Eureka Math

Engage NY Eureka Math 3rd Grade Module 1 Lesson 6 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 6 Problem Set Answer Key

Question 1.
Rick puts 15 tennis balls into cans. Each can holds 3 balls.
Circle groups of 3 to show the balls in each can.

Rick needs ____5___ cans.
___5___ × 3 = 15
15 ÷ 3 = __5____

Rick needs 5 cans,

Explanation:
Given Rick puts 15 tennis balls into cans and each can
holds 3 balls. Circled groups of 3 to show the balls in each can.
Rick needs 15 ÷ 3 = 5, 5 X 3 = 15,
Therefore Rick needs 5 cans.

Question 2.
Rick uses 15 tennis balls to make 5 equal groups.
Draw to show how many tennis balls are in each group.
There are ____3___ tennis balls in each group.
5 × __3____ = 15
15 ÷ 5 = ___3___

There are 3 tennis balls in each group.

Explanation:
Given Rick uses 15 tennis balls to make 5 equal groups.
Drawn circles to show 3 number of tennis balls are in
each group as 15 ÷ 5 = 3, 5 X 3 = 15,
Therefore, there are 3 tennis balls in each group.

Question 3.
Use an array to model Problem 1.
a. ___5___ × 3 = 15
15 ÷ 3 = ___5___
The number in the blanks represents
_____5 groups___________________.

Explanation:
Used an array to model Problem 1 as
5 X 3 = 15 or 15 ÷ 3 = 5,
The number in the blanks represents groups as
5 groups of balls.

b. 5 × __3____ = 15
15 ÷ 5 = __3____
The number in the blanks represents
____________3 balls in each group________.

The number in the blanks represents 3 balls in each group,

Explanation:
Rick uses 15 tennis balls to make 5 equal groups,
15 ÷ 5 = 3 or 5 X 3= 15, The number in the blanks
represents 3 balls in each group.

Question 4.
Deena makes 21 jars of tomato sauce. She puts 7 jars in
each box to sell at the market. How many boxes does Deena need?
21 ÷ 7 = __3____
___3___ × 7 = 21
What is the meaning of the unknown factor and quotient?
_________3______________

Deena needs 3 boxes,

Explanation:
Given Deena makes 21 jars of tomato sauce and she
puts 7 jars in each box to sell at the market.
So number of boxes Deena needs are 21 ÷ 7 = 3, (3 X 7 = 21)
Therefore, Deena needs 3 boxes.

Question 5.
The teacher gives the equation 4 × __3__ = 12. Charlie
finds the answer by writing and solving 12 ÷ 4 = __3__.
Explain why Charlie’s method works.

Charlie method is 12 ÷ 4 = 3,

Explanation:
Given The teacher gives the equation 4 × ____ = 12.
Charlie finds the answer by writing and solving as
12 ÷ 4 = 3,
because multiplication and division are closely related,
given that division is the inverse operation of multiplication.
When we divide, we look to separate into equal groups,
while multiplication involves joining equal groups.
If we divide this product by one of the factors,
we get the other factor as a result. So Charlie uses
division method to solve missing factor in the given
equation as 3. So the equation is 4 X 3 = 12.

Question 6.
The blanks in Problem 5 represent the size of the groups.
Draw an array to represent the equations.

Explanation:
Drawn an array that represents the equations
as  4 X 3 = 12.

Eureka Math Grade 3 Module 1 Lesson 6 Exit Ticket Answer Key

Cesar arranges 12 notecards into rows of 6 for his presentation.
Draw an array to represent the problem.
12 ÷ 6 = ___2_____
____2____ × 6 = 12
What do the unknown factor and quotient represent?
___2 notecards in each row____

Explanation:
Given Cesar arranges 12 notecards into rows of
6 for his presentation.
Drawn an array to represent the problem as
12 ÷ 6 = 2, 2 X 6 = 12, means 6 rows  and 2 columns
as shown above, The unknown factor and quotient 2
represents 2 notecards in each row.

Eureka Math Grade 3 Module 1 Lesson 6 Homework Answer Key

Question 1.
Mr. Hannigan puts 12 pencils into boxes. Each box holds 4 pencils.
Circle groups of 4 to show the pencils in each box.

Mr. Hannigan needs ___3____ boxes.
___3___ × 4 = 12
12 ÷ 4 = __3____

Mr. Hannigan needs 3 boxes,

Explanation:
Given Mr. Hannigan puts 12 pencils into boxes.
Each box holds 4 pencils.
Circled groups of 4 to show the pencils in each box
as shown above, So Mr. Hannigan needs 12 ÷ 4 = 3 boxes,
( 3 x 4 = 12).

Question 2.
Mr. Hannigan places 12 pencils into 3 equal groups.
Draw to show how many pencils are in each group.
There are ___4____ pencils in each group.
3 × ___4___ = 12
12 ÷ 3 = ___4___

There are 4 pencils in each group,

Explanation:
Given Mr. Hannigan places 12 pencils into 3 equal groups,
Drawn to show 4 number of pencils are in each group
as 12 ÷ 3 = 4, 3 X 4 = 12, Therefore 4 pencils are there in each group.

Question 3.
Use an array to model Problem 1.
a. ___3___ × 4 = 12
12 ÷ 4 = ___3___
The number in the blanks represents
___________3 groups______________.
a.

The number in the blanks represents 3 groups,

Explanation:
Used an array to model Problem 1 as 3 x 4 = 12,
or 12 ÷ 4 = 3, therefore the number in the blanks
represents 3 groups.

b. 3 × ___4___ = 12
12 ÷ 3 = __4____
The number in the blanks represents
___________4 pencils in each group_____________.

The number in the blanks represents 4 pencils in each group,

Explanation:
Given Mr. Hannigan places 12 pencils into 3 equal groups.
12 ÷ 3 = 4 or 3 X 4= 12, The number in the blanks
represents 4 pencils in each group.

Question 4.
Judy washes 24 dishes. She then dries and stacks the dishes
equally into 4 piles. How many dishes are in each pile?
24 ÷ 4 = ___6____
4 × ___6_____ = 24
What is the meaning of the unknown factor and quotient?
__________6 dishes are in each pile____________________

There are 6 dishes in each pile,

Explanation:
Given Judy washes 24 dishes and she then dries and
stacks the dishes equally into 4 piles.
So number of dishes in each pile are 24 ÷ 4 = 6, or
4 X 6 = 24, The meaning of the unknown factor and
quotient is 6 dishes are there in each pile.

Question 5.
Nate solves the equation _____ × 5 = 15 by writing and
solving 15 ÷ 5 = ____. Explain why Nate’s method works.

Nate solves the equation as 3 X 5 = 15,

Explanation:
Given Nate solves the equation _____ × 5 = 15 by writing and
solving as 15 ÷ 5 = 3, Nate’s method is correct because in
the given equation ___ X 5 = 15 , we bring 5 to other side it becomes as
15 ÷ 5 = 3 now we check as 3 X 5 it becomes 15 only,
So Nate’s method work.

Question 6.
The blanks in Problem 5 represent the number of groups.
Draw an array to represent the equations.

Explanation:
Drawn an array to represent the equations as
3 X 5 = 15 or 15 ÷ 5 = 3 as shown in the picture above.

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

Eureka Math Grade 3 Module 1 Answer Key

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

Question 1.
Fill in the blanks to make true statements.

a. 3 groups of five = ___15______
3 fives = ____15_____
3 × 5 = ___15______

3 groups of five = 15,
3 fives = 15,
3 X 5 = 15,

Explanation:
Given expressions as
3 groups of five means 3 X 5 = 15,
3 fives = 3 x 5 = 15 and
3 X 5 = 15 as 3 is multiplied by 5 we get 15.

b. 3 + 3 + 3 + 3 + 3 = __15___
5 groups of three = ___15______
5 × 3 = ____15_____

3 + 3 + 3 + 3 + 3 = 15,
5 groups of three = 15,
5 × 3 = 15,

Explanation:
Given expressions as
3 + 3 + 3 + 3 + 3 = 15 as 3 is added 5 times,
5 groups of three = 15 means 5  X 3 = 15,
5 X 3 = 15 as 5 is multiplied by 3 we get 15.

c. 6 + 6 + 6 + 6 = ___24________
____4___ groups of six = ___24_______
4 × __6____ = ____24______

6 + 6 + 6 + 6 = 24,
4 groups of six = 24,
4 X 6 = 24,

Explanation:
Given expressions as
6 + 6 + 6 + 6 = 24 as 6 is added 4 times,
4 groups of six = 24 means 4  X 6 = 24,
4 X 6 = 24 as 4 is multiplied by 6 we get 24.

d. 4 +__4__ + __4__ + __4__ + _4___ + __4__ = ___24______
6 groups of ____4____ = ____24_______
6 × __4____ = ____24______

4 + 4 + 4 + 4 + 4 + 4 = 24,
6 groups of 4 = 24,
6 x 4 = 24,

Explanation:
Given expressions as
4 + ___ + ___ +___ +___ +___ =  means by seeing the picture
4 is added 6 times as 4 + 4 + 4 + 4 + 4 + 4 we get 24,
6 groups of ______ =  means again by seeing the picture
6 X 4 we get 24, So 6 groups of 4 = 24,
6 X 4 = 24 as 6 is multiplied by 4 we get 24.

Question 2.
The picture below shows 2 groups of apples. Does the picture
show 2 × 3? Explain why or why not.

No, the picture does not show 2 X 3,

Explanation:
Given the picture below shows 2 groups of apples but
in the 2 groups the number of apples are not the same
as in group -1 we have 3 apples, group 2 we have 2 apples
in total there are 5 apples and 2 X 3 = 6 whose value is not
same as shown in the picture ,
So, the picture does not show 2 X 3.

Question 3.
Draw a picture to show 2 × 3 = 6.

Explanation:
Drawn a picture above to show 2 X 3 = 6 .

Question 4.
Caroline, Brian, and Marta share a box of chocolates.
They each get the same amount. Circle the chocolates below
to show 3 groups of 4. Then, write a repeated addition sentence
and a multiplication sentence to represent the picture.

Addition sentence is 4 + 4 + 4 = 12,
Multiplication sentence is 3 X 4 = 12,

Explanation:
Given Caroline, Brian, and Marta share a box of chocolates and
they each get the same amount as 12 ÷ 3 = 4 each.
Circled the chocolates above to show 3 groups of 4 and
wrote a repeated addition sentence as  4 + 4 + 4 = 12 and
a multiplication sentence to represent the picture as 3 X 4 = 12.

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

Question 1.
The picture below shows 4 groups of 2 slices of watermelon.
Fill in the blanks to make true repeated addition and
multiplication sentences that represent the picture.

True repeated addition is 2 + 2 + 2 + 2 = 8,
Multiplication sentences is 4 X 2 = 8.

Explanation:
Given the picture above shows 4 groups of
2 slices of watermelon.
Filled in the blanks to make true repeated addition
as 2 + 2 + 2 + 2 = 8 and multiplication sentences that
represent the picture as 4 X 2 = 8.

Question 2.
Draw a picture to show 3 + 3 + 3 = 9. Then,
write a multiplication sentence to represent the picture.

Addition sentence is  3 + 3 + 3 = 9,
Multiplication sentences is 3 X 3 = 9.

Explanation:
Drawn a picture above as  3 groups of 3 birds  to show
3 + 3 + 3 = 9 and writing a multiplication sentence to
represent the picture as 3 X 3 = 9 birds in total.

Eureka Math Grade 3 Module 1 Lesson 1 Homework Answer Key

Question 1.
Fill in the blanks to make true statements.

a. 4 groups of five = ____20_____
4 fives = ___20______
4 × 5 = ____20_____

4 groups of five = 20,
4 fives = 20,
4 X 5 = 20,

Explanation:
Given expressions as
4 groups of five means 4 X 5 = 20,
4 fives = 4 x 5 = 20 and
4 X 5 = 20 as 4 is multiplied by 5 we get 20.

b. 5 groups of four = _________
5 fours = _________
5 × 4 = _________

5 groups of four = 20,
5 fours = 20,
5 X 4 = 20,

Explanation:
Given expressions as
5 groups of four means 5 X 4 = 20,
5 fours = 5 x 4 = 20 and
5 X 4 = 20 as 5 is multiplied by 4 we get 20.

c. 6 + 6 + 6 = ___18________
____3___ groups of six = ____18______
3 × __6____ = ____18______

6 + 6 + 6  = 18,
3 groups of 6 = 18,
3 x 6 = 18,

Explanation:
Given expressions as 6 + 6 + 6 = 18 as
6 is added 3 times we get 18,
_____ groups of six =  means again by seeing the picture
3 groups of 6 =  3 X 6 we get 18, So 3 groups of 6 = 18,
3 X 6 = 18 as 3 is multiplied by 6 we get 18.

d. 3 + __3__ + __3__ + __3__ + __3__ + __3__ = ___18____
6 groups of ____3____ = _____18______
6 × __3__ = ____18______

3 + 3 + 3 + 3 + 3 + 3 = 18,
6 groups of 3 = 18,
6 x 3 = 18,

Explanation:
Given expressions as
3 + ___ + ___ +___ +___ +___ =  means by seeing the picture
3 is added 6 times as 3 + 3 + 3 + 3 + 3 + 3 we get 18,
6 groups of ______ =  means again by seeing the picture
6 X 3 we get 18, So 6 groups of 3 = 18,
6 X 3 = 18 as 6 is multiplied by 3 we get 18.

Question 2.
The picture below shows 3 groups of hot dogs.
Does the picture show 3 × 3? Explain why or why not.

Yes, the picture shows 3 X 3,

Explanation:
Given the picture below shows 3 groups of of hot dogs,
In the 3 groups the number of hot dogs are the same as 3,
So in total we have 3 X 3 = 9 hotdogs in the picture
which matches with the given equation as 3 X 3 whose
value is also same 9, So, the picture shows 3 X 3.

Question 3.
Draw a picture to show 4 × 2 = 8,

Explanation:
Drawn picture of flowers as shown above to show 4 X 2 = 8.

Question 4.
Circle the pencils below to show 3 groups of 6.
Write a repeated addition and a multiplication sentence
to represent the picture.


Repeated addition sentence is 6 + 6 + 6 = 18,
Multiplication sentence is 3 X 6 = 18,

Explanation:
Circled the pencils above to show 3 groups of 6.
Wrote a repeated addition as 6 + 6 + 6 = 18 and
a multiplication sentence as 3 X 6 = 18 to represent the picture.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 12 Answer Key

Eureka Math Grade 3 Module 7 Lesson 12 Pattern Sheet Answer Key

Multiply.

multiply by 7 (6–10)
Answer:

Explanation:
7 × 6 = 42
7 × 7 = 49
7 × 8 = 56
7 × 9 = 63
7 × 10 = 70.

Eureka Math Grade 3 Module 7 Lesson 12 Problem Set Answer Key

Question 1.
Measure and label the side lengths of the shapes below in centimeters. Then, find the perimeter of each shape.
a.

Perimeter = _____cm +_____cm +_____cm +_____cm
= _______ cm
Answer:
Perimeter of the ABCD given shape = Side + Side + Side + Side
= 2cm + 2cm + 2cm + 2cm
= 8cm.

Explanation:

Length of the AB side of the ABCD shape = 2cm
Length of the BC side of the ABCD shape = 2cm
Length of the CD side of the ABCD shape = 2cm
Length of the DA side of the ABCD shape = 2cm
Perimeter of the given ABCD shape = Side + Side + Side + Side
= AB + BC + CD+ DA
= 2cm + 2cm + 2cm + 2cm
= 4cm + 2cm + 2cm
= 6cm + 2cm
= 8cm.

b.

Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the given ABCDEF Hexagon shape = Side + Side + Side + Side + Side + Side
= 3cm + 3cm + 3cm + 3cm + 3cm + 3cm
= 18cm.

Explanation:

Length of the AB side of the given ABCDEF Hexagon shape = 3cm.
Length of the BC side of the given ABCDEF Hexagon shape = 3cm.
Length of the CD side of the given ABCDEF Hexagon shape = 3cm.
Length of the DE side of the given ABCDEF Hexagon shape = 3cm.
Length of the EF side of the given ABCDEF Hexagon shape = 3cm.
Length of the FA side of the given ABCDEF Hexagon shape = 3cm.
Perimeter of the given ABCDEF Hexagon shape = Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF + FA
= 3cm + 3cm + 3cm + 3cm + 3cm + 3cm
= 6cm +3cm + 3cm + 3cm + 3cm
= 9cm +3cm + 3cm + 3cm
= 12cm +3cm + 3cm
= 15cm + 3cm
= 18cm.

c.

Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the given ABCD Parallelogram shape = Side + Side + Side + Side
= 4cm + 4cm + 4cm + 4cm
= 16cm.

Explanation:

Length of the AB side of the given ABCD Parallelogram shape = 4cm
Length of the BC side of the given ABCD Parallelogram shape = 4cm
Length of the CD  side of the given ABCD Parallelogram shape = 4cm
Length of the DA side of the given ABCD Parallelogram shape = 4cm
Perimeter of the given ABCD Parallelogram shape = Side + Side + Side + Side
=AB + BC + CD +DA
= 4cm + 4cm + 4cm + 4cm
= 8cm + 4cm + 4cm
= 12cm + 4cm
=16cm.

d.

Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the ABC Triangle = Side + Side + Side
= 5cm + 5cm + 5cm
= 15cm.

Explanation:

Length of the AB side of the ABC Triangle = 5cm
Length of the BC side of the ABC Triangle = 5cm
Length of the CA side of the ABC Triangle = 5cm
Perimeter of the ABC Triangle = Side + Side + Side
= AB + BC + CA
= 5cm + 5cm + 5cm
= 10cm + 5cm
= 15cm.

e.

Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the given ABCDEF figure = Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE+ EF + FA
= 5.5cm+ 1cm + 3cm + 2cm + 3cm + 3cm
= 17.5cm

Explanation:

Length of AB side if the given ABCDEF figure = 5.5cm
Length of BC side if the given ABCDEF figure = 1cm
Length of CD side if the given ABCDEF figure = 3cm
Length of DE side if the given ABCDEF figure = 2cm
Length of EF side if the given ABCDEF figure = 3cm
Length of FA side if the given ABCDEF figure = 3cm
Perimeter of the given ABCDEF figure = Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE+ EF + FA
= 5.5cm+ 1cm + 3cm + 2cm + 3cm + 3cm
= 6.5cm + 3cm + 2cm + 3cm + 3cm
= 9.5cm + 2cm + 3cm + 3cm
= 11.5cm + 3cm + 3cm
= 14.5cm + 3cm
= 17.5cm

Question 2.
Carson draws two triangles to create the new shape shown below. Use a ruler to find the side lengths of Carson’s shape in centimeters. Then, find the perimeter.

Answer:
Perimeter of ABCD Carson’s shape = Side + Side + Side + Side
= AB + BC + CD+ DA
= 2cm + 2cm + 2cm + 2cm
=8cm.

Explanation:

Length of AB side of ABCD Carson’s shape = 2cm
Length of BC side of ABCD Carson’s shape = 2cm
Length of CD side of ABCD Carson’s shape = 2cm
Length of DA side of ABCD Carson’s shape = 2cm
Perimeter of ABCD Carson’s shape = Side + Side + Side + Side
= AB + BC + CD+ DA
= 2cm + 2cm + 2cm + 2cm
=4cm + 2cm + 2cm
= 6cm + 2cm
=8cm.

Question 3.
Hugh and Daisy draw the shapes shown below. Measure and label the side lengths in centimeters. Whose shape has a greater perimeter? How do you know?

Answer:
Perimeter of ABCDE Hugh’s shape = Side + Side + Side + Side + Side
= AB + BC + CD + DE + EA
= 3cm + 3cm + 3cm + 3cm + 3cm
= 15cm.
Perimeter of FGHI Daisy’s shape =  Side + Side + Side + Side
= FG+ GH+ HI+ IF
=  4cm + 5cm + 4cm + 3cm
= 16cm.
Perimeter of FGHI Daisy’s shape is greater than the Perimeter of ABCDE Hugh’s shape because the  measurement value is greater than the other one’s shape.

Explanation:

Length of the AB side of ABCDE Hugh’s shape = 3cm
Length of the BC side of ABCDE Hugh’s shape = 3cm
Length of the CD side of ABCDE Hugh’s shape = 3cm
Length of the DE side of ABCDE Hugh’s shape = 3cm
Length of the EA side of ABCDE Hugh’s shape = 3cm
Perimeter of ABCDE Hugh’s shape = Side + Side + Side + Side + Side
= AB + BC + CD + DE + EA
= 3cm + 3cm + 3cm + 3cm + 3cm
= 6cm + 3cm + 3cm + 3cm
= 9cm + 3cm + 3cm
= 12cm + 3cm
= 15cm.

Length of the FG side of FGHI Daisy’s shape = 4cm
Length of the GH side of FGHI Daisy’s shape = 5cm
Length of the HI side of FGHI Daisy’s shape = 4cm
Length of the IF side of FGHI Daisy’s shape = 3cm
Perimeter of FGHI Daisy’s shape =  Side + Side + Side + Side
= FG+ GH+ HI+ IF
=  4cm + 5cm + 4cm + 3cm
= 9cm + 4cm +3cm
= 13cm + 3cm
= 16cm.

Question 4.
Andrea measures one side length of the square below and says she can find the perimeter with that measurement. Explain Andrea’s thinking. Then, find the perimeter in centimeters.

Answer:
Andrea’s figure is a Square. So, her figure’s  all sides are going to be equal. Yes, she is correct in her thinking.
Perimeter of ABCD Andrea’s Square = Side + Side + Side + Side
= 4cm + 4cm + 4cm+ 4cm
= 16cm.

Explanation:

Andrea’s figure is a Square. So, her figure’s  all sides are going to be equal. Yes, she is correct in her thinking, ” measuring one side length of the square below, she can find the perimeter with that measurement.”
Length of AB side of Andrea’s Square = 4cm
Perimeter of ABCD Andrea’s Square = Side + Side + Side + Side
= 4cm + 4cm + 4cm+ 4cm
= 8cm + 4cm + 4cm
=12cm + 4cm
= 16cm.

Eureka Math Grade 3 Module 7 Lesson 12 Exit Ticket Answer Key

Measure and label the side lengths of the shape below in centimeters. Then, find the perimeter.

Perimeter = __________________________________________
= _______ cm
Answer:
Perimeter of the given ABCDEFGHIJK Shape =  Side + Side + Side + Side + Side + Side + Side + Side + Side + Side + Side + Side
= AB + BC+ CD + DE + EF + FG + GH + HI + IJ + JK +KL + LA
= 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 32cm.

Explanation:

Length of the AB side of the given ABCDEFGHIJK Shape = 4cm
Length of the BC side of the given ABCDEFGHIJK Shape = 2cm
Length of the CD side of the given ABCDEFGHIJK Shape = 2cm
Length of the DE side of the given ABCDEFGHIJK Shape = 4cm
Length of the EF side of the given ABCDEFGHIJK Shape = 2cm
Length of the FG side of the given ABCDEFGHIJK Shape = 2cm
Length of the GH side of the given ABCDEFGHIJK Shape = 4cm
Length of the HI side of the given ABCDEFGHIJK Shape = 2cm
Length of the IJ side of the given ABCDEFGHIJK Shape =  2cm
Length of the JK side of the given ABCDEFGHIJK Shape = 4cm
Length of the KL side of the given ABCDEFGHIJK Shape = 2cm
Length of the LA side of the given ABCDEFGHIJK Shape = 2cm
Perimeter of the given ABCDEFGHIJK Shape =  Side + Side + Side + Side + Side + Side + Side + Side + Side + Side + Side + Side
= AB + BC+ CD + DE + EF + FG + GH + HI + IJ + JK +KL + LA
= 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 6cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 8cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 12cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 14cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 16cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 20 cm + 2cm + 2cm + 4cm + 2cm + 2cm
= 22cm + 2cm + 4cm + 2cm + 2cm
= 24cm + 4cm + 2cm + 2cm
= 28cm + 2cm + 2cm
= 30cm + 2cm
= 32cm.

Eureka Math Grade 3 Module 7 Lesson 12 Homework Answer Key

Question 1.
Measure and label the side lengths of the shapes below in centimeters. Then, find the perimeter of each shape.
a.

Perimeter = _____cm +_____cm +_____cm
= _______ cm
Answer:
Perimeter of ABC triangle = Side + Side + Side
= AB + BC + CA
= 3cm + 5cm + 4cm
= 12cm.

Explanation:

Length of the AB side of ABC triangle = 3cm
Length of the BC side of ABC triangle = 5cm
Length of the CA side of ABC triangle = 4cm
Perimeter of ABC triangle = Side + Side + Side
= AB + BC + CA
= 3cm + 5cm + 4cm
= 8cm + 4cm
= 12cm.

b.

Perimeter = _____________________
= _______ cm
Answer:
Perimeter of ABCD rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 6cm + 4cm + 6cm + 4cm
= 20cm.

Explanation:

Length of the AB side of ABCD rectangle = 6cm
Length of the BC side of ABCD rectangle = 4cm
Length of the CD side of ABCD rectangle = 6cm
Length of the DA side of ABCD rectangle = 4cm
Perimeter of ABCD rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 6cm + 4cm + 6cm + 4cm
= 10cm + 6cm + 4cm
= 16cm + 4cm
= 20cm.

c.

Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the ABCD Quadrilateral = Side + Side + Side + Side
= AB + BC + CD + DA
= 3cm + 4cm + 5cm + 4cm
= 16cm.

Explanation:

Length of the AB side in the ABCD Quadrilateral = 3cm
Length of the BC side in the ABCD Quadrilateral = 4cm
Length of the CD side in the ABCD Quadrilateral = 5cm
Length of the DA side in the ABCD Quadrilateral = 4cm
Perimeter of the ABCD Quadrilateral = Side + Side + Side + Side
= AB + BC + CD + DA
= 3cm + 4cm + 5cm + 4cm
= 7cm + 5cm + 4cm
= 12cm + 4cm
= 16cm.

d.

Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the ABCD Parallelogram = Side + Side + Side + Side
= AB + BC + CD + DA
= 5cm + 5cm + 5cm + 5cm
= 20cm.

Explanation:

Length of the AB side of the ABCD parallelogram = 5cm
Length of the BC side of the ABCD parallelogram = 5cm
Length of the CD side of the ABCD parallelogram = 5cm
Length of the DA side of the ABCD parallelogram = 5cm
Perimeter of the ABCD Parallelogram = Side + Side + Side + Side
= AB + BC + CD + DA
= 5cm + 5cm + 5cm + 5cm
= 10cm + 5cm + 5cm
= 15cm + 5cm
= 20cm.

e.

Perimeter = _____________________
= _______ cm
Answer:
Perimeter of the given ABCDEFGH figure = Side + Side + Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF + FG + GH + HA
= 2cm + 2cm + 3.5cm + 2cm + 2cm + 2.5cm + 7.5cm + 2.5cm
= 24cm.

Explanation:

Length of AB side of the given ABCDEFGH figure = 2cm
Length of BC side of the given ABCDEFGH figure = 2cm
Length of CD side of the given ABCDEFGH figure = 3.5cm
Length of DE side of the given ABCDEFGH figure = 2cm
Length of EF side of the given ABCDEFGH figure = 2cm
Length of FG side of the given ABCDEFGH figure = 2.5cm
Length of GH side of the given ABCDEFGH figure = 7.5cm
Length of HA side of the given ABCDEFGH figure = 2.5cm
Perimeter of the given ABCDEFGH figure = Side + Side + Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF + FG + GH + HA
= 2cm + 2cm + 3.5cm + 2cm + 2cm + 2.5cm + 7.5cm + 2.5cm
= 4cm + 3.5cm + 2cm + 2cm + 2.5cm + 7.5cm + 2.5cm
= 7.5cm + 2cm + 2cm + 2.5cm + 7.5cm + 2.5cm
= 9.5cm + 2cm + 2.5cm + 7.5cm + 2.5cm
= 11.5cm + 2.5cm + 7.5cm + 2.5cm
= 14cm + 7.5cm + 2.5cm
= 21.5cm + 2.5cm
= 24cm.

Question 2.
Melinda draws two trapezoids to create the hexagon shown below. Use a ruler to find the side lengths of Melinda’s hexagon in centimeters. Then, find the perimeter.

Answer:
Perimeter of ABCDEF Melinda’s hexagon = Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF +FA
= 3cm + 3cm + 3cm + 3cm + 3cm + 3cm
= 18cm.

Explanation:

Length of the AB side of ABCDEF Melinda’s hexagon = 3cm
Length of the BC side of ABCDEF Melinda’s hexagon = 3cm
Length of the CD side of ABCDEF Melinda’s hexagon = 3cm
Length of the DE side of ABCDEF Melinda’s hexagon = 3cm
Length of the EF side of ABCDEF Melinda’s hexagon = 3cm
Length of the FA side of ABCDEF Melinda’s hexagon = 3cm
Perimeter of the ABCDEF Melinda’s hexagon = Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF +FA
= 3cm + 3cm + 3cm + 3cm + 3cm + 3cm
= 6cm + 3cm + 3cm + 3cm + 3cm
= 9cm + 3cm + 3cm + 3cm
= 12cm + 3cm + 3cm
= 15cm + 3cm
= 18cm.

Question 3.
Victoria and Eric draw the shapes shown below. Eric says his shape has a greater perimeter because it has more sides than Victoria’s shape. Is Eric right? Explain your answer.

Answer:
Yes, Eric is correct because DEFG Square has more sides than the ABC Victoria’s triangle shape.
Perimeter of DEFG Eric’s Square shape is greater than the Perimeter of ABC Victoria’s triangle shape.
Perimeter of ABC Victoria’s triangle shape = 12cm

Explanation:

Length of AB Victoria’s triangle shape = 3cm
Length of BC Victoria’s triangle shape = 5cm
Length of CA Victoria’s triangle shape =4cm
Perimeter of ABC Victoria’s triangle shape = Side + Side + Side
= 3cm +  4cm + 5cm
= 7cm + 5cm
= 12cm.

Length of DEFG Eric’s Square shape = 4cm
Perimeter of DEFG Eric’s Square shape = Side × Side
= 4cm × 4cm
= 16cm.
Perimeter of Eric’s Square shape is greater than the Perimeter of Victoria’s triangle shape.

Question 4.
Jamal uses his ruler and a right angle tool to draw the rectangle shown below. He says the perimeter of his rectangle is 32 centimeters. Do you agree with Jamal? Why or why not?

Answer:
NO, I disagree with Jamal. Jamal answer is incorrect because actual Perimeter of his ABCD rectangle is 30cm not 32cm.

Explanation:

Length of the ABCD Rectangle Jamal drawn = 7.5cm
Width of the ABCD Rectangle Jamal drawn = 4cm
Perimeter of ABCD Rectangle Jamal drawn = Length × Width
= 7.5 × 4
= 30 cm.
Jamal answer is incorrect because actual Perimeter of his ABCD rectangle is 30cm not 32cm.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 5 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 5 Problem Set Answer Key

Question 1.

Divide 6 tomatoes into groups of 3.
There are ____2_____ groups of 3 tomatoes.
6 ÷ 3 = 2

There are 2 groups of 3 tomatoes,

Explanation:
Dividing 6 tomatoes into groups of 3 we get
6 ÷ 3 = 2 groups,
So there are 2 groups of 3 tomatoes.

Question 2.

Divide 8 lollipops into groups of 2.
There are ____4___ groups.
8 ÷ 2 = ___4____

There are 4 groups,

Explanation:
Dividing 8 lollipops into groups of 2 as
8 ÷ 2 = 4 we get 4 groups,
So, there are 4 groups of 2.

Question 3.

Divide 10 stars into groups of 5.
10 ÷ 5 = ___2____

There are 2 groups,

Explanation:
Dividing 10 stars into groups of 5 as
10 ÷ 5 = 2 we get 2 groups,
So, there are 2 groups of 5.

Question 4.

Divide the shells to show 12 ÷ 3 = ____4____,
where the unknown represents the number of groups.
How many groups are there? ___4_____

There are 4 groups,

Explanation:
Dividing 12 shells into groups of 3 as
12 ÷ 3 = 4 we get 4 groups,
So, there are 4 groups of 3.

Question 5.
Rachel has 9 crackers. She puts 3 crackers in each bag.
Circle the crackers to show Rachel’s bags.

a. Write a division sentence where the answer
represents the number of Rachel’s bags.
b. Draw a number bond to represent the problem.

Division sentence 9 ÷ 3 = 3,
Number of Rachel’s bags are 3,

Explanation:
Given Rachel has 9 crackers.
She puts 3 crackers in each bag.
Circled the crackers to show Rachel’s bags,
a. Division sentence 9 ÷ 3 = 3,
therefore, number of Rachel’s bags are 3.
b.

Explanation:
Drawn a number bond to represent the problem
as shown above

Question 6.
Jameisha has 16 wheels to make toy cars.
She uses 4 wheels for each car.
a. Use a count-by to find the number of cars
Jameisha can build. Make a drawing to match
your counting.
b. Write a division sentence to represent the problem.

a.

Number of cars Jameisha builds are 4,

Explanation:
Given Jameisha has 16 wheels to make toy cars.
She uses 4 wheels for each car.
Used a count-by to find the number of cars
Jameisha can build as 16 ÷ 4 = 4 cars,
Made a drawing to match my counting as shown above.

b. Division sentence to represent the problem is 16 ÷ 4 = 4 cars,

Explanation:
Given Jameisha has 16 wheels to make toy cars.
She uses 4 wheels for each car, So the division sentence
to represent the problem is 16 ÷ 4 = 4 cars.

Eureka Math Grade 3 Module 1 Lesson 5 Exit Ticket Answer Key

Question 1.
Divide 12 triangles into groups of 6.

12 ÷ 6 = ___2____

There are 2 groups of 6,

Explanation:
Dividing 12 triangles into groups of 6 as
12 ÷ 6 = 2 we get 2 groups,
So, there are 2 groups of 6.

Question 2.
Spencer buys 20 strawberries to make smoothies.
Each smoothie needs 5 strawberries.
Use a count-by to find the number of smoothies
Spencer can make.
Make a drawing to match your counting.

