enVisionmath 2.0: Grade 6, Volume 1 – Math Book Answers

Chapter 2: Algebra Solve Equations And Inequalities Review What You Know

Schindler — Tue, 29 Apr 2025 09:17:13 +0000

Chapter 2: Algebra Solve Equations And Inequalities

Review What You Know

Page 80 Exercise 1 Answer

In 6x, x is a(n) variable.

Result

Variable

Page 80 Exercise 2 Answer

x + 5 is an example of a(n) algebraic expression.

Result

Algebraic Expression

Page 80 Exercise 3 Answer

Evaluate an expression to find its value.

Result

Evaluate

Page 80 Exercise 4 Answer

6 + 2 Given

= 2 + 6 Commutative Property of Addition

Result

6 + 2 = 2 + 6

TRUE

Page 80 Exercise 5 Answer

2.5 − 1 = 1.5 Subtract

1 − 2.5 = −1.5 Subtract

Result

2.5 − 1 ≠ 1 − 2.5

FALSE

Page 80 Exercise 6 Answer

$\frac{1}{2}$ x 3 = 3 x $\frac{1}{2}$ Commutative Property of Multiplication: a × b = b × a

Result

TRUE

Page 80 Exercise 7 Answer

$\frac{3}{4}$ ÷ 5 Given

= $\frac{3}{4}$ x $\frac{1}{5}$ Rewrite Division as Multiplication

Result

$\frac{3}{4} \div 5=\frac{3}{4} \times \frac{1}{5}$

TRUE

Page 80 Exercise 8 Answer

$5 \div \frac{1}{3}=5 \times 3=15$ Rewrite Division as Multiplication

$\frac{5}{3}=1.66$ Divide

Result

$5 \div \frac{1}{3} \neq \frac{5}{3}$

FALSE

Page 80 Exercise 9 Answer

$\frac{2}{3}$ x 5 Given

= $\frac{2}{3}$ x $\frac{5}{1}$ Write 5 as a fraction

= $\frac{10}{3}$ Multiply

Result

$\frac{2}{3}$ x 5 ≠ $\frac{10}{15}$

FALSE

Page 80 Exercise 10 Answer

x − 2 Evaluate

= 8 − 2 Substitute x = 8

= 6 Subtract

Result

Page 80 Exercise 11 Answer

2b Evaluate

= 2(9) Substitute b = 9

= 18 Multiply

Result

Page 80 Exercise 12 Answer

$3 \frac{3}{4}$ + y Evaluate

= $\frac{15}{4}$ + $\frac{5}{6}$ Substitute y = $\frac{5}{6}$

= $\frac{45+10}{12}$ LCM is 12

= $\frac{55}{12}$ Simplify

Result

$\frac{55}{12}$

Page 80 Exercise 13 Answer

$\frac{15}{x}$ Evaluate

= $\frac{15}{3}$ Substitute x = 3

= 5 Divide

Result

Page 80 Exercise 14 Answer

5.6t Evaluate

= 5.6(0.7) Substitute t = 0.7

= 3.92 Multiply

Result

3.92

Page 80 Exercise 15 Answer

4x Evaluate

= $4\left(\frac{1}{2}\right)$ Substitute x = $\frac{1}{2}$

= 2 Multiply

Result

Page 80 Exercise 16 Answer

[(33 ÷ 3) + 1] − $2^2$

Order of Operation :

1. Evaluate the parentheses and bracket.

2. Evaluate the power.

3. Subtract.

[(33 ÷ 3) + 1] − $2^2$

= [11 + 1] − $2^2$ Evaluate the parentheses

= 12 − $2^2$ Evaluate the bracket

= 12 − 4 Evaluate the power

= 8 Subtract

Result

enVisionmath 2.0 Grade 6 Volume 1 Solutions

Alekhya — Tue, 29 Apr 2025 05:46:51 +0000

enVisionmath 2.0 Grade 6 Volume 1 Chapter 1 Use Positive Rational Numbers

enVisionmath 2.0: Grade 6, Volume 1, Chapter 2, Integers and Rational Numbers

enVisionmath 2.0: Grade 6, Volume 1, Chapter 3, Numeric and Algebraic Expressions

enVisionmath 2.0: Grade 6, Volume 1, Chapter ,4 Represent and Solve Equations and Inequalities

enVisionmath 2.0: Grade 6, Volume 1 Chapter 1 Algebra: Understand Numerical And Algebraic Expressions Guided Practice 3

Schindler — Tue, 29 Apr 2025 05:32:55 +0000

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Guided Practice 3

Page 27 Exercise 1 Answer

2(3 + 4)

Yes, the expression can be described as a product of two factors.

The expression has two factors.

One factor is 2 and the other is the sum 3 + 4.

Result

One factor is 2 and other factor is the sum 3 + 4

Page 27 Exercise 2 Answer

2(3 + 4)

The part of the expression (3 + 4) is the sum of two terms.

Result

(3 + 4)

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 27 Exercise 3 Answer

$\frac{w}{4}$ + 12.5 – 7z

The expression $\frac{w}{4}$ + 12.5 − 7z has three terms.

The terms are $\frac{w}{4}$, 12.5, and 7z

Result

The expression has three terms.

Page 27 Exercise 4 Answer

$\frac{w}{4}$ + 12.5 − 7z

The third term 7z is the product of two factors.

One factor is 7 and other factor is z.

Result

7z is the product of two factors.

Page 27 Exercise 5 Answer

5 − g

The expression 5 − g has two terms.

One term is 5 and other term is g.

NOTE : Each part of expression that is separated by a plus or a minus sign is called a term.

Result

The expression 5 − g has two terms.

Page 27 Exercise 6 Answer

$3+\frac{1}{2} b$

The expression $3+\frac{1}{2} b$ has two terms.

One term is 3 and other term is $\frac{1}{2} b$

NOTE : Each part of expression that is separated by a plus or a minus sign is called a term.

Result

The expression $3+\frac{1}{2} b$ has two terms.

Page 27 Exercise 7 Answer

$\frac{v}{3}$ + 2 . 5

The expression $\frac{v}{3}$ + 2 ⋅5 has two terms.

One term is $\frac{v}{3}$ and other term is 2 ⋅ 5.

NOTE : Each part of expression that is separated by a plus or a minus sign is called a term.

Result

The expression $\frac{v}{3}$ + 2 ⋅ 5 has two terms.

Page 27 Exercise 8 Answer

16.2 − (3 ⋅ 4) + (14 ÷ 2)

The expression 16.2 − (3 ⋅ 4) + (14 ÷ 2) has three terms.

One term is 16.2, second term is 3 ⋅ 4 and third term is 14 ÷ 2.

NOTE : Each part of expression that is separated by a plus or a minus sign is called a term.

Result

The expression 16.2 − (3 ⋅ 4) + (14 ÷ 2) has three terms.

Page 27 Exercise 9 Answer

5.3t − (20 ÷ 4) + 11

The expression 5.3t − (20 ÷ 4) + 11 has three terms.

The terms are 5.3t, 20 ÷ 4 and 11.

The second term 20 ÷ 4 represent the quotient of 20 divided by 4

NOTE : Each part of expression that is separated by a plus or a minus sign is called a term.

Result

20 ÷ 4 represent the quotient of 20 divided by 4

Page 27 Exercise 10 Answer

5.3t − (20 ÷ 4) + 11

The expression 5.3t − (20 ÷ 4) + 11 has three terms.

The terms are 5.3t, 20 ÷ 4 and 11.

The first term 5.3t represent the product of two factors.

One factor is 5.3 and other factor is t

NOTE : Each part of expression that is separated by a plus or a minus sign is called a term.

Result

5.3t represent the product of two factors.

Page 27 Exercise 11 Answer

7(10 + 8) − 9

The expression 7(10 + 8) − 9 has two terms.

The part 10 + 8 represent a sum.

Result

(10 + 8) represent a sum

Page 27 Exercise 12 Answer

7(10 + 8) − 9

The expression 7(10 + 8) represents a product.

One factor is 7 and other factor is the sum 10 + 8

Result

7(10 + 8) represents a product as a factor.

One factor is 7 and other factor is 10 + 8

enVisionmath 2.0: Grade 6, Volume 1 Chapter 1 Algebra: Understand Numerical And Algebraic Expressions Homework And Practice 4

Schindler — Tue, 29 Apr 2025 05:32:19 +0000

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Homework And Practice 4

Page 29 Exercise 1 Answer

4c + $7 \frac{1}{2}$

The expression 4c + $7 \frac{1}{2}$ has two terms.

The terms are 4c and $7 \frac{1}{2}$

NOTE : Each part of the expression that is separated by a plus or minus sign is called a term.

Result

The expression 4c + $7 \frac{1}{2}$ has two terms.

Page 29 Exercise 2 Answer

80 . 6 − 3p − q

The expression 80 . 6 − 3p − q has three terms.

The terms are 80 . 6 , 3p, and q

NOTE : Each part of the expression that is separated by a plus or minus sign is called a term.

Result

The expression 80 . 6 − 3p − q has three terms.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 29 Exercise 3 Answer

(7 ⋅ 2) ÷ s

The expression (7 ⋅ 2) ÷ s has only one term.

NOTE : Each part of the expression that is separated by a plus or minus sign is called a term.

Result

The expression (7 ⋅ 2) ÷ s has one term.

Page 29 Exercise 4 Answer

100 + (8 ⋅ 6) − 50 + 2

The expression 100 + (8 ⋅ 6) − 50 + 2 has four terms.

The terms are 100, 8 ⋅ 6, 50, and 2

NOTE : Each part of the expression that is separated by a plus or minus sign is called a term.

Result

The expression 100 + (8 ⋅ 6) − 50 + 2 has four terms.

Page 29 Exercise 5 Answer

1 + $\frac{z}{3}$ + 2w

The expression 1 + $\frac{z}{3}$ + 2w has three terms.

First term is 1

Second term is $\frac{z}{3}$

Third term is 2w

The second term $\frac{z}{3}$ represent the quotient of z divided by 3

NOTE : Each part of the expression that is separated by a plus or minus sign is called a term.

Result

The second term z represent the quotient of z divided by 3

Page 29 Exercise 6 Answer

1 + $\frac{z}{3}$ + 2w

The expression 1 + $\frac{z}{3}$ + 2w has three terms.

First term is 1

Second term is $\frac{z}{3}$

Third term is 2w

The third term 2w represent the product as a factor with one factor 2 and other factor is w

NOTE : Each part of the expression that is separated by a plus or minus sign is called a term.

Result

The third term 2w represent the product as a factor with one factor 2 and other factor is w

Page 29 Exercise 7 Answer

$\frac{3}{4}$ + 3(14 − 7)

The expression $\frac{3}{4}$ + 3(14 − 7) has two terms.

First term is $\frac{3}{4}$

Second term is 3(14 − 7)

The part 14 − 7 represents a difference.

NOTE : Each part of the expression that is separated by a plus or minus sign is called a term.

Result

The part 14 − 7 represents a difference.

Page 29 Exercise 8 Answer

$\frac{3}{4}$ + 3(14 − 7)

The expression $\frac{3}{4}$ + 3(14 − 7) has two terms.

First term is $\frac{3}{4}$

Second term is 3(14 − 7)

The second term 3(14 − 7) represents a product.

NOTE : Each part of the expression that is separated by a plus or minus sign is called a term.

Result

The second term 3(14 − 7) represents a product.

Page 30 Exercise 9 Answer

Cost of one sandwich = $5

Cost of one Drink = $1

Cost of salad = $4

Expression to show the cost of 2 sandwiches, 2 drinks and a salad :

2 ⋅ 5 + 2 ⋅ 1 + 4

The expression 2 ⋅ 5 + 2 ⋅ 1 + 4 has 3 terms.

The terms are 2 ⋅ 5, 2 ⋅ 1 and 4

NOTE : Each part of expression that is separated by a plus or minus sign is called a term.

Result

Expression: 2 ⋅ 5 + 2 ⋅ 1 + 4

Terms = 3

Page 30 Exercise 10 Answer

Cost of one sandwich = $5

Cost of one Drink = $1

Number of drinks ordered = 16

Let the number of sandwiches ordered be = n

Expression for total cost of 16 drinks and some sandwiches ordered : 16 ⋅ 1 + n ⋅ 5

The expression 16 ⋅ 1 + n ⋅ 5 has two terms.

The first term is 16 ⋅ 1 represent a product as factor.

One factor is 16 and other factor is 1

The second term is n⋅5 which also represents a product as factor

One factor is n and other factor is 1

NOTE : Each part of expression that is separated by a plus or minus sign is called a term.

Result

Expression: 16 ⋅ 1 + n ⋅ 5

The first term is 16 ⋅ 1 represent a product as factor

The second term is n ⋅ 5 which also represents a product as factor

Page 30 Exercise 11 Answer

$\frac{a}{2}$

The expression $\frac{a}{2}$ has one term which represents a quotient of a divided by 2

So, Mary is not correct.

