Eureka Math Grade 3 Module 3 Lesson 3 Answer Key

Teachers and students can find this Eureka Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Eureka Math Answer Key for Grade 3 aids teachers to differentiate instruction, building and reinforcing foundational mathematics skills that alter from the classroom to real life. With the help of Eureka’s primary school Grade 3 Answer Key, you can think deeply regarding what you are learning, and you will really learn math easily just like that.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 3 Answer Key

Teachers and students can find this Eureka Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Eureka Math Answer Key for Grade 3 aids teachers to differentiate instruction, building and reinforcing foundational mathematics skills that alter from the classroom to real life. With the help of Eureka’s primary school Grade 3 Answer Key, you can think deeply regarding what you are learning, and you will really learn math easily just like that.

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

Question 1.
Each equation contains a letter representing the unknown. Find the value of the unknowns, and then write the letters that match the answers to solve the riddle.
Eureka Math Grade 3 Module 3 Lesson 3 Problem Set Answer Key 1

Answer:
e = 20, i = 1/ 6, s = 4, n = 10, k = 6, a = 2.6, l = 7, c = 3, t = 0.7, b = 2, h = 5.

Explanation:
In the above-given question,
given that,
The equation contains the letter representing unknown.
5 x 4 = e , e = 20.
24/ i = 4, i = 4/ 24 = 1/6.
32 = s x 8, s = 32/ 8, s = 4.
8 = 80 / n, n = 10.
8 = a / 3, a = 2.6.
21 / 3 = l, l = 7.
21 = c x 7, c = 3.
t / 10 = 7, t = 0.7.
24/b = 12, b = 24/ 12, b = 2.
35 = 7 x h, h = 5.
Eureka-Math-Grade-3-Module-3-Lesson-3-Answer Key-1

Which tables do you NOT have to learn?
Eureka Math Grade 3 Module 3 Lesson 3 Problem Set Answer Key 2

Answer:
The tables not have to learn is 10, 20, and 70.

Explanation:
In the above-given question,
given that,
The tables are 9, 6, 3, 5, 70, 20, 10, 24, 2, 7, and 4.
out of all the tables 10 and 20 are very easy.
so we need not to learn.

Question 2.
Lonna buys 3 t-shirts for $8 each.
a. What is the total amount Lonna spends on 3 t-shirts? Use the letter m to represent the total amount of money Lonna spends, and then solve the problem.

Answer:
The total amount of money Lonna spends = 24$.

Explanation:
In the above-given question,
given that,
Lonna buys 3 t-shirts for $8 each.
m x 3 = $8.
m = 8 x 3.
m = $24.
so the total amount of money Lonna spends = $24.

b. If Lonna hands the cashier 3 ten dollar bills, how much change will she receive? Use the letter c in an equation to represent the change, and then find the value of c.

Answer:
The change she received = $6.

Explanation:
In the above-given question,
given that,
Lonna hands the cashier 3 ten dollar bills.
c x 30 = 24.
c = 30 – 24.
c = $6.
so the change Lonna received = $6.

Question 3.
Miss Potts used a total of 28 cups of flour to bake some bread. She used 4 cups of flour for each loaf of bread. How many loaves of bread did she bake? Represent the problem using multiplication and division sentences and a letter for the unknown. Then, solve the problem.
__7___ × ___4__ = __28____
__28___ ÷ __4___ = ___7___

Answer:
The number of loaves of bread did she bake = 7.

Explanation:
In the above-given question,
given that,
Miss potts used a total of 28 cups of flour to bake some bread.
she used 4 cups of flour for each loaf of bread.
7 x 4 = 28.
28 / 4 = 7.
so the number of loaves of bread did she bake = 7.

Question 4.
At a table tennis tournament, two games went on for a total of 32 minutes. One game took 12 minutes longer than the other. How long did it take to complete each game? Use letters to represent the unknowns. Solve the problem.
Eureka Math Grade 3 Module 3 Lesson 3 Problem Set Answer Key 3

Answer:
The other game takes to complete = 10 minutes.

Explanation:
In the above-given question,
given that,
At a table tennis tournament, two games went on for a total of 32 minutes.
one game took 12 minutes longer than the other.
32 + t = 12.
t = 32 – 12.
t = 20.
so the other game takes to complete = 10 minutes.

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

Find the value of the unknown in Problems 1 – 4.

Question 1.
z = 5 × 9
z = ___45___

Answer:
z = 45.

Explanation:
In the above-given question,
given that,
the equation z = 5 x 9.
z = 45.

Question 2.
30 ÷ 6 = v
v = ___5___

Answer:
v = 5.

Explanation:
In the above-given question,
given that,
the equation 30/ 6 = v.
v = 5.

Question 3.
8 × w = 24
w = ____3__

Answer:
w = 3.

Explanation:
In the above-given question,
given that,
the equation 8 x w = 24.
w = 24/8.
w = 3.

Question 4.
y ÷ 4 = 7
y = __1.75____

Answer:
y = 1.75.

Explanation:
In the above-given question,
given that,
y / 4 = 7.
y = 7 / 4.
y = 1.75.

Question 5.
Mr. Strand waters his rose bushes for a total of 15 minutes. He waters each rose bush for 3 minutes. How many rose bushes does Mr. Strand water? Represent the problem using multiplication and division sentences and a letter for the unknown. Then, solve the problem.
__3___ × __5___ = __15___
__15___ ÷ __3___ = ___5__

Answer:
The number of rose bushes Mr. strand water = 15.

Explanation:
In the above-given question,
given that,
Mr. Strand waters his rose bushes for a total of 15 minutes.
He waters each rose bush for 3 minutes.
3 x 5 = 15.
15 / 3 = 5.
so the number of rose bushes Mr. Strand water = 15.

Eureka Math Grade 3 Module 3 Lesson 3 Homework Answer Key

Question 1.
a. Complete the pattern.
Eureka Math Grade 3 Module 3 Lesson 3 Homework Answer Key 4

Answer:
The missing numbers in the pattern are 40, 50, 70, 80, and 100.

Explanation:
In the above-given question,
given that,
the pattern with numbers.
the missing numbers are 40, 50, 70, 80, and 100.
Eureka-Math-Grade-3-Module-3-Lesson-3-Answer Key-2

b. Find the value of the unknown.

10 × 2 = d   d = __20___
3 × 10 = e   e = __30___
f = 4 × 10   f = ___40__
p = 5 × 10  p = __50___
10 × 6 = w  w = ___60__
10 × 7 = n  n = __70___
g = 8 × 10  g = __80___

Answer:
d = 20.
e = 30.
f = 40.
p = 50.
w = 60.
n = 70.
g = 80.

Explanation:
In the above-given question,
given that,
the equations 10 x 2 = d, d = 20.
3 x 10 = e, e = 30.
f = 4 x 10, f = 40.
p = 5 x 10, p = 50.
10 x 6 = w, w = 60.
10 x 7 = n, n = 70.
g = 8 x 10, g = 80.

Question 2.
Each equation contains a letter representing the unknown. Find the value of the unknown.

8 ÷ 2 = nn = _4____
3 × a = 12a = __4___
p × 8 = 40p = __5___
18 ÷ 6 = cc = __3___
d × 4= 24d = ___6__
h ÷ 7 = 5h = __1.4___
6 × 3 = ff = ___18__
32 ÷ y = 4y = __8___

Answer:
n = 4.
a = 4.
p = 5.
c = 3.
d = 6.
h = 1.4.
f = 18.
y = 8.

Explanation:
In the above-given question,
given that,
the equations 8 / 2 = n, n = 4.
3 x a = 12, a = 12 / 3, a = 4.
p x 8 = 40, p = 40 / 8, p = 5.
18 / 6 = c, c = 3.
d x 4 = 24, d = 24 / 4, d = 6.
h / 7 = 5, h = 1.4.
6 x 3 = f, f = 18.
32 / y = 4. y = 8.

Question 3.
Pedro buys 4 books at the fair for $7 each.
a. What is the total amount Pedro spends on 4 books? Use the letter b to represent the total amount Pedro spends, and then solve the problem.

Answer:
The total amount Pedro spends on 4 books = $28.

Explanation:
In the above-given question,
given that,
Pedro buys 4 books at the fair for $7 each.
b x 4 = 7.
b = 7 x 4.
b = $28.
so the total amount Pedro spends on 4 books = $28.

b. Pedro hands the cashier 3 ten dollar bills. How much change will he receive? Write an equation to solve. Use the letter c to represent the unknown.

Answer:
The change Pedro receives = $2.

Explanation:
In the above-given question,
given that,
Pedro hands the cashier 3 ten dollar bills.
c + 30 = 28.
c = 30 – 28.
c = $2.

Question 4.
On field day, the first-grade dash is 25 meters long. The third-grade dash is twice the distance of the first-grade dash. How long is the third-grade dash? Use a letter to represent the unknown and solve.

Answer:
The third-grade dash = 75 meters long.

Explanation:
In the above-given question,
given that,
On field day, the first-grade dash is 25 meters long.
The third-grade dash is twice the distance of the first-grade dash.
g = 25 + 50.
g = 75.
so the third-grade dash = 75 meters long.

Eureka Math Grade 3 Module 3 Lesson 4 Answer Key

Students of Grade 3 Module 3 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 3, students need to solve all questions and exercises included in Eureka’s Grade 3 Textbook.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 4 Answer Key

So teachers and students can find this Eureka Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Eureka Math Answer Key for Grade 3 aids teachers to differentiate instruction, building and reinforcing foundational mathematics skills that alter from the classroom to real life. With the help of Eureka’s primary school Grade 3 Answer Key, you can think deeply regarding what you are learning, and you will really learn math easily just like that.

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

Question 1.
Eureka Math Grade 3 Module 3 Lesson 4 Problem Set Answer Key 1

Answer:
The missing numbers are 12, 24, 42, and 54.

Explanation:
In the above-given question,
given that,
the numbers are 6, 18, 30, 36, 48, and 60.
the missing numbers are 12, 24, 42, and 54.
Eureka-Math-Grade-3-Module-3-Lesson-4-Answer Key-1

Question 2.
Count by six to fill in the blanks below.

6, __12_____, ___18____, ___24____

Answer:
The missing numbers are 12, 18, and 24.

Explanation:
In the above-given question,
given that,
the table of 6.
count by six.
so the missing numbers are 12, 18, and 24.

Complete the multiplication equation that represents the final number in your count-by.

6 × ___4____ = ___24____

Answer:
6 x 4 = 24.

Explanation:
In the above-given question,
given that,
the table of 6 x 4 = 24.
so the missing number is 6 x 4 = 24.

Complete the division equation that represents your count-by.

____24___ ÷ 6 = ___4____

Answer:
24 / 6 = 4.

Explanation:
In the above-given question,
given that,
24 / 6 = 4.
6 x 4 = 24.

Question 3.
Count by six to fill in the blanks below.

   6     , ___12___, ___18___, ___24___, ___30___, __36___, __42___

Answer:
The missing numbers are 12, 18, 24, 30, 36, and 42.

Explanation:
In the above-given question,
given that,
the numbers are 6, 12, 18, 24, 30, 36, and 42.
so the missing numbers are 12, 18, 24, 30, 36, and 42.

Complete the multiplication equation that represents the final number in your count-by.

6 × ___7____ = ___42____

Answer:
6 x 7 = 42.

Explanation:
In the above-given question,
given that,
the numbers are 6 x 7 = 42.
6 x 7 = 42.

Complete the division equation that represents your count-by.

___42____ ÷ 6 = ___7____

Answer:
42 / 6 = 7.

Explanation:
In the above-given question,
given that,
42 / 6 = 7.
6 x 7 = 42.

Question 4.
Mrs. Byrne’s class skip-counts by six for a group counting activity. When she points up, they count up by six, and when she points down, they count down by six. The arrows show when she changes direction.

a. Fill in the blanks below to show the group counting answers.
↑ 0, 6, __12___, 18, __24___↓ ___18__, 12 ↑ __18___, 24, 30, __36___ ↓ 30, 24, ___18__ ↑ 24, __30___, 36, __42___, 48

Answer:
The missing numbers are 12, 24, 18, 18, 36, 18, 30, and 42.

Explanation:
In the above-given question,
given that,
The numbers are 0, 6, 12, 18, 24, 18, 12, 18, 24, 30, 36, 30, 24, 18, 24, 30, 36, 42, and 48.
the missing numbers are 12, 24, 18, 36, 18, 30, and 42.

b. Mrs. Byrne says the last number that the class counts is the product of 6 and another number. Write a multiplication sentence and a division sentence to show she’s right.
6 × ____8___ = 48
48 ÷ 6 = ___8____

Answer:
6 x 8 = 48.
48 / 6 = 8.

Explanation:
In the above-given question,
given that,
Mrs. Byme says the last number that the class counts is the product of 6 and another number.
6 x 8 = 48.
48 / 6 = 8.

Question 5.
Julie counts by six to solve 6 × 7. She says the answer is 36. Is she right? Explain your answer.

Answer:
No, she was not correct.
6 x 7 = 42.

Explanation:
In the above-given question,
given that,
Julie counts by six to solve 6 x 7.
she says the answer is 36.
no she was not correct.
6 x 7 = 42.

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

Question 1.
Sylvia solves 6 × 9 by adding 48 + 6. Show how Sylvia breaks apart and bonds her numbers to complete the ten. Then, solve.

Answer:
6 x 9 = 54.
48 + 6 = 54.

Explanation:
In the above-given question,
given that,
Sylvia solves 6 x 9 by adding 48 + 6.
10 + 10 + 10 + 10 + 10 + 4 = 54.
6 x 9 = 54.
48 + 6 = 54.

Question 2.
Skip-count by six to solve the following:
a. 8 × 6 = ___48___

Answer:
8 x 6 = 48.

Explanation:
In the above-given question,
given that,
6 x 8 = 48.
8 x 6 = 48.

b. 54 ÷ 6 = ____9__

Answer:
54 / 6 = 9.

Explanation:
In the above-given question,
given that,
6 x 9 = 54.
54 / 6 = 9.

Eureka Math Grade 3 Module 3 Lesson 4 Homework Answer Key

Question 1.
Use number bonds to help you skip-count by six by either making a ten or adding to the ones.
Eureka Math Grade 3 Module 3 Lesson 4 Problem Set Answer Key 2

Answer:
10 + 8 = 18, 30 + 6 = 36, 10 + 20 = 30, 30 + 12 = 42, 40 + 8 = 48, 40 + 14 = 54, 50 + 10 = 60.

Explanation:
In the above-given question,
given that,
the numbers are 10 + 8 = 18.
30 + 6 = 36.
10 + 20 = 30.
30 + 12 = 42.
40 + 8 = 48.
40 + 14 = 54.
50 + 10 = 60.
Eureka-Math-Grade-3-Module-3-Lesson-4-Answer Key-2

Question 2.
Count by six to fill in the blanks below.
6, ___12____, ___18____, ___24____, ___30____

Answer:
The missing numbers are 6, 12, 18, 24, and 30.

Explanation:
In the above-given question,
given that,
the tables of 6.
6, 12, 18, 24, and 30.
so the missing numbers are 6, 12, 18, 24, and 30.

Complete the multiplication equation that represents the final number in your count-by.

6 × ___5____ = ___30____

Answer:
6 x 5 = 30.

Explanation:
In the above-given question,
given that,
the tables of 6.
6 x 5 = 30.

Complete the division equation that represents your count-by.

__30_____ ÷ 6 = ___5____

Answer:
30 / 6 = 5.

Explanation:
In the above-given question,
given that,
6 x 5 = 30.
30 / 6 = 5.

Question 3.
Count by six to fill in the blanks below.
6, __12____, __18____, ___24___, __30____, __36____

Answer:
The missing numbers are 12, 18, 24, 30, and 36.

Explanation:
In the above-given question,
given that,
the numbers are 6, 12, 18, 24, 30, and 36.
so the missing numbers are 12, 18, 24, 30, and 36.

Complete the multiplication equation that represents the final number in your count-by.

6 × ___6____ = ___36____

Answer:
6 x 6 = 36.

Explanation:
In the above-given question,
given that,
the table of 6.
6 x 6 = 36.

Complete the division equation that represents your count-by.

____36___ ÷ 6 = ___6____

Answer:
36 / 6 = 6.

Explanation:
In the above-given question,
given that,
6 x 6 = 36.
36 / 6 = 6.

Question 4.
Count by six to solve 48 ÷ 6. Show your work below.

Answer:
48 / 6 = 8.

Explanation:
In the above-given question,
given that,
6 x 8 = 48.
48 / 6 = 8.

 

Eureka Math Grade 3 Module 3 Lesson 5 Answer Key

Teachers and students can find this Eureka Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Eureka Math Answer Key for Grade 3 aids teachers to differentiate instruction, building and reinforcing foundational mathematics skills that alter from the classroom to real life. With the help of Eureka’s primary school Grade 3 Answer Key, you can think deeply regarding what you are learning, and you will really learn math easily just like that.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 5 Answer Key

Students of Grade 3 Module 3 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 3, students need to solve all questions and exercises included in Eureka’s Grade 3 Textbook.

Eureka Math Grade 3 Module 3 Lesson 5 Pattern Sheet Answer Key

Multiply.

Eureka Math Grade 3 Module 3 Lesson 5 Pattern Sheet Answer Key 1

Answer:
6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24, 6 x 5 = 30, 6 x 1 = 6, 6 x 2 = 12, 6 x 1 = 6, 6 x 3 = 18, 6 x 1 = 6, 6 x 4 = 24, 6 x 1 = 6, 6 x 5 = 30, 6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 2 = 12, 6 x 4 = 24, 6 x 2 = 12, 6 x 5 = 30, 6 x 2 = 12, 6 x 1 = 6,  6 x 2 = 12, 6 x 3 = 18, 6 x 1 = 6, 6 x 3 = 18, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24, 6 x 3 = 18, 6 x 5 = 30, 6 x 3 = 18, 6 x 4 = 24, 6 x 1 = 6, 6 x 4 = 24, 6 x 2 = 12, 6 x 4 = 24, 6 x 3 = 18, 6 x 4 = 24, 6 x 5 = 30, 6 x 4 = 24, 6 x 5 = 30, 6 x 1 = 6, 6 x 5 = 30, 6 x 2 = 12, 6 x 5 = 30, 6 x 3 = 18, 6 x 5 = 30, 6 x 4 = 24, 6 x 2 = 12, 6 x 4 = 24, 6 x 3 = 18, 6 x 5 = 30, 6 x 3 = 18, 6 x 2 = 12, 6 x 4 = 24, 6 x 3 = 18, 6 x 5 = 30, 6 x 2 = 12, 6 x 4 = 24.

Explanation:
In the above-given question,
given that,
the table of 6.
6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24, 6 x 5 = 30, 6 x 1 = 6, 6 x 2 = 12, 6 x 1 = 6, 6 x 3 = 18, 6 x 1 = 6, 6 x 4 = 24, 6 x 1 = 6, 6 x 5 = 30, 6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 2 = 12, 6 x 4 = 24, 6 x 2 = 12, 6 x 5 = 30, 6 x 2 = 12, 6 x 1 = 6,  6 x 2 = 12, 6 x 3 = 18, 6 x 1 = 6, 6 x 3 = 18, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24, 6 x 3 = 18, 6 x 5 = 30, 6 x 3 = 18, 6 x 4 = 24, 6 x 1 = 6, 6 x 4 = 24, 6 x 2 = 12, 6 x 4 = 24, 6 x 3 = 18, 6 x 4 = 24, 6 x 5 = 30, 6 x 4 = 24, 6 x 5 = 30, 6 x 1 = 6, 6 x 5 = 30, 6 x 2 = 12, 6 x 5 = 30, 6 x 3 = 18, 6 x 5 = 30, 6 x 4 = 24, 6 x 2 = 12, 6 x 4 = 24, 6 x 3 = 18, 6 x 5 = 30, 6 x 3 = 18, 6 x 2 = 12, 6 x 4 = 24, 6 x 3 = 18, 6 x 5 = 30, 6 x 2 = 12, 6 x 4 = 24.
Eureka-Math-Grade-3-Module-3-Lesson-5-Answer Key-1

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

Question 1.
Skip-count by seven to fill in the blanks in the fish bowls. Match each count-by to its multiplication expression. Then, use the multiplication expression to write the related division fact directly to the right.
Eureka Math Grade 3 Module 3 Lesson 5 Problem Set Answer Key 2

Answer:
The missing numbers are 14, 28, 35, 56, 64, and 70.

Explanation:
In the above-given question,
given that,
the numbers of table 7 are missing.
The missing numbers are 14, 28, 35, 56, 63, and 70.
Eureka-Math-Grade-3-Module-3-Lesson-5-Answer Key-2

Question 2.
Complete the count-by seven sequence below. Then, write a multiplication equation and a division equation to represent each blank you filled in.
7, 14, ___21____, 28, ___35____, 42, ___49____,  ___56____, 63, ____70___

a. __3____ × 7 = ____21___    ___21____ ÷ 7 = ___3____

Answer:
The missing numbers are 21, 35, 49, 56, 63, and 70.

Explanation:
In the above-given question,
given that,
the numbers of table 7 are given.
the missing numbers are 21, 35, 49, 56, 63, and 70.

b. ___5____ × 7 = ____35___    ____35___ ÷ 7 = ___5____

Answer:
5 x 7 = 35.
35 / 7 = 5.

Explanation:
In the above-given question,
given that,
the multiplication and division are given.
5 x 7 = 35.
35 / 7 = 5.

c. __7_____ × 7 = ___49____    ___49____ ÷ 7 = ___7____

Answer:
7 x 7 = 49.
49 / 7 = 7.

Explanation:
In the above-given question,
given that,
the multiplication and division are given.
7 x 7 = 49.
49 / 7 = 7.

d. ____8___ × 7 = __56_____    ___56____ ÷ 7 = ___8____

Answer:
8 x 7 = 56.
56 / 7 = 8.

Explanation:
In the above-given question,
given that,
the multiplication and division are given.
8 x 7 = 56.
56 / 7 = 8.

e. ___10____ × 7 = ___70____   ___70____ ÷ 7 = __10_____

Answer:
10 x 7 = 70.
70 / 7 = 10.

Explanation:
In the above-given question,
given that,
the multiplication and division are given.
10 x 7 = 70.
70 / 7 = 10.

Question 3.
Abe says 3 × 7 = 21 because 1 seven is 7, 2 sevens are 14, and 3 sevens are 14 + 6 + 1, which equals 21. Why did Abe add 6 and 1 to 14 when he is counting by seven?

Answer:
3 x 7 = 21.

Explanation:
In the above-given question,
given that,
Abe says 3 x 7 = 21 because 1 seven is 7, 2 sevens are 14.
3 sevens are 21.
14 + 6 + 1 = 21.
6 + 1 = 7.

