,

Explanation :
Given expression 80 ÷ 4 when 80 is divided by 4
we get 20, So quotient is 20 and remainder is 0 and
(4 X 20 = 80).

Question 33.
60 ÷ 4 =
Answer:
60 ÷ 4 = 15,

Explanation :
Given expression 60 ÷ 4 when 60 is divided by 4
we get 15, So quotient is 15 and remainder is 0 and
(4 X 15 = 60).

Question 34.
64 ÷ 4 =
Answer:
64 ÷ 4 = 16,

Explanation :
Given expression 64 ÷ 4 when 64 is divided by 4
we get 16, So quotient is 16 and remainder is 0 and
(4 X 16 = 64).

Question 35.
20 ÷ 2 =
Answer:
20 ÷ 2 = 10,

Explanation :
Given expression 20 ÷ 2 when 20 is divided by 2
we get 10, So quotient is 10 and remainder is 0 and
(2 X 10 = 20).

Question 36.
40 ÷ 2 =
Answer:
40 ÷ 2 = 20,

Explanation :
Given expression 40 ÷ 2 when 40 is divided by 2
we get 20, So quotient is 20 and remainder is 0 and
(2 X 20 = 40).

Question 37.
30 ÷ 2 =
Answer:
30 ÷ 2 = 15,

Explanation :
Given expression 30 ÷ 2 when 30 is divided by 2
we get 15, So quotient is 15 and remainder is 0 and
(2 X 15 = 30).

Question 38.
38 ÷ 2 =
Answer:
38 ÷ 2 = 19,

Explanation :
Given expression 38 ÷ 2 when 38 is divided by 2
we get 19, So quotient is 19 and remainder is 0 and
(2 X 19 = 38).

Question 39.
30 ÷ 3 =
Answer:
30 ÷ 3 = 10,

Explanation :
Given expression 30 ÷ 3 when 30 is divided by 3
we get 10, So quotient is 10 and remainder is 0 and
(3 X 10 = 30).

Question 40.
36 ÷ 3 =
Answer:
36 ÷ 3 = 12,

Explanation :
Given expression 36 ÷ 3 when 36 is divided by 3
we get 12, So quotient is 12 and remainder is 0 and
(3 X 12 = 36).

Question 41.
42 ÷ 3 =
Answer:
42 ÷ 3 = 14,

Explanation :
Given expression 42 ÷ 3 when 42 is divided by 3
we get 14, So quotient is 14 and remainder is 0 and
(3 X 14 = 42).

Question 42.
60 ÷ 3 =
Answer:
60 ÷ 3 = 20,

Explanation :
Given expression 60 ÷ 3 when 60 is divided by 3
we get 20, So quotient is 20 and remainder is 0 and
(3 X 20 = 60).

Question 43.
54 ÷ 3 =
Answer:
54 ÷ 3 = 18,

Explanation :
Given expression 54 ÷ 3 when 54 is divided by 3
we get 18, So quotient is 18 and remainder is 0 and
(3 X 18 = 54).

Question 44.
48 ÷ 3 =
Answer:
48 ÷ 3 = 16,

Explanation :
Given expression 48 ÷ 3 when 48 is divided by 3
we get 16, So quotient is 16 and remainder is 0 and
(3 X 16 = 48).

Eureka Math Grade 4 Module 3 Lesson 19 Problem Set Answer Key

Question 1.
When you divide 94 by 3, there is a remainder of 1.
Model this problem with place value disks. In the place
value disk model, how did you show the remainder?
Answer:

I showed my remainder by circling the remaining one
in the ones place.

Explanation:
When we divide 94 by 3, there is a remainder of 1.
Modeled the problem with place value disks. In the place
value disk model, showed the remainder by circling the
remaining one in the ones place.

Question 2.
Cayman says that 94 ÷ 3 is 30 with a remainder of 4.
He reasons this is correct because (3 × 30) + 4 = 94.
What mistake has Cayman made? Explain how he can
correct his work.
Answer:
Cayman mistake is that his remainder is greater than
his divisor means he can divide more.
Instead of 30 groups he can make 31 groups.

Explanation:
Given Cayman says that 94 ÷ 3 is 30 with a remainder of 4.
He reasons this is correct because (3 × 30) + 4 = 94.
Instead of 30 groups he can make 31 groups,
94 ÷ 3 when 94 is divided by 3
we get 31 as quotient and remainder is 1.

Question 3.
The place value disk model is showing 72 ÷ 3.
Complete the model. Explain what happens to the
1 ten that is remaining in the tens column.

Answer:
The 1 ten remaining gets decomposed into
10 ones column,

Explanation:
Given the place value disk model is showing 72 ÷ 3.
Completed the model. Explained what happens to the
1 ten that is remaining in the tens column gets decomposed
into10 ones column as shown above.

Question 4.
Two friends evenly share 56 dollars.
a. They have 5 ten-dollar bills and 6 one-dollar bills.
Draw a picture to show how the bills will be shared.
Will they have to make change at any stage?
Answer:
Yes, they will have to make change for 1 dollar bill.

