Go Math Answer Key

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Go Math Grade 4 Answer Key Homework FL Chapter 2 Multiply by 1-Digit Numbers Review/Test

Improve your conceptual learning through the Go Math Grade 4 Answer Key Homework FL Review/Test. This guide is very helpful to tally your answers & correct the mistakes by providing various methods to solve the questions. Learn the concepts properly and approach different techniques to solve problems in Chapter 2 Multiply by 1-Digit Numbers.

Chapter 2: Review/Test

• Review/Test – Page No. 95
• Review/Test – Page No. 96
• Review/Test – Page No. 97
• Review/Test – Page No. 98

Review/Test – Page No. 95

Vocabulary

Choose the best term from the box.

Question 1.
To find the product of a 3-digit number and a 1-digit number, you can multiply the ones, multiply the tens, multiply the hundreds, and find the sum of each.
_________

Question 1.

Question 2.
The _______________________ states that multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products.

Answer:
The distributive property states that multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products.

Concepts and Skills

Estimate. Then find the product.

Question 3.
5 5
× 2
———-
Estimate: _________
Product: _________

Answer:
Estimate: 150
Product: 110

Explanation:
5 5
× 2
———-
110

Question 4.
$2 5
×   3
———-
Estimate: $ _________
Product: $ _________

Answer:
Estimate: $ 100
Product: $ 75

Explanation:
$2 5
×   3
———-
$ 75

Question 5.
3 0 6
×    8
———-
Estimate: _________
Product: _________

Answer:
Estimate: 2,500.
Product: 2,448.

Explanation:
3 0 6
×    8
———-
2,448

Question 6.
$9 2 4
×      5
———-
Estimate: $ _________
Product: $ _________

Answer:
Estimate: $ 5,000
Product: $ 4,620.

Explanation:
$9 2 4
×      5
———-
4,620

Question 7.
3, 563
×      9
———-
Estimate: _________
Product: _________

Answer:
Estimate: 30,000.
Product: 32,067

Explanation:
3, 563
×      9
———-
32,067

Question 8.
7, 048
×      7
———-
Estimate: _________
Product: _________

Answer:
Estimate: 50,000
Product: 49,336

Explanation:
7, 048
×      7
———-
49,336

Question 9.
6, 203
×      3
————
Estimate: _________
Product: _________

Answer:
Estimate: 19,000
Product: 18,609

Explanation:
6, 203
×      3
————
18,609

Question 10.
8, 798
×      6
————
Estimate: _________
Product: _________

Answer:
Estimate: 53,000
Product: 52,788

Explanation:
8, 798
×      6
————
52,788

Review/Test – Page No. 96

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Question 11.
Which number sentence shows the Distributive Property?
Options:
a. 2 × 3 = 3 × 2
b. 5 × 0 = 0
c. 3 × (5 + 2) = (3 × 5) + (3 × 2)
d. (3 × 7) × 4 = 3 × (7 × 4)

Answer: c

Explanation:
The distributive property means solving the expression in the form of a×(b+c). So the correct option is 3 × (5 + 2) = (3 × 5) + (3 × 2).

Question 12.
Look at the pattern below. What is the missing number?

Options:
a. 8,000
b. 6,000
c. 600
d. 60

Answer: b

Explanation:
5×6,000= 30,000.

Question 13.
Which comparison sentence represents the equation?
45 = 5 × 9
a. 9 more than 5 is 45
b. 9 is 5 times as many as 45
c. 5 is 4 times as many as 45
d. 45 is 5 times as many as 9

Answer: d

Explanation:
45 is 5 times as many as 9.

Question 14.
There are 4 times as many alligators as crocodiles. If the total number of alligators and crocodiles is 40, how many alligators are there?
Options:
a. 40
b. 32
c. 24
d. 8

Answer: b

Explanation:
As there are 4 times as many alligators as crocodiles and if the total number of alligators and crocodiles is 40. To find how many alligators are there, we will put them in a group of 5 as there are 4 alligators and 1 crocodile, so 40÷5= 8 groups, in which 32 are alligators and 8 are crocodiles.

Review/Test – Page No. 97

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Question 15.
Gardeners at Seed Stop are planting seeds in 12-row seed trays. They plant 8 seeds in each row. How many plants will there be in each tray if all of the seeds germinate, or grow?
Options:
a. 84
b. 86
c. 96
d. 104

Answer: c

Explanation:
As gardeners at Seed Stop are planting seeds in 12-row seed trays, and they plant 8 seeds in each row. So the number of plants will there be in each tray are 12×8= 96.

Question 16.
Which shows the product of 4 × 15 × 25?
Options:
a. 150
b. 1,200
c. 1,500
d. 1,600

Answer: c

Explanation:
The product of 4×15×25= 1,500.

Question 17.
A Broadway musical group will have 9 performances. The theater can seat 2,518 people. If all of the seats at each performance are taken, how many people will see the show?
Options:
a. 18,592
b. 22,652
c. 22,662
d. 31,622

Answer: c

Explanation:
As a broadway musical group will have 9 performances and the theater can seat 2,518 people, so number of people will see the show are 9×2,518= 22,662.

Question 18.
The table below shows the type of film sold and the number of rolls in one pack at a local gift shop.

Hannah buys 3 packs of 36 exposure film and 2 packs of 24 exposure film. She uses 8 rolls of film. How many rolls does she have left?
Options:
a. 8
b. 12
c. 20
d. 24

Answer: b

Explanation:
Hannah buys 3 packs of 36 exposure film and 2 packs of 24 exposure film, so the total number of films are 3×4= 12, 2×4= 8 which is 12+8= 20. And she uses 8 rolls, then the number of rolls left are 20-8= 12.

Review/Test – Page No. 98

Constructed Response

Question 19.
John’s grade has 3 classrooms. Each classroom has 14 tables. Two students sit at each table. About how many students are there in all? Use pictures, words, or numbers to show how you know.
About ______ students

Answer: 84 students.

Explanation:
As John’s grade has 3 classrooms, and each classroom has 14 tables and two students sit at each other, so total number of students are 14×2= 28 as John has 3 classrooms, so 28×3= 84 students.

Performance Task

Justin has $450 to buy supplies for the school computer lab. He buys 8 boxes of printer paper that cost $49 each.

Question 20.
A. About how much money does Justin spend on the printer paper? Describe how you made your estimate.
About $ ______

Answer: $400.

Explanation:
As each printer paper costs $49 and Justin buys 8 boxes, so total costs $49×8= $392. And the estimated cost is $400.

Question 20.
B. Find the actual amount of money Justin spends on the printer paper. Explain whether your estimate is close to the actual price.
Actual price $ ______

Answer: $392.

Explanation:
The actual amount of money Justin spends on the printer paper is $49×8= $392. Yes, the estimation is close to the actual price.

Question 20.
C. Will Justin have enough money left over to buy 3 packages of blank DVDs that cost $17 each? Explain your answer.

Answer: Yes, Justin will have enough money left.

Explanation:
As Justin left some money which is $450-$392= $51, after he bought printer paper for $392, so he can buy 3 packages of blank DVDs that cost $17 each which costs $17×3= $51.

Conclusion:

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Students of 4th grade can collect chapter 3 Multiply 2-Digit Numbers Go Math Homework Review/Test Answer key from this page in pdf format. All you have to do is click on the links provided over here and practice more from the HMH Go Math Grade 4 Answer Key Homework FL Chapter 3 Multiply 2-Digit Numbers Review/Test. However, students can score good marks in the exam.

Go Math Grade 4 Answer Key Homework FL Chapter 3 Multiply 2-Digit Numbers Review/Test

Moreover, educators and instructors can also make use of this Go Math Grade 4 Answer Key Homework FL Review/Test as a test paper to keep the exam and verify their student’s knowledge. We as a team designed this guide by providing detailed solutions for each and every question from Practice Tests, Chapter Tests, Cumulative Practice. Practice regularly by using the 4th Grade Go Math Chapter 3 Answer Key Review/Test and also use it as a quick reference to assess your knowledge after preparing the concepts within it.

Chapter 3: Review/Test

• Review/Test – Page No. 131
• Review/Test – Page No. 132
• Review/Test – Page No. 133
• Review/Test – Page No. 134

Review/Test – Page No. 131

Concepts and Skills

Question 1.
Explain how to find 14 × 19 by breaking apart the factors into tens and ones and finding the sum of the four partial products.

Answer: 266.

Explanation:
We can break 14 and 19 as 10+4 and 10+9 and to get the answer we will multiply both of the first two numbers by the other two. So we will multiply 10(10+9) and 4(10+9), then the values will be (100+90) and (40+36). By adding both we will get 100+90+40+36= 266

Question 2.
Explain how to find 40 × 80 using mental math.

Answer: 3600.