Spencer makes 4 smoothies,

Explanation:
Given Spencer buys 20 strawberries to make smoothies.
Each smoothie needs 5 strawberries.
Used a count-by the number of smoothies
Spencer can make are 20 ÷ 5 = 4,
Made a drawing to match my counting as
5 X 4 = 20 strawberries as shown above in the picture.

Eureka Math Grade 3 Module 1 Lesson 5 Homework Answer Key

Question 1.

Divide 4 triangles into groups of 2.
There are _____2____ groups of 2 triangles.
4 ÷ 2 = 2

There are 2 groups of 2 triangles,

Explanation:
Dividing 4 triangles into groups of 2 as
4 ÷ 2 = 2 we get 2 groups,
So, there are 2 groups of 2 triangles.

Question 2.

Divide 9 eggs into groups of 3.
There are ___3____ groups.
9 ÷ 3 = __3_____

There are 3 groups of 3 eggs,

Explanation:
Dividing 9 eggs into groups of 3 as
9 ÷ 3 = 3 we get 3 groups,
So, there are 3 groups of 3 eggs.

Question 3.

Divide 12 buckets of paint into groups of 3.
12 ÷ 3 = __4_____

There are 4 groups of 3 paint buckets,

Explanation:
Divided 12 buckets of paint into groups of 3 as
12 ÷ 3 = 4 we get 4 groups of paint buckets,
So, there are 4 groups of 3 paint buckets.

Question 4.

Group the squares to show 15 ÷ 5 = __3___,
where the unknown represents the number of groups.
How many groups are there? ____3____

There are 3 groups of 5 squares,

Explanation:
Grouped the squares to show 15 ÷ 5 = 3,
where the unknown represents the number of groups
as 3 groups of 5 squares.

Question 5.
Daniel has 12 apples. He puts 6 apples in each bag.
Circle the apples to find the number of bags Daniel makes.

a. Write a division sentence where the answer
represents the number of Daniel’s bags.
b. Draw a number bond to represent the problem.
a.

Daniel’s has 2 bags of 6 apples each,
Division sentence : 12 ÷ 6 = 2 bags,

Explanation:
Daniel has 12 apples and he puts 6 apples in each bag.
Circled the apples to find the number of bags Daniel makes
as 12 ÷ 6 = 2 bags,
a. Writing a division sentence where the answer
represents the number of Daniel’s bags as 12 ÷ 6 = 2 bags.

b.

Explanation:
Drawn a number bond to represent the problem as
shown above in the picture.

Question 6.
Jacob draws cats. He draws 4 legs on each
cat for a total of 24 legs.
a. Use a count-by to find the number of cats
Jacob draws. Make a drawing to match your counting.
b. Write a division sentence to represent the problem.

a.

The number of cats Jacob draws are 6,

Explanation:
Given Jacob draws cats. He draws 4 legs on each
cat for a total of 24 legs.
a. Used a count-by to find the number of cats
Jacob draws as 24 ÷ 4 = 6 cats,
Made a drawing to match my counting as shown above.

b. Division sentence is 24 ÷ 4 = 6 cats,

Explanation:
Wrote a division sentence to represent the problem as
24 ÷ 4 = 6 cats.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 11 Answer Key

Eureka Math Grade 3 Module 7 Lesson 11 Problem Set Answer Key

Question 1.
Follow the directions below using the shape you created yesterday.
a. Tessellate your shape on a blank piece of paper.
b. Color your tessellation to create a pattern.
c. Outline the perimeter of your tessellation with a highlighter.
d. Use a string to measure the perimeter of your tessellation.
Answer:
a.

b.

c.

d. A string is used to measure the perimeter of the figure.

Explanation:
The figure asked to draw has been drawn considering all the given points to follow while drawing it.

Question 2.
Compare the perimeter of your tessellation to a partner’s. Whose tessellation has a greater perimeter? How do you know?
Answer:
My tessellation has the greater perimeter than my friend’s tessellation perimeter.

Explanation:
My tessellation has the greater perimeter than my friend’s tessellation perimeter because I have compared both the tessellation perimeter with the string. My tessellation string was down that compared to my friend’s tessellation.

Question 3.
How could you increase the perimeter of your tessellation?
Answer:
I could increase the perimeter of my tessellation by tessellating it by adding some more rows to my shape.

Explanation:
I could increase the perimeter of my tessellation by tessellating  more shapes. If I wish to tessellating my figure   I would tesselate by adding some more rows to my shape.

Question 4.
How would overlapping your shape when you tessellated change the perimeter of your tessellation?
Answer:
If I overlap my figure, it wouldn’t  fit together I think. If I overlap my figure, I think the perimeter will decrease of the figure.

Explanation:
Overlapping any figure means covering the figure with same sized figure or different sized figure. If I overlap my figure, I think the perimeter will decrease of my figure that of the overlapped figure.

Eureka Math Grade 3 Module 7 Lesson 11 Exit Ticket Answer Key

Estimate to draw at least four copies of the given regular hexagon to make a new shape, without gaps or overlaps. Outline the perimeter of your new shape with a highlighter. Shade in the area with a colored pencil.

Answer:

Explanation:
The area of the new shape is shaded with Yellow color.
The perimeter of the new shape is highlighted with Red color.

Eureka Math Grade 3 Module 7 Lesson 11 Homework Answer Key

Question 1.
Samson tessellates regular hexagons to make the shape below.

a. Outline the perimeter of Samson’s new shape with a highlighter.
Answer:

Explanation:
Perimeter is the distance around the outside of a shape. Perimeter of the given figure is highlighted in red highlighter.

b. Explain how Samson could use a string to measure the perimeter of his new shape.
Answer:
Samson could use a string to measure the perimeter of his new shape by placing it all-around the outer layer of the new shape.

Explanation:

Perimeter is the distance around the outside of a shape.

c. How many sides does his new shape have?
Answer:
The new shape has 18 or eighteen sides in it.

Explanation:

The sides of the new shape are counted one by one.

d. Shade in the area of his new shape with a colored pencil.
Answer:

Explanation:
The new shape area has be shaded with light blue colored pencil in it.

Question 2.
Estimate to draw at least four copies of the given triangle to make a new shape, without gaps or overlaps. Outline the perimeter of your new shape with a highlighter. Shade in the area with a colored pencil.

Answer:

Explanation:
The Area of the new shape is shaded with light grey colored pencil in it.
The Perimeter of the new shape is highlighted with red color.

Question 3.
The marks on the strings below show the perimeters of Shyla’s and Frank’s shapes. Whose shape has a greater perimeter? How do you know?

Answer:
The perimeter of Frank’s shape is having greater perimeter than Shyla’s shape because the  string measurement of Frank’s shape is lower than the Shyla’s shape string measurement.

Explanation:

The string measurement of Frank’s shape has further crossed  the string measurement of Shyla’s shape.

Question 4.
India and Theo use the same shape to create the tessellations shown below.

a. Estimate to draw the shape India and Theo used to make their tessellations.
Answer:

Explanation:
The shape India and Theo used to make their tessellations is been draw in the figure highlighted in the yellow color.

b. Theo says both tessellations have the same perimeter. Do you think Theo is right? Why or why not?
Answer:
Yes, Theo says correct because the shape used to draw the tessellation of both India and Theo’s are the same.  The number of shapes used by both are the same in their respective own  tessellation yet the position differs one is in vertical presentation and horizontal presentation .

Explanation:

The number of shapes used by both are the same in their respective own  tessellation yet the position differs one is in vertical presentation and horizontal presentation .

Engage NY Eureka Math 3rd Grade Module 1 Lesson 14 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 14 Sprint Answer Key

A
Multiply or Divide by 3

Question 1.
2 × 3 =
2 X 3 = 6,

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

Question 2.
3 × 3 =
3 X 3 = 9,

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

Question 3.
4 × 3 =
4 X 3 = 12,

Explanation:
Given 4 X 3 we multiply 4 with 3,
we get 12 as 4 X 3 = 12.

Question 4.
5 × 3 =
5 X 3 = 15,

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

Question 5.
1 × 3 =
1 X 3 = 6,

Explanation:
Given 1 X 3 we multiply 1 with 3,
we get 3 as 1 X 3 = 3.

Question 6.
6 ÷ 3 =
6 ÷ 3 = 2,

Explanation:
Given 6 ÷ 3 we divide 6 by 3,
we get 2 as 6 ÷ 3 = 2.

Question 7.
9 ÷ 3 =
9 ÷ 3 = 3,

Explanation:
Given 9 ÷ 3 we divide 9 by 3,
we get 3 as 9 ÷ 3 = 3.

Question 8.
15 ÷ 3 =
15 ÷ 3 = 5,

Explanation:
Given 15 ÷ 3 we divide 15 by 3,
we get 5 as 15 ÷ 3 = 5.

Question 9.
3 ÷ 1 =
3 ÷ 1 = 3,

Explanation:
Given 3 ÷ 1 we divide 3 by 1,
we get 3 as 3 ÷ 1 = 3.

Question 10.
12 ÷ 3 =
12 ÷ 3 = 4,

Explanation:
Given 12 ÷ 3 we divide 12 by 3,
we get 4 as 12 ÷ 3 = 4.

Question 11.
6 × 3 =
6 X 3 = 18,

Explanation:
Given 6 X 3 we multiply 6 with 3,
we get 18 as 6 X 3 = 18.

Question 12.
7 × 3 =
7 X 3 = 21,

Explanation:
Given 7 X 3 we multiply 7 with 3,
we get 21 as 7 X 3 = 21.

Question 13.
8 × 3 =
8 X 3 = 24,

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

Question 14.
9 × 3 =
9 X 3 = 27,

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

Question 15.
10 × 3 =
10 X 3 = 30,

Explanation:
Given 10 X 3 we multiply 10 with 3,
we get 30 as 10 X 3 = 30.

Question 16.
24 ÷ 3 =
24 ÷ 3 = 8,

Explanation:
Given 24 ÷ 3 we divide 24 by 3,
we get 8 as 24 ÷ 3 = 8.

Question 17.
21 ÷ 3 =
21 ÷ 3 = 7,

Explanation:
Given 21 ÷ 3 we divide 21 by 3,
we get 7 as 27 ÷ 3 = 7.

Question 18.
27 ÷ 3 =
27 ÷ 3 = 9,

Explanation:
Given 27 ÷ 3 we divide 27 by 3,
we get 9 as 27 ÷ 3 = 9.

Question 19.
18 ÷ 3 =
18 ÷ 3 = 6,

Explanation:
Given 18 ÷ 3 we divide 18 by 3,
we get 6 as 18 ÷ 3 = 6.

Question 20.
30 ÷ 3 =
30 ÷ 3 = 10,

Explanation:
Given 30 ÷ 3 we divide 30 by 3,
we get 10 as 30 ÷ 3 = 10.

Question 21.
__ × 3 = 15
5 X 3 = 15,

Explanation:
Given __ X 3 = 15, Let us take missing number
as x, So x X  3 = 15, means x = 15 ÷ 3 = 5,
therefore 5 X 3 = 15.

Question 22.
__ × 3 = 12
4 X 3 = 12,
Explanation:
Given __ X 3 = 12, Let us take missing number
as x, So x X  3 = 12, means x = 12 ÷ 3 = 4,
therefore 4 X 3 = 12.

Question 23.
__ × 3 = 30
10 X 3 = 30,

Explanation:
Given __ X 3 = 30, Let us take missing number
as x, So x X  3 = 30, means x = 30 ÷ 3 = 10,
therefore 10 X 3 = 30.

Question 24.
__ × 3 = 6
2 X 3 = 6,

Explanation:
Given __ X 3 = 6, Let us take missing number
as x, So x X  3 = 6, means x = 6 ÷ 3 = 2,
therefore 2 X 3 = 6.

Question 25.
__ × 3 = 9
3 X 3 = 9,

Explanation:
Given __ X 3 = 9, Let us take missing number
as x, So x X  3 = 9, means x = 9 ÷ 3 = 3,
therefore 3 X 3 = 9.

Question 26.
30 ÷ 3 =
30 ÷ 3 = 10,

Explanation:
Given 30 ÷ 3 we divide 30 by 3,
we get 10 as 30 ÷ 3 = 10.

Question 27.
15 ÷ 3 =
15 ÷ 3 = 5,

Explanation:
Given 15 ÷ 3 we divide 15 by 3,
we get 5 as 15 ÷ 3 = 5.

Question 28.
3 ÷ 1 =
3 ÷ 1 = 3,

Explanation:
Given 3 ÷ 1 we divide 3 by 1,
we get 3 as 3 ÷ 1 = 3.

Question 29.
6 ÷ 3 =
6 ÷ 3 = 2,

Explanation:
Given 6 ÷ 3 we divide 6 by 3,
we get 2 as 6 ÷ 3 = 2.

Question 30.
9 ÷ 3 =
9 ÷ 3 = 3,

Explanation:
Given 9 ÷ 3 we divide 9 by 3,
we get 3 as 9 ÷ 3 = 3.

Question 31.
__ × 3 = 18
6 X 3 = 18,

Explanation:
Given __ X 3 = 18, Let us take missing number
as x, So x X  3 = 18, means x = 18 ÷ 3 = 6,
therefore 6 X 3 = 18.

Question 32.
__ × 3 = 21
7 X 3 = 21,

Explanation:
Given __ X 3 = 21, Let us take missing number
as x, So x X  3 = 21, means x = 21 ÷ 3 = 7,
therefore 7 X 3 = 21.

Question 33.
__ × 3 = 27
9 X 3 = 27,

Explanation:
Given __ X 3 = 27, Let us take missing number
as x, So x X  3 = 27, means x = 27 ÷ 3 = 9,
therefore 9 X 3 = 27.

Question 34.
__ × 3 = 24
8 X 3 = 24,

Explanation:
Given __ X 3 = 24, Let us take missing number
as x, So x X  3 = 24, means x = 24 ÷ 3 = 8,
therefore 8 X 3 = 24.

Question 35.
21 ÷ 3 =
21 ÷ 3 = 7,

Explanation:
Given 21 ÷ 3 we divide 21 by 3,
we get 7 as 21 ÷ 3 = 7.

Question 36.
27 ÷ 3 =
27 ÷ 3 = 9,

Explanation:
Given 27 ÷ 3 we divide 27 by 3,
we get 9 as 27 ÷ 3 = 9.

Question 37.
18 ÷ 3 =
18 ÷ 3 = 6,

Explanation:
Given 18 ÷ 3 we divide 18 by 3,
we get 6 as 18 ÷ 3 = 6.

Question 38.
24 ÷ 3 =
24 ÷ 3 = 8,

Explanation:
Given 24 ÷ 3 we divide 24 by 3,
we get 8 as 24 ÷ 3 = 8.

Question 39.
11 × 3 =
11 X 3 = 33,

Explanation:
Given 11 X 3 we multiply 11 with 3,
we get 33 as 11 X 3 = 33.

Question 40.
33 ÷ 3 =
33 ÷ 3 = 11,

Explanation:
Given 33 ÷ 3 we divide 33 by 3,
we get 11 as 33 ÷ 3 = 11.

Question 41.
12 × 3 =
12 X 3 = 36,

Explanation:
Given 12 X 3 we multiply 12 with 3,
we get 36 as 12 X 3 = 36.

Question 42.
36 ÷ 3 =
36 ÷ 3 = 12,

Explanation:
Given 36 ÷ 3 we divide 36 by 3,
we get 12 as 36 ÷ 3 = 12.

Question 43.
13 × 3 =
13 X 3 = 39,

Explanation:
Given 13 X 3 we multiply 13 with 3,
we get 39 as 13 X 3 = 39.

Question 44.
39 ÷ 3 =
39 ÷ 3 = 13,

Explanation:
Given 39 ÷ 3 we divide 39 by 3,
we get 13 as 39 ÷ 3 = 13.

B
Multiply or Divide by 3

Question 1.
1 × 3 =
1 X 3 = 3,

Explanation:
Given 1 X 3 we multiply 1 with 3,
we get 3 as 1 X 3 = 3.

Question 2.
2 × 3 =
2 X 3 = 6,

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

Question 3.
3 × 3 =
3 X 3 = 9,

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

Question 4.
4 × 3 =
4 X 3 = 12,

Explanation:
Given 4 X 3 we multiply 4 with 3,
we get 12 as 4 X 3 = 12.

Question 5.
5 × 3 =
5 X 3 = 15,

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

Question 6.
9 ÷ 3 =
9 ÷ 3 = 3,

Explanation:
Given 9 ÷ 3 we divide 9 by 3,
we get 3 as 9 ÷ 3 = 3.

Question 7.
6 ÷ 3 =
6 ÷ 3 = 2,

Explanation:
Given 6 ÷ 3 we divide 6 by 3,
we get 2 as 6 ÷ 3 = 2.

Question 8.
12 ÷ 3 =
12 ÷ 3 = 4,

Explanation:
Given 12 ÷ 3 we divide 12 by 3,
we get 4 as 12 ÷ 3 = 4.

Question 9.
3 ÷ 1 =
3 ÷ 1 = 3,

Explanation:
Given 3 ÷ 1 we divide 3 by 1,
we get 3 as 3 ÷ 1 = 3.

