Texas Go Math

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 5.4 Answer Key Applying Addition and Subtraction of Integers.

Texas Go Math Grade 6 Lesson 5.4 Answer Key Applying Addition and Subtraction of Integers

Your Turn

Question 1.
Anna is in a cave 40 feet below the cave entrance. She descends 13 feet, then ascends 18 feet, Find her new position relative to the cave entrance.
Answer:
The expression that models Anna’s change of position is:
– 40 – 13 + 18
First find the difference:
– 40 – 13
Subtracting 13 is equal to adding its additive inverse -13 so you can find sum:
– 40 + (- 13)
Find absolute values of given integers:
|- 40| = 40 and |- 13| = 13
Sum that absolute values:
40 + 13 = 53
Result would be 53 or – 53, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so result is – 53.
Now, find the sum:
– 53 + 18
Find absolute values and subtract lesser absolute value from greater:
|- 53| – |18| = 35
Sum would be 35 or – 35, depending on signs of given integers.
Use the sign of integer with greater absolute value to find the sum:
– 53 + 18 = – 35
Therefore, Anna’s new position iS:
– 40 – 13 + 18 = – 35

Anna’s new position is – 35 feet
Find the difference – 40 – 13 using rules for subtracting integers and then add 18 using rules for adding integers.

Reflect

Question 2.
Communicate Mathematical Ideas Describe a different way to find the change in Irene’s account.
Answer:
You can find the change in Irenne’s account {without using the Commutative Property.
First, find sum:
– 160 + 125
Find absolute values and subtract lesser absolute value from greater:
|- 160| – |125| = 35
Sum would be 35 or – 35, depending on signs of given integers.
Use the sign of number with greater absolute value to find the sum:
– 160 + 125 = – 35
Then, calculate:
– 35 + (- 40)
Find absolute values of given integers:
|- 35| = 35 and |- 40| = 40
Sum that absolute values:
35 + 40 = 75
Given integers are negative so final result is – 75.

You can find the change in Irenne’s account without using the Commutative Property.
Find sum – 160 + 125 using rules for adding integers and then add – 40 to that sum.

Your Turn

Lesson 5.4 Answer Key 6th Grade Go Math Question 3.
Alex wrote checks on Tuesday for $35 and $45. He also made a deposit in his checking account of $180. Find the overall change in the amount in his checking account.
Answer:
Use negative integers to represent the checks that Alex wrote and positive integers to represent the deposit he made.
The expression that modeLs change in his account is:
– 35 + (- 45) + 180
First, find sum:
– 35 + (- 45)
Find absolute values of given integers:
|- 35| = 35 and |- 45| = 45
Sum that absolute values:
35 + 45 = 80
Given integers are negative so final result is -80.
Then, calculate:
– 80 + 180
Find absolute values and subtract lesser absolute value from greater:
|180| – |- 80| = 100
Sum would be 100 or – 100. depending on signs of given integers.
Use the sign of number with greater absoLute vaLue to find the sum:
– 80 + 180 = 100
Therefore, the change in his account is
$100

The overall change in his account is $100.
Find the difference – 35 – 45 using rules for subtracting integers and then add 180 to the result using rules for adding integers.

Your Turn

Question 4.
Jim and Carla are scuba diving. Jim started out 10 feet below the surface. He descended 18 feet, rose 5 feet, and descended 12 more feet. Then he rested. Carla started out at the surface. She descended 20 feet, rose 5 feet, and descended another 18 feet. Then she rested. Which person rested at a greater depth? Explain.
Answer:
Use negative integer to represent Jim’s start because he started out below the surface.
Subtract integer if he descended and add integer if he rose.
The expression you get is:
– 10 – 18 + 5 – 12
Subtracting 18 and 12 is equal to adding additive inverses – 18 and – 12 so you can find sum:
– 10 + (- 18) + 5 + (- 12)
Use Commutative Property and Associative Property to simplify addition:
– 10 + (- 18) + (- 12) + 5
– 10 – ((- 18) + (- 12)) + 5
Use rules for adding integers to get the sum in brackets and to get the final result:
– 10 + (-30) + 5
– 40 + 5 = – 35
Therefore, Jim’s position is:
– 10 – 18 + 5 – 12 = – 35

0 represents Carla’s start because she started out at the surface.
Subtract integer if she descended and add integer if she rose.
The expression you get is:
0 – 20 + 5 – 18
Subtracting 20 and 18 is equal to adding additive inverses – 20 and – 18 so you can find sum:
0 + (- 20) + 5 + (- 18)
Use Commutative Property to simplify addition:
o + (- 20) + (- 18) + 5
Use rules for adding integers to get the final result:
– 20 + (- 18) + 5
– 38 + 5 = – 33
therefore, Carla’s position is:
0 – 20 + 5 – 18 = – 33

– 35 < – 33
Jim rested at a greater depth because his position white resting was 35 feet below the surface.
Carla’s position while resting was 33 feet below the surface.

Texas Go Math Grade 6 Lesson 5.4 Guided Practice Answer Key

Write an expression. Then find the value of the expression.

Question 1.
Tomas works as an underwater photographer. He starts at a position that is 15 feet below sea level. He rises 9 feet, then descends 12 feet to take a photo of a coral reef. Write and evaluate an expression to find his position relative to sea level when he took the photo.
Answer:
Use negative integer to represent his position at start because he starts out below sea level.
Subtract integer if he descends and add integer if he rises.
The expression you get is:
– 15 + 9 – 12
Subtracting 12 is equal to adding its additive inverse – 12 so you can find sum:
– 15 + 9 + (- 12)
Use Commutative Property to simplify addition:
– 15 + (- 12) + 9
Use rules for adding integers to get the final result:
– 15 + (- 12) + 9
– 27 + 9 = – 18
Therefore, his position is:
– 15 + 9 – 12 = – 18

Tomas was 18 feet below sea level when he took the photo.
The expression that models his position is – 15 + 9 – 12.

Question 2.
The temperature on a winter night was – 23 °F. The temperature rose by 5 °F when the sun came up. When the sun set again, the temperature dropped by 7 °F. Write and evaluate an expression to find the temperature after the sun set.
Answer:
Add integer if temperature rose and subtract integer if temperature dropped.
The expression you get is:
– 23 + 5 – 7
Subtracting 7 is equal to adding its additive inverse -7 so you can find sum:
– 23 + 5 + (- 7)
Use Commutative Property to simplify addition:
– 23 + (- 7) + 5
Use rules for adding integers to get the final result:
– 23 + (- 7) + 5
– 30 + 5 = – 25
Therefore, the temperature after sun set was:
– 23 + 5 – 7 = – 25

The temperature after sun set was – 25 °F.
The expression that models temperature is – 23 + 5 – 7.

Question 3.
Jose earned 50 points in a video game. He lost 40 points, earned 87 points, then lost 30 more points. Write and evaluate an expression to find his final score in the video game.
Answer:
Add integer if Jose earned points and subtract integer if he lost points.
The expression you get is:
50 – 40 + 87 – 30
Subtracting 40 and 30 is equal to adding additive inverses – 40 and – 30 so you can find sum:
50 + (- 40) + 87 + (- 30)
Use Commutative Property and Associative Property to simplify addition:
50 + 87 + (- 40) + (- 30)
(50 + 87) + ((- 40) + (- 30))
Use rules for adding integers to get the final result:
137 + (- 70) = 67
Therefore, his final score is:
50 – 40 + 87 – 30 = 67

Jose’s final score is 67 points
The expression that models his score is 50 – 40 + 87 – 30.

Find the value of each expression.

Question 4.
– 6 + 15 + 15 = ____________
Answer:
The expression is:
– 6 + 15 + 15
Use Associative Property to simplify addition:
– 6 + (15 + 15)
Find the result in brackets first:
– 6 – 30
Use rules for adding integers to get the final result:
– 6 + 30 = 24

The value is 24.
Use Associative Property and rules for adding integers to find the result

Question 5.
9 – 4 – 17 = _______
Answer:
The expression is:
9 – 4 – 17
Subtracting 4 and 17 is equal to adding additive inverses – 4 and – 17 so you can find sum:
9 + (- 4) + (- 17)
Use Associative Property to simplify addition:
9 + ((- 4) + (- 17))
Find the result in brackets first:
9 + (- 21)
Use rules for adding integers to get the final result:
9 + (- 21) = – 12

The value is – 12.
Use Associative Property and rules for adding integers to find the value of expression.

Question 6.
50 – 42 + 10 = _________
Answer:
The expression is:
50 – 42 + 10
Subtracting 42 is equal to adding its additive inverse -42 so you can find sum:
50 + (- 42) + 10
Use Commutative Property to simplify addition:
50 + 10 + (- 42)
Use Associative Property and rules for adding integers to get the final result:
60 + (- 42) = 18

The value is 18.
Use Commutative Property, Associative Property and rules for adding integers to find the value of expression.

Question 7.
6 + 13 + 7 – 5 = ____________
Answer:
The expression is:
6 + 13 + 7 – 5
Subtracting 5 is equal to adding its additive inverse – 5 so you can find sum:
6 + 13 + 7 + (- 5)
Use Associative Property to simplify addition:
6 + (13 + 7) + (- 5)
Find the result in brackets first:
6 + 20 + (- 5)
Use Associative Property and rules for adding integers to get the final result:
26 + (- 5) = 21

The value is 21.
Use Associative Property and rules for adding integers to find the value of expression.

Question 8.
65 + 43 – 11 = _____________
Answer:
The expression is:
65 + 43 – 11
Subtracting 11 is equal to adding its additive inverse – 11 so you can find sum:
65 + 43 + (- 11)
Use Associative Property and rules for adding integers to get the final result:
108 + (- 11) = 97

The value is 97.
Use Associative Property and rules for adding integers to find the value of expression.

Go Math Grade 6 Lesson 5.4 Answer Key Question 9.
– 35 – 14 + 45 + 31 = _____________
Answer:
The expression is:
– 35 – 14 + 45 + 31
Subtracting 14 is equal to adding its additive inverse – 14 so you can find sum:
– 35 + (- 14) + 45 + 31
use Associative Property to simplify addition:
(- 35 + (- 14)) + (45 + 31)
Find results in brackets using rules for adding integers:
– 49 + 76
Use rules for adding integers to get the final result:
– 49 + 76 = 27

The value is 27.
Use Associative Property and rules for adding integers to find the value of expression.

Determine which expression has a greater value.

Question 10.
– 12 + 6 – 4 or – 34 – 3 + 39
Answer:
First expression is:
-12 + 6 – 4
Subtracting 4 is equal to adding its additive inverse – 4 so you can find sum:
– 12 + 6 + (- 4)
Use Commutative Property to simplify addition:
– 12 + (- 4) + 6
Use Associative Property and rules for adding integers to get the final result:
– 16 + 6 = – 10

Second expression is:
– 34 – 3 + 39
Subtracting 3 is equal to adding its additive inverse – 3 so you can find sum:
– 34 + (- 3) + 39
Use Associative Property and rules for adding integers to get the final result:
– 37 + 39 = 2

Compare values of first and second expression:
– 10 < 2
The expression – 34 – 3 + 39 has greater value.

The expression – 34 – 3 + 39 has greater value, its value is 2.
VaLue of expression – 12 + 6 – 4 is – 10.
Use Associative Property, Commutative Property and rules for adding integers to find values of expressions.

Question 11.
21 – 3 + 8 or – 14 + 31 – 6
Answer:
First expression is:
21 – 3 + 8
Subtracting 3 is equal to adding its additive inverse -3 so you can find sum:
21 + (- 3) + 8
Use Commutative Property to simplify addition:
21 + 8 + (- 3)
Use Associative Property and rules for adding integers to get the final result:
29 + (- 3) = 26

Second expression is:
– 14 + 31 – 6
Subtracting 6 is equal to adding its additive inverse – 6 so you can find sum:
– 14 + 31 + (- 6)
Use Commutative Property to simplify addition:
– 14 + (- 6) + 31
Use Associative Property and rules for adding integers to get the final result
– 20 + 31 = 11

Compare values of first and second expression:
26 > 11
The expression 21 – 3 + 8 has greater value

The expression 21 – 3 + 8 has greater value, its value is 26.
Value of expression 14 + 31 – 6 is 11.
Use Commutative Property Associative Property and rules for adding integers to find values of expressions.

Essential Question Check-In

Question 12.
Explain how you can find the value of the expression – 5 + 12 + 10 – 7.
Answer:
The expression is:
– 5 + 12 + 10 – 7
You can use Commutative Property, Associative Property and rules for adding and subtracting integers to find value of that expression.
Subtracting 7 is equal to adding its additive inverse – 7 so you can find sum:
– 5 + 12 + 10 + (- 7)
Use Associative Property:
– 5 + (12 + 10) + (- 7)
Calculate sum in brackets first:
– 5 + 22 (- 7)
Now, use Commutative Property once again:
– 5 + (- 7) + 22
Use Associative Property and rules for adding integers to get the final result:
– 12 + 22 = 10

You can use Commutative Property, Associative Property and rules for adding and subtracting integers to find value of that expression.
Its value is 10.

Question 13.
Sports Cameron is playing 9 holes of golf. He needs to score a total of at most 15 over par on the last four holes to beat his best golf score. On the last four holes, he scores 5 over par, 1 under par, 6 over par, and 1 under par.

a. Write and find the value of an expression that gives Cameron’s score for 4 holes of golf.
Answer:
Add integer if he scores over par and subtract integer if he scores under par.
The expression you get S:
5 – 1 + 6 – 1
Subtracting 1 is equal to adding its additive inverse – 1 so you can find sum:
5 + (- 1) + 6 + (- 1)
Use Commutative Property to simpLify addition:
5 + 6 + (- 1) + (- 1)
Use Associative Property and rules for adding integers to get the final result:
(5 + 6) + ((- 1) + (- 1))
11 + (- 2) = 9
His total score is 9 over par.

b. Is Cameron’s score on the last four holes over or under par?
Answer:
Cameron’s score on those four holes is over par.

c. Did Cameron beat his best golf score?
Answer:
Compare his score with given condition

15 over par
9 < 15
He did not beat his best score.

Question 14.
Herman is standing on a ladder that is partly in a hole. He starts out on a rung that is 6 feet under ground, climbs up 14 feet, then climbs down 11 feet. What is Herman’s final position, relative to ground level?
Answer:
Start with negative integer because he is under ground. Add integer if he climbs up and subtract integer if he
climbs down
The expression you get is:
– 6 + 14 – 11
Subtracting 11 is equal to adding its additive inverse -11 so you can find sum:
– 6 + 14 + (- 11)
Use Commutative Property to simplify addition:
– 6 + (- 11) + 14
Use Associative Property and rules for adding integers to get the final result:
– 17 + 14 = – 3

Herman’s final position is 3 feet under ground
Find value of expression – 6 + 14 – 11 using rules for adding integers.

Question 15.
Explain the Error Jerome tries to find the value of the expression 3 – 6 + 5 by first applying the Commutative Property. He rewrites the expression as 3 – 5 + 6. Explain what is wrong with Jerome’s approach.
Answer:
He tried to find value of expression:
3 – 6 + 5
Jerome applied the Commutative Property only on absolute values 6 and 5 without including signs of given numbers.
He was supposed to change places of – 6 and 5. In that case, he would get the expression:
3 + 5 – 6
Then he would be able to find the correct result.
To find the result, use Associative Property and rules for adding integers:
8 + (- 6) = 2

Jerome applied the Commutative Property only on absoLute vaLues 6 and 5 without including signs of given numbers.
He was supposed to change places of – 6 and 5.

Question 16.
Lee and Barry play a trivia game in which questions are worth different numbers of points. If a question is answered correctly, a player earns points. If a question is answered incorrectly, the player loses points. Lee currently has – 350 points.

a. Before the game ends, Lee answers a 275-point question correctly, a 70-point question correctly, and a 50-point question incorrectly. Write and find the value of an expression to find Lee’s final score.
Answer:
Start with – 350, add integer if lees answer is correct and subtract integer if his answer is incorrect.
The expression you get is:
– 350 + 275 – 70 – 50
Subtracting 50 is equal to adding its additive inverse – 50 so you can find sum:
– 350 + 275 + 70 + (- 50)
Use Associative Property to simplify addition:
– 350 + (275 + 70) + (- 50)
Find the sum in brackets first:
– 350 + 345 + (- 50)
Use Commutative Property
– 350 + (- 50) + 345
Use Associative Property and rules for adding integers to get the final result:
-400 + 345 = -55
Lee’s final score is – 55 points.

b. Barry’s final score is 45. Which player had the greater final score?
Answer:
Compare Lees and Barry’s scores:
– 55 < 45
Barry had greater final score.

Question 17.
Multistep Rob collects data about how many customers enter and leave a store every hour. He records a positive number for customers entering the store each hour and a negative number for customers leaving the store each hour.

a. During which hour did more customers leave than arrive?
Answer:
Sum number of customers entering and number of customers leaving the store for each hour.
First expression is:
30 + (-12)
Use rules for adding integers to find the sum:
30 + (- 12) = 18
Second expression is:
14 +(- 8)
Use rules for adding integers to find the sum:
14 + (- 8) = 6
Third expression is:
18 + (- 30)
Use rules for adding integers to find the sum:
18 + (- 30) = – 12
Negative number represents number of customers leaving the store so during the last hour more customers left store than arrived.

b. There were 75 customers in the store at 1:00. The store must be emptied of customers when it closes at 5:00. How many customers must leave the store between 4:00 and 5:00?
Answer:
Start with 75 and add numbers you got in part a for each hour to calculate number of customers that are in the store at 4 : 00.
the expression you get is:
75 + 18 + 6 + (- 12)
Use Associative Property and rules for adding integers to get the final result:
(75 + 18) + 6 + (- 12)
93 – 6 + (- 12)
99 + (- 12) = 87
87 customers must leave the store between 4 : 00 and 5 : 00.

The table shows the changes in the values of two friends savings accounts since the previous month.