Result

Mary is not correct

Page 30 Exercise 12 Answer

Multiplying the number (x+y) by $\frac{1}{3}$ is the same as dividing it by 3

$(x+y) \times \frac{1}{3}=\frac{x}{3}+\frac{y}{3}$ $(x+y) \div 3=\frac{x}{3}+\frac{y}{3}$

So, $\frac{1}{3}$ = (x + y) ÷ 3

Result

Multiplying the number (x + y) by $\frac{1}{3}$ is the same as dividing it by 3

Page 30 Exercise 13 Answer

6 + 5(12 − 8)

The expression 6 + 5(12 − 8) has two terms which is separated by a plus sign.

So, the word sum describes the entire expression.

Result

SUM

Page 30 Exercise 14 Answer

$3 t-\frac{10}{(4+1)}-2$

Result

Coefficient → 3

Quotient → 10

Sum → 4 + 1

Product → 3t

enVisionmath 2.0: Grade 6, Volume 1 Chapter 1 Algebra: Understand Numerical And Algebraic Expressions Guided Practice 4

Schindler — Tue, 29 Apr 2025 05:31:27 +0000

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Guided Practice 4

Page 33 Exercise 1 Answer

It is necessary to use Order of Operation to evaluate algebraic expression because the algebraic expression contains different operations so it is necessary to follow the Order of Operations to know which operations should

we evaluate first and there after.

If we do not use Order of Operation then the solution of the expression will be incorrect.

Result

It is necessary because we should know which operations to evaluate first.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 33 Exercise 3 Answer

3t − 8 Evaluate

= 3(8) − 8 Substitute t = 8

= 24 − 8 Multiply

= 16 Subtract

Result

Page 33 Exercise 4 Answer

6w ÷ x + 9 Evaluate

= $6\left(\frac{1}{2}\right) \div(3)+9$ Substitute w = $\frac{1}{2}$ and x = 3

= 3 ÷ 3 + 9 Evaluate the parentheses

= 1 + 9 Divide

= 10 Add

Result

Page 33 Exercise 5 Answer

$t^2$ − 12w ÷ x Evaluate

= $(8)^2-12\left(\frac{1}{2}\right) \div(3)$ Substitute t=8, w = $\frac{1}{2}$ and x = 3

= 64 − 6 ÷ 3 Evaluate the parentheses

= 64 – 2 Divide

= 62 Subtract

Result

Page 33 Exercise 6 Answer

5x − 2w + t Evaluate

= $5(3)-2\left(\frac{1}{2}\right)+(8)$ Substitute x = 3, w = $\frac{1}{2}$ and t = 8

= 15 − 1 + 8 Evaluate the parentheses

= 14 + 8 Subtract

= 22 Add

Result

Page 33 Exercise 7 Answer

9x Evaluate

= 9(3) Substitute x = 3

= 27 Multiply

Result

Page 33 Exercise 8 Answer

3w + 6 ÷ 2x Evaluate

= 3(5) + 6 ÷ 2(3) Substitute w = 5 and x = 3

= 15 + 6 ÷ 6 Evaluate the parentheses

= 15 + 1 Divide

= 16 Add

Result

Page 33 Exercise 9 Answer

$w^2$ + 2 + 48 ÷ 2x Evaluate

= $(5)^2$ + 2 + 48 ÷ 2(3) Substitute w = 5 and x = 3

= 25 + 2 + 48 ÷ 6 Evaluate the parentheses

= 25 + 2 + 8 Divide

= 35 Add

Result

Page 33 Exercise 10 Answer

$3^3$ + 5y ÷ w + z Evaluate

= $(3)^3$ + 5(4) ÷ (5) + (8) Substitute x = 3, y = 4, w = 5 and z = 8

= 27 + 20 ÷ 5 + 8 Evaluate the parentheses

= 27 + 4 + 8 Divide

= 39 Add

Result

Page 33 Exercise 11 Answer

9y ÷ x + $z^{2}$ − w Evaluate

= 9(4) ÷ (3) + $(8)^2$ − (5) Substitute y = 4, x = 3, z = 8 and w = 5

= 36 ÷ 3 + 64 − 5 Evaluate the parentheses

= 12 + 64 − 5 Divide

= 76 − 5 Add

= 71 Subtract

Result

Page 34 Exercise 12 Answer

$x^{2}$ + 4w − 2y ÷ z Evaluate

= $(3)^2$ + 4(5) − 2(4) ÷ (8)

Substitute x = 3, w = 5, y = 4 and z = 8

= 9 + 20 − 8 ÷ 8 Evaluate the parentheses

= 9 + 20 − 1 Divide

= 29 − 1 Add

= 28 Subtract

Result

Page 34 Exercise 13 Answer

Weekly fee to rent a small car for a week = $250

Cost per mile = $0.30

a) Let m = number of miles Ms. White drives during the week

Expression for the amount she will pay for the car:

250 + 0.30m

b) Evaluate the expression if she drives 100 miles:

250 + 0.30m

=250 + 0.30(100) Substitute m = 100

= 250 + 30 Multiply

= $280

Result

a) 250 + 0.30m

b) $280

Page 34 Exercise 14 Answer

325 + 120d Evaluate

= 325 + 120(11) Substitute d = 11

= 325 + 1320 Multiply

= 1645 Add

Result

Mr. Black will have to pay $1645 for 11 − day rental

Page 34 Exercise 15 Answer

Small Car:

Cost for week = $250

Cost per day = $100

Cost for 2 days = 100 ⋅ 2 = $200

Cost for 3 days = 100 ⋅ 3 = $300

So, if we rent small car for 2 day then it will cost $200 which is less expensive to rent for the week.

Result

2 days

Page 34 Exercise 16 Answer

No, we cannot evaluate the expression 5 + 3n by adding 5 + 3 first and then multiplying by the value of n.

According to Charlene :

Example:

Let n = 2

5 + 3n

= 5 + 3(2) Substitute n = 2

= 8(2) Add

= 16 Multiply

We should multiply the value of n with 3 first and then add the value with 5

Example:

Let n = 2

5 + 3n

= 5 + 3(2) Substitute n = 2

= 5 + 6 Multiply

= 11 Add

Thus Charlene is not correct

Result

Charlene is not correct

Page 34 Exercise 17 Answer

$\left(d \cdot 10^4\right)+\left(d \cdot 10^3\right)+\left(d \cdot 10^2\right)+\left(d \cdot 10^1\right)+\left(d \cdot 10^0\right)$ Evaluate

= $\left(7 \cdot 10^4\right)+\left(7 \cdot 10^3\right)+\left(7 \cdot 10^2\right)+\left(7 \cdot 10^1\right)+\left(7 \cdot 10^0\right)$ Substitute d = 7

= (7 ⋅ 10000) + (7 ⋅ 1000) + (7 ⋅ 100) + (7 ⋅ 10) + (7 ⋅ 1) Evaluate the power

= 70000 + 7000 + 700 + 70 + 7 Multiply

= 77,777 Add

Result

77,777

Page 34 Exercise 18 Answer

$a^{2}$ + 3b ÷ c − d Evaluate

= $(3)^2$ + 3(8) ÷ (6) − (1) Substitute a = 7, b = 8, c = 6 and d = 1

= 49 + 24 ÷ 6 − 1 Evaluate the parentheses

= 49 + 4 − 1 Divide

= 53 − 1 Add

= 52 Subtract

Result

A) 52

Page 34 Exercise 19 Answer

8b ÷ a − $c^{2}$ + d Evaluate

= 8(5) ÷ (2) − $(3)^2$ + (9) Substitute b = 5, a = 2, c = 3 and d = 9

= 40 ÷ 2 − 9 + 9 Evaluate the parentheses

= 20 − 9 + 9 Divide

= 11 + 9 Subtract

= 20 Add

Result

C) 20

enVisionmath 2.0: Grade 6, Volume 1 Chapter 1 Use Positive Rational Numbers Section 1.7

Schindler — Tue, 29 Apr 2025 05:30:03 +0000

Chapter 1 Use Positive Rational Numbers

Section 1.7: Solve Problems With Rational Numbers

Page 49 Exercise 1a Answer

To answer the question, we first need to calculate how much food the cat eats in one day.

Since Jenna feeds her cat twice a day, each time a $\frac{3}{4}$ can, find the product of those two numbers.

Result

To answer the question, we first need to calculate how much food the cat eats in one day by multiplying 2 and $\frac{3}{4}$.

Page 49 Exercise 1b Answer

Since Jenna feeds her cat twice a day, each time a $\frac{3}{4}$ can, find teh product of those two numbers.

$2 \times \frac{3}{4}=\frac{2}{1} \times \frac{3}{4}=\frac{2 \times 3}{1 \times 4}=\frac{6}{4}=\frac{4}{4}+\frac{2}{4}=1 \frac{1}{2}$

Jenna’s cat eats $1 \frac{1}{2}$ can of food per day. Since Jenna’s friend will take car of her cat for 5 days multiply $1 \frac{1}{2}$ by 5.

$1 \frac{1}{2} \times 5=\frac{3}{2} \times 5=\frac{3}{2} \times \frac{5}{1}=\frac{3 \times 5}{2 \times 1}=\frac{15}{2}=\frac{14}{2}+\frac{1}{2}=7 \frac{1}{2}$

Jenna’s cat will eat $7 \frac{1}{2}$ cans of food in five days, so since Jenna bought 8 cans of food she bought enough.

Result

Since Jenna feeds her cat twice a day, each time a $\frac{3}{4}$ can, we can find the product of those two numbers to find how much the cat eats per day. We can then multiply the product by 5 to find how much food the cat eats in 5 days. This gives $7 \frac{1}{2}$cans which is less than 8 so she bought enough.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 49 Exercise 1 Answer

We know Jenna feeds her cat twice a day, feeds her cat $\frac{3}{4}$ can each time, needs to buy enough for 5 days, and she bought 8 cans

If Jenna multiplied, divided, and compared to find out whether she has enough cat food, then Jenna first multiplied to find how many cans per day her cat eats:

$2 \times \frac{3}{4}=\frac{2}{1} \times \frac{3}{4}=\frac{2 \times 3}{1 \times 4}=\frac{6}{4}=\frac{3}{2}$

Then she divided the number of cans she bought by the number of cans needed per day:

$8 \div \frac{3}{2}=\frac{8}{1} \times \frac{2}{3}=\frac{8 \times 2}{1 \times 3}=\frac{16}{3}=5 \frac{1}{3}$

Jenna then bought enough cans for $5 \frac{1}{3}$ days. Comparing this to 5 days, she can then conclude that she bought enough cans.

Result

Jenna multiplied 2 and $\frac{3}{4}$ to find the number of cans needed per day. Next she divided the number of cans she bought by the number of cans needed per day. She then compared the quotient of $5 \frac{1}{3}$ with 5 to determine that she bought enough.

Page 50 Exercise 1 Answer

A farmer is building a small horse-riding arena that is built using three rows of wood planks. The farmer uses wood planks that are $8 \frac{1}{2}$ feet long and ordered 130 of these wood planks.

We need to determine if the farmer ordered enough planks for an arena that measured $115 \frac{1}{4}$ ft by $63 \frac{1}{2}$ ft.

To answer the question we must find the perimeter of the arena.

$P=2 \times 115 \frac{1}{4}+2 \times 63 \frac{1}{2}=2 \times \frac{461}{4}+2 \times \frac{127}{2}=\frac{461}{2}+127$ $=\frac{461}{2}+\frac{127 \times 2}{2}=\frac{461}{2}+\frac{254}{2}=\frac{461+254}{2}=\frac{715}{2}=357 \frac{1}{2}$

The farmer needs enough wood planks for 3 times the perimeter.

$3 \times 357 \frac{1}{2}=\frac{3}{1} \times \frac{715}{2}=\frac{3 \times 715}{1 \times 2}=\frac{2145}{2}=1072 \frac{1}{2} \mathrm{ft}$

Divide to find how many $8 \frac{1}{2}$-foot-long planks are needed.

$1072 \frac{1}{2} \div 8 \frac{1}{2}=\frac{2145}{2} \div \frac{17}{2}=\frac{2145}{2} \times \frac{2}{17}=\frac{2145 \times 2}{2 \times 17}=\frac{2145}{17}=126 \frac{3}{2}$

The farmer needs at least 127 wood planks to build the fencing. Since he ordered 130 wood planks, he has enough to build the fencing.

Result

The farmer is correct.

Page 51 Exercise 2 Answer

A 26.2-mile marathon is being planned.

The number of runners who finish the marathon is 320. Runners donate $2.50 for each mile they run. How much money is donated?

26.2 × 320 = 8384miles

320 runners finished the marathon which means they ran all 26.2 miles. Thus, together they ran 8384 miles. For each mile they donated $2.50.

8384 × $2.50 = 20960

Result

$20960 is donated.

Page 52 Exercise 1 Answer

There are two types of problems with rational numbers – problems with fractions and problems with decimals.

When solving multistep problems with fractions or decimals, first decide the steps to use to solve the problem and choose the correct operations. Then identify the information you need from the problem and use it correctly. Calculate and interpret solutions and check that the answer is reasonable.