Question 10.
15 ÷ 3 =
15 ÷ 3 = 5,

Explanation:
Given 15 ÷ 3 we divide 15 by 3,
we get 5 as 15 ÷ 3 = 5.

Question 11.
10 × 3 =
10 X 3 = 30,

Explanation:
Given 10 X 3 we multiply 10 with 3,
we get 30 as 10 X 3 = 30.

Question 12.
6 × 3 =
6 X 3 = 18,

Explanation:
Given 6 X 3 we multiply 6 with 3,
we get 18 as 6 X 3 = 18.

Question 13.
7 × 3 =
7 X 3 = 21,

Explanation:
Given 7 X 3 we multiply 7 with 3,
we get 21 as 7 X 3 = 21.

Question 14.
8 × 3 =
8 X 3 = 24,

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

Question 15.
9 × 3 =
9 X 3 = 27,

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

Question 16.
21 ÷ 3 =
21 ÷ 3 = 7,

Explanation:
Given 21 ÷ 3 we divide 21 by 3,
we get 7 as 21 ÷ 3 = 7.

Question 17.
18 ÷ 3 =
18 ÷ 3 = 6,

Explanation:
Given 18 ÷ 3 we divide 18 by 3,
we get 6 as 18 ÷ 3 = 6.

Question 18.
24 ÷ 3 =
24 ÷ 3 = 8,

Explanation:
Given 24 ÷ 3 we divide 24 by 3,
we get 8 as 24 ÷ 3 = 8.

Question 19.
30 ÷ 3 =
30 ÷ 3 = 10,

Explanation:
Given 30 ÷ 3 we divide 30 by 3,
we get 10 as 30 ÷ 3 = 10.

Question 20.
27 ÷ 3 =
27 ÷ 3 = 9,

Explanation:
Given 27 ÷ 3 we divide 27 by 3,
we get 9 as 27 ÷ 3 = 9.

Question 21.
__ × 3 = 12
4 X 3 = 12,

Explanation:
Given __ X 3 = 12, Let us take missing number
as x, So x X  3 = 12, means x = 12 ÷ 3 = 4,
therefore 4 X 3 = 12.

Question 22.
__ × 3 = 15
5 X 3 = 15,

Explanation:
Given __ X 3 = 15, Let us take missing number
as x, So x X  3 = 15, means x = 15 ÷ 3 = 5,
therefore 5 X 3 = 15.

Question 23.
__ × 3 = 6
2 X 3 = 6,

Explanation:
Given __ X 3 = 6, Let us take missing number
as x, So x X  3 = 6, means x = 6 ÷ 3 = 2,
therefore 2 X 3 = 6.

Question 24.
__ × 3 = 30
10 X 3 = 30,

Explanation:
Given __ X 3 = 30, Let us take missing number
as x, So x X  3 = 30, means x = 30 ÷ 3 = 10,
therefore 10 X 3 = 30.

Question 25.
__ × 3 = 9
3 X 3 = 9,

Explanation:
Given __ X 3 = 9, Let us take missing number
as x, So x X  3 = 9, means x = 9 ÷ 3 = 3,
therefore 3 X 3 = 9.

Question 26.
6 ÷ 3 =
6 ÷ 3 = 2,

Explanation:
Given 6 ÷ 3 we divide 6 by 3,
we get 2 as 6 ÷ 3 = 2.

Question 27.
3 ÷ 1 =
3 ÷ 1 = 3,

Explanation:
Given 3 ÷ 1 we divide 3 by 1,
we get 3 as 3 ÷ 1 = 3.

Question 28.
30 ÷ 3 =
30 ÷ 3 = 10,

Explanation:
Given 30 ÷ 3 we divide 30 by 3,
we get 10 as 30 ÷ 3 = 10.

Question 29.
15 ÷ 3 =
15 ÷ 3 = 5,

Explanation:
Given 15 ÷ 3 we divide 15 by 3,
we get 5 as 15 ÷ 3 = 5.

Question 30.
9 ÷ 3 =
9 ÷ 3 = 3,

Explanation:
Given 9 ÷ 3 we divide 9 by 3,
we get 3 as 9 ÷ 3 = 3.

Question 31.
__ × 3 = 18
6 X 3 = 18,

Explanation:
Given __ X 3 = 18, Let us take missing number
as x, So x X  3 = 18, means x = 18 ÷ 3 = 6,
therefore 6 X 3 = 18.

Question 32.
__ × 3 = 24
8 X 3 = 24,

Explanation:
Given __ X 3 = 24, Let us take missing number
as x, So x X  3 = 24, means x = 24 ÷ 3 = 8,
therefore 8 X 3 = 24.

Question 33.
__ × 3 = 27
9 X 3 = 27,

Explanation:
Given __ X 3 = 27, Let us take missing number
as x, So x X  3 = 27, means x = 27 ÷ 3 = 9,
therefore 9 X 3 = 27.

Question 34.
__ × 3 = 21
7 X 3 = 21,

Explanation:
Given __ X 3 = 21, Let us take missing number
as x, So x X  3 = 21, means x = 21 ÷ 3 = 7,
therefore 7 X 3 = 21.

Question 35.
24 ÷ 3 =
24 ÷ 3 = 8,

Explanation:
Given 24 ÷ 3 we divide 24 by 3,
we get 8 as 24 ÷ 3 = 8.

Question 36.
27 ÷ 3 =
27 ÷ 3 = 9,

Explanation:
Given 27 ÷ 3 we divide 27 by 3,
we get 9 as 27 ÷ 3 = 9.

Question 37.
18 ÷ 3 =
18 ÷ 3 = 6,

Explanation:
Given 18 ÷ 3 we divide 18 by 3,
we get 6 as 18 ÷ 3 = 6.

Question 38.
21 ÷ 3 =
21 ÷ 3 = 7,

Explanation:
Given 21 ÷ 3 we divide 21 by 3,
we get 7 as 21 ÷ 3 = 7.

Question 39.
11 × 3 =
11 X 3 = 33,

Explanation:
Given 11 X 3 we multiply 11 with 3,
we get 33 as 11 X 3 = 33.

Question 40.
33 ÷ 3 =
33 ÷ 3 = 11,

Explanation:
Given 33 ÷ 3 we divide 33 by 3,
we get 11 as 33 ÷ 3 = 11.

Question 41.
12 × 3 =
12 X 3 = 36,

Explanation:
Given 12 X 3 we multiply 12 with 3,
we get 36 as 12 X 3 = 36.

Question 42.
36 ÷ 3 =
36 ÷ 3 = 12,

Explanation:
Given 36 ÷ 3 we divide 36 by 3,
we get 12 as 36 ÷ 3 = 12.

Question 43.
13 × 3 =
13 X 3 = 39,

Explanation:
Given 13 X 3 we multiply 13 with 3,
we get 39 as 13 X 3 = 39.

Question 44.
39 ÷ 3 =
39 ÷ 3 = 13,

Explanation:
Given 39 ÷ 3 we divide 39 by 3,
we get 13 as 39 ÷ 3 = 13.

Eureka Math Grade 3 Module 1 Lesson 14 Problem Set Answer Key

Question 1.
Skip-count by fours. Match each answer to the appropriate expression.

Skipped-counts by fours.
Matched each answer to the appropriate expression as
4 = 1 X 4,
8 = 2 X 4,
12 = 3 X 4,
16 = 4 X 4,
20 = 5 X 4,
24 = 6 X 4,
28 = 7 X 4,
32 = 8 X 4,
36 = 9 X 4,
40 = 10 X 4.

Question 2.
Mr. Schmidt replaces each of the 4 wheels on 7 cars.
How many wheels does he replace? Draw and label a tape diagram to solve.
Mr. Schmidt replaces _____28______ wheels.

Mr. Schmidt replaces 28 wheels on 7 cars,

Explanation:
Given Mr. Schmidt replaces each of the 4 wheels on 7 cars.
Drawn and labeled a tape diagram to solve how many wheels
he replaced as shown above, So number of wheels replaced are
7 X 4 = 28, therefore, Mr. Schmidt replaces _____28______ wheels.

Question 3.
Trina makes 4 bracelets. Each bracelet has 6 beads.
Draw and label a tape diagram to show the
total number of beads Trina uses.

Trina uses 24 beads to make 4 bracelets,

Explanation:
Given Trina makes 4 bracelets. Each bracelet has 6 beads.
Draw and label a tape diagram to show the
total number of beads Trina uses as 4 X 6 =24 beads.

Question 4.
Find the total number of sides on 5 rectangles.

Total number of sides on 5 rectangles are 20 side,

Explanation:
We know a rectangle has 4 sides, So number of sides on
5 rectangles are 5 X 4 = 20 sides.

Eureka Math Grade 3 Module 1 Lesson 14 Exit Ticket Answer Key

Arthur has 4 boxes of chocolates. Each box has 6 chocolates inside. How many chocolates does Arthur have altogether? Draw and label a tape diagram to solve.

Eureka Math Grade 3 Module 1 Lesson 14 Homework Answer Key

Question 1.
Skip-count by fours. Match each answer to the appropriate expression.

Skipped-counts by fours.
Matched each answer to the appropriate expression as
4 = 1 X 4,
8 = 2 X 4,
12 = 3 X 4,
16 = 4 X 4,
20 = 5 X 4,
24 = 6 X 4,
28 = 7 X 4,
32 = 8 X 4,
36 = 9 X 4,
40 = 10 X 4.

Question 2.
Lisa places 5 rows of 4 juice boxes in the refrigerator.
Draw an array and skip-count to find the total number of juice boxes.
There are ____20_______ juice boxes in total.

The total number of juice boxes are 20,

Explanation:
Given Lisa places 5 rows of 4 juice boxes in the refrigerator.
Drawn an array and skipped-count for finding the total
number of juice boxes are as 5 X 4 = 20 juice boxes in total.

Question 3.
Six folders are placed on each table. How many folders are
there on 4 tables? Draw and label a tape diagram to solve.

There are 24 folders on 4 tables,

Explanation:
Given Six folders are placed on each table. So number of
folders are there on 4 tables are 4 X 6 = 24,
Drawn and labeled a tape diagram to solve as shown above,
Therefore, there are 24 folders on 4 tables.

Question 4.
Find the total number of corners on 8 squares.

Total number of corners on 8 squares is 32,

Explanation:
We know a square has 4 corners,

So 8 squares will have 8 X 4 = 32 corners,
therefore, total number of corners on 8 squares is 32.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 13 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 13 Sprint Answer Key

A
Multiply or Divide by 2

Question 1.
2 × 2 =
2 X 2 = 4,

Explanation:
Given 2 X 2 we multiply 2 with 2,
we get 4 as 2 X 2 = 4.

Question 2.
3 × 2 =
3 X 2 = 6,

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

Question 3.
4 × 2 =
4 X 2 = 8,

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

Question 4.
5 × 2 =
5 X 2 = 10,
Explanation:
Given 5 X 2 we multiply 5 with 2,
we get 10 as 5 X 2 = 10.

Question 5.
1 × 2 =
1 X 2 = 2,

Explanation:
Given 1 X 2 we multiply 1 with 2,
we get 2 as 1 X 2 = 2.

Question 6.
4 ÷ 2 =
4 ÷ 2 = 2,

Explanation:
Given 4 ÷ 2 we divide 4 by 2,
we get 2 as 4 ÷ 2 = 2.

Question 7.
6 ÷ 2 =
6 ÷ 2 = 3,
Explanation:
Given 6 ÷ 2 we divide 6 by 2,
we get 3 as 6 ÷ 2 = 3.

Question 8.
10 ÷ 2 =
10 ÷ 2 = 5,

Explanation:
Given 10 ÷ 2 we divide 10 by 2,
we get 5 as 10 ÷ 2 = 5.

Question 9.
2 ÷ 1 =
2 ÷ 1 = 2,

Explanation:
Given 2 ÷ 1 we divide 2 by 1,
we get 2 as 2 ÷ 1 = 2.

Question 10.
8 ÷ 2 =
8 ÷ 2 = 4,

Explanation:
Given 8 ÷ 2 we divide 8 by 2,
we get 4 as 8 ÷ 2 = 4.

Question 11.
6 × 2 =
6 X 2 = 12,

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

Question 12.
7 × 2 =
7 X 2 = 14,
Explanation:
Given 7 X 2 we multiply 7 with 2,
we get 14 as 7 X 2 = 14.

Question 13.
8 × 2 =
8 X 2 = 16,

Explanation:
Given 8 X 2 we multiply 8 with 2,
we get 16 as 8 X 2 = 16.

Question 14.
9 × 2 =
9 X 2 = 18,

Explanation:
Given 9 X 2 we multiply 9 with 2,
we get 18 as 9 X 2 = 18.

Question 15.
10 × 2 =
10 X 2 = 20,

Explanation:
Given 10 X 2 we multiply 10 with 2,
we get 20 as 10 X 2 = 20.

Question 16.
16 ÷ 2 =
16 ÷ 2 = 8,

Explanation:
Given 16 ÷ 2 we divide 16 by 2,
we get 8 as 16 ÷ 2 = 8.

Question 17.
14 ÷ 2 =
14 ÷ 2 = 7,

Explanation:
Given 14 ÷ 2 we divide 14 by 2,
we get 7 as 14 ÷ 2 = 7.

Question 18.
18 ÷ 2 =
18 ÷ 2 = 9,

Explanation:
Given 18 ÷ 2 we divide 18 by 2,
we get 9 as 18 ÷ 2 = 9.

Question 19.
12 ÷ 2 =
12 ÷ 2 = 6,

Explanation:
Given 12 ÷ 2 we divide 12 by 2,
we get 6 as 12 ÷ 2 = 6.

Question 20.
20 ÷ 2 =
20 ÷ 2 = 10,

Explanation:
Given 20 ÷ 2 we divide 20 by 2,
we get 10 as 20 ÷ 2 = 10.

Question 21.
__ × 2 = 10
5 X 2 = 10,

Explanation:
Given __ X 2 = 10, Let us take missing number
as x, So x X 2 = 10, means x = 10 ÷ 2 = 5,
therefore 5 X 2 = 10.

Question 22.
__ × 2 = 12
6 X 2 = 12,

Explanation:
Given __ X 2 = 12, Let us take missing number
as x, So x X 2 = 12, means x = 12 ÷ 2 = 6,
therefore 6 X 2 = 12.

Question 23.
__ × 2 = 20
10 X 2 = 20,

Explanation:
Given __ X 2 = 20, Let us take missing number
as x, So x X 2 = 20, means x = 20 ÷ 2 = 10,
therefore 10 X 2 = 20.

Question 24.
__ × 2 = 4
2 X 2 = 4,

Explanation:
Given __ X 2 = 4, Let us take missing number
as x, So x X 2 = 4, means x = 4 ÷ 2 = 2,
therefore 2 X 2 = 4.

Question 25.
__ × 2 = 6
3 X 2 = 6,

Explanation:
Given __ X 2 = 6, Let us take missing number
as x, So x X 2 = 6, means x = 6 ÷ 2 = 3,
therefore 3 X 2 = 6.

Question 26.
20 ÷ 2 =
20 ÷ 2 = 10,

Explanation:
Given 20 ÷ 2 we divide 20 by 2,
we get 10 as 20 ÷ 2 = 10.

Question 27.
10 ÷ 2 =
10 ÷ 2 = 5,

Explanation:
Given 10 ÷ 2 we divide 10 by 2,
we get 5 as 10 ÷ 2 = 5.

Question 28.
2 ÷ 1 =
2 ÷ 1 = 2,

Explanation:
Given 2 ÷ 1 we divide 2 by 1,
we get 2 as 2 ÷ 1 = 2.

Question 29.
4 ÷ 2 =
4 ÷ 2 = 2,

Explanation:
Given 4 ÷ 2 we divide 4 by 2,
we get 2 as 4 ÷ 2 = 2.

Question 30.
6 ÷ 2 =
6 ÷ 2 = 3,

Explanation:
Given 6 ÷ 2 we divide 6 by 2,
we get 3 as 6 ÷ 2 = 3.

Question 31.
__ × 2 = 12
6 X 2 = 12,

Explanation:
Given __ X 2 = 12, Let us take missing number
as x, So x X 2 = 12, means x = 12 ÷ 2 = 6,
therefore 6 X 2 = 12.

Question 32.
__ × 2 = 14
7 X 2 = 14,

Explanation:
Given __ X 2 = 14, Let us take missing number
as x, So x X 2 = 14, means x = 14 ÷ 2 = 7,
therefore 7 X 2 = 14.

Question 33.
__ × 2 = 18
9 X 2 = 18,

Explanation:
Given __ X 2 = 18, Let us take missing number
as x, So x X 2 = 18, means x = 18 ÷ 2 = 9,
therefore 9 X 2 = 18.

Question 34.
__ × 2 = 16
8 X 2 = 16,

Explanation:
Given __ X 2 = 16, Let us take missing number
as x, So x X 2 = 16, means x = 16 ÷ 2 = 8,
therefore 8 X 2 = 16.

Question 35.
14 ÷ 2 =
14 ÷ 2 = 7,

Explanation:
Given 14 ÷ 2 we divide 14 by 2,
we get 7 as 14 ÷ 2 = 7.