Lesson 5.4 Applying Addition and Subtraction of Integers Question 18.
Carla had $100 in her account in May. How much money does she have in her account in August?
Answer:
Start with 100 and add integers that represent a change in Carla’s account for each month.
The expression you get is:
100 + (- 18) + 22 + (- 53)
Use Commutative Property to simplify addition:
100 + 22 + (- 18) + (- 53)
Use Associative Property and rules for adding integers to get the final result:
(100 + 22) + ((-18) + (- 53))
122 + (- 71) = 51
Carla has $51 in her account in August.

Carla has $51 in her account in August.
Use Commutative Property, Associative Property and rules for adding integers to find the result.

Question 19.
Leta had $45 in her account in May. How much money does she have in her account in August?
Answer:
Start with 45 and add integers that represent change in Leta’s account for each month.
The expression you get is:
45 + (- 17) + (- 22) + 18
Use Associative Property to simplify addition:
45 + ((- 17) + (- 22)) + 18
Find the sum in brackets first:
45 + (- 39) + 18
Use Commutative Property:
45 + 18 + (- 39)
Use Associative Property and rules for adding integers to get the finaL resuLt:
(45 + 18) – (- 39)
63 + (- 39) = 24
Leta has $24 in her account in August

Leta has $24 in her account in August
Use Commutative Property, Associative Property and rules for adding integers to find the result.

Question 20.
Analyze Relationships Whose account had the greatest decrease in value from May to August?
Answer:
Carla had $ 100 in her account in May and she has $ 51 in August. Find the difference:
100 – 51
to calculate decrease from May to August.
Subtracting 51 is equal to adding its additive inverse – 51 so you can find sum:
100 + (- 51)
Use rules for adding integers to get the final result:
100 + (- 51) = 49
Leta bad $ 45 in her account in May and she has $ 24 in August. Find the difference:
45 – 24
to calculate decrease from May to August.
Subtracting 24 is equal to adding its additive inverse – 24 so you can find sum:
45 + (- 24)
Use rules for adding integers to get the final result:
45 +(- 24) = 21
49 > 21
Carla’s count had the greatest decrease in value.

Carla’s account had the greatest decrease in value.
Carla’s account had $49 decrease in value and Leta’s account had $21 decrease.

H.O.T Focus On Higher Order Thinking

Question 21.
Represent Real-World Problems Write and solve a word problem that matches the diagram shown.

Answer:
Tom played a game in which a player gets or loses points in each turn. His score after first turn was – 1 point. He lost 6 points in his second turn and got 3 points in last, third turn. What was his final score after third turn?
Start with – 1, subtract integer if he lost points and add integer if he got points.
The expression you get is:
– 1 – 6 + 3
Subtracting 6 is equal to adding its additive inverse – 6 so you can find sum:
– 1 + (- 6) + 3
Use Associative Property to simplify addition:
(- 1 + (- 6)) + 3
Find the sum in brackets and then add 3. Use number line to find the result.
Start at – 1 and move 6 units to the left to represent adding – 6.
Then, move 3 units to the right to represent adding 3.
Read the result from number line below.

Tom’s final score after third turn was – 4 points.

Tom played a game in which a player gets or loses points in each turn. His score after first turn was – 1 point. He lost 6 points in his second turn and got 3 points in last, third run. What was his final score after three turns?
His final score after third turn was 4 points.

Question 22.
Critical Thinking Mary has $10 in savings. She owes her parents $50. She does some chores and her parents pay her $12. She also gets $25 for her birthday from her grandmother. Does Mary have enough money to pay her parents what she owes them? If not, how much more money does she need? Explain.
Answer:
Calculate the sum of amount Mary has in savings and amounts she gets from parents and grandmother and compare that sum with the amount she owes her parents.
The sum you need to find is:
10 + 12 + 25
Use Associative Property and rules for adding integers to get the final result:
(10 + 12) + 25
22 + 25 = 47
Compare that sum with $50:
47 < 50
Miry doesn’t have enough money to pay her parents.
She needs $3 more to pay her parents because she has $3 less than $50, the amount she owes.

Mary doesn’t have enough money to pay her parents because she has $3 less than $50, the amount she owes.
She needs $3 more to pay her parents.

Question 23.
Draw Conclusions An expression involves subtracting two numbers from a given first number. Under what circumstances will the value of the expression be negative? Give an example.
Answer:
The expression is:
a – b – c
The value of this expression will be negative if the sum:
b + c
is greater than first number, a.
For example, find the value of an expression:
4 – 5 – 1
Subtracting 5 and 1 is equal to adding additive inverses – 5 – 1 so you can find sum:
4 + (- 5) + (- 1)
Use Associative Property and rules for adding integers to find that sum:
4 + (- 5 + (- 1))
4 + (- 6) = – 2
Compare 4 and sum 5 + 1:
4 < 6

The value of the expression will be negative if the sum b + c is greater than first number, a.
For example, the value of an expression 4 – 5 – 1 is – 2 where 4 is lesser than 5 + 1.

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 5.1 Answer Key Adding Integers with the Same Sign.

Texas Go Math Grade 6 Lesson 5.1 Answer Key Adding Integers with the Same Sign

Texas Go Math Grade 6 Lesson 5.1 Explore Activity Answer Key

Modeling Sums of Integers with the Same Sign

You can use colored counters to add positive integers and to add negative integers.

Model with two-color counters.

(A) 3 + 4

How many counters are there in total? __________
What is the sum and how do you find it?

(B) – 5 + (- 3)

How many counters are there in total? __________
Since the counters are negative integers, what is the sum? __________

Reflect

Question 1.
Communicate Mathematical Ideas When adding two numbers with the same sign, what sign do you use for the sum?
Answer:
The sum of two positive numbers is a positive number
The sum of two negative numbers is a negative number.
So, when you add two numbers with the same sign the sum has exactly that sign

Go Math Grade 6 Lesson 5.1 Answer Key Question 2.
What If? Suppose the temperature is – 1°F and drops by 3°F. Explain how to use the number line to find the new temperature.
Answer:
First graph number line with units.
Start at 0 and move 1 unit left to represent number – 1.
Then, start at – 1 move 3 units left (because the temperature drops by 3), and read the new temperature from the number line.

Graph:

New temperature would be – 4 because it is 4 units left from 0 on the number line.

Question 3.
Communicate Mathematical Ideas How would using a number line to find the sum 2 + 5 be different from using a number line to find the sum – 2 + (- 5)?
Answer:
To find 2 + 5 on number line you start at 0, then move 2 units right and then move 5 units also right

To find – 2 + (- 5) on number line you start at 0, then move 2 units left and then move 5 units also left
So, only difference is where you move, left or right.

It is different direction of moving on number line.
For positive right, for negative left.

Question 4.
Analyze Relationships What are two other negative integers that have the same sum as – 2 and – 5?
Answer:
First, graph number line with units.
Start at 0 and move 2 unit left to represent number – 2.
Then, start at – 2 and move 5 units left to add – 5 and read sum from number Line. It would be – 7.
So, it is 7 units left from 0, and you also could move 3 units left and then 4 units left and to be again 7 units left from 0.
Thus:
– 2 + (-5) = – 3 + (- 4)

Two other negative integers that have the same sum as – 2 and – 5 are -3 and -4.

Question 5.
Communicate Mathematical Ideas Does the Commutative Property of Addition apply when you add two negative integers? Explain.
Answer:
It does.
You can apply Commutative Property when you add two negative integers because in both cases sum would be negative and equal minus sum of absolute values of that integer.

Example:
– 4 + (- 8) = – (|- 4| + |- 8|) = – (4 + 8)
– 8 + (- 4) = – (|- 8| + |- 4|) = – (8 + 4)
– (4 + 8) = – (8 + 4) → – 4 + (- 8) = – 8 + (- 4)

Adding Integers Lesson 5.1 Answer Key 6th Grade Question 6.
Critical Thinking Choose any two negative integers, Is the sum of the integers less than or greater than the value of either of the integers? Will this be true no matter which integers you choose? Explain.
Answer:
The sum of two negative integers would always be less than the value of either of those integers.

This is because the sum of two

The sum of two negative integers would always be less than the value of either of that integers.
Use a + b = – (|a| + |b|) for negative a and b.

Your Turn

Find each sum.

Question 7.
– 8 + (- 1) = ____________
Answer:
First, find absolute values of given integers:
|- 8| = 8 and |- 1| = 1

Then, sum that absolute values:
8 + 1 = 9
Final result would be 9 or – 9, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 9.

Result is – 9.
Find absolute values, sum them and add minus if both were negative.

Question 8.
– 3 + (- 7) = __________
Answer:
First find absolute values of given integer:
| – 3| = 3 and |- 7| = 7
Then, sum that absolute values:
3 + 7 = 10
Final result would be 10 or – 10, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 10.

Question 9.
– 48 + (- 12) = __________
Answer:
First find absolute values of given integers:
|- 48| = 48 and |- 12| = 12
Then, sum that absolute values
48 + 12 = 60
Final result would be 60 or – 60, depending on sign of given integers (both positive then +, both negative then – ).
Given integers are negative so final result is – 60.

Question 10.
– 32 + (- 38) = _________
Answer:
First find absolute values of given integers:
|- 32| = 32 and |- 38| = 38
Then, sum that absolute values
32 + 38 = 60
Final result would be 70 or – 70, depending on sign of given integers (both positive then +, both negative then – ).
Given integers are negative so final result is – 70.

Adding Integers with the Same Sign Lesson 5.1 Answer Key Question 11.
109 + 191 = ___________
Answer:
First find the absolute values of given integers:
|109| = 109 and |191| = 191
Then, sum that absolute values
109 + 191 = 300
The final result would be 300 or – 300, depending on the sign of given integers (both positive then +, both negative then – ).
Given integers are positive the final result is 300.
Note:
You could simply add them, but this is another (harder) way that shows that adding integers with the same sign as it was explained is correct

Question 12.
– 40 + (- 105) = ________
Answer:
First find absolute values of given integers:
|- 40| = 40 and |- 105| = 105
Then, sum that absolute values
40 + 105 = 145
Final result would be 145 or – 145, depending on sign of given integers (both positive then +, both negative then – ).
Given integers are negative so final result is – 145.

Question 13.
– 150 + (- 1,500) = ________
Answer:
First find absolute values of given integers:
|- 150| = 150 and |- 1,500| = 1,500
Then, sum that absolute values
150 + 1,500 = 1,650
Final result would be 1,650 or – 1,650 depending on sign of given integers (both positive then +, both negative then – ).
Given integers are negative so final result is – 1,650.

Question 14.
– 200 + (- 800) = _______
Answer:
First find absolute values of given integers:
|- 200| = 200 and |- 800| = 800
Then, sum that absolute values
200 + 800 = 1,000
Final result would be 1,000 or – 1,000, depending on sign of given integers (both positive then +, both negative then – ).
Given integers are negative so final result is – 1,000.

Texas Go Math Grade 6 Lesson 5.1 Guided Practice Answer Key

Find each sum.

Question 1.
– 5 + (- 1)

Answer:
In the first row there are 5 negative counters which represent – 5 and in second row there is 1 negative counter which represent – 1.
So, on the given picture there are 6 counters and all of them are negative.
(which represent – 6)
Thus, sum would be:
– 5 – (- 1) = – 6
On the picture there are 6 negative counters and thus – 5 + (- 1) = – 6.

a. How many counters are there? ________
Answer:
6

b. Do the counters represent positive or negative numbers? ____________
Answer:
negative

c. – 5 + (- 1) = _____________
Answer:
– 6

Go Math Lesson 5.1 Answer Key 6th Grade Question 2.
– 2 + (- 7)

Answer:
In the first row there are 2 negative counters which represent – 2 and in second row there is 7 negative counter which represent – 7.
So, on the given picture there are 9 counters and all of them are negative.
(which represent – 9)
Thus, sum would be:
– 2 + (- 7) = – 9
On the picture there are 6 negative counters and thus – 2 + (- 7) = – 9.

a. How many counters are there? ________
Answer:
9

b. Do the counters represent positive or negative numbers? ____________
Answer:
negative

c. – 2 + (- 7) = _____________
Answer:
– 9

Model each addition problem on the number line to find each sum.

Question 3.
– 5 + (- 2) = _____________

Answer:
First, graph number line with units.
Start at 0 and move 5 unit left to represent number – 5.
Then, start at – 5 and move 2 units left (to add – 2) and read sum from number line

Number line:

Sum is – 7 because it is 7 units left from 0 on the number line.

Question 4.
– 1 + (- 3) = _____________

Answer:
First, graph number line with units.
Start at 0 and move 1 unit left to represent number – 1.
Then, start at – 1 and move 3 units left (to add – 3) and read sum from number line

Number line:

Sum is – 4 because it is 4 units left from 0 on the number line.

Question 5.
– 3 + (- 7) = _____________

Answer:
First, graph number line with units.
Start at 0 and move 3 unit left to represent number – 3.
Then, start at – 3 and move 7 units left (to add – 7) and read sum from number line

Number line:

Sum is – 10 because it is 10 units left from 0 on the number line.

Question 6.
– 4 + (- 1) = _____________

Answer:
First, graph number line with units.
Start at 0 and move 4 unit left to represent number – 4.
Then, start at – 4 and move 1 units left (to add – 1) and read sum from number line

Number line:

Sum is – 5 because it is 5 units left from 0 on the number line.

Question 7.
– 2 + (- 2) = _____________

Answer:
First, graph number line with units.
Start at 0 and move 2 unit left to represent number – 2.
Then, start at – 2 and move 2 units left (to add – 2) and read sum from number line

Number line:

Sum is – 4 because it is 4 units left from 0 on the number line.

Adding Integers with the Same Sign Lesson 5.1 Go Math 6th Grade Question 8.
– 6 + (- 8) = _____________

Answer:
First, graph number line with units.
Start at 0 and move 6 unit left to represent number – 6.
Then, start at – 6 and move 8 units left (to add – 8) and read sum from number line

Number line:

Sum is – 14 because it is 14 units left from 0 on the number line.

Find each sum.

Question 9.
– 5 + (- 4) = _____________
Answer:
First, find absolute values of given integers:
| – 5| = 5 and |- 4| = 4
Then, sum that absolute values:
5 + 4 = 9
Final result would be 9 or – 9, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 9.

Result is – 9
Find absolute values, sum them and add minus if both were negative.

Question 10.
– 1 + (- 10) = __________________
Answer:
First, find absolute values of given integers:
| – 1| = 1 and |- 10| = 10
Then, sum that absolute values:
1 + 10 = 11
Final result would be 11 or – 11, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 11.

Result is – 11
Find absolute values, sum them and add minus if both were negative.

Question 11.
– 9 + (- 1) = ____________________
Answer:
First, find absolute values of given integers:
| – 9| = 9 and |- 1| = 1
Then, sum that absolute values:
9 + 1 = 10
Final result would be 10 or – 10, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 10.

Result is – 10
Find absolute values, sum them and add minus if both were integers were negative.

Question 12.
– 90 + (- 20) = __________________
Answer:
First, find absolute values of given integers:
| – 90| = 90 and |- 20| = 20
Then, sum that absolute values:
90 + 20 = 110
Final result would be 110 or – 110, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 110.

Result is – 110
Find absolute values, sum them and add minus if both were integers were negative.

Question 13.
– 52 + (- 48) = __________________
Answer:
First, find absolute values of given integers:
| – 52| = 52 and |- 48| = 48
Then, sum that absolute values:
52 + 48 = 100
Final result would be 100 or – 100, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 100.

Result is – 100
Find absolute values, sum them and add minus if both were integers were negative.

Question 14.
5 + 198 = __________________
Answer:
First, find absolute values of given integers:
| 5 | = 5 and |198| = 198
Then, sum that absolute values:
5 + 198 = 203
Final result would be 203 or – 203, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is 203.

Note: You could simply add them, but this is another way.

Result is – 203
Find absolute values, sum them and add minus if both were integers were negative.

Question 15.
– 4 + (- 5) + (- 6) = _____________
Answer:
First, find absolute values of given integers:
|- 4| = 4 |- 5| = 5 and | – 6| = 6
Then, sum that absolute values:
4 + 5 + 6 = 15
Final result would be 15 or – 15, depending on sign of given integers (all positive then +, all negative then -).
Given integers are negative so final result is – 15.

Result is – 15
Find absolute values, sum them and add minus if given integers were negative.

Question 16.
– 50 + (- 175) + (- 345) = __________________
Answer:
First, find absolute values of given integers:
|- 50| = 50 |- 175| = 175 and | – 345| = 345
Then, sum that absolute values:
50 + 175 + 345 = 570
Final result would be 570 or – 570, depending on sign of given integers (all positive then +, all negative then -).
Given integers are negative so final result is – 570.

Result is – 570
Find absolute values, sum them and add minus if given integers were negative.

Essential Question Check-In

Question 17.
How do you add integers with the same sign?
Answer:
First, find absolute values of given integers;
Then, sum that absolute values;
If given integers were negative add minus.

Find absolute values, sum them and add minus if given integers were negative.

Go Math Grade 6 Lesson 5.1 Homework Answer Key Question 18.
Represent Real-World Problems Jane and Sarah both dive down from the surface of a pool. Jane first dives down 5 feet, and then dives down 3 more feet. Sarah first dives down 3 feet and then dives down 5 more feet.

a. Multiple Representations Use the number line to model the equation
– 5 + (- 3) = – 3 + (- 5).

Answer:
First graph number Line with units.
Representing – 5 + (- 3):
Start at 0 and move 5 unit left to represent number – 5.
Then, start at – 5 and move 3 units left (to add – 3).
Now, you are 8 units Left from 0 on the number Line.
Representing – 3 + (- 5):
Start at 0 and move 3 unit left to represent number – 3.
Then, start at – 3 and move 5 units left (to add – 5).
Again, you are 8 units left from 0 on the number line.
So, it is equal.

Number line:

b. Does the order in which you add two integers with the same sign affect the sum? Explain.
Answer:
The order does not affect the sum because you move on number line in the same direction and get to the same value in both cases.

Question 19.
A golfer has the following scores for a 4-day tournament.

What was the golfer’s total score for the tournament?
Answer:
Total score equals:
– 3 + (- 1) + (- 5) + (- 2)
First find absolute values of given integers:
|- 3| = 3
|- 1| = 1
|- 5| = 5
|- 2| = 2
Then, sum that absolute values:
3 + 1 + 5 + 2 = 11
Final result would be 11 or – 11, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 11.