Result

Page 52 Exercise 2 Answer

Meghan has $5 \frac{1}{4}$ yards of fabric and plans to use $\frac{2}{3}$ of the fabric to make 4 identical backpacks. Meghan multiplies $5 \frac{1}{4}$ by $\frac{2}{3}$ to find how much fabric she will use to make the backpacks:

$5 \frac{1}{4} \times \frac{2}{3}=\frac{21}{4} \times \frac{2}{3}=\frac{21 \times 2}{4 \times 3}=\frac{42}{12}=\frac{36}{12}+\frac{6}{12}=3 \frac{6}{12}=3 \frac{1}{2}$

Meghan will use $3 \frac{1}{2}$ yards of fabric to make 4 backpacks.

To find out how much fabric she needs for each backpack she must divide $3 \frac{1}{2}$ by 4.

$3 \frac{1}{2} \div 4=\frac{7}{2} \div \frac{4}{1}=\frac{7}{2} \times \frac{1}{4}=\frac{7 \times 1}{2 \times 4}=\frac{7}{8}$

Meghan needs $\frac{7}{8}$ yard of fabric for each backpack.

Result

She must divide $3 \frac{1}{2}$ by 4 to get $\frac{7}{8}$ which means she needs $\frac{7}{8}$ yard of fabric for each backpack.

Page 52 Exercise 3 Answer

Each side of a square patio is 10.5 feet. The patio is made up of 1.5 foot by 1.5 foot square stones. What is the number of stones in the patio?

The solution which is given:

10.5 x 10.5 = 110.25, 110.25 ÷ 1.5 = 73.5

To solve the problem we need to divide the area of the patio by the area of the stones. In the solution, the area of the patio is correctly calculated, 10.5 x 10.5 = 110.25. However, the area is then divided by 1.5 which is not the area of the stone, it is the length of the side of the stone. To find the solution we first need to find the area of the stone.

The area of the stone is 2.25 square feet.

To calculate how many stones are in the patio we must divide the area of the patio, 110.25, by the area of the stones.

There are 49 stones in the patio.

Result

The solution does not include all the steps needed to solve the problem. The find the solution, we need to find the area of the stone. There are 49 stones in the patio.

Page 52 Exercise 4a Answer

Devon records 4 hours of reality shows on her DVR and records comedy shows for $\frac{3}{8}$ of that amount of time.

To find the number of hours of comedy shows that Devon records, we need to multiply 4 by $\frac{3}{8}$.

$\frac{3}{8} \times 4=\frac{3}{8} \times \frac{4}{1}=\frac{3 \times 4}{8 \times 1}=\frac{12}{8}=\frac{3}{2}=1 \frac{1}{2}$

Result

Devon recorded $1 \frac{1}{2}$ hour of comedy shows.

Page 52 Exercise 4b Answer

Devon records 4 hours of reality shows on her DVR and $1 \frac{1}{2}$ hours of comedy shows, which we calculated in Exercise 4a.

To find the total number of hours of reality and comedy shows that Devon records, we need to add the number of hours of reality and the number of hours of comedy shows.

$4+1 \frac{1}{2}=\frac{4}{1}+\frac{3}{2}=\frac{8}{2}+\frac{3}{2}=\frac{8+3}{2}=\frac{11}{2}=\frac{10}{2}+\frac{1}{2}=5 \frac{1}{2}$

Result

The total number of hours of reality and comedy shows that Devon recored is $5 \frac{1}{2}$.

Page 52 Exercise 4c Answer

The total number of hours of reality and comedy shows that Devon recorded is $5 \frac{1}{2}$, which we calculated in Exercise 4b. Devon watches all the reality and comedy shows in half-hour sittings.

To find the number of half-hour sittings needed to watch all the shows we must divide the total number of hours of reality and comedy shows that Devon recorded, which is $5 \frac{1}{2}$, by $\frac{1}{2}$.

$5 \frac{1}{2} \div \frac{1}{2}=\frac{11}{2} \div \frac{1}{2}=\frac{11}{2} \times \frac{2}{1}=\frac{11 \times 2}{2 \times 1}=\frac{22}{2}=11$

Result

The number of half-hour sittings needed to watch all the shows is 11.

Page 52 Exercise 5a Answer

An auto mechanic earns $498.75 in 35 hours during the week, his pay is $2.50 more per hour on weekends, and he works 6 hours on the weekend in addition to 35 hours during the week. We need to find how much he earns.

The questions we need to answer to solve the problem are:

1. How much does he earn per hour during the week?

2. How much does he earn per hour during the weekend?

3. How much does he earn during the whole weekend?

Page 52 Exercise 5b Answer

The questions we need to answer to solve the problem are:

1. How much does he earn per hour during the week?

2. How much does he earn per hour during the weekend?

3. How much does he earn during the whole weekend?

To answer the first question, we must divide $498.75 by 35, since he earns $498.75 during the whole week and he works 35 hours during the whole week.

During the week, the auto mechanic earns $14.25 per hour.

To answer the second question, we must add $2.50 to $14.25, since he earns $2.50 more per hour on weekends.

During the weekend, the auto mechanic earns $16.75 per hour. To answer the third question, we must multiply $16.75 by 6, since he earns $16.75 per hour and works 6 hours during the whole weekend.

During the whole weekend, the auto mechanic earns $100.50.

To determine how much the auto mechanic earns, we must add how much he earns during the whole week and how much he earns during the weekend.

Result

The auto mechanic earns $599.25.

Page 53 Exercise 6 Answer

First, we must calculate how much we would pay separately for apples, pears, and oranges we bought.

The dollar amounts must be rounded to the nearest cent so the oranges cost $2.79, the pears cost $1.49, and the apples cost $3.14.

Second, we add the cost of apples, pears, and oranges to get a total bill.

Result

Total bill rounded to the nearest cent is $7.42.

Page 53 Exercise 7a Answer

First, to solve the problem, we must calculate how much 8.9 pounds of apples cost. Since apples cost $0.99 per lb we multiply 8.9 and 0.99.

8.9 x 0.99 = $8.81

8.9 pounds of apples cost $8.81.

Result

First, we must calculate how much 8.9 pounds of apples cost, which is $8.81.

Page 53 Exercise 7b Answer

First, to solve the problem, we must calculate how much does 8.9 pounds of apples cost, which we did in Exercise 7a.

8.9 pounds of apples cost $8.81.

Next, we have to calculate the difference between the money the student payed with and the cost of 8.9 pounds of apples.

10.00 – 8.81 = 1.19

The student receives $1.19 of change.

Result

Next, we have to calculate the difference between the money the student paid with and the cost of the 8.9 pounds of apples. The student receives $1.19 in change.

Page 53 Exercise 8a Answer

To solve the problem we must first calculate how many oranges and how many pears the customer bought. Since he paid $3.27 for oranges and they cost $1.09 lb, we must divide 3.27 by 1.09. We use the same logic to find how many pears he bought, so we divide 4.76 by 1.19.

The customer bought 3 pounds of oranges and 4 pounds of pears.

Result

We must first calculate how many oranges and how many pears the customer bought. The customer bought 3 pounds of oranges and 4 pounds of pears.

Page 53 Exercise 8b Answer

To solve the problem we must first calculate how many oranges and how many pears the customer bought, which we did in Exercise 8a.

The customer bought 3 pounds of oranges and 4 pounds of pears.

To find how many pounds of fruit the customer bought we now must add the number of pounds of oranges and the number of pounds of pears he bought.

3 + 4 = 7

The customer bought 7 pounds of fruit.

Result

Next, we must add the number of pounds of oranges and the number of pounds of pears he bought. The customer bought 7 pounds of fruit.

Page 53 Exercise 9 Answer

It is given that students put $2 \frac{1}{4}$ pounds of trail mix into bags that each weigh $\frac{3}{8}$ pound and bring $\frac{2}{3}$ of the bags of trail mix on a hiking trip.

To find how many bags are left, we would need to perform the following steps:

Find how many bags the students made by dividing $2 \frac{1}{4}$ pounds by $\frac{3}{8}$.

Multiply the number of bags by $\frac{2}{3}$ to find how many they brought on the trip.

Subtract the total number of bags and the number of bags they brought to find how many are left.

We would then need 3 steps so no we cannot complete just one step to determine how many bags are left.

Performing step 1 we get:

$2 \frac{1}{4} \div \frac{3}{8}=\frac{9}{4} \times \frac{8}{3}=\frac{72}{12}=6$

Performing step 1 gives:

$6 \times \frac{2}{3}=\frac{12}{3}=4$

Performing step 3 then gives:

6 − 4 = 2

There are then 2 bags left.

Result

No, we cannot complete just one step. We need to find how many bags they made, how many they took on the hike, and then we can find how many are left, which was 2 bags.

Page 53 Exercise 10 Answer

It is given that $\frac{3}{5}$ of the T-shirts are blue and $\frac{5}{8}$ of the blue shirts are on sale. The fraction of T-shirts that are blue and on sale is then:

$\frac{3}{5} \times \frac{5}{8}=\frac{3 \times 5}{5 \times 8}=\frac{15}{40}=\frac{3}{8}$

If $\frac{1}{3}$ of the blue shirts that are on sale are size medium, then the fraction of blue T-shirts that are medium and on sale is:

$\frac{1}{3} \times \frac{3}{8}=\frac{1 \times 3}{3 \times 8}=\frac{3}{24}=\frac{1}{8}$

Result

$\frac{1}{8}$

Page 54 Exercise 11 Answer

The area of the community garden is given as the product 6.2 × 4.5 and the area of the vegetable garden is 0.4 times the area of the community garden.

6.2 x 4.5

0.4 x 27.9

Result

The area of the vegetable garden is 11.16 square meters.

Page 54 Exercise 12 Answer

The area of the herb garden is given by the product 2.2 x 1.6 and the area of the flower garden is 9.7 square meters greater than the herb garden.

Result

The area of the flower garden is 13.22 square meters.

Page 54 Exercise 13 Answer

It is given that $\frac{3}{4}$ cup is left and that Jim will divide $\frac{4}{5}$ of the leftover dip equally between 2 friends.

Since $\frac{3}{4}$ cup is left and he divides up $\frac{4}{5}$ of the leftover dip, then the amount of dip he divides between his friends is:

$\frac{3}{4} \times \frac{4}{5}=\frac{3 \times 4}{4 \times 5}=\frac{12}{20}=\frac{3}{5}$

Since he divides it equally between 2 friends, then each friend gets half of the $\frac{3}{5}$ cup of dip:

$\frac{1}{2} \times \frac{3}{5}=\frac{1 \times 3}{2 \times 5}=\frac{3}{10}$

Each friend will then get $\frac{3}{10}$ cup.

Result

$\frac{3}{10}$ cup

Page 54 Exercise 14 Answer

On the first day of a 3 day hiking trip the students hike 0.28 of the total distance which is 18.5 kilometers.

On the first day the students hiked 5.18 kilometers.

The remaining distance is equally split between the second and the third day.

The remaining distance is 13.32 kilometers. It is divided by two.

Result

On day 3 they will hike 6.66 kilometers.

Page 54 Exercise 15 Answer

From the picture, we can see that the three containers of potato salad that Kelly bought are different weights 1.03 lb, 1.12 lb, and 1.6 lb. First, we must calculate the sum 1.03 + 1.12 + 1.6 to find how many pounds she bought and then multiply that by $\frac{4}{5}$ to find how many she brought to the picnic.

$(1.03+1.12+1.6) \times \frac{4}{5}=3.75 \times \frac{4}{5}=3.75 \times 0.8=3$

Another way to find the answer is to write 3.75 as a fraction and than multiply that fraction by $\frac{4}{5}$.

$3.75=\frac{375}{100}=\frac{15}{4}$ $3.75 \times \frac{4}{5}=\frac{15}{4} \times \frac{4}{5}=\frac{15 \times 4}{4 \times 5}=\frac{60}{20}=3$

Result

Kelly brought 3 pounds of potato salad to the picnic.

Page 54 Exercise 16 Answer

To calculate how much did the students earn, we must first calculate how man bottles they sold. 84.5 ounces od liquid soap was divided into 6.5 ounce bottles.

The students sold bottles of soap for $5.50 and they sold 13 of them.

Result

The students earned $71.50.

Page 54 Exercise 17 Answer

Last week it took Clair $\frac{7}{12}$ hour to mow one lawn.

$\frac{7}{12} \times 5=\frac{7}{12} \times \frac{5}{1}=\frac{7 \times 5}{12 \times 1}=\frac{35}{12}=\frac{24}{12}+\frac{11}{12}=2 \frac{11}{12}$

Last week it took her $2 \frac{11}{12}$ to mow 5 lawns. This week she mowed the same lawns but it took her $\frac{5}{12}$ hour because she used her new lawn mower.

$\frac{5}{12} \times 5=\frac{5}{12} \times \frac{5}{1}=\frac{5 \times 5}{12 \times 1}=\frac{25}{12}=\frac{24}{12}+\frac{1}{12}=2 \frac{1}{12}$

This week it took Claire $2 \frac{1}{12}$ hour to mow 5 lawns.

