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## Bridges in Mathematics Grade 5 Student Book Answer Key Unit 5 Module 1

**Bridges in Mathematics Grade 5 Student Book Unit 5 Module 1 Session 1 Answer Key**

**Mike’s Measurements**

Mike is moving to a new house. He is measuring his furniture to see which items will fit in his new room. Help Mike determine the measurements of his furniture.

Question 1.

Mikeâ€™s bed has a base of 192 cm by 96 cm. What is the perimeter of the base of Mikeâ€™s bed in meters? Show your work.

Answer: 576 cm

Mikeâ€™s bed has a base of 192 cm by 96 cm.

perimeter of the base of Mikeâ€™s bed = ?

perimeter = 2 x (192 + 96)Â =Â 2 x 288 = 576

Question 2.

Mikeâ€™s wooden storage box is 25 inches by 36 inches by 39 inches. What is the volume of Mikeâ€™s box? Show your work.

Answer:Â 35100 in^{3}

Mikeâ€™s wooden storage box is 25 inches by 36 inches by 39 inches.

volume = 25 * 36 * 39 = 35100 in^{3}

Question 3.

Mikeâ€™s room is rectangular. One wall of Mikeâ€™s new room is 3.96 meters long.

a. How long is this wall in centimeters?

Answer:Â 396 centimeters

1 meter = 100 centimeters

3.96 meters = ?

Therefore, 3.96 meters = 3.96 * 100 = 396 centimeters

b. How long is this wall in millimeters?

Answer: 3690 millimeters

1 meter = 1000 millimeters

3.96 meters = ?

Therefore, 3.96 * 1000 = 3960 millimeters

Question 4.

Another wall in Mikeâ€™s room is 2.51 meters long.

a. How long is this wall in centimeters?

Answer: 251 centimeters

1 meter = 100 centimeters

2.51 meters = ?

Therefore, 2.51 * 100 = 251 centimeters

b. How long is this wall in millimeters?

Answer: 2510 millimeters

1 meter = 1000 millimeters

2.51 metersÂ = ?

Therefore, 2.51 meters =2.51 * 1000 = 2510

Question 5.

What is the area of Mikeâ€™s room in square centimeters? Show your work.

Answer: 99369 cm^{2}

Area of rectangle = l * b

l = 396 centimeters

b = 251 centimeters

Area = 396 * 251 =Â 99369 cm^{2}

**Bridges in Mathematics Grade 5 Student Book Unit 5 Module 1 Session 2 Answer Key**

**Work Place Instructions 5A Target One Fractions**

Each pair of partners needs:

- 2 Target One Fractions Record Sheets
- 1 deck of Number Cards, with the 0s, 7s, 9s, and wild cards removed
- math journals

1. Players shuffle the deck of cards and decide who will be the dealer. The dealer gives 5 cards to each player.

2. Player 1 chooses three cards to make a whole number and a fraction that will result in a product that is as close to 1 as possible. Each card is only used once.

For the fraction, one card is the numerator and the other is the denominator. For example, if the cards 2, 5, and 8 are used, players could choose 5 Ã— \(\frac{2}{8}\), 5 Ã— \(\frac{8}{2}\), 2 Ã— \(\frac{5}{8}\), etc.

Player 1 0K, I got two 5s, a 2, a 6, and an 8. Hmm… I think Iâ€™m going to use one of the 5s as my whole number, and the 2 and the 8 for the fraction, like this.

3. Player 1 multiplies the two numbers and shows her work in her math journal. Then players discuss how to multiply the two numbers.

Player 1 I think that 5 Ã— \(\frac{2}{8}\) is \(\frac{10}{8}\) which is 1\(\frac{2}{8}\), and thatâ€™s the same as 1\(\frac{1}{4}\).

Player 2 I thought of it as 5 Ã— \(\frac{2}{8}\) = 5 Ã— \(\frac{1}{4}\), so itâ€™s \(\frac{5}{4}\). Thatâ€™s also 1\(\frac{1}{4}\).