The golfer’s total score was – 11.
Find absolute values, sum them and add minus if all integers were negative.

Question 20.
A football team loses 3 yards on one play and 6 yards on another play. Write a sum of negative integers to represent this situation. Find the sum and explain how it is related to the problem.
Answer:
Let represent the team’s score:
(-3) + (-6)
First, find absolute vaLues of given integers:
|- 3| = 3 and |- 6| = 6
Then, sum that absolute values
3 + 6 = 9
Final result would be 9 or – 9, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 9.
That means that the team loses 9 yards on two plays

Sum is – 9.
Find absolute values, sum them and add minus if both integers were negative.
That means that the team loses 9 yards on two plays.

Question 21.
When the quarterback is sacked, the team loses yards. In one game, the quarterback was sacked four times. What was the total sack yardage?

Answer:
The total sack yardage equals:
– 14 + (- 5) + (- 12) + (- 23)
First find absolute values of given integers:
|- 14| = 14
|- 5| = 5
|- 12| = 12
|- 23| = 23
Then, sum that absolute values
14 + 5 + 12 + 23 = 54
Final result would be 54 or – 54, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 54.

The total sack yardage was – 54.
Find absolute values, sum them and add minus if all integers were negative.

Question 22.
Multistep The temperature in Jonestown and Cooperville was the same at 1:00. By 2:00, the temperature in Jonestown dropped 10 degrees, and the temperature in Cooperville dropped 6 degrees. By 3:00, the temperature in Jonestown dropped 8 more degrees, and the temperature in Cooperville dropped 2 more degrees.

a. Write an equation that models the change to the temperature in Jonestown since 1:00.
Answer:
Let calculate the change to the temperature in Jonestown:
(- 10) + (- 8)
First find absolute values of given integers:
|- 10| =10 and |- 8| = 8
Then, sum that absolute values
10 + 8 = 18
Final result would be 18 or – 18, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 18.
The temperature in Jonestown dropped 18 degrees since 1:00.

b. Write an equation that models the change to the temperature in Cooperville since 1:00.
Answer:
Let calculate the change to the temperature in Cooperville:
(- 6) + (- 2)
First, find absolute values of given integers:
|- 6| = 6 and |- 2| = 2
Then, sum that absolute values
6 + 2 = 8
Final result would be – 8 or 8, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 8.
The temperature in Cooperville dropped 8 degrees since 1:00.

c. Where is it colder at 3:00, Jonestown or Cooperville?
Answer:
Since the temperature was the same in both cities at 1:00, Jonestown is colder at 3:00, because – 18 is less than – 8.

Question 23.
Represent Real-World Problems Julio is playing a trivia game. On his first turn, he lost 100 points. On his second turn, he lost 75 points. On his third turn, he lost 85 points. Write a sum of three negative integers that models the change to Julio’s score after his first three turns.
Answer:
Show how the score changes after three turns:
– 100 + (- 75) + (- 85)
First, find absoLute vaLues of given integers:
|- 100| = 100
|- 75| = 75
|- 85| = 85
Then, sum that absolute values
100 + 75 + 85 = 260
Final result would be 260 or – 260, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 260
Julio’s score got – 260 points after three turns.

The sum is
– 100 + (- 75) + (- 85)
and the change to his score is – 260.
Find absolute values, sum them and add minus if all integers were negative.

H.O.T. Focus on Higher Order Thinking

Question 24.
Multistep On Monday, Jan made withdrawals of $25, $45, and $75 from her savings account. On the same day, her twin sister Julie made withdrawals of $35, $55, and $65 from her savings account.

a. Write a sum of negative integers to show Jan’s withdrawals on Monday. Find the total amount Jan withdrew.
Answer:
Let calculate Jan’s withdrawals:
– 25 + (- 45) + (- 75)
First find absolute values of given integers:
|- 25| = 25
|- 45| = 45
|- 75| = 75
then, sum that absoLute vaLues:
25 + 45 + 75 = 145
Final resuLt would be 145 or – 145, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 145
Jan withdrew $145.

b. Write a sum of negative integers to show Julie’s withdrawals on Monday. Find the total amount Julie withdrew.
Answer:
Let calculate Julie’s withdrawals:
– 35 + (- 55) + (- 65)
First find absolute values of given integers:
|- 35| = 35
|- 55| = 55
|- 65| = 65
Then, sum that absolute values
35 + 55 + 65 = 155
Final result would be 155 or – 155, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 155.
Julie withdrew $155.

c. Julie and Jan’s brother also withdrew money from his savings account on Monday. He made three withdrawals and withdrew $10 more than Julie did. What are three possible amounts he could have withdrawn?
Answer:
Their brother also made three withdrawals, and his total amount is $ 10 bigger then Julies.
He could have withdrawn first amount $10 bigger. Let calculate the first with drawal:
– 35 + (- 10)
First, find absolute values of given integers :
|- 35| = 35
|- 10| = 10
Then, sum that absolute values:
35 + 10 = 45
Final result would be 45 or – 45, depending on sign of given integers
(both positive then +. both negative then -).
Given integers are negative so final result is – 45.
He could have withdrawn amounts $ 45, $ 55 and $65.

Go Math Grade 6 Lesson 5.1 Integers Same Sign Answer Key Question 25.
Communicate Mathematical Ideas Why might you want to use the Commutative Property to change the order of the integers in the following sum before adding?
– 80 + (- 173) + (- 20)
Answer:
First, write down the sum:
– 80 + (- 173) + (- 20)
You might want to use the Commutative Property before adding because it makes adding easier
If you change the order of the integers – 173 and – 20, you will get the sum:
– 80 + (- 20) + (- 173)
Let’s calculate the sum
– 80 + (- 20)
First, find the absolute values of given integers:
|- 80| = 80 and |- 20| = 20
Then, sum that absolute values
80 + 20 = 100
Final result would be 100 or – 100, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 100.
Now it is easy to calculate
– 100 + (- 173)

Using the Commutative Property to change the order of the second and third integers before adding makes adding easier.

Question 26.
Critique Reasoning The absolute value of the sum of two different integers with the same sign is 8. Pat says there are three pairs of integers that match this description. Do you agree? Explain.
Answer:
First pair of negative integers is – 1 and 7.
Find absolute values of given integers:
|- 1| = 1 and |- 7| = 7
Sum that absolute values:
1 + 7 = 8
The absolute value of the sum is 8.
Second pair is – 2 and – 6.
Find absolute values of given integers:
|- 2| = 2 and |- 6| = 6
Sum that absolute values:
2 + 6 = 8
The absolute value of the sum is 8.
Third pair is – 3 and – 5.
Find absolute values of given integers:
– 3 = 3 and – 5 = 5
Sum that absolute values:
3 + 5 = 8
The absolute value of the sum is 8.

First pair of positive integers is 1 and 7.
Find absolute values of given integers:
|1| = 1 and |7| = 7
Sum that absolute values:
1 + 7 = 8
The absolute value of the sum is 8
Second pair is 2 and 6.
Find absolute values of given integers:
|2| = 2 and |6| = 6
Sum that absolute values
2 + 6 = 8
The absolute value of the sum ¡s 8
Third pair is 3 and 5.
Find absolute values of given integers:
|3| = 3 and |5| = 5
Sum that absolute values:
3 + 5 = 8
The absolute value of the sum is 8.

There are six pairs of integers with the same sign: – 1 and – 7, – 2 and – 6, – 3 and – 5, 1 and 7, 2 and 6, 3 and 5.

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Module 5 Answer Key Adding and Subtracting integers.

Texas Go Math Grade 6 Module 5 Answer Key Adding and Subtracting integers

Texas Go Math Grade 6 Module 5 Are You Ready? Answer Key

Write an integer to represent each situation.

Question 1.
an elevator ride down 27 stories
Answer:
First, you need to decide whether the integer is positive or negative
ride down → it would be negative
So, unknown integer is – 27 (minus is for negative and 27 is given number).

Question 2.
a $ 700 profit
Answer:
First, you need to decide whether the integer is positive or negative
Profit → it would be positive
So, unknown integer is 700 (it is positive 700 is given number).

Go Math Grade 6 Module 5 Answer Key Pdf Question 3.
46 degrees below zero
Answer:
First, you need to decide whether the integer is positive or negative
below zero → it would be negative
So, unknown integer is – 46 (minus is for negative and 46 is the given number).

Question 4.
a gain of 12 yards
Answer:
First, you need to decide whether the integer is positive or negative
gain → it would be positive.
So, unknown integer is 12 (it is positive and 12 is given number).

Find the sum or difference.

Question 5.

Answer:
First, add ones:
3 + 8 = 11
Now, 11 ones equals 1 ten plus 1.

Then, add tens:
8 + 7 + 1 = 16
Now, 16 tens equals 1 hundred plus 6 tens.

Then, add hundreds:
1 + 0 + 1 = 2 (last 1 is hundred from adding tens)
Finally, write 2 for hundreds:

Final result is 261.

First add ones, then tens and then hundreds Result is 261.

Grade 6 Module 5 Answer Key Go Math Question 6.

Answer:
First, Subtract ones.
Note that 7 < 8, so regroup 1 ten as 10 ones
7 ones + 10 ones = 17 ones (and 6 tens left)
Now, 17 – 8 = 9

Then, Subtract tens.
Note that 6 < 8, so regroup 1 hundred as 10 tens.
6 ones + 10 ones = 16 ones
Now, 16 – 8 = 8:

Then, subtract hundreds. Because 5 > 2 only subtract them:
5 – 2 = 3
Finally:

Final result is 389.

First add ones, then tens and then hundreds Result is 389.

Question 7.

Answer:
First, add ones:
8 + 2 = 10
Now, 10 ones equals 1 ten plus 0.

Then, add tens:
8 + 0 + 1 = 0 (1 is ten from adding ones)
Now, write 9 for tens.

Then, add hundreds:
1 + 9 = 10
Now, 10 hundreds equals 1 thousand plus 0 hundred

Finally, add thousand:
1 + 0 + 1 = 2 (1 is thousand from adding hundreds)

Now, write 2 for thousand:

Final result is 2090.

First add ones, then tens, then hundreds and finally thousands Result is 2090.

Question 8.

Answer:
First, Subtract ones. Because 7 > 5 only subtract them:
7 – 5 = 2
Now:

Then, subtract tens
Note that 4 < 8, so regroup 1 hundred as 10 tens
4 ones + 10 ones = 14 ones (and 5 hundreds left)
Now, 14 – 8 = 6:

Then, subtract hundreds.
Note that 5 < 8, so regroup 1 thousand as 10 hundreds.
5 hundreds + 10 hundreds = 15 hundreds (and 1 thousand left)
Now, 15 – 8 = 7

First, subtract thousands:
1 – 1 = 0
Now:

Final result is 762.

First subtract ones, then tens, then hundreds and then thousands.
Regroup if necessary. Result is 762.

Graph each number on the number line.

Question 9.
7
Answer:
First, graph number line with units.
To graph 7 start at 0 and move 7 units to the right.

Graph:

Go Math Grade 6 Answer Key Module 5 Question 10.
– 4
Answer:
First, graph the number line with units.
To graph – 4 start at 0 and move 4 units to the left.

Graph:

Question 11.
– 9
Answer:
First, graph number line with units.
To graph – 9 start at 0 and move 9 units to the left.

Graph:

Question 12.
4
Answer:
First, graph number line with units.
To graph 4 start at 0 and move 4 units to the right.

Graph:

Texas Go Math Grade 6 Module 5 Reading Start-Up Answer Key

Visualize Vocabulary
Use the ✓ words to fill in the ovals on the graphic. You may put more than one word in each oval.

Understand Vocabulary

Complete the sentences using the preview words.

Question 1.
The ____________________ of a number gives its distance from zero.
Answer:
Correct answer is “absolute value”
This is because:
|4| = 4, which is its distance from 0.
|- 9| = 9, which is its distance from 0.

Module 5 Grade 6 Answer Key Go Math Question 2.
The sum of a number and its _________________ is zero.
Answer:
Correct answer is “additive inverse”
For example:
Additive inverse of 10 is – 10.
Now:
10 + (- 10) = 10 – 10 = 0

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 5.2 Answer Key Adding Integers with Different Signs.

Texas Go Math Grade 6 Lesson 5.2 Answer Key Adding Integers with Different Signs

Reflect

Question 1.
Make a Prediction Predict the sum of – 2 + 2. Explain your prediction and check it using the number line.

Answer:
The sum of 2 + 2 is 0 because you start at – 2 and move |2| = 2 units right on the number line.
First graph number line with units.
Start at – 2 and move 2 units to the right.
You are 0 units left from 0 on the number line.
The sum is 0.

Number line:

The sum of – 2 + 2 is 0 because you start at – 2 and move 2 units right, or in the positive direction.

Go Math Grade 6 Lesson 5.2 Answer Key Question 2.
Make a Prediction Kyle models a sum of two integers. He uses more negative (red) counters than positive (yellow) counters. What do you predict about the sign of the sum? Explain.
Answer:
The sign of the sum that Kyle models is – (minus).

One positive (yellow) counter and one negative (red) counter form a zero pair, so when you eliminate alt zero pairs, there are some negative (red) counters left.

The sign of the sum that Kyle models is minus, or – because he uses more negative counters.

Your Turn

Model and find each sum using counters.

Question 3.
5 + (- 1) ____________
Answer:
The sum you want to find is:
5 + (- 1)
Begin with 5 yellow counters to represent 5 (see the picture bellow).
Then, add 1 red counter to represent adding – 1.
Eliminate zero pairs.
When you remove the zero pairs, there are 4 yellow counters left.

The final result is:
5 + (- 1) = 4

There are 5 positive counters and 1 negative and thus 5 + (-1) = 4.

Question 4.
4 + (- 6) ____________
Answer:
The sum you want to find is:
4 + (- 6)
Start with 4 yellow counters to represent 4.
After 6 red counter to represent adding – 6.
Eliminate zero pairs.
When you remove the zero pairs, there are 2 red counters left.

The final result is:
4 + (- 6) = – 2

There are 4 positive and 6 negative counters and thus 4 + (-6) = – 2.

Question 5.
1 + (- 7) ____________
Answer:
The sum you want to find is:
1 + (- 7)
Start with 1 yellow counters to represent 1.
After 7 red counter to represent adding – 7.
Eliminate zero pairs.
When you remove the zero pairs, there are 6 red counters left.

The final result is:
1 + (- 7) = – 6

There are 1 positive and 5 negative counters and thus 1 + (-7) = – 6.

Lesson 5.2 Answer Key 6th Grade Go Math Question 6.
3 + (- 4) ____________
Answer:
The sum you want to find is:
3 + (- 4)
Start with 3 yellow counters to represent 3.
After 4 red counter represents adding – 4.
Eliminate zero pairs.
When you remove the zero pairs, there are 1 red counters left.

The final result is:
3 + (-4) = – 1

There are 3 positive and 4 negative counters and thus 3 + (-4) = – 1.

Your Turn

Find each sum.

Question 7.
– 51 + 23 = ____________
Answer:
Find absolute values of both integers and subtract lesser absolute value from greater:
|- 51| – |23| = 28
The final result would be 28 or – 28, depending on signs of given integers.
To find the final result, use the sign of number with greater absolute value:
– 51 + 23 = – 28

The sum is – 28.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.

Question 8.
10 + (- 18) = ____________
Answer:
Find absolute values of both integers and subtract lesser absolute value from greater:
|- 18| – |10| = 8
The final result would be 8 or – 8, depending on signs of given integers.
To find the final result, use the sign of number with greater absolute value:
10 + (- 18) = – 8

The sum is – 8.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.

Question 9.
13 + (- 13) = ____________
Answer:
The opposite of an integer is called its additive inverse.
Sum of integer and its additive inverse is 0
so final result is:
13 + (- 13) = 0

The sum is 0 because the sum of an integer and its additive inverse is 0.

Question 10.
25 + (- 26) = ____________
Answer:
Find absolute values of both integers and subtract lesser absolute value from greater:
|- 26| – |25| = 1
The final result would be 1 or – 1, depending on signs of given integers.
To find the final result, use the sign of number with greater absolute value:
25 + (- 26) = – 1

The sum is – 1.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.

Texas Go Math Grade 6 lesson 5.2 Guided Practice Answer Key

Use a number line to find each sum.

Question 1.
9 + (- 3) = __________

Answer:
First graph number line with units.
Start at 9 and move |- 3| = 3 units in the negative direction (to the left) on the number line.
To find the final result read sum from number line.

Number line:

The sum is 6 because it is 6 units right from 0 on the number line.

Question 2.
– 2 + 7 = ____________

Answer:
First graph number line with units.
Start at – 2 and move |7| = 7 units in the negative direction (to the left) on the number line.
To find the final result, read sum from number line.

Number line:

The sum is 5 because it is 5 units right from 0 on the number line.

Go Math Lesson 5.2 6th Grade Answer Key Question 3.
– 15 + 4 = _____________

Answer:
First graph number line with units.
Start at – 15 and move |4| = 4 units in the negative direction (to the left) on the number line.
To find the final result, read sum from number line.

Number line:

The sum is – 11 because it is 11 units right from 0 on the number line.

Question 4.
1 + (- 4) = _____________

Answer:
First graph number line with units.
Start at 1 and move |-4| = 4 units in the negative direction (to the left) on the number line.
To find the final result, read sum from number line.

Number line:

The sum is – 3 because it is 3 units right from 0 on the number line.

Circle the zero pairs in each model. Find the sum.

Question 5.
– 4 + 5 = _____________

Answer:
The sum you want to find is:
– 4 + 5
Start with 4 red counters to represent – 4 and add 5 yellow counters to represent 5.
Find zero pairs and circle them.
When you remove the zero pairs, there is 1 yellow counter left so the sum is:
– 4 + 5 = 1

There are 4 negative counters and 5 positive counters and thus – 4 + 5 = 1.

Question 6.
– 6 + 6 = _____________

Answer:
The sum you want to find is:
– 6 + 6
Start with 6 red counters to represent – 6 and add 6 yellow counters to represent 6.
Find zero pairs and circle them.
When you remove the zero pairs, there is 0 yellow counter left so the sum is:
– 6 + 6 = 0

There are 6 negative counters and 6 positive counters and thus – 6 + 6 = 0.