Question 4.
Molly says she can count by seven 6 times to solve 7 × 6. James says he can count by six 7 times to solve this problem. Who is right? Explain your answer.

Answer:
Molly and James both are correct.

Explanation:
In the above-given question,
given that,
Molly says she can count by seven 6 times to solve.
7 x 6 = 42.
James says he can count by six 7 times to solve.
6 x 7 = 42.
so both of them are correct.

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

Complete the count-by seven sequence below. Then, write a multiplication equation and a division equation to represent each number in the sequence.
7, 14, __21_____, 28, ___35____, 42, ___49____,  ___56____, 63, ___70____

a. ___3____ × 7 = __21_____     ____21___ ÷ 7 = ___3____

Answer:
3 x 7 = 21.
21 / 7 = 3.

Explanation:
In the above-given question,
given that,
the multiplication and division are given.
3 x 7 = 21.
21 / 7 = 3.

b. ___5____ × 7 = __35_____    ____35___ ÷ 7 = ___5____

Answer:
5 x 7 = 35.
35 / 7 = 5.

Explanation:
In the above-given question,
given that,
the multiplication and division are given.
5 x 7 = 35.
35 / 7 = 5.

c. ____7___ × 7 = ___49____    ___49____ ÷ 7 = ___7____

Answer:

7 x 7 = 49.
49 / 7 = 7.

Explanation:
In the above-given question,
given that,
the multiplication and division are given.
7 x 7 = 49.
49 / 7 = 7.

d. ___8____ × 7 = __56_____    ___56____ ÷ 7 = ___8____

Answer:
8 x 7 = 56.
56 / 7 = 8.

Explanation:
In the above-given question,
given that,
the multiplication and division are given.
8 x 7 = 56.
56 / 7 = 8.

e. ___9____ × 7 = ___63____    ___63____ ÷ 7 = ___9____

Answer:
9 x 7 = 63.
63 / 7 = 9.

Explanation:
In the above-given question,
given that,
the multiplication and division are given.
9 x 7 = 63.
63 / 7 = 9.

f. __10_____ × 7 = __70_____     ___70____ ÷ 7 = ___10____

Answer:
10 x 7 = 70.
70 / 7 = 10.

Explanation:
In the above-given question,
given that,
the multiplication and division are given.
10 x 7 = 70.
70 / 7 = 10.

g. ___6____ × 7 = __42_____    ___42____ ÷ 7 = ____6___

Answer:
6 x 7 = 42.
42 / 7 = 6.

Explanation:
In the above-given question,
given that,
the multiplication and division are given.
6 x 7 = 42.
42 / 7 = 6.

h. ___4____ × 7 = ____28___     ___28____ ÷ 7 = __4_____

Answer:
2 x 7 = 14.
14 / 7 = 2.

Explanation:
In the above-given question,
given that,
the multiplication and division are given.
2 x 7 = 14.
14 / 7 = 2.

i. ____1___ × 7 = ___7____      ___7____ ÷ 7 = __1_____

Answer:
1 x 7 = 7.
7 / 7 = 1.

Explanation:
In the above-given question,
given that,
the multiplication and division are given.
1 x 7 = 7.
7 / 7 = 1.

j. ___10____ × 7 = __7=_____      ___70____ ÷ 7 = __10_____

Answer:
10 x 7 = 70.
70 / 7 = 10.

Explanation:
In the above-given question,
given that,
the multiplication and division are given.
10 x 7 = 70.
70 / 7 = 10.

Eureka Math Grade 3 Module 3 Lesson 5 Homework Answer Key

Question 1.
Use number bonds to help you skip-count by seven by making ten or adding to the ones.
Eureka Math Grade 3 Module 3 Lesson 5 Homework Answer Key 3

Answer:
10 + 4 = 14, 10 + 11 = 21, 10 + 18 = 28, 10 + 25 = 35, 10 + 32 = 42, 10 + 39 = 49, 10 + 46 = 56, 10 + 53 = 63.

Explanation:
In the above-given question,
given that,
Table 6 is given.
10 + 4 = 14, 10 + 11 = 21, 10 + 18 = 28, 10 + 25 = 35, 10 + 32 = 42, 10 + 39 = 49, 10 + 46 = 56, and 10 + 53 = 63.
Eureka-Math-Grade-3-Module-3-Lesson-5-Answer Key-3

Question 2.
Skip-count by seven to fill in the blanks. Then, fill in the multiplication equation, and use it to write the related division fact directly to the right.
Eureka Math Grade 3 Module 3 Lesson 5 Homework Answer Key 4

Answer:
The missing numbers are 14, 21, 35, 42, 56, 63, and 70.

Explanation:
In the above-given question,
given that,
table 7 is given.
the table of 7 is given.
The missing numbers are 14, 21, 35, 42, 56, 63, and 70.
Eureka-Math-Grade-3-Module-3-Lesson-5-Answer Key-4

Eureka Math Grade 3 Module 3 Lesson 6 Answer Key

Students of Grade 3 Module 3 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 3, students need to solve all questions and exercises included in Eureka’s Grade 3 Textbook.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 6 Answer Key

So teachers and students can find this Eureka Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Eureka Math Answer Key for Grade 3 aids teachers to differentiate instruction, building and reinforcing foundational mathematics skills that alter from the classroom to real life. With the help of Eureka’s primary school Grade 3 Answer Key, you can think deeply regarding what you are learning, and you will really learn math easily just like that.

Eureka Math Grade 3 Module 3 Lesson 6 Pattern Sheet Answer Key

Multiply
Eureka Math Grade 3 Module 3 Lesson 6 Pattern Sheet Answer Key 1

Answer:
6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24, 6 x 5 = 30, 6 x 6 = 36, 6 x 7 = 42, 6 x 8 = 48, 6 x 9 = 54, 6 x 10 = 60, 6 x 5 = 30, 6 x 6 = 36, 6 x 5 = 30, 6 x 7 = 35, 6 x 5 = 30, 6 x 8 = 40, 6 x 5 = 30, 6 x 9 = 54, 6 x 5 = 30, 6 x 10 = 60, 6 x 6 = 36, 6 x 5 = 30, 6 x 6 = 36, 6 x 7 = 42, 6 x 6 = 36, 6 x 8 = 48, 6 x 6 = 36, 6 x 9 = 54, 6 x 6 = 36, 6 x 7 = 42, 6 x 6 = 36, 6 x 7 = 42, 6 x 8 = 48, 6 x 7 = 42, 6 x 9 = 54, 6 x 7 = 42, 6 x 8 = 48, 6 x 6 = 36, 6 x 8 = 48, 6 x 7 = 42, 6 x 8 = 48, 6 x 9 = 54, 6 x 9 = 54, 6 x 6 = 36, 6 x 9 = 54, 6 x 8 = 48, 6 x 6 = 36, 6 x 9 = 54, 6 x 7 = 42, 6 x 9 = 54, 6 x 6 = 36, 6 x 8 = 48, 6 x 9 = 54, 6 x 7 = 42, 6 x 6 = 36, 6 x 8 = 48.

Explanation:
In the above-given question,
given that,
the table of 6.
6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24, 6 x 5 = 30, 6 x 6 = 36, 6 x 7 = 42, 6 x 8 = 48, 6 x 9 = 54, 6 x 10 = 60, 6 x 5 = 30, 6 x 6 = 36, 6 x 5 = 30, 6 x 7 = 35, 6 x 5 = 30, 6 x 8 = 40, 6 x 5 = 30, 6 x 9 = 54, 6 x 5 = 30, 6 x 10 = 60, 6 x 6 = 36, 6 x 5 = 30, 6 x 6 = 36, 6 x 7 = 42, 6 x 6 = 36, 6 x 8 = 48, 6 x 6 = 36, 6 x 9 = 54, 6 x 6 = 36, 6 x 7 = 42, 6 x 6 = 36, 6 x 7 = 42, 6 x 8 = 48, 6 x 7 = 42, 6 x 9 = 54, 6 x 7 = 42, 6 x 8 = 48, 6 x 6 = 36, 6 x 8 = 48, 6 x 7 = 42, 6 x 8 = 48, 6 x 9 = 54, 6 x 9 = 54, 6 x 6 = 36, 6 x 9 = 54, 6 x 8 = 48, 6 x 6 = 36, 6 x 9 = 54, 6 x 7 = 42, 6 x 9 = 54, 6 x 6 = 36, 6 x 8 = 48, 6 x 9 = 54, 6 x 7 = 42, 6 x 6 = 36, 6 x 8 = 48.
Eureka-Math-Grade-3-Module-3-Lesson-6-Answer Key-1

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

Question 1.
Label the tape diagrams. Then, fill in the blanks below to make the statements true.
a. 6 × 6 = __36___
Eureka Math Grade 3 Module 3 Lesson 6 Problem Set Answer Key 2
(6 × 6) = (5 + 1) × 6
= (5 × 6) + (1 × 6)
  30    + __6____
= ____6__

Answer:
5 x 6 = 30.
1 x 6 = 6.
30 + 6 = 36.

Explanation:
In the above-given question,
given that,
6 x 6 = (5 + 1) x 6.
(5 x 6) + (1 x 6).
30 + 6.
36.
Eureka-Math-Grade-3-Module-3-Lesson-6-Answer Key-2

b. 7 × 6 = __42___
Eureka Math Grade 3 Module 3 Lesson 6 Problem Set Answer Key 3
(7 × 6) = (5 + 2) × 6
= (5 × 6) + (2 × 6)
 30     + __12____
= ____42__

Answer:
5 x 6 = 30.
2 x 6 = 12.
30 + 12 = 42.

Explanation:
In the above-given question,
given that,
7 x 6 = (5 + 2) x 6.
(5 x 6) + (2 x 6).
30 + 12.
42.
Eureka-Math-Grade-3-Module-3-Lesson-6-Answer Key-3

c. 8 × 6 = ___48__
Eureka Math Grade 3 Module 3 Lesson 6 Problem Set Answer Key 4
8 × 6 = (5 + ___3__ ) × 6
= (5 × 6) + ( __3__ × 6)
 30    + __18____
= __48____

Answer:
5 x 6 = 30.
3 x 6 = 18.
30 + 18 = 48.

Explanation:
In the above-given question,
given that,
8 x 6 = (5 + 3) x 6.
(5 x 6) + (3 x 6).
30 + 18.
48.
Eureka-Math-Grade-3-Module-3-Lesson-6-Answer Key-4

d. 9 × 6 = __54___
Eureka Math Grade 3 Module 3 Lesson 6 Problem Set Answer Key 5
9 × 6 = (5 + __4___ ) × 6
= (5 × 6) + ( _4___ × 6)
 30    + __24____
= __54____

Answer:
5 x 6 = 30.
4 x 6 = 24.
30 + 24 = 54.

Explanation:
In the above-given question,
given that,
9 x 6 = (5 + 4) x 6.
(5 x 6) + (4 x 6).
30 + 24.
54.
Eureka-Math-Grade-3-Module-3-Lesson-6-Answer Key-5

Question 2.
Break apart 54 to solve 54 ÷ 6.
Eureka Math Grade 3 Module 3 Lesson 6 Problem Set Answer Key 6
54 ÷ 6 = (30 ÷ 6) + ( __24________ ÷ 6)
= 5 + ______4______
= ____9_______

Answer:
54 / 6 = 9.

Explanation:
In the above-given question,
given that,
54 / 6 is divided into two parts.
30 / 6 and 24 / 6.
30 / 6 = 5 and 24 / 6 = 4.
5 + 4 = 9.

Question 3.
Break apart 49 to solve 49 ÷ 7.
Eureka Math Grade 3 Module 3 Lesson 6 Problem Set Answer Key 7
49 ÷ 7 = (35 ÷ 7) + ( ___14_______ ÷ 7)
= 5 + _______2_____
= _____7______

Answer:
49 / 7 = 7.

Explanation:
In the above-given question,
given that,
49 / 7 is divided into two parts.
35 / 7 and 14 / 7.
35 / 7 = 5 and 14 / 7 = 2.
5 + 2 = 7.
Eureka-Math-Grade-3-Module-3-Lesson-6-Answer Key-6

Question 4.
Robert says that he can solve 6 × 8 by thinking of it as (5 × 8) + 8. Is he right? Draw a picture to help explain your answer.

Answer:
Yes, he was right.

Explanation:
In the above-given question,
given that,
Robert says that he can solve 6 x 8 by thinking of it as (5 x 8) + 8.
6 x 8 = 48.
(5 x 8) + 8 = 40 + 8 = 48.
so he was correct.
Eureka-Math-Grade-3-Module-3-Lesson-6-Answer Key-7

Question 5.
Kelly solves 42 ÷ 7 by using a number bond to break apart 42 into two parts. Show what her work might look like below.

Answer:
42 / 7 = 6.

Explanation:
In the above-given question,
given that,
42 / 7 is divided into two parts.
28 / 7 and 14 / 7.
28 / 7 = 4 and 14 / 7 = 2.
4 + 2 = 6.

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

Question 1.
A parking lot has space for 48 cars. Six cars can park in 1 row. Break apart 48 to find how many rows there are in the parking lot.
Eureka Math Grade 3 Module 3 Lesson 6 Exit Ticket Answer Key 8

Answer:
48 / 6 = 8.

Explanation:
In the above-given question,
given that,
48 / 6 is divided into two parts.
30 / 6 and 18 / 6.
30 / 6 = 5 and 18 / 6 = 3.
5 + 3 = 8.
Eureka-Math-Grade-3-Module-3-Lesson-6-Answer Key-8

Question 2.
Malia solves 6 × 7 using (5 × 7) + 7. Leonidas solves 6 × 7 using (6 × 5) + (6 × 2). Who is correct? Draw a picture to help explain your answer.

Answer:
Yes, both of them are correct.

Explanation:
In the above-given question,
given that,
6 x 7 = 42.
(5 x 7) + 7 = 35 + 7 = 42.
(6 x 5) + (6 x 2) = 30 + 12 = 42.

Eureka Math Grade 3 Module 3 Lesson 6 Homework Answer Key

Question 1.
Label the tape diagrams. Then, fill in the blanks below to make the statements true.
a. 6 × 7 = __42___
Eureka Math Grade 3 Module 3 Lesson 6 Homework Answer Key 9
(6 × 7) = (5 + 1) × 7
= (5 × 7) + (1 × 7)
 35     + ___7___
= ___42___

Answer:
5 x 7 = 35.
1 x 7 = 7.
35 + 7 = 42.

Explanation:
In the above-given question,
given that,
6 x 7 = (5 + 1) x 7.
(5 x 7) + (1 x 7).
35 + 7.
42.
Eureka-Math-Grade-3-Module-3-Lesson-6-Answer Key-9

b. 7 × 7 = __49___
Eureka Math Grade 3 Module 3 Lesson 6 Homework Answer Key 10
(7 × 7) = (5 + 2) × 7
= (5 × 7) + (2 × 7)
 35     + __14___
= ___49___

Answer:
5 x 7 = 35.
2 x 7 = 14.
35 + 14 = 49.

Explanation:
In the above-given question,
given that,
6 x 7 = (5 + 2) x 7.
(5 x 7) + (2 x 7).
35 + 14.
49.
Eureka-Math-Grade-3-Module-3-Lesson-6-Answer Key-10

c. 8 × 7 = __56___
Eureka Math Grade 3 Module 3 Lesson 6 Homework Answer Key 11
8 × 7 = (5 + ___3__ ) × 7
= (5 × 7) + ( __3__ × 7)
 35     + ___21___
= __56____

Answer:
5 x 7 = 35.
3 x 7 = 21.
35 + 21 = 56.

Explanation:
In the above-given question,
given that,
8 x 7 = (5 + 3) x 7.
(5 x 7) + (3 x 7).
35 + 21.
56.
Eureka-Math-Grade-3-Module-3-Lesson-6-Answer Key-11

d. 9 × 7 = __63___
Eureka Math Grade 3 Module 3 Lesson 6 Homework Answer Key 12
9 × 7 = (5 + __4___ ) × 7
= (5 × 7) + ( __4__ × 7)
 35    + ___28___
= ___63___

Answer:
5 x 7 = 35.
4 x 7 = 28.
35 + 28 = 63.

Explanation:
In the above-given question,
given that,
9 x 7 = (5 + 4) x 7.
(5 x 7) + (4 x 7).
35 + 28.
63.
Eureka-Math-Grade-3-Module-3-Lesson-6-Answer Key-12

Question 2.
Break apart 54 to solve 54 ÷ 6.
Eureka Math Grade 3 Module 3 Lesson 6 Homework Answer Key 13
54 ÷ 6 = (30 ÷ 6) + ( ____24_______ ÷ 6)
= 5 + _______4_____
= _____9______

Answer:
54 / 6 = 9.

Explanation:
In the above-given question,
given that,
54 / 6 is divided into two parts.
30 / 6 and 24 / 6.
30 / 6 = 5 and 24 / 6 = 4.
5 + 4 = 9.
Eureka-Math-Grade-3-Module-3-Lesson-6-Answer Key-13

Question 3.
Break apart 56 to solve 56 ÷ 7
Eureka Math Grade 3 Module 3 Lesson 6 Homework Answer Key 14
56 ÷ 7 = ( __35___ ÷ ___7__ ) + ( __21___ ÷ ___7__ )
= 5 + ____3____
= ____8______

Answer:
56 / 7 = 8.

Explanation:
In the above-given question,
given that,
56 / 7 is divided into two parts.
35 / 7 and 21 / 7.
35 / 7 = 5 and 21 / 7 = 3.
5 + 3 = 8.
Eureka-Math-Grade-3-Module-3-Lesson-6-Answer Key-14

Question 4.
Forty-two third grade students sit in 6 equal rows in the auditorium. How many students sit in each row? Show your thinking.

Answer:
The number of students sits in each row = 7.

Explanation:
In the above-given question,
given that,
42 third-grade students sit in 6 equal rows in the auditorium.
42 / 6 = 7.
7 x 6 = 42.

Question 5.
Ronaldo solves 7 × 6 by thinking of it as (5 × 7) + 7. Is he correct? Explain Ronaldo’s strategy.

Answer:
Yes, he was correct.

Explanation:
In the above-given question,
given that,
Ronaldo solves 7 x 6 = 42.
(5 x 7) + 7 = 42.
so he was correct.

Eureka Math Grade 3 Module 3 Lesson 7 Answer Key

Teachers and students can find this Eureka Book Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Learning activities are the best option to educate elementary school kids and make them understand the basic mathematical concepts like addition, subtraction, multiplication, division, etc. Grade 3 elementary school students can find these fun-learning exercises for all math concepts through the Eureka Book.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 7 Answer Key

Students of Grade 3 can get a strong foundation 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 3, students need to solve all questions and exercises included in Eureka Grade 3 Book.

Eureka Math Grade 3 Module 3 Lesson 7 Pattern Sheet Answer Key

Multiply
Eureka Math Grade 3 Module 3 Lesson 7 Pattern Sheet Answer Key 1

Answer:
7 x 1 = 7, 7 x 2 = 14, 7 x 3 = 21, 7 x 4 = 28, 7 x 5 = 35, 7 x 1 = 7, 7 x 2 = 14, 7 x 1 = 7, 7 x 3 = 21, 7 x 1 = 7, 7 x 4 = 28, 7 x 1 = 7, 7 x 5 = 35, 7 x 1 = 7, 7 x 2 = 14, 7 x 3 = 21, 7 x 2 = 14, 7 x 4 = 28, 7 x 5 = 35, 7 x 2 = 14, 7 x 2 = 14, 7 x 1 = 7, 7 x 2 = 14, 7 x 3 = 21, 7 x 1 = 7, 7 x 3 = 21, 7 x 2 = 14, 7 x 3 = 21, 7 x 4 = 28, 7 x 3 = 21, 7 x 5 = 35, 7 x 3 = 21, 7 x 4 = 28, 7 x 1 = 7, 7 x 4 = 28, 7 x 2 = 14, 7 x 4 = 28, 7 x 3 = 21, 7 x 4 = 28, 7 x 5 = 35, 7 x 4 = 28, 7 x 5 = 35, 7 x 1 = 7, 7 x 5 = 35, 7 x 2 = 14, 7 x 5 = 35, 7 x 3 = 21, 7 x 5 = 35, 7 x 4 = 28, 7 x 2 = 14, 7 x 4 = 28, 7 x 3 = 21, 7 x 5 = 35, 7 x 3 = 21, 7 x 2 = 14, 7 x 4 = 28, 7 x 3 = 21, 7 x 5 = 35, 7 x 2 = 14, 7 x 4 = 28.

Explanation:
In the above-given question,
given that,
the table of 7 and multiply.
7 x 1 = 7, 7 x 2 = 14, 7 x 3 = 21, 7 x 4 = 28, 7 x 5 = 35, 7 x 1 = 7, 7 x 2 = 14, 7 x 1 = 7, 7 x 3 = 21, 7 x 1 = 7, 7 x 4 = 28, 7 x 1 = 7, 7 x 5 = 35, 7 x 1 = 7, 7 x 2 = 14, 7 x 3 = 21, 7 x 2 = 14, 7 x 4 = 28, 7 x 5 = 35, 7 x 2 = 14, 7 x 2 = 14, 7 x 1 = 7, 7 x 2 = 14, 7 x 3 = 21, 7 x 1 = 7, 7 x 3 = 21, 7 x 2 = 14, 7 x 3 = 21, 7 x 4 = 28, 7 x 3 = 21, 7 x 5 = 35, 7 x 3 = 21, 7 x 4 = 28, 7 x 1 = 7, 7 x 4 = 28, 7 x 2 = 14, 7 x 4 = 28, 7 x 3 = 21, 7 x 4 = 28, 7 x 5 = 35, 7 x 4 = 28, 7 x 5 = 35, 7 x 1 = 7, 7 x 5 = 35, 7 x 2 = 14, 7 x 5 = 35, 7 x 3 = 21, 7 x 5 = 35, 7 x 4 = 28, 7 x 2 = 14, 7 x 4 = 28, 7 x 3 = 21, 7 x 5 = 35, 7 x 3 = 21, 7 x 2 = 14, 7 x 4 = 28, 7 x 3 = 21, 7 x 5 = 35, 7 x 2 = 14, 7 x 4 = 28.
Eureka-Math-Grade-3-Module-3-Lesson-7-Answer Key-1

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

Question 1.
Match the words to the correct equation.
Eureka Math Grade 3 Module 3 Lesson 7 Problem Set Answer Key 2
Answer:
n x 7 = 21.
n x 6 = 30.
6 x 7 = n.
7 x n = 42,
36 / n = 6.
63 / n = 9.