Explanation:
In order to share its the ten dollar bill needs to be decomposed
into one dollar bills.
Drawn a picture to show how the bills will be shared
and have to make changes as shown above.

b. Explain how they share the money evenly.
Answer:
Each friend gets 2 ten dollar bills and 8 one dollar bills,

Explanation:
Given two friends evenly share 56 dollars,
they have 5 ten-dollar bills and 6 one-dollar bills,
they shared the money evenly as 2 ten dollar bills and
8 one dollar bills.

Question 5.
Imagine you are filming a video explaining the
blem 45 ÷ 3 to new fourth graders. Create a script to
explain how you can keep dividing after getting a
remainder of 1 ten in the first step.
Answer:

Explanation:
Watch as I solve 45 ÷ 3 using a place value chart.
First I divide my tens, Each of the 3 groups can equally
have 1 ten. There is 1 ten remaining, We can continue
dividing by decomposing the 1 ten into 10 ones. Watch as
I show this on my chart. Now I have 15 ones that can be
equally distributed into our 3 groups. Each group will get
5 ones. Now we can see that 45 ÷ 3 is 1 ten 5 ones or 15.

Eureka Math Grade 4 Module 3 Lesson 19 Exit Ticket Answer Key

Question 1.
Molly’s photo album has a total of 97 pictures.
Each page of the album holds 6 pictures.
How many pages can Molly fill? Will there be any
pictures left? If so, how many? Use place value disks to solve.
Answer:
Molly can fill 16 pages,
Yes, there will be 1 picture left,

Explanation:
Given Molly’s photo album has a total of 97 pictures.
Each page of the album holds 6 pictures, So number of
pages can Molly fill are 97 ÷ 6 = 16 quotient and 1 remainder
means Molly can fill 16 pages and there will be 1 picture left
as shown above using place value disks.

Question 2.
Marti’s photo album has a total of 45 pictures.
Each page holds 4 pictures. She said she can only
fill 10 pages completely. Do you agree? Explain why or why not.
Answer:
No, I don’t agree, why because she can fill 11 pages completly
with 1 picture left,

Explanation:
Marti’s photo album has a total of 45 pictures.
Each page holds 4 pictures. She said she can only
fill 10 pages completely, I don’t agree because
if we divide 45 pictures by 4 we get 11 pages but she
said she can only fill 10 pages completely so disagree.

Eureka Math Grade 4 Module 3 Lesson 19 Homework Answer Key

Question 1.
When you divide 86 by 4, there is a remainder of 2.
Model this problem with place value disks. In the
place value disk model, how can you see that there is a remainder?
Answer:
I showed my remainder by circling the remaining two
in the ones place,

Explanation:
When we divide 86 by 4, there is a remainder of 2.
Modeled the problem with place value disks. In the place
value disk model, showed the remainder by circling the
remaining two in the ones place.

Question 2.
Francine says that 86 ÷ 4 is 20 with a remainder of 6.
She reasons this is correct because (4 × 20) + 6 = 86.
What mistake has Francine made? Explain how she can
correct her work.
Answer:
Francine mistake is that her remainder is greater than
his divisor means he can divide more.
Instead of 20 groups he can make 21 groups,

Explanation:
Given Francine says that 86 ÷ 4 is 20 with a remainder of 6.
Her reasons this is correct because (4 X 20) + 6 = 86.
Instead of 20 groups he can make 21 groups,
86 ÷ 4 when 86 is divided by 4
we get 21 as quotient and remainder is 2.

Question 3.
The place value disk model is showing 67 ÷ 4.
Complete the model. Explain what happens to
the 2 tens that are remaining in the tens column.

Answer:
The 2 tens remaining gets decomposed into
20 ones column,

Explanation:
Given the place value disk model is showing 67 ÷ 4.
Completed the model. Explained what happens to the
2 tens that is remaining in the tens column gets decomposed
into 20 ones column as shown above.

Question 4.
Two friends share 76 blueberries.
a. To count the blueberries, they put them into small
bowls of 10 blueberries. Draw a picture to show how the
blueberries can be shared equally. Will they have to split
apart any of the bowls of 10 blueberries when they share them?
Answer:
Yes, they will have to make change for 1 bowl,

Explanation:
In order to share it the 10 blueberries needs to be decomposed
into ones blueberries.
Drawn a picture to show how the blueberries will be shared
and have to make changes as shown above.

b. Explain how the friends can share the blueberries fairly.
Answer:
Each friend gets 3 ten blueberries and 8 blueberries,

Explanation:
Given two friends can share 76 blueberries,
they have 3 ten-blueberries and 8 one-blueberries,
Each friend shared 3 ten blueberries and 8 blueberries.

Question 5.
Imagine you are drawing a comic strip showing how to
solve the problem 72 ÷ 4 to new fourth graders. Create a
script to explain how you can keep dividing after getting a
remainder of 3 tens in the first step.
Answer:

Explanation:
Created a script as I solve 72 ÷ 4 using a place value chart.
First I divide my tens into 4 groups I get 4 tens into 1 group,
There are 3 tens remaining, We can continue
dividing by decomposing the tens into 30 ones. Watch as
I show this on my chart. Now I have 32 ones that can be
equally distributed into our 4 groups. Each group will get
8 ones. Now we can see that 72 ÷ 4 is 1 ten 8 ones or 18.

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 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 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 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.