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Bridges in Mathematics Grade 5 Home Connections Answer Key Unit 1 Module 2
Bridges in Mathematics Grade 5 Home Connections Unit 1 Module 2 Session 1 Answer Key
Better Boxes
Brad found some additional boxes in his storeroom. He is wondering if these boxes will work for packaging some of his hand-stitched baseballs.
For each of the following problems, assume that a ball fits into a 1 × 1 × 1 space. Use numbers, labeled sketches, or words to find the answers. Show your work.
Question 1.
How many baseballs would fit in a box that has the dimensions (3 × 5) × 2?
Answer:
(3 × 5) × 2
= 15 × 2
= 30 cu. units
Question 2.
How many baseballs would fit in a box that has the dimensions (2 × 4) × 5?
Answer:
(2 × 4) × 5
= 8 × 5
= 40 cu. units
Question 3.
How many baseballs would fit in a box that has the dimensions 4 × (3 × 6)?
Answer:
4 × (3 × 6)
= 4 × 18
= 72 cu. units
Question 4.
How many baseballs would fit in this box?
Answer:
Given,
l = 2
w = 2
h = 6
We know that,
V = lwh
V = 2 × 2 × 6
V = 24 cu. units
Question 5.
How many baseballs would fit in this box?
Answer:
Given,
l = 4
w = 5
h = 4
We know that,
V = lwh
V = 4 × 5 × 4
V = 80 cu. units
Question 6.
How many baseballs would fit in this box?
Answer:
Given,
l = 8
w = 2
h = 6
We know that,
V = lwh
V = 8 × 2 × 6
V = 96 cu. units
Question 7.
Brad is hoping to package exactly 48 baseballs together. He sees the side of a box in his storeroom that is 2 × 3. What is he hoping the other dimension is?
Answer:
Given,
Brad is hoping to package exactly 48 baseballs together.
He sees the side of a box in his storeroom that is 2 × 3.
Let h be the height of the box.
The shape of baseball is spherical and one baseball fits in a cubical box having the minimum dimensions equal to the diameter of the spherical ball.
The volume occupied by one baseball = volume of the cubical box having the side d units = d³ cu. units
So, the volume required to pack 48 baseball = 48 d³
So, the length, 3 units and the width 2 units must be the integral multiple of the diameter d units of the baseball
lenth 3 = ad
width 2 = bd
h = cd
2 × 3 × h = 48d³
6d²h = 48d³
h = 48d³/6d²
h = 8d
So, the other dimension of the box is 8 units corresponding to the unity diameter of the baseball.
The height of the box depends on the diameter of the baseball, so it will have infinite possible value for different values of the diameter.
h = 2 × 0.5 = 1 unit
The height of the box is 2 units corresponding to 0.5 unit diameter of the baseball.
Question 8.
Brad is hoping to package exactly 64 baseballs together. He sees the side of a box in his storeroom that is 4 × 2. What is he hoping the other dimension is?
Answer:
Given,
Brad is hoping to package exactly 64 baseballs together.
He sees the side of a box in his storeroom that is 4 × 2.
Let h be the height of the box.
The shape of baseball is spherical and one baseball fits in a cubical box having the minimum dimensions equal to the diameter of the spherical ball.
The volume occupied by one baseball = volume of the cubical box having the side d units = d³ cu. units
So, the volume required to pack baseball = 64 d³
So, the length, 4 units and the width 2 units must be the integral multiple of the diameter d units of the baseball
lenth 4 = ad
width 2 = bd
h = cd
4 × 2 × h = 48d³
8d²h = 64d³
h = 64d³/8d²
h = 8d
So, the other dimension of the box is 8 units corresponding to the unity diameter of the baseball.
Question 9.
CHALLENGE Harris said that 15 × 9 is equivalent to 45 × 3 because you can multiply one dimension by 3 and divide the other dimension by 3. His partner said that only works when you double one number and halve the other. Who is right? Explain.
Answer:
Given,
Harris said that 15 × 9 is equivalent to 45 × 3 because you can multiply one dimension by 3 and divide the other dimension by 3.
15 × 9 = 15 × (3 × 3) = (15 × 3) × 3 = 45 × 3
Bridges in Mathematics Grade 5 Home Connections Unit 1 Module 2 Session 3 Answer Key
Samantha’s Strategies
Samantha has been working with a variety of multiplication strategies.
Question 1.
Write an expression to describe each of the statements Samantha made.
a. To solve 18 × 20, I find double 18 and then multiply by 10.
Answer:
18 × 20 = 360
(18 × 2) × 10 = 36 × 10 = 360
b. To solve 16 × 18, I double and halve.