$2 \frac{11}{12} \div 2 \frac{1}{12}=\frac{35}{12} \div \frac{25}{12}=\frac{35}{12} \times \frac{12}{25}=\frac{35 \times 12}{12 \times 25}=\frac{35}{25}=\frac{7}{5}=\frac{5}{5}+\frac{2}{5}=1 \frac{2}{5}$

Result

Claire’s time to mow all the lawns last week was $1 \frac{2}{5}$ times longer than this week.

enVisionmath 2.0: Grade 6, Volume 1 Chapter 1 Use Positive Rational Numbers Section 1.6

Schindler — Tue, 29 Apr 2025 05:29:24 +0000

Chapter 1 Use Positive Rational Numbers

Section 1.6: Divide Mixed Numbers

Page 43 Exercise 1 Answer

A $5 \frac{1}{2}$ inch strip of silver wire is cut into $1 \frac{3}{8}$ inch pieces so we must divide $5 \frac{1}{2}$ by $1 \frac{3}{8}$.

$5 \frac{1}{2} \div 1 \frac{3}{8}=\frac{11}{2} \div \frac{11}{8}=\frac{11}{2} \times \frac{8}{11}=\frac{11 \times 8}{2 \times 11}=\frac{88}{22}=4$

Result

She can make 4 pieces.

Page 43 Exercise 1 Answer

Use estimation to check whether the answer is reasonable, that is round up or round down the fractions and divide.

Round down both fractions, $5 \frac{1}{2}$ to 5 and $1 \frac{3}{8}$ to 1. The quotient 5 ÷ 1 is 5.

Round up both fractions, $5 \frac{1}{2}$ to 6 and $1 \frac{3}{8}$ to 2. The quotient 6 ÷ 2 is 3.

Thus, the answer must be greater than 3 and less than 5.

Since the answer is 4, it is reasonable.

Result

Rounding both fractions down gives $5 \frac{1}{2} \div 1 \frac{3}{8} \approx 5 \div 1=5$ and rounding both up gives $5 \frac{1}{2} \div 1 \frac{3}{8} \approx$ 6 ÷ 2 = 3 so the answer must be greater than 3 and less than 5. Since the answer is 4, it is reasonable.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 44 Exercise 1 Answer

To divide mixed numbers, rewrite the mixed numbers as improper fractions, rewrite as multiplication, and then multiply:

$37 \frac{1}{2} \div 10 \frac{3}{4}=\frac{75}{2} \div \frac{43}{4}=\frac{75}{2} \times \frac{4}{43}=\frac{75 \times 4}{2 \times 43}=\frac{300}{86}=\frac{258}{86}+\frac{42}{86}=3 \frac{42}{86}=3 \frac{21}{43}$

The number of stickers must be a whole number so Damon can fit 3 medium bumper stickers on his car bumper.

Result

Damon can fit 3 medium bumper stickers on his car bumper.

Page 45 Exercise 2a Answer

20 ÷ $2 \frac{2}{3}$

First, estimate using compatible numbers.

21 ÷ 3 = 7

Write the whole number and mixed number as fractions and than multiply by the reciprocal of the divisor.

$20 \div 2 \frac{2}{3}=\frac{20}{1} \div \frac{8}{3}=\frac{20}{1} \times \frac{3}{8}=\frac{20 \times 3}{1 \times 8}=\frac{60}{8}=\frac{56}{8}+\frac{4}{8}=7 \frac{4}{8}=7 \frac{1}{2}$

The estimate, 7, is close to the quotient, $7 \frac{1}{2}$. The answer is reasonable.

Result

$7 \frac{1}{2}$

Page 45 Exercise 2b Answer

$12 \frac{1}{2}$ ÷ 6

First, estimate using compatible numbers.

12 ÷ 6 = 2

Write the whole number and mixed number as fractions and than multiply by the reciprocal of the divisor.

$12 \frac{1}{2} \div 6=\frac{25}{2} \div \frac{6}{1}=\frac{25}{2} \times \frac{1}{6}=\frac{25 \times 1}{2 \times 6}=\frac{25}{12}=\frac{24}{12}+\frac{1}{12}=2 \frac{1}{12}$

The estimate, 2, is close to the quotient, $2 \frac{1}{12}$. The answer is reasonable.

Result

$2 \frac{1}{12}$

Page 46 Exercise 1 Answer

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem and then multiply. You can use an estimate to check whether the answer is reasonable. Use compatible numbers to estimate.

For example:

$4 \frac{1}{5} \div 1 \frac{2}{3}=\frac{21}{5} \div \frac{5}{3}=\frac{21}{5} \times \frac{3}{5}=\frac{21 \times 3}{5 \times 5}=\frac{63}{25}=\frac{50}{25}+\frac{13}{25}=2 \frac{13}{25}$

Estimate: 4 ÷ 2 = 2. The estimate, 2, is close to the quotient, $2 \frac{13}{25}$. The answer is reasonable.

Result

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem and then multiply.

Page 46 Exercise 2 Answer

When dividing mixed numbers, it is important to estimate the quotient to see if the answer is reasonable.

Page 46 Exercise 3 Answer

In Example 1, the length of Damon’s bumper was $37 \frac{1}{2}$ inches and the length of each long bumper sticker is 15 inches.

To find how many long bumper stickers can fit on his bumper, we must then divide $37 \frac{1}{2}$ and 15.

When dividing mixed numbers, first convert the mixed number to an improper fraction. Next, rewrite the division as multiplying by the reciprocal of the divisor and then multiply:

$37 \frac{1}{2} \div 15=\frac{75}{2} \div \frac{15}{1}=\frac{75}{2} \times \frac{1}{15}=\frac{75 \times 1}{2 \times 15}=\frac{75}{30}=\frac{5}{2}=2 \frac{1}{2}$

The number of bumper stickers must be a whole number so we need to round the quotient down to the nearest whole number. He can then fit 2 long stickers on his bumper.

Since the quotient was not a whole number and needed to be rounded down, then there is uncovered space on his bumper.

Result

2 long stickers and there is uncovered space.

Page 46 Exercise 4 Answer

When dividing by fractions less than 1, the quotient is always greater than the dividend. For example, $\frac{2}{7}$ ÷ $\frac{1}{3}=\frac{2}{7} \times \frac{3}{1}=\frac{6}{7} \text { and } \frac{6}{7}>\frac{2}{7}$

When dividing by a mixed number, the quotient is always less than the dividend. For example $\frac{5}{7} \div 2 \frac{1}{3}=$$\frac{5}{7} \div \frac{7}{3}=\frac{5}{7} \times \frac{3}{7}=\frac{15}{49} \text { and } \frac{15}{49}<\frac{5}{7} .$

Result

When dividing by fractions less than 1, the quotient is always greater than the dividend. When dividing by a mixed number, the quotient is always less than the dividend.

Page 46 Exercise 5 Answer

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem. Finally, multiply and use the estimate to check whether the answer is reasonable. Use compatible numbers to estimate.

$2 \frac{5}{8} \div 2 \frac{1}{4}=\frac{21}{8} \div \frac{9}{4}=\frac{21}{8} \times \frac{4}{9}=\frac{21 \times 4}{8 \times 9}=\frac{84}{72}=\frac{7}{6}=\frac{6}{6}+\frac{1}{6}=1 \frac{1}{6}$

Estimate: 2 ÷ 2 = 1. The estimate, 1, is close to the quotient, $1 \frac{1}{6}$. The answer is reasonable.

Result

$1 \frac{1}{6}$

Page 46 Exercise 6 Answer

$3 \div 4 \frac{1}{2}=\frac{3}{1} \div \frac{9}{2}=\frac{3}{1} \times \frac{2}{9}=\frac{3 \times 2}{1 \times 9}=\frac{6}{9}=\frac{2}{3}$

Estimate: 3 ÷ 5 = $\frac{3}{5}$. The estimate, $\frac{3}{5}$, is close to the quotient, $\frac{2}{3}$. The answer is reasonable.

Result

$\frac{2}{3}$

Page 46 Exercise 7 Answer

$18 \div 3 \frac{2}{3}=\frac{18}{1} \div \frac{11}{3}=\frac{18}{1} \times \frac{3}{11}=\frac{18 \times 3}{1 \times 11}=\frac{54}{11}=\frac{44}{11}+\frac{10}{11}=4 \frac{10}{11}$

Estimate: 18 ÷ 3 = 6. The estimate, 6, is close to the quotient, $4 \frac{10}{11}$, which is actually closer to 5 then 4 since $\frac{10}{11}$ is very close to 1 The answer is reasonable.

Result

$4 \frac{10}{11}$

Page 46 Exercise 8 Answer

$1 \frac{2}{5} \div 7=\frac{7}{5} \div \frac{7}{1}=\frac{7}{5} \times \frac{1}{7}=\frac{7 \times 1}{5 \times 7}=\frac{7}{35}=\frac{1}{5}$

Estimate: 1 ÷ 7 = $\frac{1}{7}$. The estimate, $\frac{1}{7}$, is close to the quotient, $\frac{1}{5}$, which is less than one so the answer is reasonable.

Result

$\frac{1}{5}$

Page 46 Exercise 9 Answer

$5 \div 6 \frac{2}{5}=\frac{5}{1} \div \frac{32}{5}=\frac{5}{1} \times \frac{5}{32}=\frac{5 \times 5}{1 \times 32}=\frac{25}{32}$

Estimate: 5 ÷ 6 = $\frac{5}{6}$. The estimate, $\frac{5}{6}$, is close to the quotient, $\frac{25}{32}$, which is less than one so the answer is reasonable.

Result

$\frac{25}{32}$

Page 46 Exercise 10 Answer

$8 \frac{1}{5} \div 3 \frac{3}{4}=\frac{41}{5} \div \frac{15}{4}=\frac{41}{5} \times \frac{4}{15}=\frac{41 \times 4}{5 \times 15}=\frac{164}{75}=\frac{150}{75}+\frac{14}{75}=2 \frac{14}{75}$

Estimate: 8 ÷ 4 = 2. The estimate, 2, is close to the quotient, $2 \frac{14}{75}$. The answer is reasonable.

Result

$2 \frac{14}{75}$

Page 46 Exercise 11 Answer

$2 \frac{1}{2} \div 4 \frac{1}{10}=\frac{5}{2} \div \frac{41}{10}=\frac{5}{2} \times \frac{10}{41}=\frac{5 \times 10}{2 \times 41}=\frac{50}{82}=\frac{25}{41}$

Estimate: 2 ÷ 4 = $\frac{1}{2}$. The estimate, $\frac{1}{2}$, is close to the quotient, $\frac{25}{41}$, which is less than one so the answer is reasonable.

Result

$\frac{25}{41}$

Page 46 Exercise 12 Answer

$2 \frac{2}{3} \div 6=\frac{8}{3} \div \frac{6}{1}=\frac{8}{3} \times \frac{1}{6}=\frac{8 \times 1}{3 \times 6}=\frac{8}{18}=\frac{4}{9}$

Estimate: 3 ÷ 6 = $\frac{1}{2}$. The estimate, $\frac{1}{2}$, is close to the quotient, $\frac{4}{9}$, which is less than one so the answer is reasonable.

Result

$\frac{4}{9}$

Page 46 Exercise 13 Answer

$6 \frac{5}{9} \div 1 \frac{7}{9}=\frac{59}{9} \div \frac{16}{9}=\frac{59}{9} \times \frac{9}{16}=\frac{59 \times 9}{9 \times 16}=\frac{531}{144}=\frac{59}{16}=\frac{48}{16}+\frac{11}{16}=3 \frac{11}{16}$

Estimate: 6 ÷ 2 = 3. The estimate, 3, is close to the quotient, $3 \frac{11}{16}$. The answer is reasonable.