4. Player 1 writes an equation to represent her work on her record sheet.

5. Player 1 figures her score by finding the difference between the product and 1. Both Player 1 and Player 2 both record Player 1’s score on their record sheets.

A product of 1\(\frac{1}{4}\) has a score of \(\frac{1}{4}\).

A product of \(\frac{7}{8}\) has a score of \(\frac{1}{8}\).

A product of 1 has a score of 0.

At the end of each turn, players keep the 2 cards they did not use. The dealer passes out 3 new cards to each player, so each player has a total of 5 cards to begin the next round.

6. Player 2 takes a turn. Players continue to take turns until they have played 5 rounds of the game.

7. Players add the scores of all 5 rounds. The lower total score wins.

Game Variations

A. Include the wild cards in the deck. A wild card can be any numeral 1-6. Players put a star above the number made from a wild card in the equation on their record sheets.

B. Products below 1 get a positive score. Products above 1 get a negative score. Players add those scores together and the final score closest to 0 wins.

For example, a product of \(\frac{7}{8}\) would be scored as +\(\frac{1}{8}\), and a product of 1\(\frac{1}{4}\) would be scored as –\(\frac{1}{4}\).

C. Players play Target Two Fractions, trying to get as close as possible to 2 instead of 1.

**Target One Fractions**

In Target One Fractions, players choose 3 numbers to create a whole number and a fraction that have a product close to 1. Their score is the difference between their product and 1.

Question 1.

Claudia is playing Target One Fractions. She has these cards: 2, 3, 5, 5, 8. Help Claudia by choosing 3 cards and writing an equation she will solve.

Answer: The numbers that will yield a result closest to 1 are 2, 3 and 5 used in one of the following equations:

2 x \(\frac{3}{5}\)Â = \(\frac{6}{5}\)Â or 3 x \(\frac{2}{5}\)Â = \(\frac{6}{5}\)

Question 2.

Ernesto is playing Target One Fractions. He chose these cards: 2, 3, 8. He made the problem 8 Ã— \(\frac{2}{3}\).

a What is 8 Ã— \(\frac{2}{3}\) your work.

Answer: \(\frac{16}{3}\)

8 x \(\frac{2}{3}\)Â = \(\frac{16}{3}\)

b. What is Ernestoâ€™s score?

Answer: 4 \(\frac{1}{3}\)

Ernestos’s Score is 4 \(\frac{1}{3}\)

c. How else could you arrange the numbers?

Answer: The arrangement that will yield a product closest to 1 is 2 x \(\frac{3}{8}\)

d. What would your product be? Show your work.

Answer: The cards which I have chosen are 5, 8, 6, 2, 5. I haver chosen 5, 2, 8 as my three cards.

e. What is your score?

Answer:

5 Ã— \(\frac{2}{8}\) = 5 Ã— \(\frac{1}{4}\), so itâ€™s \(\frac{5}{4}\). Thatâ€™s also 1\(\frac{1}{4}\).

f. What is the difference between your score and Ernestoâ€™s? Show your work.

Answer:

Ernestos’s Score is 4 \(\frac{1}{3}\)

My score is 1\(\frac{1}{4}\)

Difference between my score and Ernesto’s is 4 \(\frac{1}{3}\) – 1\(\frac{1}{4}\)

= Â \(\frac{13}{3}\) – \(\frac{5}{4}\)

= \(\frac{52 – 15}{12}\)

= \(\frac{37}{12}\)

=3 \(\frac{1}{12}\)

**Bridges in Mathematics Grade 5 Student Book Unit 5 Module 1 Session 3 Answer Key**

**A Fraction of a Whole**

**Use numbers and labeled sketches to solve the problems on this page. Show your work.**

Question 1.

Nathan participated in a 5K (5 kilometer) race. He walked \(\frac{1}{4}\) of the way, jogged \(\frac{3}{5}\) of the way, and ran \(\frac{3}{20}\) of the way.

a. How many kilometers did Nathan walk?

