Eureka Math

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.

Object	Perimeter (to the nearest quarter inch)
Cap of my jam jar	13\(\frac{1}{4}\) inches
Bangle	17\(\frac{1}{4}\) inches
Plate of my Dog	34\(\frac{1}{4}\) inches
Plastic glass Mouth	21\(\frac{3}{4}\) inches
Lid of my Lunch Box	29\(\frac{3}{2}\) inches
Ear ring	2\(\frac{2}{4}\) inches
Water Bottle cap	7\(\frac{2}{4}\) inches
Pencil Mouth	1\(\frac{2}{4}\) inches
Pen Cap	3\(\frac{1}{4}\) inches
Perfume Bottom surface	12\(\frac{3}{4}\) inches

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.

Object	Perimeter (to the nearest quarter inch)
Example:  Peanut Butter Jar Cap	9\(\frac{1}{2}\) inches
Cap of my jam jar	13\(\frac{1}{4}\) inches
Plate of my Dog	34\(\frac{1}{4}\) inches
Lid of my Lunch Box	29\(\frac{3}{2}\) inches
Water Bottle cap	7\(\frac{2}{4}\) inches
Pen Cap	3\(\frac{1}{4}\) inches

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.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 4 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 4 Sprint Answer Key

A
Repeated Addition as Multiplication

Question 1.
5 + 5 + 5 =
5 + 5 + 5 = 15,

Explanation:
Given 5 + 5 + 5 we add 5 by 3 times
we get 15, So 5 + 5 + 5 = 15.

Question 2.
3 × 5 =
3 X 5 = 15,

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

Question 3.
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 4.
2 + 2 + 2 =
2 + 2 + 2 = 6,

Explanation:
Given 2 + 2 + 2 we add 2 by 3 times
we get 6, So 2 + 2 + 2 = 6.

Question 5.
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 6.
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 7.
5 + 5 =
5 + 5 = 10,

Explanation:
Given 5 + 5 we add 5 by 2 times
we get 10, So 5 + 5 = 10.

Question 8.
2 × 5 =
2 X 5 = 10,

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

Question 9.
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 10.
2 + 2 + 2 + 2 =
2 + 2 + 2 + 2 = 8,

Explanation:
Given 2 + 2 + 2 + 2 we add 2 by 4 times
we get 8, So 2 + 2 + 2 + 2= 8.

Question 11.
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 12.
2 × 4 =
2 X 4 = 8,

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

Question 13.
2 + 2 + 2 + 2 + 2 =
2 + 2 + 2 + 2 + 2 = 10,

Explanation:
Given 2 + 2 + 2 + 2 + 2 we add 2 by 5 times
we get 10, So 2+ 2 + 2 + 2 + 2= 10.

Question 14.
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 15.
2 × 5 =
2 X 5 = 10,

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

Question 16.
3 + 3 =
3 + 3 = 6,

Explanation:
Given 3 + 3 we add 3 by 2 times
we get 6, So 3 + 3 = 6.

Question 17.
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 18.
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 19.
5 + 5 + 5+ 5 =
5 + 5 + 5 + 5 = 20,

Explanation:
Given 5 + 5 + 5 + 5 we add 5 by 4 times
we get 20, So 5 + 5 + 5 + 5 = 20.

Question 20.
4 × 5 =
4 X 5 = 20,

Explanation:
Given 4 X 5 we multiply 4 with 5,
we get 20 as 4 X 5 = 20.

Question 21.
5 × 4 =
5 X 4 = 20,

Explanation:
Given 5 X 4 we multiply 5 with 4,
we get 20 as 5 X 4 = 20.

Question 22.
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 23.
3 + 3 + 3 + 3 =
3 + 3 + 3 + 3 = 12,

Explanation:
Given 3 + 3 + 3 + 3 we add 3 by 4 times
we get 12, So 3 + 3 + 3 + 3 = 12.

Question 24.
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 25.
3 × 4 =
3 X 4 = 12,

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

Question 26.
3 + 3 + 3 =
3 + 3 + 3 = 9,

Explanation:
Given 3 + 3 + 3 we add 3 by 3 times
we get 9, So 3 + 3 + 3 = 9.

Question 27.
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 28.
3 + 3 + 3 + 3 + 3 =
3 + 3 + 3 + 3 + 3 = 15,

Explanation:
Given 3 + 3 + 3 + 3 + 3 we add 3 by 5 times
we get 15, So 3 + 3 + 3 + 3 + 3 = 15.

Question 29.
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 30.
3 × 5 =
3 X 5 = 15,

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

Question 31.
7 + 7 =
7 + 7 = 14,

Explanation:
Given 7 + 7 we add 7 by 2 times
we get 14, So 7 + 7 = 14.

Question 32.
2 × 7 =
2 X 7 = 14,

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

Question 33.
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 34.
9 + 9 =
9 + 9 = 18,

Explanation:
Given 9 + 9 we add 9 by 2 times
we get 18, So 9 + 9 = 18.

Question 35.
2 × 9 =
2 X 9 = 18,

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

Question 36.
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 37.
6 + 6 =
6 + 6 = 12,

Explanation:
Given 6 + 6 we add 6 by 2 times
we get 12, So 6 + 6 = 12.

Question 38.
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 39.
2 × 6 =
2 X 6 = 12,

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

Question 40.
8 + 8 =
8 + 8 = 16,

Explanation:
Given 8 + 8 we add 8 by 2 times
we get 16, So 8 + 8 = 16.

Question 41.
2 × 8 =
2 X 8 = 16,

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

Question 42.
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 43.
7 + 7 + 7 + 7 =
7 + 7 + 7 + 7 = 28,

Explanation:
Given 7 + 7 + 7 + 7 we add 7 by 4 times
we get 28, So 7 + 7 + 7 + 7 = 28.

Question 44.
4 × 7 =
4 X 7 = 28,

Explanation:
Given 4 X 7 we multiply 4 with 7,
we get 28 as 4 X 7 = 28.

B
Repeated Addition as Multiplication

Question 1.
2 + 2 + 2 =
2 + 2 + 2 = 6,

Explanation:
Given 2 + 2 + 2 we add 2 by 3 times
we get 6, So 2 + 2 + 2 = 6.

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.
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 4.
5 + 5 + 5 =
5 + 5 + 5 = 15,
Explanation:
Given 5 + 5 + 5 we add 5 by 3 times
we get 15, So 5 + 5 + 5 = 15.

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

Question 6.
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 7.
2 + 2 + 2 + 2 =
2 + 2 + 2 + 2 = 8,

Explanation:
Given 2 + 2 + 2 + 2 we add 2 by 4 times
we get 8, So 2 + 2 + 2 + 2= 8.

Question 8.
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 9.
2 × 4 =
2 X 4 = 8,
Explanation:
Given 2 X 4 we multiply 2 with 4,
we get 8 as 2 X 4 = 8.

Question 10.
5 + 5 =
5 + 5 = 10,
Explanation:
Given 5 + 5 we add 5 by 2 times
we get 10, So 5 + 5 = 10.

Question 11.
2 × 5 =
2 X 5 = 10,
Explanation:
Given 2 X 5 we multiply 2 with 5,
we get 10 as 2 X 5 = 10.

Question 12.
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 13.
3 + 3 =
3 + 3 = 6,

Explanation:
Given 3 + 3 we add 3 by 2 times
we get 6, So 3 + 3 = 6.

Question 14.
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 15.
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 16.
2 + 2 + 2 + 2 + 2 =
2 + 2 + 2 + 2 + 2 = 10,
Explanation:
Given 2 + 2 + 2 + 2 + 2 we add 2 by 5 times
we get 10, So 2+ 2 + 2 + 2 + 2= 10.

Question 17.
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 18.
2 × 5 =
2 X 5 = 10,
Explanation:
Given 2 X 5 we multiply 2 with 5,
we get 10 as 2 X 5 = 10.

Question 19.
5 + 5 + 5 + 5 =
5 + 5 + 5 + 5 = 20,

Explanation:
Given 5 + 5 + 5 + 5 we add 5 by 4 times
we get 20, So 5 + 5 + 5 + 5 = 20.

Question 20.
4 × 5 =
4 X 5 = 20,
Explanation:
Given 4 X 5 we multiply 4 with 5,
we get 20 as 4 X 5 = 20.