Result

$3 \frac{11}{16}$

Page 47 Exercise 14 Answer

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem. Finally, multiply:

$10 \div 2 \frac{1}{4}=\frac{10}{1} \div \frac{9}{4}=\frac{10}{1} \times \frac{4}{9}=\frac{10 \times 4}{1 \times 9}=\frac{40}{9}=\frac{36}{9}+\frac{4}{9}=4 \frac{4}{9}$

Result

$4 \frac{4}{9}$

Page 47 Exercise 15 Answer

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem. Finally, multiply:

$9 \frac{1}{3} \div 6=\frac{28}{3} \div \frac{6}{1}=\frac{28}{3} \times \frac{1}{6}=\frac{28 \times 1}{3 \times 6}=\frac{28}{18}=\frac{18}{18}+\frac{10}{18}=1 \frac{10}{18}=1 \frac{5}{9}$

Result

$1 \frac{5}{9}$

Page 47 Exercise 16 Answer

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem. Finally, multiply:

$1 \frac{3}{8} \div 4 \frac{1}{8}=\frac{11}{8} \div \frac{33}{8}=\frac{11}{8} \times \frac{8}{33}=\frac{11 \times 8}{8 \times 33}=\frac{88}{264}=\frac{1}{3}$

Result

$\frac{1}{3}$

Page 47 Exercise 17 Answer

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem. Finally, multiply:

$2 \frac{2}{3} \div 8=\frac{8}{3} \div \frac{8}{1}=\frac{8}{3} \times \frac{1}{8}=\frac{8 \times 1}{3 \times 8}=\frac{8}{24}=\frac{1}{3}$

Result

$\frac{1}{3}$

Page 47 Exercise 18 Answer

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem. Finally, multiply:

$4 \frac{1}{3} \div 3 \frac{1}{4}=\frac{13}{3} \div \frac{13}{4}=\frac{13}{3} \times \frac{4}{13}=\frac{13 \times 4}{3 \times 13}=\frac{52}{39}=\frac{4}{3}=\frac{3}{3}+\frac{1}{3}=1 \frac{1}{3}$

Result

$1 \frac{1}{3}$

Page 47 Exercise 19 Answer

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem. Finally, multiply:

$1 \div 8 \frac{5}{9}=\frac{1}{1} \div \frac{77}{9}=\frac{1}{1} \times \frac{9}{77}=\frac{1 \times 9}{1 \times 77}=\frac{9}{77}$

Result

$\frac{9}{77}$

Page 47 Exercise 20 Answer

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem. Finally, multiply:

$3 \frac{5}{6} \div 9 \frac{5}{6}=\frac{23}{6} \div \frac{59}{6}=\frac{23}{6} \times \frac{6}{59}=\frac{23 \times 6}{6 \times 59}=\frac{138}{354}=\frac{23}{59}$

Result

$\frac{23}{59}$

Page 47 Exercise 21 Answer

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem. Finally, multiply:

$16 \div 2 \frac{2}{3}=\frac{16}{1} \div \frac{8}{3}=\frac{16}{1} \times \frac{3}{8}=\frac{16 \times 3}{1 \times 8}=\frac{48}{8}=6$

Result

Page 47 Exercise 22 Answer

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem. Finally, multiply:

$2 \frac{5}{8} \div 13=\frac{21}{8} \div \frac{13}{1}=\frac{21}{8} \times \frac{1}{13}=\frac{21 \times 1}{8 \times 13}=\frac{21}{104}$

Result

$\frac{21}{104}$

Page 47 Exercise 23 Answer

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem. Finally, multiply:

$3 \frac{6}{7} \div 6 \frac{3}{4}=\frac{27}{7} \div \frac{27}{4}=\frac{27}{7} \times \frac{4}{27}=\frac{27 \times 4}{7 \times 27}=\frac{108}{189}=\frac{4}{7}$

Result

$\frac{4}{7}$

Page 47 Exercise 24 Answer

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem. Finally, multiply:

$2 \frac{1}{3} \div 1 \frac{1}{3}=\frac{7}{3} \div \frac{4}{3}=\frac{7}{3} \times \frac{3}{4}=\frac{7 \times 3}{3 \times 4}=\frac{21}{12}=\frac{7}{4}=\frac{4}{4}+\frac{3}{4}=1 \frac{3}{4}$

Result

$1 \frac{3}{4}$

Page 47 Exercise 25 Answer

To divide with mixed numbers, write mixed numbers and any whole numbers as fractions. Use the reciprocal of the divisor to rewrite the problem as a multiplication problem. Finally, multiply:

$3 \frac{3}{4} \div 1 \frac{1}{2}=\frac{15}{4} \div \frac{3}{2}=\frac{15}{4} \times \frac{2}{3}=\frac{15 \times 2}{4 \times 3}=\frac{30}{12}=\frac{10}{4}=\frac{8}{4}+\frac{2}{4}=2 \frac{2}{4}=2 \frac{1}{2}$

Result

$2 \frac{1}{2}$

Page 47 Exercise 26 Answer

If each step of the ladder is $2 \frac{1}{3}$ feet wide and Beth has a rope that is 21 feet long, then the number of steps she can make is:

$21 \div 2 \frac{1}{3}=\frac{21}{1} \div \frac{7}{3}=\frac{21}{1} \times \frac{3}{7}=\frac{21 \times 3}{1 \times 7}=\frac{63}{7}=9$

Result

Beth can make 9 steps from the rope.

Page 47 Exercise 27 Answer

The area of the rectangle is 257$\frac{1}{4}$in2 and the length of one side is 10$\frac{1}{2}$in. The width is then the area divided by the length:

$w=257 \frac{1}{4} \div 10 \frac{1}{2}$

= $\frac{1029}{4} \div \frac{21}{2}$ (Rewrite as improper fractions)

= $\frac{1029}{4} \times \frac{2}{21}$ (Rewrite as a multiplication problem)

= $\frac{1029 \times 2}{4 \times 21}$ (Multiply)

= $\frac{2058}{84}$

= $\frac{2016}{84}+\frac{42}{84}$

= $24 \frac{42}{84}$

= $24 \frac{1}{2}$

Result

w = $24 \frac{1}{2}$

Page 48 Exercise 28 Answer

The smaller room is $20 \frac{4}{5}$ feet long and the longer room is twice as long, so we must find the product $20 \frac{4}{5} \times 2$.

$20 \frac{4}{5} \times 2=\frac{104}{5} \times 2=\frac{104}{5} \times \frac{2}{1}=\frac{104 \times 2}{5 \times 1}=\frac{208}{5}=\frac{205}{5}+\frac{3}{5}=41 \frac{3}{5}$

Result

The larger room is $41 \frac{3}{5}$ feet long.

Page 48 Exercise 29 Answer

To find how long is each part if the smaller room is divided into four equal parts we need to find the quotient $20 \frac{4}{5} \div 4$

$20 \frac{4}{5} \div 4=\frac{104}{5} \div \frac{4}{1}=\frac{104}{5} \times \frac{1}{4}=\frac{104 \times 1}{5 \times 4}=\frac{104}{20}=\frac{100}{20}+\frac{4}{20}=5 \frac{4}{20}=5 \frac{1}{5}$

Result

Each part is $5 \frac{1}{5}$ feet long.

Page 48 Exercise 30 Answer

First, we must calculate how many pounds of meat he used to make those 6 burgers.

$\frac{3}{8} \times 6=\frac{3}{8} \times \frac{6}{1}=\frac{3 \times 6}{8 \times 1}=\frac{18}{8}=\frac{9}{4}=\frac{8}{4}+\frac{1}{4}=2 \frac{1}{4}$

He used $2 \frac{1}{4}$ pounds to make 6 burgers. We need to calculate how many pounds of meat he has left.

$3-2 \frac{1}{4}=\frac{3}{1}-\frac{9}{4}=\frac{3 \times 4}{1 \times 4}-\frac{9}{4}=\frac{12}{4}-\frac{9}{4}=\frac{12-9}{4}=\frac{3}{4}$

He has left $\frac{3}{4}$ pound of meat. Obviously, he can make three $\frac{1}{4}$ pound burgers.

Result

Luis can make three $\frac{1}{4}$ pound burgers.

Page 48 Exercise 31 Answer

$9 \times \frac{n}{5}=9 \div \frac{n}{5}$

The only number by which we can multiply and divide the same number, in this case 9, and get the equal result is one. Thus, $\frac{n}{5}$ must be equal to one. This is true when n is 5.

Result

n = 5

Page 48 Exercise 32 Answer

First, we must calculate the quotient $1 \frac{3}{4} \div 12$ to find out how many teaspoons of vanilla Margaret puts in one cupcake.

$1 \frac{3}{4} \div 12=\frac{7}{4} \div \frac{12}{1}=\frac{7}{4} \times \frac{1}{12}=\frac{7 \times 1}{4 \times 12}=\frac{7}{48}$

Margaret uses $\frac{7}{48}$ teaspoon of vanilla to make one cupcake.

$\frac{7}{48} \times 30=\frac{7}{48} \times \frac{30}{1}=\frac{7 \times 30}{48 \times 1}=\frac{210}{48}=\frac{192}{48}+\frac{18}{48}=4 \frac{18}{48}=4 \frac{3}{8}$

Result

Margaret will use $4 \frac{3}{8}$ teaspoons to make 30 cupcakes.

Page 48 Exercise 33 Answer

If the diamond was $1 \frac{1}{2}$ pounds and was cut into 6 equal pieces, then the weight of each piece was:

$1 \frac{1}{2} \div 6=\frac{3}{2} \div \frac{6}{1}=\frac{3}{2} \times \frac{1}{6}=\frac{3 \times 1}{2 \times 6}=\frac{3}{12}=\frac{1}{4}$

Result

Each piece would weigh $\frac{1}{4}$ pound.

Page 48 Exercise 34 Answer

To find out how many aquariums did the owner fill we must calculate the quotient $17 \frac{1}{2} \div 5 \frac{5}{6}$

$17 \frac{1}{2} \div 5 \frac{5}{6}=\frac{35}{2} \div \frac{35}{6}=\frac{35}{2} \times \frac{6}{35}=\frac{35 \times 6}{2 \times 35}=\frac{6}{2}=3$

Result

He filled 3 aquariums.

Page 48 Exercise 35 Answer

To estimate the quotient $17 \frac{1}{5} \div 3 \frac{4}{5}$ we must round both numbers to closest whole compatible numbers. For $17 \frac{1}{5}$ we can use 18 and for $3 \frac{4}{5}$ we can use 3, thus the estimate would be 18 ÷ 3 which is equal to 6.

$17 \frac{1}{5} \div 3 \frac{4}{5}=\frac{86}{5} \div \frac{19}{5}=\frac{86}{5} \times \frac{5}{19}=\frac{86 \times 5}{5 \times 19}=\frac{86}{19}=\frac{76}{19}+\frac{10}{19}=4 \frac{10}{19}$

The estimate was correct, the answer $4 \frac{10}{19}$ is close to 4.

Result

Page 48 Exercise 36a Answer

Since the restaurant has $15 \frac{1}{5}$ pounds of ground beef and each meat loaf requires $2 \frac{3}{8}$ pounds of meat, to answer the question how many meat loaves can be made find the quotient $15 \frac{1}{5} \div 2 \frac{3}{8}$

$15 \frac{1}{5} \div 2 \frac{3}{8}=\frac{76}{5} \div \frac{19}{8}=\frac{76}{5} \times \frac{8}{19}=\frac{76 \times 8}{5 \times 19}=\frac{608}{95}=\frac{570}{95}+\frac{38}{95}=6 \frac{2}{5}$

The restaurant can make 6 meat loaves.

Result

The restaurant can make 6 meat loaves.

Page 48 Exercise 36b Answer

$15 \frac{1}{5} \div 1 \frac{3}{5}=\frac{76}{5} \div \frac{8}{5}=\frac{76}{5} \times \frac{5}{8}=\frac{76 \times 5}{5 \times 8}=\frac{380}{40}=\frac{360}{40}+\frac{20}{40}=9 \frac{1}{2}$

The restaurant can make 9 smaller size meat loaves. Since the restaurant can make 6 bigger meat loaves, thus they can make 3 more meat laoves more if they choose to make them smaller.

Result

The restaurant can make 3 more meat loaves.

enVisionmath 2.0: Grade 6, Volume 1 Chapter 1 Use Positive Rational Numbers Section 1.5

Schindler — Tue, 29 Apr 2025 05:28:47 +0000

Chapter 1 Use Positive Rational Numbers

Section 1.5: Divide Fractions by Fractions

Page 37 Exercise 1 Answer

To answer the question write the fraction $\frac{2}{3}$ as a fraction with the denominator 6.

$\frac{2}{3}=\frac{2 \times 2}{3 \times 2}$

= $\frac{4}{6}$

Since $\frac{2}{3}$ of the original bar remains, and $\frac{2}{3}$ is equal to $\frac{4}{6}$, which means there are four $\frac{1}{6}$ parts left.

$\frac{2}{3} \div \frac{1}{6}=\frac{2}{3} \times \frac{6}{1}$

= $\frac{12}{3}$

= 4

Result

4 parts

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 37 Exercise 1 Answer

In the Solve & Discuss It! problem, we found that $\frac{2}{3} \div \frac{1}{6}=4$.

To check our answer using multiplication, we can use a related multiplication fact. Recall that a division statement a ÷ b = c has a related multiplication fact of a = b × c. For the division statement $\frac{2}{3}$ ÷ $\frac{1}{6}$ = 4, the related multiplication fact is $\frac{2}{3}$ = $\frac{1}{6}$ × 4.

Our answer is then correct if $\frac{1}{6}$ × 4 is equal to $\frac{2}{3}$. Multiplying $\frac{1}{6}$ and 4 gives:

$\frac{1}{6} \times 4=\frac{1}{6} \times \frac{4}{1}=\frac{1 \times 4}{6 \times 1}=\frac{4}{6}=\frac{2}{3}$

Our answer is then correct.

Result

We can use the related multiplication fact $\frac{2}{3}$ = $\frac{1}{6}$ x 4 to check our answer.

Page 38 Exercise 1 Answer

To use the number line to represent $\frac{1}{6}$ x 3 = $\frac{1}{2}$, we first need to divide the number line into sixths since the first factor of the multiplication sentence is $\frac{1}{6}$.

Since the second factor is 3, we need to mark off three $\frac{1}{6}$ parts, as shown below by the red ovals.

Note that the last oval ends at $\frac{3}{6}$ = $\frac{1}{2}$ which is why the product $\frac{1}{6}$ × 3 is equal to $\frac{1}{2}$.

To write an equivalent division sentence for the number line, we need to look at the total width of the ovals to get the dividend, the number of ovals to get the divisor, and the width of each oval to get the quotient.