Answer:
16 × 18 = 16 × 8 × 10 = (16 × 8) × 10 = 128 × 10 = 1280
c. To solve 31 × 8, I find 3 × 8, multiply by 10, and then add 1 group of 8.
Answer:
31 × 8
3 × 8 = 24
24 × 10 = 240
240 + (1 × 8) = 240 + 8 = 248
31 × 8 = 248
Question 2.
Evaluate the three expressions above (in other words, solve the problems).
a.
Answer:
b.
Answer:
c.
Answer:
Question 3.
Fill in the blanks.
a. (5 × 3) × _______ = 60
Answer:
Let the unknown number be x.
(5 × 3) × x = 60
15 × x = 60
x = 60/15
x = 4
So, the missing number is 4.
b. 4 × (____ × 9) = 72
Answer:
4 × (____ × 9) = 72
Let the unknown number be x.
4 × (x × 9) = 72
x × 9 = 72/4
9x = 18
x = 18/9
x = 2
So, the missing number is 2.
c. 7 × (2 × ___) = 42
Answer:
7 × (2 × ___) = 42
Let the unknown number be x.
7 × (2 × x) = 42
2x = 42/7
2x = 6
x = 6/2
x = 3
So, the missing number is 3.
d. (___ × 5) × 5 = 75
Answer:
Let the unknown number be x.
(x × 5) × 5 = 75
5x = 75/5
5x = 15
x = 3
So, the missing number is 3.
e. (3 × 3) × ____ = 27
Answer:
Let the unknown number be x.
(3 × 3) × x = 27
9x = 27
x = 27/9
x = 3
So, the missing number is 3.
Question 4.
True or False?
a. ____ 9 × 9 = (10 × 10) – 1
Answer:
9 × 9 = 81
(10 × 10) – 1 = 100 – 1 = 99
The statement is false.
b. ____ 7 × 21 = (20 × 7) + (1 × 7)
Answer:
7 × 21 = 147
(20 × 7) + (1 × 7) = 140 + 7 = 147
The statement is true.
c. ____ 16 × 20 = 10 × (16 × 2)
Answer:
16 × 2 = 320
10 × (16 × 2) = 10 × 32 = 320
320 = 320
The statement is true.
d. _____ 8 × 13 = 2 × 52
Answer:
8 × 13 = 104
2 × 52 = 104
104 = 104
The statement is true.
e. _____ 6 × 18 = (6 × 20) – (6 × 2)
Answer:
6 × 18 = (6 × 20) – (6 × 2)
6 × 18 = 108
(6 × 20) – (6 × 2) = 120 – 12 = 108
108 = 108
The statement is true.
Question 5.
William needs a box to hold his golf ball collection. He found a box that can fit 8 layers with 14 balls in each layer. How many golf balls can this box hold? Show your work.
Answer:
Given,
William needs a box to hold his golf ball collection. He found a box that can fit 8 layers with 14 balls in each layer.
8 × 14 = 112
Thus the box can hold 112 balls.
Question 6.
William found a different box in his garage. The label outside said it would hold 120 golf balls. If 24 balls fit in each layer, how many layers tall is the box? Show your work.
Answer:
Given,
William found a different box in his garage. The label outside said it would hold 120 golf balls.
24 balls fit in each layer.
120/24 = 5
Thus the box is 5 layers tall.
Question 7.
CHALLENGE William has a total of 292 golf balls in his collection.
a. Write an equation to show how many golf balls are left if William fills the two boxes described in problems 5 and 6 above.
Answer:
112 + 120 + b = 292
b = 292 – 232
b = 60
So, 60 golf balls are left if William fills the two boxes.
b. What are the dimensions of two different boxes that William could use to store the rest of his collection? Show your work.
Answer:
The two dimensions of the boxes are
l = 14
w = 8
l = 24
w = 5
Bridges in Mathematics Grade 5 Home Connections Unit 1 Module 2 Session 5 Answer Key
Finding Factors page
Question 1.
Find all the factors of each of the numbers below.
ex 15: 1, 3, 5, 15
a. 21: ____, ___, ___, _____
Answer:
The factors of 21 are 1, 3, 7 and 21
1 × 21 = 21
3 × 7 = 21
7 × 3 = 21
21 × 1 = 21
b. 28: ___, ___, ____, ___, ____, ____
Answer:
The factors of 28 are 1, 2, 4, 7, 14, 28
1 × 28 = 28
2 × 14 = 28
4 × 7 = 28
7 × 4 = 28
14 × 2 = 28
28 × 1 = 28
c. 42: ____, ____, ___, ___, ___, ___, ___, ____
Answer:
The factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42.