Answer: \(\frac{5}{4}\) K = 1.25 K

Total kilometers Nathan walked = 5

He walked \(\frac{1}{4}\) of the way

Therefore, number of kilometers he walked = 5 x \(\frac{1}{4}\) = \(\frac{5}{4}\) = 1.25 K

b. How many meters did Nathan walk?

Answer: 1250 meters

1 kilometer = 1000 meters

1.25 kilometers = ?

Therefore, 1.25 x 1000 = 1250 meters

c. How many kilometers did Nathan jog?

Answer: 3 k

Total kilometers Nathan walked = 5

Total kilometers Nathan joggedÂ = \(\frac{3}{5}\)

Therefore, number of kilometers he jogged = 5 x \(\frac{3}{5}\) = 3 K

d. How many meters did Nathan jog?

Answer: 3000 meters

1 kilometer = 1000 meters

3 kilometers = ?

Therefore, 3 x 1000 = 3000 meters

e. How far in kilometers did Nathan run?

Answer: \(\frac{3}{4}\)K = 0.75 K

Total kilometers Nathan walked = 5

Total kilometers Nathan runÂ = \(\frac{3}{20}\)

Therefore, number of kilometers he run = 5 x \(\frac{3}{20}\) = \(\frac{3}{4}\)K = 0.75 K

f. How many meters did Nathan run?

Answer: 750 meters

1 kilometer = 1000 meters

0.75 kilometers = ?

Therefore, 0.75 x 1000 = 750 meters

Question 2.

Dejaâ€™s bedroom is 9 square meters. She just got a new rug that covers \(\frac{3}{5}\) of her bedroom floor. How big is Dejaâ€™s new rug?

Answer:

Dejaâ€™s bedroom is 9 square meters.

She just got a new rug that covers \(\frac{3}{5}\) of her bedroom floor.

9 – \(\frac{3}{5}\)

Therefore, Dejaâ€™s new rug is

Question 3.

Write a story problem for \(\frac{2}{3}\) Ã— 4. Then solve your own problem.

Answer: Vijay is having 4 apples. He wants to sell each apple at $ \(\frac{2}{3}\) . Find the cost of total apples after seeling?

Solution: total apples = 4

cost of each apple = \(\frac{2}{3}\)

total cost = \(\frac{2}{3}\)Â x 4Â = \(\frac{8}{3}\)

Question 4.

Solve the following combinations. Show your work for each.

Hint: Use one or more of the strategies on the chart you made with your classmates.

3 Ã— \(\frac{3}{4}\) = ______________

6 Ã— \(\frac{2}{5}\) = ______________

8 Ã— \(\frac{2}{3}\) = ______________

Answer:

3 Ã— \(\frac{3}{4}\) = \(\frac{9}{4}\)

6 Ã— \(\frac{2}{5}\) = \(\frac{12}{5}\)

8 Ã— \(\frac{2}{3}\) = \(\frac{16}{3}\)

Question 5.

Fill in the blanks.

\(\frac{1}{4}\) Ã— ______________ = 9

\(\frac{2}{4}\) Ã— ______________ = 9

\(\frac{3}{4}\) Ã— ______________ = 9

\(\frac{4}{4}\) Ã— ______________ = 9

Find and describe at least one pattern in the 4 combinations above.

Answer:

\(\frac{1}{4}\) Ã— **36** = 9

\(\frac{2}{4}\) Ã— **17** = 9

\(\frac{3}{4}\) Ã— **12** = 9

\(\frac{4}{4}\) Ã—** 1**= 9

Question 6.

Fill in the blanks.

\(\frac{1}{5}\) Ã— ______________ = 15

\(\frac{1}{5}\) Ã— ______________ = 30

\(\frac{1}{5}\) Ã— ______________ = 60

\(\frac{1}{5}\) Ã— ______________ = 120

Find and describe at least one pattern in the 4 combinations above.

Answer:

\(\frac{1}{5}\) Ã— **75** = 15

\(\frac{1}{5}\) Ã— **150** = 30

\(\frac{1}{5}\) Ã— **300** = 60

\(\frac{1}{5}\) Ã— **600** = 120

Question 7.