The ovals have a total width of $\frac{1}{2}$ so the dividend is $\frac{1}{2}$. The width of each oval is $\frac{1}{6}$ so the divisor is $\frac{1}{6}$. There are 3 ovals so the quotient is 3.

An equivalent division sentence is then $\frac{1}{2} \div \frac{1}{6}=3$

Result

To represent $\frac{1}{6}$ x 3 = $\frac{1}{2}$ on the number line, draw 3 ovals with a width of $\frac{1}{6}$, starting from 0. The equivalent division statement for the number line is $\frac{1}{2}$ ÷ $\frac{1}{6}$ = 3.

The dividend is represented in the number line as the total width of the ovals, the divisor is the width of each oval, and the quotient is the number of ovals.

Page 39 Exercise 2 Answer

To find $\frac{1}{4} \div \frac{3}{8}$ using an area model, start by drawing two rectangles. Divide the first rectangle into fourths and shade 1 of them and divide the second rectangle into eights and shade 3 of them, as shown below in Step 1.

Next, we need to divide the two rectangles so they have the same number of parts. Divide each fourth into halves so the top rectangle will have 8 parts like the bottom rectangle, as shown below in Step 2.

Since $\frac{1}{4}$ is divided into 2 equal parts while $\frac{3}{8}$ is divided into 3 equal parts, then $\frac{1}{4} \div \frac{3}{8}=\frac{2}{3}$.

Result

$\frac{2}{3}$

Page 40 Exercise 1 Answer

To divide a fraction by a fraction,

rewrite the division problem as a multiplication problem, that is, multiply the dividend by the reciprocal of the divisor.

For example,

$\frac{3}{4} \div \frac{1}{5}=\frac{3}{4} \times \frac{5}{1}$

Result

To divide a fraction by a fraction, rewrite the division problem as a multiplication problem, that is, multiply the dividend by the reciprocal of the divisor.

Page 40 Exercise 2 Answer

Division is not commutative that is a dividend and a dividor can’t switch places. When rewriting the division problem as a mulitplication problem multiply the dividend by the reciprocal of a divisor.

In Corey’s problem $\frac{2}{5}$ is the dividend and $\frac{8}{5}$ is the divisor. So the correct answer would be multiplying $\frac{2}{5}$ by the reciprocal of $\frac{8}{5}$ which is $\frac{5}{8}$.

$\frac{2}{5} \div \frac{8}{5}=\frac{2}{5} \times \frac{5}{8}$

Corey’s mistake is that he multiplied the divisor by a reciprocal of a dividend.

Result

Corey’s mistake is that he multiplied the divisor by a reciprocal of a dividend. The correct expression

$\frac{2}{5} \div \frac{8}{5}=\frac{2}{5} \times \frac{5}{8}$

Page 40 Exercise 3 Answer

The quotient of $\frac{3}{5}$ ÷ $\frac{6}{7}$ is greater than $\frac{3}{5}$ since the divisor is less than 1.

Result

The quotient is greater than $\frac{3}{5}$.

Page 40 Exercise 4 Answer

To divide a fraction by a whole number, we must first rewrite the whole number as a fraction and then rewrite the division as multiplying by the reciprocal of the divisor. For example:

$\frac{3}{7} \div 5=\frac{3}{7} \div \frac{5}{1}=\frac{3}{7} \times \frac{1}{5}$

Also, when dividing a fraction by a whole number, the quotient is always smaller than the dividend.

To divide a fraction by a fraction, we go directly to rewriting the division as multiplying by the reciprocal of the divisor. For example:

$\frac{3}{7} \div \frac{2}{3}=\frac{3}{7} \times \frac{3}{2}$

When dividing a fraction by a fraction, the quotient is smaller than the dividend if the divisor is a fraction greater than 1 and the quotient is bigger than the dividend if the divisor is a fraction smaller than 1.

Result

To divide a fraction by a whole number, we must first rewrite the whole number as a fraction and then rewrite the division as multiplying. The quotient is always smaller than the dividend.

To divide a fraction by a fraction, we go directly to rewriting the division as multiplying. The quotient is smaller than the dividend if the divisor is a fraction greater than 1 and the quotient is bigger than the dividend if the divisor is a fraction smaller than 1.

Page 40 Exercise 5 Answer

To write a division sentence to represent the model, we need to determine the total width of the ovals, the number of ovals, and the width of each oval.

The dividend is the total width of the ovals, which is $\frac{4}{5}$.

The divisor is the width of each oval, which is $\frac{1}{5}$.

The quotient is the number of ovals, which is 4.

The division sentence is then $\frac{4}{5} \div \frac{1}{5}=4$

Result

$\frac{4}{5} \div \frac{1}{5}=4$

Page 40 Exercise 6 Answer

To write a division sentence to represent the model, we need to see what fraction of each rectangle is shown and how many equal parts are in each fraction.

The top rectangle is labeled as $\frac{1}{2}$ so the dividend is $\frac{1}{2}$.

The bottom rectangle is labeled as $\frac{2}{3}$ so the divisor is $\frac{2}{3}$

$\frac{1}{2}$ is divided into 3 equal parts while $\frac{2}{3}$ is divided into 4 equal parts. The quotient is then $\frac{3}{4}$.

Therefore, the division sentence is $\frac{1}{2} \div \frac{2}{3}=\frac{3}{4}$

Result

$\frac{1}{2} \div \frac{2}{3}=\frac{3}{4}$

Page 40 Exercise 7 Answer

To write a division sentence for the model, we need to determine what fraction of the whole rectangle is shaded, how many parts the rectangle is divided into, and how many parts are shaded.

The rectangle is initially divided into 4 parts and 3 parts are shaded to represent $\frac{3}{4}$ so the dividend is $\frac{3}{4}$.

The rectangle is then divided into eighths so the divisor is $\frac{1}{8}$.

There are 6 eighths shaded so the quotient is 6.

The division sentence is then $\frac{3}{4} \div \frac{1}{8}=6$.

Result

$\frac{3}{4} \div \frac{1}{8}=6$

Page 40 Exercise 8 Answer

$\frac{3}{4} \div \frac{2}{3}$ = $\frac{3}{4}$ x $\frac{3}{2}$ (Rewrite as a multiplication problem)

= $\frac{3 \times 3}{4 \times 2}$

= $\frac{9}{8}$ (Multiply the fractions)

= $\frac{8}{8}+\frac{1}{8}$

= $1 \frac{1}{8}$

Result

= $1 \frac{1}{8}$

Page 40 Exercise 9 Answer

$\frac{3}{12} \div \frac{1}{8}=\frac{1}{4} \div \frac{1}{8}$ (Rewrite $\frac{3}{12}$ as $\frac{1}{4}$)

= $\frac{1}{4}$ x $\frac{8}{1}$ (Rewrite as a multiplication problem)

= $\frac{8}{4}$ (Multiply the fractions)

= 2

Result

Page 40 Exercise 10 Answer

$\frac{1}{2} \div \frac{4}{5}$ = $\frac{1}{2}$ x $\frac{5}{4}$ (Rewrite as a multiplication problem)

= $\frac{5}{8}$ (Multiply the fractions)

Result

$\frac{5}{8}$

Page 40 Exercise 11 Answer

$\frac{7}{10} \div \frac{2}{5}$ = $\frac{7}{10}$ x $\frac{5}{2}$ (Rewrite as a multiplication problem)

= $\frac{35}{20}$ (Multiply the fractions)

= $\frac{7}{4}$

Result

$\frac{7}{4}$

Page 41 Exercise 12 Answer

When representing a division sentence using a number line, the number of ovals is the quotient.

From the given number line, there are 4 ovals so $\frac{1}{3} \div \frac{1}{2}=4$.

Result

Page 41 Exercise 13 Answer

When using an area model to represent a division sentence, the number of shaded parts is the quotient.

From the given diagram, the rectangle has 4 shaded parts so $\frac{2}{5} \div \frac{1}{10}=4$.

Result

Page 41 Exercise 14 Answer

To divide a fraction by a fraction, rewrite the division equation as a multiplication equation. To divide by a fraction, a multiply by the reciprocal of the divisor.

$\frac{2}{3} \div \frac{1}{3}=\frac{2}{3} \times \frac{3}{1}=\frac{2 \times 3}{3 \times 1}=\frac{6}{3}=2$

Result

Page 41 Exercise 15 Answer

To divide a fraction by a fraction, rewrite the division equation as a multiplication equation. To divide by a fraction, a multiply by the reciprocal of the divisor.

$\frac{1}{2} \div \frac{1}{16}=\frac{1}{2} \times \frac{16}{1}=\frac{1 \times 16}{2 \times 1}=\frac{16}{2}=8$

Result

Page 41 Exercise 16 Answer

To divide a fraction by a fraction, rewrite the division equation as a multiplication equation. To divide by a fraction, a multiply by the reciprocal of the divisor.

$\frac{1}{4} \div \frac{1}{12}=\frac{1}{4} \times \frac{12}{1}=\frac{1 \times 12}{4 \times 1}=\frac{12}{4}=3$

Result

Page 41 Exercise 17 Answer

To divide a fraction by a fraction, rewrite the division equation as a multiplication equation. To divide by a fraction, a multiply by the reciprocal of the divisor.

$\frac{6}{7} \div \frac{3}{7}=\frac{6}{7} \times \frac{7}{3}=\frac{6 \times 7}{7 \times 3}=\frac{42}{21}=2$

Result

Page 41 Exercise 18 Answer

To divide a fraction by a fraction, rewrite the division equation as a multiplication equation. To divide by a fraction, a multiply by the reciprocal of the divisor.

$\frac{5}{14} \div \frac{4}{7}=\frac{5}{14} \times \frac{7}{4}=\frac{5 \times 7}{14 \times 4}=\frac{35}{56}=\frac{5}{8}$

Result

$\frac{5}{8}$

Page 41 Exercise 19 Answer

To divide a fraction by a fraction, rewrite the division equation as a multiplication equation. To divide by a fraction, a multiply by the reciprocal of the divisor.

$\frac{5}{8} \div \frac{1}{2}=\frac{5}{8} \times \frac{2}{1}=\frac{5 \times 2}{8 \times 1}=\frac{10}{8}=\frac{5}{4}=\frac{4}{4}+\frac{1}{4}=1 \frac{1}{4}$

Result

$1 \frac{1}{4}$

Page 41 Exercise 20 Answer

To divide a fraction by a fraction, rewrite the division equation as a multiplication equation. To divide by a fraction, a multiply by the reciprocal of the divisor.

$\frac{7}{12} \div \frac{3}{4}=\frac{7}{12} \times \frac{4}{3}=\frac{7 \times 4}{12 \times 3}=\frac{28}{36}=\frac{7}{9}$

Result

$\frac{7}{9}$

Page 41 Exercise 21 Answer

To divide a fraction by a fraction, rewrite the division equation as a multiplication equation. To divide by a fraction, a multiply by the reciprocal of the divisor.

$\frac{2}{7} \div \frac{1}{2}=\frac{2}{7} \times \frac{2}{1}=\frac{2 \times 2}{7 \times 1}=\frac{4}{7}$

Result

$\frac{4}{7}$

Page 41 Exercise 22 Answer

To divide a fraction by a fraction, rewrite the division equation as a multiplication equation. To divide by a fraction, a multiply by the reciprocal of the divisor.

$\frac{4}{9} \div \frac{2}{3}=\frac{4}{9} \times \frac{3}{2}=\frac{4 \times 3}{9 \times 2}=\frac{12}{18}=\frac{2}{3}$

Result

$\frac{2}{3}$

Page 41 Exercise 23 Answer

To divide a fraction by a fraction, rewrite the division equation as a multiplication equation. To divide by a fraction, a multiply by the reciprocal of the divisor.

$\frac{7}{12} \div \frac{1}{8}=\frac{7}{12} \times \frac{8}{1}=\frac{7 \times 8}{12 \times 1}=\frac{56}{12}=\frac{14}{3}=\frac{12}{3}+\frac{2}{3}=4 \frac{2}{3}$

Result

$4 \frac{2}{3}$

Page 41 Exercise 24 Answer

To divide a fraction by a fraction, rewrite the division equation as a multiplication equation. To divide by a fraction, a multiply by the reciprocal of the divisor.

$\frac{3}{10} \div \frac{3}{5}=\frac{3}{10} \times \frac{5}{3}=\frac{3 \times 5}{10 \times 3}=\frac{15}{30}=\frac{1}{2}$

Result

$\frac{1}{2}$

Page 41 Exercise 25 Answer

To divide a fraction by a fraction, rewrite the division equation as a multiplication equation. To divide by a fraction, a multiply by the reciprocal of the divisor.

$\frac{2}{5} \div \frac{1}{8}=\frac{2}{5} \times \frac{8}{1}=\frac{2 \times 8}{5 \times 1}=\frac{16}{5}=\frac{15}{5}+\frac{1}{5}=3 \frac{1}{5}$

Result

$3 \frac{1}{5}$

Page 41 Exercise 26 Answer

We need to determine how many $\frac{1}{3}$ pound bags can be filled with $\frac{12}{15}$ pounds of granola. To find the number of bags, we must then divide $\frac{12}{15}$ and $\frac{1}{3}$.