Explanation:
By using mental math we will multiply 4×8= 36 and then we will add zeros, so the answer will be 3600.

Estimate the product. Choose a method.

Question 3.
80 × 26

Answer: 2,080.

Explanation:
80
×26
———
480
+160
———–
2,080

Question 4.
19 × $67
$ ____

Answer: $1,273.

Explanation:
By breaking apart the factors into tens and ones and we can find the sum of the four partial products.
19×67= (10+9)×(60+7)
= (10×60)+(10×7)+(9×60)+(9×7)
= 600+70+540+63
= $1,273.

Question 5.
43 × 25

Answer: 1,075.

Explanation:
By breaking apart the factors into tens and ones and we can find the sum of the four partial products.
43 × 25= (40+3)×(20+5)
= (40×20)+(40×5)+(3×20)+(3×5)
= 800+200+60+15
= 1,075.

Question 6.
54 × 83

Answer: 4,482.

Explanation:
By breaking apart the factors into tens and ones and we can find the sum of the four partial products.
54 × 83= (50+4)×(80+3)
= (50×80)+(50×3)+(4×80)+(4×3)
= 4000+150+320+12
= 4,482.

Estimate. Then find the product.

Question 7.
$ 2 4
× 9 6
———–
Estimate: $ ________
Product: $ _________

Answer:
Estimate: $ 2,300
Product: $ 2,304

Explanation:
$ 2 4
× 9 6
———–
14 4
+ 216
———–
2,304

Question 8.
4 4
× 6 0
————
Estimate: _________
Product: _________

Answer:
Estimate: 2,600.
Product: 2,640.

Explanation:
4 4
× 6 0
————
00
+264
———–
2640

Question 9.
9 9
× 1 4
————
Estimate: _________
Product: _________

Answer:
Estimate: 1,400
Product: 1,386

Explanation:
9 9
× 1 4
————
396
+99
———–
1,386.

Question 10.
6 7
× 2 5
————
Estimate: _________
Product: _________

Answer:
Estimate: 1,700
Product: 1,675

Explanation:
6 7
× 2 5
————
335
+134
————
1,675

Question 11.
3 6
× 5 7
————
Estimate: _________
Product: _________

Answer:
Estimate: 2,000.
Product: 2,052.

Explanation:
3 6
× 5 7
————
252
+180
———–
2,052

Question 12.
$ 5 4
× 2 9
————
Estimate: $ _________
Product: $ _________

Answer:
Estimate: 1,600.
Product: 1,566.

Explanation:
$ 5 4
× 2 9
————
486
+108
———–
1,566

Question 13.
7 6
× 3 8
————
Estimate: _________
Product: _________

Answer:
Estimate: 2,900.
Product: 2,888.

Explanation:
7 6
× 3 8
————
608
+228
———–
2,888.

Question 14.
8 5
× 4 6
————
Estimate: _________
Product: _________

Answer:
Estimate: 3,900.
Product: 3,910

Explanation:
8 5
× 4 6
————
510
+340
———-
3,910.

Review/Test – Page No. 132

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Question 15.
Each month Sid’s parents put $75 into his college fund. How much do his parents put in the fund during 2 years?
Options:
a. $150
b. $450
c. $1,800
d. $15,300

Answer: c.

Explanation:
As Sid’s parents put $75 into his college fund, during two years Sid’s parent’s fund $75×24= $1800.

Question 16.
Mrs. Jenks wrote the correct answer to a homework problem on the board below. Which of the following could have been the homework problem?

Options:
a. 5 × 4,000
b. 50 × 400
c. 50 × 40
d. 50 × 4,000

Answer: c.

Explanation:
Mrs. Jenks’s homework problem is 50 × 40 because 50 × 40= 2,000.

Question 17.
George buys 30 cartons of 18 eggs for the Community Pancake Breakfast. How many eggs does he buy?
Options:
a. 340
b. 354
c. 460
d. 540

Answer: d

Explanation:
As George bought 30 cartons of 18 eggs for the Community Pancake Breakfast, the number of eggs George bought is 30×18= 540.

Review/Test – Page No. 133

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Question 18.
Mrs. Sampson donated a carton of pencils for each of the 35 classes at Lancet Elementary School. Each carton holds 64 pencils. Which is the best estimate for the number of pencils Mrs. Sampson donated?
Options:
a. A 99
b. B 1,800
c. C 2,400
d. D 2,800

Answer:  c.

Explanation:
As Mrs. Sampson donated a carton of pencils for each of the 35 classes at Lancet Elementary School, and each carton holds 64 pencils, so Mrs. Sampson donates 35×64= 2,240. And the estimated value is 2,400.

Question 19.
The school’s athletic department ordered 95 dozen badminton feather shuttles. How many feather shuttles were ordered?
Options:
a. A 2,280
b. B 1,140
c. C 1,030
d. D 114

Answer: b

Explanation:
One dozen is equal to 12. As school’s athletic department ordered 95 dozen badminton feather shuttles, so 95 dozens means
95×12= 1,140 badminton feather shuttles.

Question 20.
Jill sold 35 adult tickets and 48 child tickets for a fund-raising dinner. An adult ticket costs $18 and a child ticket costs $14. How much did Jill collect for the tickets?
Options:
a. A $1,354
b. B $1,302
c. C $1,232
d. D $1,102

Answer: b

Explanation:
As Jill sold 35 adult tickets and 48 child tickets for a fund-raising dinner and each adult ticket costs $18 and a child ticket costs $14, so total amount Jill collected is 35×$18= 630 and 48×$18= 672 by adding 630+672= $1,302.

Question 21.
Which shows a way to find 35 × 74?
Options:
a. A (30 × 7) + (30 × 4) + (70 × 3) + (70 × 5)
b. B (30 × 70) + (30 × 4) + (50 × 70) + (50 × 4)
c. C (30 + 70) + (30 + 4) + (70 + 30) + (70 + 5)
d. D (30 × 70) + (30 × 4) + (5 × 70) + (5 × 4)

Answer: d

Explanation:
By breaking apart the factors into tens and ones and finding the sum of the four partial products,
35 × 74= (30 × 70) + (30 × 4) + (5 × 70) + (5 × 4)

Question 22.
New seats are being delivered to the theater. There are 45 new seats for each row in a 15-row section. How many seats are being delivered?
Options:
a. A 60
b. B 400
c. C 675
d. D 1,000

Answer: c

Explanation:
As new seats are being delivered to the theater and there are 45 new seats for each row in a 15-row section, so the total number of new seats is 45×15= 675.

Review/Test – Page No. 134

Constructed Response

Question 23.
Gulfside Gifts has 48 boxes of postcards to sell. There are 24 postcards in each box. If the shop sells 3 boxes of postcards, how many postcards does the shop have left to sell? Explain how you found the answer.
______ remaining cards

Answer: 1,080 remaining cards.

Explanation:
As Gulfside Gifts has 48 boxes of postcards to sell and there are 24 postcards in each box. So total number of post cards are
48×24= 1,152. And the shop sold 3 boxes of postcards i.e 3×24= 72, so shop has left 1,152-72= 1,080 cards are remaining to sell.

Question 24.
Several steps in finding the product of 68 and 34 are shown below. Describe the remaining steps. Use pictures, words, or numbers. Then complete the multiplication.

_____

Answer:  2,312.

Explanation:
68
× 34
———-
272
+ 204
———
2,312

Performance Task

Question 25.
A city is having a festival in a local park. Alison’s Bakery has agreed to donate $1,200 worth of baked goods for the event. The city wants to order 12 loaves of holiday bread, 18 dozen biscuits, 12 dozen bagels, and 14 dozen multigrain rolls.
A. Is the cost of the baked goods under the $1,200 donation limit? Use pictures, numbers, or words to explain how you found your answer.

Answer: Yes, the donation is under $1,200.

Explanation:
As the city ordered 12 loaves of holiday bread, 18 dozen biscuits, 12 dozen bagels, and 14 dozen multigrain rolls. And holiday bread costs $20, one dozen busicuits costs $12, and one dozen bagels costs $28, 1 dozen multigrain rolls costs $22. So by adding them
(12×20)+(12×18)+(12×28)+(14×22) we will get $1,100 which is less than $1,200

Question 25.
B. If yes, what could the city add to the order? If no, what could the city remove from the order?

Answer: The city can add whatever they want with the remaining $100. As $1,200-$1,100= $100.

Conclusion:

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Test your knowledge with the subject experts provided Go Math Grade 4 Answer Key Homework FL Chapter 4 Divide by 1-Digit Numbers Review/Test & Score higher grades in your exams. By referring to the Go Math Grade 4 Solution Key Homework FL Chapter 4 Divide by 1-Digit Numbers Review/Test you will have strong command over fundamentals. Hence, Download the HMH Go Math 4th Grade Solution Key Chapter 4 for free and ace up your preparation directly.