**CHALLENGE** Fill in the blanks.

\(\frac{5}{4}\) Ã— ______________ = 10

\(\frac{8}{4}\) Ã— ______________ = 9

\(\frac{3}{5}\) Ã— ______________ = 30

\(\frac{4}{5}\) Ã— ______________ = 60

Answer:

\(\frac{5}{4}\) Ã— **8** = 10

\(\frac{8}{4}\) Ã— **4.5 **= 9

\(\frac{3}{5}\) Ã— **50** = 30

\(\frac{4}{5}\) Ã— **75** = 60

**Fractions of Wholes**

Question 1.

Find the product.

a. \(\frac{1}{4}\) of 7 = _____________

Answer:Â \(\frac{7}{4}\)

\(\frac{1}{4}\) of 7 = \(\frac{1}{4}\) x 7

\(\frac{1}{4}\) of 7 = \(\frac{7}{4}\)

b. \(\frac{1}{5}\) Ã— 25 = _____________

Answer: 5

\(\frac{1}{5}\) Ã— 25 = 5

c. \(\frac{1}{3}\) of 36 = ____________

Answer: 12

\(\frac{1}{3}\) of 36 = \(\frac{1}{3}\) x 36

\(\frac{1}{3}\) of 36 = 12

d. \(\frac{3}{4}\) of 7 = _______________

Answer: \(\frac{21}{4}\)

\(\frac{3}{4}\) of 7 = \(\frac{3}{4}\) x 7

\(\frac{3}{4}\) of 7 = \(\frac{21}{4}\)

e. \(\frac{4}{5}\) Ã— 25 = _______________

Answer: 20

\(\frac{4}{5}\) Ã— 25 = 4 x 5

\(\frac{4}{5}\) Ã— 25 = 20

f. \(\frac{2}{3}\) Ã— 36 = _______________

Answer: 24

\(\frac{2}{3}\) Ã— 36 = 2 x 12

\(\frac{2}{3}\) Ã— 36 = 24

Question 2.

True or False?

a. \(\frac{1}{2}\) Ã— 11 = 8 \(\frac{1}{4}\)

True

False

Answer: False

\(\frac{1}{2}\) Ã— 11 = \(\frac{11}{2}\)Â = 5.5

8 \(\frac{1}{4}\) = Â \(\frac{33}{4}\) = 8.25

Therefore both are not equal

b. \(\frac{3}{5}\) of 20 = 15

True

False

Answer: False

\(\frac{3}{5}\) of 20 = \(\frac{3}{5}\) x 20 =Â 3 x 4 = 12

12 is not equal to 15.

therefore, both are not equal

c. \(\frac{2}{5}\) of 30 = 18

True

False

Answer: False

\(\frac{2}{5}\) of 30 = \(\frac{2}{5}\) x 30 = 12

12 is not equal to 18

Therefore, both are not equal

d. 16 Ã— \(\frac{1}{5}\) = \(\frac{16}{5}\)

True

False

Answer: True

16 Ã— \(\frac{1}{5}\) = \(\frac{16}{5}\)

both are equal

e. \(\frac{2}{6}\) Ã— 21 = 7

True

False

Answer: True

\(\frac{2}{6}\) Ã— 21 = \(\frac{1}{3}\) Ã— 21 = 7

Therefore, both are equal

f. 24 Ã— \(\frac{2}{3}\) = \(\frac{48}{3}\)

True

False

Answer: True

24 Ã— \(\frac{2}{3}\) = 8 x 2 = 16

\(\frac{48}{3}\) = 16

Therefore, both are equal

Question 3.

Madeline read \(\frac{2}{3}\) of her favorite book on the car ride to her grandparentsâ€™ house. If the book had 225 pages, how many pages of the book has she read?

Answer: 150

Total number of pages in the book = 225

Madeline read \(\frac{2}{3}\) of the book

Therefore, total number of pages she read =Â 225Â x \(\frac{2}{3}\)

= 75 x 2 = 150

Question 4.