To divide by a fraction, rewrite the division as multiplying by the reciprocal of the divisor:

$\frac{12}{15} \div \frac{1}{3}=\frac{12}{15} \times \frac{3}{1}=\frac{12 \times 3}{15 \times 1}=\frac{36}{15}=\frac{12}{5}=2 \frac{2}{5}$

We can then completely fill 2 bags and can fill $\frac{2}{5}$ of another bag.

To find how much granola is left, we need to find how much granola is $\frac{2}{5}$ of a bag. Since each bag is $\frac{1}{3}$ pound, we can multiply $\frac{2}{5}$ and $\frac{1}{3}$:

$\frac{2}{5} \times \frac{1}{3}=\frac{2 \times 1}{5 \times 3}=\frac{2}{15}$

The amount of leftover granola is then $\frac{2}{15}$ pound.

Result

2 bags can be filled and $\frac{2}{15}$ pound is left over.

Page 41 Exercise 27 Answer

To divide by a fraction, rewrite the division as multiplying by the reciprocal of the divisor:

$\frac{3}{4} \div \frac{2}{3}=\frac{3}{4} \times \frac{3}{2}=\frac{3 \times 3}{4 \times 2}=\frac{9}{8}=1 \frac{1}{8}$

To use a model to find the quotient, we can start by drawing two rectangles. Divide the first rectangle into fourths and shade three of them to represent $\frac{3}{4}$. Divide the second rectangle into thirds and shade two of them to represent $\frac{2}{3}$.

Next, divide each rectangle into the same number of parts. Since 4 × 3 = 12, we can divide each rectangle into 12 equal parts.

$\frac{3}{4}$ is divided into 9 equal parts while $\frac{2}{3}$ is divided into 8 equal parts. Therefore, $\frac{3}{4} \div \frac{2}{3}=\frac{9}{8}=1 \frac{1}{8}$

Result

$1 \frac{1}{8}$

Page 41 Exercise 28 Answer

Using the formula A = l x w

A = $\frac{1}{6}$ W = $\frac{2}{3}$

$A=l \times w \rightarrow \frac{1}{6}=l \times \frac{2}{3} \rightarrow l=\frac{1}{6} \div \frac{2}{3}=\frac{1}{6} \times \frac{3}{2}=\frac{1 \times 3}{6 \times 2}=\frac{3}{12}=\frac{1}{4}$

Thus, the length of the painting is $\frac{1}{4}$ yard.

Result

l = $\frac{1}{4}$

Page 41 Exercise 29 Answer

Solve the equation $\frac{13}{16} \div \frac{1}{6}=n$ or n by rewriting the division as multiplying by the reciprocal of the divisor:

$n=\frac{13}{16} \div \frac{1}{6}=\frac{13}{16} \times \frac{6}{1}=\frac{13 \times 6}{16 \times 1}=\frac{78}{16}=\frac{39}{8}=\frac{32}{8}+\frac{7}{8}=4 \frac{7}{8}$

Result

n = $4 \frac{7}{8}$

Page 42 Exercise 30a Answer

It is given that the cafeteria has $\frac{2}{3}$ pound of coffee and uses $\frac{1}{6}$ pound to fill the dispenser. To find how many times the dispenser can be filled, we then need to find the quotient $\frac{2}{3} \div \frac{1}{6}$.

To find the quotient using a model, start by representing the dividend. Divide the rectangle into 3 parts and shade 2 of them to represent the dividend $\frac{2}{3}$.

Next, divide the rectangle into sixths to represent the divisor $\frac{1}{6}$.

Since 4 parts of the rectangle are shaded, then the quotient is 4 so $\frac{2}{3} \div \frac{1}{6}=4$

The cafeteria can then fill 4 dispensers.

Result

4 dispensers

Page 42 Exercise 30b Answer

From part (a), we know the division sentence is $\frac{2}{3} \div \frac{1}{6}=4$

Since the quotient is 4, the cafeteria can fill 4 dispensers.

Result

$\frac{2}{3} \div \frac{1}{6}=4$ 4 dispensers

Page 42 Exercise 31a Answer

We know that the truck can haul $\frac{2}{3}$ ton and is currently hauling $\frac{1}{2}$ ton.

To find how much of a full load the truck is hauling, we must then find the quotient $\frac{1}{2} \div \frac{2}{3}$

To find the quotient using the model, start by dividing the first rectangle into halves and shade 1 half to represent the dividend $\frac{1}{2}$. Shade the second rectangle into thirds and shade 2 thirds to represent the divisor $\frac{2}{3}$.

Next, we need to divide the two rectangles into the same number of parts. Since 2 × 3 = 6, we can divide both rectangles into 6 equal parts.

$\frac{1}{2}$ is divided into 3 equal parts while $\frac{2}{3}$ is divided into 4 equal parts. The quotient is then $\frac{1}{2} \div \frac{2}{3}=\frac{3}{4}$.

The truck is then currently hauling $\frac{3}{4}$ of a full load.

Result

$\frac{3}{4}$ of a full load

Page 42 Exercise 31b Answer

From part (a), we know the division sentence is $\frac{1}{2} \div \frac{2}{3}=\frac{3}{4}$

Since the quotient is $\frac{3}{4}$, then the truck is hauling $\frac{3}{4}$ of a full load

Result

$\frac{1}{2} \div \frac{2}{3}=\frac{3}{4}$

$\frac{3}{4}$ of a full load

Page 42 Exercise 32 Answer

To answer the question find the quotient $\frac{5}{8} \div \frac{1}{4}$, since the piece of metal is $\frac{5}{8}$ inch long and is cut into $\frac{1}{4}$ inch pieces.

$\frac{5}{8} \div \frac{1}{4}$ = $\frac{5}{8}$ x $\frac{4}{1}$ (Rewrite as a multiplication problem)

= $\frac{5 \times 4}{8 \times 1}$

= $\frac{20}{8}$ (Multiply the fractions)

= $\frac{5}{2}$

= $\frac{4}{2}+\frac{1}{2}$

= $2 \frac{1}{2}$ (Rewrite as a mixed number)

The number of $\frac{1}{4}$ inch long pieces must be a whole number. A piece of metal $\frac{5}{8}$ inch long can then be divided into 2 pieces that are each $\frac{1}{4}$ inch long.

Result

2 pieces that are $\frac{1}{4}$ inch long

Page 42 Exercise 33 Answer

Margaret has a rope $\frac{5}{8}$ inch long. She wants to cut it into smaller pieces, each the same length of $\frac{2}{5}$ inch. How many $\frac{2}{5}$ inch pieces of rope will she have and will there be any left over rope?

$\frac{5}{8} \div \frac{2}{5}$ = $\frac{5}{8}$ x $\frac{5}{2}$ (Rewrite as a multiplication problem)

= $\frac{5 \times 5}{8 \times 2}$

= $\frac{25}{16}$ (Multiply the fractions)

= $\frac{16}{16}+\frac{9}{16}$

= $1 \frac{9}{16}$

The quotient is a mixed number but the number of $\frac{2}{5}$ inch pieces must be a whole number. This means she will have 1 piece that is $\frac{2}{5}$ inch and will have left over rope.

Result

Possible solution: Margaret has a rope $\frac{5}{8}$ inch long. She wants to cut it into smaller pieces, each the same length of $\frac{2}{5}$ inch. How many $\frac{2}{5}$ inch pieces of rope will she have and will there be any left over rope?

She will have 1 piece that is $\frac{2}{5}$ inch and will have left over rope.

Page 42 Exercise 34 Answer

The colored in rectangle represents $\frac{2}{3}$. Since each of the smallest rectangles in the model represent $\frac{1}{9}$, the model represents the quotient $\frac{2}{3} \div \frac{1}{9}$

This answer is marked A, that is $\frac{2}{3} \div \frac{1}{9}$ = 6.

Result

A $\frac{2}{3} \div \frac{1}{9}$ = 6

envisionmath 2.0: Grade 6, Volume 1 Chapter 1 Use Positive Rational Numbers Section 1.4

Schindler — Tue, 29 Apr 2025 05:28:11 +0000

Chapter 1 Use Positive Rational Numbers

Section 1.4: Understand Division With Fractions

Page 31 Exercise 1a Answer

Each point represents the position of each runner after his turn:

Result

Plot points which represent the position of each runner after their turn.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 31 Exercise 1b Answer

Step 1

We can find each point (points represent distances of each runner from the start) by multiplying $\frac{2}{5}$ by the turn number of each runner.

So, the points which represent each runner after his turn can be obtained by the following procedure:

$\frac{2}{5} \times 1=\frac{2}{5}$ $\frac{2}{5} \times 2=\frac{4}{5}$ $\frac{2}{5} \times 3=\frac{6}{5}$ $\frac{2}{5} \times 4=\frac{8}{5}$ $\frac{2}{5} \times 5=2$ $\frac{2}{5} \times 6=\frac{12}{5}$ $\frac{2}{5} \times 7=\frac{14}{5}$ $\frac{2}{5} \times 8=\frac{16}{5}$ $\frac{2}{5} \times 9=\frac{18}{5}$ $\frac{2}{5} \times 10=4$

Result

Multiply $\frac{2}{5}$ by the turn number of each runner.

Page 31 Exercise 1 Answer

The line would have values between 0 and 5, and the distance between adjacent points is $\frac{1}{2}$. Each point can be found by multiplying $\frac{1}{2}$ by turn number of each runner.

Result

Each point can be found by multiplying $\frac{1}{2}$ by turn number of each runner.

Page 32 Exercise 1 Answer

We need to determine how many $\frac{2}{3}$ foot long pieces can be made from a 6-foot long board. We must find $6 \div \frac{2}{3}$ using the given number line.

Label the given number line from 0 to 6 and then divide each segment into thirds.

Next, divide the number line into $\frac{2}{3}$ foot parts, as shown by the red arcs below.

Since the number line from 0 to 6 can be divided into 9 parts, then $6 \div \frac{2}{3}=9$. Therefore, 9 pieces can be cut from the board.

When dividing by a number that is less than 1, the quotient is always greater than the dividend. Since the divisor is $\frac{2}{3}$, which is less than 1, then the quotient must be greater than the dividend of 6.

Result

9 pieces

Since the divisor is $\frac{2}{3}$, which is less than 1, then the quotient must be greater than the dividend of 6.

Page 33 Exercise 2 Answer

To make a diagram to find $\frac{2}{3} \div 4$, start by drawing a rectangle and dividing it into thirds. Shade 2 of the thirds, as shown in step 1 below.

The divide the rectangle into 4 equal parts, as shown in step 2 below.

Each part is $\frac{1}{12}$ of the whole and each fourth has two shaded parts. We then get:

$\frac{2}{3} \div 4=\frac{2}{12}=\frac{1}{6}$

Result

$\frac{1}{6}$

Page 33 Exercise 3 Answer

To divide a whole number rewrite the problem as a multiplication problem using the reciprocal of the divisor.

$8 \div \frac{3}{4}=8 \times \frac{4}{3}=\frac{8}{1} \times \frac{4}{3}=\frac{8 \times 4}{1 \times 3}=\frac{32}{3}=\frac{30}{3}+\frac{2}{3}=10 \frac{2}{3}$

Result

$10 \frac{2}{3}$

Page 34 Exercise 4 Answer

When the divisor is a fraction less than 1, the quotient is greater than the dividend.

Page 34 Exercise 5 Answer

When the divisor is a fraction less than 1, the quotient is greater than the dividend.

Page 34 Exercise 6 Answer

To find the division equation represented by the diagram, we need to look to see how many parts the rectangle is divided up into and how many parts are shaded.

The rectangle is divided into 4 columns and 3 of the columns are shaded so the dividend is $\frac{3}{4}$.

The rectangle is divided into 2 rows so the divisor is 2.

Since the rectangle is divided up into 8 parts and 3 of those parts are circled, then the quotient is $\frac{3}{8}$.

The division equation is then $\frac{3}{4} \div 2=\frac{3}{8}$

Result

$\frac{3}{4} \div 2=\frac{3}{8}$

Page 34 Exercise 7 Answer

Two numbers whose product is one are called reciprocals of each other. If a nonzero number is named as a fraction $\frac{a}{b}$, then its reciprocal is $\frac{b}{a}$.

The reciprocal of $\frac{3}{5}$ is $\frac{5}{3}$.

Result

$\frac{5}{3}$

Page 34 Exercise 8 Answer

Two numbers whose product is one are called reciprocals of each other. If a nonzero number is named as a fraction $\frac{a}{b}$, then its reciprocal is $\frac{b}{a}$.

The reciprocal of $\frac{1}{6}$ is $\frac{6}{1}$, which is 6.

Result

Page 34 Exercise 9 Answer

Two numbers whose product is one are called reciprocals of each other. If a nonzero number is named as a fraction $\frac{a}{b}$, then its reciprocal is $\frac{b}{a}$.

The reciprocal of 9, which can be written as $\frac{9}{1}$ is $\frac{1}{9}$.