Question 36.
18 ÷ 2 =
18 ÷ 2 = 9,

Explanation:
Given 18 ÷ 2 we divide 18 by 2,
we get 9 as 18 ÷ 2 = 9.

Question 37.
12 ÷ 2 =
12 ÷ 2 = 6,

Explanation:
Given 12 ÷ 2 we divide 12 by 2,
we get 6 as 12 ÷ 2 = 6.

Question 38.
16 ÷ 2 =
16 ÷ 2 = 8,

Explanation:
Given 16 ÷ 2 we divide 16 by 2,
we get 8 as 16 ÷ 2 = 8.

Question 39.
11 × 2 =
11 X 2 = 22,

Explanation:
Given 11 X 2 we multiply 11 with 2,
we get 22 as 11 X 2 = 22.

Question 40.
22 ÷ 2 =
22 ÷ 2 = 11,

Explanation:
Given 22 ÷ 2 we divide 22 by 2,
we get 11 as 22 ÷ 2 = 11.

Question 41.
12 × 2 =
12 X 2 = 24,

Explanation:
Given 12 X 2 we multiply 12 with 2,
we get 24 as 12 X 2 = 24.

Question 42.
24 ÷ 2 =
24 ÷ 2 = 12,

Explanation:
Given 24 ÷ 2 we divide 24 by 2,
we get 12 as 24 ÷ 2 = 12.

Question 43.
14 × 2 =
14 X 2 = 28,

Explanation:
Given 14 X 2 we multiply 14 with 2,
we get 28 as 14 X 2 = 28.

Question 44.
28 ÷ 2 =
28 ÷ 2 = 14,

Explanation:
Given 28 ÷ 2 we divide 28 by 2,
we get 14 as 28 ÷ 2 = 14.

B
Multiply or Divide by 2

Question 1.
1 × 2 =
1 X 2 = 2,

Explanation:
Given 1 X 2 we multiply 1 with 2,
we get 2 as 1 X 2 = 2.

Question 2.
2 × 2 =
2 X 2 = 4,

Explanation:
Given 2 X 2 we multiply 2 with 2,
we get 4 as 2 X 2 = 4.

Question 3.
3 × 2 =
3 X 2 = 6,

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

Question 4.
4 × 2 =
4 X 2 = 8,

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

Question 5.
5 × 2 =
5 X 2 = 10,

Explanation:
Given 5 X 2 we multiply 5 with 2,
we get 10 as 5 X 2 = 10.

Question 6.
6 ÷ 2 =
6 ÷ 2 = 3,

Explanation:
Given 6 ÷ 2 we divide 6 by 2,
we get 3 as 6 ÷ 2 = 3.

Question 7.
4 ÷ 2 =
4 ÷ 2 = 2,

Explanation:
Given 4 ÷ 2 we divide 4 by 2,
we get 2 as 4 ÷ 2 = 2.

Question 8.
8 ÷ 2 =
8 ÷ 2 = 4,

Explanation:
Given 8 ÷ 2 we divide 8 by 2,
we get 4 as 8 ÷ 2 = 4.

Question 9.
2 ÷ 1 =
2 ÷ 1 = 2,

Explanation:
Given 2 ÷ 1 we divide 2 by 1,
we get 2 as 2 ÷ 1 = 2.

Question 10.
10 ÷ 2 =
10 ÷ 2 = 5,

Explanation:
Given 10 ÷ 2 we divide 10 by 2,
we get 5 as 10 ÷ 2 = 5.

Question 11.
10 × 2 =
10 X 2 = 20,

Explanation:
Given 10 X 2 we multiply 10 with 2,
we get 20 as 10 X 2 = 20.

Question 12.
6 × 2 =
6 X 2 = 12,

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

Question 13.
7 × 2 =
7 X 2 = 14,

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

Question 14.
8 × 2 =
8 X 2 = 16,

Explanation:
Given 8 X 2 we multiply 8 with 2,
we get 16 as 8 X 2 = 16.

Question 15.
9 × 2 =
9 X 2 = 18,

Explanation:
Given 9 X 2 we multiply 9 with 2,
we get 18 as 9 X 2 = 18.

Question 16.
14 ÷ 2 =
14 ÷ 2 = 7,

Explanation:
Given 14 ÷ 2 we divide 14 by 2,
we get 7 as 14 ÷ 2 = 7.

Question 17.
12 ÷ 2 =
12 ÷ 2 = 6,

Explanation:
Given 12 ÷ 2 we divide 12 by 2,
we get 6 as 12 ÷ 2 = 6.

Question 18.
16 ÷ 2 =
16 ÷ 2 = 8,

Explanation:
Given 16 ÷ 2 we divide 16 by 2,
we get 8 as 16 ÷ 2 = 8.

Question 19.
20 ÷ 2 =
20 ÷ 2 = 10,

Explanation:
Given 20 ÷ 2 we divide 20 by 2,
we get 10 as 20 ÷ 2 = 10.

Question 20.
18 ÷ 2 =
18 ÷ 2 = 9,

Explanation:
Given 18 ÷ 2 we divide 18 by 2,
we get 9 as 18 ÷ 2 = 9.

Question 21.
__ × 2 = 12
6 X 2 = 12,

Explanation:
Given __ X 2 = 12, Let us take missing number
as x, So x X 2 = 12, means x = 12 ÷ 2 = 6,
therefore 6 X 2 = 12.

Question 22.
__ × 2 = 10
5 X 2 = 10,

Explanation:
Given __ X 2 = 10, Let us take missing number
as x, So x X 2 = 10, means x = 10 ÷ 2 = 5,
therefore 5 X 2 = 10.

Question 23.
__ × 2 = 4
2 X 2 = 4,

Explanation:
Given __ X 2 = 4, Let us take missing number
as x, So x X 2 = 4, means x = 4 ÷ 2 = 2,
therefore 2 X 2 = 4.

Question 24.
__ × 2 = 20
10 X 2 = 20,

Explanation:
Given __ X 2 = 20, Let us take missing number
as x, So x X 2 = 20, means x = 20 ÷ 2 = 10,
therefore 10 X 2 = 20.

Question 25.
__ × 2 = 6
3 X 2 = 6,

Explanation:
Given __ X 2 = 6, Let us take missing number
as x, So x X 2 = 6, means x = 6 ÷ 2 = 3,
therefore 3 X 2 = 6.

Question 26.
4 ÷ 2 =
4 ÷ 2 = 2,

Explanation:
Given 4 ÷ 2 we divide 4 by 2,
we get 2 as 4 ÷ 2 = 2.

Question 27.
2 ÷ 1 =
2 ÷ 1 = 2,

Explanation:
Given 2 ÷ 1 we divide 2 by 1,
we get 2 as 2 ÷ 1 = 2.

Question 28.
20 ÷ 2 =
20 ÷ 2 = 10,

Explanation:
Given 20 ÷ 2 we divide 20 by 2,
we get 10 as 20 ÷ 2 = 10.

Question 29.
10 ÷ 2 =
10 ÷ 2 = 5,

Explanation:
Given 10 ÷ 2 we divide 10 by 2,
we get 5 as 10 ÷ 2 = 5.

Question 30.
6 ÷ 2 =
6 ÷ 2 = 3,

Explanation:
Given 6 ÷ 2 we divide 6 by 2,
we get 3 as 6 ÷ 2 = 3.

Question 31.
__ × 2 = 12
6 X 2 = 12,

Explanation:
Given __ X 2 = 12, Let us take missing number
as x, So x X 2 = 12, means x = 12 ÷ 2 = 6,
therefore 6 X 2 = 12.

Question 32.
__ × 2 = 16
8 X 2 = 16,

Explanation:
Given __ X 2 = 16, Let us take missing number
as x, So x X 2 = 16, means x = 16 ÷ 2 = 8,
therefore 8 X 2 = 16.

Question 33.
__ × 2 = 18
9 X 2 = 18,

Explanation:
Given __ X 2 = 18, Let us take missing number
as x, So x X 2 = 18, means x = 18 ÷ 2 = 9,
therefore 9 X 2 = 18.

Question 34.
__ × 2 = 14
7 X 2 = 14,

Explanation:
Given __ X 2 = 14, Let us take missing number
as x, So x X 2 = 14, means x = 14 ÷ 2 = 7,
therefore 7 X 2 = 14.

Question 35.
16 ÷ 2 =
16 ÷ 2 = 8,

Explanation:
Given 16 ÷ 2 we divide 16 by 2,
we get 8 as 16 ÷ 2 = 8.

Question 36.
18 ÷ 2 =
18 ÷ 2 = 9,

Explanation:
Given 18 ÷ 2 we divide 18 by 2,
we get 9 as 18 ÷ 2 = 9.

Question 37.
12 ÷ 2 =
12 ÷ 2 = 6,

Explanation:
Given 12 ÷ 2 we divide 12 by 2,
we get 6 as 12 ÷ 2 = 6.

Question 38.
14 ÷ 2 =
14 ÷ 2 = 7,

Explanation:
Given 14 ÷ 2 we divide 14 by 2,
we get 7 as 14 ÷ 2 = 7.

Question 39.
11 × 2 =
11 X 2 = 22,

Explanation:
Given 11 X 2 we multiply 11 with 2,
we get 22 as 11 X 2 = 22.

Question 40.
22 ÷ 2 =
22 ÷ 2 = 11,

Explanation:
Given 22 ÷ 2 we divide 22 by 2,
we get 11 as 22 ÷ 2 = 11.

Question 41.
12 × 2 =
12 X 2 = 24,

Explanation:
Given 12 X 2 we multiply 12 with 2,
we get 24 as 12 X 2 = 24.

Question 42.
24 ÷ 2 =
24 ÷ 2 = 12,

Explanation:
Given 24 ÷ 2 we divide 24 by 2,
we get 12 as 24 ÷ 2 = 12.

Question 43.
13 × 2 =
13 X 2 = 26,

Explanation:
Given 13 X 2 we multiply 13 with 2,
we get 23 as 13 X 2 = 26.

Question 44.
26 ÷ 2 =
26 ÷ 2 = 13,

Explanation:
Given 26 ÷ 2 we divide 26 by 2,
we get 26 as 26 ÷ 2 = 13.

Eureka Math Grade 3 Module 1 Lesson 13 Problem Set Answer Key

Question 1.
Fill in the blanks to make true number sentences.

Explanation:
Filled the blanks to make true number sentences as
1 X 3 = 3, 3 ÷ 3 = 1,
2 X 3 = 6, 6 ÷ 3 = 2,
3 X 3 = 9, 9 ÷ 3 = 3,
4 X 3 = 12, 12 ÷ 3 = 4,
5 X 3 = 15, 15 ÷ 3 = 5,
6 X 3 = 18, 18 ÷ 3 = 6,
7 X 3 = 21, 21 ÷ 3 = 7,
8 X 3 = 24, 24 ÷ 3 = 8,
9 X 3 = 27, 27 ÷ 3 = 9,
10 x 3 = 30, 30 ÷ 3 = 10.

Question 2.
Mr. Lawton picks tomatoes from his garden.
He divides the tomatoes into bags of 3.

a. Circle to show how many bags he packs.
Then, skip-count to show the total number of tomatoes.


Circled the bags as 4,
The total number of tomatoes are 12,

Explanation:
Given Mr. Lawton picks tomatoes from his garden.
He divides the tomatoes into bags of 3.
a. Circled and showed number of bags he packs
as 12 ÷ 3 = 4 bags,
Then, skipped-count and showed the total number of
tomatoes are 4 X 3 = 12.

b. Draw and label a tape diagram to represent the problem.
____12____ ÷ 3 = ___4 bags__________
Mr. Lawton packs ___4____ bags of tomatoes.

Mr. Lawton packs  4 bags of tomatoes,

Explanation:
Drawn and labeled a tape diagram to represent
the problem as shown above 12 ÷ 3 = 4 bags.

Question 3.
Camille buys a sheet of stamps that measures 15 centimeters long.
Each stamp is 3 centimeters long.
How many stamps does Camille buy?
Draw and label a tape diagram to solve.
Camille buys ____5_____ stamps.

Camille buy’s 5 stamps,

Explanation:
Given Camille buys a sheet of stamps that measures
15 centimeters long and each stamp is 3 centimeters long.
So number of  stamps Camille buy’s is 15 ÷ 3 = 5 stamps,
Drawn and labeled a tape diagram to solve as shown above.

Question 4.
Thirty third-graders go on a field trip. They are equally
divided into 3 vans. How many students are in each van?

In each van there are 11 students,

Explanation:
Given thirty third-graders go on a field trip and they are equally
divided into 3 vans. So number of  students in each van are
33 ÷ 3 = 11, Therefore in each van there are 11 students.

Question 5.
Some friends spend $24 altogether on frozen yogurt.
Each person pays $3. How many people buy frozen yogurt?

8 people buy’s frozen yogurt,

Explanation:
Given some friends spend $24 altogether on frozen yogurt
and each person pays $3, So number of people buy’s frozen
yogurt is $24 ÷ $3 = 8, Therefore 8 people buy’s frozen yogurt.

Eureka Math Grade 3 Module 1 Lesson 13 Exit Ticket Answer Key

Question 1.
Andrea has 21 apple slices. She uses 3 apple slices to
decorate 1 pie. How many pies does Andrea make?
Draw and label a tape diagram to solve.

Andrea makes 7 pies,

Explanation:
Given Andrea has 21 apple slices and she uses 3 apple slices to
decorate 1 pie. So number of pies Andrea makes are 21 ÷ 3 = 7,
Drawn and labeled a tape diagram to solve as shown above.

Question 2.
There are 24 soccer players on the field. They form 3 equal teams.
How many players are on each team?

Number of players in each team are 8,

Explanation:
Given there are 24 soccer players on the field and
they form 3 equal teams, So number of players in
each team are 24 ÷ 3 = 8 players.

Eureka Math Grade 3 Module 1 Lesson 13 Homework Answer Key

Question 1.
Fill in the blanks to make true number sentences.

Explanation:
Filled in the blanks to make true number sentences as
2 X 3 = 6, 6 ÷ 3 = 2, 1 x 3 = 3, 3 ÷ 3 =1,
7 X 3 = 21, 21 ÷ 3 = 7 and 9 X 3 = 27, 27 ÷ 3 = 9.

Question 2.
Ms. Gillette’s pet fish are shown below.
She keeps 3 fish in each tank.

a. Circle to show how many fish tanks she has.
Then, skip-count to find the total number of fish.

Circled the fish tanks as 5,
The total number of fishes are 15,

Explanation:
Given Ms. Gillette’s pet fishes, She keeps 3 fish in each tank,
Circled to show number of fish tanks she has.
Then, skipped-count to find the total number of fishes as
5 X 3 = 15.

b. Draw and label a tape diagram to represent the problem.
_____15______ ÷ 3 = ____5______
Ms. Gillette has ___5____ fish tanks.

Ms. Gillette has 5 fish tanks.

Explanation:
Drawn and labeled a tape diagram to represent
the problem as shown above 15 ÷ 3 = 5 fish tanks.

Question 3.
Juan buys 18 meters of wire. He cuts the wire into
pieces that are each 3 meters long. How many pieces
of wire does he cut?

Juan cuts 6 pieces of wire.

Explanation:
Given Juan buys 18 meters of wire and he cuts the wire into
pieces that are each 3 meters long So number of pieces
of wire he cuts is 18 ÷ 3 = 6 pieces.

Question 4.
A teacher has 24 pencils. They are divided equally
among 3 students. How many pencils does each student get?

Each student will get 8 pencils,

Explanation:
Given a teacher has 24 pencils and they are divided equally
among 3 students, So number of pencils each student gets is
24 ÷ 3 = 8 pencils.

Question 5.
There are 27 third-graders working in groups of 3.
How many groups of third-graders are there?

There are 9 groups of third-graders working,

Explanation:
Given there are 27 third-graders working in groups of 3,
So, number of groups of third-graders working are
27 ÷ 3 = 9 groups.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 18 Answer Key

Eureka Math Grade 3 Module 7 Lesson 18 Problem Set Answer Key

Question 1.
Use unit squares to build as many rectangles as you can with an area of 24 square units. Shade in squares on your grid paper to represent each rectangle that you made with an area of 24 square units.
a. Estimate to draw and label the side lengths of each rectangle you built in Problem 1. Then, find the perimeter of each rectangle. One rectangle is done for you.

P = 24 units + 1 unit + 24 units + 1 unit = 50 units
b. The areas of the rectangles in part (a) above are all the same. What do you notice about the perimeters?
Answer:
a. Perimeter of ABCD rectangle = 22units.
Perimeter of EFGH rectangle = 28units.
Perimeter of IJKL rectangle = 20units.

b. All rectangles drawn in the above1.a are not the same sided figures. They are not having same perimeters because their lengths are different compared to one another.

Explanation:
a.



Perimeter of ABCD rectangle =  Side + Side + Side + Side
= AB + BC + CD + DA
= 8units + 3units + 8units + 3units
= 11units + 8units + 3units
= 19units + 3units
= 22units.
Perimeter of EFGH rectangle = Side + Side + Side + Side
= EF + FG + GH + HE
= 12units + 2units + 12units + 2units
= 14units + 12units + 2units
= 26units + 2units
= 28units.
Perimeter of IJKL rectangle = Side + Side + Side + Side
= IJ + JK + KL + KI
= 6units + 4 units + 6units + 4 units
= 10units + 6units + 4units
= 16units + 4units
= 20units.

b. All rectangles drawn in the above1a are not the same figures. They are not having same perimeters.