Explanation:
In the above-given question,
given that,
the equations of the products and divisions.
n x 7 = 21.
n x 6 = 30.
6 x 7 = n.
7 x n = 42,
36 / n = 6.
63 / n = 9.
Eureka-Math-Grade-3-Module-3-Lesson-7-Answer Key-2

Question 2.
Write an equation to represent the tape diagram below, and solve for the unknown.
Eureka Math Grade 3 Module 3 Lesson 7 Problem Set Answer Key 3
Equation: __48__/___k___=_8._____

Answer:
48 / k = 8.
k = 6.

Explanation:
In the above-given question,
given that,
There are 6 eights.
6 x 8 = 48.
48 / k = 8.
k = 6.

Question 3.
Model each problem with a drawing. Then, write an equation using a letter to represent the unknown, and solve for the unknown.

a. Each student gets 3 pencils. There are a total of 21 pencils. How many students are there?

Answer:
The number of students = 7.

Explanation:
In the above-given question,
given that,
Each student gets 3 pencils.
There are a total of 21 pencils.
21 x p = 3.
21 / 3 = p.
so p = 7.
Eureka-Math-Grade-3-Module-3-Lesson-7-Answer Key-3

b. Henry spends 24 minutes practicing 6 different basketball drills. He spends the same amount of time on each drill. How much time does Henry spend on each drill?

Answer:
Henry spends on each drill = 4 minutes.

Explanation:
In the above-given question,
given that,
Henry spends 24 minutes practicing 6 different basketball drills.
He spends the same amount of time on each drill.
24 / m = 6.
m = 24 / 6 .
m = 4 minutes.
Eureka-Math-Grade-3-Module-3-Lesson-7-Answer Key-4

c. Jessica has 8 pieces of yarn for a project. Each piece of yarn is 6 centimeters long. What is the total length of the yarn?

Answer:
The total length of the yarn = 48 centimeters.

Explanation:
In the above-given question,
given that,
Jessica has 8 pieces of yarn for a project.
Each piece of yarn is 6 centimeters long.
8 x y = 6.
y = 6 x 8.
y = 48.
Eureka-Math-Grade-3-Module-3-Lesson-7-Answer Key-5

d. Ginny measures 6 milliliters of water into each beaker. She pours a total of 54 milliliters. How many beakers does Ginny use?

Answer:
The number of beakers does Ginny uses = 9.

Explanation:
In the above-given question,
given that,
Ginny measures 6 millimeters of water into each beaker.
she pours a total of 54 milliliters.
6 / b = 54.
b = 54 / 6.
b = 9.
Eureka-Math-Grade-3-Module-3-Lesson-7-Answer Key-6

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

Model each problem with a drawing. Then, write an equation using a letter to represent the unknown, and solve for the unknown.

Question 1.
Three boys and three girls each buy 7 bookmarks. How many bookmarks do they buy all together?

Answer:
The total number of bookmarks = 42.

Explanation:
In the above-given question,
given that,
Three boys and three girls each buy 7 bookmarks.
3 + 3 = 6.
6 x b = 7.
b = 7 x 6.
b = 42.
Eureka-Math-Grade-3-Module-3-Lesson-7-Answer Key-7

Question 2.
Seven friends equally share the cost of a $56 meal. How much does each person pay?

Answer:
The number of dollars each person pays = $8.

Explanation:
In the above-given question,
given that,
seven friends equally share the cost of a $56 meal.
7 / p = 56.
p = 56 / 7.
p = 8.
Eureka-Math-Grade-3-Module-3-Lesson-7-Answer Key-8

Eureka Math Grade 3 Module 3 Lesson 7 Homework Answer Key

Question 1.
Match the words on the arrow to the correct equation on the target.
Eureka Math Grade 3 Module 3 Lesson 7 Homework Answer Key 4

Answer:
n x 7 = 21 = A number times 7 equals 21.
7 x n = 42 = 7 times a number equals 42.
63 / n = 9 = 63 divided by a number equals 9.
36 / n = 6 = 36 divided by a number equals 6.

Explanation:
In the above-given question,
given that,
the equations that multiply and divide.
n x 7 = 21 = A number times 7 equals 21.
7 x n = 42 = 7 times a number equals 42.
63 / n = 9 = 63 divided by a number equals 9.
36 / n = 6 = 36 divided by a number equals 6.
Eureka-Math-Grade-3-Module-3-Lesson-7-Answer Key-12

Question 2.
Ari sells 6 boxes of pens at the school store.
a. Each box of pens sells for $7. Draw a tape diagram, and label the total amount of money he makes as m. Write an equation, and solve for m.

Answer:
The total amount of money = 42.

Explanation:
In the above-given question,
given that,
Ari sells 6 boxes of pens at the school store.
Each box of pens sells for $7.
6 x m = $7.
m = 6 x 7.
m = 42.
Eureka-Math-Grade-3-Module-3-Lesson-7-Answer Key-9

b. Each box contains 6 pens. Draw a tape diagram, and label the total number of pens as p. Write an equation, and solve for p.

Answer:
The total number of pens = 36.

Explanation:
In the above-given question,
given that,
Each box contains 6 pens.
6 x p = 6.
p = 6 x 6.
p = 36.
Eureka-Math-Grade-3-Module-3-Lesson-7-Answer Key-10

Question 3.
Mr. Lucas divides 28 students into 7 equal groups for a project. Draw a tape diagram, and label the number of students in each group as n. Write an equation, and solve for n.

Answer:
The number of students in each group = 4.

Explanation:
In the above-given question,
given that,
Mr. Lucas divides 28 students into 7 equal groups for a project.
28 / s = 7.
s = 28 / 7.
s = 4.
Eureka-Math-Grade-3-Module-3-Lesson-7-Answer Key-11

Eureka Math Grade 3 Module 3 Lesson 8 Answer Key

Students of Grade 3 can get a strong foundation 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 3, students need to solve all questions and exercises included in Eureka Grade 3 Book.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 8 Answer Key

Teachers and students can find this Eureka Book Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Learning activities are the best option to educate elementary school kids and make them understand the basic mathematical concepts like addition, subtraction, multiplication, division, etc. Grade 3 elementary school students can find these fun-learning exercises for all math concepts through the Eureka Book.

Eureka Math -Grade 3 Module 3 Lesson 8 Pattern Sheet Answer Key

Multiply
Eureka Math Grade 3 Module 3 Lesson 8 Pattern Sheet Answer Key 1

Answer:
7 x 1 = 7, 7 x 2 = 14, 7 x 3 = 21, 7 x 4 = 28, 7 x 5 = 35, 7 x 6 = 42, 7 x 7 = 49, 7 x 8 = 56, 7 x 9 = 63, 7 x 10 = 70, 7 x 5 = 35, 7 x 6 = 42, 7 x 5 = 35, 7 x 7 = 49, 7 x 5 = 35, 7 x 8 = 56, 7 x 5 = 35, 7 x 9 = 63, 7 x 5 = 35, 7 x 10 = 70, 7 x 6 = 42, 7 x 5 = 35, 7 x 6 = 42, 7 x 7 = 49, 7 x 6 = 42, 7 x 8 = 56, 7 x 6 = 42, 7 x 9 = 63, 7 x 6 = 42, 7 x 7 = 49, 7 x 6 = 42, 7 x 7 = 49, 7 x 8 = 56, 7 x 7 = 49, 7 x 9 = 63, 7 x 7 = 49, 7 x 8 = 56, 7 x 6 = 42, 7 x 8 = 56, 7 x 7 = 49, 7 x 8 = 56, 7 x 9 = 63, 7 x 9 = 63, 7 x 6 = 42, 7 x 9 = 63, 7 x 7 = 49, 7 x 9 = 63, 7 x 8 = 56, 7 x 9 = 63, 7 x 8 = 56, 7 x 6 = 42, 7 x 9 = 63, 7 x 7 = 49, 7 x 9 = 63, 7 x 6 = 42, 7 x 8 = 56, 7 x 9 = 63, 7 x 7 = 49, 7 x 6 = 42, 7 x 8 = 56.

Explanation:
In the above-given question,
given that,
the table of 6 is given.
7 x 1 = 7, 7 x 2 = 14, 7 x 3 = 21, 7 x 4 = 28, 7 x 5 = 35, 7 x 6 = 42, 7 x 7 = 49, 7 x 8 = 56, 7 x 9 = 63, 7 x 10 = 70, 7 x 5 = 35, 7 x 6 = 42, 7 x 5 = 35, 7 x 7 = 49, 7 x 5 = 35, 7 x 8 = 56, 7 x 5 = 35, 7 x 9 = 63, 7 x 5 = 35, 7 x 10 = 70, 7 x 6 = 42, 7 x 5 = 35, 7 x 6 = 42, 7 x 7 = 49, 7 x 6 = 42, 7 x 8 = 56, 7 x 6 = 42, 7 x 9 = 63, 7 x 6 = 42, 7 x 7 = 49, 7 x 6 = 42, 7 x 7 = 49, 7 x 8 = 56, 7 x 7 = 49, 7 x 9 = 63, 7 x 7 = 49, 7 x 8 = 56, 7 x 6 = 42, 7 x 8 = 56, 7 x 7 = 49, 7 x 8 = 56, 7 x 9 = 63, 7 x 9 = 63, 7 x 6 = 42, 7 x 9 = 63, 7 x 7 = 49, 7 x 9 = 63, 7 x 8 = 56, 7 x 9 = 63, 7 x 8 = 56, 7 x 6 = 42, 7 x 9 = 63, 7 x 7 = 49, 7 x 9 = 63, 7 x 6 = 42, 7 x 8 = 56, 7 x 9 = 63, 7 x 7 = 49, 7 x 6 = 42, 7 x 8 = 56.
Eureka-Math-Grade-3-Module-3-Lesson-8-Answer Key-1

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

Question 1.
Solve.
a. (12 – 4) + 6 = __14____

Answer:
(12 – 4) + 6 = 14.

Explanation:
In the above-given question,
given that,
(12 – 4) = 8.
8 + 6 = 14.

b. 12 – (4 + 6) = ___2___

Answer:
12 – (4 + 6) = 2.

Explanation:
In the above-given question,
given that,
4 + 6 = 10.
12 – 10 = 2.

c. __5____ = 15 – (7 + 3)

Answer:
15 – (7 + 3) = 5.

Explanation:
In the above-given question,
given that,
7 + 3 = 10.
15 – 10 = 5.

d. __11____ = (15 – 7) + 3

Answer:
(15 – 7) + 3 = 11.

Explanation:
In the above-given question,
given that,
15 – 7 = 8.
8 + 3 = 11.

e. ___30___ = (3 + 2) × 6

Answer:
(3 + 2 ) x 6 = 30.

Explanation:
In the above-given question,
given that,
3 + 2 = 5.
5 x 6 = 30.

f. __15____ = 3 + (2 × 6)

Answer:
3 + ( 2  x 6 ) = 15.

Explanation:
In the above-given question,
given that,
2 x 6 = 12.
3 + 12 = 15.

g. 4 × (7 – 2) = __20____

Answer:
4 x ( 7  – 2 ) = 20.

Explanation:
In the above-given question,
given that,
7 – 2 = 5.
5 x 4 = 20.

h. (4 × 7) – 2 = __26____

Answer:
(4 x 7) – 2 = 26.
Explanation:
In the above-given question,
given that,
4 x 7 = 28.
28 – 2 = 26.

i. ___10___ = (12 ÷ 2) + 4

Answer:
4 + ( 12  / 2 ) = 10.

Explanation:
In the above-given question,
given that,
12 / 2 = 6.
4 + 6 = 10.

j. __2____ = 12 ÷ (2 + 4)

Answer:
12 / ( 2  + 4 ) = 2.

Explanation:
In the above-given question,
given that,
12 / 6 = 2.
2 + 4 = 6.

k. 9 + (15 ÷ 3) = __14____

Answer:
9 + ( 15  / 3 ) = 14.

Explanation:
In the above-given question,
given that,
15 / 3 = 5.
9 + 5 = 14.

l. (9 + 15) ÷ 3 = ___8___

Answer:
(9 + 15) / 3 = 8.

Explanation:
In the above-given question,
given that,
9 + 15 = 24.
24 / 3 = 8.

m. 60 ÷ (10 – 4) = __10____

Answer:
60 / (10 – 4) = 10.
Explanation:
In the above-given question,
given that,
10 – 4 = 6.
60 / 6 = 10.

n. (60 ÷ 10) – 4 = __2____

Answer:
(60 / 10) – 4 = 2.

Explanation:
In the above-given question,
given that,
(60 / 10) = 6.
6 – 4 = 2.

o. __37____ = 35 + (10 ÷ 5)

Answer:
35 + ( 10 / 5 ) = 37.

Explanation:
In the above-given question,
given that,
10 / 5 = 2.
35 + 2 = 37.

p. ___9___ = (35 + 10) ÷ 5

Answer:
(35 + 10) / 5 = 9.

Explanation:
In the above-given question,
given that,
35 + 10 = 45.
45 / 5 = 9.

Question 2.
Use parentheses to make the equations true.
Answer:

Question 3.
The teacher writes 24 ÷ 4 + 2 = __4 , __8__ on the board. Chad says it equals 8. Samir says it equals 4. Explain how placing the parentheses in the equation can make both answers true.

Answer:
Yes, both the answers are true.

Explanation:
In the above-given question,
given that,
The teacher writes 24 / 4 + 2 = 8 and 4 on the board.
24 / 4 = 6.
6 + 2 = 8.
4 + 2 = 6.
24 / 6 = 4.
so both answers are true.

Question 4.
Natasha solves the equation below by finding the sum of 5 and 12. Place the parentheses in the equation to show her thinking. Then, solve.
12 + 15 ÷ 3 = __17______

Answer:
12 + 15 / 3 = 17.

Explanation:
In the above-given question,
given that,
Natasha solves the equation below by finding the sum of 5 and 12.
15 / 3 = 5.
12 + 5 = 17.

Question 5.
Find two possible answers to the expression 7 + 3 × 2 by placing the parentheses in different places.

Answer:
The two possible answers are 20 and 13.

Explanation:
In the above-given question,
given that,
7 + 3 x 2.
7 + 3 = 10.
10 x 2 = 20.
3 x 2 = 6.
7 + 6 = 13.

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

Question 1.
Use parentheses to make the equations true.
a. 24 = 32 – 14 + 6

Answer:
24 = (32 – 14) + 6.

Explanation:
In the above-given question,
given that,
24 = 32 – 14 + 6.
32 – 14 = 18.
18 + 6 = 24.

b. 12 = 32 – 14 + 6

Answer:
12 = 32 – (14 + 6).

Explanation:
In the above-given question,
given that,
12 = 32 – 14 + 6.
14 + 6 = 20.
32 – 20 = 12.

c. 2 + 8 × 7 = 70

Answer:
70 = 70.

Explanation:
In the above-given question,
given that,
2 + 8 x 7.
2 + 8 = 10.
10 x 7 = 70.

d. 2 + 8 × 7 = 58

Answer:
58 = 58.

Explanation:
In the above-given question,
given that,
2 + 8 x 7.
7 x 8 = 56.
56 + 2 = 58.

Question 2.
Marcos solves 24 ÷ 6 + 2 = ___6 and 3___. He says it equals 6. Iris says it equals 3. Show how the position of parentheses in the equation can make both answers true.

Answer:
Yes, both the answers are true.

Explanation:
In the above-given question,
given that,
Marcos solves 24 / 6 + 2.
24 / 6 = 4.
4 + 2 = 6.
6 + 2 = 8.
24 / 8 = 3.
so both answers are true.

Eureka Math Grade 3 Module 3 Lesson 8 Homework Answer Key

Question 1.
Solve.
a. 9 – (6 + 3) = __0____

Answer:
9 – ( 6 + 3) = 0.

Explanation:
In the above-given question,
given that,
6 + 3 = 9.
9 – 9 = 0.

b. (9 – 6) + 3 = __6____

Answer:
(9 – 6) + 3 = 6.

Explanation:
In the above-given question,
given that,
(9 – 6) = 3.
3 + 3 = 6.

c. __8___ = 14 – (4 + 2)

Answer:
14 – (4 + 2) = 8.

Explanation:
In the above-given question,
given that,
(4 + 2) = 6.
14 – 6 = 8.

d. ___12__ = (14 – 4) + 2

Answer:
(14 – 4) + 2 = 12.

Explanation:
In the above-given question,
given that,
(14 – 4) = 10.
10 + 2 = 12.

e. _42____ = (4 + 3) × 6

Answer:
(4 + 3) x 6 = 42.

Explanation:
In the above-given question,
given that,
(4 + 3) = 7.
7 x 6 = 42.

f. __22___ = 4 + (3 × 6)

Answer:
(3 x 6) + 4 = 22.

Explanation:
In the above-given question,
given that,
(3 x 6) = 18.
18 +  4 = 22.

g. (18 ÷ 3) + 6 = __12___

Answer:
(18 / 3) + 6 = 12.

Explanation:
In the above-given question,
given that,
(18 / 3) = 6.
6 + 6 = 12.

h. 18 ÷ (3 + 6)= __2___

Answer:
18 / (3 + 6) = 2.

Explanation:
In the above-given question,
given that,
(3 + 6) = 9.
18 / 9 = 2.

Question 2.
Use parentheses to make the equations true
a. 14 – 8 + 2 = 4

Answer:
14 – (8 + 2) = 4.
Explanation:
In the above-given question,
given that,
4 = 14 – 8 + 2.
8 + 2 = 10.
14 – 10 = 4.

b. 14 – 8 + 2 = 8

Answer:
8 = 8.

Explanation:
In the above-given question,
given that,
14 – 8 + 2 = 8.
14 – 8 = 6.
6 + 2 = 8.

c. 2 + 4 × 7 = 30

Answer:
30 = 30.

Explanation:
In the above given question,
given that,
30 = 2 + 4 x 7.
4 x 7 = 28.
28 + 2 = 30.

d. 2 + 4 × 7 = 42

Answer:
42 = 42.

Explanation:
In the above given question,
given that,
42 = 2 + 4 x 7.
2 + 4 = 6.
6 x 7 = 42.

e. 12 = 18 ÷ 3 × 2

Answer:
12 = 12.

Explanation:
In the above-given question,
given that,
12 = 18 / 3 x 2.
18 / 3 = 6.
6 x 2 = 12.

f. 3 = 18 ÷ 3 × 2

Answer:
3 = 3.

Explanation:
In the above-given question,
given that,
3 = 18 / 3 x 2.
3 x 2 = 6.
18 / 6 = 3.

g. 5 = 50 ÷ 5 × 2

Answer:
5 = 5.

Explanation:
In the above-given question,
given that,
5 = 50 / 5 x 2.
5 x 2 = 10.
50 / 10 = 5.

h. 20 = 50 ÷ 5 × 2

Answer:
20 = 20.

Explanation:
In the above-given question,
given that,
20 = 50 / 5 x 2.
50 / 5 = 10.
10 x 2 = 20.

Question 3.
Determine if the equation is true or false.

a. (15 – 3) ÷ 2 = 6Example: True
b. (10 – 7) × 6 = 18True
c. (35 – 7) ÷ 4 = 8 False
d. 28 = 4 × (20 – 13) True
e. 35 = (22 – 8) ÷ 5False 

Answer:
a. true.
b. true.
c. false.
d. true.
e. false.

Explanation:
In the above-given question,
given that,
(15 – 3) / 2 = 6.
(10 – 7) x 6 = 18.
(35 – 7) / 4 = 7.
28 = 4 x (20 – 13).
2.8 = (22 – 8) / 5.

Question 4.
Jerome finds that (3 × 6) ÷ 2 and 18 ÷ 2 are equal. Explain why this is true.

Answer:
Yes, it is true.

Explanation:
In the above-given question,
given that,
Jerome finds that (3 x 6) / 2 and 18 / 2 are equal.
3 x 6 = 18.
so both of them are true.

Question 5.
Place parentheses in the equation below so that you solve by finding the difference between 28 and 3. Write the answer
4 × 7 – 3 = __25______

Answer:
The difference between 28 and 3 is 25.

Explanation:
In the above-given question,
given that,
4 x 7 – 3 = 25.
4 x 7 = 28.
28 – 3 = 25.

Question 6.
Johnny says that the answer to 2 × 6 ÷ 3 is 4 no matter where he puts the parentheses. Do you agree? Place parentheses around different numbers to help you explain his thinking.

Answer:
( 2 x 6 ) / 3 = 4.

Explanation:
In the above-given question,
given that,
2 x 6 = 12.
12 / 3 = 4.

Eureka Math Grade 3 Module 3 Lesson 9 Answer Key

Teachers and students can find this Eureka Book Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Learning activities are the best option to educate elementary school kids and make them understand the basic mathematical concepts like addition, subtraction, multiplication, division, etc. Grade 3 elementary school students can find these fun-learning exercises for all math concepts through the Eureka Book.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 9 Answer Key

Students of Grade 3 can get a strong foundation 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 3, students need to solve all questions and exercises included in Eureka Grade 3 Book.

Eureka Math Grade 3 Module 3 Lesson 9 Application Problems Answer Key

Solve the following pairs of problems. Circle the pairs where both problems have the same answer.

Question 1.
a. 7 + (6 + 4)
b. (7 + 6) + 4

Answer:
a. 17.
b. 17.

Explanation:
In the above-given question,
given that,
a. 7 + (6 + 4)
6 + 4 = 10.
7 + 10 = 17.
b. (7 + 6) + 4.
7 + 6 = 13.
13 + 4 = 17.
Eureka-Math-Grade-3-Module-3-Lesson-9-Answer Key-1

Question 2.
a. (3 × 2) × 4
b. 3 × (2 × 4)

Answer:
a. 24.
b. 24.

Explanation:
In the above-given question,
given that,
a. 4 x (3 x 2)
3 x 2 = 6.
6 x 4 = 24.
b. (2 x 4) x 3.
2 x 4 = 8.
8 x 3 = 24.
Eureka-Math-Grade-3-Module-3-Lesson-9-Answer Key-2

Question 3.
a. (2 × 1) × 5
b. 2 × (1 × 5)
Answer:
a. 10.
b. 10.

Explanation:
In the above-given question,
given that,
a. 5 x (2 x 1)
2 x 1 = 2.
5 x 2 = 10.
b. (1 x 5) x 2.
5 x 1 = 5.
5 x 2 = 10.
Eureka-Math-Grade-3-Module-3-Lesson-9-Answer Key-3

Question 4.
a. (4 × 2) × 2
b. 4 × (2 × 2)

Answer:
a. 16.
b. 16.