Go Math Grade 4 Answer Key Homework FL Chapter 4 Divide by 1-Digit Numbers Review/Test

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Chapter 4: Review/Test

• Review/Test – Page No. 187
• Review/Test – Page No. 188
• Review/Test – Page No. 189
• Review/Test – Page No. 190

Review/Test – Page No. 187

Choose the best term from the box.

Question 1.
1. When a number cannot be divided evenly, the amount left over is called the:

Answer:
When a number cannot be divided evenly, the amount left over is called the remainder.

Question 2.
You use the _______________ method of dividing when multiples of the divisor are subtracted from the dividend and then the quotients are added together.

Answer:
You use the compatible numbers method of dividing when multiples of the divisor are subtracted from the dividend and then the quotients are added together.

Use grid paper or base-ten blocks to model the quotient.

Then record the quotient.

Question 3.
96 ÷ 6 = ____

Answer: 16

Explanation:

Question 4.
86 ÷ 2 = ____

Answer: 43

Explanation:

Question 5.
155 ÷ 5 = ____

Answer: 31

Explanation:

Find two numbers the quotient is between.
Then estimate the quotient.

Question 6.
787 ÷ 2
Estimate: ____

Answer: 400.

Explanation:
787 ÷ 2= 393.5
Estimate: 800 ÷ 2= 400.

Question 7.
391 ÷ 6
Estimate: ____

Answer: 65.

Explanation:
391 ÷ 6= 65.157
Estimate: 390 ÷ 6= 65

Question 8.
789 ÷ 8
Estimate: ____

Answer: 100.

Explanation:
789 ÷ 8= 98.62
Explanation: 800 ÷ 8= 100.

Divide.

Question 9.
3)\(\overline { 987 } \)
____

Answer: 329.

Explanation:
3)\(\overline { 987 } \)
= 987÷3
= 329.

Question 10.
7)\(\overline { 501 } \)
____ R ____

Answer: 71 R 4.

Explanation:
7)\(\overline { 501 } \)
= 501÷7
= 71 R 4.

Question 11.
5)\(\overline { 153 } \)
____ R ____

Answer: 30 R 3.

Explanation:
5)\(\overline { 153 } \)
= 153÷5
= 30 R 3.

Question 12.
4)\(\overline { 808 } \)
____ R ____

Answer: 202 R 0.

Explanation:
4)\(\overline { 808 } \)
= 808÷4
= 202 R 0.

Question 13.
6)\(\overline { 8,348 } \)
____ R ____

Answer: 1391 R 2.

Explanation:
6)\(\overline { 8,348 } \)
= 8348÷6
= 1391 R 2.

Question 14.
8)\(\overline { 4,897 } \)
____ R ____

Answer: 612 R 1.

Explanation:
8)\(\overline { 4,897 } \)
= 4897÷8
= 612 R 1.

Review/Test – Page No. 188

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Question 15.
There are 96 tourists who have signed up to tour the island. The tourists are assigned to 6 equal-size groups. How many tourists are in each group?
Options:
a. 1 r3
b. 1 r6
c. 11
d. 16

Answer: 16.

Explanation:
As there are 96 tourists who have signed up to tour the island and the tourists are assigned to 6 equal-size groups. So the number of tourists are in each group is 96÷6= 16.

Question 16.
Maria needs to share the base-ten blocks equally among 4 equal groups.

Which model shows how many are in each equal group?
Options:
a.
b.
c.
d.

 

Question 17.
Manny has 39 rocks. He wants to put the same number of rocks in each of 7 boxes. Which sentence shows how many rocks will be in each box?
Options:
a. He will need 6 boxes.
b. There will be 6 rocks in each box.
c. There will be 5 rocks in each box.
d. There will be 5 rocks left over.

Answer: c

Explanation:
As Manny has 39 rocks. He wants to put the same number of rocks in each of the 7 boxes, so there will be 5 rocks in each box.

Review/Test – Page No. 189

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Question 18.
There are 176 students in the marching band. They are arranged in equal rows of 8 students for a parade. How many rows of students are there?
Options:
a. 220 rows
b. 120 rows
c. 22 rows
d. 21 rows

Answer: c

Explanation:
As there are 176 students in the marching band and they arranged in equal rows of 8 students for a parade, so 176÷8= 22 rows of students are there.

Question 19.
Naomi wants to plant 387 tulip bulbs in 9 equal rows. She uses division to find the number of tulips in each row. In which place is the first digit of the quotient?
Options:
a. ones
b. tens
c. hundreds
d. thousands

Answer: b

Explanation:
Naomi wants to plant 387 tulip bulbs in 9 equal rows and she uses division to find the number of tulips in each row, so 387÷9= 43. And the first digit of the quotient is tens place.

Question 20.
Kevin and 2 friends are playing a game of cards. There are 52 cards in the deck to be shared equally. Kevin wants each player to receive the same number of cards. How many cards will each player receive? How many cards will be left over?
Options:
a. 16 cards and 4 cards left over
b. 17 cards and 1 card left over
c. 25 cards and 2 cards left over
d. 26 cards and no cards left over

Answer: d.

Explanation:
Kevin and 2 friends are playing a game of cards and there are 52 cards in the deck to be shared equally, as Kevin wants each player to receive the same number of cards, each player will receive 52÷2= 26. So 26 cards each player receives and no cards left over.

Question 21.
Which number is the quotient?
1,125 ÷ 5 = ■
Options:
a. 25
b. 105
c. 125
d. 225

Answer: d

Explanation:
1,125 ÷ 5 =225.

Review/Test – Page No. 190

Constructed Response

Question 22.
Mrs. Valdez bought 6 boxes of roses. Each box had 24 roses. She divided all the roses into 9 equal bunches. How many roses were in each bunch? Explain how to use a diagram to help solve the problem. Show your diagrams.
______ roses

Answer: 16 roses.

Explanation:
As Mrs. Valdez bought 6 boxes of roses and each box had 24 roses, so the total number of roses are 6×24= 144 then she divided all the roses into 9 equal bunches. So each bunch will have 144÷9= 16 roses.

Performance Task

Question 23.
Mr. Owens plans to rent tables for a spaghetti fundraiser. He needs to seat 184 people.

A. If Mr. Owens wants all rectangular tables, how many tables should he rent? Explain.
______ tables

Answer: 31 tables.

Explanation:
The number of rectangular tables Mr. Owens should rent is 184÷6= 30.67. We will round off 30.67 to 31, so 31 tables he should rent.

Question 23.
B. Square tables rent for $12 each. Circular tables rent for $23 each. Mr. Owens says it would cost him less to rent square tables instead of circular tables. Is he right? Explain.

Answer: Yes, Mr. Owens is wrong.

Explanation:
As square tables rent for $12 each and circular tables rent for $23 each, so if Mr. Owens chooses square tables to rent and it has only 4 chairs, so 184÷4= 46 square tables should he rent which costs 46×$12= $552. And if Mr. Owens chooses circular tables he should rent 184÷8= 23 circular tables and which costs 23×$23= $529. So if he chooses circular tables he can pay less rent.

Conclusion:

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Go Math Grade 4 Answer Key Homework FL Chapter 5 Factors, Multiples, and Patterns Review/Test includes all the topics of chapter 5. In Practice Test, Chapter Test, Cumulative Practice, etc., we have compiled detailed Solutions for all the Questions. So, check out the 4th Grade Go math Answer Key Homework FL Chapter 5 Factors, Multiples, and Patterns Review/Test & cross-check your knowledge & math skills.

Go Math Grade 4 Answer Key Homework FL Chapter 5 Factors, Multiples, and Patterns Review/Test

Improve your subject Skills with the help of Go Math Grade 4 Answer Key Homework FL Review/Test and score better grades in your exams. Also, you can take help from the 4th Grade Go Math Ch 5 Review/Test Solution Key for effective preparation and better practice sessions. Moreover, it let students know where they went wrong and clear their doubts. Hence, this guide is so helpful for 4th standard students to score high in exams.

Chapter 5: Review/Test

• Review/Test – Page No. 219
• Review/Test – Page No. 220
• Review/Test – Page No. 221
• Review/Test – Page No. 222

Review/Test – Page No. 219

Choose the best term from the box.

Question 1.
The product of two numbers is a _______________ of both numbers.

Answer:
The product of two numbers is a multiple of both numbers.

Question 2.
A _______________ has exactly two factors.

Answer:
A prime has exactly two factors.

Question 3.
A number is always a multiple of its ____________ .

Answer:
A number is always a multiple of its multiple.

List all the factor pairs in the table.

Question 4.

Answer:
Factors of 48 are 1,2,3,4,6.

Explanation:
1×48= 48    1,48.
2×24= 48    2,24.
3×16= 48     3,16.
4×12= 48     4,12.
6×8= 48       6,8

Question 5.