Theo entered a race that required him to ride his bike 54 kilometers and run \(\frac{1}{6}\) as far as he bikes.

a. How many kilometers will Theo run?

Answer:Â 9 kilometers

Bike rides 54 kilometers

Theo runs \(\frac{1}{6}\) as far as he bikes

Therefore, number of kilometers will Theo runÂ = 54 x \(\frac{1}{6}\)

= 9 kilometers

b. How many meters will he run?

Answer: 9000 meters

1 kilometer = 1000 meters

9 kilometers = ?

Therefore, 9 x 1000 = 9000 meters

**Bridges in Mathematics Grade 5 Student Book Unit 5 Module 1 Session 4 Answer Key**

**Thinking About Strategy**

Question 1.

Anna is playing Target One Fractions. She has these cards: 4, 3, 6, 4, 2.

a. What three cards should she choose to make a whole number and fraction whose product is close to 1?

Answer: The 3 numbers that will result in the product closest to 1 are 2, 3 and 6 used in one of the two equations are 2 x \(\frac{3}{6}\)Â or 3 x \(\frac{2}{6}\)

b. Write an equation for the problem Anna will solve.

Answer:

Anna’s problem equation would be the following two:

2 x \(\frac{3}{6}\)Â or 3 x \(\frac{2}{6}\)

c. Solve the problem.

Answer:

There are two possibilities.

1) 2 x \(\frac{3}{6}\) = Â \(\frac{6}{6}\) = 1

2) 3 x \(\frac{2}{6}\) = Â \(\frac{6}{6}\) = 1

d. What is Annaâ€™s score for this round?

Answer: 1

Whatever equation she would choose, her score would be 1

Question 2.

Multiply.

Hint: Use one of the strategies on the class poster from todayâ€™s session.

\(\frac{3}{5}\) Ã— 4 = _______________

\(\frac{4}{5}\) Ã— 3 = _______________

\(\frac{4}{5}\) Ã— 16 = _______________

Answer:

\(\frac{3}{5}\) Ã— 4 = **\(\frac{12}{5}\)Â **

\(\frac{4}{5}\) Ã— 3 = **\(\frac{12}{5}\)**

\(\frac{4}{5}\) Ã— 16 = **\(\frac{64}{5}\)**

Question 3.

**CHALLENGE **Morgan thought she would be able to sell 75 plants for the school fundraiser. So far, she has sold \(\frac{2}{3}\) of her goal. Morganâ€™s friend Billy has sold \(\frac{4}{5}\) of his goal of 60 plants.

a. Make a prediction: which of the two students has sold more plants?

Answer:Â Morgan sold more plants than Billy

Morgan goal = 75 plants

Total plants Morgan sold = \(\frac{2}{3}\) of her goal

= \(\frac{2}{3}\) x 75

= 2 x 25

= 50

Therefore, Morgan sold 50 plants.

Billy goal = 60 plants

Total plants Billy sold = \(\frac{4}{5}\) of his goal of 60 plants.

= \(\frac{4}{5}\) x 60

= 4 x 12

= 48

Therefore, Billy sold 48 plants

So, Morgan sold more plants than Billy

b. How many plants has each student sold? Show your work.

Answer:Â Morgan sold 50 plants and Billy sold 48 plants.

Morgan goal = 75 plants

Total plants Morgan sold = \(\frac{2}{3}\) of her goal

= \(\frac{2}{3}\) x 75

= 2 x 25

= 50

Therefore, Morgan sold 50 plants.

Billy goal = 60 plants

Total plants Billy sold = \(\frac{4}{5}\) of his goal of 60 plants.

= \(\frac{4}{5}\) x 60

= 4 x 12

= 48

Therefore, Billy sold 48 plants

**Bridges in Mathematics Grade 5 Student Book Unit 5 Module 1 Session 5 Answer Key**

**Ryan’s Baseball Cards**

Question 1.

Ryan has 48 baseball cards. He gives some of them away to his friends. Help Ryan figure out how many cards each of his friends will get. Show your work.

a. If Ryan gives \(\frac{1}{4}\) of his 48 cards to Anna, how many cards does he give Anna?