Lesson 5.2 Go Math 6th Grade Answer Key Question 7.
2 + (- 5) = _____________

Answer:
The sum you want to find is:
2 + (- 5)
Start with 2 yellow counters to represent 2 and add 5 red counters to represent adding – 5.
Find zero pairs and circle them.
When you remove the zero pairs, there are 3 red counters left so the sum is:
2 + (- 5) = – 3

There are 2 positive counters and 5 negative and thus 2 + (- 5) = – 3.

Question 8.
– 3 + 7 = ______________

Answer:
The sum you want to find is:
– 3 + 7
Start with 3 red counters to represent – 3 and add 7 yellow counters to represent adding 7.
Find zero pairs and circle them.
When you remove the zero pairs, there are 4 yellow counters left so the sum is:
– 3 + 7 = 4

There are 3 negative and 7 positive counters and thus – 3 + 7 = 4

Find each sum.

Question 9.
– 8 + 14 = ______________
Answer:
Find absolute values and subtract lesser absolute value from greater:
|14| – |- 8| = 6
The final result would be 6 or – 6, depending on signs of given integers.
To find the final result, use the sign of number with greater absolute value:
– 8 + 14 = 6

The sum is 6.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.

Question 10.
7 + (- 5) = ______________
Answer:
Find absolute values and subtract lesser absolute value from greater:
|7| – |- 5| = 2
The final result would be 2 or – 2, depending on signs of given integers.
To find the final result, use the sign of number with greater absolute value:
7 + (- 5) = 2

The sum is 2.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.

Lesson 5.2 Adding Integers with Opposite Signs Answer Key Question 11.
5 + (- 21) = ______________
Answer:
Find absolute values and subtract lesser absolute value from greater:
|- 21| – |5| = 16
The final result would be 16 or – 16, depending on signs of given integers.
To find the final result, use the sign of a number with greater absolute value:
5 + (- 21) = – 16

The sum is 16.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.

Question 12.
14 + (- 14) = ______________
Answer:
The opposite of an integer is called its additive inverse.
Sum of integer and its additive inverse is 0

So final result is:
14 + (- 14) = 0

The sum is 0 because the sum of an integer and its additive inverse is 0.

Question 13.
0 + (- 5) = ______________
Answer:
Find absolute values and subtract lesser absolute value from greater:
|- 5| – |0| = 5
The final result would be 5 or – 5, depending on signs of given integers.
To find the final result, use the sign of number with greater absolute value:
0 + (- 5) = – 5

The sum is – 5.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.

Question 14.
32 + (- 8) = ______________
Answer:
Find absolute values and subtract lesser absolute value from greater:
|32| – |- 8| = 24
The final result would be 24 or – 24, depending on signs of given integers.
To find the final result, use the sign of number with greater absolute value:
32 + (- 8) = 24

The sum is 24.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.

Essential Question Check-In

Question 15.
Describe how to find the sums – 4 + 2 and – 4 + (- 2) on a number line.
Answer:
First sum is:
– 4 + 2
Graph number line with units.
Start at – 4 and move |2| = 2 units to the right to represent adding 2.
Read the sum from number line
Second sum is:
– 4 + (-2)
Start at – 4 and move |- 2| = 2 units to the left to represent adding – 2.
Again, read the sum from number line

Number line:

First sum is – 2 because it is 2 units left from 0 on the number line.
Second sum is – 6 because it is 6 units left from 0 on the number line.

To find these sums, start from – 4 on the number line and move 2 units to the right to represent the first sum, and 2 unit to the left to represent the second sum.
The first sum is – 2 and the second sum is – 6.

Find each sum.

Question 16.
– 15 + 71 = ______________
Answer:
Find absolute values and subtract lesser absolute value from greater:
|71| – |- 15| = 56
The final result would be 56 or – 56, depending on signs of given integers.
To find the final result use the sign of a number with greater absolute value:
– 15 + 71 = 56

The sum is 56.
Find absolute values, subtract them, and add minus if the integer with the greater absolute value was negative.

Adding Integers with Counters Lesson 5.2 6th Grade Question 17.
– 53 + 45 = ______________
Answer:
Find absolute values and subtract lesser absolute value from greater:
|- 53| – |- 45| = 8
The final result would be 8 or – 8, depending on signs of given integers.
To find the final result use the sign of number with greater absolute value:
– 53 + 45 = – 8

The sum is – 8.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.

Question 18.
– 79 + 79 = ______________
Answer:
The opposite of an integer is called its additive inverse.
The sum of an integer and its additive inverse is 0
so the final result is
– 79 + 79 = 0

The sum is 0 because the sum of an integer and its additive inverse is 0.

Question 19.
– 25 + 50 = ______________
Answer:
Find absolute values and subtract lesser absolute value from greater:
|50| – |- 25| = 25
The final result would be 25 or – 25, depending on signs of given integers.
To find the final result use the sign of number with greater absolute value:
– 25 + 50 = 25

The sum is 25.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.

Question 20.
18 + (- 32) = ______________
Answer:
Find absolute values and subtract lesser absolute value from greater:
|- 32| – |18| = 14
The final result would be 14 or – 14, depending on signs of given integers.
To find the final result use the sign of number with greater absolute value:
18 + (- 32) = – 14

The sum is – 14.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.

Question 21.
5 + (- 100) = ______________
Answer:
Find absolute values and subtract lesser absolute value from greater:
|- 100| – | 5 | = 95
The final result would be 95 or – 95, depending on signs of given integers.
To find the final result use the sign of number with greater absolute value:
5 + (- 100) = – 95

The sum is – 95.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.

Question 22.
– 12 + 8 + 7 = ______________
Answer:
First find the sum:
– 12 + 8
Find absolute values and subtract lesser absolute value from greater:
|- 12| – 8 = 4
The result would be 4 or – 4, depending on signs of given integers.
To find the result, use the sign of number with greater absolute value:
– 12 + 8 = – 4
Now, find the sum:
– 4 + 7
Find absolute values and subtract lesser absolute value from greater:
|7| – |- 4| = 3
The final result would be 3 or – 3, depending on signs of given integers.
To find the result, use the sign of number with greater absolute value:
– 4 + 7 = 3
The final result is 3.
The sum is 3.
First, find the sum – 12 + 8 (find absolute values, subtract them and minus if the integer with the greater absolute value was negative) and then add 7 to that sum.

Question 23.
– 8 + (- 2) + 3 = ______________
Answer:
First find the sum:
– 8 + (- 2)
Find absolute values of given integers:
|- 8| = 8 and |- 2| = 2
Then, sum that absolute values
8 + 2 = 10
The result would be 10 or – 10, depending on sign of given integers (both positive then +, both negative then -)
Given integers are negative so the result is – 10.
Now, find the sum:
– 10 + 3
Subtract the lesser absolute value from the greater:
|- 10| – |3| = 7
The final result would be 7 or – 7, depending on signs of given integers.
To find the result, use the sign of the number with the greater absolute value:
– 10 + 3 = – 7
The final result is – 7.
The sun is – 7
First, find the sum – 8 + (- 2) (find absolute values, sum them and add minus if all integers were negative) and then add 3 to that sum (find absolute values, subtract them and add minus if the integer with the greater absolute value was negative).

Go Math Lesson 5.2 6th Grade Adding Integers with Different Signs Question 24.
15 + (- 15) + 200 = ______________
Answer:
First, find the sum:
15 + (- 15)
The sum is 0 because
the sum of a number and its additive inverse is 0
Now, find the sum:
0 + 200
The final result is 200.
The sum is 200
First, find the sum 15 + (-15) (the sum of 15 and its additive inverse is 0) and then add 200 to that sum.

Question 25.
– 500 + (- 600) + 1200 = ______________
Answer:
First, find the sum:
– 500 + (- 600)
Find absolute values of given integers:
|- 500| = 500 and |- 600| = 600
Then, sum that absolute values
500 + 600 = 1100
The result would be 1100 or – 1100, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative the result is – 1100
Now, find the sum:
– 1100 + 1200
Subtract the lesser absolute value from the greater:
|1200| – |- 1100| = 100
The final result would be 100 or – 100, depending on signs of given integers
To find the result, use the sign of the number with the greater absolute value:
– 1100 + 1200 = 100
The final result is 100.
The sun is 100
First, find the sum – 500 + (- 600) (find absolute values, sum them and add minus if all integers were negative) and then add 1200 to that sum (find absolute values, subtract them and add minus if the integer with the greater absolute value was negative).

Question 26.
A football team gained 9 yards on one play and then lost 22 yards on the next. Write a sum of integers to find the overall change in field position. Explain your answer.
Answer:
You can represent the change in field position by writing the sum:
9 + (- 22)
9 shows gain on one play and adding – 22 shows loss on the next
Now, to find the overall change, subtract lesser absolute value from greater:
|- 22| – |9| = 13
Final result would be 13 or – 13, depending on signs of integers.
Since the number with the greater absolute value is – 22, final result is:
9 + (- 22) = – 13

Gain on one play is represented with positive integer 9 and loss on the next game is represented with negative integer – 22 so the sum is 9 + (- 22).
The overall change in field position after two plays is – 13.

Question 27.
A soccer team is having a car wash. The team spent $55 on supplies. They earned $275, including tips. The team’s profit is the amount the team made after paying for supplies. Write a sum of integers that represents the team’s profit.
Answer:
A soccer team spout $55 on supplies and earned $275 so the sum that represents the team’s profit is:
– 55 + 275
Find absolute values and subtract lesser absolute value from greater:
|275| – |- 55|= 220
Final result would be 220 or – 220. depending on sighs of given integers.
Since 275 ix integer with greater absolute value, the sum is:
– 55 + 275 = 220

The sum that represents soccer team’s profit is – 55 + 275 and the profit is $220.
Find absolute values, subtract them and add minus if integer with greater absolute value was negative.

Question 28.
As shown in the illustration, Alexa had a negative balance in her checking account before depositing a $47.00 check. What is the new balance of Alexa’s checking account? What property did you use to find the sum?

Answer:
The new balance in Alexa’s account is the sum of:
– 47 + 47
Since 47 is additive inverse of – 47 and the sum of integer and its additive inverse is 0, new balance in her account is:
– 47 + 47 = 0

New balance of Alexa’s account is 0.
A property that is used to find the sum is the Inverse Property of Addition.

Question 29.
The sum of two integers with different signs is 8. Give two possible integers that fit this description.
Answer:
Two possible integers with different signs whose sum is 8 are

You can check it:
Find absolute values and subtract lesser absolute value from greater:
|20| – |- 12| = 8
Final result would be 8 or – 8, depending on signs of given integers.
To find the sum, use the sign of number with greater absolute value:
20 + (- 12) = 8

Two possible integers with different signs whose sum is 8 are 20 and – 12.
Find absolute values, subtract them and add minus it the integer with the greater absolute value was negative.

Question 30.
Multistep Bart and Sam played a game in which each player earns or loses points in each turn. A player’s total score after two turns is the sum of his points earned or lost. The player with the greater score after two turns wins. Bart earned 123 points and lost 180 points. Sam earned 185 points and lost 255 points. Use a problem-solving model to find which person won the game. Explain.
Answer:
Bait’s total score is the sum:
123 + (- 180)
To find his score, subtract lesser absolute value from greater:
|- 180| – |123| = 57
Final result would be 57 or – 57, depending on signs of given integers.
Use the sign of number with greater absolute value to find the result:
123 + (- 180) = – 57
Bait’s total score after two turns is – 57 points

Sam’s total score is the sum:
185 + (- 255)
To find his score, subtract lesser absolute value from greater:
|- 255| – |185| = 70
Final result would be 70 or – 70, depending on signs of given integers.
Use the sign of number with greater absolute value to find the result:
185 + (- 255) = – 70
Sam’s total score after two turns is – 70 points.

Since both Bart and Sam have negative score after two turns, winner is the player who lost less points.
Bart lost 5 points and Sam lost 70 points so Bart won the game.

H.O.T. FOCUS ON HIGHER ORDER THINKING

Question 31.
Critical Thinking Explain how you could use a number line to show that – 4 + 3 and 3 + (- 4) have the same value. Which property of addition states that these sums are equivalent?
Answer:
Use number line to represent sums:
– 4 + 3 and 3 + (- 4)
First graph number line with units.
To find first sum, start at – 4 and move 3 units to the right to represent adding 3 (yellow arrow).
To find second sum, start at 3 and move 4 units to the left to represent adding – 4 (red arrow).
Read both sums from number line
The Commutative Property states that these sums are equivalent

Number line:

First sum is -1 because it is 1 unit left from 0 on the number line.
Second sum is -1 because it is 1 unit left from 0 on the number line.

Question 32.
Represent Real-World Problems Jim is standing beside a pool. He drops a weight from 4 feet above the surface of the water in the pool. The weight travels a total distance of 12 feet down before landing on the bottom of the pool. Explain how you can write a sum of integers to find the depth of the water.
Answer:
Jim drops a weight from 4 feet above the surface and it travels 12 feet down until landing on the bottom so the sum that represents the depth of the water is:
4 + (- 12)
To calculate the sum, subtract lesser absolute value from greater:
|- 12| – |4| = 8
Result would be 8 or – 8, depending on signs of given integers
Use the sign of number with greater absolute value to find result:
4 + (- 12) = – 8
Since you want to find the depth of the water, use the absolute value of the result because the depth of the water is positive number. The depth of the water is 8 feet.

Sum that represents the depth is 4 + (-12).
The absolute value of this sum is the depth of the water and that would be 8 feet

Question 33.
Communicate Mathematical Ideas Use counters to model two integers with different signs whose sum is positive. Explain how you know the sum is positive.
Answer:
Yellow counters represent positive integer and red counters represent negative integer.
To model positive sum of two integers with different sign, take some yellow and some red counters and let there be more yellow counters.
Form and eliminate zero pairs.
When you remove zero pairs, there are some yellow counters left so the sum is positive.

For example:

There are more yellow than red counters When you remove zero pairs, there would be some yellow counters left so the sum is positive.

Question 34.
Analyze Relationships You know that the sum of – 5 and another integer is a positive integer. What can you conclude about the sign of the other integer? What can you conclude about the value of the other integer? Explain.
Answer:
Since the sum of – 5 and another integer is positive, that other integer is positive.
In case second integer is negative, the sum would be negative, too, which is not correct.
To find the sum of – 5 and another integer, you subtract the lesser absolute value from the greater and then you use the sign of number with greater absolute value.
Therefore, you would use the sign of that other integer so the absolute value of that integer is greater than absolute value of – 5

The sign of second integer is positive because the sum is positive
Absolute value of second integer is greater than absolute value of – 5 because you use the sign of number with greater absolute value to find the sum of integers with different signs.

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 4.3 Answer Key Applying Multiplication and Division of Rational Numbers.

Texas Go Math Grade 6 Lesson 4.3 Answer Key Applying Multiplication and Division of Rational Numbers

Your Turn

Question 1.
Some friends went out to dinner. The bill was $14.50. Charles paid for \(\frac{3}{5}\) of the entire bill. How much did Charles pay? Show how to solve the problem two different ways.
Answer:
The first way:
We will first convert decimal to a fraction and then multiply.
After that we will write the product in simplest form:

Charlic paid is $ 8\(\frac{7}{10}\)
The second way:
We will convert fraction to a decimal and then multiply.
\(\frac{3}{5}\) × 14.50 = 0.6 × 14.50 = 8.7
We can notice that Charlie paid $ 8.7.

Go Math Grade 6 Lesson 4.3 Answer Key Question 2.
Shaneeka is saving up to buy a camera that costs $65.60. She has already saved \(\frac{3}{4}\) of the amount she needs for the camera. How much has Shaneeka saved so far? Explain your solution.
Answer:
In order to calculate how much Shaneeka saved, we need to multiply 65.60 by \(\frac{3}{4}\). We will convert the decimal to a fraction and then multiply.
After that we will write the product in simplest form.

So, Shaneeka saved 49\(\frac{1}{5}\)

Question 3.
Serena charges $6.50 an hour to walk dogs. She worked 5 hours last week and 9.5 hours this week. How much did she make altogether in the last two weeks? Show your work.
Answer:
We actually need to calculate the value of the following expression to find how much Serena made altogether in the last two weeks.
(5 ÷ 9.5) ÷ 6.50
First we will add to find the total working hours in the last two weeks.
5 + 9.5 = 14.5
Now, we will multiply 14.5 by 6.50:
14.5 × 6.50 = 94.25
We can see that Serena made 94.25 in the last two weeks.

Texas Go Math Grade 6 Lesson 4.3 Guided Practice Answer Key

Solve. Show your work.

Question 1.
Chanasia has 8.75 gallons of paint. She wants to use \(\frac{2}{5}\) of the paint to paint her living room. How many gallons of paint will Chanasia use?
Answer:
Convert the fraction to a decimal
\(\frac{2}{5}\) × 8.75 = 0.4 × 8.75
Multiply
0.4 × 8.75 = 3.5
Chanasia use 3.5 gallons of point

Question 2.
Bob and Cheryl are taking a road trip that is 188.3 miles. Bob drove \(\frac{5}{7}\) of the total distance. How many miles did Bob drive?
Answer:
Convert the fraction to a decimal
\(\frac{5}{7}\) × 188.3 = 0.71 × 188.3
Multiply
0.71 × 188.3 = 133.7
Bob drives 133.7 miles

Lesson 4.3 Go Math Answer Key Grade 6 Question 3.
The winner of a raffle will receive \(\frac{3}{4}\) of the money raised from raffle ticket sales. The raffle raised $530.40. How much money will the winner of the raffle get?
Answer:
We need to multiply \(\frac{3}{4}\) by 530.40 in order to find how much money the winner of the ruffle will get. We will convert the decimal to a fraction and then multiply. After that we will write the product in simplest form.

So, the winner of the raffle will get $397\(\frac{4}{5}\).

Question 4.
Multistep An object that weighs 1 pound on Earth weighs only 0.17 pound on the Moon. Zach’s dog weighs 17\(\frac{1}{2}\) pounds on Earth. How much more does his dog weigh on Earth than on the Moon?
Answer:
We will calculate how much Zach’s dog weighs on the Moon multiplying 17\(\frac{1}{2}\) by 0.17 We will first convert the decimal to a fraction and then multip1y After that we will write the product in the simplest form.