Explanation:
In the above-given question,
given that,
a. 2 x (4 x 2)
4 x 2 = 8.
8 + 2 = 16.
b. (2 x 2) x 4.
2 x 2 = 4.
4 x 4 = 16.
Eureka-Math-Grade-3-Module-3-Lesson-9-Answer Key-4

Question 5.
a. (3 + 2) × 5
b. 3 + (2 × 5)

Answer:
a. 25.
b. 13.

Explanation:
In the above-given question,
given that,
a. 5 x (3 + 2)
3 + 2 = 5.
5 x 5 = 25.
b. (2 x 5) + 3.
2 x 5 = 10.
10 + 3 = 13.

Question 6.
a. (8 ÷ 2) × 2
b. 8 ÷ (2 × 2)

Answer:
a. 8.
b. 2.

Explanation:
In the above-given question,
given that,
a. 2 x (8 / 2)
8 / 2 = 4.
2 x 4 = 8.
b. (2 x 2) / 8.
2 x 2 = 4.
8 / 4 = 2.

Question 7.
a. (9 – 5) + 3
b. 9 – (5 + 3)

Answer:
a. 7.
b. 1.

Explanation:
In the above-given question,
given that,
a. 3 + (9 – 5)
9 – 5 = 4.
4 + 3 = 7.
b. (5 + 3) – 9.
5 + 3 = 8.
9 – 8 = 1.

Question 8.
a. (8 × 5) – 4
b. 8 × (5 – 4)

Answer:
a. 36.
b. 8.

Explanation:
In the above-given question,
given that,
a. 4 – (8 x 5)
8 x 5 = 40.
40 – 4 = 36.
b. (5 – 4) x 8.
5 – 4 = 1.
1 x 8 = 8.

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

Question 1.
Use the array to complete the equation.
a.
Eureka Math Grade 3 Module 3 Lesson 9 Problem Set Answer Key 1
3 × 12 =___36______

Answer:
The number of rows = 3.
the number of columns = 12.

Explanation:
In the above-given question,
given that,
The number of rows = 3.
the number of columns = 12.
3 x 12 = 36.

b.
Eureka Math Grade 3 Module 3 Lesson 9 Problem Set Answer Key 2
(3 × 3) × 4
= __9___ × 4
= __36___

Answer:
The number of rows = 3.
the number of columns = 12.

Explanation:
In the above-given question,
given that,
The number of rows = 3.
the number of columns = 12.
(3 x 3) x 4.
9 x 4 = 36.

c.
Eureka Math Grade 3 Module 3 Lesson 9 Problem Set Answer Key 3
3 × 14 = __42___

Answer:
The number of rows = 3.
the number of columns = 14.

Explanation:
In the above-given question,
given that,
The number of rows = 3.
the number of columns = 14.
3 x 14 = 42.

d.
Eureka Math Grade 3 Module 3 Lesson 9 Problem Set Answer Key 4
(___2__ × __3___) × 7
= ___6__ × __7___
= __42___

Answer:
The number of rows = 3.
the number of columns = 14.

Explanation:
In the above-given question,
given that,
The number of rows = 3.
the number of columns = 14.
(2 x 3) x 7.
6 x 7 = 42.

Question 2.
Place parentheses in the equations to simplify. Then, solve. The first one has been done for you.
a.
Eureka Math Grade 3 Module 3 Lesson 9 Problem Set Answer Key 5

Answer:
3 x 16 = 3 x (2 x 8)
(3 x 2) x 8.
6 x 8 = 48.

Explanation:
In the above-given question,
given that,
the equation 3 x 16.
3 x ( 2 x 8 )
( 3 x 2 ) x 8.
6 x 8 = 48.

b.
Eureka Math Grade 3 Module 3 Lesson 9 Problem Set Answer Key 6

Answer:
2 x 14 = 2 x (2 x 7)
(2 x 2) x 7.
4 x 7 = 28.

Explanation:
In the above-given question,
given that,
the equation 2 x 14.
2 x ( 2 x 7 )
( 2 x 2 ) x 7.
4 x 7 = 28.
Eureka-Math-Grade-3-Module-3-Lesson-9-Answer Key-5

c.
Eureka Math Grade 3 Module 3 Lesson 9 Problem Set Answer Key 7

Answer:
3 x 12 = 3 x (3 x 4)
(3 x 3) x 4.
9 x 4 = 36.

Explanation:
In the above-given question,
given that,
the equation 3 x 12.
3 x ( 3 x 4 )
( 3 x 3 ) x 4.
9 x 4 = 36.
Eureka-Math-Grade-3-Module-3-Lesson-9-Answer Key-6

d.
Eureka Math Grade 3 Module 3 Lesson 9 Problem Set Answer Key 8

Answer:
3 x 14 = 3 x (2 x 7)
(3 x 2) x 7.
6 x 7 = 42.

Explanation:
In the above-given question,
given that,
the equation 3 x 14.
3 x ( 2 x 7 )
( 3 x 2 ) x 7.
6 x 7 = 42.
Eureka-Math-Grade-3-Module-3-Lesson-9-Answer Key-7

e.
Eureka Math Grade 3 Module 3 Lesson 9 Problem Set Answer Key 9

Answer:
15 x 3 = 5 x 3 x 3
(5 x 3) x 3.
15 x 3 = 45.

Explanation:
In the above-given question,
given that,
the equation 3 x 15.
3 x ( 5 x 3 )
( 3 x 5 ) x 3.
15 x 3 = 45.
Eureka-Math-Grade-3-Module-3-Lesson-9-Answer Key-8

f.
Eureka Math Grade 3 Module 3 Lesson 9 Problem Set Answer Key 10

Answer:
2 x 15 = 2 x (5 x 3)
(5 x 3) x 2.
15 x 2 = 30.

Explanation:
In the above-given question,
given that,
the equation 2 x 15.
2 x ( 5 x 3)
( 5 x 3 ) x 2.
15 x 2 = 30.
Eureka-Math-Grade-3-Module-3-Lesson-9-Answer Key-9

Question 3.
Charlotte finds the answer to 16 × 2 by thinking about 8 × 4. Explain her strategy.

Answer:
16 x 2 = 32.
8 x 4 = 32.

Explanation:
In the above-given question,
given that,
16 x 2 = 8 x 4.
16 x 2 = 32.
8 x 4 = 32.

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

Simplify to find the answer to 18 × 3. Show your work, and explain your strategy.

Answer:
18 x 3 = 54.

Explanation:
In the above-given question,
given that,
18 x 3 = 54.
3 x 18 = 54.

Eureka Math Grade 3 Module 3 Lesson 9 Worksheet Answer Key

Question 1.
Use the array to complete the equation.
a.
Eureka Math 3rd Grade Module 3 Lesson 9 Worksheet Answer Key 11
3 × 16 = ____48_____

Answer:
The number of rows = 3.
the number of columns = 16.

Explanation:
In the above-given question,
given that,
The number of rows = 3.
the number of columns = 16.
3 x 16 = 48.

b.
Eureka Math 3rd Grade Module 3 Lesson 9 Worksheet Answer Key 12
(3 × __2__) × 8
= __6___ × ___8__
= __48___

Answer:
The number of rows = 3.
the number of columns = 16.

Explanation:
In the above-given question,
given that,
The number of rows = 3.
the number of columns = 16.
(3 x 2) x 8.
6 x 8 = 48.

c.
Eureka Math 3rd Grade Module 3 Lesson 9 Worksheet Answer Key 13
4 × 18 = __72___

Answer:
The number of rows = 4.
the number of columns = 18.

Explanation:
In the above-given question,
given that,
The number of rows = 4.
the number of columns = 18.
4 x 18 = 72.

d.
Eureka Math 3rd Grade Module 3 Lesson 9 Worksheet Answer Key 14
(4 × __2__) × 9
= ___8__ × ___9__
= __72___

Answer:
The number of rows = 4.
the number of columns = 18.

Explanation:
In the above-given question,
given that,
The number of rows = 4.
the number of columns = 18.
(4 x 2) x 9.
8 x 9 = 72.

Question 2.
Place parentheses in the equations to simplify and solve.
a.
Eureka Math 3rd Grade Module 3 Lesson 9 Worksheet Answer Key 15

Answer:
14 x 3 = 2 x 7 x 3
(3 x 2) x 7.
6 x 7 = 42.

Explanation:
In the above-given question,
given that,
the equation 3 x 14.
3 x ( 2 x 7 )
( 3 x 2 ) x 7.
6 x 7 = 42.
Eureka-Math-Grade-3-Module-3-Lesson-9-Answer Key-10

b.
Eureka Math 3rd Grade Module 3 Lesson 9 Worksheet Answer Key 16

Answer:
12 x 3 = 4 x 3 x 3
(3 x 3) x 4.
9 x 4 = 36.

Explanation:
In the above-given question,
given that,
the equation 3 x 12.
3 x ( 3 x 4 )
( 3 x 3 ) x 4.
9 x 4 = 36.
Eureka-Math-Grade-3-Module-3-Lesson-9-Answer Key-11
Question 3.
Solve. Then, match the related facts.
Eureka Math 3rd Grade Module 3 Lesson 9 Worksheet Answer Key 17

Answer:
20 x 2 = 40.
30 x 2 = 60.
35 x 2 = 70.
40 x 2 = 80.

Explanation:
In the above-given question,
given that,
the equations are 20 x 2 = 40.
30 x 2 = 60.
35 x 2 = 70.
40 x 2 = 80.
Eureka-Math-Grade-3-Module-3-Lesson-9-Answer Key-12

Eureka Math Grade 3 Module 3 Lesson 10 Answer Key

Students of Grade 3 can get a strong foundation 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 3, students need to solve all questions and exercises included in Eureka Grade 3 Book.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 10 Answer Key

Teachers and students can find this Eureka Book Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Learning activities are the best option to educate elementary school kids and make them understand the basic mathematical concepts like addition, subtraction, multiplication, division, etc. Grade 3 elementary school students can find these fun-learning exercises for all math concepts through the Eureka Book.

Eureka Math Grade 3 Module 3 Lesson 10 Problem Set Answer Key

Question 1.
Label the arrays. Then, fill in the blanks below to make the statements true.
a. 8 × 8 = __64___
Eureka Math Grade 3 Module 3 Lesson 10 Problem Set Answer Key 1
8 × 8 = 8 × (5 + __3___)
= (8 × 5) + (8 × __3___)
 40     + ___24___
= ____64__

Answer:
8 x 8 = 64.

Explanation:
In the above-given question,
given that,
the number of rows = 8.
the number of columns = 8.
8 x 8 = 8 x ( 5 + 3 ).
(8 x 5 ) + ( 8 x 3)
40 + 24.
64.
Eureka-Math-Grade-3-Module-3-Lesson-10-Answer Key-1

b. 8 × 9 = 9 × 8 = __72___
Eureka Math Grade 3 Module 3 Lesson 10 Problem Set Answer Key 2
9 × 8 = 8 × (5 + __4___)
= (8 × 5) + (8 × __4__ )
 40    + __32____
= __72____

Answer:
9 x 8 = 72.

Explanation:
In the above-given question,
given that,
the number of rows = 8.
the number of columns = 9.
8 x 9 = 8 x ( 5 + 4 ).
(8 x 5 ) + ( 8 x 4)
40 + 32.
72.
Eureka-Math-Grade-3-Module-3-Lesson-10-Answer Key-2

Question 2.
Break apart and distribute to solve 56 ÷ 8.
Eureka Math Grade 3 Module 3 Lesson 10 Problem Set Answer Key 3
56 ÷ 8 = (40 ÷ 8) + ( ____16_____ ÷ 8)
= 5 + _____2_______
= _____7_______

Answer:
56 / 8 = 7.

Explanation:
In the above-given question,
given that,
56 / 8 = (40 / 8 ) + ( 16 / 8 )
40 / 8 = 5.
16 / 8 = 2.
5 + 2 = 7.

Question 3.
Break apart and distribute to solve 72 ÷ 8.
Eureka Math Grade 3 Module 3 Lesson 10 Problem Set Answer Key 4
72 ÷ 8 = (40 ÷ 8) + ( ___32______ ÷ 8)
= 5 + ______4______
= _______9_____

Answer:
72 / 8 = 9.

Explanation:
In the above-given question,
given that,
72 / 8 = (40 / 8 ) + ( 32 / 8 )
40 / 8 = 5.
32 / 8 = 4.
5 + 4 = 9.
Eureka-Math-Grade-3-Module-3-Lesson-10-Answer Key-3

Question 4.
An octagon has 8 sides. Skip-count to find the total number of sides on 9 octagons.
Eureka Math Grade 3 Module 3 Lesson 10 Problem Set Answer Key 5
Nine octagons have a total of _____72____ sides.

Answer:
8 x 9 = 72.

Explanation:
In the above-given question,
given that,
An octagon has 8 sides.
the total number of sides o 9 octagons.
8 x 9 = 72.
Eureka-Math-Grade-3-Module-3-Lesson-10-Answer Key-4

Question 5.
Multiply.
Eureka Math Grade 3 Module 3 Lesson 10 Problem Set Answer Key 6

Answer:
8 x 6 = 48, 3 x 8 = 24, 8 x 10 = 80, 8 x 8 = 64, 7 x 8 = 56.

Explanation:
In the above-given question,
given that,
the tables of 7 and 8.
8 x 6 = 48.
3 x 8 = 24.
8 x 10 = 80.
8 x 8 = 64.
7 x 8 = 56.
Eureka-Math-Grade-3-Module-3-Lesson-10-Answer Key-5

Question 6.
Match
Eureka Math Grade 3 Module 3 Lesson 10 Problem Set Answer Key 7

Answer:
24 / 8 = 3.
32 / 8 = 4.
16 / 8 = 2.
64 / 8 = 8.
48 / 8 = 6.
72 / 8 = 9.

Explanation:
In the above-given question,
given that,
8 x 3 = 24, 8 x 4 = 32, 8 x 2 = 16, 8 x 8 = 64, 8 x 6 = 48, 8 x 9 = 72.
24 / 8 = 3.
32 / 8 = 4.
16 / 8 = 2.
64 / 8 = 8.
48 / 8 = 6.
72 / 8 = 9.
Eureka-Math-Grade-3-Module-3-Lesson-10-Answer Key-6

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

Use the break apart and distribute strategy to solve the following problem. You may choose whether or not to draw an array.
7 × 8 =__56___

Answer:
7 x 8 = 56.

Explanation:
In the above-given question,
given that,
the number of rows = 7.
the number of columns = 8.
7 x 8 = 7 x ( 5 + 3 ).
(7 x 5 ) + ( 7 x 3)
35 + 21.
56.
Eureka-Math-Grade-3-Module-3-Lesson-10-Answer Key-7

Eureka Math Grade 3 Module 3 Lesson 10 Homework Answer Key

Question 1.
Label the array. Then, fill in the blanks to make the statements true.
8 × 7 = 7 × 8 =__56___
Eureka Math Grade 3 Module 3 Lesson 10 Homework Answer Key 8
8 × 7 = 7 × (5 + __3___)
= (7 × 5) + (7 × __3__)
  35     + ___21___
= __56____

Answer:
7 x 8 = 56.

Explanation:
In the above-given question,
given that,
the number of rows = 7.
the number of columns = 8.
7 x 8 = 7 x ( 5 + 3 ).
(7 x 5 ) + ( 7 x 3)
35 + 21.
56.
Eureka-Math-Grade-3-Module-3-Lesson-10-Answer Key-8

Question 2.
Break apart and distribute to solve 72 ÷ 8.
Eureka Math Grade 3 Module 3 Lesson 10 Homework Answer Key 9
72 ÷ 8 = (40 ÷ 8) + ( ____32_____ ÷ 8)
= 5 + ______4______
= ______9______

Answer:
72 / 8 = 9.

Explanation:
In the above-given question,
given that,
72 / 8 = (40 / 8 ) + ( 32 / 8 )
40 / 8 = 5.
32 / 8 = 4.
5 + 4 = 9.

Question 3.
Count by 8. Then, match each multiplication problem with its value.
8 , ___16____, ___24____, ____32___, ____40___, ___48____, ___56____, ____64___, ___72____, ___80____
Eureka Math Grade 3 Module 3 Lesson 10 Homework Answer Key 10

Answer:
8 x 2 = 16, 8 x 3 = 24, 8 x 4 = 32, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64, 8 x 9 = 72, and 8 x 10 = 8.

Explanation:
In the above-given question,
given that,
the table of 8 and match it with the given equations.
8 x 2 = 16, 8 x 3 = 24, 8 x 4 = 32, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64, 8 x 9 = 72, and 8 x 10 = 8.
Eureka-Math-Grade-3-Module-3-Lesson-10-Answer Key-9

Question 4.
Divide
Eureka Math Grade 3 Module 3 Lesson 10 Homework Answer Key 11

Answer:
16 / 8 = 2.
40 / 8 = 5.
32 / 8 = 4.
48 / 8 = 6.
56 / 8 = 7.
72 / 8 = 9.

Explanation:
In the above-given question,
given that,
the division of 8.
16 / 8 = 2.
40 / 8 = 5.
32 / 8 = 4.
48 / 8 = 6.
56 / 8 = 7.
72 / 8 = 9.
Eureka-Math-Grade-3-Module-3-Lesson-10-Answer Key-10

 

Eureka Math Grade 3 Module 3 Lesson 11 Answer Key

Teachers and students can find this Eureka Book Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Learning activities are the best option to educate elementary school kids and make them understand the basic mathematical concepts like addition, subtraction, multiplication, division, etc. Grade 3 elementary school students can find these fun-learning exercises for all math concepts through the Eureka Book.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 11 Answer Key

Students of Grade 3 can get a strong foundation 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 3, students need to solve all questions and exercises included in Eureka Grade 3 Book.

Eureka Math Grade 3 Module 3 Lesson 11 Pattern Sheet Answer Key

Multiply.

Answer:Eureka Math Grade 3 Module 3 Lesson 11 Pattern Sheet Answer Key 1
8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 4 = 8 x 5 = 40, 8 x 1 = 8, 8 x 2 = 16, 8 x 1 = 8, 8 x 3 = 24, 8 x 1 = 8,  8 x 4 = 32, 8 x 1 = 8, 8 x 5 = 40, 8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 2 = 16, 8 x 4 = 32, 8 x2 = 16, 8 x 5 = 40, 8 x 2 = 16, 8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 1 = 8, 8 x 3 = 24, 8 x 2 = 16, 8 x 3 = 24, 8 x 4 = 32, 8 x 3 = 24, 8 x 5 = 40, 8 x 3 = 24, 8 x 4 = 32, 8 x 1 = 8, 8 x 4 = 32, 8 x 2 = 16, 8 x 4 = 32, 8 x 3 = 24, 8 x 4 = 32, 8 x 5 = 40, 8 x 4 = 32, 8 x 5 = 40, 8 x 1 = 8, 8 x 5 = 40, 8 x 2 = 16, 8 x 5 = 40, 8 x 3 = 24, 8 x 5 = 40, 8 x 4 = 32, 8 x 2 = 16, 8 x 4 = 32, 8 x 3 = 24, 8 x 5 = 40, 8 x 3 = 24, 8 x 2 = 16, 8 x 4 = 32, 8 x 3 = 24, 8 x 5 = 40, 8 x 2 = 16, 8 x 4 = 32.

Explanation:
In the above-given question,
given that,
the tables of 8.
8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 4 = 8 x 5 = 40, 8 x 1 = 8, 8 x 2 = 16, 8 x 1 = 8, 8 x 3 = 24, 8 x 1 = 8,  8 x 4 = 32, 8 x 1 = 8, 8 x 5 = 40, 8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 2 = 16, 8 x 4 = 32, 8 x2 = 16, 8 x 5 = 40, 8 x 2 = 16, 8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 1 = 8, 8 x 3 = 24, 8 x 2 = 16, 8 x 3 = 24, 8 x 4 = 32, 8 x 3 = 24, 8 x 5 = 40, 8 x 3 = 24, 8 x 4 = 32, 8 x 1 = 8, 8 x 4 = 32, 8 x 2 = 16, 8 x 4 = 32, 8 x 3 = 24, 8 x 4 = 32, 8 x 5 = 40, 8 x 4 = 32, 8 x 5 = 40, 8 x 1 = 8, 8 x 5 = 40, 8 x 2 = 16, 8 x 5 = 40, 8 x 3 = 24, 8 x 5 = 40, 8 x 4 = 32, 8 x 2 = 16, 8 x 4 = 32, 8 x 3 = 24, 8 x 5 = 40, 8 x 3 = 24, 8 x 2 = 16, 8 x 4 = 32, 8 x 3 = 24, 8 x 5 = 40, 8 x 2 = 16, 8 x 4 = 32.
Eureka-Math-Grade-3-Module-3-Lesson-11-Answer Key-1

Eureka Math Grade 3 Module 3 Lesson 11 Problem Set Answer Key

Question 1.
Ms. Santor divides 32 students into 8 equal groups for a field trip. Draw a tape diagram, and label the number of students in each group as n. Write an equation, and solve for n.

Answer:
32 / n = 8.
n = 4.

Explanation:
In the above-given question,
given that,
Ms. Santor divides 32 students into 8 equal groups for a field trip.
the number of students in each group as n.
32 / n = 8.
n = 4.
Eureka-Math-Grade-3-Module-3-Lesson-11-Answer Key-2

Question 2.
Tara buys 6 packs of printer paper. Each pack of paper costs $8. Draw a tape diagram, and label the total amount she spends as m. Write an equation, and solve for m.

Answer:
The total amount she spends = $48.

Explanation:
In the above-given question,
given that,
Tara buys 6 packs of printer paper.
Each pack of paper costs $8.
6 x m = $8.
m = 6 x 8 = $48.
Eureka-Math-Grade-3-Module-3-Lesson-11-Answer Key-3

Question 3.
Mr. Reed spends $24 on coffee beans. How many kilograms of coffee beans does he buy? Draw a tape diagram, and label the total amount of coffee beans he buys as c. Write an equation, and solve for c.
Eureka Math Grade 3 Module 3 Lesson 11 Problem Set Answer Key 2

Answer:
The total amount of coffee beans he buys = 3 kg.

Explanation:
In the above-given question,
given that,
Mr. Reed spends $24 on coffee beans.
$24 / c = 8.
c = 24 / 8.
c = 3.
Eureka-Math-Grade-3-Module-3-Lesson-11-Answer Key-4

Question 4.
Eight boys equally share 4 packs of baseball cards. Each pack contains 10 cards. How many cards does each boy get?

Answer:
The number of cards each boy gets = 40.

Explanation:
In the above-given question,
given that,
Eight boys equally share 4 packs of baseball cards.
each pack contains 10 cards.
10 x 4 = 40.
so the number of cards each boy gets = 40.

Question 5.
There are 8 bags of yellow and green balloons. Each bag contains 7 balloons. If there are 35 yellow balloons, how many green balloons are there?

Answer:
The number of green balloons = 21.