Answer: 12 cards

Total cards = 48 cards

Ryan gives \(\frac{1}{4}\) of his 48 cards to Anna

Therefore, number of cards he gives to Anna =\(\frac{1}{4}\) x 48

= 12

So, Ryan gives 12 cards to Anna

b. If Ryan gives \(\frac{3}{8}\) of his 48 cards to Josiah, how many cards does he give Josiah?

Answer: 18 cards

Total cards = 48 cards

Ryan gives \(\frac{3}{8}\) of his 48 cards to Josiah

Therefore, number of cards he gives to Josaih =\(\frac{3}{8}\) x 48

= 18

So, Ryan gives 18 cards to Josaih

c. If Ryan gives \(\frac{2}{6}\) of his 48 cards to Max, how many cards does he give Max?

Answer: 16 cards

Total cards = 48 cards

Ryan gives \(\frac{2}{6}\) of his 48 cards to Max

Therefore, number of cards he gives to Anna =\(\frac{2}{6}\) x 48

= 16

So, Ryan gives 16 cards to Max.

d. How many cards does Ryan have left? What fraction of his original 48 cards is this?

Answer: 2 cards

Total cards = 48 cards

Therefore, number of cards Ryan haveÂ = total cards – (Anna cards + Josiah cards + Max cards)

= 48 – (12 + 18 +16)

= 2

So, Ryan have 2 cards

Question 2.

Solve the problems below.

\(\frac{2}{3}\) + \(\frac{5}{6}\) = ________________

1\(\frac{1}{3}\) – \(\frac{7}{8}\) = ________________

1\(\frac{4}{5}\) + 1\(\frac{3}{10}\) = _______________

16 Ã— \(\frac{7}{8}\) = _______________

\(\frac{4}{5}\) Ã— 24 = _______________

27 Ã— \(\frac{4}{9}\) = _______________

16\(\frac{1}{8}\) – 15\(\frac{3}{5}\) = _______________

208\(\frac{4}{7}\) + 201\(\frac{3}{4}\) = ________________

20\(\frac{1}{6}\) – 15\(\frac{3}{5}\) = ______________

Answer:

\(\frac{2}{3}\) + \(\frac{5}{6}\) = \(\frac{4}{6}\) + \(\frac{5}{6}\)Â = \(\frac{9}{6}\) = **\(\frac{3}{2}\)**

1\(\frac{1}{3}\) – \(\frac{7}{8}\) = \(\frac{4}{3}\) – \(\frac{7}{8}\) = \(\frac{32}{24}\) – \(\frac{21}{24}\) = **\(\frac{11}{24}\)**

1\(\frac{4}{5}\) + 1\(\frac{3}{10}\) = \(\frac{9}{5}\) + \(\frac{13}{10}\) = \(\frac{18}{10}\) + \(\frac{13}{10}\)Â = **\(\frac{31}{10}\)**

16 Ã— \(\frac{7}{8}\) = 2 x 7 = **14**

\(\frac{4}{5}\) Ã— 24 =\(\frac{96}{5}\)

27 Ã— \(\frac{4}{9}\) =Â 3 x 4 = 12

16\(\frac{1}{8}\) – 15\(\frac{3}{5}\) = \(\frac{129}{8}\) – \(\frac{78}{5}\) = \(\frac{645}{40}\) – \(\frac{624}{40}\) = **\(\frac{21}{40}\)**

208\(\frac{4}{7}\) + 201\(\frac{3}{4}\) = \(\frac{1460}{7}\) + \(\frac{807}{4}\) = \(\frac{5840}{28}\) + \(\frac{5649}{28}\) = **\(\frac{11489}{40}\)**

20\(\frac{1}{6}\) – 15\(\frac{3}{5}\) = \(\frac{121}{6}\) – \(\frac{78}{5}\) = \(\frac{605}{30}\) – \(\frac{468}{30}\) = **\(\frac{137}{30}\)**