So, Zach’s dog weights 2\(\frac{39}{40}\) pounds on the Moon.
Now we will calculate the difference between Zach’s dog weight on Earth and on the Moon.

Conclusion is that Zach’s dog weights for 14\(\frac{21}{40}\) pounds on the Earth more than on the Moon.

Question 5.
Naomi has earned $54 mowing lawns the past two days. She worked 2.5 hours yesterday and 4.25 hours today. If Naomi is paid the same amount for every hour she works, how much does she earn per hour to mow lawns?
Answer:
We actually need to calculate the value of the following expression to find how much earns per hour to mow lawns:
51 ÷ (2.5 + 4.25)
First, we will add to find the total working hours of Naomi:
2.5 + 1.25 = 6.75
Now, we will divide 51 by 6.75. but first we need to change the divisor to a whole number by multiplying by a power of 100, and then multiply the dividend by the same power of 100 and get:

Conclusion is that Naomi earns $8 per hour.

Lesson 4.3 Answer Key 6th Grade Go Math Question 6.
Harold bought 3 pounds of red apples and 4.2 pounds of green apples from Blue Star Grocery, where both kinds of apples are $1.75 a pound. How much did Harold spend on apples?
Answer:
The total weight of all apples is:
3 + 4.2 = 7.2
Multiply
7.2 × 1.75 = 12.6
Harold spent $12.6 on apples

Essential Question Check-In

Question 7.
How can you solve a problem that involves multiplying or dividing a fraction by a decimal?
Answer:
The first way is that we can convert the decimal to a fraction and then multiply.
The second way is that we can convert fraction to a decimal and then multiply
Convert the decimal to a fraction or convert fraction to a decimal

Samuel and Jason are brothers. They sell cans to a recycling center that pays $0.40 per pound of cans. The table shows the number of pounds of cans that Samuel and Jason sold for several days.

Question 8.
Samuel wants to use his earnings from Monday Wednesday 12.5 and Tuesday to buy new batteries for his video game device. The batteries cost $5.60 each. How many batteries can Samuel buy? Show your work.
Answer:
We actually need to calculate the value of the following expression to find how many bateries Samuel can buy:
(16.2 + 11.8) ÷ 5.60
First, we will add to find how much money Samuel has:
16.2 + 11.8 = 28
Now, we will divide 28 by 5.60. but first we need to change the divisor to a whole number by multiplying by a power of 100, and then multiply the dividend by the same power of 100 and get:

Conclusion is that Samuel can buy 5 batteries.

Question 9.
Jason wants to use his earnings from Monday and Tuesday for online movie rentals. The movies cost $2.96 each to rent. How many movies can Jason rent? Show your work.
Answer:
We actually need to calculate the value of the following expression to find how many movies Jason can rent:
(11.5 + 10.7) ÷ 2.96
First, we will add to find how much money Jason has:
11.5 + 10.7 = 22.2
Now, we will divide 22.2 by 2.96, but first we need to change the divisor to a whole number by multiplying by a power of 100. amid then multiply the dividend by the same power of 100 and get:

The result is 7.5, so, the conclusion is that Jason can rent 7 movies.

Go Math Lesson 4.3 Answer Key 6th Grade Question 10.
Multistep Samuel and Jason combine their earnings from Wednesday to buy their mother a gift. They spend \(\frac{3}{4}\) of their shared earnings on the gift. How much do they spend? Use a problem-solving model to determine if there is enough left over from Wednesday’s earnings to buy a greeting card that costs $3.25. Explain.
Answer:
If Samuel and Jason combine their earnings from Wednesday, they will have the following sum of money:
12.5 + 7.1 = 19.6
Now, we will calculate how much they spent on the gift by multiplying \(\frac{3}{4}\) by 1 9.6. We will convert a decimal to a fraction and multiply.

So, they spent $ 14\(\frac{7}{10}\) on the gift.
In order to determine if there is enough money to buy a greeting card, we need to subtract the sum of money they spent on a gift, 14\(\frac{7}{10}\), from Wednesday’s earnings, which is 19.6.
19.6 – 14\(\frac{7}{10}\) = 19.6 – 14.7 = 4.9
They have $4.9 left over from Wednesday’s earnings and the greeting card costs $3.25, so. they have enough money to buy greeting cards.
$14\(\frac{7}{10}\); they have enough money to buy greeting card

Question 11.
Eva wants to make two pieces of pottery. She needs \(\frac{3}{5}\) pound of clay for one piece and \(\frac{7}{10}\) pound of clay for the other piece. She has three bags of clay that weigh \(\frac{4}{5}\) pound each. How many bags of clay will Eva need to make both pieces of pottery? How many pounds of clay will she have left over?
Answer:
First, we will calculate how much clay she needs. We will sum \(\frac{3}{5}\) and \(\frac{7}{10}\) and get:
\(\frac{3}{5}+\frac{7}{10}=\frac{6+7}{10}=\frac{13}{10}\)
So, she needs \(\frac{13}{10}\) pounds of clay.
Now, we will calculate how much clay she has, we will multiply 3 by \(\frac{4}{5}\) and get:
3 × \(\frac{4}{5}=\frac{3 \times 4}{1 \times 5}=\frac{12}{5}\)
In order to find how many bags of clay Eva will need to make both pieces of pottery, we will divide \(\frac{13}{10}\) by \(\frac{4}{5}\) and get:

So, she will need 2 bags of clay to make two pieces of pottery.
Now, we will calculate how many pounds of clay she will have left over subtracting \(\frac{13}{8}\) from \(\frac{12}{5}\) and get:
\(\frac{12}{5}-\frac{13}{8}=\frac{96-65}{40}=\frac{31}{40}\)
So, she will have \(\frac{31}{10}\) pounds of clay left over.

Question 12.
Multiple Representations Give an example of a problem that could be solved using the expression 9.5 × (8 + 12.5).
Answer:
Samanta is paid 9.5 per hour for her work. This week she worked 12.5 hours and last week she worked 8 hours.
How much money she earned in past two weeks?

Tony and Alice are trying to reduce the amount of television they watch. For every hour they watch television, they have to put $2.50 into savings. The table shows how many hours of television Tony and Alice have watched in the past two months.

Question 13.
Tony wants to use his savings at the end of March to buy video games.
The games cost $35.75 each. How many games can Tony buy?
Answer:
We will first calculate how much money Tony has ax. the cud of March. We will add 35.4 and 18.2 and get:
35.4 + 18.2 = 53.6
So, Tony has $ 53.6. Now, we will divide 53.6 by 35.75 to calculate how many games Tony can buy, but find we need to change the divisor to a whole number by multiplying by a power of 100. and then multiply the dividend by the same power of 100 and get:

So, Tony has money for one video game.

Go Math 5th Grade Lesson 4.3 Answer Key Question 14.
Alice wants to use her savings at the end of March to buy fabric for a crafts project, The fabric costs $17.50 a yard. How many full yards of fabric can Alice buy?
Answer:
We will first calculate how much money Alice has at the end of March. We will add 21.8 and 26.6 and get:
21.8 + 26.6 = 18.1
Alice has $ 48.4. Now, we will divide 48.4 by 17.50 to find how many full yards of fabric she can buy. but first, we need to change the divisor to a whole number 1w multiplied by a power of 10, and then multiply the dividend by the same power of 10:

So, Alice has money to buy 2 full yards of fabric.

H.O.T. Focus On Higher Order Thinking

Question 15.
Multiple Representations You are measuring walnuts for banana walnut oatmeal and a spinach and walnut salad. You need \(\frac{3}{8}\) cup of walnuts for the oatmeal and \(\frac{3}{4}\) cup of walnuts for the salad. You have a \(\frac{1}{4}\)-cup scoop. Describe two different ways to find how many scoops of walnuts you will need.
Answer:
First way is to add \(\frac{3}{4}\) and \(\frac{3}{8}\) in order to find how many cups of walnut we need.
\(\frac{3}{4}+\frac{3}{8}=\frac{6+3}{8}=\frac{9}{8}\)
Now, we will divide \(\frac{9}{8}\) by \(\frac{1}{4}\) and get:

The other way is to convert fractions to a decimals and divide like that. Here, \(\frac{9}{8}\) = 1.125 and \(\frac{1}{25}\) = 0.25. but first we need to change the divisor to a whole number by multiplying by a power of 100. and then multiply the dividend by the same power of 100.  In this case we have the following:

So, we will need 4.5 or 4\(\frac{1}{2}\) scoops of walnuts.

Nadia charges $7.50 an hour for babysitting. She babysits 18.5 hours the first week of the month and 20 hours the second week of the month.

Question 16.
Explain the Error To find her total earnings for those two weeks, Nadia writes 7.5 × 18.5 + 20 = $158.75. Explain her error. Show the correct solution.
Answer:
The error is this:
7.5 × 18.5 + 7.5 × 20 = 288.75, she should multiply the rate she charges with both the set of hours she worked and not just one. It can also be done using bracket operation : (20 + 18.5) × 7.5 = 288.75

Go Math Lesson 4.3 6th Grade Multiplication and Division of Rational Numbers Question 17.
What If? Suppose Nadia raises her rate by $0.75 an hour. How many hours would she need to work to earn the same amount of money she made in the first two weeks of the month? Explain.
Answer:
Nadia earned in the past two weeks the following sum of money:
7.50 × (18.5 + 20) = 7.50 × 38.5 = 288.75
Now, Nadia raised her rate by $0.75 an hour. so, now she charges:
0.75 × 7.50 = 5.625
we will divide 288.75 by 5.625 in order to find how many hours she will need to work, but first, we need to change the divisor to a whole number by multiplying by a power of 1000, and then multiplying the dividend by the same power of 1000 and get:

So, she will need to work 51.3 hours in the first two weeks of the month to earn the same amount of money as before.

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Module 4 Answer Key Multiplying and Dividing Decimals.

Texas Go Math Grade 6 Module 4 Answer Key Multiplying and Dividing Decimals

Texas Go Math Grade 6 Module 4 Are You Ready? Answer Key

Write the decimals represented by the shaded square.

Question 1.

Answer:
We have that 10 squares are 10 of 100 equal parts, or \(\frac{10}{100}\) or 0.1.
Here we have 70 shaded squares, so, they represent 7 × 0.1, or 0.7.

Grade 6 Module 4 Answer Key Go Math Question 2.

Answer:
We have that 10 squares are 10 of 100 equal parts, or \(\frac{10}{100}\) or 0.1.
Here we have 70 shaded squares, so, they represent 4 × 0.1, or 0.4.

Question 3.

Answer:
We have that 1 square is 1 of 100 equal parts, or \(\frac{1}{100}\), or 0.01.
Also, 10 squares are 10 of 100 equal parts, or \(\frac{10}{100}\), or 0.1.
Here we have 53 shaded squares, we can rewrite them as 503.
So, 3 squares represent 3 × 0.01, or 0.03 and 50 squares represent 50 × 0.1, or 0.5.
Finally, 53 squares represent 0.03 + 0.5 = 0.53

Grade 6 Module 4 Answer Key Texas Go Math Question 4.

Answer:
Here, we have 100 shaded squares, they are 100 of 100 equal parts, or \(\frac{100}{100}\), or 1.

Find the product.

Question 5.
0.49 × 10 ___________
Answer:
Number of zeros in 10 is 1, so, we have to move the decimal point 1 place to the right and we get that the result is 4.9.

Question 6.
25.34 × 1,000 ___________
Answer:
The number of zeros in 1000 is 3, so, we have to move the decimal point 3 places to the right and we get that the result is 25, 340
Here, because we had two decimal places, we added 0 at the third place because we moved the decimal point to 3 places.

Texas Go Math Grade 6 Answer Key Module 4 Question 7.
87 × 100 ___________
Answer:
Here, the number of zeros in 100 is 2. In this case we will add two zeros to 87 and get that the result is 8,700

Write a numerical expression for the word expression.

Question 8.
20 decreased by 8 _______________
Answer:
Decreased means “to subtract 8 from 20”, so, required numerical expression would be 20 – 8.

Question 9.
the quotient of 14 and 7 _______
Answer:
The quotient means the answer in a division problem, so, the required numerical expression would be \(\frac{14}{7}\)

Question 10.
the difference between 72 and 16 _________
Answer:
The difference between two numbers means to subtract them, so, the required expression would be 72 – 16.

Go Math Grade 6 Module 4 Answer Key Question 11.
the sum of 19 and 3 __________
Answer:
The sum of two numbers means to add them, so, the required expression would be 19 + 3.

Texas Go Math Grade 6 Module 4 Reading Start-Up Answer Key

Visualize Vocabulary

Use the ✓ words to complete the chart. You may put more than one word in each section.

Answer:

Understand Vocabulary

Match the term on the left to the definition on the right.

Answer:
We have the following:
Division is split something into equal groups, so, the conclusion is that 1 matches to C.
The denominator is the bottom number in a fraction, so, 2 matches to A.
The quotient is the answer in a division problem, so, the conclusion is that 3 matches to D.
Numerator is the top number in a fraction, so, the conclusion is that 4 matches to B.

1 – C
2 – A
3 – D
4 – B

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 4.2 Answer Key Dividing Decimals.

Texas Go Math Grade 6 Lesson 4.2 Answer Key Dividing Decimals

Reflect

Question 1.
Multiple Representations When using models to divide decimals, when might you want to use grids divided into tenths instead of hundredths?
Answer:
We can use grids divided into tenths if we divide decimals with one decimal place.

Question 2.
Check for Reasonableness How can you estimate to check that your quotient in A is reasonable?
Answer:
We can estimate to check that our quotient is reasonable dividing the whole number which is closest to given decimal by already given whole number.

Your Turn

Divide

Question 3.

Answer:
Solution to this example is given below

Go Math Dividing Decimals Grade 6 Lesson 4.2 Question 4.

Answer:

Question 5.

Answer:
The divisor has one decimal, place, so multiply both the dividend and the divisor by 10 so that the divisor is a whole number

0.5 × 10 = 5
4.25 × 10 = 42.5

Question 6.

Answer:
The divisor has two decimal place, so mutliply both the dividend and the divisor by 100 so that the divisor is a whole number

0.84 × 100 = 84
15.12 × 100 = 1512

Texas Go Math Grade 6 Lesson 4.2 Guided Practice Answer Key

Divide

Question 1.

Answer:
Here we will divide using long division as with whole numbers. will place a decimal point in the quotient directly above the decimal point in the dividend. So, we have the following:

So, the quotient is 7.4

Dividing Decimals Answer Key Go Math Lesson 4.2 Grade 6 Question 2.

Answer:
Solution to this example is given below

Question 3.

Answer:
The divisor has one decimal place, so multiply both the dividend and the divisor by 10 so that the divisor is a whole number

2.4 × 10 = 24
16.8 × 10 = 168

Question 4.

Answer:
The divisor has two decimal place, so multiply both the dividend and the divisor by 100 so that the divisor is a whole number

0.96 × 100 = 96
0.144 × 10 = 14.4

Question 5.
38.5 ÷ 0.5 = _____________
Answer:
The divisor has one decimal place, so multiply both the dividend and the divisor by 10 so that the divisor is a whole number

0.5 × 10 = 5
38.5 × 10 = 385

Question 6.
23.85 ÷ 9 _____________
Answer:
Solution to this example is given below

Go Math Grade 6 Lesson 4.2 Answer Key Question 7.
5.6375 ÷ 0.71 _____________
Answer:
The divisor has two decimal places, so multiply both the dividend and the divisor by 100 so that the divisor is a whole number

0.17 × 100 = 17
5.6372 × 100 = 563.72

Question 8.
8.19 ÷ 4.2 _____________
Answer:
The divisor has one decimal place, so multiply both the dividend and the divisor by 10 so that the divisor is a whole number

4.2 × 10 = 42
8.19 × 10 = 81.9

Question 9.
66.5 ÷ 3.5 _____________
Answer:
The divisor has one decimal place, so multiply both the dividend and the divisor by 10 so that the divisor is a whole number

3.5 × 10 = 35
66.5 × 10 = 665

Question 10.
0.234 ÷ 0.78 _____________
Answer:
The divisor has two decimal place, so multiply both the dividend and the divisor by 100 so that the divisor is a whole number

0.78 × 100 = 78
0.234 × 100 = 23.4

Divide

Question 11.
78.74 ÷ 12.7 _____________
Answer:
The divisor has one decimal place, so multiply both the dividend and the divisor by 10 so that the divisor is a whole number

12.7 × 10 = 127
78.74 × 10 = 787.4

Divide

Question 12.
36.34 ÷ 0.09 _____________
Answer:
The divisor has two decimal place, so multiply both the dividend and the divisor by 100 so that the divisor is a whole number

0.09 × 100 = 9
36.45 × 100 = 3645

Dividing Decimals for 6th Grade Go Math Question 13.
90 ÷ 0.36 _____________
Answer:
The divisor has two decimal places, so multiply both the dividend and the divisor by 100 so that the divisor is a whole number

0.36 × 100 = 36
90 × 100 = 9000

Divide

Question 14.
18.88 ÷ 1.6 _____________
Answer:
The divisor has one decimal place, so multiply both the dividend and the divisor by 10 so that the divisor is a whole number

1.6 × 10 = 16
18.88 × 10 = 188.8

Question 15.
Corrine has 9.6 pounds of trail mix to divide into 12 bags. How many pounds of trail mix will go in each bag?
Answer:
The divisor has one decimal place, so multiply both the dividend and the divisor by 10 so that the divisor is a whole number

12 × 10 = 120
9.6 × 10 = 96

Question 16.
Michael paid $1 1.48 for sliced cheese at the deli counter. The cheese cost $3.28 per pound. How much cheese did Michael buy?
Answer:
The divisor has two decimal place, so multiply both the dividend and the divisor by 100 so that the divisor is a whole number

3.28 × 100 = 328
11.48 × 100 = 1148

Question 17.
A four-person relay team completed a race in 72.4 seconds. On average, what was each runner’s time?
Answer:
The divisor has one decimal place, so multiply both the dividend and the divisor by 10 so that the divisor is a whole number

4 × 10 = 40
72.4 × 10 = 724

Each person’s time 18.1 seconds

Question 18.
Elizabeth has a piece of ribbon that is 4.5 meters long. She wants to cut it into pieces that are 0.25 meter long. How many pieces of ribbon will she have?
Answer:
The divisor has two decimal place, so multiply both the dividend and the divisor by 100 so that the divisor is a whole number

0.25 × 100 = 25
4.5 × 100 = 450

Divide

Elizabeth will have 18 pieces of ribbon

Question 19.
Lisa paid $43.95 for 16.1 gallons of gasoline. What was the cost per gallon, rounded to the nearest hundredth?
Answer:
The divisor has one decimal place, so multiply both the dividend and the divisor by 10 so that the divisor is a whole number

16.1 × 10 = 161
43.95 × 10 =439.5

Cost per gallon 2.73 dollars

Question 20.
One inch is equivalent to 2.54 centimeters. How many inches are there in 50.8 centimeters?
Answer:
The divisor has one decimal place, so multiply both the dividend and the divisor by 10 so that the divisor is a whole number

2.54 × 100 = 254
50.08 × 100 = 5008

In 50.8 centimeters there is 20 inches

Essential Question Check-In

Question 21.
How can you determine if you divided the numbers correctly?
Answer:
We can check it if we muLtipLy the resuLt by divisor. Multiplying those two numbers, we suppose to get the dividend.
If we get the dividend, that means that we divided the number correctly.
Multiplying the result by divisor.