Explanation:
In the above-given question,
given that,
There are 8 bags of yellow and green balloons.
each bag contains 7 balloons.
if there are 35 yellow balloons.
8 x 7 = 56.
56 – 35 = 21.
so the number of green balloons = 21.

Question 6.
The fruit seller packs 72 oranges into bags of 8 each. He sells all the oranges at $4 a bag. How much money did he receive?

Answer:
The amount of money he receives = 40.

Explanation:
In the above-given question,
given that,
The fruit seller packs 72 oranges into bags of 8 each.
He sells all the oranges at $4 a bag.
8 x 4 = 32.
72 – 32 = 40.
so the amount of money he receives = 40.

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

Erica buys some packs of rubber bracelets. There are 8 bracelets in each pack.

a. How many packs of rubber bracelets does she buy if she has a total of 56 bracelets? Draw a tape diagram, and label the total number of packages as p. Write an equation, and solve for p.

Answer:
The total amount of packages = 7.

Explanation:
In the above-given question,
given that,
Erica buys some packs of rubber bracelets.
There are 8 bracelets in each pack.
8 / p = 56.
p = 56 / 8.
p = 7.
so the total amount of packages = 7.
Eureka-Math-Grade-3-Module-3-Lesson-11-Answer Key-5

b. After giving some bracelets away, Erica has 18 left. How many bracelets did she give away?

Answer:
The number of bracelets did she give away = 9.

Explanation:
In the above-given question,
given that,
After giving some bracelets away, Erica has 18 left.
18 / 2 = 9.
so the number of bracelets did she give away = 9.

Eureka Math Grade 3 Module 3 Lesson 11 Homework Answer Key

Question 1.
Jenny bakes 10 cookies. She puts 7 chocolate chips on each cookie. Draw a tape diagram, and label the total amount of chocolate chips as c. Write an equation, and solve for c.

Answer:
The total amount of chocolate chips = 70.

Explanation:
In the above-given question,
given that,
Jenny bakes 10 cookies.
she puts 7 chocolate chips on each cookie.
10 x 7 = c.
c = 70.
Eureka-Math-Grade-3-Module-3-Lesson-11-Answer Key-6

Question 2.
Mr. Lopez arranges 48 dry erase markers into 8 equal groups for his math stations. Draw a tape diagram, and label the number of dry erase markers in each group as v. Write an equation, and solve for v.

Answer:
The number of dry erase markers in each group = 6.

Explanation:
In the above-given question,
given that,
Mr. Lopez arranges 48 dry erase markers into 8 equal groups for his math stations.
48 / v = 8.
v = 48 / 8.
v = 6.
Eureka-Math-Grade-3-Module-3-Lesson-11-Answer Key-7

Question 3.
There are 35 computers in the lab. Five students each turn off an equal number of computers.
How many computers does each student turn off? Label the unknown as m, and then solve.

Answer:
The number of computers does each student turn off = 7.

Explanation:
In the above-given question,
given that,
There are 35 computers in the lab.
Five students each turn off an equal number of computers.
5 / m = 35.
m = 35 / 5.
m = 7.
Eureka-Math-Grade-3-Module-3-Lesson-11-Answer Key-8

Question 4.
There are 9 bins of books. Each bin has 6 comic books. How many comic books are there altogether?

Answer:
The number of comic books is there altogether = 54.

Explanation:
In the above-given question,
given that,
There are 9 bins of books.
Each bin has 6 comic books.
9 x 6 = 54.
so the number of comic books is there altogether = 54.

Question 5.
There are 8 trail mix bags in one box. Clarissa buys 5 boxes. She gives an equal number of bags of trail mix to 4 friends. How many bags of trail mix does each friend receive?

Answer:
The bags of trail mix does each friend receives = 40.

Explanation:
In the above-given question.
given that,
There are 8 trail mix bags in one box.
Clarissa buys 5 boxes.
she gives an equal number of bags of trail mix to 4 friends.
8 x 5 = 40.

Question 6.
Leo earns $8 each week for doing chores. After 7 weeks, he buys a gift and has $38 left. How much money does he spend on the gift?

Answer:
The amount he spends on the gift = $18.

Explanation:
In the above-given question,
given that,
Leo earns $8 each week for doing chores.
After 7 weeks, he buys a gift and has $38 left.
7 x 8 = 56.
56 – 38 = $18.
so the amount he spends on the gift = $18.

Eureka Math Grade 3 Module 3 Lesson 12 Answer Key

Students of Grade 3 can get a strong foundation 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 3, students need to solve all questions and exercises included in Eureka Grade 3 Book.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 12 Answer Key

Teachers and students can find this Eureka Book Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Learning activities are the best option to educate elementary school kids and make them understand the basic mathematical concepts like addition, subtraction, multiplication, division, etc. Grade 3 elementary school students can find these fun-learning exercises for all math concepts through the Eureka Book.

Eureka Math Grade 3 Module 3 Lesson 12 Pattern Sheet Answer Key

Multiply

Eureka Math Grade 3 Module 3 Lesson 12 Pattern Sheet Answer Key 1

Answer:
8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64, 8 x 9 = 72, 8 x 10 = 80, 8 x 5 = 40, 8 x 6 = 48, 8 x 5 = 40, 8 x 7 = 56, 8 x 5 = 40, 8 x 6 = 48, 8 x 5 = 40, 8 x 7 = 56, 8 x 5 = 40, 8 x 8 = 64, 8 x 5 = 40, 8 x 9 = 72, 8 x 5 = 40, 8 x 10 = 80, 8 x 6 = 48, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 6 = 48, 8 x 8 = 64, 8 x 6 = 48, 8 x 9 = 72, 8 x 6 = 48, 8 x 7 = 56, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64, 8 x 7 = 56, 8 x 9 = 72, 8 x 7 = 56, 8 x 8 = 64, 8 x 6 48, 8 x 8 = 64, 8 x 7 = 56, 8 x 8 = 64, 8 x 9 = 72, 8 x 9 = 72, 8 x 6 = 48, 8 x 9 = 72, 8 x 7 = 56, 8 x 9 = 72, 8 x 8 = 64, 8 x 9 = 72, 8 x 8 = 64, 8 x 6 = 48, 8 x 9 = 72, 8 x 7 = 56, 8 x 9 = 72, 8 x 6 = 48, 8 x 8 = 64, 8 x 9 = 72, 8 x 7 = 56, 8 x 6 = 48, 8 x 8 = 64.

Explanation:
In the above-given question,
given that,
the table of 8 and multiply.
8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64, 8 x 9 = 72, 8 x 10 = 80, 8 x 5 = 40, 8 x 6 = 48, 8 x 5 = 40, 8 x 7 = 56, 8 x 5 = 40, 8 x 6 = 48, 8 x 5 = 40, 8 x 7 = 56, 8 x 5 = 40, 8 x 8 = 64, 8 x 5 = 40, 8 x 9 = 72, 8 x 5 = 40, 8 x 10 = 80, 8 x 6 = 48, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 6 = 48, 8 x 8 = 64, 8 x 6 = 48, 8 x 9 = 72, 8 x 6 = 48, 8 x 7 = 56, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64, 8 x 7 = 56, 8 x 9 = 72, 8 x 7 = 56, 8 x 8 = 64, 8 x 6 48, 8 x 8 = 64, 8 x 7 = 56, 8 x 8 = 64, 8 x 9 = 72, 8 x 9 = 72, 8 x 6 = 48, 8 x 9 = 72, 8 x 7 = 56, 8 x 9 = 72, 8 x 8 = 64, 8 x 9 = 72, 8 x 8 = 64, 8 x 6 = 48, 8 x 9 = 72, 8 x 7 = 56, 8 x 9 = 72, 8 x 6 = 48, 8 x 8 = 64, 8 x 9 = 72, 8 x 7 = 56, 8 x 6 = 48, 8 x 8 = 64.
Eureka-Math-Grade-3-Module-3-Lesson-12-Answer Key-1

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

Question 1.
Each Eureka Math Grade 3 Module 3 Lesson 12 Problem Set Answer Key 2 has a value of 9. Find the value of each row. Then, add the rows to find the total.

a. 6 × 9 = ___54____
Eureka Math Grade 3 Module 3 Lesson 12 Problem Set Answer Key 3
6 × 9 = (5 + 1) × 9
= (5 × 9) + (1 × 9)
= 45 + ___9__
= __54___

Answer:
6 x 9 = 54.

Explanation:
In the above-given question,
given that,
6 x 9 = ( 5 + 1 ) x 9.
( 5 x 9 ) + ( 1 x 9 )
45 + 9
54.
Eureka-Math-Grade-3-Module-3-Lesson-12-Answer Key-2

b. 7 × 9 = ____63___
Eureka Math Grade 3 Module 3 Lesson 12 Problem Set Answer Key 4
7 × 9 = (5 + __2__) × 9
= (5 × 9) + (__2__ × 9)
= 45 + __18___
= ___63__

Answer:
7 x 9 = 63.

Explanation:
In the above-given question,
given that,
7 x 9 = ( 5 + 2 ) x 9.
( 5 x 9 ) + ( 2 x 9 )
45 + 18.
63.
Eureka-Math-Grade-3-Module-3-Lesson-12-Answer Key-3

c. 8 × 9 = ___72____
Eureka Math Grade 3 Module 3 Lesson 12 Problem Set Answer Key 5
8 × 9 = (5 + ___3__) × 9
= (5 × 9) + (___3__ × __9___)
= 45 + __27___
= __72___

Answer:
8 x 9 = 72.

Explanation:
In the above-given question,
given that,
8 x 9 = ( 5 + 3 ) x 9.
( 5 x 9 ) + ( 3 x 9 )
45 + 27.
72.
Eureka-Math-Grade-3-Module-3-Lesson-12-Answer Key-4

d. 9 × 9 = ____81___
Eureka Math Grade 3 Module 3 Lesson 12 Problem Set Answer Key 6
9 × 9 = (5 + ___4__) × 9
= (5 × 9) + (__4___ × ___9__)
= 45 + __36___
= __81___

Answer:
9 x 9 = 81.

Explanation:
In the above-given question,
given that,
9 x 9 = ( 5 + 4 ) x 9.
( 5 x 9 ) + ( 4 x 9 )
45 + 36.
81.
Eureka-Math-Grade-3-Module-3-Lesson-12-Answer Key-5

Question 2.
Find the total value of the shaded blocks.

a. 9 × 6 = 54
Eureka Math Grade 3 Module 3 Lesson 12 Problem Set Answer Key 7
9 sixes = 10 sixes – 1 six
=__60___ – 6
= __54___

Answer:
9 x 6 = 54.

Explanation:
In the above-given equation,
given that,
9 sixes = 10 sixes – 1 six.
60 – 6.
54.

b. 9 × 7 = 63
Eureka Math Grade 3 Module 3 Lesson 12 Problem Set Answer Key 8
9 sevens = 10 sevens – 1 seven
= __70___ – 7
= __63___

Answer:
9 x 7 = 63.

Explanation:
In the above-given equation,
given that,
9 sevens = 10 sevens – 1 seven.
70 – 7.
63.

c. 9 × 8 = 72.
Eureka Math Grade 3 Module 3 Lesson 12 Problem Set Answer Key 9
9 eights = 10 eights – 1 eight
= __80___ – 8
= __72___

Answer:
9 x 8 = 72.

Explanation:
In the above-given equation,
given that,
9 eights = 10 eights – 1 eight.
80 – 8.
72.

d. 9 × 9 = 81.
Eureka Math Grade 3 Module 3 Lesson 12 Problem Set Answer Key 10
9 nines = 10 nines – 1 nine
= __90___ – __9___
= __81___

Answer:
9 x 9 = 81.

Explanation:
In the above-given equation,
given that,
9 nines = 10 nines – 1 nine.
90 – 9.
81.

Question 3.
Matt buys a pack of postage stamps. He counts 9 rows of 4 stamps. He thinks of 10 fours to find the total number of stamps. Show the strategy that Matt might have used to find the total number of stamps.

Answer:
9 x 4 = 36.

Explanation:
In the above-given equation,
given that,
9 fours = 10 fourss – 1 four.
40 – 4.
36.

Question 4.
Match
Eureka Math Grade 3 Module 3 Lesson 12 Problem Set Answer Key 11

Answer:
3 x 9 = 27, 45 / 9 = 5.
9 x 9 = 81, 9 / 9 = 1.
8 x 9 = 72, 90 / 9 = 10.
9 x 4 = 36, 72 / 9 = 8.
2 x 9 = 18, 54 / 9 = 6.

Explanation:
In the above-given question,
given that,
the multiplication and division of 9.
3 x 9 = 27, 45 / 9 = 5.
9 x 9 = 81, 9 / 9 = 1.
8 x 9 = 72, 90 / 9 = 10.
9 x 4 = 36, 72 / 9 = 8.
2 x 9 = 18, 54 / 9 = 6.
Eureka-Math-Grade-3-Module-3-Lesson-12-Answer Key-6

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

Question 1.
Each Eureka Math 3rd Gradde Module 3 Lesson 12 Exit Ticket Answer Key 12 has a value of 9. Complete the equations to find the total value of the tower of blocks.
Eureka Math 3rd Grade Module 3 Lesson 12 Exit Ticket Answer Key 13
___5__ × 9 = (5 + __1__ ) × 9
= (5 × ___9__ ) + ( __1___ × __9___ )
= 45 + __9___
= __54___

Answer:
6 x 9 = 54.

Explanation:
In the above-given question,
given that,
5 x 9 = ( 5 + 1 ) x 9.
( 5 x 9 ) + ( 1 x 9 )
45 + 9.
54.

Question 2.
Hector solves 9 × 8 by subtracting 1 eight from 10 eights. Draw a model, and explain Hector’s strategy.

Answer:
9 x 8 = 72.

Explanation:
In the above-given question,
given that,
9 x 8 = 72.
10 eights – 1 eight.
80 – 8.
72.

Eureka Math Grade 3 Module 3 Lesson 12 Homework Answer Key

Question 1.
Find the value of each row. Then, add the rows to find the total.

a. Each Eureka Math 3rd Grade Module 3 Lesson 12 Homework Answer Key 15 has a value of 6.
9 × 6 = ___54_____
Eureka Math 3rd Grade Module 3 Lesson 12 Homework Answer Key 14
9 × 6 = (5 + 4) × 6
= (5 × 6) + (4 × 6)
= 30 + __24___
= __54___

Answer:
6 x 9 = 54.

Explanation:
In the above-given question,
given that,
6 x 9 = ( 5 + 4 ) x 6.
( 5 x 6 ) + ( 4 x 6 )
30 + 24.
54.
Eureka-Math-Grade-3-Module-3-Lesson-12-Answer Key-7

b. Each Eureka Math 3rd Grade Module 3 Lesson 12 Homework Answer Key 15 has a value of 7.
9 × 7 = ___63_____
Eureka Math 3rd Grade Module 3 Lesson 12 Homework Answer Key 16
9 × 7 = (5 + __4___) × 7
= (5 × 7) + (__4___ × 7)
= 35 + __28___
= ___63__

Answer:
9 x 7 = 63.

Explanation:
In the above-given question,
given that,
9 x 7 = ( 5 + 4 ) x 7.
( 5 x 7 ) + ( 4 x 9 )
35 + 28.
63.
Eureka-Math-Grade-3-Module-3-Lesson-12-Answer Key-8

c. Each Eureka Math 3rd Grade Module 3 Lesson 12 Homework Answer Key 15 has a value of 8.
9 × 8 = ____72____
Eureka Math 3rd Grade Module 3 Lesson 12 Homework Answer Key 17
9 × 8 = (5 + __4___) × 8
= (5 × 8) + (__4___ × ___8__)
= 40 + ___32__
= __72___

Answer:
9 x 8 = 72.

Explanation:
In the above-given question,
given that,
9 x 8 = ( 5 + 4 ) x 8.
( 5 x 8 ) + ( 4 x 8 )
40 + 32.
72.
Eureka-Math-Grade-3-Module-3-Lesson-12-Answer Key-9

d. EachEureka Math 3rd Grade Module 3 Lesson 12 Homework Answer Key 15 has a value of 9.
9 × 9 = ____81____
Eureka Math 3rd Grade Module 3 Lesson 12 Homework Answer Key 18
9 × 9 = (5 + _____) × 9
= (5 × 9) + (__4___ × __9___)
= 45 + ___36__
= __81___

Answer:
9 x 9 = 81.

Explanation:
In the above-given question,
given that,
9 x 9 = ( 5 + 4 ) x 9.
( 5 x 9 ) + ( 4 x 9 )
45 + 36.
81.
Eureka-Math-Grade-3-Module-3-Lesson-12-Answer Key-10

Question 2.
Match
Eureka Math 3rd Grade Module 3 Lesson 12 Homework Answer Key 19
Answer:
50 – 5 = 45, 9 x 5 = 45.
60 – 6 = 54, 9 x 6 = 54.
70 – 7 = 63, 9 x 7 = 63.
80 – 8 = 72, 9 x 8 = 72.
90 – 9 = 81, 9 x 9 = 81.
40 – 4 = 36, 9 x 4 = 36.

Explanation:
In the above-given question,
given that,
the multiplication and division of 9 is given.
50 – 5 = 45, 9 x 5 = 45.
60 – 6 = 54, 9 x 6 = 54.
70 – 7 = 63, 9 x 7 = 63.
80 – 8 = 72, 9 x 8 = 72.
90 – 9 = 81, 9 x 9 = 81.
40 – 4 = 36, 9 x 4 = 36.

Eureka-Math-Grade-3-Module-3-Lesson-12-Answer Key-11

Eureka Math Grade 3 Module 3 Lesson 1 Answer Key

Teachers and students can find this Eureka Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Eureka Math Answer Key for Grade 3 aids teachers to differentiate instruction, building and reinforcing foundational mathematics skills that alter from the classroom to real life. With the help of Eureka’s primary school Grade 3 Answer Key, you can think deeply regarding what you are learning, and you will really learn math easily just like that.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 1 Answer Key

Students of Grade 3 Module 3 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 3, students need to solve all questions and exercises included in Eureka’s Grade 3 Textbook.

Eureka Math Grade 3 Module 3 Lesson 1 Sprint Answer Key

Mixed Multiplication

Eureka Math Grade 3 Module 3 Lesson 1 Sprint Answer Key 1

Answer:
2 x 1 = 2, 2 x 2 = 4, 2 x 3 = 6, 4 x 1 = 4, 4 x 2 = 8, 4 x 3 = 12, 1 x 6 = 6, 2 x 6 = 12, 1 x 8 = 8, 2 x 8 = 16, 3 x 1 = 3, 3 x 2 = 6, 3 x 3 = 9, 5 x 1 = 5, 5 x 2 = 10, 5 x 3 = 15, 1 x 7 = 7, 2 x 7 = 14, 1 x 9 = 9, 2 x 9 = 18, 2 x 5 = 10, 2 x 6 = 12,
2 x 7 = 14, 5 x 5 = 25, 5 x 6 = 30, 5 x 7 = 35, 4 x 5 = 20, 4 x 6 = 24, 4 x 7 = 28, 3 x 5 = 15, 3 x 6 = 18, 3 x 7 = 21, 2 x 7 = 14, 2 x 8 = 16, 2 x 9 = 18, 5 x 7 = 35, 5 x 8 = 40, 5 x 9 = 45, 4 x 7 = 28, 4 x 8 = 32, 4 x 9 = 36, 3 x 7 = 21, 3 x 8 = 24, 3 x 9 = 27.

Explanation:
In the above-given question,
given that,
The tables of 1, 2, 3, 4, and 5.

2 x 1 = 2, 2 x 2 = 4, 2 x 3 = 6, 4 x 1 = 4, 4 x 2 = 8, 4 x 3 = 12, 1 x 6 = 6, 2 x 6 = 12, 1 x 8 = 8, 2 x 8 = 16, 3 x 1 = 3, 3 x 2 = 6, 3 x 3 = 9, 5 x 1 = 5, 5 x 2 = 10, 5 x 3 = 15, 1 x 7 = 7, 2 x 7 = 14, 1 x 9 = 9, 2 x 9 = 18, 2 x 5 = 10, 2 x 6 = 12,
2 x 7 = 14, 5 x 5 = 25, 5 x 6 = 30, 5 x 7 = 35, 4 x 5 = 20, 4 x 6 = 24, 4 x 7 = 28, 3 x 5 = 15, 3 x 6 = 18, 3 x 7 = 21, 2 x 7 = 14, 2 x 8 = 16, 2 x 9 = 18, 5 x 7 = 35, 5 x 8 = 40, 5 x 9 = 45, 4 x 7 = 28, 4 x 8 = 32, 4 x 9 = 36, 3 x 7 = 21, 3 x 8 = 24, 3 x 9 = 27.
Eureka-Math-Grade-3-Module-3-Lesson-1-Answer Key-1

Eureka Math Grade 3 Module 3 Lesson 1 Sprint Answer Key 2

Answer:
5 x 1 = 5, 5 x 2 = 10, 5 x 3 = 15, 3 x 1 = 3, 3 x 2 = 6, 3 x 3 = 9, 1 x 7 = 7, 2 x 7 = 14, 1 x 9 = 9, 2 x 9 = 18, 2 x 1 = 2, 2 x 2 = 4, 2 x 3 = 6, 4 x 1 = 4, 4 x 2 = 8, 4 x 3 = 12, 1 x 6 = 6, 2 x 6 = 12, 1 x 8 = 8, 2 x 8 = 16, 5 x 5 = 25, 5 x 6 = 30,
5 x 7 = 35, 2 x 5 = 10, 2 x 6 = 12, 2 x 7 = 14, 3 x 5 = 15, 3 x 6 = 18, 3 x 7 = 21, 4 x 5 = 20, 4 x 6 = 24, 4 x 7 = 28, 5 x 7 = 35, 5 x 8 = 40, 5 x 9 = 45, 2 x 7 = 14, 2 x 8 = 16, 2 x 9 = 18, 3  x 7 = 21, 3 x 8 = 24, 3 x 9 = 27, 4 x 7 = 28, 4 x 8 = 32, 4 x 9 = 36.