Answer:
Factors of 81 are 1,3,9.

Explanation:
1×81= 81     1,81
3×27= 81     3,27
9×9= 81       9,9

Is the number a multiple of 9? Write yes or no.

Question 6.
3 _____

Answer: No

Explanation:
The number 3 is a factor of 9 but not a multiple of 9.

Question 7.
39 _____

Answer: No

Explanation:
The number 39 is not a multiple of 39.

Question 8.
45 _____

Answer: Yes

Explanation:
9×5= 45, so the number 45 is a multiple of 9.

Question 9.
93 _____

Answer: No.

Explanation:
The number 93 is not a multiple of 9.

Tell whether the number is prime or composite.

Question 10.
65 _________

Answer: Composite number.

Explanation:
As the number 65 factors are 1,5,13,65. So the number 65 is a composite number as it has more than two factors.

Question 11.
37 _________

Answer: Prime number.

Explanation:
The number 37 has only two factors 1 and 37, so the number is a prime number.

Question 12.
77 _________

Answer: Composite number.

Explanation:
The factors of 77 are 1,7,11 and 77, so the number 77 is a composite number.

Use the rule to write the first twelve terms in the pattern.
Describe another pattern in the numbers.

Question 13.
Rule: Add 10, subtract 5.

Answer:
1,6,11,16,21,26,31,36,41,46,51,56.

Explanation:
1
(1+10)-5= 11-5= 6
(6+10)-5= 16-5= 11
(11+10)-5= 21-5= 16
(16+10)-5= 26-5= 21
(21+10)-5= 31-5= 26
(26+10)-5= 36-5= 31
(31+10)-5= 41-5= 36
(36+10)-5= 46-5= 41
(41+10)-5= 51-5= 46
(46+10)-5= 56-5= 51
(51+10)-5= 61-5= 56.

Review/Test – Page No. 220

Question 14.
Erica knits 18 squares on Monday. She knits 7 more squares each day for the rest of the week. How many squares does Erica have on Friday?
Options:
a. 36
b. 46
c. 54
d. 90

Answer: b

Explanation:
As Erica knits 18 squares on Monday and she knits 7 more squares each day for the rest of the week, so on Friday Erica have 18+7+7+7+7= 46.

Question 15.
James works in a flower shop. He will put 36 tulips in vases for a wedding. He must use the same number of tulips in each vase. How many tulips could be in each vase?
Options:
a. 1, 2, 8
b. 2, 4, 8
c. 2, 4, 9
d. 6, 12, 16

Answer: c

Explanation:
As James put 36 tulips in vases for a wedding and he must use the same number of tulips in each vase, so we must find the factors of 36 to find how many tulips could be in each vase. So the factors of 36 are 1,2,3,4,6,9,12,18,36. So
2 tulips in 18 vases each
4 tulips in 9 vases each
9 tulips in 4 vases each.

Question 16.
What multiple of 7 is a factor of 7?
Options:
a. 0
b. 1
c. 7
d. 14

Answer: c

Explanation:
The number 7 is multiple and a factor of 7.

Question 17.
Hot dogs come in packages of 6. Hot dog buns come in packages of 8. Antonio will buy the same number of hot dogs as hot dog buns. How many hot dogs could he buy?
Options:
a. 6
b. 8
c. 18
d. 24

Answer: 24.

Explanation:
As hot dogs come in packages of 6, and hot dog buns come in packages of 8. So to find how many hot dogs could Antonio bought we must find the multiples of 6 and 8. So multiples of 6 and 8 are
Multiples of 6 are 6, 12, 18, 24, 30
Multiples of 8 are 8, 16, 24, 32, 40.
So Antonio bought 24 hot dogs.

Question 18.
Sean has 54 flower bulbs. He planted all the bulbs in rows. Each row has the same number of bulbs. How many bulbs could be in each row?
Options:
a. 6
b. 8
c. 12
d. 26

Answer: a

Explanation:
As Sean has 54 flower bulbs and planted all the bulbs in rows and each row has the same number of bulbs, so we will find the factors of 54. And the factors of 54 are 1,2,3,6,9,18,27, and 54. So Sean will plant 6 bulbs in each row.

Review/Test – Page No. 221

Question 19.
An ice-cream truck visits Julio’s street every 3 days and Lara’s street every 4 days. The truck visits both streets on April 12. When will the truck visit both streets next?
Options:
a. April 15
b. April 16
c. April 19
d. April 24

Answer: d

Explanation:
As an ice-cream truck visits Julio’s street every 3 days and Lara’s street every 4 days, and the truck visits both streets on April 12, so the next visit will be on April 24. By finding the multiples of 3 and 4 we will get the answer.
Multiples of 3 are 3,6,9,12,15,18,21,24
Multiples of 6 are 6,12,18,24.

Question 20.
The factors of a number include 2, 3, 4, 6, 8, 12, 16, 32, and 48. Which could be the number?
Options:
a. 32
b. 64
c. 96
d. 98

Answer: 96

Explanation:
As the number 96 is divisible by all the given numbers.

Question 21.
Ms. Booth has 16 red buttons and 24 blue buttons. She is making finger puppets. Each puppet has the same number of blue buttons and red buttons. How many puppets can she make if she uses all of the buttons?
Options:
a. 1, 2, 4, or 8
b. 1, 2, 4, 8, or 16
c. 1, 2, 4, 8, or 24
d. 1, 2, 4, 8, 16, or 24

Answer: a

Explanation:
As Ms. Booth has 16 red buttons and 24 blue buttons and she is making finger puppets and each puppet has the same number of blue buttons and red buttons, so to find how many puppets can she make if she uses all of the buttons we will find the factors of 16 and 24
so the factors of 16 are 1,2,4,8,16
Factors of 24 are 1,2,3,4,6,8,12,24.
So the common factors in both 16 and 24 are 1,2,4,8,16.

Review/Test – Page No. 222

Question 22.
I am a number between 60 and 100. My ones digit is two less than my tens digit. I am a prime number. What number am I?
Explain.
_____

Answer: 97.

Explanation:
Let’s name the digit:
X be one’s digit and y be tens digit
we know that X=Y-2. Now, Y can be 6,7,8,9 the number is between 60 and 100
As the possibilities with x=y-2, the numbers would be 64,75,86,97.
And 64 and 86 are even, so they can’t be prime. 75 is a composite number as there are more than two factors. So the remaining number is 97.

Question 23.
The number of pieces on display at an art museum is shown in the table.

A. The museum’s show for July features 30 oil paintings by different artists. All artists show the same number of paintings and each artist shows more than 1 painting. How many artists could be featured in the show?

Answer:
15 artists with 2 paintings per artist.
10 artists with 3 paintings per artist.
6 artists with 5 paintings per artist.
5 artists with 6 paintings per artist.
3 artists with 10 paintings per artist.
2 artists with 15 paintings per artist.

Explanation:
As the museum’s show for July features 30 oil paintings by different artists and all artists show the same number of paintings and each artist shows more than 1 painting, so the number of artists are
15 artists with 2 paintings per artist.
10 artists with 3 paintings per artist.
6 artists with 5 paintings per artist.
5 artists with 6 paintings per artist.
3 artists with 10 paintings per artist.
2 artists with 15 paintings per artist.

Question 23.
B. The museum wants to display all the art pieces in rows. Each row has the same number of pieces and the same type of pieces. How many pieces could be in each row?

Answer: 3

Explanation:
Given that 30 oil paintings, 24 photographs, and 21 sketches that a museum wants a display. The arrangement of all these art pieces must be in rows such that each row has the same number and same type of art piece displayed. And the greatest common factor of 30,24,21 is 3. So 3 pieces could be in each row.

Question 23.
C. The museum alternates between adding 3 new pieces one month and retiring one piece the following month. If the museum starts with 75 pieces and the pattern continues, write the numbers in the pattern for the next 8 months. Describe other patterns in the numbers.

Answer: 78, 77, 80, 79, 82, 81, 84, 83.

Explanation:
As the museum alternates between adding 3 new pieces one month and retiring one piece the following month and if the museum starts with 75 pieces and the pattern continues, so the numbers are 78, 77, 80, 79, 82, 81, 84, 83. Here the pattern is every other number differs by 2 and the numbers alternate between even and odd.

Conclusion:

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Go Math Grade 4 Answer Key Homework FL Chapter 6 Fraction Equivalence and Comparison Review/Test

Students who are searching for the best review/test guide can refer to this helpful Go Math Grade 4 Solution Key Homework FL Chapter 6 Fraction Equivalence and Comparison Review/Test. As it includes all the concepts in Chapter 6 Fraction Equivalence and Comparison. So, students can discover the methods to solve the problems easily and can explore the knowledge by giving the question from Review/Test. Click on the respective link and download it for better practice.