Use the table for 22 and 23.

Question 22.
What is the price per mug for 25 coffee mugs?
Answer:
The solution to this example is given below

The price per mug is 4.29 dollars

Question 23.
Find the price per T-shirt for 75 T-shirts.
Answer:
Solution to this example is given below

The price per T-shirt is 8.5 dollars

A movie rental website charges $5.00 per month for membership and $1.25 per movie.

Texas Go Math Dividing Decimals Lesson 4.2 Grade 6 Question 24.
How many movies did Andrew rent this month if this month’s bill was $16.25?
Answer:
In order to calculate how many movies Andrew rented this month, we will first need to subtract 5.00 from the month’s bill, which is 16.25, and get:
16.25 – 5.00 = 11.25
Now, we will divide 11.25 by 1.25, but first, we need to change the divisor to a whole number by multiplying by a power of 100, and then multiply the dividend by the same power of 100 and get:

So, Andrew rented 9 movies this month.

Question 25.
Marissa has $18.50 this month to spend on movie rentals.
a. How many movies can she view this month? ___________________
Answer:
In order to find how many movies Marissa can view, we will first subtract $ 5.00 for membership from $ 18.50 and get:
18.50 – 5.00 = 13.50
Now, we will divide 13.50 by 1.25, but first, we need to change the divisor to a whole number by multiplying by a power of 100, and then multiply the dividend by the same power of 100 and get:

We get that she can rent 10 movies.

b. Critique Reasoning Marisa thinks she can afford 11 movies in one month. What mistake could she be making?
Answer:
In part (a) WP gets that She can rent 10.8 movies, but it is not a whole number. it is true that 11 is the closest whole number to 10.8. but Marissa does not have enough money for 11 movies, only for 10.

Victoria went shopping for ingredients to make a stew. The table shows the weight and the cost of each of the ingredients that she bought.

Question 26.
What is the price for one pound of bell peppers?
Answer:
The divisor has two decimal place, so multiply both the dividend and the divisor by 100 so that the divisor is a whole number
2.50 × 100 = 250
1.25 × 100 = 125

Price for one pound of bell peppers is 0.5 dollars

Question 27.
Which ingredient costs the most per pound?
Answer:
Solution to this example is given below

Beef costs the most per pound.

Question 28.
What If? If carrots were $0.50 less per pound, how much would Victoria have paid for 8.5 pounds of carrots?
Answer:
New cost of carrots
15.30 – 0.50 = 14.8

Total cost of carrots for Vistoria is 125.80 dollars

Question 29.
Brenda is planning her birthday party. She wants to have 10.92 liters of punch, 6.5 gallons of ice cream, 3.9 pounds of fudge, and 25 guests at the birthday party.
a. Brenda and each guest drink the same amount of punch. How many liters of punch will each person drink? ______________________
Answer:
Because Brenda has 10.92 liters of punch and 25 guests, we need to divide 10.92 by 26 in order to get how many liters of punch each person will drink, but first we need to change the divisor to a whole number by multiplying by a power of 100, and then multiply the dividend by the same power of 100.

So, we can notice that each person, including Brenda will drink 0.42 liters of punch.

b. Brenda and each guest eat the same amount of ice cream. How many gallons of ice cream will each person eat? ___________________
Answer:
Brenda has 6.5 gallons of ice cream and 25 guests. so. we need to divide 6.5 by 26 in order to get how many gallons of ice cream each person will eat, but first we need to change the divisor to a whole number by multiplying by a power of 10, and then multiply the dividend by the same power of 10

So, each person, including Brenda will eat 0.25 gallons of ice cream.

c. Brenda and each guest eat the same amount of fudge. How many pounds of fudge will each person eat? ____________________
Answer:
Brenda has 3.9 pounds of fudge and 25 guests, so we need to divide 3.9 by 26 in order to get how many pounds of fudge each person will eat, but first We need to change the divisor to a whole number by multiplying by a power of 10. and then multiply the dividend by the same power of 10.

So, each person including Brenda, will eat 0.15 pounds of fudge.

To make costumes for a play, Cassidy needs yellow and white fabric that she will cut into strips. The table shows how many yards of each fabric she needs, and how much she will pay for those yards.

Question 30.
Which costs more per yard, the yellow fabric or the white fabric?
Answer:
We will first calculate how much costs the yellow fabric per yard dividing 86.40 by 12.8, hut first we need to change the divisor to a whole number by multiplying by a power of 10. and then multiply the dividend by the same power of 10

So, the yellow fabric costs $ 6.75 per yard.
Now, we will calculate how much costs the white fabric per yard dividing 45.60 by 9.5, but first we need to change the divisor to a whole number by multiplying by a power of 10, and then multiply the dividend by the same power of 10

So, the white fabric costs $ 4.8 per yard.
Conclusion is that the yellow fabric costs more per yard.

Question 31.
Cassidy wants to cut the yellow fabric into strips that are 0.3 yards wide.
How many whole strips of yellow fabric can Cassidy make?
Answer:
The divisor has one decimal place, so multiply both the dividend and the divisor by 10 so that the divisor is a whole number

0.3 × 10 = 3
12.8 × 10 = 128

Cassidy make 42 strips of yellow fabric.

H.O.T. Focus On Higher Order Thinking

Question 32.
Problem Solving Eight friends purchase various supplies for a camping trip and agree to share the total cost equally. They spend $85.43 on food, $32.75 on water, and $239.66 on other items. How much does each person owe?
Answer:
First, we will sum all their costs and get:
85.43 + 32.75 + 239.66 = 357.84
Now, we will divide total cost by 8 to calculate how much each person owes:

So, each person owes $ 44.73

Question 33.
Analyze Relationships Constance ¡s saving money to buy a new bicycle that costs $195.75. She already has $40 saved and plans to save $8 each week. How many weeks will it take her to save enough money to purchase the bicycle?
Answer:
First we will subtract 10 form 195.75 and get:
195.75 – 40 = 155.75
Now, we will divide 155.75 by 8 in order to find how ninny weeks it will take Constance to save enough money to purchase the bicycle.

So, it will take her about 20 weeks to save enough money for bicycle.

Question 34.
Represent Real-World Problems A grocery store sells twelve bottles of water for $13.80. A convenience store sells ten bottles of water for $1 1.80. Which store has the better buy? Explain.
Answer:
We will first calculate how much costs one bottle of water in the first store dividing 13.80 by 12 and get:

So, in the first store. bottle of water costs S 1.15.
Now we will calculate how much costs one bottle of water in the second store dividing 11.80 by 10 and get:

In the second store, one bottle of water costs $ 1.18.
We can conclude that the better buy has the first store.

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Module 3 Quiz Answer Key.

Texas Go Math Grade 6 Module 3 Quiz Answer Key

Texas Go Math Grade 6 Module 3 Ready to Go On? Answer Key

3.1 Multiplying Fractions

Multiply.

Question 1.
\(\frac{4}{5} \times \frac{3}{4}\)
Answer:
First we need to write the problem as a single fraction.
After that, we need to multiply numerators and denominators and then simplify by dividing by the GCF. Here, the GCF of 12 and 20 is 4.
Finally, we will write the answer in the simplest form. So, we have the following:

So, the product is \(\frac{3}{5}\)

Go Math Grade 6 Module 3 Answer Key Question 2.
\(\frac{5}{7} \times \frac{9}{10}\)
Answer:
First, we need to write the problem as a single fraction.
After that, we need to multiply numerators and denominators and then simplify by dividing by the GCF. Here, the GCF of 45 and 70 is 5.
Finally, we will write the answer in the simplest form. So, we have the following:

So, the product is \(\frac{9}{14}\).

Question 3.
Fred had 264 books in his personal library. He donated \(\frac{2}{11}\) of these books to the public library. How many books did he donate?
Answer:
In order to find how many books Fred donated, we have to multiply 264 by \(\frac{2}{11}\) so, we have the following:

The conclusion is that Fred donated 48 books to the public library.

3.2 Multiplying Mixed Numbers

Multiply.

Question 4.
\(\frac{3}{8} \times 2 \frac{1}{2}\)
Answer:
First we need to rewrite the mixed number as a fraction greater than 1. After that we have to simplify before multiplying using the GCF and rewrite the fraction. Finally, multiply numerators and denominators and write the result in the simplest form.
So, we have the following:

The result is \(\frac{15}{16}\).

Grade 6 Module 3 Quiz Answer Key Question 5.
\(3 \frac{3}{5} \times \frac{5}{6}\)
Answer:
First, we need to rewrite the mixed number as a fraction greater than 1. After that, we have to simplify before multiplying using the GCF and rewrite the fraction. Finally, multiply numerators and denominators and write the result in the simplest form.
So, we have the following:

The result is 3.

Question 6.
Jamal and Dorothy were hiking and had a choice between two trails. One was 5\(\frac{1}{3}\) miles long, and the other was 1\(\frac{3}{4}\) times as long. How long was the longer trail?
Answer:
In order to calculate how long was the longer trail, we have to multiply 5\(\frac{1}{3}\) and 1\(\frac{3}{4}\).
We have to rewrite each mixed number as a fraction greater than 1.
Then multiply numerators and denominators, but before that we need to simplify using the GCF and rewrite the fraction.
At the end, we will write the result in the simplest form.
We have the following:

So, the longer trail was 9\(\frac{1}{3}\) miles long.

3.3 Dividing Fractions

Divide.

Question 7.
\(\frac{7}{8} \div \frac{3}{4}\)
Answer:
First, we will rewrite division as multiplication, using the reciprocal of the division. After that we will multiply the numerators and denominators and finally write the answer in the simplest form. So, we have the following:

The result is \(\frac{4}{3}\).

Module 3 Test Answers Go Math Grade 6 Question 8.
\(\frac{4}{5} \div \frac{6}{7}\)
Answer:
First, we will rewrite division as multiplication, using the reciprocal of the division. After that, we will multiply the numerators and denominators and finally write the answer in the simplest form. So, we have the following:

The result is \(\frac{14}{15}\).

Question 9.
\(\frac{1}{3} \div \frac{7}{9}\)
Answer:
First, we will rewrite division as multiplication, using the reciprocal of the division. After that we will multiply the numerators and denominators and finally write the answer in the simplest form. So, we have the following:

The result is \(\frac{3}{7}\).

Question 10.
\(\frac{1}{3} \div \frac{5}{8}\)
Answer:
First, we will rewrite division as multiplication, using the reciprocal of the division. After that we will multiply the numerators and denominators and finally write the answer in the simplest form. So, we have the following:

The result is \(\frac{8}{15}\).

3.4 Dividing Mixed Numbers

Divide.

Question 11.
\(3 \frac{1}{3} \div \frac{2}{3}\)
Answer:
First we need to rewrite the mixed numbers as a fractions grater than 1.
Then rewrite the problem as multiplication using the reciprocal of the second fraction. After that, we will simplify using the GCF and multiply the numerators and denominators. Finally, we will write the result as a mixed number if it is possible. So, we have the following:

The result is 5.

Module 3 Quiz Answers Go Math Grade 6 Question 12.
\(1 \frac{7}{8} \div 2 \frac{2}{5}\)
Answer:
First, we need to rewrite the mixed numbers as fractions grater than 1.
Then rewrite the problem as multiplication using the reciprocal of the second fraction. After that, we will simplify using the GCF and multiply the numerators and denominators. Finally, we will write the result as a mixed number if it is possible. So, we have the following:

The result is \(\frac{25}{32}\)

Question 13.
\(4 \frac{1}{4} \div 4 \frac{1}{2}\)
Answer:
First we need to rewrite the mixed numbers as fractions grater than 1.
Then rewrite the problem as multiplication using the reciprocal of the second fraction. After that, we will simplify using the GCF and multiply the numerators and denominators. Finally, we will write the result as a mixed number if it is possible. So, we have the following:

The result is \(\frac{17}{18}\)

Question 14.
\(8 \frac{1}{3} \div 4 \frac{2}{7}\)
Answer:
First we need to rewrite the mixed numbers as a fractions grater than 1.
Then rewrite the problem as multiplication using the reciprocal of the second fraction. After that, we will simplify using the GCF and multiply the numerators and denominators. Finally, we will write the result as a mixed number if it is possible. So, we have the following:

The result is \(\frac{35}{18}\) = 1\(\frac{17}{18}\).

Essential Question

Go Math Grade 6 Module 3 Answer Key Pdf Question 15.
Describe a real-world situation that is modeled by multiplying two fractions or mixed numbers.
Answer:
For, example, a real-world situation that would be appropriate to be modeled by multiplying two fractions or mixed numbers is the following.
Jane has 150\(\frac{1}{2}\) grams of chocolate and she wants to split it into three friends.
How much chocolate will each friend get?

Jane has 150\(\frac{1}{2}\) grams of chocolate and she wants to split it into three friends.
How much chocolate will each friend get?

Texas Go Math Grade 6 Module 3 Mixed Review Texas Test Prep Answer Key

Texas Test Prep

Question 1.
Which of the following statements is correct?
(A) The product of \(\frac{3}{4}\) and \(\frac{7}{8}\) is less than \(\frac{7}{8}\).
(B) The product of 1\(\frac{1}{3}\) and \(\frac{9}{10}\) is less than \(\frac{9}{10}\).
(C) The product of \(\frac{3}{4}\) and \(\frac{7}{8}\) is greater than \(\frac{7}{8}\).
(D) The product of \(\frac{7}{8}\) and \(\frac{9}{10}\) is greater than \(\frac{9}{10}\).
Answer:
(A) The product of \(\frac{3}{4}\) and \(\frac{7}{8}\) is less than \(\frac{7}{8}\).

Explaination:
The correct answer is A. Really:
\(\frac{3}{4} \times \frac{7}{8}\) = \(\frac{3 \times 7}{4 \times 8}\)
= \(\frac{21}{32}\)

And
\(\frac{7}{8}\) = \(\frac{7 \times 4}{8 \times 4}\)
= \(\frac{28}{32}\)

We can notice that:
\(\frac{3}{4} \times \frac{7}{8}=\frac{21}{32}<\frac{28}{32}=\frac{7}{8}\)

Question 2.
Which shows the GCF of 18 and 24 with \(\frac{18}{24}\) in simplest form?
(A) GCF: 3; \(\frac{3}{4}\)
(B) GCF: 3; \(\frac{6}{8}\)
(C) GCF: 6; \(\frac{3}{4}\)
(D) GCF: 6; \(\frac{6}{8}\)
Answer:
(C) GCF: 6; \(\frac{3}{4}\)

Explaination:
GCF for 48 and 24 is 6, so, we have the following form of this fraction:
\(\frac{18}{24}=\frac{18 \div 6}{24 \div 6}=\frac{3}{4}\)
Conclusion is that correct answer is C.

Grade 6 Module 3 Test Form A Answer Key Question 3.
A jar contains 133 pennies. A bigger jar contains 1\(\frac{2}{7}\) times as many pennies. What is the value of the pennies in the bigger jar?
(A) $ 1.49
(B) $ 1.52
(C) $ 1.68
(D) $ 1.71
Answer:
(D) $ 1.71

Explaination:
The value of pennies in the bigger jar is product of 133 and 1\(\frac{2}{7}\), so we have the following:

So, the value of pennies in the bigger jar is 171.
Conclusion is that correct answer is D.

Question 4.
Which of these is the same as \(\frac{3}{5}\) ÷ \(\frac{4}{7}\)?
(A) \(\frac{3}{5} \div \frac{7}{4}\)
(B) \(\frac{4}{7} \div \frac{3}{5}\)
(C) \(\frac{3}{5} \times \frac{4}{7}\)
(D) \(\frac{3}{5} \times \frac{7}{4}\)
Answer:
(D) \(\frac{3}{5} \times \frac{7}{4}\)

Explaination:
The first step at dividing two factors is to rewrite the problem as multiplication
Using the reciprocal of the second fraction.
Here, reciprocal of \(\frac{4}{7}\) is \(\frac{7}{4}\) and equivalent is expression \(\frac{3}{5} \times \frac{7}{4}\)
Conclusion is that correct answer is D.

Question 5.
What is the reciprocal of 3\(\frac{3}{7}\)?
(A) \(\frac{7}{24}\)
(B) \(\frac{3}{7}\)
(C) \(\frac{7}{3}\)
(D) \(\frac{24}{7}\)
Answer:
(A) \(\frac{7}{24}\)

Explaination:
First we will find mixed number as a fraction greater than 1:
3\(\frac{3}{7}\) = \(\frac{24}{7}\)
The reciprocal of \(\frac{24}{7}\) is \(\frac{7}{24}\). So, conclusion is that correct answer is A.