Explanation:
In the above-given question,
given that,
the tables of 1, 2, 3, 4, and 5.
5 x 1 = 5, 5 x 2 = 10, 5 x 3 = 15, 3 x 1 = 3, 3 x 2 = 6, 3 x 3 = 9, 1 x 7 = 7, 2 x 7 = 14, 1 x 9 = 9, 2 x 9 = 18, 2 x 1 = 2, 2 x 2 = 4, 2 x 3 = 6, 4 x 1 = 4, 4 x 2 = 8, 4 x 3 = 12, 1 x 6 = 6, 2 x 6 = 12, 1 x 8 = 8, 2 x 8 = 16, 5 x 5 = 25, 5 x 6 = 30,
5 x 7 = 35, 2 x 5 = 10, 2 x 6 = 12, 2 x 7 = 14, 3 x 5 = 15, 3 x 6 = 18, 3 x 7 = 21, 4 x 5 = 20, 4 x 6 = 24, 4 x 7 = 28, 5 x 7 = 35, 5 x 8 = 40, 5 x 9 = 45, 2 x 7 = 14, 2 x 8 = 16, 2 x 9 = 18, 3  x 7 = 21, 3 x 8 = 24, 3 x 9 = 27, 4 x 7 = 28, 4 x 8 = 32, 4 x 9 = 36.

Eureka-Math-Grade-3-Module-3-Lesson-1-Answer Key-2

Eureka Math Grade 3 Module 3 Lesson 1 Problem Set Answer Key

Question 1.
a. Solve. Shade in the multiplication facts that you already know. Then, shade in the facts for sixes, sevens, eights, and nines that you can solve using the commutative property
Eureka Math Grade 3 Module 3 Lesson 1 Problem Set Answer Key 3

Answer:
The facts of 6 are 2, 3, and 12.
the facts of 7 are 7, 1, 14.
the facts of 8 are 8, 16.
the facts of 9 are 18.

Explanation:
In the above-given question,
given that,
the facts of 6, 7, 8, and 9.
The facts of 6 are 2, 3, and 12.
the facts of 7 are 7, 1, 14.
the facts of 8 are 8, 16.
the facts of 9 are 18.

Eureka-Math-Grade-3-Module-3-Lesson-1-Answer Key-3

b. Complete the chart. Each bag contains 7 apples.
Eureka Math Grade 3 Module 3 Lesson 1 Problem Set Answer Key 4

Answer:
The number of bags is 2, the total number of apples = 14.
The number of bags is 3, the total number of apples = 21.
The number of bags is 4, the total number of apples = 28.
The number of bags is 5, the total number of apples = 35.
The number of bags is 6, the total number of apples = 42.

Explanation:
In the above-given question,
given that,
Each bag contains 7 apples.
The number of bags is 2, the total number of apples = 14.
The number of bags is 3, the total number of apples = 21.
The number of bags is 4, the total number of apples = 28.
The number of bags is 5, the total number of apples = 35.
The number of bags is 6, the total number of apples = 42.
Eureka-Math-Grade-3-Module-3-Lesson-1-Answer Key-4

Question 2.
Use the array to write two different multiplication sentences.
Eureka Math Grade 3 Module 3 Lesson 1 Problem Set Answer Key 5

Answer:
The number of rows is 4, the number of columns = 6.
6 x 4 = 24.
4 x 6 = 24.

Explanation:
In the above-given question,
given that,
the array contains the number of rows and number of columns.
The number of rows is 4, the number of columns = 6.
6 x 4 = 24.
4 x 6 = 24.
Eureka-Math-Grade-3-Module-3-Lesson-1-Answer Key-5

Question 3.
Complete the equations.

a. 2 sevens = __seven_____ twos
= ____14____

Answer:
2 sevens = seven twos.
14 = 14.

Explanation:
In the above-given question,
given that,
2 sevens = seven twos.
2 x 7 = 7 x 2.
14 = 14.

b. 3 ___sixs_____ = 6 threes
= ___18_____

Answer:
3 sixs = 6 threes.
18 = 18.

Explanation:
In the above-given question,
given that,
3 sixs = 6 threes.
3 x 6 = 6 x 3.
18 = 18.

c. 10 eights = 8 __tens______
= ___80_____

Answer:
10 eights = 8 tens.
80 = 80.

Explanation:
In the above-given question,
given that,
10 eights = 8 tens.
10 x 8 = 80 and 8 x 10 = 80.

d. 4 × __six_____ = 6 × 4
= ___24_____

Answer:
4 x six = 6 x 4.
24 = 24.

Explanation:
In the above-given question,
given that,
4 x six = six x 4.
4 x 6 = 6 x 4.
24 = 24.

e. 8 × 5 = ____five____ × 8
= __40______

Answer:
8 x 5 = five x 8.
40 = 40.

Explanation:
In the above-given question,
given that,
8 x 5 = 5 x 8.
40 = 40.

f. ___5____ × 7 = 7 × ___five____
= __35______

Answer:
5 x 7 = 7 x five.
35 = 35.

Explanation:
In the above-given question,
given that,
5 x 7 = 7 x five.
35 = 35.

g. 3 × 9 = 10 threes – ___1____ three
= __27______

Answer:
3 x 9 = 10 threes – 1 three.
27 = 30 – 3.

Explanation:
In the above-given question,
given that,
3 x 9 = 10 threes – 1 three.
27 = 30 – 3.
27 = 27.

h. 10 fours – 1 four = __9_____ × 4
= ____36____

Answer:
10 fours – 1 four = 9 x 4.
40 – 4 = 36.

Explanation:
In the above-given question,
given that,
10 fours – 1 four = 9 x 4.
40 – 4 = 36.
36 = 36.

i. 8 × 4 = 5 fours + __3_____ fours
= ____32____

Answer:
8 x 4 = 5 fours + 3 fours.
32 = 20 + 12.

Explanation:
In the above-given question,
given that,
8 x 4 = 5 fours + 3 fours.
32 = 20 + 12.
32 = 32.

j. __six_____ fives + 1 five = 6 × 5
= ____35____

Answer:
six fives + 1 five = 6 x 5.
30 + 5 = 35.

k. 5 threes + 2 threes = __7___ × __3___
= ___21_____

Answer:
5 threes + 2 threes = 7 x 3.
21 = 21.

Explanation:
In the above-given question,
given that,
5 threes + 2 threes = 7 x 3.
15 + 6 = 21.
21 = 21.

l. ____5__ twos + __five____ twos = 10 twos
= ___20_____

Answer:
5 twos + 5 twos.
20 = 20.

Explanation:
In the above-given question,
given that,
5 twos + five twos = 10 twos.
10 + 10 = 10 twos.
20 = 20.

Eureka Math Grade 3 Module 3 Lesson 1 Exit Ticket Answer Key

Question 1.
Use the array to write two different multiplication facts.
Eureka Math Grade 3 Module 3 Lesson 1 Exit Ticket Answer Key 6

Answer:
The number of rows is 4, the number of columns = 7.
7 x 4 = 28.
4 x 7 = 28.

Explanation:
In the above-given question,
given that,
the array contains the number of rows and number of columns.
The number of rows is 4, the number of columns = 7.
7 x 4 = 28.
4 x 7 = 28.

Eureka-Math-Grade-3-Module-3-Lesson-1-Answer Key-6
Question 2.
Karen says, “If I know 3 × 8 = 24, then I know the answer to 8 × 3.” Explain why this is true.

Answer:
3 x 8 = 24.
8 x 3 = 24.

Explanation:
In the above-given question,
given that,
3 x 8 = 24.
8 x 3 = 24.
24 = 24.

Eureka Math Grade 3 Module 3 Lesson 1 Homework Answer Key

Question 1.
Complete the charts below.
a. A tricycle has 3 wheels.
Eureka Math 3rd Module 3 Lesson 1 HomeWork Answer Key 7

Answer:
The number of tricycles is 3, the total number of wheels = 9.
The number of tricycles is 4, the total number of wheels = 12.
The number of tricycles is 5, the total number of wheels = 15.
The number of tricycles is 6, the total number of wheels = 18.
The number of tricycles is 7, the total number of wheels = 21.

Explanation:
In the above-given question,
given that,
A tricycle has 3 wheels.
The number of tricycles is 3, the total number of wheels = 9.
The number of tricycles is 4, the total number of wheels = 12.
The number of tricycles is 5, the total number of wheels = 15.
The number of tricycles is 6, the total number of wheels = 18.
The number of tricycles is 7, the total number of wheels = 21.
Eureka-Math-Grade-3-Module-3-Lesson-1-Answer Key-7

b. A tiger has 4 legs.
Number of Tigers
Eureka Math 3rd Module 3 Lesson 1 HomeWork Answer Key 8

Answer:
The number of tigers is 5, the total number of legs = 20.
The number of tigers is 6, the total number of legs = 24.
The number of tigers is 7, the total number of legs = 28.
The number of tigers is 8, the total number of legs = 32.
The number of tigers is 9, the total number of legs = 36.

Explanation:
In the above-given question,
given that,
A tiger has 4 legs.
The number of tigers is 5, the total number of legs = 20.
The number of tigers is 6, the total number of legs = 24.
The number of tigers is 7, the total number of legs = 28.
The number of tigers is 8, the total number of legs = 32.
The number of tigers is 9, the total number of legs = 36.

Eureka-Math-Grade-3-Module-3-Lesson-1-Answer Key-8

c. A package has 5 erasers.
Number of Packages 6
Eureka Math 3rd Module 3 Lesson 1 HomeWork Answer Key 9

Answer:
The number of packages is 6, the total number of Erasers = 30.
The number of packages is 7, the total number of Erasers = 35.
The number of packages is 8, the total number of Erasers = 40.
The number of packages is 9, the total number of Erasers = 45.
The number of packages is 10, the total number of Erasers = 50.

Explanation:
In the above-given question,
given that,
A package has 5 erasers.
the number of packages = 6.
The number of packages is 6, the total number of Erasers = 30.
The number of packages is 7, the total number of Erasers = 35.
The number of packages is 8, the total number of Erasers = 40.
The number of packages is 9, the total number of Erasers = 45.
The number of packages is 10, the total number of Erasers = 50.
Eureka-Math-Grade-3-Module-3-Lesson-1-Answer Key-9

Question 2.
Write two multiplication facts for each array.
Eureka Math 3rd Module 3 Lesson 1 HomeWork Answer Key 10

Answer:
The number of rows is 4, the number of columns = 6.
6 x 4 = 24.
4 x 6 = 24.

Explanation:
In the above-given question,
given that,
the array contains the number of rows and number of columns.
The number of rows is 4, the number of columns = 6.
6 x 4 = 24.
4 x 6 = 24.

Eureka-Math-Grade-3-Module-3-Lesson-1-Answer Key-10

Eureka Math 3rd Module 3 Lesson 1 HomeWork Answer Key 11

Answer:
The number of rows is 3, the number of columns = 8.
3 x 8 = 24.
8 x 3 = 24.

Explanation:
In the above-given question,
given that,
the array contains the number of rows and number of columns.
The number of rows is 3, the number of columns = 8.
3 x 8 = 24.
8 x 3 = 24.

Eureka-Math-Grade-3-Module-3-Lesson-1-Answer Key-11

Question 3.
Match the expressions.

3 × 67 threes
3 sevens2 × 10
2 eights9 × 5
5 × 98 × 2
10 twos6 × 3

Answer:
The expressions 3 x 6 = 6 x 3.
3 sevens = 7 threes.
2 eights = 8 x 2.
5 x 9 = 9 x 5.
10 twos = 2 x 10.

Explanation:
In the above-given question,
given that,
The expressions 3 x 6 = 6 x 3.
3 sevens = 7 threes.
2 eights = 8 x 2.
5 x 9 = 9 x 5.
10 twos = 2 x 10.
Eureka-Math-Grade-3-Module-3-Lesson-1-Answer Key-12

Question 4.
Complete the equations.

a. 2 sixes = ___six___ twos
= ____12___

Answer:
2 sixes = six twos.
12 = 12.

Explanation:
In the above-given question,
given that,
2 sixes = six twos.
12 = 12.

b. ___3__ × 6 = 6 threes
= ____18___

Answer:
3 x 6 = 6 x 3.
18 = 18.

Explanation:
In the above-given question,
given that,
3 x 6 = 6 x3.
18 = 18.

c. 4 × 8 = _8____ × 4
= __32_____

Answer:
4 x 8 = 8 x 4.
32 = 32.

Explanation:
In the above-given question,
given that,
4 x 8 = 8 x 4.
32 = 32.

d. 4 × _8____ = _8____ × 4
= ___32____

Answer:
4 x 8 = 8 x 4.
32 = 32.

Explanation:
In the above-given question,
given that,
4 x 8 = 8 x 4.
32 = 32.

e. 5 twos + 2 twos = _7____ × ___2__
= __14_____

Answer:
5 twos + 2 twos = 14.

Explanation:
In the above-given question,
given that,
5 twos + 2 twos = 7 x 2.
10 + 4 = 14.
14 = 14.

f. __5___ fives + 1 five = 6 × 5
= ____30___

Answer:
5 fives + 1 five.
30 = 30.

Explanation:
In the above-given question,
given that,
5 fives + 1 five = 6 x 5.
25 + 5 = 30.
30 = 30.

Decimal Numbers – Definition, Types, Properties, Facts & Examples

Decimal Numbers

Decimal numbers are the part of numbers that have two parts. The decimal numbers are in the standard form representing integer and non-integer numbers. Generally, the decimal points are written in a fraction which consists of 10, 100, 1000 in the denominators. The numbers that are expressed in the decimal form are called decimal numbers. Let us check the complete concept on decimal in the below article.

Also, Read:

Decimals – Definition

A decimal number is a number that has two parts. One part has a whole number and the other part is a fractional part. Both parts are separated by decimal points. If 2.48 is a decimal number, then 2 is the whole number and 48 is the fractional part. “.” is the decimal point.

  • The digits having in the whole number part are called ones, then tens, then hundreds, then thousands, and so on.
  • The places after the decimal point begin with tenths, then hundredths, then thousandths, and so on………

Examples:

(i) In the decimal number 47.25, the whole number part is 47 and the decimal part is .25
(ii) In the decimal number 89.063, the whole number part is 89 and the decimal part is .063
(iii) Take the decimal number 11.056 where the whole number part is 11 and the decimal part is .056

Types of Decimal Numbers

Decimal numbers are classifieds into different types. They are given with definitions and examples explained in detail.

Recurring Decimal Numbers – Recurring Decimal Numbers are Repeating or Non-Terminating Decimals.
Examples of Recurring Decimal Numbers are 2.123123 (Finite) and 4.252525252525… (Infinite)

Non-Recurring Decimal Numbers – Non-Recurring Decimal Numbers are Non Repeating or Terminating Decimals.
Examples of Non-Recurring Decimal Numbers are 5.14812 (Finite) and 2.5428454845…. (Infinite)

Decimal Fraction – Decimal Fraction is the fraction that consists of the denominator as powers of ten.
Examples are 72.66 = 7266/100 and 43.536 = 43536/1000.

Converting a Decimal Number into Decimal Fraction
To convert the Decimal Number into Decimal Fraction, place the 1 in the denominator and remove the decimal point from the given number. The 1 is followed by the number of zeros that are equal to the number of digits given after the decimal point.

Examples:
1. 47.59
The given decimal number is 47.59
47.59 = 4759/100
4 represents the power of 101 that is the tenths position.
7 represents the power of 100 that is the unit’s position.
5 represents the power of 10-1 that is the one-tenth position.
9 represents the power of 10-2 that is the one-hundredths position.
So that is how each digit is represented by a particular power of 10 in the decimal number.
2. 61.27
The given decimal number is 61.27
61.27 = 6127/100
6 represents the power of 101 that is the tenths position.
1 represents the power of 100 that is the unit’s position.
2 represents the power of 10-1 that is the one-tenth position.
7 represents the power of 10-2 that is the one-hundredths position.
So that is how each digit is represented by a particular power of 10 in the decimal number.

Place Value in Decimals

Place value of a number in decimals is the position of every digit that helps to find its value. The position of each digit that before and after the decimal point is different. Check out the below examples to know each digit’s place value.

Examples:
Let us take a number 296.

  • The position of “2” is in One’s place, which means 2 ones (i.e. 2).
  • The position of “9” is in the Ten’s place, which means 9 tens (i.e. ninety).
  • The position of “6” is in the Hundred’s place, which means 6 hundred.
  • As we go left, each position becomes ten times greater.
  • Hence, we read it as “two hundred ninety-six”.

As each digit, we move to the left side the value becomes 10 times greater than the previous value.

  • The tens place digit is 10 times bigger than Ones.
  • The hundreds place digit is 10 times bigger than Tens.

If we consider a decimal number, the digits after the decimal points will become 10 times smaller than other digits. The digits present on the left side of the decimal are multiplied with the positive powers of ten in increasing order from right to left. The digits present in the right of the decimal point are multiplied with the negative powers of 10 in increasing order from left to right.

Example:
1. 61.28
The decimal expansion of the given number 61.28 is
[(6 * 10) + (1 * 1)] + [(2 * 0.1) + (5 * 0.01)]

Properties of Decimals

We have given the main properties of the decimal numbers below those are under multiplication and division operations. Check out all the properties of decimals given below.

  • When two decimal numbers are multiplied with each other, then the result will be a decimal number.
  • When a decimal number and a whole number are multiplied with each other, then the result will be a decimal number.
  • If any decimal fraction is multiplied by 1, the product remains the same decimal fraction by itself.
  • If any decimal fraction is multiplied by 0, then the product becomes 0.
  • When a decimal number divided by 1, then the quotient must be a decimal number.
  • Also, when a decimal number is divided by the same number, then the quotient becomes 1.
  • If in case, 0 divided by any decimal number, the quotient becomes 0.
  • The division of a decimal number by 0 is not applicable and possible as the reciprocal of 0 does not exist.

Arithmetic Operations on Decimals

We can perform addition, subtraction, multiplication, and division operations on Decimals easily. Check out the below concepts to understand different Arithmetic Operations on Decimals.

Addition Operation on Decimals
When you add two decimal numbers, line up the decimal points of the given numbers and add them. If you don’t see a decimal point, then that is only a whole number.

Subtraction Operation on Decimals
Subtraction Operation on Decimals is also similar to the Addition Operation on Decimals. You need to line up the decimal point of the given numbers and subtract the values.

Multiplication Operation on Decimals
Multiplication Operation on Decimals is like integers as if the decimal point not present. Firstly, find out the product and count the number after the decimal point in the given numbers. The count will let you know how many numbers present after the decimal point in the result.

Division Operation on Decimals
The Division Operation on Decimals is simply dividing the given two decimal numbers. Move the decimal points to make them whole numbers. Then, perform the division operation like normal integers.

Decimal to Fraction Conversion

We consider the digits after the decimal point as the tenths, hundredths, thousandths, and so on. Write down the decimal numbers in the expanded form and simplify the values.
Example:
Let us consider a decimal number 5.21
Conver it into a fraction number.
The expanded form of 5.21 is 521 x (1/100) = 521/100.

Fraction to Decimal Conversion

To convert the fraction number into a decimal number, divide the numerator by denominator.
Example: 9/7 is a fraction. If it is divided, we get 1.285714

Decimal Problems with Solutions

Example 1:

Convert 6/10 in decimal form?
Solution:
Given fraction number is 6/10.
To convert fraction to decimal, divide 6 by 10, we get the decimal form.
Thus, 6/10 = 0.6
Hence, the decimal form of 6/10 is 0.6.

Example 2:

Express 2.36 in fraction form?
Solution:
The given decimal number is 2.36
The expanded form of 2.36 is
= 236 x (1/100)
= 236 /100
= 118/50
= 59/25
Hence, the equivalent fraction for 2.36 is 59/25.

Frequently Asked Questions on Decimals

1. What is meant by Decimal?

A decimal is a number that mainly has two parts named as a whole number part and a fractional part separated by a decimal point.

2. What are the different types of decimals?

There are two types of decimals considered. They are

  • Terminating decimals (or) Non-recurring decimals
  • Non-terminating decimals (or) Recurring decimals

3. Write the expanded form of 85.3?

The expanded form of 85.3 is 80 + 5 + (3/10)

4. How to convert fractions to decimals?

Factions to decimals are converted by dividing the numerator by the denominator value.

Different Types of Polygons – Definition, Properties, Shapes, Examples

Different Types of Polygons

A polygon is a two-dimensional figure that consists of a finite number of sides. The sides of a polygon are connected by line segments. The end connected by the line segments is called the vertex. There are Different Types of Polygons available in geometry. Poly means many and gon means an angle. The area and perimeter of the polygon completely depend on their shape. Any polygon can be classified depends on its sides and vertices. If a polygon has four sides and four vertices, then that polygon is named a quadrilateral.

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Types of Polygons and their Properties

Based on the sides and vertices, the polygons are classified into different types. Check out the below polygons and their properties along with the examples and images.

  • Regular Polygons
  • Irregular Polygons
  • Concave Polygons
  • Convex Polygons
  • Trigons
  • Quadrilateral Polygons
  • Pentagon Polygons
  • Hexagon Polygons
  • Equilateral Polygons
  • Equiangular Polygons

Regular Polygon

A Polygon is said to be a Regular Polygon when all of its sides are equal. Also, a regular polygon has all the interior angles are the same. For example, a regular hexagon consists of equal six sides, and also its interior angles give a total of 120 degrees.
square 2 CBSE Class 7 Maths The Triangle and Its Properties Worksheets 3

Examples:

  • A square consists of all its sides equal to 4cm, and also all the angles are at 90°.
  • A regular pentagon that has 5 equal sides. All the interior angles measure 108 degrees.
  • An equilateral triangle has all three sides equal to 8cm and angles measure to 60°.

Irregular Polygon

A Polygon is said to be an irregular polygon when all of its sides are not equal. They have irregular shapes. Also, the angles of irregular polygons are not equal.
trapezium

Examples:

  • A triangle with unequal sides.
  • A quadrilateral with unequal sides.

Convex Polygon

A Polygon is said to be a convex polygon when the measure of the interior angle is less than 180 degrees. The corners of a convex polygon are always outwards. Convex Polygon is completely opposite to the concave polygon.

Example:

An irregular hexagon that vertices are completely outwards.

Concave Polygon

A Polygon is said to be a concave polygon when there are at least one angle measures more than 180 degrees. The corners of a concave polygon present inwards and also outwards.
concave-polygon

Trigons

A Polygon is said to be a trigon that has three sides. Trigons are also known as triangles. The triangles are classifieds into different categories. They are
Scalene Triangle: It has all sides that are unequal.
Isosceles Triangle: It has two sides that are equal.
Equilateral Triangle: All sides are equal in length and also all angles are 60 degrees.

Quadrilateral Polygon

A Polygon is said to be a Quadrilateral Polygon when it has four sides. Examples of Quadrilateral Polygon are square, rectangle, rhombus, and parallelogram.
quadrilateral-polygon

Pentagon Polygon

A Polygon is said to be a Pentagon Polygon when it has five sides. If all the sides of a pentagon polygon are equal, then it is called a regular pentagon. If not it is called an irregular pentagon.
pentagon

Hexagons

A polygon is called a hexagon when that has six sides and six vertices. If all the six sides of a hexagon are equal, then it is called a regular hexagon. Also, in a regular hexagon, all the interior and exterior angles are equal.