Chapter 6: Review/Test

• Review/Test – Page No. 261
• Review/Test – Page No. 262
• Review/Test – Page No. 263
• Review/Test – Page No. 264

Review/Test – Page No. 261

Choose the best term from the box.

Question 1.
A ________________ is a common multiple of two or more denominators.
________

Answer:
A common denominator is a common multiple of two or more denominators..

Question 2.
A fraction is in _________________ when the numerator and denominator have only 1 as a common factor
________

Answer:
A fraction is in simplest form when the numerator and denominator have only 1 as a common factor.

Question 3.
A ________________ is a known size or amount that helps you understand another size or amount.
________

Answer:
A benchmark is a known size or amount that helps you understand another size or amount.

Write two equivalent fractions.

Question 4.
\(\frac{4}{6}\)

Answer: \(\frac{6}{9}\) and \(\frac{8}{12}\).

Explanation:
To find equivalent fractions we will multiply its numerator and denominator by the same number. Firstly we will calculate GCF for the given fraction i.e \(\frac{4}{6}\), the GCF for (4,6) is 2. As GCF is not equal to 1, we will divide the numerator and denominator by 2. By dividing with 2 we will get the fraction as \(\frac{2}{3}\). Now we will multiply the numerator and denominator with 3,
So the fraction will be 3(\(\frac{2}{3}\))
= \(\frac{6}{9}\). For the second equivalent fraction, we will multiply numerator and denominator with 4,
So the fraction will be 4(\(\frac{2}{3}\))
= \(\frac{8}{12}\).
So, the two equivalent fractions of \(\frac{4}{6}\) are \(\frac{6}{9}\) and \(\frac{8}{12}\).

Question 5.
\(\frac{6}{10}\)

Answer: \(\frac{9}{15}\) and \(\frac{12}{20}\).

Explanation:
To find equivalent fractions we will multiply its numerator and denominator by the same number. Firstly we will calculate GCF for the given fraction i.e \(\frac{6}{10}\), the GCF for (6,10) is 2. As GCF is not equal to 1, we will divide the numerator and denominator by 2. By dividing with 2 we will get the fraction as \(\frac{3}{5}\). Now we will multiply the numerator and denominator with 3,
So the fraction will be 3(\(\frac{3}{5}\))
= \(\frac{9}{15}\). For the second equivalent fraction, we will multiply numerator and denominator with 4,
So the fraction will be 4(\(\frac{3}{5}\))
= \(\frac{12}{20}\).
So, the two equivalent fractions of \(\frac{3}{5}\) are \(\frac{9}{15}\) and \(\frac{12}{20}\).

Question 6.
\(\frac{2}{8}\)

Answer: \(\frac{3}{12}\) and \(\frac{4}{16}\).

Explanation:
To find equivalent fractions we will multiply its numerator and denominator by the same number. Firstly we will calculate GCF for the given fraction i.e \(\frac{2}{8}\), the GCF for (2,8) is 2. As GCF is not equal to 1, we will divide the numerator and denominator by 2. By dividing with 2 we will get the fraction as \(\frac{1}{4}\). Now we will multiply the numerator and denominator with 3,
So the fraction will be 3(\(\frac{1}{4}\))
= \(\frac{3}{12}\). For the second equivalent fraction, we will multiply numerator and denominator with 4,
So the fraction will be 4(\(\frac{1}{4}\))
= \(\frac{4}{16}\).
So, the two equivalent fractions of \(\frac{2}{8}\) are \(\frac{3}{12}\) and \(\frac{4}{16}\).

Write each pair of fractions as a pair of fractions with a common denominator.

Question 7.
\(\frac{3}{4} \text { and } \frac{7}{8}\)

Answer: \(\frac{6}{8}\) , \(\frac{7}{8}\).

Explanation:
To get the common denominators we will multiply \(\frac{3}{4}\) with 2, so that the fraction will be \(\frac{6}{8}\). As the other fraction is \(\frac{7}{8}\). So the denominators are the same.

Question 8.
\(\frac{2}{3} \text { and } \frac{1}{4}\)

Answer: \(\frac{8}{12}\) and \(\frac{3}{12}\).

Explanation:
To get the common denominators we will multiply \(\frac{2}{3}\) with 4 and \(\frac{1}{4}\) with 3, so that the fractions will be \(\frac{8}{12}\) and \(\frac{3}{12}\). So the denominators are same.

Question 9.
\(\frac{7}{10} \text { and } \frac{4}{5}\)

Answer: \(\frac{7}{10}\) and \(\frac{8}{10}\).

Explanation:
To get the common denominators we will multiply \(\frac{4}{5}\) with 2, so that the fraction will be \(\frac{8}{10}\). As the other fraction is \(\frac{7}{10}\). And the denominators are same.

Compare. Write <, >, or 5.

Question 10.
\(\frac{5}{8}\) _____ \(\frac{5}{12}\)

Answer: \(\frac{5}{8}\) > \(\frac{5}{12}\).

Explanation:
To compare \(\frac{5}{8}\) and \(\frac{5}{12}\) first we will find LCM of 8 and 12.
And the LCM of (8,12) is 24. Now we will multiply \(\frac{5}{8}\) with 3 and \(\frac{5}{12}\) with 2, so the fraction will be \(\frac{15}{24}\) and the other fraction is \(\frac{10}{24}\).
So \(\frac{15}{24}\) is greater than \(\frac{10}{24}\).

Question 11.
\(\frac{10}{12}\) _____ \(\frac{5}{6}\)

Answer: \(\frac{10}{12}\) = \(\frac{5}{6}\).

Explanation:
To compare \(\frac{10}{12}\) and \(\frac{5}{6}\),first we will find LCM of 12 and 6.
And the LCM of (12,6) is 12. Now we will multiply \(\frac{5}{6}\) with 2, so the fraction will be \(\frac{10}{12}\) and the other fraction is \(\frac{10}{12}\).
So \(\frac{10}{12}\) is equal to \(\frac{10}{12}\).

Question 12.
\(\frac{1}{2}\) _____ \(\frac{3}{10}\)

Answer: \(\frac{1}{2}\) > \(\frac{3}{10}\).

Explanation:
To compare \(\frac{1}{2}\) and \(\frac{3}{10}\) first we will find LCM of 2 and 10.
And the LCM of (2,10) is 10. Now we will multiply \(\frac{1}{2}\) with 5, so the fraction will be \(\frac{5}{10}\) and the other fraction is \(\frac{3}{10}\).
So \(\frac{5}{10}\) is greater than \(\frac{3}{10}\).

Question 13.
\(\frac{1}{4}\) _____ \(\frac{1}{3}\)

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

Explanation:
To compare \(\frac{1}{4}\) and \(\frac{1}{3}\) first we will find LCM of 4 and 3.
And the LCM of (4,3) is 12. Now we will multiply \(\frac{1}{4}\) with 3 and \(\frac{1}{3}\) with 4, so the fraction will be \(\frac{3}{12}\) and the other fraction is \(\frac{4}{12}\).
So \(\frac{3}{12}\) is less than \(\frac{4}{12}\).

Write the fractions in order from least to greatest.

Question 14.
\(\frac{2}{3}, \frac{3}{4}, \frac{1}{6}\)

Answer: \(\frac{1}{6}\) < \(\frac{2}{3}\)< \(\frac{3}{4}\).

Explanation:
To write the fraction from least to greatest we will find LCM of 3,4,6. And the LCM of (3,4,6) is 12. Now we will multiply
\(\frac{2}{3}\) with 4 and \(\frac{3}{4}\) with 3 and \(\frac{1}{6}\) with 2, so the fraction will be
\(\frac{8}{12}\) and \(\frac{9}{12}\), \(\frac{2}{12}\)
So \(\frac{2}{12}\) is less than \(\frac{8}{12}\) is less than \(\frac{9}{12}\).

Question 15.
\(\frac{7}{10}, \frac{4}{5}, \frac{1}{2}, \frac{4}{12}\)

Answer: \(\frac{4}{12}\) < \(\frac{1}{2}\)< \(\frac{7}{10}\)< \(\frac{4}{5}\).

Explanation:
To write the fraction from least to greatest we will find LCM of 10,5,2,12. And the LCM of (10,5,2,12) is 60. Now we will multiply
\(\frac{7}{10}\) with 6 and \(\frac{4}{5}\) with 12 and \(\frac{1}{2}\) with 30 and \(\frac{4}{12}\) with 5 , so the fraction will be
\(\frac{42}{60}\) and \(\frac{48}{60}\), \(\frac{30}{60}\), \(\frac{20}{60}\)
So \(\frac{20}{60}\) is less than \(\frac{30}{60}\) is less than \(\frac{42}{60}\) is less than
\(\frac{48}{60}\).

Review/Test – Page No. 262

Fill in the bubble completely to show your answer.