Question 6.
A rectangular patio has a length of 12\(\frac{1}{2}\) feet and an area of 103\(\frac{1}{8}\) square feet. What is the width of the patio?
(A) 4\(\frac{1}{8}\) feet
(B) 8\(\frac{1}{4}\) feet
(C) 16\(\frac{1}{2}\) feet
(D) 33 feet
Answer:
(B) 8\(\frac{1}{4}\) feet

Explaination:
The width of the patio we will dividing the area of patio, 103\(\frac{1}{8}\), by its length, 12\(\frac{1}{2}\)
So, we have the following:

So, the width of the patio is 8\(\frac{1}{4}\) feet. The correct answer is B.

Gridded Response

Module 3 Test Answers Go Math Grade 6 Question 7.
Jodi is cutting out pieces of paper that measure 8\(\frac{1}{2}\) inches by 11 inches from a large sheet that has an area of 1,000 square inches. What is the area of each piece of paper that Jodi is cutting out written as a decimal?

Answer:
The area of each piece of paper is product of 8\(\frac{1}{2}\) and 11:
8\(\frac{1}{2}\) × 11 = \(\frac{17}{2} \times \frac{11}{1}\)
= \(\frac{17 \times 11}{2}\)
= \(\frac{187}{2}\)
= 93.5
So, the area of each piece of paper that Jodi is cutting out is 93.5 square inches.

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 3.2 Answer Key Multiplying Mixed Numbers.

Texas Go Math Grade 6 Lesson 3.2 Answer Key Multiplying Mixed Numbers

Reflect

Question 1.
Make a Conjecture Will the product 1\(\frac{1}{2}\) × \(\frac{1}{2}\) be greater than or less than \(\frac{1}{2}\)? Explain.
Answer:
We can solve this task using logic, or simply by doing the multiplication and seeing the result!

(1) Let’s think logically:
As we said before, when we multiply some number by a number that is greater than 1, our output will be greater than our original value.
Now, when we know which way to think, let’s check if 1\(\frac{1}{2}\) is greater than 1.
1\(\frac{1}{2}\) = 1 + \(\frac{1}{2}\)
Now it is clear, given mixed number is surely greater than 1.
Therefore, we can make our conclusion:
Product 1\(\frac{1}{2}\) × \(\frac{1}{2}\) will be greater than \(\frac{1}{2}\).

(2) Using exact calculation to prove facts:
Let’s calculate the product of given expressions!
\(\frac{3}{2} \times \frac{1}{2}=\frac{3 \times 1}{2 \times 2}\)
= \(\frac{3}{4}\)
All that is left to do, is to compare our result with the original value!
Number 3 is greater than 2.
So we can conclude that:
Number \(\frac{3}{4}\) is greater than \(\frac{1}{2}\).

Exam stuff:
1\(\frac{1}{2}\) = 1 + \(\frac{1}{2}\)
= \(\frac{2}{2}+\frac{1}{2}\)
= \(\frac{3}{2}\)

\(\frac{1}{2}\) = \(\frac{1 \times 2}{2 \times 2}\)
= \(\frac{2}{4}\)

Your Turn

Multiply. Write each product in simplest form.

Question 2.
3\(\frac{1}{3}\) × \(\frac{3}{4}\) ________________
Answer:
Process of multiplying fractions with mixed numbers
(1) Step:
Estimate the product Round the mixed number to the nearest whole number Find the nearest benchmark for the fraction.
3\(\frac{1}{3}\) is close to 3, so multiply \(\frac{3}{4}\) times 3.
3 × \(\frac{3}{4}\) = \(\frac{3 \times 3}{4}\)
= \(\frac{9}{4}\)

(2) Step:
We transform mixed number into traction, and then multiply those tractions as in previous examples.

Note that we will not do a transformation from mixed numbers into fractions as we can do it automatically now.
One more thing to add, the First step may not make sense at first, but it is crucial to find out whether our precise solution is correct!

Multiplying Two Mixed Numbers Lesson 3.2 Answer Key Question 3.
1\(\frac{4}{5}\) × \(\frac{1}{2}\) ________________
Answer:
Process of multiplying fractions with mixed numbers
(1) Step:
Estimate the product Round the mixed number to the nearest whole number Find the nearest benchmark for the fraction.
1\(\frac{4}{5}\) is close to 2, so multiply \(\frac{1}{2}\) times 2.
\(\frac{1}{2}\) × 2 = \(\frac{1 \times 2}{2}\)
= \(\frac{2}{2}\) = 1

(2) Step:
We transform mixed numbers into traction and then multiply those tractions as in previous examples.

Note that we will not do transformation from mixed number into fraction as we can do it automatically now.
One more thing to add, First step may not make sense at first, but it is crucial to find out whether our precise solution is correct!

Question 4.
\(\frac{5}{6}\) × 2\(\frac{3}{4}\) ________________
Answer:
Process of multiplying fractions with mixed numbers
(1) Step:
Estimate the product Round the mixed number to the nearest whole number Find the nearest benchmark for the fraction.

(2) Step:
We transform mixed number into traction, and then multiply those tractions as in previous examples.

Note that we will not do transformation from mixed number into fraction as we can do it automatically now.
One more thing to add, First step may not make sense at first, but it is crucial to find out whether our precise solution is correct!

Question 5.
\(\frac{3}{5}\) × 2\(\frac{1}{5}\) ________________
Answer:
Process of multiplying fractions with mixed numbers
(1) Step:
Estimate the product Round the mixed number to the nearest whole number Find the nearest benchmark for the fraction.
2\(\frac{1}{5}\) is close to 2, so multiply \(\frac{3}{5}\) times 2.
\(\frac{3}{5}\) × 2 = \(\frac{3 \times 2}{5}\)
= \(\frac{6}{5}\)

(2) Step:
We transform mixed number into traction, and then multiply those tractions as in previous examples.

Note that we will not do a transformation from mixed numbers into fractions as we can do it automatically now.
One more thing to add, the First step may not make sense at first, but it is crucial to find out whether our precise solution is correct!

Lesson 3.2 Go Math 6th Grade Go Math Question 6.
\(\frac{9}{10}\) × 4\(\frac{1}{3}\) ________________
Answer:
Process of multiplying fractions with mixed numbers
(1) Step:
Estimate the product Round the mixed number to the nearest whole number Find the nearest benchmark for the fraction.
4\(\frac{1}{3}\) is close to 4, so multiply \(\frac{9}{10}\) times 4.

(2) Step:
We transform mixed number into fraction, and then multiply those tractions as in previous examples.

Note that we will not do transformation from mixed number into fraction as we can do it automatically now.
One more thing to add, First step may not make sense at first, but it is crucial to find out whether our precise solution is correct!

Question 7.
5\(\frac{1}{6}\) × \(\frac{1}{8}\) ________________
Answer:
Process of multiplying fractions with mixed numbers
(1) Step:
Estimate the product Round the mixed number to the nearest whole number Find the nearest benchmark for the fraction.
5\(\frac{1}{6}\) is close to 5, so multiply \(\frac{1}{8}\) times 5.
\(\frac{1}{8}\) × 5 = \(\frac{1 \times 5}{8}\)
= \(\frac{5}{8}\)

(2) Step:
We transform mixed number into traction, and then multiply those tractions as in previous examples.

Note that we will not do transformation from mixed number into fraction as we can do it automatically now.
One more thing to add, First step may not make sense at first, but it is crucial to find out whether our precise solution is correct!

Reflect

Question 8.
Analyze Relationships When you multiply two mixed numbers, will the product be less than or greater than the factors? Use an example to explain.
Answer:
When we say that some number is mixed, it automatically gives us a clue.
Every mixed number is greater than 1, or at least its absolute value(we want to pay attention onLy on positive mixed numbers here).
So product of mixed numbers will be Greater than the factors.

Let’s show this on example:

From here, we can see that 4 is greater than both factors!

We would like to add one more side note here.
In previous tasks, we were rounding our mixed numbers to the nearest whole number
We skipped that part in our solutions, but we will write the explanation here.
So, how can we know whether we should round some mixed number to greater whole number or to less whole number?
Take a look at fraction, check if it is greater than or equal to a half, if it is, then round it to the greater whole number.
Otherwise, round it to the same whole number as it is given in mixed number.

Your Turn

Multiply. Write each product in simplest form.

Question 9.
2\(\frac{2}{3}\) × 1\(\frac{1}{7}\)
Answer:
First, we need to rewrite each mixed number as a fraction greater than 1. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:

the result is \(\frac{64}{21}\) or 3\(\frac{1}{21}\).

Lesson 3.2 Answer Key 6th Grade Go Math Question 10.
2\(\frac{3}{8}\) × 1\(\frac{3}{5}\)
Answer:
First, we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:

So, the result is \(\frac{19}{5}\) or 3\(\frac{4}{5}\).

Question 11.
4\(\frac{1}{2}\) × 3\(\frac{3}{7}\)
Answer:
First, we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:

So, the result is \(\frac{108}{7}\) or 15\(\frac{3}{7}\).

Question 12.
5\(\frac{1}{4}\) × 4\(\frac{2}{3}\)
Answer:
First, we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:

So, the result is \(\frac{49}{2}\) or 24\(\frac{1}{2}\).

Texas Go Math Grade 6 Lesson 3.2 Guided Practice Answer Key

Question 1.
Mr. Martin’s yard is 1\(\frac{1}{3}\) acres. He wants to plant grass on \(\frac{1}{6}\) of his yard.

a. Draw a model to show how many acres will be covered by grass.

Answer:
On the following picture there is shown how many acres will be covered by grass.

b. How many acres will be covered by grass?
Answer:
It will be covered \(\frac{2}{9}\) acres by grass.

c. Write the multiplication shown by the model.
Answer:
First, we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:

d. Will the mixed number that represents the original size of Mr. Martin’s yard increase or decrease when multiplied by \(\frac{1}{6}\)? Explain.
Answer:
We can notice that Mr. Martin’s yard will decrease when multiply by \(\frac{1}{6}\) because 0 < \(\frac{1}{6}\) < 1 and when multiplying some number greater than 1 by number which is from interval (0, 1), product will decrease.

Multiply. Write each product in simplest form.

Question 2.
1\(\frac{1}{5}\) × \(\frac{3}{5}\)
Answer:
First, we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite the fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:
\(\frac{6}{5} \times \frac{3}{5}\) = \(\frac{6 \times 3}{5 \times 5}\)
= \(\frac{18}{25}\)
So, the result is \(\frac{18}{25}\)

Lesson 3.2 Answer Key 6th Grade Multiplying Mixed Numbers Question 3.
1\(\frac{3}{4}\) × \(\frac{4}{7}\)
Answer:
First, we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite the fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:

So, the result is 1

Question 4.
1\(\frac{5}{6}\) × \(\frac{2}{5}\)
Answer:
First, we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:

So, the result is \(\frac{11}{15}\)

Question 5.
1\(\frac{7}{10}\) × \(\frac{4}{5}\)
Answer:
First, we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:

So, the result is \(\frac{34}{25}\) = 1\(\frac{9}{25}\)

Question 6.
\(\frac{5}{9}\) × 3\(\frac{9}{10}\)
Answer:
First, we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:

So, the result is \(\frac{13}{6}\) = 2\(\frac{1}{6}\)

Question 7.
\(\frac{7}{8}\) × 3\(\frac{1}{3}\)
Answer:
First, we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:

So, the result is \(\frac{35}{12}\) = 2\(\frac{11}{12}\)

Question 8.
2\(\frac{1}{5}\) × 2\(\frac{3}{5}\)
Answer:
First, we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:

So, the result is \(\frac{143}{25}\) = 5\(\frac{18}{25}\)

Question 9.
4\(\frac{3}{4}\) × 3\(\frac{4}{5}\)
Answer:
First, we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:

So, the result is \(\frac{361}{20}\) = 18\(\frac{1}{20}\)

Essential Question Check-In

Question 10.
How can you multiply two mixed numbers?
Answer:
We can first rewrite the mixed number as a fraction grater than 1, if we are given mixed number Then simplify before multiplying using the GCF, if there are possibilities for this, after that rewrite fraction.

Finally, we multiply numerators and denominators and get the result which we can rewrite as a mixed number if it is grater than 1 and of course, if it is a fraction.

Rewrite mixed number as a fraction, simplify, multiply numerators and denominators.

Estimate. Then Solve.

Question 11.
Carly is making 3\(\frac{1}{2}\) batches of biscuits. If one batch calls for 2\(\frac{1}{3}\) cups of flour, how much flour will she need?
Answer:
In order to find how much flour Carly will need, we have to multiply those mixed numbers.

First, we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:

So, conclusion is that Carly will need 12\(\frac{1}{4}\) cups of flour.

Go Math Grade 6 Lesson 3.2 Multiplying Mixed Numbers Question 12.
Bashir collected 4\(\frac{1}{3}\) baskets of peaches at an orchard. If each basket holds 21 peaches, how many peaches did he collect in all?
Answer:
In order to find how much flour Carly will need, we have to multiply those mixed numbers.

First, we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1. So, applying all this, we have the following:

Conclusion is that

Bashir collected 91 peach.

Question 13.
Jared used 1\(\frac{2}{5}\) bags of soil for his garden. He is digging another garden that will need \(\frac{1}{5}\) as much soil as the original. How much will he use total?
Answer:
In order to calculate how much bags of soil Jared used total we have to multiply 1\(\frac{2}{5}\) and \(\frac{1}{5}\).
First we need to rewrite each mixed number as a fraction greater than 1 and simplify before multiplying using the GCF and then rewrite fraction. Then, we will multiply numerators and denominators. Final result we can write as a mixed number if it is greater than 1.

So, applying all this, we have the following:

So, he used total \(\frac{7}{25}\) bags of soil.

Question 14.
Critical Thinking is the product of two mixed numbers less than, between, or greater than the two factors? Explain.
Answer:
If both fractions are greater than 1, or, if they are mixed numbers, then their product is greater than both fractions.
For example, 1\(\frac{1}{4}\) = \(\frac{5}{4}\) auch 1\(\frac{1}{5}\) = \(\frac{6}{5}\) are both greater than 1 and their product is following:

And \(\frac{5}{4}\) < \(\frac{6}{4}\) = \(\frac{3}{2}\), \(\frac{6}{5}\) = \(\frac{12}{10}\) < \(\frac{15}{10}\) = \(\frac{3}{2}\)
Really, their product, \(\frac{3}{2}\), is greater than both fractions.
If both fractions are less than 1, their product is less than both fractions.
For example, \(\frac{1}{4}\) and \(\frac{3}{5}\) are both less than 1 and their product is following:

And \(\frac{1}{4}=\frac{5}{20}>\frac{3}{20}, \frac{3}{5}=\frac{12}{20}>\frac{3}{20}\)

Really, their product, \(\frac{3}{20}\), is less than both tractions.
And it one traction is greater than 1 or if it is mixed number and the other is less than 1, their product will be between those two tractions.
For example, \(\frac{1}{5}\) is less than 1 and 1\(\frac{2}{5}\) = \(\frac{7}{5}\) is greater than 1, their product is following:
\(\frac{1}{5} \times \frac{7}{5}\) = \(\frac{1}{5} \times \frac{7}{5}\)
= \(\frac{7}{25}\)
And \(\frac{5}{25}=\frac{1}{5}<\frac{7}{25}, \frac{30}{25}=\frac{6}{5}>\frac{7}{25}\)
Really, their product, \(\frac{7}{25}\), is between those two fractions.

If both fractions are mixed numbers, fractions greater than 1, their product will be mixed number greater than both mixed numbers.
If both fractions are less than 1, their product will be less than both fractions.
If one fraction is less than 1 and the other is mixed number, greater than 1, their product wilL be between those two fractions.

Question 15.
There are approximately 402\(\frac{1}{4}\) meters around a typical running track. Sandra has challenged herself to run 10 laps a day for 5 days. How many meters will Sandra run if she meets her challenge?
Answer:
First we will calculate how many meters Sandra will run for a day.

Now, we will calculate how many meters Sandra wilt run for 5 days muLtiplying previous result by 5

So, Sandra will run 20112\(\frac{1}{2}\) meters if she meets her challenge.

Question 16.
Ron wants to make a rectangular basketball court. What is the area of Ron’s court?

Answer:
The area of rectangular is product of length of its sides
Here, at this rectangular basketball court, one side is 4\(\frac{3}{4}\) yd long and the other is 3\(\frac{2}{5}\) yd long.
The area of this basketball court is the following:

So, the area of Rou’s Court is 16\(\frac{3}{20}\)yd²

Question 17.
Each of 15 students will give a 1\(\frac{1}{2}\)-minute speech in English class.
a. How long will it take to give the speeches?
Answer:
In order to calculate how long it will take to give the speeches, we have to multiply those two mixed numbers and get:

So, it will take 22\(\frac{1}{2}\) minutes to give the speeches.

b. If the teacher begins recording on a digital camera with an hour available, is there enough time to record everyone if she gives a 15-minute introduction at the beginning of class and every student takes a minute to get ready? Explain.
Answer:
Because we have 15 students and every student takes a minute to get ready, on result from (a) we have to add this 15 × 1 = 15 minutes.
Also, we have to add 15 minutes more because that time is needed for introduction at the begining of class.
So, the whole ceremony with speeches wilL take:

Because 1 hour has 60 minutes and whole ceremony with speaches wiLl take 52\(\frac{1}{2}\) minutes, which is less than 60 minutes, conclusion is that there will be enough time to record everyone on a digital camera.

c. How much time is left on the digital camera?
Answer:
Now we will calculate how much time is left on the digital camera subtracting 52\(\frac{1}{2}\), which is time needed for recording everyone, from 60 minutes, which is time for recording on digital camera.
So, we have the following:

So, there is 7\(\frac{1}{2}\) minutes left on digital camera.

Question 18.
Communicate Mathematical Ideas How is multiplying a whole number by a mixed number the same as multiplying two mixed numbers?
Answer:

H.O.T. Focus on Higher Order Thinking

Question 19.
Critique Reasoning To find the product 3\(\frac{3}{8}\) x 4\(\frac{1}{9}\), Tara rewrote 3\(\frac{3}{8}\) as \(\frac{17}{8}\) and 4\(\frac{1}{9}\) as \(\frac{13}{2}\). Then she multiplied the fractions to find the product \(\frac{221}{72}\). What were her errors?
Answer:
It is the same because whole number we need to rewrite as a fraction as welt as mixed numbers in order to multiply them.
We need to rewrite whole number as a fraction as well as mixed number.