Question 21.
5 × 4 =
5 X 4 = 20,

Explanation:
Given 5 X 4 we multiply 5 with 4,
we get 20 as 5 X 4 = 20.

Question 22.
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 23.
4 + 4 + 4 =
4 + 4 + 4 = 12,

Explanation:
Given 4 + 4 + 4 we add 4 by 3 times
we get 12, So 4 + 4 + 4 = 12.

Question 24.
3 × 4 =
3 X 4 = 12,

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

Question 25.
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 26.
4 + 4 + 4 + 4 =
4 + 4 + 4 + 4 = 16,

Explanation:
Given 4 + 4 + 4 + 4 we add 4 by 4 times
we get 16, So 4 + 4 + 4 +4 = 16.

Question 27.
4 × 4 =
4 X 4 = 16,

Explanation:
Given 4 X 4 we multiply 4 with 4,
we get 14 as 4 X 4 = 16.

Question 28.
4 + 4 + 4 + 4 + 4 =
4 + 4 + 4 + 4 + 4 = 20,

Explanation:
Given 4 + 4 + 4 + 4 + 4 we add 4 by 5 times
we get 20, So 4 + 4 + 4 + 4 + 4 = 20.

Question 29.
4 × 5 =
4 X 5 = 20,

Explanation:
Given 4 X 5 we multiply 4 with 5,
we get 20 as 4 X 5 = 20.

Question 30.
5 × 4 =
5 X 4 = 20,

Explanation:
Given 5 X 4 we multiply 5 with 4,
we get 20 as 5 X 4 = 20.

Question 31.
6 + 6 =
6 + 6 = 12,

Explanation:
Given 6 + 6 we add 6 by 2 times
we get 12, So 6 + 6 = 12.

Question 32.
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 33.
2 × 6 =
2 X 6 = 12,

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

Question 34.
8 + 8 =
8 + 8 = 16,

Explanation:
Given 8 + 8 we add 8 by 2 times
we get 16, So 8 + 8 = 16.

Question 35.
2 × 8 =
2 X 8 = 16,

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

Question 36.
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 37.
7 + 7 =
7 + 7 = 14,

Explanation:
Given 7 + 7 we add 7 by 2 times
we get 14, So 7 + 7 = 14.

Question 38.
2 × 7 =
2 X 7 = 14,

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

Question 39.
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 40.
9 + 9 =
9 + 9 = 18,

Explanation:
Given 9 + 9 we add 9 by 2 times
we get 18, So 9 + 9 = 18.

Question 41.
2 × 9 =
2 X 9 = 18,

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

Question 42.
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 43.
6 + 6 + 6 + 6 =
6 + 6 + 6 + 6 = 24,

Explanation:
Given 6 + 6 + 6 + 6 we add 6 by 4 times
we get 24, So 6 + 6 + 6 + 6 = 24.

Question 44.
4 × 6 =
4 X 6 = 24,

Explanation:
Given 4 X 6 we multiply 4 with 6,
we get 24 as 4 X 6 = 24.

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

Question 1.

14 flowers are divided into 2 equal groups.
There are ____7_____ flowers in each group.

There are 7 flowers in each group,

Explanation:
Given 14 flowers are divided into 2 equal groups,
So there are 14 ÷ 2 = 7 flowers in 2 equal groups.

Question 2.

28 books are divided into 4 equal groups.
There are _____7____ books in each group.

There are 7 books in each group.

Explanation:
Given 28 books are divided into 4 equal groups,
So there are 28 ÷ 4 = 7 books in 4 equal groups.

Question 3.

30 apples are divided into ___3___ equal groups.
There are ____10_____ apples in each group.

30 apples are divided into 3 equal groups.
There are 10 apples in each group.

Explanation:
Given in the picture there are30 apples divided into
3 equal groups. So there are 30 ÷ 3 = 10 apples
in each group.

Question 4.

___12____ cups are divided into ___2____ equal groups.
There are ____6_____ cups in each group.
12 ÷ 2 = ___6______

12 cups are divided into 2 equal groups.
There are 6 cups in each group. 12 ÷ 2 = 6 cups,

Explanation:
As given in the picture there are 12 cups divided
into 2 equal groups, There are 6 cups  in each group
as 12 ÷ 2 = 6 cups.

Question 5.

There are ____15_____ toys in each group.
15 ÷ 3 = ____5_____

There are ____15_____ toys in each group,

Explanation:
As given in the picture there are 15 toys divided
as 15 ÷ 3 = 5 toys in each group, So, there are
15 toys in 3 equal groups.

Question 6.

9 ÷ 3 = ____3______

There are 3 cars in each group,

Explanation:
As given in the picture there are 9 cars divided
as 9 ÷ 3 = 3 cars in each group, So, there are
3 cars in 3 equal groups.

Question 7.
Audrina has 24 colored pencils. She puts them in
4 equal groups. How many colored pencils are in each group?

There are ___6____ colored pencils in each group.
24 ÷ 4 = ___6____

There are 6 colored pencils in each group,

Explanation:
Given Audrina has 24 colored pencils. She puts them in
4 equal groups. So number of  colored pencils in each group
are 24 ÷ 4 = 6 pencils in 4 equal groups.

Question 8.
Charlie picks 20 apples. He divides them equally
between 5 baskets. Draw the apples in each basket.

There are _____4______ apples in each basket.
20 ÷ ____5____ = ____4______

There are 4 apples in each basket,

Explanation:
Given Charlie picks 20 apples. He divides them equally
between 5 baskets. Drawn the apples in each basket as
20 ÷ 5 = 4 apples in 5 equal groups.

Question 9.
Chelsea collects butterfly stickers. The picture shows
how she placed them in her book. Write a division sentence
to show how she equally grouped her stickers.
There are ______3______ butterflies in each row.
____15______ ÷ _____5_____ = ____3______

Division sentence : 15 ÷ 5 = 3,
Chelsea equally grouped 3 butterflies
in her stickers.

Explanation:
Given Chelsea collects butterfly stickers.
The picture is showing 15 butterflies she placed
them in her book. Wrote a division sentence as
15 ÷ 5 = 3 butterflies to show how she equally
grouped 3 butterflies in her stickers.

Eureka Math Grade 3 Module 1 Lesson 4 Exit Ticket Answer Key

Question 1.
There are 16 glue sticks for the class. The teacher divides them into 4 equal groups. Draw the number of glue sticks in each group.

There are _____16______ glue sticks in each group.
16 ÷ ___4_____ = ____4______

There are 16 glue sticks in each group.

Explanation:
Given there are 16 glue sticks for the class.
The teacher divides them into 4 equal groups.
Drawn the number of glue sticks in each group as
16 ÷ 4 =  4 glue sticks in 4 equal groups.

Question 2.
Draw a picture to show 15 ÷ 3. Then, fill in the blank to make a true division sentence.
15 ÷ 3 = ____5______

Drawn a picture to show division sentence as
15 ÷ 3 = 5,
Filled in the blank to make a true division sentence as
15 ÷ 3 = ____5____,

Explanation:
Drawn 15 dogs and wrote division sentence as
15 ÷ 3 = 5 as shown above and filled in the blank to
make a true division sentence as 15 ÷ 3 = ____5__ or 5 X 3 = 15.

Eureka Math Grade 3 Module 1 Lesson 4 Homework Answer Key

Question 1.

12 chairs are divided into 2 equal groups.
There are ____6_____ chairs in each group.

There are 6 chairs in each group,

Explanation:
As given in the picture there are 12 chairs divided
as 12 ÷ 2 = 6 chairs in each group, So, there are
6 chairs in 2 equal groups.

Question 2.

21 triangles are divided into 3 equal groups.
There are ____7_____ triangles in each group.

There are 7 triangles in each group,

Explanation:
As given in the picture there are 21 triangles divided
as 21 ÷ 3 =7 triangles in each group, So, there are
7 triangles in 3 equal groups.

Question 3.

25 erasers are divided into ___5___ equal groups.
There are ____5_____ erasers in each group.

25 erasers are divided into 5 equal groups.
There are 5 erasers in each group as 25 ÷ 5 = 5 erasers,

Explanation:
As given in the picture there are 25 erasers divided
into 5 equal groups, There are 5 erasers in each group
as 25 ÷ 5 = 25 erasers.

Question 4.