Question 16.
Paco needs at least \(\frac{3}{8}\) yard of twine to build a model ship. How much twine could he buy?
Options:
a. \(\frac{3}{10}\) yard
b. \(\frac{1}{4}\) yard
c. \(\frac{3}{5}\) yard
d. \(\frac{1}{8}\) yard

Answer: c.

Explanation:

a) 3/10 yard. As we know that for two rational numbers with the same numerator but with different denominators the number whose denominator is smaller is a greater quantity.
Hence 3/10 < 3/8. And option a is incorrect.

b) 1/4 yard. As to compare to rational numbers we have to either make the numerator equal or their denominator equal. Hence here we multiply and divide 1/4 by 2 to get 8 in the denominator. As 2/8 < 3/8
since the denominator is the same and the number with the same denominator but with different numerators are compared as whose numerator is greater is a greater quantity. And the option b is incorrect.

c) 3/4 yard. As both the numbers have the same numerator but different denominator and we know that for two rational numbers with the same numerator but with different denominators the number whose denominator is smaller is a greater quantity. As 3/8 < 3/4, so option c is correct.

d) 1/8 yard. As both the numbers have the same denominator and we know that for two rational numbers with the same denominator but with the different numerators, the number whose numerator is smaller is a smaller quantity. So 1/8 < 3/8 and the option d is incorrect.

Question 17.
Rachel, Nancy, and Diego were in a fishing competition. Rachel’s fish was \(\frac{7}{8}\) foot long, Nancy’s fish was \(\frac{1}{4}\) foot long, and Diego’s fish was \(\frac{1}{2}\) foot long. What are the lengths of the fish in order from least to greatest?
Options:
a. \(\frac{7}{8}\) foot, \(\frac{1}{2}\) foot, \(\frac{1}{4}\) foot
b. \(\frac{1}{2}\) foot, \(\frac{7}{8}\) foot, \(\frac{1}{4}\) foot
c. \(\frac{7}{8}\) foot, \(\frac{1}{4}\) foot, \(\frac{1}{2}\) foot
d. \(\frac{1}{4}\) foot, \(\frac{1}{2}\) foot, \(\frac{7}{8}\) foot

Answer: d

Explanation:
As Rachel’s fish was \(\frac{7}{8}\) foot long, Nancy’s fish was \(\frac{1}{4}\) foot long, Diego’s fish was \(\frac{1}{2}\) foot long, so to find the lengths of the fish in order from least to greatest we will find the LCM of (8,4,2), so the LCM of (8,4,2) is 8 and we will multiply \(\frac{1}{4}\) with 2 and \(\frac{1}{2}\) with 4, so the fraction will be \(\frac{2}{8}\) and \(\frac{4}{8}\). The lengths of the fish in order from least to greatest are  \(\frac{2}{8}\), latex]\frac{4}{8}[/latex], latex]\frac{7}{8}[/latex]

Question 18.
Amy needs \(\frac{6}{8}\) gallon of fruit juice to make punch. She needs an equal amount of sparkling water. How much sparkling water does she need?
Options:
a. \(\frac{2}{8}\) gallon
b. \(\frac{1}{2}\) gallon
c. \(\frac{2}{3}\) gallon
d. \(\frac{3}{4}\) gallon

Answer: d

Explanation:
Amy needs \(\frac{6}{8}\) gallon of fruit juice to make punch and she needs an equal amount of sparkling water, so Amy needs \(\frac{6}{8}\) or \(\frac{3}{4}\) gallon.

Question 19.
Gavin is building a model of a kitchen. In the model, \(\frac{2}{5}\) of the floor tiles are white, \(\frac{1}{2}\) of the floor tiles are yellow, and \(\frac{1}{10}\) of the floor tiles are brown. How many floor tiles could be in the model?
Options:
a. 2
b. 5
c. 10
d. 17

Answer: c

Explanation:
As Gavin is building a model of a kitchen and \(\frac{2}{5}\) of the floor tiles are white, \(\frac{1}{2}\) of the floor tiles are yellow, and \(\frac{1}{10}\) of the floor tiles are brown. To find the total number of tiles we will add up all color tiles. For that, we will multiply \(\frac{1}{2}\) with 5 and \(\frac{2}{5}\) with 2 to set the denominators equal. Then the fractions will be \(\frac{5}{10}\) and \(\frac{4}{10}\). Now add all three
\(\frac{5}{10}\)+\(\frac{4}{10}\)+\(\frac{1}{10}\)
= 10.
So the number of floor tiles modeled is 10

Review/Test – Page No. 263

Fill in the bubble completely to show your answer.

Question 20.
Bill has enough money to buy no more than \(\frac{1}{2}\) pound of cheese. How much cheese could he buy?
Options:
a. \(\frac{1}{3}\) pound
b. \(\frac{4}{6}\) pound
c. \(\frac{5}{8}\) pound
d. \(\frac{3}{4}\) pound

Answer: a

Explanation:
As Bill has enough money to buy no more than \(\frac{1}{2}\) pound of cheese, so he needs to buy \(\frac{1}{3}\) pounds.

Question 21.
Students planted 6 equal-size gardens on Earth Day. They divided each garden into 3 equal sections and planted herbs in 2 of the 3 sections. What fraction of the gardens did the students plant with herbs?
Options:
a. \(\frac{3}{6}\)
b. \(\frac{2}{6}\)
c. \(\frac{6}{18}\)
d. \(\frac{12}{18}\)

Answer: d

Explanation:
As students planted 6 equal-size gardens on Earth Day, and they divided each garden into 3 equal sections and planted herbs in 2 of the 3 sections, so the fraction of the gardens did the students plant with herbs are we need to multiply 6×3 and will get 18 sections in all gardens, then we need to multiply 2×6 and get 12 sections are herbs. So, 12 out of 18 are herbs i.e \(\frac{12}{18}\).

Question 22.
Noah and Leslie live the same distance from school. Which could be the distances they live from school?
Options:
a. \(\frac{7}{100}\) kilometer and \(\frac{7}{10}\) kilometer
b. \(\frac{5}{10}\) kilometer and \(\frac{1}{5}\) kilometer
c. \(\frac{80}{100}\) kilometer and \(\frac{8}{10}\) kilometer
d. \(\frac{6}{10}\) kilometer and \(\frac{2}{5}\) kilometer

Answer: c.

Explanation:
The option c is correct, as \(\frac{80}{100}\) km is equal to \(\frac{8}{10}\) when it is reduced.

Question 23.
Keiko needs \(\frac{8}{12}\) yard of fabric to finish her quilt. What is \(\frac{8}{12}\) written in simplest form?
Options:
a. \(\frac{4}{6}\)
b. \(\frac{2}{3}\)
c. \(\frac{3}{4}\)
d. \(\frac{1}{2}\)

Answer: b

Explanation:
As Keiko needs \(\frac{8}{12}\) yard of fabric to finish her quilt and the simplest form of \(\frac{8}{12}\) is \(\frac{2}{3}\).

Review/Test – Page No. 264

Question 24.
Sam needs \(\frac{4}{6}\) cup of laundry detergent for his laundry. The cap on top of the laundry detergent holds \(\frac{1}{3}\) cup. He has 1 capful of detergent. Does he have enough? Explain.

Answer: Sam does not have enough.

Explanation:
As Sam needs \(\frac{4}{6}\) cup of laundry detergent for his laundry and the cap holds only \(\frac{1}{3}\) and Sam has 1 capful of detergent, and Sam needs 2 cups instead of 1 cup because \(\frac{4}{6}\) is equivalent to \(\frac{2}{3}\) and Sam only has \(\frac{1}{3}\) cup, so he needs 2 cups.

Question 25.
The table shows the distances of some places in town from the school.

A. Are any of the places shown in the table closer than \(\frac{1}{2}\) mile to school? Explain how you know.

Answer: Library \(\frac{3}{5}\) mile.

Explanation:
To find which place is closer, we will find the LCM of the denominators i.e (5,2,4,10). And the LCM of (5,2,4,10) is 20, so we will divide \(\frac{3}{5}\) with 4, \(\frac{1}{2}\) with 10, \(\frac{3}{4}\) with 5 and \(\frac{8}{10}\) with 2. So that the fractions will have same denominators and we can find easily which place is closer. And the fractions after multiplying are \(\frac{12}{20}\), \(\frac{10}{20}\), \(\frac{15}{20}\) and \(\frac{16}{20}\). So the places closer than \(\frac{1}{2}\) mile to school are post office which is \(\frac{10}{20}\) mile and next place is library which is \(\frac{12}{20}\) mile.

Question 25.
B. Are any of the places shown in the table the same distance from school? Explain how you know.

Answer: Yes.

Question 25.
C. Which place is farthest from school? Explain.

Answer: Townhall.

Explanation:
Townhall is the farthest from the school as it’s distance is \(\frac{8}{10}\) mile.