Equilateral Polygons

The polygons are called equilateral polygons when it has all sides are equal. An equilateral triangle, a square, etc., are examples of Equilateral Polygons.

Equiangular Polygons

The polygons are called Equiangular Polygons when all of their interior angles are equal. An example is a rectangle.

Types of Polygons with Sides 3-20

Below is the Classification of Polygons according to the number of sides and the angle measure. They are as follows

Name of the PolygonsSidesVerticesAngle
Triangle (also called Trigon)3360°
Quadrilateral (also called Tetragon)4490°
Pentagon55108°
Hexagon66120°
Heptagon77128.571°
Octagon88135°
Nonagon (also called Enneagon)99140°
Decagon1010144°
Hendecagon1111147.27
Dodecagon1212150
Tridecagon or triskaidecagon1313152.3
Tetradecagon or tetrakaidecago1414154.28
Pendedecagon1515156
Hexdecagon1616157.5
Heptdecagon1717158.82
Octdecagon1818160
Enneadecagon1919161.05
Icosagon2020162
n-gonnn(n-2)× 180° / n

FAQs on Different Types of Polygons?

1. What is a Polygon?

A polygon is a two-dimensional figure that consists of a finite number of sides.

2. How many different types of Polygons are there?

There are two different types of Polygons namely Regular and Irregular. However, depending on the sides and vertices they are furthermore classified as Concave Polygons, Convex Polygons, Trigons, Quadrilateral Polygons, Pentagon Polygons, Hexagon Polygons, Equilateral Polygons, Equiangular Polygons, etc.

3. What is a 9 sided shape called?

9 Sided Shape is Called an Enneagon or Nonagon.

Worksheet on Decimal Word Problems | Decimal Word Problems with Answers

Worksheet on Decimal Word Problems

Worksheet on Decimal Word Problems will help the students to explore their knowledge of decimal word problems. Solve all the problems to learn the depth concept of Decimal Word Problems. Know the definition, properties, different operations performed on decimals by visiting our website. We have given the complete decimal concepts along with examples. Check out the Decimal Word Problems Worksheet and know the various strategies to solve problems in an easy way.

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What are Decimals?

The decimal is a number that has the whole number and the fractional part separated by a decimal point. The point between the whole number and fractions part is known as the decimal point. For example, 24.9 is the decimal number where 24 is a whole number and 9 is the fractional part after the decimal. We have included addition, subtraction, multiplication, and division decimal word problems below.

Word Problems on Decimals

Have a look at all the problems given below and get a grip on all types of problems on decimals.

1. Anshika scored 521 marks out of 600 in the final examination. How many marks did she lose?

Solution:

Given that Anshika scored 521 marks out of 600 in the final examination.
So, the total marks are 600.
She scored 521 marks.
To get the marks she loses in an examination, we have to subtract 521 marks from 600.
Now, subtract 521 from 600.
600 – 521 = 79 marks.

Anshika lost 79 marks in the final examination.


2. Arun had 0.24 liters of cold drink. Kiran had 0.68 liters more cold drink. How much cold drink did they have together?

Solution:

Given that Arun had 0.87 liters of cold drink. Kiran had 0.92 liters more cold drink.
Arun = 0.24 litres of cold drink
Kiran = 0.68 liters more cold drink
To find the cold drink they have together we have to add both cold drink liters.
So, add 0.24 liters of cold drink and 0.68 liters more cold drink
0.24 liters of cold drink + 0.68 liters more cold drink = 0.92 liters of cold drink.

They have 0.92 liters of cold drink together.


3. The weight of a baby lion was 169.24 kg. After two years, his weight increased by 107.64 kg. Find the weight of the lion after two years?

Solution:

Given that the weight of a baby lion was 169.24 kg. After two years, his weight increased by 107.64 kg.
The lion’s present age is 169.24 kg.
After two years, his weight increased by 107.64 kg.
Now, add both weights to get the lion’s age after two years.
So, add 169.24 kg and 107.64 kg
169.24 kg + 107.64 kg = 276.88 kg

After two years, the lion weight is 276.88 kg.


4. Ganesh had a rope of 54.28 m. He cut the rope into two pieces. If the length of one piece was 13.26 m, what was the length of the other piece?

Solution:

Given that Ganesh had a rope of 54.28 m. He cut the rope into two pieces.
The total length of the rope is 54.28 m.
The rope is cut into two pieces.
If the length of one piece was 13.26 m, let the other piece’s length is x.
Add two pieces 13.26 m and x m.
13.26 m + x m = 54.28 m
x = 54.28 m – 13.26 m
x = 41.02 m

The length of the other piece is 41.02 m.


5. Each side of a regular polygon is 21.7 m and its perimeter is 108.5 m. Find the number of sides of the polygon?

Solution:

Given that each side of a regular polygon is 21.7 m and its perimeter is 108.5 m.
The perimeter of a polygon is the sum of all sides of a polygon.
To find the number of sides of the polygon, divide the perimeter with each side of a regular polygon is 21.7 m.
Divide 108.5 m with 21.7 m
108.5 m/21.7 m = 5.

The number of sides of the polygon is 5.


6. Fariha took 2.4 minutes to complete the race and Anil took 2.1 minutes to complete the race. Who won the race?

Solution:

Given that Fariha took 2.4 minutes to complete the race and Anil took 2.1 minutes to complete the race.
To find the winner of the race, we need to compare the time taken by each of them. The one who traveled in less time is the winner of the race.
Fariha took 2.4 minutes.
Anil took 2.1 minutes.
2.1 minutes is less time compared to 2.4 minutes.
Therefore, Anil took less time to complete the race.

Anil won the race.


7. The annual rainfall received by Arunachal Pradesh is 261.5 cm and that by Assam is 297.4 cm. Who received less rainfall?

Solution:

Given that the annual rainfall received by Arunachal Pradesh is 261.5 cm and that by Assam is 297.4 cm.
To find the less rainfall received, we need to compare the rainfall in two states. The place where the less cm rainfall occurred will be the answer.
Arunachal Pradesh = 261.5 cm rainfall
Assam = 297.4 cm rainfall
261.5 cm rainfall is less time compared to 297.4 cm rainfall.
Therefore, Arunachal Pradesh received less rainfall.

The answer is Arunachal Pradesh.


8. Rishu’s height is 134.51 cm. She stands on a tool of height 9.40 cm. What is the combined height now?

Solution:

Given that Rishu’s height is 134.51 cm. She stands on a tool of height 9.40 cm.
To find the combined height, add both given heights.
Add 134.51 cm and 9.40 cm
134.51 cm + 9.40 cm = 143.91 cm.

The combined height is 143.91 cm.


9. The milkman delivers 6.03 liter of milk to a house in the morning and 3.230 liters in the evening. What is the total quantity of milk delivered by the milkman?

Solution:

Given that the milkman delivers 6.03 liter of milk to a house in the morning and 3.230 liters in the evening.
To get the total quantity of milk delivered by the milkman, combine the milk he delivers in the morning and also in the evening.
Milkman delivers 6.03 liter of milk in the morning
Milkman delivers 3.230 liters of milk in the evening
Add 6.03 liter of milk and 3.230 liters of milk.
6.03 liter of milk + 3.230 liters of milk = 9.26 liters of milk

The total quantity of milk delivered by the milkman = 9.26 liters of milk.


10. Rebecca‘s kite is flying at a height of 28.3 m and Shelly’s at a height of 32.6 m from the ground. Whose kite is flying high and by how much?

Solution:

Given that Rebecca‘s kite is flying at a height of 28.3 m and Shelly’s at a height of 32.6 m from the ground.
To find whose kite is flying high, compare the two heights. The largest height of the kite considered to be flying high.
Rebecca‘s kite is flying at a height of 28.3 m
Shelly’s at a height of 32.6 m
32.6 m is greater than the 28.3 m.
Shelly’s kite is flying high.
To find how much the kite is flying high, subtract the heights.
Subtract 28.3 m from 32.6 m.
32.6 m – 28.3 m = 4.3 m

Shelly’s kite is flying high of 4.3 m than Rebecca‘s kite.


11. A bike travels 39.2 km in 7 hours. How much distance will it travel in 1 hour?

Solution:

Given that a bike travels 39.2 km in 7 hours.
Now, to find the distance it will travel in 1 hour, we have to divide the 39.2 km by 7 hours.
Divide 39.2 km by 7 hours.
39.2 km/7 hours = 5.6km/hr

The bike travels 5.6km in one hour.


12. Ron jogged 3.3 km, Mike jogged 4.8 times more distance than Ron. Find the distance covered by Mike?

Solution:

Given that Ron jogged 3.3 km, Mike jogged 4.8 times more distance than Ron.
The distance covered by Mike, multiply the Ron jogged km with the Mike jogged 4.8 times more distance than Ron.
The distance covered by Ron 3.3 kilometers.
Mike jogged 4.8 times more distance than Ron.
Multiply 3.3 kilometers and 4.8 kilometers.
The distance jogged is 3.3 kilometers × 4.8 kilometers = 15.84 kilometers.

The distance covered by mike is 15.84 kilometers.


13. The daily consumption of milk in a house is 4.36 liters. How much milk will be consumed in 30 days?

Solution:

Given that the daily consumption of milk in a house is 4.36 liters.
To find the milk that will be consumed in 30 days, we need to multiply 30 with 4.36 liters.
Multiply 4.36 liters with 30.
4.36 liters × 30 = 130.8 liters

130.8 liters of milk will be consumed in 30 days.


14. A tin contains 13.4 liters of oil. How many such tin contains 80.4 liters of oil?

Solution:

Given that a tin contains 13.4 liters of oil.
To find the number of tins that contain 80.4 liters of oil, we need to divide the 80.4 liters of oil by 13.4 liters of oil.
Divide 80.4 liters of oil by 13.4 liters of oil
80.4 liters of oil/13.4 liters of oil = 6.

6 tins contain 80.4 liters of oil.


15. Find the cost of 58.3 m cloth if the cost of 1 m cloth is $44.80.

Solution:

Given that the cost of 1 m cloth is $44.80
To find the cost of 58.3 m cloth, multiply 58.3 m cloth with $44.80.
Multiply 58.3 m cloth with $44.80
58.3 m cloth × $44.80 = $2611.84

The cost of 58.3 m cloth is $2611.84.


16. Shruti bought a bag for $387.05. She gave the shopkeeper 2 notes of $200. How much money will she get back?

Solution:

Given that Shruti bought a bag for $387.05. She gave the shopkeeper 2 notes of $200.
she gave 2 notes of 200
Add 2 notes cost = 200 + 200 = 400.
Now, subtract $400 from $387.05 to get the amount she gets back from the shopkeeper.
Subtract $400 from $387.05.
$400 – $387.05 = $12.95

Shruti gets back $12.95 from the shopkeeper.


17. A tailor needs 46.36 m of cloth for the shirts and 56.90 m for trousers. How much cloth does the tailor need in all?

Solution:

Given that a tailor needs 46.36 m of cloth for the shirts and 56.90 m for trousers.
The tailor needs all will get by adding the cloth for the shirts and cloth for the trousers.
46.36 m of cloth for the shirts
56.90 m of cloth for trousers.
Add 46.36 m and 56.90 m
46.36 m + 56.90 m = 103.26 m

The tailor needs 103.26 m cloth in total.


18. A spool of thread has a thread measuring 97.60 m. If 53.44 m thread has been cut, what length of thread is still left in the spool?

Solution:

Given that a spool of thread has a thread measuring 97.60 m.
If 53.44 m thread has been cut, subtract the 53.44 m thread from the thread measuring 97.60 m.
The total amount of thread is 97.60 m.
The thread that has been cut is 53.44 m.
Subtract 53.44 m from 97.60 m.
97.60 m – 53.44 m = 44.16 m.

The length of thread is still left in the spool is 44.16 m.


19. The cost of a chair is $3056.94. Tania wants to buy 7 chairs for her house. How much money will she pay to the shopkeeper?

Solution:

Given that the cost of a chair is $3056.94. Tania wants to buy 7 chairs for her house.
To find the money she will pay to the shopkeeper, multiply the number of chairs by the cost of the chair.
The cost of a chair is $3056.94.
The number of chairs is 7.
Multiply $3056.94 with 7.
$3056.94 × 7 = $21398.58

She will pay $21398.58 to the shopkeeper.


20. David has a jug full of milk. He pours the complete milk in 5 glasses, each glass of capacity 0.6 l. How much milk was there in the jug?

Solution:

Given that David has a jug full of milk. He pours the complete milk in 5 glasses, each glass of capacity 0.6 l.
To find the total milk in the jug, multiply 5 glasses of milk with 0.6 l capacity.
The total number of glasses = 5.
Each glass capacity is 0.6 l
Multiply 0.6 l with 5.
0.6 l × 5 = 3 l.

The total milk in the jug is 3 l.


21. Find the area of a square whose side is 2.50 m.

Solution:

Given that the side of a square is 2.50 m.
To find the area of a square, multiply 2.50 m with 2.50 m.
2.50 m × 2.50 m = 6.25 m.

The area of a square whose side is 2.50 m is 6.25 m.


22. The weight of 1 bag of sugar is 11.4 kg. What is the weight of 14 such bags?

Solution:

Given that the weight of 1 bag of sugar is 11.4 kg.
To find the weight of 14 such bags, multiply the weight of 1 bag of sugar with 14 bags.
Multiply the weight of 1 bag of sugar 11.4 kg with 14.
11.4 kg × 14 = 159.6 kg

The weight of 14 such bags is 159.6 kg.


23. A vehicle covers a distance of 10.2 km in 3.4 liters of petrol. How much distance will it cover in 1 liter of petrol?

Solution:

Given that a vehicle covers a distance of 10.2 km in 3.4 liters of petrol.
To find the distance that covers 1 liter of petrol, divide the total distance by given liters of petrol.
The total distance covered is 10.2 km.
The total liters of petrol is 3.4 liters.
Divide 10.2 km by 3.4 liters.
10.2 km/3.4 liters = 3 km/liter.

3 km covered in 1 liter of petrol.


24. Ron has 5.60 l of juice. He pours it into 8 glasses equally. How much juice is there in each glass?

Solution:

Given that Ron has 5.60 l of juice. He pours it into 8 glasses equally.
The total amount of juice is 5.60 l.
The number of glasses is 8.
To find out the juice in each glass, divide the 5.60 l by 8.
divide the 5.60 l by 8
5.60 l/8 = 0.7 l.

0.7 l juice is there in each glass.


25. Shelly has a ribbon of length 44.44 m. She cuts it into 11 equal parts. What is the length of each equal part?

Solution:

Given that Shelly has a ribbon of length 44.44 m. She cuts it into 11 equal parts.
To find the length of each equal part, divide the length of the ribbon by the number of equal parts.
The ribbon length is 44.44 m.
The number of equal parts is 11.
Divide 44.44 m by 11.
44.44 m/11 = 4.4 m.

4.4 m is the length of each equal part.


26. The cost of 4 pens is $49.24. What is the cost of 1 pen?

Solution:

Given that the cost of 4 pens is $49.24.
To find the cost of each pen, divide the total amount by the number of pens.
The total cost of the pens is $49.24.
The total number of pens is 4.
Divide $49.24 by 4.
$49.24/4 = $12.31

The cost of 1 pen is $12.31.


27. The weight of a box is 141.029 kg. What will be the weight of 23 such boxes?

Solution:

Given that the weight of a box is 141.029 kg.
To find the weight of 23 such boxes, multiply 141.029 kg by 23.
Divide 141.029 kg by 23.
141.029 kg × 23 = 3243.667 kg.

The weight of 23 such boxes is 3243.667 kg.


28. Sonia has 18.48 l of juice. She pours it into 8 jars equally. How much juice is there in each jar?

Solution:

Given that Sonia has 18.48 l of juice. She pours it into 8 jars equally.
To find the amount of juice in each jar, divide the total liters of juice by the number of jars.
The total liters of juice is 18.48 l.
The number of jars is 8.
Divide 18.48 l of juice by 8 jars.
18.48 l/8 = 2.31 liters

2.31 liters of juice present in each jar.


Multiplying Fractions – Definition, Examples | How to Multiply Fractions?

Multiplying Fractions

Multiplying numbers is so easy compared to Multiplying Fractions. The fraction is represented as the division of the whole. The fraction is in the form of “x/y” where “x” is the numerator and “y” is the denominator. One can apply the fraction concept to real-time examples easily after reading the entire concept here. If you have an apple and you made it into 4 equal parts, then it can represent as \(\frac { 1 }{ 4 } \). Or else if it cut into 7 pieces then it can represented as \(\frac { 1 }{ 7 } \).

How to Multiply Fractions?

We have provided simple steps to multiply fractions and find the solution of multiplying fractions. It is easy to find out the solution using the below procedure. Multiplying fractions can be defined as the product of a fraction with another fraction or with the variables or with an integer. Follow the below process to multiply fractions

  • Multiply the numerator with numerator
  • Multiply the denominator with the denominator
  • Simplify the fractions, if needed

Example:
1. Multiply \(\frac { 2 }{ 3 } \) × \(\frac { 1 }{ 5 } \)

Solution:
Given that \(\frac { 2 }{ 3 } \) × \(\frac { 1 }{ 5 } \)
To multiply the above fractions, firstly multiply the numerators
2 × 1 = 2
multiply the denominators
3 × 5 = 15
Now, simplify the fraction, we get \(\frac { 2 }{ 15 } \)
If \(\frac { x }{ y } \) and \(\frac { m }{ n } \) are the multiplicand and multiplier, then the output is \(\frac { xm }{ yn } \)
Product of Fraction = Product of Numerator/Product of Denominator

Fractions Parts and Types

A fraction consists of two parts. One is the numerator and another one is the denominator. If \(\frac { a }{ b } \) is a fraction, then the two parts are a and b where a is the numerator and b is the denominator. Or else if \(\frac { 3 }{ 4 } \) is a fraction, then the two parts are 3 and 4 where 3 is the numerator and 4 is the denominator.

Mainly, there are three types of fractions considered. They are proper fractions, improper fractions, and mixed fractions.
Proper fractions: A fraction is said to be a proper fraction when the numerator of a fraction is less than the denominator.
Examples: \(\frac { 1 }{ 4 } \), \(\frac { 5 }{ 6 } \), \(\frac { 7 }{ 11 } \)
Improper fractions: A fraction is said to be an improper fraction when the numerator is greater than the denominator.
Examples: \(\frac { 5 }{ 4 } \), \(\frac { 7 }{ 6 } \), \(\frac { 13 }{ 11 } \)
Mixed Fraction: A fraction is said to be a mixed fraction when we write the improper fraction in the combination of a whole number and a fraction.
Examples: 1 \(\frac { 5 }{ 4 } \), 3 \(\frac { 9 }{ 6 } \), 2 \(\frac { 6 }{ 7 } \)

Also, Read:

Fractional Simplification

Generally, the multiplication of fractions can be finished by multiplying numerators with numerators and multiplying denominators with denominators. To make the fractional multiplication simpler, we can reduce the fraction by canceling the common factors. By canceling out the common factors from the given factor, it becomes easier to find the exact output.

Example: \(\frac { 9 }{ 4 } \) and \(\frac { 2 }{ 3 } \)
\(\frac { 9 }{ 4 } \) can written as \(\frac { 3 × 3 }{ 2×2 } \)
\(\frac { 3 × 3 }{ 2×2 } \) × \(\frac { 2 }{ 3 } \) = \(\frac { 3 }{ 2 } \)
If there is no common factors, then the numerators and denominators are multiplied directly.

Types of Fraction Multiplication

There are different types of Fraction Multiplication available. They are

  • Multiplication of Fraction with Whole Numbers
  • Multiplication of Fraction with another Fraction
  • Multiplication of Fraction with Variables

Multiplication of Fractional Number by a Whole Number

In Multiplication of Fractional Number by a Whole Number, we multiply the numerator with the numerator and the denominator remains the same. Before you multiply, reduce the fraction to the lowest terms. Check out different problems on the Multiplication of Fractional Number by a Whole Number below.

1. Multiply 2 \(\frac { 2 }{ 3 } \) by 9

Solution:
Given that multiply 2 \(\frac { 2 }{ 3 } \) by 9.
Firstly, convert given mixed fraction 2 \(\frac { 2 }{ 3 } \) to fraction.
2 \(\frac { 2 }{ 3 } \) = \(\frac { 8 }{ 3 } \)
Now, multiply \(\frac { 8 }{ 3 } \) by 9.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
8 × 9 = 72.
So, the fraction is \(\frac { 72 }{ 3 } \)
Simplify the fraction to get the final answer.
\(\frac { 72 }{ 3 } \) = 24.

The final answer is 24.

(ii) Multiply \(\frac { 3 }{ 4 } \) by 6

Solution:
Given that multiply \(\frac { 3 }{ 4 } \) by 6
multiply \(\frac { 3 }{ 4 } \) by 6.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
3 × 6 = 18.
So, the fraction is \(\frac { 18 }{ 4 } \)
Simplify the fraction to get the final answer.
\(\frac { 18 }{ 4 } \) = \(\frac { 9 }{ 2 } \).

The final answer is \(\frac { 9 }{ 2 } \).

(iii) Multiply 3 \(\frac { 3 }{ 2 } \) by 6

Solution:
Given that multiply 3 \(\frac { 3 }{ 2 } \) by 6.
Firstly, convert given mixed fraction 3 \(\frac { 3 }{ 2 } \) to fraction.
3 \(\frac { 3 }{ 2 } \) = \(\frac { 9 }{ 2 } \)
Now, multiply \(\frac { 9 }{ 2 } \) by 6.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
9 × 6 = 54.
So, the fraction is \(\frac { 54 }{ 2 } \)
Simplify the fraction to get the final answer.
\(\frac { 54 }{ 2 } \) = 27.

The final answer is 27.

Multiplication of Fractional Number by Another Fractional Number

Multiplying Fraction with Another Fraction is explained with the below examples.

(i). Multiply \(\frac { 3 }{ 5 } \) by \(\frac { 6 }{ 5 } \)

Solution:
Given that Multiply \(\frac { 3 }{ 5 } \) by \(\frac { 6 }{ 5 } \).
Firstly, multiply the numerators with numerators.
3 × 6 = 18.
Next, multiply the denominators with denominators.
5 × 5 = 25.
Finally, write the fraction in the simplest form.
\(\frac { 18 }{ 25 } \)

The final answer is \(\frac { 18 }{ 25 } \)

(ii) Multiply \(\frac { 3 }{ 4 } \) by \(\frac { 7 }{ 2 } \)

Solution:
Given that Multiply \(\frac { 3 }{ 4 } \) by \(\frac { 7 }{ 2 } \).
Firstly, multiply the numerators with numerators.
3 × 7 = 21.
Next, multiply the denominators with denominators.
4 × 2 = 8.
Finally, write the fraction in the simplest form.
\(\frac { 21 }{ 8 } \)

The final answer is \(\frac { 21 }{ 8 } \).