___9____ chickens are divided into ___3____ equal groups.
There are ____3_____ chickens in each group.
9 ÷ 3 = ____3______

9 chickens are divided into 3 equal groups.
There are 3 chickens in each group as 9 ÷ 3 = 3 chickens,

Explanation:
As given in the picture there are 9 chickens divided
into 3 equal groups, There are 3 chickens in each group
as 9 ÷ 3 = 3 chickens.

Question 5.

There are ____3_____ buckets in each group.
12 ÷ 4 = ____3____

12 buckets are divided into 4 equal groups.
There are 3 buckets in each group as 12 ÷ 4 = 3 buckets,

Explanation:
As given in the picture there are 12 buckets divided
into 4 equal groups, There are 3 buckets in each group
as 12 ÷ 4 = 3 buckets.

Question 6.

16 ÷ 4 = _4_

16 bricks are divided into 4 equal groups.
There are 4 bricks in each group as 16 ÷ 4 = 4 bricks,

Explanation:
As given in the picture there are 16 bricks divided
into 4 equal groups, There are 4 bricks in each group
as 16 ÷ 4 = 4 bricks.

Question 7.
Andrew has 21 keys. He puts them in 3 equal groups.
How many keys are in each group?

There are ___7____ keys in each group.
21 ÷ 3 = ____7______

21 keys are divided into 3 equal groups.
There are 7 keys in each group.

Explanation:
Given Andrew has 21 keys. He puts them in 3
equal groups. So, number of keys in each group
are 7 keys as 21 ÷ 3 = 7 keys in each group.

Question 8.
Mr. Doyle has 20 pencils. He divides them equally
between 4 tables. Draw the pencils on each table.

There are _____5_____ pencils on each table.
20 ÷ ___4_____ = ____5______

There are 5 pencils on each table,

Explanation:
Given Mr. Doyle has 20 pencils and he divides them equally
between 4 tables. Drawn the pencils on each table as
20 ÷ 4= 5 pencils on each table as shown above.

Question 9.
Jenna has markers. The picture shows how she placed
them on her desk. Write a division sentence to represent
how she equally grouped her markers.
There are ______4______ markers in each row.

____20______ ÷ ___5_______ = ____4____

Division sentence to represent Jenna is
20 ÷ 5 = 4 markers equally grouped in each row,

Explanation:
Given Jenna has markers and the picture shows
how she placed them on her desk. Wrote a
division sentence 20 ÷ 5 = 4 markers to represent
how Jenna equally grouped 4 markers in each row.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 10 Answer Key

Eureka Math Grade 3 Module 7 Lesson 10 Pattern Sheet Answer Key

Multiply

multiply by 7 (1–5)
Answer:

 

Explanation:
7 × 1 = 7
7 × 2 = 14
7 × 3 = 21
7 × 4 = 28
7 × 5 = 35.

Eureka Math Grade 3 Module 7 Lesson 10 Problem Set Answer Key

Question 1.
Use a 2-inch square to answer the questions below.
a. Trace the square in the space below with a red crayon.

Answer:

Explanation:
ABCD is a Square of having side as 2cm each.

 

b. Trace the new shape you made with the square in the space below with a red crayon.
Answer:

Explanation:
ABCD is a square transformed into a new shape with red crayon.

 

c. Which shape has a greater perimeter? How do you know?
Answer:
My new shape has a greater perimeter than the square because when I used my string to mark the both perimeters of the shapes, new shape was down the string.

Explanation:
My new shape has a greater perimeter than the square because when I used my string to mark the both perimeters of the shapes, new shape was down the string. So, new shape perimeter is greater than the square.

d. Color the inside of the shapes in Problem 1 (a) and (b) with a blue crayon.
Answer:

Explanation:
First, shape is square.
Second, shape is the new shape.

 

e. Which color represents the perimeters of the shapes? How do you know?
Answer:
Red color represents the Perimeter of the square.

Explanation:
The perimeter of a square is the length that its boundary covers. The perimeter of a square is obtained by adding all the sides together.

f. What does the other color represent? How do you know?
Answer:
Blue color represents the Area of the square.
Red color represents the Perimeter of the square.

Explanation:
The area is defined as the region occupied inside the boundary of a flat object or 2d figure.
Area of the square = Side × Side.
The perimeter of a square is the length that its boundary covers. The perimeter of a square is obtained by adding all the sides together.
Perimeter of the square = 4 × side.

g. Which shape has a greater area? How do you know?
Answer:
Neither, shape has greater area because the look of the shape of the square has been changed not the area.

Explanation:
Neither shapes have the greater area among themselves. the look of the square has just changed yet not the area.

 

Question 2.
a. Outline the perimeter of the shapes below with a red crayon.

Answer:

Explanation:
Rhombus ABCD, Triangle EFG, Parallelogram HIJK shapes are outlined the perimeter with red crayon.

b. Explain how you know you outlined the perimeters of the shapes above.
Answer:
The perimeter of any figure is the length that its boundary covers.

Explanation:

Question 3.
Outline the perimeter of this piece of paper with a highlighter.
Answer:

Explanation:
The Perimeter of this all figures is highlighted with a light blue crayon.

Eureka Math Grade 3 Module 7 Lesson 10 Exit Ticket Answer Key

Jason paints the outside edges of a rectangle purple. Celeste paints the inside of the rectangle yellow.
Question 1.
Use your crayons to color the rectangle that Jason and Celeste painted.

Answer:

Explanation:
ABCD is a rectangle.
Jason paints the outside edges of a rectangle purple. Celeste paints the inside of the rectangle yellow.

Question 2.
Which color represents the perimeter of the rectangle? How do you know?
Answer:
Purple Color represents the perimeter of the rectangle because it is the boundary color of the rectangle.

Explanation:

The perimeter of any figure is the length that its boundary covers.

Eureka Math Grade 3 Module 7 Lesson 10 Homework Answer Key

Question 1.
Trace the perimeter of the shapes below.

a. Explain how you know you traced the perimeters of the shapes above.

Answer:
The perimeter of any figure is the length that its boundary covers. The color used to trace the perimeter of the figures is Red color.

Explanation:

b. Explain how you could use a string to figure out which shape above has the greatest perimeter.
Answer:
Perimeter of any figure is calculated by adding its all sides. Place the string on all the sides of each figure and later add all sides of each figure to calculate the Perimeter of the figures to know which above figure as the greatest perimeter among all.

Explanation:
The perimeter of any figure is the length that its boundary covers.
Perimeter of any figure is calculated by adding its all sides.

Question 2.
Draw a rectangle on the grid below.

a. Trace the perimeter of the rectangle.

Answer:

Explanation:
ABCD is the rectangle drawn in the grid.
The perimeter of any figure is the length that its boundary covers.

b. Shade the area of the rectangle.

Answer:

Explanation:
The area is defined as the region occupied inside the boundary of a flat object or 2d figure.
Area of the rectangle is the inside part, which is shaded with yellow color.

c. How is the perimeter of the rectangle different from the area of the rectangle?
Answer:
The perimeter and the Area of the rectangle are two different topics.
They have different units for perimeter and area. The perimeter has the same units as its the length of the sides whereas the area’s unit are squared.

Explanation:
The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle, you have to add the lengths of all the four sides.
Perimeter of the rectangle = 2(Length + Width) = 2(L + W)

The area is measurement of the surface of a shape. To find the area of a rectangle  you need to multiply the length and the width of a rectangle.
Area of the rectangle = Length × Width = L × W.

 

Question 3.
Maya draws the shape shown below. Noah colors the inside of Maya’s shape as shown. Noah says he colored the perimeter of Maya’s shape. Maya says Noah colored the area of her shape. Who is right? Explain your answer.

Answer:
Maya is  correct not Noah because she colored inside part of the figure, which is area not perimeter.

Explanation:
The area is defined as the region occupied inside the boundary of a flat object or 2d figure.
Area of the figure is the inside color, which Noah colored.
The perimeter of a figure is the length that its boundary covers.
Perimeter of the figure is the boundaries.

 

Engage NY Eureka Math 3rd Grade Module 7 Lesson 9 Answer Key

Eureka Math Grade 3 Module 7 Lesson 9 Pattern Sheet Answer Key

Multiply.

multiply by 6 (6─10)
Answer:

Explanation:
6 × 5 = 30
6 × 6 = 36
6 × 7 = 42
6 × 8 = 48
6 × 9 = 54
6 × 10 = 60

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

Question 1.
Use at least two tangram pieces to make and draw two of each of the following shapes. Draw lines to show where the tangram pieces meet.
a. A rectangle that does not have all equal sides.
Answer:

Explanation:
ABCD is an rectangle of different length size sides, folded half pointing EF.
After folding, they meet at BGC.