Conclusion:

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Go Math Grade 4 Answer Key Homework FL Chapter 7 Add and Subtract Fractions Review/Test

Go Math Grade 4 Answer Key Homework FL Review/Test holds all the topics in ch 7 Add and Subtract Fractions you might require as a part of preparation. Following this Go Math Grade 4 Review/Test Answer guide of Ch 7 Add and Subtract Fractions helps you to secure better marks in exams. Get a good grip on the Add and Subtract Fractions concepts & solve the sums within no time.

Chapter 7: Review/Test

• Review/Test – Page No. 309
• Review/Test – Page No. 310
• Review/Test – Page No. 311
• Review/Test – Page No. 312

Review/Test – Page No. 309

Choose the best term from the box.

Question 1.
A number represented by a whole number and a fraction is a _________________ .
_________

Answer:
A number represented by a whole number and a fraction is a Mixed number.

Question 2.
A fraction that always has a numerator of 1 is a _______________ .
_________

Answer:
A fraction that always has a numerator of 1 is a Unit Fraction.

Write the fraction as a sum of unit fractions.

Question 3.
\(\frac{4}{5}\) =

Answer:
\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)

Explanation:
For a unit fraction the numerator should be 1, here we can see the numerator as 4 so we will add \(\frac{1}{5}\) four times. And the fraction can be written as the sum of a unit fraction as
\(\frac{1+1+1+1}{5}\)
= \(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\).

Question 4.
\(\frac{5}{10}\) =

Answer:
\(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)

Explanation:
For a unit fraction the numerator should be 1, here we can see the numerator as 4 so we will add \(\frac{1}{5}\) four times. And the fraction can be written as the sum of a unit fraction as
\(\frac{1+1+1+1}{10}\)
= \(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\).

Write the mixed number as a fraction.

Question 5.
1 \(\frac{3}{8}\) =
\(\frac{□}{□}\)

Answer: So the answer is \(\frac{11}{8}\).

Explanation:
To convert a mixed number as a fraction, we will multiply the whole number by the fraction’s denominator, and then we will add to the numerator and the result will be on the top of the denominator.
1 \(\frac{3}{8}\)
= (1×8)+3
= 8+3
= 11
So the answer is \(\frac{11}{8}\).

Question 6.
4 \(\frac{2}{3}\) =
\(\frac{□}{□}\)

Answer: \(\frac{14}{3}\).

Explanation:
To convert a mixed number as a fraction, we will multiply the whole number by the fraction’s denominator, and then we will add to the numerator and the result will be on the top of the denominator.
4 \(\frac{2}{3}\)
= 4×3
= 12
= 12+2
= 14.
The answer is \(\frac{14}{3}\).

Question 7.
2 \(\frac{3}{5}\) =
\(\frac{□}{□}\)

Answer: \(\frac{13}{5}\).

Explanation:
To convert a mixed number as a fraction, we will multiply the whole number by the fraction’s denominator, and then we will add to the numerator and the result will be on the top of the denominator.
2 \(\frac{3}{5}\)
= 2×5
= 10
= 10+3
= 13.
The answer is \(\frac{13}{5}\).

Write the fraction as a mixed number.

Question 8.
\(\frac{12}{10}\) =
_____ \(\frac{□}{□}\)

Answer: 1 \(\frac{1}{5}\).

Explanation:
To convert the fraction to a mixed number we will divide the numerator with denominator and write the whole number, then the remainder will be written above the denominator.
\(\frac{12}{10}\)
= 12÷10
= 1 \(\frac{2}{10}\)
= 1 \(\frac{1}{5}\).

Question 9.
\(\frac{10}{3}\) =
_____ \(\frac{□}{□}\)

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

Explanation:
To convert the fraction to a mixed number we will divide the numerator with denominator and write the whole number, then the remainder will be written above the denominator.
\(\frac{10}{3}\)
= 10÷3
= 3 \(\frac{1}{3}\).

Question 10.
\(\frac{15}{6}\) =
_____ \(\frac{□}{□}\)

Answer: 2 \(\frac{1}{2}\).

Explanation:
To convert the fraction to a mixed number we will divide the numerator with denominator and write the whole number, then the remainder will be written above the denominator.
\(\frac{15}{6}\)
= 15÷6
= 2 \(\frac{3}{6}\)
= 2 \(\frac{1}{2}\).

Find the sum or difference.

Question 11.
\(2 \frac{3}{8}+1 \frac{6}{8}\) =
_____ \(\frac{□}{□}\)

Answer: \(\frac{33}{8}\).

Explanation:
\(2 \frac{3}{8}+1 \frac{6}{8}\)
= \(\frac{19}{8}\)+\(\frac{14}{8}\)
= \(\frac{33}{8}\).

Question 12.
\(\frac{9}{12}-\frac{2}{12}\) =
_____ \(\frac{□}{□}\)

Answer: \(\frac{7}{12}\).

Explanation:
\(\frac{9}{12}-\frac{2}{12}\)
= \(\frac{7}{12}\).

Question 13.
\(5 \frac{7}{10}-4 \frac{5}{10}\) =
_____ \(\frac{□}{□}\)

Answer: \(\frac{6}{5}\).

Explanation:
\(5 \frac{7}{10}-4 \frac{5}{10}\)
= \(\frac{57}{10}\)–\(\frac{45}{10}\)
= \(\frac{12}{10}\)
= \(\frac{6}{5}\).

Question 14.
\(4 \frac{1}{6}-2 \frac{5}{6}\) =
_____ \(\frac{□}{□}\)

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

Explanation:
\(4 \frac{1}{6}-2 \frac{5}{6}\)
= \(\frac{25}{6}\)–\(\frac{17}{6}\)
= \(\frac{8}{6}\)
= \(\frac{4}{3}\).

Question 15.
\(3 \frac{2}{5}-1 \frac{4}{5}\) =
_____ \(\frac{□}{□}\)

Answer: \(\frac{8}{5}\).

Explanation:
\(3 \frac{2}{5}-1 \frac{4}{5}\)
= \(\frac{17}{5}\)–\(\frac{9}{5}\)
= \(\frac{8}{5}\).

Question 16.
\(\frac{4}{12}+\frac{6}{12}\) =
\(\frac{□}{□}\)

Answer: \(\frac{5}{6}\).

Explanation:
\(\frac{4}{12}+\frac{6}{12}\)
= \(\frac{10}{12}\)
= \(\frac{5}{6}\).

Use the properties and mental math to find the sum.

Question 17.
(1 \(\frac{2}{5}\) + \(\frac{1}{5}\)) + 2 \(\frac{3}{5}\) =
_______ \(\frac{□}{□}\)

Answer: \(\frac{21}{5}\).

Explanation:
(1 \(\frac{2}{5}\) + \(\frac{1}{5}\)) + 2 \(\frac{3}{5}\)
= ( \(\frac{7}{5}\) + \(\frac{1}{5}\)) + \(\frac{13}{5}\)
= \(\frac{21}{5}\).

Question 18.
2 \(\frac{4}{6}\) + (2 \(\frac{3}{6}\) + 2 \(\frac{2}{6}\)) =
_______ \(\frac{□}{□}\)

Answer: \(\frac{45}{6}\).

Explanation:
2 \(\frac{4}{6}\) + (2 \(\frac{3}{6}\) + 2 \(\frac{2}{6}\))
= \(\frac{16}{6}\) + (\(\frac{15}{6}\)) + \(\frac{14}{6}\))
= \(\frac{16}{6}\) +(\(\frac{29}{6}\))
= \(\frac{45}{6}\).

Question 19.
\(\frac{3}{10}\) + (2 \(\frac{4}{10}\) + \(\frac{7}{10}\)) =
_______ \(\frac{□}{□}\)

Answer: \(\frac{34}{10}\).

Explanation:
\(\frac{3}{10}\) + (2 \(\frac{4}{10}\) + \(\frac{7}{10}\))
= \(\frac{3}{10}\) + (\(\frac{24}{10}\) + \(\frac{7}{10}\))
= \(\frac{3}{10}\) + ( \(\frac{31}{10}\))
= \(\frac{34}{10}\).

Review/Test – Page No. 310

Fill in the bubble completely to show your answer.

Question 20.
Eddie cut 2 \(\frac{2}{4}\) feet of balsa wood for the length of a kite. He cut \(\frac{3}{4}\) foot for the width of the kite. How much longer is the length of the kite than the width?
Options:
a. 1 \(\frac{1}{4}\) feet
b. 1 \(\frac{3}{4}\) feet
c. 2 feet
d. 3 \(\frac{1}{4}\) feet

Answer: b

Explanation:
The length of Eddie cut is 2 \(\frac{2}{4}\) feet and the width is \(\frac{3}{4}\) feet, so the difference in the length and width is 2 \(\frac{2}{4}\)– \(\frac{3}{4}\)
= \(\frac{10}{4}\)–\(\frac{3}{4}\)
= \(\frac{7}{4}\)
= 1 \(\frac{3}{4}\) feet.

