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

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 Problem Set Answer Key

Question 1.
Label the tape diagrams. Then, fill in the blanks below to make the statements true.
a. 6 Ã— 6 = __36___

(6 Ã— 6) = (5 + 1) Ã— 6
= (5 Ã— 6) + (1 Ã— 6)
=Â  Â  30Â  Â  + __6____
= ____6__

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.

b. 7 Ã— 6 = __42___

(7 Ã— 6) = (5 + 2) Ã— 6
= (5 Ã— 6) + (2 Ã— 6)
=Â  Â 30Â  Â  Â + __12____
= ____42__

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.

c. 8 Ã— 6 = ___48__

8 Ã— 6 = (5 + ___3__ ) Ã— 6
= (5 Ã— 6) + ( __3__ Ã— 6)
=Â  Â 30Â  Â  + __18____
= __48____

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.

d. 9 Ã— 6 = __54___

9 Ã— 6 = (5 + __4___ ) Ã— 6
= (5 Ã— 6) + ( _4___ Ã— 6)
=Â  Â 30Â  Â  + __24____
= __54____

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.

Question 2.
Break apart 54 to solve 54 Ã· 6.

54 Ã· 6 = (30 Ã· 6) + ( __24________ Ã· 6)
= 5 + ______4______
= ____9_______

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 Ã· 7.

49 Ã· 7 = (35 Ã· 7) + ( ___14_______ Ã· 7)
= 5 + _______2_____
= _____7______

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.

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

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

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.

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.

Question 2.
Malia solves 6 Ã— 7 using (5 Ã— 7) + 7. Leonidas solves 6 Ã— 7 using (6 Ã— 5) + (6 Ã— 2). Who is correct? Draw a picture to help explain your 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 Ã— 7 = __42___

(6 Ã— 7) = (5 + 1) Ã— 7
= (5 Ã— 7) + (1 Ã— 7)
=Â  Â 35Â  Â  Â + ___7___
= ___42___

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.

b. 7 Ã— 7 = __49___

(7 Ã— 7) = (5 + 2) Ã— 7
= (5 Ã— 7) + (2 Ã— 7)
=Â  Â 35Â  Â  Â + __14___
= ___49___

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.

c. 8 Ã— 7 = __56___

8 Ã— 7 = (5 + ___3__ ) Ã— 7
= (5 Ã— 7) + ( __3__ Ã— 7)
=Â  Â 35Â  Â  Â + ___21___
= __56____

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.

d. 9 Ã— 7 = __63___

9 Ã— 7 = (5 + __4___ ) Ã— 7
= (5 Ã— 7) + ( __4__ Ã— 7)
=Â  Â 35Â  Â  + ___28___
= ___63___

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.

Question 2.
Break apart 54 to solve 54 Ã· 6.

54 Ã· 6 = (30 Ã· 6) + ( ____24_______ Ã· 6)
= 5 + _______4_____
= _____9______

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 56 to solve 56 Ã· 7

56 Ã· 7 = ( __35___ Ã· ___7__ ) + ( __21___ Ã· ___7__ )
= 5 + ____3____
= ____8______

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.

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.

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 Ã— 6 by thinking of it as (5 Ã— 7) + 7. Is he correct? Explain Ronaldoâ€™s strategy.