 

b. A triangle.
Answer:

Explanation:
ABC is a triangle, folded half pointing D.
After the folding EF is where they meet.

c. A parallelogram.
Answer:

Explanation:
ABCD is a parallelogram, half folded.
After folding, at BE and FD they meet.

d. A trapezoid.
Answer:

Explanation:
ABCD is a trapezium, folded at CE.
After the folding, they meet at BF and CG.

Question 2.
Use your two smallest triangles to create a square, a parallelogram, and a triangle. Show how you created them below.
Answer:

Explanation:
ABCD  is a square formed by joining two small triangles.
EFGH is a parallelogram formed by joining two small triangles.
IJK is a triangle formed by joining two small triangles.

Question 3.
Create your own shape on a separate sheet of paper using all seven pieces. Describe its attributes below.
Answer:

Explanation:
The figure which I have drawn using the seven pieces is having six sides known as a Hexagon. It has a pair of parallel sides. My figure does not have any right angles in it. It is not a regular hexagon because it does not have any equal sides in it.

Question 4.
Trade your outline with a partner to see if you can re-create her shape using your tangram pieces. Reflect on your experience below. What was easy? What was challenging?
Answer:

Explanation:
My partner has recreated the figure of mine into a trapezium. I have found it to be easy in identifying the figure by its outer size. It was challenging for me to identify how the pieces are kept to figure out the trapezium.

Eureka Math Grade 3 Module 7 Lesson 9 Exit Ticket Answer Key

Nancy uses her tangram pieces to make a trapezoid without using the square piece. Below, sketch how she might have created her trapezoid.
Answer:

Explanation:
Nancy uses her tangram pieces to make a trapezoid without using the square piece. She uses her five triangles to make a shape that has four sides with different size of sides, named as trapezium ADEG.

Eureka Math Grade 3 Module 7 Lesson 9 Homework Answer Key

Question 1.
Use at least two tangram pieces to make and draw each of the following shapes. Draw lines to show where the tangram pieces meet.
a. A triangle.
Answer:

Explanation:
ABC is a triangle formed by joining two triangle and they meet at CD.

b. A square.
Answer:

Explanation:
ABCD is a square, formed by joining two triangles joining at BD.

c. A parallelogram.
Answer:

Explanation:
ABCD is a parallelogram, formed by joining two triangles meeting at BD.

d. A trapezoid.
Answer:

Explanation:
ABCD is a trapezium formed by joining two triangles meeting at AC.

 

Question 2.
Use your tangram pieces to create the cat below. Draw lines to show where the tangram pieces meet.

Answer:

Explanation:
ABGE and  FQDC are two rectangles joined together to form the shape of the cat, they meet at B.
CHIP  Trapezium and PGKL  rectangle and RJK triangle are used to form the shape of the body, meeting at P.
NMOL is a combination of two triangle used together to form the shape of the cat’s tail meeting at OM.

Question 3.
Use the five smallest tangram pieces to make a square. Sketch your square below, and draw lines to show where the tangram pieces meet.
Answer:

Explanation:
ABCD is a square formed by combining five different sizes of triangles, meeting at A,B,C,D.

 

Engage NY Eureka Math 3rd Grade Module 1 Lesson 2 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 2 Sprint Answer Key

A
Add or Subtract Using 2

Question 1.
0 + 2 =
0 + 2 = 2,

Explanation:
Given 0 + 2 we add 0 with 2,
we get 2 as 0 + 2 = 2.

Question 2.
2 + 2 =
2 + 2 = 4,

Explanation:
Given 2 + 2 we add 2 with 2,
we get 4 as 2 + 2 = 4.

Question 3.
4 + 2 =
4 + 2 = 6,

Explanation:
Given 4 + 2 we add 4 with 2,
we get 6 as 4 + 2 = 6.

Question 4.
6 + 2 =
6 + 2 = 8,

Explanation:
Given 6 + 2 we add 6 with 2,
we get 8 as 6 + 2 = 8.

Question 5.
8 + 2 =
8 + 2 = 10,

Explanation:
Given 8 + 2 we add 8 with 2,
we get 10 as 8 + 2 = 10.

Question 6.
10 + 2 =
10 + 2 = 12,

Explanation:
Given 10 + 2 we add 10 with 2,
we get 12 as 10 + 2 = 12.

Question 7.
12 + 2 =
12 + 2 = 14,

Explanation:
Given 12 + 2 we add 12 with 2,
we get 14 as 12 + 2 = 14.

Question 8.
14 + 2 =
14 + 2 = 16,

Explanation:
Given 14 + 2 we add 14 with 2,
we get 16 as 14 + 2 = 16.

Question 9.
16 + 2 =
16 + 2 = 18,

Explanation:
Given 16 + 2 we add 16 with 2,
we get 18 as 16 + 2 = 18.

Question 10.
18 + 2 =
18 + 2 = 20,

Explanation:
Given 18 + 2 we add 18 with 2,
we get 20 as 18 + 2 = 20.

Question 11.
20 – 2 =
20 – 2 = 18,

Explanation:
Given 20 – 2 we subtract 2 from 20,
we get 18 as 20 – 2 = 18.

Question 12.
18 – 2 =
18 – 2 = 16,

Explanation:
Given 18 – 2 we subtract 2 from 18,
we get 16 as 18 – 2 = 16.

Question 13.
16 – 2 =
16 – 2 = 14,

Explanation:
Given 16 – 2 we subtract 2 from 16,
we get 14 as 16 – 2 = 14.

Question 14.
14 – 2 =
14 – 2 = 12,

Explanation:
Given 14 – 2 we subtract 2 from 14,
we get 12 as 14 – 2 = 12.

Question 15.
12 – 2 =
12 – 2 = 10,

Explanation:
Given 12 – 2 we subtract 2 from 12,
we get 10 as 12 – 2 = 10.

Question 16.
10 – 2 =
10 – 2 = 8,

Explanation:
Given 10 – 2 we subtract 2 from 10,
we get 8 as 10 – 2 = 8.

Question 17.
8 – 2 =
8 – 2 = 6,

Explanation:
Given 8 – 2 we subtract 2 from 8,
we get 6 as 8 – 2 = 6.

Question 18.
6 – 2 =
6 – 2 = 4,

Explanation:
Given 6 – 2 we subtract 2 from 6,
we get 4 as 6 – 2 = 4.

Question 19.
4 – 2 =
4 – 2 = 2,

Explanation:
Given 4 – 2 we subtract 2 from 4,
we get 2 as 4 – 2 = 2.

Question 20.
2 – 2 =
2 – 2 = 0,

Explanation:
Given 2 – 2 we subtract 2 from 2,
we get 0 as 2 – 2 = 0.

Question 21.
2 + 0 =
2 + 0 = 2,

Explanation:
Given 2 + 0 we add 2 with 0,
we get 2 as 2 + 0 = 2.

Question 22.
2 + 2 =
2 + 2 = 4,

Explanation:
Given 2 + 2 we add 2 with 2,
we get 4 as 2 + 2 = 4.

Question 23.
2 + 4 =
2 + 4 = 6,

Explanation:
Given 2 + 4 we add 2 with 4,
we get 6 as 2 + 4 = 6.

Question 24.
2 + 6 =
2 + 6 = 8,

Explanation:
Given 2 + 6 we add 2 with 6,
we get 8 as 2 + 6 = 8.

Question 25.
2 + 8 =
2 + 8 = 10,

Explanation:
Given 2 + 8 we add 2 with 8,
we get 10 as 2 + 8 = 10.

Question 26.
2 + 10 =
2 + 10 = 12,

Explanation:
Given 2 + 10 we add 2 with 10,
we get 12 as 2 + 10 = 12.

Question 27.
2 + 12 =
2 + 12 = 14,

Explanation:
Given 2 + 12 we add 2 with 12,
we get 14 as 2 + 12 =14.

Question 28.
2 + 14 =
2 + 14 = 16,

Explanation:
Given 2 + 14 we add 2 with 14,
we get 16 as 2 + 14 =16.