Question 21.
On a trip to the art museum, Lily rode the subway for \(\frac{7}{10}\) mile and walked for \(\frac{3}{10}\) mile. How much farther did she ride on the subway than walk?
Options:
a. \(\frac{3}{10}\) mile
b. \(\frac{4}{10}\) mile
c. \(\frac{7}{10}\) mile
d. 1 mile

Answer: d

Explanation:
As Lily rode \(\frac{7}{10}\) mile and walked for \(\frac{3}{10}\) mile, so she ride total of
\(\frac{7}{10}\)+ \(\frac{3}{10}\)
= 1 mile.

Question 22.
Pablo is training for a marathon. He ran 5 \(\frac{4}{8}\) miles on Friday, 6 \(\frac{5}{8}\) miles on Saturday, and 7 \(\frac{4}{8}\) miles on Sunday. How many miles did he run on all three days ?
Options:
a. 1 \(\frac{5}{8}\) miles
b. 12 \(\frac{1}{8}\) miles
c. 19 \(\frac{4}{8}\) miles
d. 19 \(\frac{5}{8}\) miles

Answer: d

Explanation:
Pablo ran 5 \(\frac{4}{8}\) miles on Friday and 6 \(\frac{5}{8}\) miles on Saturday, 7 \(\frac{4}{8}\) miles on Sunday. So total he ran on three days is
5 \(\frac{4}{8}\)+ 6 \(\frac{5}{8}\)+7 \(\frac{4}{8}\)
= \(\frac{44}{8}\)+ \(\frac{53}{8}\)+ \(\frac{60}{8}\)
= \(\frac{157}{8}\)
= 19 \(\frac{5}{8}\) miles.

Question 23.
Cindy has two jars of paint.

Which fraction below represents how much paint Cindy has?
Options:
a. \(\frac{1}{8}\)
b. \(\frac{4}{8}\)
c. \(\frac{5}{8}\)
d. \(\frac{7}{8}\)

Answer: c

Explanation:
The first jar contains \(\frac{3}{8}\) and in the second jar \(\frac{2}{8}\) of paint. So total paint Cindy contains
\(\frac{3}{8}\)+\(\frac{2}{8}\)
= \(\frac{5}{8}\).

Review/Test – Page No. 311

Question 24.
Cole grew 2 \(\frac{3}{4}\) inches last year. Kelly grew the same amount. Which fraction below represents the number of inches that Kelly grew last year?
Options:
a. \(\frac{3}{4}\)
b. \(\frac{5}{4}\)
c. \(\frac{11}{4}\)
d. \(\frac{14}{4}\)

Answer: c

Explanation:
As Cole grew 2 \(\frac{3}{4}\) inches and Kelly has same amount which is 2 \(\frac{3}{4}\) inches, so the fraction is
\(\frac{11}{4}\) inches.

Question 25.
Olivia’s dog is 4 years old. Her cat is 1 \(\frac{1}{2}\) years younger. How old is Olivia’s cat?
Options:
a. 5 \(\frac{1}{2}\) years old
b. 3 \(\frac{1}{2}\) years old
c. 2 \(\frac{1}{2}\) years old
d. 1 \(\frac{1}{2}\) years old

Answer: c

Explanation:
Olivia’s dog is 4 years old and her cat is 1 \(\frac{1}{2}\) years younger, so Olivia’s cat is
= 4- 1 \(\frac{1}{2}\)
= \(\frac{8}{2}\) – \(\frac{3}{2}\)
= \(\frac{5}{2}\)
= 2 \(\frac{1}{2}\) years old.

Question 26.
Lisa mixed 4 \(\frac{4}{6}\) cups of orange juice with 3 \(\frac{1}{6}\) cups of milk to make a health shake. She drank 3 \(\frac{3}{6}\) cups of the health shake. How much of the health shake did Lisa not drink?
Options:
a. \(\frac{2}{6}\) cup
b. 4 \(\frac{2}{6}\) cups
c. 7 \(\frac{5}{6}\) cups
d. 11 \(\frac{2}{6}\) cups

Answer: b

Explanation:
Lisa mixed 4 \(\frac{4}{6}\) cups of orange juice with 3 \(\frac{1}{6}\) cups of milk to make a health shake, so total health shake is 4 \(\frac{4}{6}\)+3 \(\frac{1}{6}\)
= \(\frac{28}{6}\)+ \(\frac{19}{6}\)
= \(\frac{47}{6}\) cups of health shake. As she drank 3 \(\frac{3}{6}\) cups of health shake, so
= \(\frac{47}{6}\)– 3 \(\frac{3}{6}\)
= \(\frac{47}{6}\)– \(\frac{21}{6}\)
= \(\frac{26}{6}\)
= 4 \(\frac{2}{6}\) cups.

Question 27.
Keiko entered a contest to design a new school flag. Five twelfths of her flag has stars and \(\frac{3}{12}\) has stripes. What fraction of Keiko’s flag has stars and stripes?
Options:
a. \(\frac{8}{12}\)
b. \(\frac{8}{24}\)
c. \(\frac{2}{12}\)
d. \(\frac{2}{24}\)

Answer: a

Explanation:
As Keiko’s flag has Five-twelfths of stars and \(\frac{3}{12}\) of strips, so the fraction of Keiko’s flag has stars and stripes is
\(\frac{5}{12}\)+\(\frac{3}{12}\)
= \(\frac{8}{12}\).

Review/Test – Page No. 312

Constructed Response

Question 28.
Ela is knitting a scarf from a pattern. The pattern calls for 4 \(\frac{2}{12}\) yards of yarn. She has only 2 \(\frac{11}{12}\) yards of yarn. How much more yarn does Ela need to finish knitting the scarf? Explain how you found your answer.
_____ \(\frac{□}{□}\) yards

Answer: 1 \(\frac{3}{12}\) yards.

Explanation:
Ela’s pattern calls for 4 \(\frac{2}{12}\) yards of yarn and Ela has 2 \(\frac{11}{12}\) yards of yarn only, so she needs
4 \(\frac{2}{12}\)– 2 \(\frac{11}{12}\)
= \(\frac{50}{12}\) – \(\frac{35}{12}\)
= \(\frac{15}{12}\)
= 1 \(\frac{3}{12}\) yards more.

Performance Task

Question 29.
Miguel’s class went to the state fair. The fairground is divided into sections. Rides are in \(\frac{6}{10}\) of the fairground. Games are in \(\frac{2}{10}\) of the fairground. Farm exhibits are in \(\frac{1}{10}\) of the fairground.
A. How much greater is the fraction of the fairground with rides than the fraction with farm exhibits? Draw a model to prove your answer is correct.
\(\frac{□}{□}\)

Answer: \(\frac{5}{10}\).

Explanation:
As the fairground is divided into sections, rides are in \(\frac{6}{10}\) of the fairground, games are in \(\frac{2}{10}\) of the fairground and Farm exhibits are in \(\frac{1}{10}\) of the fairground. So the fraction of the fairground with rides than the fraction with farm exhibits is \(\frac{6}{10}\)– \(\frac{1}{10}\)
= \(\frac{5}{10}\) greater than farm exhibits.

Question 29.
B. What fraction of the fairground has games and farm exhibits?
Write an equation to show your answer.

Answer: \(\frac{3}{10}\).

Explanation:
The fraction of the fairground has games and farm exhibits is \(\frac{2}{10}\)+\(\frac{1}{10}\)
= \(\frac{3}{10}\).

Question 29.
C. The rest of the fairground is refreshment booths. What fraction of the fairground is refreshment booths? Describe the steps you follow to solve the problem.

Answer: 9 \(\frac{1}{10}\).

Explanation:
As the fairground is divided into sections, rides are in \(\frac{6}{10}\) of the fairground, games are in \(\frac{2}{10}\) of the fairground and Farm exhibits are in \(\frac{1}{10}\) of the fairground. So the fraction of the fairground is refreshment booths \(\frac{6}{10}\)+\(\frac{2}{10}\)+\(\frac{1}{10}\)
= \(\frac{9}{10}\).
To find a fraction of the fairground is refreshment booths we will subtract \(\frac{9}{10}\) with 10, so
10- \(\frac{9}{10}\)
= \(\frac{100-9}{10}\)
= \(\frac{91}{10}\)
= 9 \(\frac{1}{10}\).

Conclusion:

Hoping the data gave above on Go Math Grade 4 Answer Key Homework FL Chapter 7 Add and Subtract Fractions Review/Test has benefited you a lot. For solving your doubts and need more questions related to the Ch 7 Add and Subtract Fraction refer to Go Math Grade 4 Solution Key & apply them in the real world.