Question 2.
Use unit square tiles to build as many rectangles as you can with an area of 16 square units. Estimate to draw each rectangle below. Label the side lengths.
a. Find the perimeters of the rectangles you built.
b. What is the perimeter of the square? Explain how you found your answer.
Answer:
a. Perimeter of OPQR Rectangle = 14units.
Perimeter of EFGH Rectangle = 34units.

b. Perimeter of ABCD Square= 16units. As, the sides in a square are equal, we can multiple the number of sides into the side value to get the Perimeter of square instead of adding them separately.

Explanation:
a.


Perimeter of OPQR Rectangle = Side + Side + Side + Side
= OP + PQ + QR + RO
= 5units + 2units + 5units + 2units
= 7units + 5units + 2units
= 12units + 2units
= 14units.
Perimeter of EFGH Rectangle = Side + Side + Side + Side
= EF + FG+ GH + HE
= 16units + 1unit + 16units + 1unit
= 17units + 16units + 1unit
= 33units + 1unit
= 34unit.

b.

Perimeter of ABCD Square= 4 × Side
= 4 × 4units
= 16units.

Question 3.
Doug uses square unit tiles to build rectangles with an area of 15 square units. He draws the rectangles as shown below but forgets to label the side lengths. Doug says that Rectangle A has a greater perimeter than Rectangle B. Do you agree? Why or why not?

Answer:
Yes, Doug is correct, the perimeter of rectangle A is greater than the perimeter of the rectangle B because in the appearance itself we notice that rectangle A is bigger in size than that of rectangle B.

Explanation:

Well, comparing the rectangles drawn by Doug its easy to say rectangle A is going to have the greater perimeter that compared to the perimeter of the rectangle B because the size of the rectangle A is bigger than that of rectangle B.

Eureka Math Grade 3 Module 7 Lesson 18 Exit Ticket Answer Key

Tessa uses square-centimeter tiles to build rectangles with an area of 12 square centimeters. She draws the rectangles as shown below. Label the unknown side lengths of each rectangle. Then, find the perimeter of each rectangle.

P = _____

P = _____

P = _____
Answer:
Perimeter of the ABCD Rectangle = 26cm.
Perimeter of the EFGH Rectangle = 12cm.
Perimeter of the IJKL Rectangle = 16cm.

Explanation:

The unknown side are measured by using a ruler by me.
Perimeter of the ABCD Rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 12cm + 1cm + 12cm + 1cm
= 13cm + 12cm + 1cm
= 25cm + 1cm
= 26cm.
Perimeter of the EFGH Rectangle = Side + Side + Side + Side
= EF + FG + GH+ HE
= 3cm + 3cm + 3cm + 3cm
= 6cm + 3cm + 3cm
= 9cm + 3cm
= 12cm.
Perimeter of the IJKL Rectangle = Side + Side + Side + Side
IJ + JK + KL + KI
= 6cm + 2cm + 6cm + 2cm
= 8cm + 6cm + 2cm
= 14cm + 2cm
= 16cm.

Eureka Math Grade 3 Module 7 Lesson 18 Homework Answer Key

Question 1.
Shade in squares on the grid below to create as many rectangles as you can with an area of 18 square centimeters.

Answer:
ABCD rectangle
EFGH rectangle
IJKL rectangle.

Explanation:

Question 2.
Find the perimeter of each rectangle in Problem 1 above.
Answer:
Perimeter of the ABCD rectangle = 20units.
Perimeter of the EFGH rectangle = 12units.
Perimeter of the IJKL rectangle = 18units.

Explanation:

Perimeter of the ABCD rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 8units + 2units + 8units + 2units
= 10units + 8units + 2units
= 18units + 2units
= 20units.
Perimeter of the EFGH rectangle = Side + Side + Side + Side
= EF + FG + GH+ HE
= 4units + 2units + 4units + 2units
= 6units + 4units + 2units
= 10units + 2units
= 12units.
Perimeter of the IJKL rectangle = Side + Side + Side + Side
= IJ + JK + KL + LI
=  5units + 4units + 5units + 4units
= 9units + 5units + 4units
= 14units + 4units
= 18units.

Question 3.
Estimate to draw as many rectangles as you can with an area of 20 square centimeters. Label the side lengths of each rectangle.
a. Which rectangle above has the greatest perimeter? How do you know just by looking at its shape?
b. Which rectangle above has the smallest perimeter? How do you know just by looking at its shape?
Answer:
a. Among all the three rectangles drawn, through looks  IJKL Rectangles is having greater perimeter compared to other rectangles. We can say that by seeing the length of the rectangles, which length is high that going to have greater perimeter.

b. Among all the three rectangles drawn, through looks  EFGH Rectangles is having smallest perimeter compared to other rectangles. We can say that by seeing the length of the rectangles.

Explanation:
a.

Perimeter of the ABCD Rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 7units + 4units + 7units + 4units
= 11units + 7units + 4units
= 18units + 4units
= 22units.
Perimeter of the EFGH Rectangle = Side + Side + Side + Side
= EF + FG + GH + HE
= 5units + 3units + 5units + 3units
= 8units + 5units + 3units
= 13units + 3units
= 16units.
Perimeter of the IJKL Rectangle = Side + Side + Side + Side
= IJ + JK + KL + LI
= 11units + 2units + 11units + 2units
= 13units + 11units + 2units
= 24units + 2units
= 26 units.

b. Among all the three rectangles drawn, through looks  EFGH Rectangles is having smallest perimeter compared to other rectangles. We can say that by seeing the length of the rectangles, which length is small that going to have smallest perimeter.

Engage NY Eureka Math 3rd Grade Module 7 Mid Module Assessment Answer Key

Eureka Math Grade 3 Module 7 Mid Module Assessment Task Answer Key

Question 1.
Three shapes are shown below.
a. Circle the shape(s) with only one pair of parallel sides.
b. Cross out the shape(s) with two pairs of parallel sides.

c. Which of the three shapes are quadrilaterals? Explain how you know.
Answer:

Question 2.
Use your ruler and right angle tool to draw the following shapes.
a. Draw and name a shape with four right angles.
b. Draw a four-sided shape with no right angles and no equal sides. Label the side lengths.
c. Draw triangles to create a rhombus. Label the side lengths.
Answer:

Question 3.
Mr. Cooper builds a fence to make a rectangular horse stall. The stall is 5 meters long and 7 meters wide. How many meters of fence does Mr. Cooper use? Draw a picture and write an equation to show your thinking.
Answer:

Question 4.
Jamal wants to put wood trim around his rectangular bedroom and square closet. His bedroom is 10 feet wide and 8 feet long. His closet is 3 feet wide and 3 feet long.

a. Wood trim is sold by the foot. How many feet of wood trim does Jamal need to go around his bedroom and closet? Show your work.
b. How much more wood trim does Jamal need for his bedroom than his closet? Write and solve an equation. Use a letter to represent the unknown.
Answer:

Question 5.
The figure below is composed of rectangles. Use the picture and the descriptions to find the perimeter of the shape. Show your work.
Each side labeled with A is 6 inches.
Each side labeled with B is 3 inches.
Each side labeled with C is 8 inches.

Answer:

Question 6.
Mrs. Gomez builds a fence around her backyard. Her plan shows the fence as a dotted line below.

Together, the garage and backyard make a rectangle. The fence goes only where there is a dotted line. How many feet of fence does Mrs. Gomez need to build? Show your work.
Answer:

Engage NY Eureka Math 3rd Grade Module 1 Lesson 12 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

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

Multiply.

multiply by 3 (6–10)

Multiplied by 3 (6–10) as shown below

Eureka Math Grade 3 Module 1 Lesson 12 Problem Set Answer Key

Question 1.
There are 8 birds at the pet store. Two birds are in each cage.
Circle to show how many cages there are.

8 ÷ 2 = ____4______
There are ___4____ cages of birds.

There are 4 cages of birds,

Explanation:
Given there are 8 birds at the pet store and two
birds are in each cage.
Number of cages are 8 ÷ 2 = 4 cages,
Circled to show 4 number of cages of birds as shown above.

Question 2.
The pet store sells 10 fish. They equally divide the fish into 5 bowls.
Draw fish to find the number in each bowl.

5 × ___2____ = 10
10 ÷ 5 = ___2_____
There are ____2____ fish in each bowl.

There are 2 fish in each bowl,

Explanation:
Given the pet store sells 10 fish and they are equally
divided the fish into 5 bowls.
Means each bowl has 10 ÷ 5 = 2 fish ( 5 X 2 = 10),
Drawn fish and found the number in each bowl as 2 fish,

Question 3.
Match.

Explanation:
Matched expressions as
10 ÷ 2 = 5,
16 ÷ 2 = 8,
18 ÷ 2 = 9,
14 ÷ 2 = 7 and
12 ÷ 2 = 6.

Question 4.
Laina buys 14 meters of ribbon. She cuts her ribbon into
2 equal pieces. How many meters long is each piece?
Label the tape diagram to represent the problem, including the unknown.

Each piece is ____7______ meters long.

Laina’s each piece of ribbon is 7 meters long,

Explanation:
Given Laina buys 14 meters of ribbon and she cuts
her ribbon into 2 equal pieces. So number of meters
long each piece is 14 ÷ 2 = 7 meters long,
Labeled the tape diagram to represent the problem,
including the unknown as shown in the above picture.

Question 5.
Roy eats 2 cereal bars every morning. Each box has a total
of 12 bars. How many days will it take Roy to finish 1 box?

It will take Roy to  finish 1 box in 6 days,

Explanation:
Given Roy eats 2 cereal bars every morning and each
box has a total of 12 bars.
So, Number of days it will take Roy to finish 1 box is
12 ÷ 2 = 6 days.

Question 6.
Sarah and Esther equally share the cost of a present.
The present costs $18. How much does Sarah pay?

Sarah pay’s $9 for the present,

Explanation:
Given Sarah and Esther equally share the cost of a present.
The present costs $18, As 2 persons have equally paid for the
present each paid $18 ÷ 2 = $9,
Therefore  Sarah pay’s $9 for the present.

Eureka Math Grade 3 Module 1 Lesson 12 Exit Ticket Answer Key

There are 14 mints in 1 box. Cecilia eats 2 mints each day.
How many days does it take Cecilia to eat 1 box of mints?
Draw and label a tape diagram to solve.

It takes Cecilia ____7___ days to eat 1 box of mints.

Cecilia eats 1 box of mints in 7 days,

Explanation:
Given there are 14 mints in 1 box and Cecilia eats
2 mints each day. So, number of days it will take Cecilia
to eat 1 box of mints are 14 ÷ 2 = 7 days,
Drawn and labeled a tape diagram to solve as shown above
in the picture.

Eureka Math Grade 3 Module 1 Lesson 12 Homework Answer Key

Question 1.
Ten people wait in line for the roller coaster.
Two people sit in each car. Circle to find the total number of cars needed.

10 ÷ 2 = ____5______
There are ___5____ cars needed.

Number of cars needed are 5,

Explanation:
Given Ten people wait in line for the roller coaster,
Two people sit in each car. So, number of cars needed are
10 ÷ 2 = 5, Circled the total 5 number of cars needed as
shown above in the picture.

Question 2.
Mr. Ramirez divides 12 frogs equally into 6 groups for
students to study. Draw frogs to find the number in each group.
Label known and unknown information on the tape diagram
to help you solve.

6 × ___2____ = 12
12 ÷ 6 = ___2____
There are ____2____ frogs in each group.

The number of frogs in each group are 2,

Explanation:
Given Mr. Ramirez divides 12 frogs equally into 6 groups for
students to study. Drawn frogs to find the number in each group
as 12 ÷ 6 = 2 frogs, Labeled known and unknown information
on the tape diagram and solved as shown above in the picture.

Question 3.
Match.

Explanation:
Matched expressions as
10 ÷ 2 = 5,
16 ÷ 2 = 8,
18 ÷ 2 = 9 and
14 ÷ 2 = 7 respectively.

Question 4.
Betsy pours 16 cups of water to equally fill 2 bottles.
How many cups of water are in each bottle?
Label the tape diagram to represent the problem, including the unknown.
There are ____8_____ cups of water in each bottle.

There are 8 cups of water in each bottle,

Explanation:
Given Betsy pours 16 cups of water to equally fill 2 bottles,
So, number of cups of water in each bottle are 16 ÷ 2 = 8 cups,
Labeled the tape diagram to represent the problem
including the unknown as shown above in the picture.

Question 5.
An earthworm tunnels 2 centimeters into the ground each day.
The earthworm tunnels at about the same pace every day.
How many days will it take the earthworm to tunnel 14 centimeters?

It will take 7 days for the earthworm to tunnel
14 centimeters into the ground,

Explanation:
Given an earthworm tunnels 2 centimeters into the
ground each day and the earthworm tunnels at about
the same pace every day, It will take the earthworm to
tunnel 14 centimeters is 14 ÷ 2 = 7 days.

Question 6.
Sebastian and Teshawn go to the movies.
The tickets cost $16 in total. The boys share the cost equally.
How much does Teshawn pay?

Teshawn pays $8 cost for the ticket,

Explanation:
Given Sebastian and Teshawn go to the movie and
the tickets cost $16 in total. Both the boys share the
cost equally means each had cost of $16 ÷ 2 = $8,
Therefore Teshawn pays $8 cost for the ticket.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 17 Answer Key

Eureka Math Grade 3 Module 7 Lesson 17 Problem Set Answer Key

Question 1.
The shapes below are made up of rectangles. Label the unknown side lengths. Then, write and solve an equation to find the perimeter of each shape.
a.

P =
Answer:
Perimeter of the given figure = 16 cm.

Explanation:

Length of the side AB in the given figure = 4 cm
Length of the side BC in the given figure = 2 cm
Length of the side CD in the given figure = 2 cm
Length of the side ED in the given figure = 1 cm
Length of the side EF in the given figure = 2 cm
Length of the side FA in the given figure = 3 cm
Length of the side GD in the given figure = 2 cm
Perimeter of the given figure = Length of the side AB + Length of the side BC + Length of the side CD + Length of the side ED + Length of the side EF + Length of the side FA + Length of the side GD
= 4 cm + 2 cm + 2 cm + 1 cm + 2 cm + 3 cm + 2cm
= 6 cm + 2 cm + 1 cm + 2 cm + 3 cm + 2cm
= 8 cm + 1 cm + 2 cm + 3 cm + 2cm
= 9 cm + 2 cm + 3 cm + 2cm
= 11 cm + 3 cm + 2cm
= 14 cm + 2 cm
= 16 cm.

b.

P =
Answer:
Perimeter of the given figure = 16ft.

Explanation:

Length of the side AB in the given figure = 2ft
Length of the side BC in the given figure = 1ft
Length of the side CD in the given figure = 1ft
Length of the side DE in the given figure = 1ft
Length of the side EF in the given figure = 2ft
Length of the side GF in the given figure = 2ft
Length of the side GH in the given figure = 5ft
Length of the side HA in the given figure = 2ft
Perimeter of the given figure = Length of the side AB + Length of the side BC + Length of the side CD + Length of the side DE + Length of the side EF + Length of the side GF + Length of the side GH + Length of the side HA
= 2ft + 1ft + 1ft + 1ft  + 2ft + 2ft+ 5ft + 2ft
= 3ft + 1ft + 1ft  + 2ft + 2ft+ 5ft + 2ft
= 4ft + 1ft  + 2ft + 2ft+ 5ft + 2ft
= 5ft + 2ft + 2ft+ 5ft + 2ft
= 7ft + 2ft + 5ft + 2ft
= 9ft + 5ft + 2ft
= 14ft + 2ft
= 16ft.

c.

P =
Answer:
Perimeter of the given figure = 24m.

Explanation:

Length of the side AB in the given figure = 2m
Length of the side BC in the given figure = 2m
Length of the side CD in the given figure = 4m
Length of the side DE in the given figure = 2m
Length of the side EF in the given figure = 4m
Length of the side GF in the given figure = 2m
Length of the side GH in the given figure = 2m
Length of the side HA in the given figure = 6m
Perimeter of the given figure = Length of the side AB + Length of the side BC + Length of the side CD + Length of the side DE + Length of the side EF + Length of the side GF + Length of the side GH + Length of the side HA
= 2m + 2m + 4m + 2m + 4m + 2m + 2m + 6m
= 4m + 4m + 2m + 4m + 2m + 2m + 6m
= 8m + 2m + 4m + 2m + 2m + 6m
= 10m + 4m + 2m + 2m + 6m
= 14m + 2m + 2m + 6m
= 16m + 2m + 6m
= 18m + 6m
= 24m.

d.

P =
Answer:
Perimeter of the given figure = 26yd.