Question 29.
2 + 16 =
2 + 16 = 18,

Explanation:
Given 2 + 16 we add 2 with 16,
we get 16 as 2 + 16 =18.

Question 30.
2 + 18 =
2 + 18 = 20,

Explanation:
Given 2 + 18 we add 2 with 18,
we get 20 as 2 + 18 =20.

Question 31.
0 + 22 =
0 + 22 = 22,

Explanation:
Given 0 + 22 we add 0 with 22,
we get 22 as 0 + 22 =22.

Question 32.
22 + 22 =
22 + 22 = 44,

Explanation:
Given 22 + 22 we add 22 with 22,
we get 44 as 22 + 22 =44.

Question 33.
44 + 22 =
44 + 22 = 66,

Explanation:
Given  44 + 22 we add 44 with 22,
we get 66 as 44 + 22 =66.

Question 34.
66 + 22 =
66 + 22 = 88,

Explanation:
Given 66 + 22 we add 66 with 22,
we get 88 as 66 + 22 = 88.

Question 35.
88 – 22 =
88 – 22 = 66,

Explanation:
Given 88 – 22 we subtract 22 from 88,
we get 66 as 88 – 22 = 66.

Question 36.
66 – 22 =
66 – 22 = 44,

Explanation:
Given 66 – 22 we subtract 22 from 66,
we get 44 as 66 – 22 = 44.

Question 37.
44 – 22 =
44 – 22 = 22,

Explanation:
Given 44 – 22 we subtract 22 from 44,
we get 22 as 44 – 22 = 22.

Question 38.
22 – 22 =
22 – 22 = 0,

Explanation:
Given 22 – 22 we subtract 22 from 22,
we get 0 as 22 – 22 = 0.

Question 39.
22 + 0 =
22 + 0 = 22,

Explanation:
Given 22 + 0 we add 22 with 0,
we get 22 as 22 + 0 = 22.

Question 40.
22 + 22 =
22 + 22 = 44,

Explanation:
Given 22 + 22 we add 22 with 22,
we get 44 as 22 + 22 =44.

Question 41.
22 + 44 =
22 + 44 = 66,

Explanation:
Given 22 + 44 we add 22 with 44,
we get 66 as 22 + 44 = 66.

Question 42.
66 + 22 =
66 + 22 = 88,

Explanation:
Given 66 + 22 we add 66 with 22,
we get 88 as 66 + 22 = 88.

Question 43.
888 – 222 =
888 – 222 = 666,

Explanation:
Given 888 – 222 we subtract 222 from 888,
we get 666 as 888 – 222 = 666.

Question 44.
666 – 222 =
666 – 222 = 444,

Explanation:
Given 666 – 222 we subtract 222 from 666,
we get 444 as 666 – 222 = 444.

B
Add or Subtract Using 2

Question 1.
2 + 0 =
2 + 0 = 2,

Explanation:
Given 2 + 0 we add 2 with 0,
we get 2 as 2 + 0 = 2.

Question 2.
2 + 2 =
2 + 2 = 4,

Explanation:
Given 2 + 2 we add 2 with 2,
we get 4 as 2 + 2 = 4.

Question 3.
2 + 4 =
2 + 4 = 6,

Explanation:
Given 2 + 4 we add 2 with 4,
we get 6 as 2 + 4 = 6.

Question 4.
2 + 6 =
2 + 6 = 8,

Explanation:
Given 2 + 6 we add 2 with 6,
we get 8 as 2 + 6 = 8.

Question 5.
2 + 8 =
2 + 8 = 10,

Explanation:
Given 2 + 8 we add 2 with 8,
we get 10 as 2 + 8 = 10.

Question 6.
2 + 10 =
2 + 10 = 12,

Explanation:
Given 2 + 10 we add 2 with 10,
we get 12 as 2 + 10 = 12.

Question 7.
2 + 12 =
2 + 12 = 14,

Explanation:
Given 2 + 12 we add 2 with 12,
we get 14 as 2 + 12 =14.

Question 8.
2 + 14 =
2 + 14 = 16,

Explanation:
Given 2 + 14 we add 2 with 14,
we get 16 as 2 + 14 =16.

Question 9.
2 + 16 =
2 + 16 = 18,

Explanation:
Given 2 + 16 we add 2 with 16,
we get 16 as 2 + 16 =18.

Question 10.
2 + 18 =
2 + 18 = 20,

Explanation:
Given 2 + 18 we add 2 with 18,
we get 20 as 2 + 18 =20.

Question 11.
20 – 2 =
20 – 2 = 18,

Explanation:
Given 20 – 2 we subtract 2 from 20,
we get 18 as 20 – 2 = 18.

Question 12.
18 – 2 =
18 – 2 = 16,

Explanation:
Given 18 – 2 we subtract 2 from 18,
we get 16 as 18 – 2 = 16.

Question 13.
16 – 2 =
16 – 2 = 14,

Explanation:
Given 16 – 2 we subtract 2 from 16,
we get 14 as 16 – 2 = 14.

Question 14.
14 – 2 =
14 – 2 = 12,

Explanation:
Given 14 – 2 we subtract 2 from 14,
we get 12 as 14 – 2 = 12.

Question 15.
12 – 2 =
12 – 2 = 10,

Explanation:
Given 12 – 2 we subtract 2 from 12,
we get 10 as 12 – 2 = 10.

Question 16.
10 – 2 =
10 – 2 = 8,

Explanation:
Given 10 – 2 we subtract 2 from 10,
we get 8 as 10 – 2 = 8.

Question 17.
8 – 2 =
8 – 2 = 6,

Explanation:
Given 8 – 2 we subtract 2 from 8,
we get 6 as 8 – 2 = 6.

Question 18.
6 – 2 =
6 – 2 = 4,

Explanation:
Given 6 – 2 we subtract 2 from 6,
we get 4 as 6 – 2 = 4.

Question 19.
4 – 2 =
4 – 2 = 2,

Explanation:
Given 4 – 2 we subtract 2 from 4,
we get 2 as 4 – 2 = 2.

Question 20.
2 – 2 =
2 – 2 = 0,

Explanation:
Given 2 – 2 we subtract 2 from 2,
we get 0 as 2 – 2 = 0.

Question 21.
0 + 2 =
0 + 2 = 2,
Explanation:
Given 0 + 2 we add 0 with 2,
we get 2 as 0 + 2 = 2.

Question 22.
2 + 2 =
2 + 2 = 4,
Explanation:
Given 2 + 2 we add 2 with 2,
we get 4 as 2 + 2 = 4.

Question 23.
4 + 2 =
4 + 2 = 6,
Explanation:
Given 4 + 2 we add 4 with 2,
we get 6 as 4 + 2 = 6.

Question 24.
6 + 2 =
6 + 2 = 8,

Explanation:
Given 6 + 2 we add 6 with 2,
we get 8 as 6 + 2 = 8.

Question 25.
8 + 2 =
8 + 2 = 10,
Explanation:
Given 8 + 2 we add 8 with 2,
we get 10 as 8 + 2 = 10.

Question 26.
10 + 2 =
10 + 2 = 12,
Explanation:
Given 10 + 2 we add 10 with 2,
we get 12 as 10 + 2 = 12.

Question 27.
12 + 2 =
12 + 2 = 14,
Explanation:
Given 12 + 2 we add 12 with 2,
we get 14 as 12 + 2 = 14.

Question 28.
14 + 2 =
14 + 2 = 16,
Explanation:
Given 14 + 2 we add 14 with 2,
we get 16 as 14 + 2 = 16.

Question 29.
16 + 2 =
16 + 2 = 18,
Explanation:
Given 16 + 2 we add 16 with 2,
we get 18 as 16 + 2 = 18.

Question 30.
18 + 2 =
18 + 2 = 20,
Explanation:
Given 18 + 2 we add 18 with 2,
we get 20 as 18 + 2 = 20.

Question 31.
0 + 22 =
0 + 22 = 22,
Explanation:
Given 0 + 22 we add 0 with 22,
we get 22 as 0 + 22 =22.

Question 32.
22 + 22 =
22 + 22 = 44,

Explanation:
Given 22 + 22 we add 22 with 22,
we get 44 as 22 + 22 =44.

Question 33.
22 + 44 =
22 + 44 = 66,

Explanation:
Given 22 + 44 we add 22 with 44,
we get 66 as 22 + 44 = 66.