(iii) Multiply \(\frac { 2 }{ 3 } \), \(\frac { 2 }{ 5 } \), and \(\frac { 2 }{ 7 } \)

Solution:
Given that Multiply \(\frac { 2 }{ 3 } \), \(\frac { 2 }{ 5 } \), and \(\frac { 2 }{ 7 } \).
Firstly, multiply the numerators with numerators.
2 × 2 × 2 = 8.
Next, multiply the denominators with denominators.
3 × 5 × 7 = 105.
Finally, write the fraction in the simplest form.
\(\frac { 8 }{ 105 } \)

The final answer is \(\frac { 8 }{ 105 } \).

Multiplication of a Mixed number by Another Mixed Number

To find the multiplication of a mixed number with another mixed number, we need to change the mixed fractions to fractions and multiply them.

(i) Multiply 3 \(\frac { 2 }{ 5 } \) and 2 \(\frac { 3 }{ 7 } \)

Solution:
Given that multiply 3 \(\frac { 2 }{ 5 } \) and 2 \(\frac { 3 }{ 7 } \).
Firstly, convert given mixed fractions 3 \(\frac { 2 }{ 5 } \) and 2 \(\frac { 3 }{ 7 } \) to fractions.
3 \(\frac { 2 }{ 5 } \) = \(\frac { 17 }{ 5 } \)
2 \(\frac { 3 }{ 7 } \) = \(\frac { 17 }{ 7 } \)
Now, multiply \(\frac { 17 }{ 5 } \) by \(\frac { 17 }{ 7 } \).
Firstly, multiply the numerators with numerators.
17 × 17 = 289.
Next, multiply the denominators with denominators.
5 × 7 = 35.
Finally, write the fraction in the simplest form.
\(\frac { 289 }{ 35 } \)

The final answer is \(\frac { 289 }{ 35 } \).

(ii) Multiply 2 \(\frac { 4 }{ 3 } \) and 1 \(\frac { 6 }{ 5 } \)

Solution:
Given that multiply 2 \(\frac { 4 }{ 3 } \) and 1 \(\frac { 6 }{ 5 } \).
Firstly, convert given mixed fractions 2 \(\frac { 4 }{ 3 } \) and 1 \(\frac { 6 }{ 5 } \) to fractions.
2 \(\frac { 4 }{ 3 } \) = \(\frac { 10 }{ 3 } \)
1 \(\frac { 6 }{ 5 } \) = \(\frac { 11 }{ 5 } \)
Now, multiply \(\frac { 10 }{ 3 } \) by \(\frac { 11 }{ 5 } \).
Firstly, multiply the numerators with numerators.
10 × 11 = 110.
Next, multiply the denominators with denominators.
5 × 3 = 15.
Finally, write the fraction in the simplest form.
\(\frac { 110 }{ 15 } \) = \(\frac { 22 }{ 3 } \)

The final answer is \(\frac { 22 }{ 3 } \).

(i) Multiply 4 \(\frac { 2 }{ 9 } \) and 5 \(\frac { 1 }{ 4 } \)

Solution:
Given that multiply 4 \(\frac { 2 }{ 9 } \) and 5 \(\frac { 1 }{ 4 } \).
Firstly, convert given mixed fractions 4 \(\frac { 2 }{ 9 } \) and 5 \(\frac { 1 }{ 4 } \) to fractions.
4 \(\frac { 2 }{ 9 } \) = \(\frac { 38 }{ 9 } \)
5 \(\frac { 1 }{ 4 } \) = \(\frac { 21 }{ 4 } \)
Now, multiply\(\frac { 38 }{ 9 } \) by \(\frac { 21 }{ 4 } \).
Firstly, multiply the numerators with numerators.
38 × 21 = 798.
Next, multiply the denominators with denominators.
9 × 4 = 36.
Finally, write the fraction in the simplest form.
\(\frac { 798 }{ 36 } \)

The final answer is \(\frac { 798 }{ 36 } \).

Multiplying Fractions Examples

I. Find the product
(i) \(\frac { 5 }{ 4 } \) × 1
(ii) \(\frac { 3 }{ 5 } \) × 6
(iii) \(\frac { 10 }{ 15 } \) × 7
(iv) \(\frac { 2 }{ 3 } \) × 0
(v) \(\frac { 1 }{ 4 } \) × \(\frac { 2 }{ 7 } \)
(vi) 2\(\frac { 9 }{ 13 } \) × 4
(vii) \(\frac { 1 }{ 6 } \) × \(\frac { 7 }{ 1 } \)
(viii) \(\frac { 1 }{ 4 } \) × \(\frac { 8 }{ 6 } \) × \(\frac { 3 }{ 10 } \)
(ix) \(\frac { 5 }{ 16 } \) × \(\frac { 11 }{ 23 } \)
(x) \(\frac { 1 }{ 2 } \) of 50
(xi) \(\frac { 1 }{ 3 } \) of 90
(xii) \(\frac { 5 }{ 6 } \) of \(\frac { 9 }{ 12 } \)

(i) \(\frac { 5 }{ 4 } \) × 1
Solution:
Given that \(\frac { 5 }{ 4 } \) × 1
Now, multiply \(\frac { 5 }{ 4 } \) by 1.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
5 × 1 = 5.
So, the fraction is \(\frac { 5 }{ 4 } \)

The final answer is \(\frac { 5 }{ 4 } \).

(ii) \(\frac { 3 }{ 5 } \) × 6
Solution:
Given that \(\frac { 3 }{ 5 } \) × 6
multiply \(\frac { 3 }{ 5 } \) by 6
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
3 × 6 = 18.
So, the fraction is \(\frac { 18 }{ 5 } \)
Simplify the fraction to get the final answer.
\(\frac { 18 }{ 5 } \).

The final answer is \(\frac { 18 }{ 5 } \).

(iii) \(\frac { 10 }{ 15 } \) × 7
Solution:
Given that \(\frac { 10 }{ 15 } \) × 7
multiply \(\frac { 10 }{ 15 } \) by 7
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
10 × 7 = 70.
So, the fraction is \(\frac { 70 }{ 15 } \)
Simplify the fraction to get the final answer.
\(\frac { 70 }{ 15 } \) = \(\frac { 14 }{ 3 } \).

The final answer is \(\frac { 14 }{ 3 } \).

(iv) \(\frac { 2 }{ 3 } \) × 0
Solution:
Any fraction that multiplies with 0 gives 0.
Therefore, the answer is 0.

(v) \(\frac { 1 }{ 4 } \) × \(\frac { 2 }{ 7 } \)
Solution:
Given that Multiply \(\frac { 1 }{ 4 } \) by \(\frac { 2 }{ 7 } \).
Firstly, multiply the numerators with numerators.
1 × 2 = 2.
Next, multiply the denominators with denominators.
4 × 7 = 28.
Finally, write the fraction in the simplest form.
\(\frac { 2 }{ 28 } \)
Simplify the fraction to get the final answer.
\(\frac { 1 }{ 14 } \).

The final answer is \(\frac { 1 }{ 14 } \).

(vi) 2\(\frac { 9 }{ 13 } \) × 4
Solution:
Given that multiply 2\(\frac { 9 }{ 13 } \) × 4.
Firstly, convert given mixed fraction 2\(\frac { 9 }{ 13 } \) to fraction.
2\(\frac { 9 }{ 13 } \) = \(\frac { 35 }{ 13 } \)
Now, multiply \(\frac { 35 }{ 13 } \) by 4.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
35 × 4 = 140.
So, the fraction is \(\frac { 140 }{ 13 } \)
Simplify the fraction to get the final answer.
\(\frac { 140 }{ 13 } \).

The final answer is \(\frac { 140 }{ 13 } \).

(vii) \(\frac { 1 }{ 6 } \) × \(\frac { 7 }{ 1 } \)
Solution:
Given that Multiply \(\frac { 1 }{ 6 } \) × \(\frac { 7 }{ 1 } \).
Firstly, multiply the numerators with numerators.
1 × 7 = 7.
Next, multiply the denominators with denominators.
6 × 1 = 6.
Finally, write the fraction in the simplest form.
\(\frac { 7 }{ 6 } \)
Simplify the fraction to get the final answer.
\(\frac { 7 }{ 6 } \).

The final answer is \(\frac { 7 }{ 6 } \).

(viii) \(\frac { 1 }{ 4 } \) × \(\frac { 8 }{ 6 } \) × \(\frac { 3 }{ 10 } \)
Solution:
Given that Multiply \(\frac { 1 }{ 4 } \) × \(\frac { 8 }{ 6 } \) × \(\frac { 3 }{ 10 } \).
Firstly, multiply the numerators with numerators.
1 × 8 × 3 = 24.
Next, multiply the denominators with denominators.
4 × 6 × 10 = 240.
Finally, write the fraction in the simplest form.
\(\frac { 24 }{ 240 } \)
Simplify the fraction to get the final answer.
\(\frac { 1 }{ 10 } \).

The final answer is \(\frac { 1 }{ 10 } \).

(ix) \(\frac { 5 }{ 16 } \) × \(\frac { 11 }{ 23 } \)
Solution:
Given that Multiply \(\frac { 5 }{ 16 } \) × \(\frac { 11 }{ 23 } \).
Firstly, multiply the numerators with numerators.
5 × 11 = 55.
Next, multiply the denominators with denominators.
16 × 23 = 368.
Finally, write the fraction in the simplest form.
\(\frac { 55 }{ 368 } \)
Simplify the fraction to get the final answer.
\(\frac { 55 }{ 368 } \).

The final answer is \(\frac { 55 }{ 368 } \).

(x) \(\frac { 1 }{ 2 } \) of 50
Solution:
Given that \(\frac { 1 }{ 2 } \) of 50
multiply \(\frac { 1 }{ 2 } \) by 50
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
1 × 50 = 50.
So, the fraction is \(\frac { 50 }{ 2 } \)
Simplify the fraction to get the final answer.
\(\frac { 50 }{ 2 } \) = 25.

The final answer is 25.

(xi) \(\frac { 1 }{ 3 } \) of 90

Solution:
Given that \(\frac { 1 }{ 3 } \) of 90
multiply \(\frac { 1 }{ 3 } \) by 90
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
1 × 90 = 90.
So, the fraction is \(\frac { 90 }{ 3 } \)
Simplify the fraction to get the final answer.
\(\frac { 90 }{ 3 } \) = 30.

The final answer is 30.

II. Multiply and write the product in the lowest terms.

(i) \(\frac { 1 }{ 2 } \) × 60
(ii) \(\frac { 1 }{ 3 } \) × 18
(iii) \(\frac { 2 }{ 5 } \) × 25
(iv) \(\frac { 4 }{ 3 } \) × 0
(v) \(\frac { 7 }{ 29 } \) × 1
(vi) 6 × \(\frac { 7 }{ 36 } \)
(vii) \(\frac { 5 }{ 34 } \) × \(\frac { 34 }{ 8 } \)
(viii) \(\frac { 12 }{ 25 } \) × \(\frac { 5 }{ 6 } \)
(ix) \(\frac { 6 }{ 14 } \) × \(\frac { 56 }{ 7 } \)
(x) \(\frac { 1 }{ 3 } \) × \(\frac { 4 }{ 5 } \) × \(\frac { 5 }{ 8 } \)
(xi) \(\frac { 6 }{ 3 } \) × \(\frac { 1 }{ 2 } \) × \(\frac { 3 }{ 3 } \)
(xii) 3\(\frac { 8 }{ 5 } \) × \(\frac { 5 }{ 4 } \)

(i) \(\frac { 1 }{ 2 } \) × 60
Solution:
Given that \(\frac { 1 }{ 2 } \) × 60
Now, multiply \(\frac { 1 }{ 2 } \) by 60.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
1 × 60 = 60.
So, the fraction is \(\frac { 60 }{ 2 } \)
Simplify the fraction to get the final answer.
\(\frac { 60 }{ 2 } \) = 30.

The final answer is 30.

(ii) \(\frac { 1 }{ 3 } \) × 18
Solution:
Given that \(\frac { 1 }{ 3 } \) × 18
Now, multiply \(\frac { 1 }{ 3 } \) by 18.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
1 × 18 = 18.
So, the fraction is \(\frac { 18 }{ 3 } \)
Simplify the fraction to get the final answer.
\(\frac { 18 }{ 3 } \) = 6.

The final answer is 6.

(iii) \(\frac { 2 }{ 5 } \) × 25
Solution:
Given that \(\frac { 2 }{ 5 } \) × 25
Now, multiply \(\frac { 2 }{ 5 } \) by 25.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
2 × 25 = 50.
So, the fraction is \(\frac { 50 }{ 5 } \)
Simplify the fraction to get the final answer.
\(\frac { 50 }{ 5 } \) = 10.

The final answer is 10.

(iv) \(\frac { 4 }{ 3 } \) × 0
Solution:
Any fraction that multiplies with 0 gives 0.
Therefore, the answer is 0.

(v) \(\frac { 7 }{ 29 } \) × 1
Solution:
Any fraction that multiplies with 1 gives the same output.
Therefore, the answer is \(\frac { 7 }{ 29 } \).

(vi) 6 × \(\frac { 7 }{ 36 } \)
Solution:
Given that 6 × \(\frac { 7 }{ 36 } \)
Now, multiply 6 by \(\frac { 7 }{ 36 } \).
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
6 × 7 = 42.
So, the fraction is \(\frac { 42 }{ 36 } \)
Simplify the fraction to get the final answer.
\(\frac { 42 }{ 36 } \) = \(\frac { 21 }{ 18 } \).

The final answer is \(\frac { 21 }{ 18 } \).

(vii) \(\frac { 5 }{ 34 } \) × \(\frac { 34 }{ 8 } \)
Solution:
Given that Multiply \(\frac { 5 }{ 34 } \) × \(\frac { 34 }{ 8 } \).
Firstly, multiply the numerators with numerators.
5 × 34 = 170.
Next, multiply the denominators with denominators.
34 × 8 = 272.
Finally, write the fraction in the simplest form.
\(\frac { 170 }{ 272 } \)
Simplify the fraction to get the final answer.
\(\frac { 5 }{ 8 } \).

The final answer is \(\frac { 5 }{ 8 } \).

(viii) \(\frac { 12 }{ 25 } \) × \(\frac { 5 }{ 6 } \)
Solution:
Given that Multiply \(\frac { 12 }{ 25 } \) × \(\frac { 5 }{ 6 } \).
Firstly, multiply the numerators with numerators.
12 × 5 = 60.
Next, multiply the denominators with denominators.
25 × 6 = 150.
Finally, write the fraction in the simplest form.
\(\frac { 60 }{ 150 } \)
Simplify the fraction to get the final answer.
\(\frac { 60 }{ 150 } \) = \(\frac { 2 }{ 5 } \).

The final answer is \(\frac { 2 }{ 5 } \).

(ix) \(\frac { 6 }{ 14 } \) × \(\frac { 56 }{ 7 } \)
Solution:
Given that Multiply \(\frac { 6 }{ 14 } \) × \(\frac { 56 }{ 7 } \).
Firstly, multiply the numerators with numerators.
6 × 56 = 336.
Next, multiply the denominators with denominators.
14 × 7 = 98.
Finally, write the fraction in the simplest form.
\(\frac { 336 }{ 98 } \)
Simplify the fraction to get the final answer.
\(\frac { 336 }{ 98 } \) = 168.

The final answer is 168.

(x) \(\frac { 1 }{ 3 } \) × \(\frac { 4 }{ 5 } \) × \(\frac { 5 }{ 8 } \)
Solution:
Given that Multiply \(\frac { 1 }{ 3 } \) × \(\frac { 4 }{ 5 } \) × \(\frac { 5 }{ 8 } \).
Firstly, multiply the numerators with numerators.
1 × 4 × 5 = 20.
Next, multiply the denominators with denominators.
3 × 5 × 8 = 120.
Finally, write the fraction in the simplest form.
\(\frac { 20 }{ 120 } \)
Simplify the fraction to get the final answer.
\(\frac { 20 }{ 120 } \) = \(\frac { 1 }{ 6 } \).

The final answer is \(\frac { 1 }{ 6 } \).

(xi) \(\frac { 6 }{ 3 } \) × \(\frac { 1 }{ 2 } \) × \(\frac { 3 }{ 3 } \)
Solution:
Given that Multiply \(\frac { 6 }{ 3 } \) × \(\frac { 1 }{ 2 } \) × \(\frac { 3 }{ 3 } \).
Firstly, multiply the numerators with numerators.
6 × 1 × 3 = 18.
Next, multiply the denominators with denominators.
3 × 2 × 3 = 18.
Finally, write the fraction in the simplest form.
\(\frac { 18 }{ 18 } \)
Simplify the fraction to get the final answer.
\(\frac { 18 }{ 18 } \) = 1.

The final answer is 1.

(xii) 3\(\frac { 8 }{ 5 } \) × \(\frac { 5 }{ 4 } \)
Solution:
Given that multiply 3\(\frac { 8 }{ 5 } \) × \(\frac { 5 }{ 4 } \).
Firstly, convert given mixed fraction 3\(\frac { 8 }{ 5 } \) to fraction.
3\(\frac { 8 }{ 5 } \) = \(\frac { 23 }{ 5 } \)
Multiply \(\frac { 23 }{ 5 } \) × \(\frac { 5 }{ 4 } \).
Firstly, multiply the numerators with numerators.
23 × 5 = 115.
Next, multiply the denominators with denominators.
5 × 4 = 20.
Finally, write the fraction in the simplest form.
\(\frac { 115 }{ 20 } \)
Simplify the fraction to get the final answer.
\(\frac { 115 }{ 20 } \) = \(\frac { 23 }{ 4 } \).

The final answer is \(\frac { 23 }{ 4 } \).

III. Find the given quantity.

(i) \(\frac { 1 }{ 6 } \) of 48 kg apples
Solution:
Given that \(\frac { 1 }{ 6 } \) of 48 kg apples.
\(\frac { 1 }{ 6 } \) × 48 kg
8 kg

The answer is 8 kg.

(ii) \(\frac { 1 }{ 7 } \) of $280
Solution:
Given that \(\frac { 1 }{ 7 } \) of $280.
\(\frac { 1 }{ 7 } \) × $280
$40

The answer is $40.

(iii) \(\frac { 6 }{ 3 } \) of 54 km
Solution:
Given that \(\frac { 6 }{ 3 } \) of 54 km.
\(\frac { 6 }{ 3 } \) × 54 km
108 km

The answer is 108 km.

(iv) \(\frac { 2 }{ 8 } \) of 40 chairs
Solution:
Given that \(\frac { 2 }{ 8 } \) of 40 chairs.
\(\frac { 2 }{ 8 } \) × 40 chairs
10 chairs

The answer is 10 chairs.

Word problems on Multiplying Fractions

1. 3\(\frac { 5 }{ 8 } \) m of cloth is required to make a shirt. Sam wants to make 32 shirts, what length of cloth does he need?

Solution:
Given that 3\(\frac { 5 }{ 8 } \) m of cloth is required to make a shirt.
Sam wants to make 32 shirts.
3\(\frac { 5 }{ 8 } \) m of 32.
Firstly, convert given mixed fraction 3\(\frac { 5 }{ 8 } \) to fraction.
3\(\frac { 5 }{ 8 } \) = \(\frac { 29 }{ 8 } \)
\(\frac { 29 }{ 8 } \) × 32
Now, multiply \(\frac { 29 }{ 8 } \) by 32.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
29 × 32 = 928.
So, the fraction is \(\frac { 928 }{ 8 } \)
Simplify the fraction to get the final answer.
\(\frac { 928 }{ 8 } \) = 116.

The final answer is 116.

2. \(\frac { 5 }{ 2 } \) cups of milk is required to make a cake of 1 kg. How many cups of milk is required to make a cake of \(\frac { 2 }{ 5 } \) kg?

Solution:
Given that \(\frac { 5 }{ 2 } \) cups of milk is required to make a cake of 1 kg.
To make a cake of \(\frac { 2 }{ 5 } \) kg, multiply \(\frac { 5 }{ 2 } \) with \(\frac { 2 }{ 5 } \).
Firstly, multiply the numerators with numerators.
2 × 5 = 10.
Next, multiply the denominators with denominators.
5 × 2 = 10.
Finally, write the fraction in the simplest form.
\(\frac { 10 }{ 10 } \)
Simplify the fraction to get the final answer.
\(\frac { 10 }{ 10 } \) = 1.

The final answer is 1.

3. Shelly bought \(\frac { 11 }{ 9 } \) liters of juice. If the cost of 1-liter juice is $36, find the total cost of juice?

Solution:
Given that Shelly bought \(\frac { 11 }{ 9 } \) liters of juice.
If the cost of 1-liter juice is $36, multiply \(\frac { 11 }{ 9 } \) with $36.
\(\frac { 11 }{ 9 } \) × $36 = $44.

The final answer is $44.

4. The weight of each bag is 7\(\frac { 1 }{ 9 } \) Kg. What would be the weight of 36 such bags?

Solution:
Given that the weight of each bag is 7\(\frac { 1 }{ 9 } \) Kg.
If the weight of 36 such bags is 7\(\frac { 1 }{ 9 } \) Kg × 36.
Firstly, convert given mixed fraction 7\(\frac { 1 }{ 9 } \) to fraction.
7\(\frac { 1 }{ 9 } \) = \(\frac { 64 }{ 9 } \)
Now, multiply \(\frac { 64 }{ 9 } \) by 36.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
64 × 36 = 2304.
So, the fraction is \(\frac { 2304 }{ 9 } \)
Simplify the fraction to get the final answer.
\(\frac { 2304 }{ 9 } \) = 256.

The final answer is 256.

5. Sam works for 1 \(\frac { 5 }{ 6 } \) hours each day. For how much time will she work in a month if she works for 24 days in a month?

Solution:
Given that Sam works for 1 \(\frac { 5 }{ 6 } \) hours each day.
If she works for 24 days in a month, 1 \(\frac { 5 }{ 6 } \) hours each day × 24
Firstly, convert given mixed fraction 1 \(\frac { 5 }{ 6 } \) to fraction.
1 \(\frac { 5 }{ 6 } \) = \(\frac { 11 }{ 6 } \)
Now, multiply \(\frac { 11 }{ 6 } \) by 24.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
11 × 24 = 264.
So, the fraction is \(\frac { 264 }{ 6 } \)
Simplify the fraction to get the final answer.
\(\frac { 264 }{ 6 } \) = 44.

The final answer is 44.