Explanation:

Length of the side AB in the given figure = 7yd
Length of the side BC in the given figure = 2yd
Length of the side CD in the given figure = 2yd
Length of the side DE in the given figure = 4yd
Length of the side EF in the given figure = 2yd
Length of the side FG in the given figure = 2yd
Length of the side GH in the given figure = 1yd
Length of the side HI in the given figure = 2yd
Length of the side IJ in the given figure = 2yd
Length of the side JA in the given figure = 2yd
Perimeter of the given figure = Length of the side AB + Length of the side BC + Length of the side CD + Length of the side DE + Length of the side EF + Length of the side FG + Length of the side GH + Length of the side HI + Length of the side IJ + Length of the side JA
= 7yd + 2yd + 2yd + 4yd + 2yd + 2yd + 1yd + 2yd + 2yd + 2yd
= 9yd + 2yd + 4yd + 2yd + 2yd + 1yd + 2yd + 2yd + 2yd
= 11yd + 4yd + 2yd + 2yd + 1yd + 2yd + 2yd + 2yd
= 15yd + 2yd + 2yd + 1yd + 2yd + 2yd + 2yd
= 17yd + 2yd + 1yd + 2yd + 2yd + 2yd
= 19yd + 1yd + 2yd + 2yd + 2yd
= 20yd + 2yd + 2yd + 2yd
= 22yd + 2yd + 2yd
= 24yd + 2yd
= 26yd.

Question 2.
Nathan draws and labels the square and rectangle below. Find the perimeter of the new shape.

Answer:
Perimeter of the  ACDF  new shape = 48cm.

Explanation:

Length of the side of AB in the given figure = 6cm
Length of the side of BC in the given figure = 12cm
Length of the side of CD in the given figure = 6cm
Length of the side of DE in the given figure = 12cm
Length of the side of EF in the given figure = 6cm
Length of the side of FA in the given figure = 6cm
Perimeter of the ACDF new shape = Length of the side of AB + Length of the side of BC + Length of the side of CD + Length of the side of DE + Length of the side of EF + Length of the side of FA
= 6cm + 12cm + 6cm + 12cm + 6cm + 6cm
= 18cm + 6cm + 12cm + 6cm + 6cm
= 24cm + 12cm + 6cm + 6cm
= 36cm + 6cm + 6cm
= 42cm + 6cm
= 48cm.

Question 3.
Label the unknown side lengths. Then, find the perimeter of the shaded rectangle.

Answer:
Perimeter of the DFGC shaded rectangle = 26in.

Explanation:

Shaded rectangle = DFGC
Length of the side of AB of the given figure  = 16in
Length of the side of BG of the given figure = 2in
Length of the side of GC of the given figure = EA – BG = 7in – 2in = 5in.
Length of the side of CD of the given figure= AB – DE = 16in – 8in = 8in.
Length of the side of DE of the given figure= 8in
Length of the side of EA of the given figure= 7in
Length of the side of DF of the given figure= 5in
Length of the side of FG of the given figure= 8in
Perimeter of the DFGC shaded rectangle = Length of the side of FG + Length of the side of GC + Length of the side of CD + Length of the side of DF
= 8in + 5in + 8in + 5in
= 13in + 8in + 5in
= 21in + 5in
= 26in.

Eureka Math 3rd Grade Module 7 Lesson 17 Exit Ticket Answer Key

Label the unknown side lengths. Then, find the perimeter of the shaded rectangle.

Answer:
Perimeter of the FGDE shaded rectangle = 30m.

Explanation:

Shaded rectangle = FGDE
Length of the side AB in the given figure = 12m
Length of the side BC in the given figure = 14m
Length of the side CD in the given figure = 5m
Length of the side DE in the given figure = AB – CD = 12m – 5m = 7m.
Length of the side EF in the given figure = BC – FA = 14m – 6m = 8m.
Length of the side FA in the given figure = 6m
Length of the side FG in the given figure = 7m
Length of the side GD in the given figure = 8m
Perimeter of the FGDE shaded rectangle = Length of the side FG  + Length of the side GD + Length of the side DE + Length of the side EF
= 7m + 8m + 7m + 8m
= 15m + 7m + 8m
= 22m + 8m
= 30m.

Eureka Math Grade 3 Module 7 Lesson 17 Homework Answer Key

Question 1.
The shapes below are made up of rectangles. Label the unknown side lengths. Then, write and solve an equation to find the perimeter of each shape.
a.

P =
Answer:
Perimeter of the given figure = 32m.

Explanation:

Length of the side of AB in the given figure = 4m
Length of the side of BC in the given figure = 9m
Length of the side of CD in the given figure = 7m
Length of the side of DE in the given figure = 2m
Length of the side of EF in the given figure = 3m
Length of the side of FA in the given figure = 7m
Perimeter of the given figure = Length of the side of AB + Length of the side of BC + Length of the side of CD + Length of the side of DE + Length of the side of EF + Length of the side of FA
= 4m + 9m + 7m + 2m + 3m + 7m
= 13m + 7m + 2m + 3m + 7m
= 20m + 2m + 3m + 7m
= 22m + 3m + 7m
= 25m + 7m
= 32m.

b.

P =
Answer:
Perimeter of the given figure = 34 cm.

Explanation:

Length of the side of AB in the given figure = 2 cm
Length of the side of BC in the given figure = 4 cm
Length of the side of CD in the given figure = 4 cm
Length of the side of DE in the given figure = 3 cm
Length of the side of EF in the given figure = 2 cm
Length of the side of FG in the given figure = 5 cm
Length of the side of GH in the given figure = 8 cm
Length of the side of HA in the given figure = 6 cm
Perimeter of the given figure = Length of the side of AB + Length of the side of BC + Length of the side of CD + Length of the side of DE + Length of the side of EF + Length of the side of FG + Length of the side of GH + Length of the side of HA
= 2 cm + 4 cm + 4 cm + 3 cm + 2 cm + 5 cm + 8 cm + 6 cm
= 6 cm + 4 cm + 3 cm + 2 cm + 5 cm + 8 cm + 6 cm
= 10 cm + 3 cm + 2 cm + 5 cm + 8 cm + 6 cm
= 13 cm + 2 cm + 5 cm + 8 cm + 6 cm
= 15 cm + 5 cm + 8 cm + 6 cm
= 20 cm + 8 cm + 6 cm
= 28 cm + 6 cm
= 34 cm.

c.

P =
Answer:
Perimeter of the given figure = 40in.

Explanation:

Length of the side of AB in the given figure = 12in
Length of the side of BC in the given figure = 2in
Length of the side of CD in the given figure = 4in
Length of the side of DE in the given figure = 6in
Length of the side of EF in the given figure = 4in
Length of the side of FG in the given figure = 6in
Length of the side of GH in the given figure = 4in
Length of the side of HA in the given figure = 2in
Perimeter of the given figure = Length of the side of AB  + Length of the side of BC + Length of the side of CD + Length of the side of DE + Length of the side of EF + Length of the side of FG + Length of the side of GH + Length of the side of HA
= 12in + 2in + 4in + 6in + 4in + 6in + 4in + 2in
= 14in + 4in + 6in + 4in + 6in + 4in + 2in
= 18in + 6in + 4in + 6in + 4in + 2in
= 24in + 4in + 6in + 4in + 2in
= 28in + 6in + 4in + 2in
= 34in + 4in + 2in
= 38in + 2in
= 40in.

d.

P =
Answer:
Perimeter of the given figure = 30ft.

Explanation:

Length of the side of AB in the given figure = 8ft
Length of the side of BC in the given figure = 3ft
Length of the side of CD in the given figure = 3ft
Length of the side of DE in the given figure = 1ft
Length of the side of EF in the given figure = 3ft
Length of the side of FG in the given figure = 3ft
Length of the side of GH in the given figure = 2ft
Length of the side of HA in the given figure = 7ft
Perimeter of the given figure = Length of the side of AB + Length of the side of BC + Length of the side of CD  + Length of the side of DE + Length of the side of EF + Length of the side of FG + Length of the side of GH + Length of the side of HA
= 8ft + 3ft + 3ft + 1ft + 3ft + 3ft + 2ft + 7ft
= 11ft + 3ft + 1ft + 3ft + 3ft + 2ft + 7ft
= 14ft + 1ft + 3ft + 3ft + 2ft + 7ft
= 15ft + 3ft + 3ft + 2ft + 7ft
= 18ft + 3ft + 2ft + 7ft
= 21ft + 2ft + 7ft
= 23ft + 7ft
= 30ft.

Question 2.
Sari draws and labels the squares and rectangle below. Find the perimeter of the new shape.

Answer:
Perimeter of the new shape = 72cm.

Explanation:

Length of the side of AB in the given figure = 6cm
Length of the side of BC in the given figure = 18cm
Length of the side of CD in the given figure = 6cm
Length of the side of DE in the given figure = 6cm
Length of the side of EF in the given figure = 6cm
Length of the side of FG in the given figure = 18cm
Length of the side of GH in the given figure = 6cm
Length of the side of HA in the given figure = 6cm
Perimeter of the new shape = Length of the side of AB + Length of the side of BC + Length of the side of CD + Length of the side of DE + Length of the side of EF + Length of the side of FG + Length of the side of GH + Length of the side of HA
= 6cm + 18cm + 6cm + 6cm + 6cm + 18cm + 6cm + 6cm
= 24cm + 6cm + 6cm + 6cm + 18cm + 6cm + 6cm
= 30cm + 6cm + 6cm + 18cm + 6cm + 6cm
= 36cm + 6cm + 18cm + 6cm + 6cm
= 42cm + 18cm + 6cm + 6cm
= 60cm + 6cm + 6cm
= 66cm + 6cm
= 72cm.

Question 3.
Label the unknown side lengths. Then, find the perimeter of the shaded rectangle.

Answer:
Perimeter of the shaded rectangle = 37in.

Explanation:

Shaded rectangle =  BCDG
Length of the side of AB in the given figure = 5in
Length of the side of BC in the given figure =  EF – AB = 18in – 5in = 13in
Length of the side of CD in the given figure = FA – DE = 8in – 2in = 6in
Length of the side of DE in the given figure = 2in
Length of the side of EF in the given figure = 18in
Length of the side of FA in the given figure = 8in
Length of the side of GB in the given figure = 6in
Length of the side of GD in the given figure = 13in
Perimeter of the shaded rectangle = Length of the side of BC+ Length of the side of CD + Length of the side of GD + Length of the side of GB
= 13in + 6in + 13in + 6in
= 19in + 13in + 6in
= 31in + 6in
= 37in.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 11 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 11 Pattern Sheet Answer Key

Multiply.

multiply by 3 (1–5)


Explanation:
Multiplied by 3 (1–5) as shown above.

Eureka Math Grade 3 Module 1 Lesson 11 Problem Set Answer Key

Question 1.
Mrs. Prescott has 12 oranges. She puts 2 oranges in each bag.
How many bags does she have?
a. Draw an array where each column shows a bag of oranges.
___12___ ÷ 2 = ___6_____.

Mrs. Prescott have 6 bags,

Explanation:
Given Mrs. Prescott has 12 oranges and she puts
2 oranges in each bag, So number of bags she have
are 12 ÷ 2 = 6 bags,

a. Drawn an array where each column shows a bag
of oranges as shown above in the picture.

b. Redraw the oranges in each bag as a unit in the tape diagram.
The first unit is done for you. As you draw,
label the diagram with known and unknown information from the problem.

b.

Explanation:
Redrawn the oranges in each bag as a unit in the tape diagram,
labeled the diagram with known and unknown information
from the problem as 6 X 2 = 12 oranges, or 12 ÷ 2 = 6 bags.

Question 2.
Mrs. Prescott arranges 18 plums into 6 bags. How many plums
are in each bag? Model the problem with both an array and
a labeled tape diagram. Show each column as the number
of plums in each bag.
There are ____3_____ plums in each bag.

Mrs. Prescott arranges 3 plums in each bag,

Explanation:
Given Mrs. Prescott arranges 18 plums into 6 bags.
So number of  plums in each bag are 18 ÷ 6 = 3 bags
Modeled the problem with both an array and
labeled tape diagram as shown each column as the
number of plums in each bag.

Question 3.
Fourteen shopping baskets are stacked equally in 7 piles.
How many baskets are in each pile? Model the problem
with both an array and a labeled tape diagram.
Show each column as the number of baskets in each pile.

There are 2 baskets in each pile,

Explanation:
Given Fourteen shopping baskets are stacked equally in 7 piles.
So number of baskets in each pile are 14 ÷ 7 = 2 baskets,
Modeled the problem with both an array and
labeled tape diagram. Shown each column as the
number of baskets in each pile.

Question 4.
In the back of the store, Mr. Prescott packs 24 bell peppers
equally into 8 bags. How many bell peppers are in each bag?
Model the problem with both an array and labeled tape diagram.
Show each column as the number of bell peppers in each bag.

There are 3 bell peppers in each bag,

Explanation:
Given In the back of the store, Mr. Prescott packs 24 bell peppers
equally into 8 bags. So number of  bell peppers in each bag are
24 ÷ 8 = 3 bell peppers,
Modeled the problem with both an array and labeled tape diagram and
shown each column as the number of bell peppers in each bag.

Question 5.
Olga saves $2 a week to buy a toy car. The car costs $16.
How many weeks will it take her to save enough to buy the toy?

It will take 8 weeks to buy a toy car,

Explanation:
Given Olga saves $2 a week to buy a toy car.
The car costs $16. So number of weeks will it take
her to save enough to buy the toy is $16 ÷ $2 = 8 weeks.

Eureka Math Grade 3 Module 1 Lesson 11 Exit Ticket Answer Key

Ms. McCarty has 18 stickers. She puts 2 stickers on
each homework paper and has no more left.
How many homework papers does she have?
Model the problem with both an array and a
labeled tape diagram.

Ms. McCarty has 9 homework papers,

Explanation:
Given Ms. McCarty has 18 stickers and she puts 2 stickers
on each homework paper and has no more left.
So number of homework papers does she have are
18 ÷ 2 = 9 homework papers,
Modeled the problem with both an array and
labeled tape diagram as shown above in the picture.

Eureka Math Grade 3 Module 1 Lesson 11 Homework Answer Key

Question 1.
Fred has 10 pears. He puts 2 pears in each basket.
How many baskets does he have?
a. Draw an array where each column represents the
number of pears in each basket.
___10____ ÷ 2 = ___5_____

Fred has 5 baskets ,

Explanation:
Given Fred has 10 pears and he puts 2 pears in each basket.
So number of baskets does he have are 10 ÷ 2 = 5 baskets,

a. Drawn an array where each column represents the
number of pears in each basket as shown in the picture above.

b. Redraw the pears in each basket as a unit in the tape diagram.
Label the diagram with known and unknown information from the problem.

Redrawn the pears in each basket as a unit in the tape diagram,

Explanation:
Redrawn the pears in each basket as a unit in the tape diagram.
Labeled the diagram with known and unknown information
from the problem as 10 ÷ 2 = 5 baskets or 5 X 2 = 10 pears.

Question 2.
Ms. Meyer organizes 15 clipboards equally into 3 boxes.
How many clipboards are in each box? Model the problem
with both an array and a labeled tape diagram. Show each
column as the number of clipboards in each box.
There are ____5_____ clipboards in each box.

Ms. Meyer organizes 5 clipboards in each box,

Explanation:
Given Ms. Meyer organizes 15 clipboards equally into 3 boxes.
So, Number of clipboards in each box are 15 ÷ 3 = 5,
Modeled the problem with both an array and
labeled tape diagram as shown above each
column has 5 number of clipboards in each box.

Question 3.
Sixteen action figures are arranged equally on 2 shelves.
How many action figures are on each shelf?
Model the problem with both an array and a
labeled tape diagram. Show each column as the number
of action figures on each shelf.

There are 8 action figures on each shelf,

Explanation:
Given Sixteen action figures are arranged equally on 2 shelves.
So number of action figures on each shelf are 16 ÷ 2 = 8,
Modeled the problem with both an array and
labeled tape diagram as show each column has 8 number
of action figures on each shelf.

Question 4.
Jasmine puts 18 hats away. She puts an equal number of
hats on 3 shelves. How many hats are on each shelf?
Model the problem with both an array and a labeled
tape diagram. Show each column as the number of
hats on each shelf.

On each shelf there are 6 hats,

Explanation:
Given Jasmine puts 18 hats away and she puts an equal
number of hats on 3 shelves. So number of hats on each
shelf are 18 ÷ 3 = 6 hats,
Modeled the problem with both an array and labeled
tape diagram as shown each column has 6 number of
hats on each shelf.

Question 5.
Corey checks out 2 books a week from the library.
How many weeks will it take him to check out a
total of 14 books?

Corey will take 7 weeks to check out total 14 books,

Explanation:
Given Corey checks out 2 books a week from the library.
So, number of weeks it will take Corey to check out a
total of 14 books are 14 ÷ 2 = 7 weeks.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 10 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 10 Pattern Sheet Answer Key

Multiply.

multiply by 2 (6–10)


Explanation:
Multiplied  by 2 (6–10) as shown above.

Eureka Math Grade 3 Module 1 Lesson 10 Problem Set Answer Key

Question 1.
7 × 3 = (5 × 3) + (2 × 3) = ___21_______

(5 × 3) + (2 × 3) = 15 + ___6___
15 + ___6___ = _____21________

7 × 3 = (5 × 3) + (2 × 3) =
15 + 6 = 21 or 7 X 3 = 21.