Question 34.
66 + 22 =
66 + 22 = 88,

Explanation:
Given 66 + 22 we add 66 with 22,
we get 88 as 66 + 22 = 88.

Question 35.
88 – 22 =
88 – 22 = 66,

Explanation:
Given 88 – 22 we subtract 22 from 88,
we get 66 as 88 – 22 = 66.

Question 36.
66 – 22 =
66 – 22 = 44,

Explanation:
Given 66 – 22 we subtract 22 from 66,
we get 44 as 66 – 22 = 44.

Question 37.
44 – 22 =
44 – 22 = 22,
Explanation:
Given 44 – 22 we subtract 22 from 44,
we get 22 as 44 – 22 = 22.

Question 38.
22 – 22 =
22 – 22 = 0,

Explanation:
Given 22 – 22 we subtract 22 from 22,
we get 0 as 22 – 22 = 0.

Question 39.
22 + 0 =
22 + 0 = 22,

Explanation:
Given 22 + 0 we add 22 with 0,
we get 22 as 22 + 0 = 22.

Question 40.
22 + 22 =
22 + 22 = 44,
Explanation:
Given 22 + 22 we add 22 with 22,
we get 44 as 22 + 22 =44.

Question 41.
22 + 44 =
22 + 44 = 44,

Explanation:
Given 22 + 44 we add 22 with 44,
we get 66 as 22 + 44 = 66.

Question 42.
66 + 22 =
66 + 22 = 88,

Explanation:
Given 66 + 22 we add 66 with 22,
we get 88 as 66 + 22 = 88.

Question 43.
666 – 222 =
666 – 222 = 444,

Explanation:
Given 666 – 222 we subtract 222 from 666,
we get 444 as 666 – 222 = 444.

Question 44.
888 – 222 =
888 – 222 = 666,

Explanation:
Given 888 – 222 we subtract 222 from 888,
we get 666 as 888 – 222 = 666.

Eureka Math Grade 3 Module 1 Lesson 2 Problem Set Answer Key

Use the arrays below to answer each set of questions.

Question 1.
a. How many rows of cars are there? ___4_______
b. How many cars are there in each row? ___2_______

a. There are 4 number of rows of cars,

Explanation:
As shown in the picture there are
4 number of rows of cars.

b. There are 2 cars are there in each row,

Explanation:
As shown in the picture there are
2 cars are there in each row.

Question 2.
a. What is the number of rows? ____3______
b. What is the number of objects in each row? _____6_____

a. There are 3 number of rows of objects,

Explanation:
As shown in the picture there are
3 number of rows of objects.

b. There are 6 objects are there in each row,

Explanation:
As shown in the picture there are
6 objects are there in each row.

Question 3.
a. There are 4 spoons in each row. How many spoons are in 2 rows? ____8______
b. Write a multiplication expression to describe the array.
_____4 x 2 = 8___

a. There are 8 spoons in 2 rows,

Explanation:
Given there are 4 spoons in each row.
Number of spoons in 2 rows are 4 X 2 = 8 spoons,

b. Multiplication expression : 4 X 2 = 8 spoons,

Explanation:
Given there are 4 spoons in each row.
Multiplication expression to describe the array is
4 X 2 = 8.

Question 4.
a. There are 5 rows of triangles.
How many triangles are in each row? ___4______
b. Write a multiplication expression to describe the total number of triangles. _____5 x 4 = 20 _________________

a. There are 4 triangles in each row,

Explanation:
Given there are 5 rows of triangles,
As shown in the picture there are 4
number of triangles in each row.

b. Multiplication expression : 5 X 4 = 20 triangles,

Explanation:
Given there are 5 rows of triangles,
Wrote a multiplication expression to
describe the total number of triangles as
5 X 4 = 20 triangles.

Question 5.
The dots below show 2 groups of 5.
a. Redraw the dots as an array that shows 2 rows of 5.

b. Compare the drawing to your array.
Write at least 1 reason why they are the same and
1 reason why they are different.

Redrawn the dots as an array that shows 2 rows of 5.

Explanation:
Given the dots below show 2 groups of 5.
So, redrawn the dots as an array that shows
2 rows of 5 as shown above in the picture.

b. Same : Both have the same number of dots,
Different : Drawing is not in any order but my array
has an order of 2 rows of 5,

Explanation:
Comparing the drawing to my array.
Wrote at least 1 reason why they are the same and
1 reason why they are different as
Same : Both have the same number of dots,
Different : Drawing is not in some order but my array
has an order of 2 rows of 5.

Question 6.
Emma collects rocks. She arranges them in 4 rows of 3.
Draw Emma’s array to show how many rocks
she has altogether. Then, write a multiplication
equation to describe the array.

Emma has 12 number of rocks altogether,
Multiplication equation : 4 X 3 = 12 rocks,

Explanation:
Given Emma collects rocks. She arranges them in 4 rows of 3.
Drawn Emma’s array to show how many rocks
she has altogether as shown above in the picture,
Then, wrote a multiplication equation to describe
the array as multiplication equation : 4 X 3 = 12 rocks.

Question 7.
Joshua organizes cans of food into an array.
He thinks, “My cans show 5 × 3!” Draw Joshua’s
array to find the total number of cans he organizes.

The total number of cans Joshua organizes are 15,

Explanation:
Given Joshua organizes cans of food into an array.
He thinks, “My cans show 5 × 3!” Drawn Joshua’s
array to find the total number of cans he organizes is
5 X 3 = 15 food cans.

Eureka Math Grade 3 Module 1 Lesson 2 Exit Ticket Answer Key

Question 1.
a. There are 4 rows of stars. How many
stars are in each row? _____3_____
b. Write a multiplication equation to
describe the array. ______4 x 3= 12_________

a. There are 3 number of stars in each row,

Explanation:
As shown in the picture there are
3 number of stars in each row.

b. Multiplication equation : 4 X 3 = 12,

Explanation:
As shown in the picture multiplication equation to
describe the array is 4 X 3 = 12 stars.

Question 2.
Judy collects seashells. She arranges them in 3 rows of 6.
Draw Judy’s array to show how many seashells
she has altogether. Then, write a multiplication equation
to describe the array.

Judy’s has 18 seashells altogether,
Multiplication Equation : 3 X 6 = 18,

Explanation:
Given Judy collects seashells. She arranges them
in 3 rows of 6.
Drawn Judy’s array to show 18 seashells
she has altogether. Then, wrote a multiplication equation
to describe the array as 3 X 6 = 18 seashells.

Eureka Math Grade 3 Module 1 Lesson 2 Homework Answer Key

Use the arrays below to answer each set of questions.

Question 1.
a. How many rows of erasers are there? ____3______
b. How many erasers are there in each row? ____2______

a. There are 3 rows of erasers,

Explanation:
In the given picture there are 3 rows are erasers.

b. There are 2 erasers in each row,

Explanation:
In the given picture there are 2 erasers in each row.

Question 2.
a. What is the number of rows? _____4_____
b. What is the number of objects in each row? ____3______

a. There are 4 rows of ducks,

Explanation:
In the given picture there are 4 rows of ducks.

b. There are 3 objects in each row,

Explanation:
In the given picture there are 3 objects in each row.

Question 3.
a. There are 3 squares in each row. How many squares are in 5 rows? ____15____
b. Write a multiplication expression to describe the array.
_5 X 3 = 15___

a. There are 15 squares in 5 rows,

Explanation:
Given there are 3 squares in each row,
So, number of squares in 5 rows are 5 X 3 = 15 squares.

b. Multiplication expression : 5 X 3 = 15,

Explanation:
Given there are 3 squares in each row,
So, multiplication expression to describe the array is
5 X 3 = 15 squares.

Question 4.
a. There are 6 rows of stars. How many stars are
in each row? ____4______
b. Write a multiplication expression to describe
the array. ____6 x 4 = 24______

a. There are 4 stars in each row,

Explanation:
In the picture there are 6 rows of stars and
in each row there are 4 stars.

b. Multiplication expression : 6 X 4 = 24,

Explanation:
In the picture there are 6 rows and each row
has 4 stars, So, multiplication expression to describe
the array is 6 X 4 = 24 stars.