1 × 42 = 42
2 × 21 = 42
3 × 14 = 42
6 × 7 = 42
7 × 6 = 42
14 × 3 = 42
21 × 2 = 42
42 × 1 = 42
d. 60: ___, ___, ___, ___, ___, ___, ___, ___, ___, ___, ____
Answer:
The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.
1 × 60 = 60
2 × 30 = 60
3 × 20 = 60
4 × 15 = 60
5 × 12 = 60
6 × 10 = 60
10 × 6 = 60
12 × 5 = 60
15 × 4 = 60
20 × 3 = 60
30 × 2 = 60
60 × 1 = 60
Question 2.
Find at least three multiples for each number below.
ex 15: 30, 45, 60
a.
21: ___, ___, ____
Answer:
The next three multiples of 21 are 42, 63, 84
b.
25: ___, ___, ____
Answer:
The next three multiples of 25 are 50, 75 and 100.
c.
35: ___, ___, ____
Answer:
The next three multiples of 35 are 70, 105, 140
d.
42: ___, ___, ____
Answer:
The next three multiples of 42 are 84, 126, 168
Question 3.
a. List all the factors of 36.
Answer:
The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.
b. How do you know you have listed them all?
Answer:
1 × 36 = 36
2 × 18 = 36
3 × 12 = 36
4 × 9 = 36
6 × 6 = 36
9 × 4 = 36
12 × 3 = 36
18 × 2 = 36
36 × 1 = 36
Question 4.
Milo is talking to his sister Lisa about factors. He said he thinks that any even number always has more factors than any odd number. Lisa said she doesn’t agree with him. Explain who you agree with and why.
Answer:
Factors of 12 are 1, 2, 3, 4, 6, 12
1 × 12 = 12
2 × 6 = 12
3 × 4 = 12
6 × 2 = 12
12 × 1 = 12
1 is a factor every number
The number is a factor itself.
Milo said he thinks that any even number always has more factors than any odd number.
Lisa said she doesn’t agree with him.
Even numbers: 2, 4, 14, 18
Odd numbers: 3, 9, 15, 21
The factors of 2 are 1, 2
2 has two factors
The factors of 4 are 1, 2, 4
4 has three factors
The factors of 14 are 1, 2, 7, 14
14 has four factors
The factors of 18 are 1, 2, 3, 6, 9, and 18.
18 has six factors.
The factors of 3 are 1 and 3.
3 has 2 factors
The factors of 9 are 1, 3, 9
9 has 3 factors
The factors of 15 are 1, 3, 5, 15
15 has 4 factors
The factors of 21 are 1, 3, 7, 21
21 has 4 factors
The factors of an even number are greater than the factors of an odd number like 18 the number of factors of 18 is greater than the number of the factors of all odd numbers above.
Thus not always any even number has more factors than any odd number.
I agree with Lisa.
ex What factors do 12 and 24 have in common? 1, 2, 3, 4, 6, 12
Question 5.
What factors do 8 and 12 have in common?
Answer:
The common factors of 8 and 12 are 1, 2, 4.
multiples of 2 are 2, 4, 6, 8, 10, 12, …
multiples of 4 are 4, 8, 12, 16, 20, ….
Question 6.
What factors do 6 and 4 have in common?
Answer:
The factors 6 and 4 have in common are 1, 2.
1 × 6 = 6
2 × 3 = 6
1 × 4 = 4
2 × 2 = 4
multiples of 2 are 2, 4, 6, 8, 10,…
ex What are two multiples that 5 and 6 have in common? 30, 60
Question 7.
What are two multiples that 4 and 8 have in common?
Answer:
The two multiples that 4 and 8 have in common are 16 and 24.
4 × 4 = 16
4 × 6 = 24
8 × 3 = 24
8 × 2 = 16
Question 8.
What are two multiples that 5 and 7 have in common?
Answer:
The two multiples that 5 and 7 have in common are 35 and 70
5 × 7 = 35
5 × 14 = 70
Question 9.
CHALLENGE Huan is redesigning his bedroom, which is the shape of a rectangle.
a. Huan knows the area of his bedroom is 180 square feet. What are all the possible whole number dimensions of Huan’s bedroom?
Answer:
Huan knows the area of his bedroom is 180 square feet.
9 × 20 = 180
3 × 60 = 180
15 × 12 = 180
45 × 4 = 180
5 × 36 = 180
2 × 90 = 180
6 × 30 = 180
10 × 18 = 180
b. Which dimensions are the most likely dimensions for Huan’s bedroom? Why?
Answer:
likely dimensions:
length 18 feet
width 10 feet
length: 15 feet
width: 12 feet
we observe that the length is close to the width.