Explanation:
Given 7 X 3 wrote 7 as (5 + 2) X 3 =
(5 × 3) + (2 × 3) =
15 + 6 = 21 or 7 X 3 = 21.

Question 2.
8 × 3 = (4 × 3) + (4 × 3) = __24____

(4 × 3) + (4 × 3) = ____12_____ + _____12____ = 24,
____8_____ × 3 = ____24______

8 × 3 = (4 × 3) + (4 × 3) =
12 + 12 = 24 or 8 X 3 = 24,

Explanation:
Given 8 X 3 wrote 8 as (4 + 4) X 3 =
(4 × 3) + (4 × 3) =
12 + 12 = 24 or 8 X 3 = 24.

Question 3.
Ruby makes a photo album. One page is shown below.
Ruby puts 3 photos in each row.
a. Fill in the equations on the right.
Use them to help you draw arrays that show the photos
on the top and bottom parts of the page.


Explanation:
Filled in the equations on the right as 2 X 3 = 6, 3 X 3 =9,
Used them to help to draw arrays that showed the photos
on the top and bottom parts of the page as shown above
in the picture.

b. Ruby calculates the total number of photos as shown below.
Use the array you drew to help explain Ruby’s calculation.

Explanation:
Ruby calculates the total 15 number of photos as shown above,
Used the array to drew to help explain Ruby’s calculation as
(2 X 3) + (3 X 3) = 6 + 9 = 15.

Eureka Math Grade 3 Module 1 Lesson 10 Exit Ticket Answer Key

Question 1.
6 × 3 = ___18___

(4 × 3) + (2 × 3) = __12__ + ___6___
6 × 3 = _12__ + _6__
_6_ × 3 = _18__

6 X 3  = (4 × 3) + (2 × 3) = 12 + 6 = 18 or 6 X 3 = 18,

Explanation:
Given 6 X 3 wrote 6 as (4 + 2) X 3 =
(4 × 3) + (2 × 3) =
12 + 6 = 18 or 6 X 3 = 18.

Question 2.
7 × 3 = _21_

(5 × 3) + (2 × 3) = __ + __
7 × 3 = __ + __
__ × 3 = ___

7 X 3  = (5 × 3) + (2 × 3) = 15 + 6 = 21 or 7 X 3 = 21,

Explanation:
Given 7 X 3 wrote 7 as (5 + 2) X 3 =
(5 × 3) + (2 × 3) =
15 + 6 = 21 or 7 X 3 = 21.

Eureka Math Grade 3 Module 1 Lesson 10 Homework Answer Key

Question 1.
6 × 3 = ____18______

6 X 3 = (4 × 3) + (2 × 3) = 12 + 6 = 18 or 6 X 3 = 18,

Explanation:
Given 6 X 3 wrote 6 as (4 + 2) X 3 =
(4 × 3) + (2 × 3) =
12 + 6 = 18 or 6 X 3 = 18.

Question 2.
8 × 2 = _16_

8 X 2 = (4 × 2) + (4 × 2) = 8 + 8 = 16 or 8 X 2 = 16,

Explanation:
Given 8 X 2 wrote 8 as (4 + 4) X 2 =
(4 × 2) + (4 × 2) =
8 + 8 = 16 or 8 X 2 = 16.

Question 3.
Adriana organizes her books on shelves. She puts 3 books in each row.
a. Fill in the equations on the right. Use them to draw arrays
that show the books on Adriana’s top and bottom shelves.


Explanation:
Given Adriana organizes her books on shelves.
She puts 3 books in each row.
a. Filled in the equations on the right as (5 X 3), (1 X 3)
Used them to draw arrays that show the books on
Adriana’s top and bottom shelves as shown above.

b. Adriana calculates the total number of books as shown below.
Use the array you drew to help explain Adriana’s calculation.


Explanation:
Adriana calculates the total 18 number of books as shown above,
Used the array to drew to help explain Adriana’s calculation as
6 X 3 = (5 X 3) + (1 X 3) = 15 + 3 = 18 or 6 X 3 = 18.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 16 Answer Key

Eureka Math Grade 3 Module 7 Lesson 16 Pattern Sheet Answer Key

Multiply.

multiply by 9 (6–10)
Answer:

Explanation:
9 × 5 = 45
9 × 6 = 54
9 × 7 = 63
9 × 8 = 72
9 × 9 = 81
9 × 10 = 90.

Eureka Math Grade 3 Module 7 Lesson 16 Problem Set Answer Key

Question 1.
Find the perimeter of 10 circular objects to the nearest quarter inch using string. Record the name and perimeter of each object in the chart below.

a. Explain the steps you used to find the perimeter of the circular objects in the chart above.

Answer:
Step 1: Took a string and wrapped around the object.
Step 2: Marked the string met.
Step 3: Measured the length of the string.

Explanation:
First I rolled the String around the object. Later, I marked the string met. Afterwards, I took a ruler to measure the length of the string.

b. Could the same process be used to find the perimeter of the shape below? Why or why not?

Answer: Yes, the same steps would be followed to find the perimeter of the given shape because I use the string to find the perimeter.

Explanation:
The same steps would be followed to find the perimeter of the given shape because I use the string to find the perimeter. First I rolled the String around the object. Later, I marked the string met. Afterwards, I took a ruler to measure the length of the string.

Question 2.
Can you find the perimeter of the shape below using just your ruler? Explain your answer.

Answer:
No, I cant find the perimeter of this given shape because it got curve line in it as which a ruler cant measure it.

Explanation:
A tool used to rule straight lines and measure distances is called as ruler.
No, I cant find the perimeter of this given shape because it got curve line in it as a ruler measures only straight lines.

Question 3.
Molly says the perimeter of the shape below is 6 \(\frac{1}{4}\) inches. Use your string to check her work. Do you agree with her? Why or why not?

Answer:
No, she is not correct has I have used my string and found the perimeter of the shape as 5 \(\frac{3}{4}\) inches.

Explanation:
Molly says the perimeter of the shape below is 6 \(\frac{1}{4}\) inches.
No, she is not correct has I have used my string and found the perimeter of the shape as 5 \(\frac{3}{4}\) inches.

Question 4.
Is the process you used to find the perimeter of a circular object an efficient method to find the perimeter of a rectangle? Why or why not?
Answer:
No, I don’t think this process to find the perimeter of a circular object an efficient method to find the perimeter of a rectangle, because I can just use a ruler to measure length of the straight lines.

Explanation:
No, I don’t think this process to find the perimeter of a circular object an efficient method to find the perimeter of a rectangle. A ruler is used to find the lengths of the straight lines which help in finding the perimeter of the shape.

Eureka Math Grade 3 Module 7 Lesson 16 Exit Ticket Answer Key

Use your string to the find the perimeter of the shape below to the nearest quarter inch.

Answer:
The perimeter of this circular shape is 26 3/4 inches.

Explanation:
I used my string to measure this circular shape. The perimeter of this circular shape is 26 3/4 inches.

Eureka Math Grade 3 Module 7 Lesson 16 Homework Answer Key

Question 1.
a. Find the perimeter of 5 circular objects from home to the nearest quarter inch using string. Record the name and perimeter of each object in the chart below.

b. Explain the steps you used to find the perimeter of the circular objects in the chart above.
Answer:
Well, as previously discussed I have used a string to measure the perimeter of the circular shapes. I wrapped string around the circular bodies and  noted the values where they met. Later I have used a ruler to measured the values.

Explanation:
Used string to measure the perimeter of the circular shapes. I wrapped string around the circular bodies and  noted the values where they met. Later I have used a ruler to measured the values.

Question 2.
Use your string and ruler to find the perimeter of the two shapes below to the nearest quarter inch.

a. Which shape has a greater perimeter?
b. Find the difference between the two perimeters.

Answer:
a. The perimeter of the given shape B is greater by 3inches than The perimeter of the given shape A.
b. The perimeter of the shape B given using string – The perimeter of the shape A given using string
= 17 1/4 inches – 14 1/4 inches
= 3 inches

Explanation:
a. The perimeter of the given shape A using string = 14 1/4 inches.
The perimeter of the given shape B using string = 17 1/4 inches.

b. Difference:
The perimeter of the shape  B given using string – The perimeter of the shape  A given using string
= 17 1/4 inches – 14 1/4 inches
= 3 inches.

Question 3.
Describe the steps you took to find the perimeter of the objects in Problem 2. Would you use this method to find the perimeter of a square? Explain why or why not.
Answer:
Step 1: Took a string and wrapped around the given shape A.
Step 2: Marked the string met.
Step 3: Measured the length of the string using ruler.
Step 4: Took a string and wrapped around the given shape B.
Step 5: Marked the string met.
Step 6: Measured the length of the string using ruler.

No, I cant use this process of finding perimeter for Square because Square’s perimeter can be found easily using ruler directly. For finding the perimeter of circular shapes we use string n later the ruler.

Explanation:
First I rolled the String around the given shape A. Later, I marked the string met. Afterwards, I took a ruler to measure the length of the string. same I did with the given  shape B.

For finding the perimeter of circular shapes we use string n later the ruler, to know their measurement value.  Square is a straight line shape, we can use ruler directly to find its perimeter no need of string.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 9 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 9 Pattern Sheet Answer Key

multiply by 2 (1–5)


Explanation:
Multiply by 2 (1–5) as shown above.

Eureka Math Grade 3 Module 1 Lesson 9 Problem Set Answer Key

Question 1.
The team organizes soccer balls into 2 rows of 5.
The coach adds 3 rows of 5 soccer balls.
Complete the equations to describe the total array.


Explanation:
Given the team organizes soccer balls into 2 rows of 5 as
2 X 5 = 10 and the coach adds 3 rows of 5 soccer balls as
3 X 5 = 15, Completed the equations to described the
total array as 5 x 5 = (2 + 3) X 5 = (2 X 5) + (3 X 5) =
10 + 15 = 25 or 5 X 5 = 25.

Question 2.
7 × 2 = __14___

7 X 2 = 14,

Explanation:
Given 7 X 2, we wrote 7 X 2 as (5 + 2) X 2 =
(5 X 2) + ( 2 x 2) = 10 + 4 = 14 or
7 X 2 = 14.

Question 3.
9 × 2 = __18___

9 X 2 = 18,

Explanation:
Given 9 X 2  we wrote 9 X 2 as
(10 – 1) X 2 = (10 X 2) – (1 X 2) =
20 – 2 = 18 or 9 X 2 = 18.

Question 4.
Matthew organizes his baseball cards in 4 rows of 3.
a. Draw an array that represents Matthew’s cards using
an x to show each card.
b. Solve the equation to find Matthew’s total number of cards.
4 × 3 = __12__,
a.

Explanation:
Given Matthew organizes his baseball cards in 4 rows of 3,
So array shown as 4 X 3,
Drawn an array that represents Matthew’s cards using
an x to show each card as shown above.

b. Total number of  Matthew’s cards are 12,

Explanation:
Solved the equation to find Matthew’s total number of
cards as 4 X 3 = 12.

Question 5.
Matthew adds 2 more rows. Use circles to show his new
cards on the array in Problem 4(a).
a. Write and solve a multiplication equation to represent
the circles you added to the array.
___2___ × 3 = __6____

Explanation:
Given Matthew adds 2 more rows.
Used circles to show his new cards on the array in
Problem 4(a) as shown above.

b. Add the totals from the equations in Problems 4(b)
and 5(a) to find Matthew’s total cards.
__12____ + ___6___ = 18

Explanation:
Added the totals from the equations in Problems 4(b)
and 5(a) to find Matthew’s total cards as
(4 X 3) + (2 X 3) = 12 + 6 = 18.

c. Write the multiplication equation that shows Matthew’s
total number of cards.
___6___ × ___3___ = 18,

The multiplication equation to show total number of cards
is 6 X 3 = 18,

Explanation:
The multiplication equation that shows Matthew’s
total number of cards is 6 X 3 = (4 X 3) + (2 X 3) =
12 + 6 = 18 or 6 X 3 = 18.

Eureka Math Grade 3 Module 1 Lesson 9 Exit Ticket Answer Key

Question 1.
Mrs. Stern roasts cloves of garlic. She places 10 rows of
two cloves on a baking sheet. Write an equation to describe
the number of cloves Mrs. Stern bakes.
___10____ × ___2____ = __20____ cloves

The number of cloves Mrs. Stern bakes is 20 cloves,

Explanation:
Given Mrs. Stern roasts cloves of garlic. She places 10 rows of
two cloves on a baking sheet. The equation to describe
the number of cloves Mrs. Stern bakes is 10 X 2 = 20 cloves.

Question 2.
When the garlic is roasted, Mrs. Stern uses some for a recipe.
There are 2 rows of two garlic cloves left on the pan.
a. Complete the equation below to show how many garlic cloves Mrs. Stern uses.
____10____ twos – ____2____ twos = ____8___twos,

10 twos – 2 twos = 8 twos,

Explanation:
Completed the equation below to show how many
garlic cloves Mrs. Stern uses as 10 twos – 2 twos = 8 twos.

b. 20 – ________ = 16
20 – 4 = 16,

Explanation:
Subtracted 4 from 20 we get 16 as 20 – 4 =16 cloves.

c. Write an equation to describe the number of garlic cloves
Mrs. Stern uses. ___8____ × 2 = ___16_____,

The number of garlic cloves Mrs. Stern uses are 16 cloves,

Explanation:
Given Mrs. Stern roasts cloves of garlic,
She places 10 rows of two cloves on a baking sheet and
Mrs. Stern uses some for a recipe there are 2 rows of
two garlic cloves left on the pan, 10 x 2 = 20 cloves and left are
2 X 2 = 4 cloves, Means used cloves are ( 10 X 2) – (2 X 2) =
20 – 4 = 16 cloves or (10 – 2) X 2 = 8 X 2= 16 cloves.
Therefore, the equation for number of garlic cloves Mrs. Stern
uses are 8 X 2 = 16 cloves.

Eureka Math Grade 3 Module 1 Lesson 9 Homework Answer Key

Question 1.
Dan organizes his stickers into 3 rows of four.
Irene adds 2 more rows of stickers. Complete the
equations to describe the total number of stickers in the array.

a. (4 + 4 + 4) + (4 + 4) = __12 + 8 = ____20_____
b. 3 fours + ___2___ fours = _____5______ fours
c. ___5_____ × 4 = ____20____

a. (4 + 4 + 4) + (4 + 4) = 12 + 8 = 20,
b. 3 fours + 2 fours = 5 fours,
c. 5 X 4 = 20,

Explanation:
Given Dan organizes his stickers into 3 rows of four.
Irene adds 2 more rows of stickers.
Completed the equations to describe the
total number of stickers in the array as 5 X 5 =
(3 X 5) + (2 X 5) = 15 + 10 = 25 or 5 X 5 = 25,
a. We wrote 5 X 5 as (4 + 4 + 4) + (4 + 4) = 12 + 8 = 20,
b. We wrote 5 X 5 as 3 fours + 2 fours = 5 fours,
c. We wrote 5 x 4 as 20, 5 X 4 = 20.

Question 2.
7 × 2 = __14____

7 x 2 = 14,

Explanation:
Given 7 X 2 we wrote 7 x 2 as (6 + 1) X 2 =
(6 x 2) + (1 x 2) = 12 + 2 = 14 or 7 x 2 = 14.

Question 3.
9 × 3 = __27___

9 X 3 = 27,

Explanation:
Given 9 X 3 we wrote 9 x 3 as (10 – 1) X 3 =
(10 x 3) – (1 x 3) = 30 – 3 = 27 or 9 x 2 = 27.

Question 4.
Franklin collects stickers. He organizes his stickers in 5 rows of four.
a. Draw an array to represent Franklin’s stickers.
Use an x to show each sticker.
b. Solve the equation to find Franklin’s total number of
stickers. 5 × 4 = ___20___
a.

b. Franklin’s total number of stickers are 20.

Explanation:
Given Franklin collects stickers. He organizes his
stickers in 5 rows of four.
a. Drawn an array to represent Franklin’s stickers,
Used an x to show each sticker as shown above.
b. Solved the equation to find Franklin’s total number of
stickers as 5 × 4 = 20.

Question 5.
Franklin adds 2 more rows. Use circles to show his new stickers
on the array in Problem 4(a).
a. Write and solve an equation to represent the circles you added to the array.
___2___ × 4 = ___8___

Explanation:
Given Franklin adds 2 more rows. Used circles to show
his new stickers on the array in Problem 4(a).
a. Wrote and solved an equation to represent the
circles I added to the array as 2 X 4 = 8.

b. Complete the equation to show how you add the
totals of 2 multiplication facts to find Franklin’s
total number of stickers. __20____ + ___8___ = 28,

Equation of totals of 2 multiplication facts to find Franklin’s
total number of stickers are 20 + 8 = 28,

Explanation:
Completed the equation to show how I added the
totals of 2 multiplication facts to find Franklin’s
total number of stickers as (5 X 4)  + (2 X 4) = 20 + 8 = 28.

c. Complete the unknown to show Franklin’s total number of stickers.
__5___ × 4 = 20,

The unknown Franklin’s total number of stickers are 20,

Explanation:
To show unknown Franklin’s total number of stickers
as given Franklin collects stickers. He organizes his
stickers in 5 rows of four, So, Franklin’s total number
of stickers are 5 X 4 = 20.