Question 5.
The triangles below show 3 groups of four.
a. Redraw the triangles as an array that shows 3 rows of four.
b. Compare the drawing to your array. How are they the same? How are they different?

a.

Shown triangles as an array that shows 3 rows of four.

Explanation:
Given the triangles as 3 groups of four.
a. Redrawn the triangles as an array that
shows 3 rows of four as shown in the picture above.

b. Same : Both have same number of triangles,
Different : Drawing is not in any order but my array
has an order of 3 rows of 4.

Question 6.
Roger has a collection of stamps.
He arranges the stamps into 5 rows of four.
Draw an array to represent Roger’s stamps.
Then, write a multiplication equation to
describe the array.

Drawn an array to represent Roger’s stamps.
Multiplication equation to describe the array is 5 X 4 = 20 stamps,

Explanation:
Given Roger has a collection of stamps.
He arranges the stamps into 5 rows of four.
Drawn an array to represent Roger’s stamps
as shown above then wrote a multiplication equation
to describe the array as 5 X 4 = 20 stamps.

Question 7.
Kimberly arranges her 18 markers as an array.
Draw an array that Kimberly might make.
Then, write a multiplication equation to describe your array.

Drawn an array that Kimberly might make,
Multiplication equation to describe the array is 3 X 6 = 18,

Explanation:
Given Kimberly arranges her 18 markers as an array,
may be of 3 rows of 6 markers each, Drawn an array
that Kimberly might make as shown in the picture above,
Then, wrote a multiplication equation to describe
my array as 3 X 6 = 18 markers.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 8 Answer Key

Eureka Math Grade 3 Module 7 Lesson 8 Pattern Sheet Answer Key

Multiply.

multiply by 6 (1─5)
Answer:

Explanation:
6 × 1 = 6
6 × 2 = 12
6 × 3 = 18
6 × 4 = 24
6 × 5 = 30.

Eureka Math Grade 3 Module 7 Lesson 8 Problem Set Answer Key

Question 1.
Fold and cut the square on the diagonal. Draw and label your 2 new shapes below.
Answer:

Explanation:
ABCD is a Square.
ABC is the half part of the square when folded and cut.
DEF  is the other half part of the square when folded and cut.

 

Question 2.
Fold and cut one of the triangles in half. Draw and label your 2 new shapes below.
Answer:

Explanation:
ABC is an Triangle.
ABD is the half part of of the Triangle when folded and cut.
CEF is the other half part of of the Triangle when folded and cut.

 

Question 3.
Fold twice, and cut your large triangle. Draw and label your 2 new shapes below.
Answer:

Explanation:
ABC Triangle is twice folded and cut into two halves.
ADEC is first half part of twice folded Triangle.
CFG is the second half part of twice folded Triangle.

 

Question 4.
Fold and cut your trapezoid in half. Draw and label your 2 new shapes below.
Answer:

Explanation:
ABCD is a Trapezium, folded and cut.
ABFE and CDHG are two parts of them formed.
We can call ABFE and CDHG  as Quadrilaterals.

 

Question 5.
Fold and cut one of your trapezoids. Draw and label your 2 new shapes below.
Answer:

Explanation:
GDCH is a half part of Trapezium.
When folded the Trapezium, we get a  GDIJ Square and a CHK Triangle  are formed.

 

Question 6.
Fold and cut your second trapezoid. Draw and label your 2 new shapes below.
Answer:

Explanation:
AEFB is the second Trapezium.
When the AEFB Trapezium is folded and cut, we get a AEOP Parallelogram  and a BFN Triangle .

Question 7.
Reconstruct the original square using the seven shapes.
a. Draw lines inside the square below to show how the shapes go together to form the square. The first one has been done for you.

Answer:

Explanation:
ABCD is the Square.
AC is the first line used to divide the Square.
EB is the second line drawn, AEB and BEC Triangles are formed.
GF is the third line drawn, GFC Triangle is formed.
GI is the fourth line drawn, DGI Triangle is formed.
IJ is the fifth line drawn, IJA Triangle is formed.
HE is the sixth line drawn, IJEH Rectangle  and HEFG Rectangle are formed.
We can reconstruct this seven shapes together to form the original square.

 

b. Describe the process of forming the square. What was easy, and what was challenging?
Answer:
I first put the two big triangles to form the Square, which was quiet easy. Later one by one shape I joined to put on the square, this process was challenging .

Explanation:
Firstly, dividing the square into two Triangle was the easy one. Later on forming other shapes which are smaller in size than the first step, was quiet difficult one because while reconstructing it back to original shape was little messy and challenging. Overall, it made me to go back to my problem and check the shapes in the list one by one, which finally gave me my original shape of square to rejoin with all seven different shapes.

 

Eureka Math Grade 3 Module 7 Lesson 8 Exit Ticket Answer Key

Choose three shapes from your tangram puzzle. Trace them below. Label the name of each shape, and describe at least one attribute that they have in common.
Answer:

Explanation:
ABJ, BJC, CIF, FED, KHG are the triangles in the ABCD square.
AEKG is a Parallelogram in the ABCD square.
KFIH is the Rectangle in the ABCD square.
They all are having one right angle in common in KHG, KFIH and AEKG shapes.

Eureka Math Grade 3 Module 7 Lesson 8 Homework Answer Key

Question 1.
Draw a line to divide the square below into 2 equal triangles.

Answer:

Explanation:
EFGH is a square.
EG is the line drawn in the square, making EFG and GHE two triangles into equal halves.

Question 2.
Draw a line to divide the triangle below into 2 equal, smaller triangles.

Answer:

Explanation:
ABC is the Triangle.
CD is the line drawn, which divides the triangle into two equal triangle ADC and BDC triangles.

Question 3.
Draw a line to divide the trapezoid below into 2 equal trapezoids.

Answer:

Explanation:
ABCD is the trapezoids
EF is the line drawn, dividing the trapezoids into two equal trapezoids AEFD and BEFC. trapezoids.

 

Question 4.
Draw 2 lines to divide the quadrilateral below into 4 equal triangles.

Answer:

Explanation:
EFGH is the quadrilateral.
Lines are drawn in the quadrilateral joining EG and HF center O, making EOF, FOG, GOH and HOE four equal triangles.

Question 5.
Draw 4 lines to divide the square below into 8 equal triangles.

Answer:

Explanation:
ABCD is a Square.
AC and BD are joined by drawing two lines.
EF and GH are another two lines drawn making center O.
AOH, HOB, BOF, FOC, COG, GOD, DOE, EOA  equal triangles are formed.

Question 6.
Describe the steps you took to divide the square in Problem 5 into 8 equal triangles.
Answer:
I have drawn two lines dividing the square into two equal triangles.
I  have later drawn two lines in between the two triangles formed.

Explanation:
I have drawn two lines dividing the square into two equal triangles.
ABC, ABD, BCD, CDA  Triangles are formed.
I  have later drawn two lines in between the two triangles formed.
AOH, HOB, BOF, FOC, COG, GOD, DOE, EOA eight equal triangles are formed having Center O.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 1 Answer Key

Eureka Math Grade 3 Module 7 Answer Key

Eureka Math 3 Module 7 Lesson 1 Pattern Sheet Answer Key

Multiply

multiply by 3(1 – 5)
Answer:

Explanation:
3 × 1 = 3
3 × 2 = 6
3  × 3 = 9
3 × 4 = 12
3  × 5 = 15.

Eureka Math Grade 3 Module 7 Lesson 1 Problem Set Answer Key

Lena’s family visits Little Tree Apple Orchard. Use the RDW process to solve the problems about Lena’s visit to the orchard. Use a letter to represent the unknown in each problem.

Question 1.
The sign below shows information about hayrides at the orchard.

a. Lena’s family buys 2 adult tickets and 2 child tickets for the hayride. How much does it cost Lena’s family to go on the hayride?
Answer:
Total cost of tickets for Lena’s family to go on the hayride = $22.

Explanation:
Cost of each adults tickets = $7
=> Cost of Two adults tickets = $7 × 2 = $14.
Cost of Each child ticket = $4
=> Cost of Two Child tickets = $4 × 2 = $8
Total cost of tickets for Lena’s family to go on the hayride = Cost of Two adults tickets + Cost of Two Child tickets
= $14 + $8
= $22.

b. Lena’s mom pays for the tickets with $5 bills. She receives $3 in change. How many $5 bills does Lena’s mom use to pay for the hayride?