Go Math Lesson 3.2 Grade 6 Multiply Mixed Fractions Question 20.
Represent Real-World Problems Ian is making his special barbecue sauce for a party. His recipe makes 3\(\frac{1}{2}\) cups of barbecue sauce and uses 2\(\frac{1}{4}\) tablespoons of soy sauce. He wants to increase his recipe to make five times as much barbecue sauce. He checks his refrigerator and finds that he has 8 tablespoons of soy sauce. Will he have enough soy sauce? Explain.
Answer:
She did not rewrite mixed numbers properly as fractions, that was Tara’s mistake
The right way is the following:

So, there product is actually \(\frac{111}{8}\), or, 13\(\frac{7}{8}\).

Question 21.
Analyze Relationships Is it possible to find the product of two mixed numbers by multiplying the whole number parts together, then multiplying the two fractional parts together, and finally adding the two products? Use an example to support your answer.
Answer:
If he wants to make five times as much barbecue sauce, we need all ingredients five times as much, so, and soy sauce.
So, now we will calculate how much soy sauce he needs if he wants to make five times as much barbecue sauce. We have the following:
5 × 2\(\frac{1}{4}\) = 5 × \(\frac{9}{4}\)
= \(\frac{5 \times 9}{1 \times 4}\)
= \(\frac{45}{4}\)
= 9
So, he needs 9 tablespoons of soy sauce to make five times as much barbecue sauce, but he has only 8 tablespoons of soy sauce in his refrigerator.
The conclusion is that he does not have enough soy sauce to make five times as much barbecue sauce.

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 3.4 Answer Key Dividing Mixed Numbers.

Texas Go Math Grade 6 Lesson 3.4 Answer Key Dividing Mixed Numbers

Reflect

Question 1.
Communicate Mathematical Ideas Which mathematical operation could you use to find the number of sushi rolls that Antoine can make? Explain.
Answer:
We can use dividing to find the number of sushi rolls that Antonie can make Actually, to find that number, we need to divide 2\(\frac{1}{2}\) by \(\frac{1}{4}\).

Question 2.
Multiple Representations Write the division shown by the model.
Answer:
The division shown by the previous model is the following:
2\(\frac{1}{2}\) ÷ \(\frac{1}{4}\)

Question 3.
What If? Suppose Antoine instead uses \(\frac{1}{8}\) cup of rice for each sushi roll. How would his model change? How many rolls can he make? Explain.
Answer:
Instead dividing 2\(\frac{1}{2}\) by \(\frac{1}{8}\), he will divide 2\(\frac{1}{2}\) by \(\frac{1}{8}\) and in that case he will make more sushi rolls.

Go Math Lesson 3.4 Answer Key Dividing Mixed Numbers Question 4.
Analyze Relationships Explain how you can check the answer.
Answer:
We can check the answer by dividing the result by the first of the fractions we were dividing or multiplying the result by the second of the fractions we were dividing.

Question 5.
What If? Harold serves himself 1\(\frac{1}{2}\)-ounce servings of cereal each morning. How many servings does he get from a box of his favorite cereal? Show your work.
Answer:
We need to divide 1 by 1\(\frac{1}{2}\) in order to get how many servings Harold gets from a box. But first, we need to rewrite 1\(\frac{1}{2}\) as a fraction and find its reciprocal.
So, we have the following:
1\(\frac{1}{2}\) = \(\frac{3}{2}\)
Reciprocal of \(\frac{3}{2}\) is \(\frac{2}{3}\)
Now, we have to multiply 1 by \(\frac{2}{3}\) and get:
1 × \(\frac{2}{3}=\frac{1 \times 2}{1 \times 3}=\frac{2}{3}\)
So, Harold gets \(\frac{2}{3}\) servings of cereal from a box.

Your Turn

Question 6.
Sheila has 10\(\frac{1}{2}\) pounds of potato salad. She wants to divide the potato salad into containers, each of which holds 1\(\frac{1}{4}\) pounds. How many containers does she need? Explain.
Answer:
In order to find how many containers Sheila wilt need, we need to divide 10\(\frac{1}{2}\) by 1\(\frac{1}{4}\). But first we need to rewrite those mixed numbers as a fraction, than find reciprocal of the second one and then multiply them.
So, we get:

She will need 8\(\frac{2}{5}\) containers.

Reflect

Question 7.
Check for Reasonableness How can you determine if your answer is reasonable?
Answer:
If we have to divide two fractions, we can check our result on two ways:
First is to multiply result by the second fraction and if product is the first fraction, then our original solution is correct.
The second way is to divide the first fraction by result and if we get as a result of this dividing the second fraction, then our original solution is correct.
Multiply result by the second fraction; Divide the fraction by the result.

Your Turn

Question 8
The area of a rectangular patio is 12\(\frac{3}{8}\) square meters.
The width of the patio is 2\(\frac{3}{4}\) meters. What is the length? ___________
Answer:
First we will write given situation as a division problem, it is the following:
12\(\frac{3}{8}\) ÷ 2\(\frac{3}{4}\)
Now, we will rewrite the mixed numbers as fractions greater than 1.
12\(\frac{3}{8}\) ÷ 2\(\frac{3}{4}\) = \(\frac{99}{8}\) ÷ \(\frac{11}{4}\)
In the next step we will rewrite the problem as multiplication using the reciprocal of the second fraction:
\(\frac{99}{8} \div \frac{11}{4}=\frac{99}{8} \times \frac{4}{11}\)
Now, we will multiply previous fractions:

So, the length of a rectangular patio 4\(\frac{1}{2}\) meters.

Go Math Grade 6 Answers Dividing With Mixed Numbers Question 9.
The area of a rectangular rug is 14\(\frac{1}{2}\) square yards.
The length of the rug is 4\(\frac{1}{3}\) yards. What is the width? ___________
Answer:
First we wilt write given situation as a division problem, it is the following:
14\(\frac{1}{12}\) ÷ 4\(\frac{1}{3}\)
Now, we will rewrite the mixed numbers as fractions greater than 1.
14\(\frac{1}{12}\) ÷ 4\(\frac{1}{3}\) = \(\frac{169}{12} \div \frac{13}{3}\)
In the next step we will rewrite the problem as multiplication using the reciprocal of the second fraction:
\(\frac{169}{12} \div \frac{13}{3}=\frac{169}{12} \times \frac{3}{13}\)
Now, we will multiply previous fractions:

So, the width of a rectangular rug is 3\(\frac{1}{4}\) yards.

Texas Go Math Grade 6 Lesson 3.4 Guided Practice Answer Key

Divide. Write each answer in simplest form.

Question 1.
4\(\frac{1}{4}\) ÷ \(\frac{3}{4}\)

Answer:
First we will rewrite the mixed numbers as fractions greater than 1, then again rewrite the problem as multiplication
using the reciprocal ot the second fractions.
We will simplify using the GCF in order to get the solution in simplest form and multiply numerators and denominators. At the end, we will rewrite the solution as a mixed number if it is greater than 1.

So, we have the following:

Conclusion is that the solution is 3.

Question 2.
1\(\frac{1}{2}\) ÷ 2\(\frac{1}{4}\)

Answer:
First we will rewrite the mixed numbers as fractions greater than 1, then again rewrite the problem as multiplication
using the reciprocal ot the second fractions.
We will simplify using the GCF in order to get the solution in simplest form and multiply numerators and denominators. At the end, we will rewrite the solution as a mixed number if it is greater than 1.

So, we have the following:

Conclusion is that the solution is \(\frac{2}{3}\).

Question 3.
4 ÷ 1\(\frac{1}{8}\) ____________
Answer:
First we will rewrite the mixed numbers as fractions greater than 1, then again rewrite the problem as multiplication
using the reciprocal ot the second fractions.
We will simplify using the GCF in order to get the solution in simplest form and multiply numerators and denominators. At the end, we will rewrite the solution as a mixed number if it is greater than 1.

So, we have the following:

Conclusion is that the solution is \(\frac{32}{9}\), or 3\(\frac{5}{9}\).

Question 4.
3\(\frac{1}{5}\) ÷ 1\(\frac{1}{7}\) ____________
Answer:
First we will rewrite the mixed numbers as fractions greater than 1, then again rewrite the problem as multiplication
using the reciprocal ot the second fractions.
We will simplify using the GCF in order to get the solution in simplest form and multiply numerators and denominators. At the end, we will rewrite the solution as a mixed number if it is greater than 1.

So, we have the following:

Conclusion is that the solution is \(\frac{63}{40}\), or 1\(\frac{23}{40}\).

Question 5.
8\(\frac{1}{3}\) ÷ 2\(\frac{1}{2}\) ____________
Answer:
First we will rewrite the mixed numbers as fractions greater than 1, then again rewrite the problem as multiplication
using the reciprocal ot the second fractions.
We will simplify using the GCF in order to get the solution in simplest form and multiply numerators and denominators. At the end, we will rewrite the solution as a mixed number if it is greater than 1.

So, we have the following:

Conclusion is that the solution is \(\frac{10}{3}\), or 3\(\frac{1}{3}\).

Question 6.
15\(\frac{1}{3}\) ÷ 3\(\frac{5}{6}\) ____________
Answer:
Solution to this example is given below
First we will rewrite the mixed numbers as fractions greater than 1, then again rewrite the problem as multiplication
using the reciprocal ot the second fractions.
We will simplify using the GCF in order to get the solution in simplest form and multiply numerators and denominators. At the end, we will rewrite the solution as a mixed number if it is greater than 1.

So, we have the following:

Conclusion is that the solution is 4.

Write each situation as a division problem. Then solve.

Question 7.
A sandbox has an area of 26 square feet, and the length is 5\(\frac{1}{2}\) feet. What is the width of the sandbox?
Answer:
First we will write given situation as a division problem, it is the following:
26 ÷ 5\(\frac{1}{2}\)
Now, we will rewrite the mixed numbers as fractions greater than 1.
26 ÷ 5\(\frac{1}{2}\) = \(\frac{26}{1}\) ÷ \(\frac{11}{2}\)
In the next step we will rewrite the problem as multiplication using the reciprocal of the second fraction:
\(\frac{26}{1} \div \frac{11}{2}=\frac{26}{1} \times \frac{2}{11}\)
Now, we will multiply previous fractions:

So, the width of the sandbox is 4\(\frac{8}{11}\) feet

Lesson 3.4 Divide Mixed Numbers Go Math Grade 6 Question 8.
Mr. Webster is buying carpet for an exercise room in his basement. The room will have an area of 230 square feet. The width of the room is 12\(\frac{1}{2}\) feet. What is the length?
Answer:
First we will write given situation as a division problem, it is the following:
230 ÷ 12\(\frac{1}{2}\)
Now, we will rewrite the mixed numbers as fractions greater than 1.
230 ÷ 12\(\frac{1}{2}\) = \(\frac{230}{1}\) ÷ \(\frac{25}{2}\)
In the next step we will rewrite the problem as multiplication using the reciprocal of the second fraction:
\(\frac{230}{1} \div \frac{25}{2}=\frac{230}{1} \times \frac{2}{25}\)
Now, we will multiply previous fractions:

So, the length of the room is 18\(\frac{2}{5}\) feet.

Essential Question Check-In

Question 9.
How does dividing mixed numbers compare with dividing fractions?
Answer:
We have to rewrite mixed numbers as a fractions greater than 1 and then the steps of dividing are the same as dividing two fractions because we actually get two fractions.
It is equivalent with dividing fractions, but first rewrite mixed numbers as mixed numbers.

Question 10.
Jeremy has 4\(\frac{1}{2}\) CUPS of iced tea. He wants to divide the tea into \(\frac{3}{4}\)-cup servings. Use the model to find the number of servings he can make.

Answer:
We will use the following model in order to find the number of servings Jeremy can make:
4\(\frac{1}{2}\) ÷ \(\frac{3}{4}\)
We will rewrite mixed number as a fraction and rewrite the model as multiplication using the reciprocal of the
second fractions.
And then multiply fractions:

So, Jeremy can make 6 servings

Question 11.
A ribbon is 3\(\frac{2}{3}\) yards long. Mae needs to cut the ribbon into pieces that are \(\frac{2}{3}\) yard long. Use the model to find the number of pieces she can cut.

Answer:
We will use the following model in order to find the number of servings Jeremy can make:
3\(\frac{2}{3}\) ÷ \(\frac{2}{3}\)
We will rewrite mixed number as a fraction and rewrite the model as multiplication using the reciprocal of the
second fractions.
And then multiply fractions:

So, Mae can cut 11 pieces of a ribbon.

Question 12.
Dao has 2\(\frac{3}{8}\)pounds of hamburger meat. He is making \(\frac{1}{4}\)-pound hamburgers. Does Dao have enough meat to make 10 hamburgers? Explain.
Answer:
In order to find the number of hamburgers Dao can make of 2\(\frac{3}{8}\) pounds of hamburger meat, we have to divide 2\(\frac{3}{8}\) by \(\frac{1}{4}\).

Dao can make 9\(\frac{1}{4}\) hamburgers. Conclusion is that Dao has not enough meat for making 10 hamburgers.

Question 13.
Multistep Zoey made 5\(\frac{1}{2}\) cups of trail mix fora camping trip. She wants to divide the trail mix into \(\frac{3}{4}\)-cup servings.
a. Ten people are going on the camping trip. Can Zoey make enough \(\frac{3}{4}\)-cup servings so that each person on the trip has one serving?
Answer:
We will find number of serving Zoey can made dividing 5\(\frac{1}{2}\) by \(\frac{3}{4}\).

So, we can notice that there are 7\(\frac{1}{3}\) sevings.
Conclusion is that there is not enough servings for 10 persons.

b. What size would the servings need to be for everyone to have a serving? Explain.
Answer:
If we want to share 2\(\frac{3}{8}\) cups of trail mix to 10 persons, we need to divide 2\(\frac{3}{8}\) by 10 in order to get size of each serving:

So, size of each serving would be \(\frac{19}{80}\)-cup.

c. If Zoey decides to use the \(\frac{3}{4}\)-cup servings, how much more trail mix will she need? Explain.
Answer:
To calculate how much more trail mix Zoey will need for 10 persons, we need to subtract 7\(\frac{1}{3}\), result we get at (a), from 10:

So, Zoey will need \(\frac{8}{3}\)-cup of trail mix moreS

Question 14.
The area of a rectangular picture frame is 3o\(\frac{1}{3}\) square inches. The length of the frame is 6\(\frac{1}{2}\) inches. Find the width of the frame.
Answer:
First we will write given situation as a division problem, it is the following:
30\(\frac{1}{3}\) ÷ 6\(\frac{1}{2}\)
Now, we will rewrite the mixed numbers as fractions greater than 1.
20\(\frac{1}{3}\) ÷ 6\(\frac{1}{2}\) = \(\frac{91}{3}\) ÷ \(\frac{13}{2}\)
In the next step we will rewrite the problem as multiplication using the reciprocal of the second fraction:
\(\frac{91}{3} \div \frac{13}{2}=\frac{91}{3} \times \frac{2}{13}\)
Now, we will multiply previous fractions:

So, the width of the frame is 4\(\frac{2}{3}\) inches.

Question 15.
The area of a rectangular mirror is 11\(\frac{11}{16}\) square feet. The width of the mirror is 2\(\frac{3}{4}\)feet. If there is a 5 foot tall space on the wall to hang the mirror, will it fit? Explain.
Answer:
We need to find the length of the mirror dividing 11\(\frac{11}{16}\) by 2\(\frac{3}{4}\). So, we have the following:

We got that the Length of mirror is 4\(\frac{1}{4}\) feet and there is a 5 feet total space on the wall for hanging the mirror, so, 4\(\frac{1}{4}\) < 5, there is enough space on the wall for hanging mirror, it will fit.

Question 16.
Ramon has a rope that is 25\(\frac{1}{2}\) feet long. He wants to cut it into 6 pieces that are equal in length. How long will each piece be?
Answer:
In order to find length of each peace, we have to divide 25\(\frac{1}{2}\) by 6, so we have following:

So, each piece will be long 4\(\frac{1}{4}\) feet.

Question 17.
Eleanor and Max used two rectangular wooden boards to make a set for the school play. One board was 6 feet long, and the other was 5\(\frac{1}{2}\) feet long. The two boards had equal widths. The total area of the set was 60\(\frac{3}{8}\) square feet. What was the width?
Answer:
We can conclude that total length of the set is:

In order to total width of the set, we have to divide 60\(\frac{3}{8}\) by \(\frac{23}{3}\) and get:

So, the total width of the set 7\(\frac{7}{8}\) feet.

H.O.T. Focus On Higher Order Thinking

Question 18.
Draw Conclusions Micah divided 11\(\frac{2}{3}\) by 2\(\frac{5}{6}\) and got 4\(\frac{2}{17}\) for an answer.
Does his answer seem reasonable? Explain your thinking. Then check Micah’s answer.
Answer:
Yes, Micah’s answer seems reasonabLe because 4\(\frac{2}{17}\) is smaller than 11\(\frac{2}{3}\), now we will also check it:

We can see that Micah’s answer is correct.

Question 19.
Explain the Error To divide 14\(\frac{2}{3}\) ÷ 2\(\frac{3}{4}\), Erik multiplied 14\(\frac{2}{3}\) × \(\frac{4}{3}\). Explain Erik’s error.
Answer:
Erik’s error was that he did not first rewrite mixed numbers as a fractions and then multiply them using the reciprocal of the second fraction.
So, the right way is:

He did not first rewrite mixed numbers as a fractions.

Question 20.
Analyze Relationships Explain how you can find the missing number in 3\(\frac{4}{5}\) ÷ = 2\(\frac{5}{7}\). Then find the missing number.
Answer:
We will find the missing number dividing 3\(\frac{4}{5}\) by 2\(\frac{5}{7}\)
First, we will rewrite those mixed numbers as a fractions greater than 1.
Then rewrite the problem. First we need to rewrite the mixed numbers as a fractions grater than 1.

So, the missing number is \(\frac{7}{5}\), or 1\(\frac{2}{5}\).