Answer:
Number of $5 bills she uses to pay the bill = 5.

Explanation:
Total cost of tickets for Lena’s family to go on the hayride = $22
Amount of change she receives = $3
Total amount paid for tickets = Total cost of tickets for Lena’s family to go on the hayride + Amount of change she receives
= $22 + $3
= $25
Number of $5 bills she uses to pay the bill = Total amount paid for tickets  ÷ 5
= $25 ÷ 5
= 5.

c. Lena’s family wants to go on the fourth hayride of the day. It’s 11:38 now. How many minutes do they have to wait for the fourth hayride?

Answer:
Time left for Lena’s family to wait for their fourth hayride = 7 minutes.

Explanation:
Time of their hayride of the day now = 11:38
Leaves every 15minutes hayride starting from 11:00.
Number of hayride lens family going = Fourth of the day
=> Time of first hayride = 11:00 to 11:15
=> Time of Second hayride = 11:15 to 11.30
=> Time of Third hayride = 11:30 to 11:45
=> Time of Fourth hayride = 11:45 to 12:00.
Time left for Lena’s family to wait for their fourth hayride = Time of Fourth hayride  – Time of their hayride of the day now
= 11:45 – 11:38
= 7 minutes.

 

Question 2.
Lena picked 17 apples, and her brother picked 19. Lena’s mom has a pie recipe that requires 9 apples. How many pies can Mom make with the apples that Lena and her brother picked?
Answer:
Number of pies Mom can make with the apples that Lena and her brother picked = 4.

Explanation:
Number of Apples Lena picked = 17
Number of Apples Lena’s brother picked = 19
Number of Apples Lena’s mom needs to prepare a pie = 9
Number of Apples Lena and her brother picked = Number of Apples Lena picked + Number of Apples Lena’s brother picked
= 19 + 17
= 36
Number of pies Mom can make with the apples that Lena and her brother picked = Number of Apples Lena and her brother picked  ÷ Number of Apples Lena’s mom needs to prepare a pie
= 36 ÷ 9
= 4.

 

Question 3.
Lena’s dad gives the cashier $30 to pay for 6 liters of apple cider. The cashier gives him $6 in change. How much does each liter of apple cider cost?
Answer:
Cost  for the a liter of apple cider = $4.

Explanation:
Amount of money Lena’s dad gives to the cashier to pay for 6 liters of apple cider = $30
Amount of money Lena’s dad gets in change = $6
Cost  for the 6 liters of apple cider = Amount of money Lena’s dad gives to the cashier to pay for 6 liters of apple cider  – Amount of money Lena’s dad gets in change
= $30 – $6
= $24.
Number of liters of apple cider purchased = 6
Cost  for the a liter of apple cider = Cost  for the 6 liters of apple cider  ÷ Number of liters of apple cider purchased
= $24 ÷ 6
= $4.

 

Question 4.
The apple orchard has 152 apple trees. There are 88 trees with red apples. The rest of the trees have green apples. How many more trees have red apples than green apples?
Answer:
Number of Green apple trees = 64.
Number of Red apples more than Green apples = 24.

Explanation:
Number of Apple trees the Apple orchard has = 152
Number of Red apple trees = 88
Number of Green apple trees = Number of Apple trees the Apple orchard has – Number of Red apple trees
= 152 – 88
= 64.
Number of Red apples more than Green apples = Number of Red apple trees – Number of Green apple trees
= 88 – 64
= 24.

Eureka Math Grade 3 Module 7 Lesson 1 Exit Ticket Answer Key

Use the RDW process to solve the problem below. Use a letter to represent the unknown.
Sandra keeps her sticker collection in 7 albums. Each album has 40 stickers in it. She starts a new album that has 9 stickers in it. How many total stickers does she have in her collection?
Answer:
Total number of stickers Sandra has in each album = 640.

Explanation:
Number of stickers Sandra has in each album of 7albums = 40 × 7 = 280.
Number of stickers Sandra has in each album of 9albums = 40 × 9 = 360
Total number of stickers Sandra has in each album = Number of stickers Sandra has in each album of 7albums  + Number of stickers Sandra has in each album of 9albums
= 280 + 360
= 640.

Eureka Math Grade 3 Module 7 Lesson 1 Homework Answer Key

Max’s family takes the train to visit the city zoo. Use the RDW process to solve the problems about Max’s trip to the zoo. Use a letter to represent the unknown in each problem.
Question 1.
The sign below shows information about the train schedule into the city.

a. Max’s family buys 2 adult tickets and 3 child tickets. How much does it cost Max’s family to take the train into the city?
Answer:
Total amount for the Max’s family to take the train into the city = $34.

Explanation:
Amount of money for Adult ticket = $8
=> Amount of money for 2 Adult ticket = $8 × 2 = $16.
Amount of money for Child ticket = $6
=> Amount of money for 3 Child ticket =$6 × 3 = $18.
Total amount for the Max’s family to take the train into the city = Amount of money for 2 Adult ticket + Amount of money for 3 Child ticket
= $16 + $18
= $34.

b. Max’s father pays for the tickets with $10 bills. He receives $6 in change. How many $10 bills does Max’s father use to pay for the train tickets?
Answer:
Number of $10 bills Max’s father used to make the payment for the tickets = 4.

Explanation:
Amount of money paid by Max’s father for the tickets with $10 bills.
Amount of money received as change = $6
Total amount for the Max’s family to take the train into the city = $34
Number of $10 bills Max’s father used to make the payment for the tickets = Total amount for the Max’s family to take the train into the city + Amount of money received as change  ÷ 6
= $34 + $6 ÷ 10
= $40 ÷ 10
=4.

c. Max’s family wants to take the fourth train of the day. It’s 6:38 a.m. now. How many minutes do they have to wait for the fourth train?
Answer:
Time left for Max’s family to wait for their fourth train = 7 minutes.

Explanation:
Time of their Fourth train of the day now = 6:38 a.m.
Leaves every 15minutes train starting from 6:00 a.m.
Number of the train Max’s family to go = Fourth of the day
=> Time of first train  = 6:00 to 6:15
=> Time of Second train  = 6:15 to 6.30
=> Time of Third train  = 6:30 to 6:45
=> Time of Fourth train = 6:45 to 7:00.
Time left for Max’s family to wait for their fourth train = Time of Fourth train – Time of their train of the day now
= 6:45 – 6:38
= 7 minutes.

Question 2.
At the city zoo, they see 17 young bats and 19 adult bats. The bats are placed equally into 4 areas. How many bats are in each area?
Answer:
Number of Bats in each area = 9.

Explanation:
Number of Young bats in the zoo = 17
Number of Adults bats in the zoo = 19
The bats are placed equally into 4 areas.
=> Number of areas bats placed equally = 4
Number of Bats in each area = (Number of Adults bats in the zoo + Number of Young bats in the zoo) ÷ Number of areas bats placed  equally
= (19 + 17) ÷ 4
= 36 ÷ 4
= 9.

Question 3.
Max’s father gives the cashier $20 to pay for 6 water bottles. The cashier gives him $8 in change. How much does each water bottle cost?
Answer:
Cost of each water bottle = $2.

Explanation:
Amount of money given to cashier by Max’s father to pay for 6 water bottles = $20
Amount of money received as change = $8
Total number of water bottles purchased  = 6
Cost of each water bottle = Amount of money given to cashier by Max’s father to pay for 6 water bottles  – Amount of money received as change ÷ Total number of water bottles purchased
= $20 – $8 ÷ 6
= $12 ÷ 6
= $2.

Question 4.
The zoo has 112 types of reptiles and amphibians in their exhibits. There are 72 types of reptiles, and the rest are amphibians. How many more types of reptiles are there than amphibians in the exhibits?
Answer:
Number of Amphibians in the zoo = 40.
Number of Reptiles more than Amphibians in the exhibits in the zoo = 32.

Explanation:
Number of Reptiles in the zoo = 72
Number of Reptiles and Amphibians in their exhibits in the zoo = 112
Number of Amphibians in the zoo = Number of Reptiles and Amphibians in their exhibits in the zoo – Number of Reptiles in the zoo
= 112 – 72
= 40.
Number of Reptiles more than Amphibians in the exhibits in the zoo = Number of Reptiles in the zoo  – Number of Amphibians in the zoo
= 72 – 40
= 32.