Result

$\frac{1}{9}$

Page 34 Exercise 10 Answer

Two numbers whose product is one are called reciprocals of each other. If a nonzero number is named as a fraction $\frac{a}{b}$, then its reciprocal is $\frac{b}{a}$.

The reciprocal of $\frac{7}{4}$ is $\frac{4}{7}$.

Result

$\frac{4}{7}$

Page 34 Exercise 11 Answer

Two numbers whose product is one are called reciprocals of each other. If a nonzero number is named as a fraction $\frac{a}{b}$, then its reciprocal is $\frac{b}{a}$.

The reciprocal of $\frac{5}{8}$ is $\frac{8}{5}$.

Result

$\frac{8}{5}$

Page 34 Exercise 12 Answer

Two numbers whose product is one are called reciprocals of each other. If a nonzero number is named as a fraction $\frac{a}{b}$, then its reciprocal is $\frac{b}{a}$.

The reciprocal of 16, which can be written as $\frac{16}{1}$ is $\frac{1}{16}$.

Result

$\frac{1}{16}$

Page 34 Exercise 13 Answer

Two numbers whose product is one are called reciprocals of each other. If a nonzero number is named as a fraction $\frac{a}{b}$, then its reciprocal is $\frac{b}{a}$.

The reciprocal of $\frac{7}{12}$ is $\frac{12}{7}$.

Result

$\frac{12}{7}$

Page 34 Exercise 14 Answer

Two numbers whose product is one are called reciprocals of each other. If a nonzero number is named as a fraction $\frac{a}{b}$, then its reciprocal is $\frac{b}{a}$.

The reciprocal of $\frac{11}{5}$ is $\frac{5}{11}$.

Result

$\frac{5}{11}$

Page 34 Exercise 15 Answer

Divide 6 by $\frac{2}{3}$.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$6 \div \frac{2}{3}=\frac{6}{1} \div \frac{2}{3}=\frac{6}{1} \times \frac{3}{2}=\frac{6 \times 3}{1 \times 2}=\frac{18}{2}=9$

Result

Page 34 Exercise 16 Answer

Divide 6 by $\frac{3}{8}$.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$12 \div \frac{3}{8}=\frac{12}{1} \div \frac{3}{8}=\frac{12}{1} \times \frac{8}{3}=\frac{12 \times 8}{1 \times 3}=\frac{96}{3}=32$

Result

Page 34 Exercise 17 Answer

Divide $\frac{1}{4}$ by 3.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$\frac{1}{4} \div 3=\frac{1}{4} \div \frac{3}{1}=\frac{1}{4} \times \frac{1}{3}=\frac{1 \times 1}{4 \times 3}=\frac{1}{12}$

Result

$\frac{1}{12}$

Page 34 Exercise 18 Answer

Divide $\frac{2}{5}$ by 3.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$\frac{2}{5} \div 2=\frac{2}{5} \div \frac{2}{1}=\frac{2}{5} \times \frac{1}{2}=\frac{2 \times 1}{5 \times 2}=\frac{2}{10}=\frac{1}{5}$

Result

$\frac{1}{5}$

Page 34 Exercise 19 Answer

Divide 2 by $\frac{1}{2}$.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$2 \div \frac{1}{2}=\frac{2}{1} \div \frac{1}{2}=\frac{2}{1} \times \frac{2}{1}=\frac{2 \times 2}{1 \times 1}=\frac{4}{1}=4$

Result

Page 34 Exercise 20 Answer

Divide 3 by $\frac{1}{4}$.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$3 \div \frac{1}{4}=\frac{3}{1} \div \frac{1}{4}=\frac{3}{1} \times \frac{4}{1}=\frac{3 \times 4}{1 \times 1}=\frac{12}{1}=12$

Result

Page 34 Exercise 21 Answer

Divide 9 by $\frac{3}{5}$.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$9 \div \frac{3}{5}=\frac{9}{1} \div \frac{3}{5}=\frac{9}{1} \times \frac{5}{3}=\frac{9 \times 5}{1 \times 3}=\frac{45}{3}=15$

Result

Page 34 Exercise 22 Answer

Divide 5 by $\frac{2}{7}$.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$5 \div \frac{2}{7}=\frac{5}{1} \div \frac{2}{7}=\frac{5}{1} \times \frac{7}{2}=\frac{5 \times 7}{1 \times 2}=\frac{35}{2}$

Result

$\frac{35}{2}$

Page 35 Exercise 23 Answer

When using a number line to represent division by a fraction, the width of each oval is the divisor.

From the diagram, we can see there are two ovals between each pair of consecutive whole numbers. This means that each oval in the diagram has a width of $\frac{1}{2}$ so the divisor must then be $\frac{1}{2}$.

The division sentence is then completed as: $6 \div \frac{1}{2}=12$

Result

$6 \div \frac{1}{2}=12$

Page 35 Exercise 24 Answer

The number of columns in the diagram tells us the dividend and the number of rows tells us the divisor.

From the diagram, the rectangle is dividend into 3 rows so the divisor must be 3.

The completed division sentence is then: $\frac{2}{3} \div 3=\frac{2}{9} .$

Result

$\frac{2}{3} \div 3=\frac{2}{9}$

Page 35 Exercise 25 Answer

Divide $\frac{3}{5}$ by 3.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$\frac{3}{5} \div 3=\frac{3}{5} \div \frac{3}{1}=\frac{3}{5} \times \frac{1}{3}=\frac{3 \times 1}{5 \times 3}=\frac{3}{15}=\frac{1}{5}$

Result

$\frac{1}{5}$

Page 35 Exercise 26 Answer

Divide 2 by $\frac{2}{5}$.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$2 \div \frac{2}{5}=\frac{2}{1} \div \frac{2}{5}=\frac{2}{1} \times \frac{5}{2}=\frac{2 \times 5}{1 \times 2}=\frac{10}{2}=5$

Result

Page 35 Exercise 27 Answer

Two numbers whose product is one are called reciprocals of each other. If a nonzero number is named as a fraction $\frac{a}{b}$, then its reciprocal is $\frac{b}{a}$.

The reciprocal of $\frac{3}{10}$ is $\frac{10}{3}$.

Result

$\frac{10}{3}$

Page 35 Exercise 28 Answer

Two numbers whose product is one are called reciprocals of each other. If a nonzero number is named as a fraction $\frac{a}{b}$, then its reciprocal is $\frac{b}{a}$.

The reciprocal of 6, which can be written as $\frac{6}{1}$ is $\frac{1}{6}$.

Result

$\frac{1}{6}$

Page 35 Exercise 29 Answer

Two numbers whose product is one are called reciprocals of each other. If a nonzero number is named as a fraction $\frac{a}{b}$, then its reciprocal is $\frac{b}{a}$.

The reciprocal of $\frac{1}{15}$ is $\frac{15}{1}$, which can be written as 15.

Result

Page 35 Exercise 30 Answer

Two numbers whose product is one are called reciprocals of each other. If a nonzero number is named as a fraction $\frac{a}{b}$, then its reciprocal is $\frac{b}{a}$.

The reciprocal of 3, which can be written as $\frac{3}{1}$ is $\frac{1}{3}$.

Result

$\frac{1}{3}$

Page 35 Exercise 31 Answer

Divide 36 by $\frac{3}{4}$.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$36 \div \frac{3}{4}=\frac{36}{1} \div \frac{3}{4}=\frac{36}{1} \times \frac{4}{3}=\frac{36 \times 4}{1 \times 3}=\frac{144}{3}=48$

Result

Page 35 Exercise 32 Answer

Divide 2 by $\frac{3}{8}$.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$2 \div \frac{3}{8}=\frac{2}{1} \div \frac{3}{8}=\frac{2}{1} \times \frac{8}{3}=\frac{2 \times 8}{1 \times 3}=\frac{16}{3}$

Result

$\frac{16}{3}$

Page 35 Exercise 33 Answer

Divide 18 by $\frac{2}{3}$.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$18 \div \frac{2}{3}=\frac{18}{1} \div \frac{2}{3}=\frac{18}{1} \times \frac{3}{2}=\frac{18 \times 3}{1 \times 2}=\frac{54}{2}=27$

Result

Page 35 Exercise 34 Answer

Divide 9 by $\frac{4}{5}$.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$9 \div \frac{4}{5}=\frac{9}{1} \div \frac{4}{5}=\frac{9}{1} \times \frac{5}{4}=\frac{9 \times 5}{1 \times 4}=\frac{45}{4}$

Result

$\frac{45}{4}$

Page 35 Exercise 35 Answer

Divide $\frac{1}{6}$ by 2.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$\frac{1}{6} \div 2=\frac{1}{6} \div \frac{2}{1}=\frac{1}{6} \times \frac{1}{2}=\frac{1 \times 1}{6 \times 2}=\frac{1}{12}$

Result

$\frac{1}{12}$

Page 35 Exercise 36 Answer

Divide $\frac{2}{3}$ by 3.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$\frac{2}{3} \div 3=\frac{2}{3} \div \frac{3}{1}=\frac{2}{3} \times \frac{1}{3}=\frac{2 \times 1}{3 \times 3}=\frac{2}{9}$

Result

$\frac{2}{9}$

Page 35 Exercise 37 Answer

Divide $\frac{3}{5}$ by 2.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$\frac{3}{5} \div 2=\frac{3}{5} \div \frac{2}{1}=\frac{3}{5} \times \frac{1}{2}=\frac{3 \times 1}{5 \times 2}=\frac{3}{10}$

Result

$\frac{3}{10}$

Page 35 Exercise 38 Answer

Divide $\frac{1}{4}$ by 4.

To divide a whole number by a fraction, first write the whole number as a fraction and than multiply the fraction by the reciprocal of the divisor.

$\frac{1}{4} \div 4=\frac{1}{4} \div \frac{4}{1}=\frac{1}{4} \times \frac{1}{4}=\frac{1 \times 1}{4 \times 4}=\frac{1}{16}$

Result

$\frac{1}{16}$

Page 35 Exercise 39 Answer

A worker is pouring 3 quart of liquid into $\frac{3}{8}$ quart containers.

$3 \div \frac{3}{8}=\frac{3}{1} \div \frac{3}{8}=\frac{3}{1} \times \frac{8}{3}=\frac{3 \times 8}{1 \times 3}=\frac{24}{3}=8$

Result

A worker can fill 8 containers.

Page 36 Exercise 40 Answer

A snail can move 120 ft in $\frac{3}{4}$ h. A tortoise can move 600 ft in $\frac{2}{3}$ h. A sloth can move 250 ft in $\frac{5}{8}$ h.

The animal which moves the fastest is the one that moves more feet in a shorter period of time.

Comparing the snail and tortoise, $\frac{2}{3}$ of an hour is less than $\frac{3}{4}$ and 120 ft is less than 600 ft, so a tortoise moves faster than a snail since it moves more feet in a shorter period of time.

Comparing the sloth and tortoise, $\frac{5}{8}$ is close to $\frac{2}{3}$ but 250 ft is less than 600 ft, so a tortoise moves faster than a sloth since it moves more feet in about the same amount of time.

Result

The animal which moves the fastest is the one that moves more feet in a shorter period of time so the tortoise moves the fastest.

Page 36 Exercise 41 Answer

Since the quotient 250 ÷ $\frac{5}{8}$ tells us about how far a sloth moves in 1 hour and 90 minutes = $1 \frac{1}{2}$ hour, we can multiply the quotient by $1 \frac{1}{2}$ to find how far teh sloth can go in 90 minutes.

First, let’s find the quotient. To divide by a fraction, rewrite the expressions as multiplying by the reciprocal of the divisor and then multiply:

$250 \div \frac{5}{8}=\frac{250}{1} \times \frac{8}{5}=\frac{250 \times 8}{1 \times 5}=\frac{2000}{5}=400$

The sloth will then move about 400 feet in 1 hour. Multiplying this distance by $1 \frac{1}{2}$ gives:

$400 \times 1 \frac{1}{2}=\frac{400}{1} \times \frac{3}{2}=\frac{400 \times 3}{1 \times 2}=\frac{1200}{2}=600$

Therefore, the sloth can move 600 ft in 90 minutes.

Result

600 ft

Page 36 Exercise 42 Answer

A tortoise can move 600 ft in $\frac{2}{3}$h.

The quotient 600 ÷ $\frac{2}{3}$ tells about how far a tortoise may move in one hour.

$600 \div \frac{2}{3}=\frac{600}{1} \div \frac{2}{3}=\frac{600}{1} \times \frac{3}{2}=\frac{600 \times 3}{1 \times 2}=\frac{1800}{2}=900$

Result

A tortoise may move 900 ft in one hour.

Page 36 Exercise 43 Answer

A snail can move 120 ft in $\frac{3}{4}$ h, so the quotient 120 ÷ $\frac{3}{4}$ how far a snail can move in one hour.

$120 \div \frac{3}{4}=\frac{120}{1} \div \frac{3}{4}=\frac{120}{1} \times \frac{4}{3}=\frac{120 \times 4}{1 \times 3}=\frac{480}{3}=160$

Result

A snail can move 160 ft in one hour.

Page 36 Exercise 44 Answer

$\frac{3}{4}$ gallon of juice is being divided equally into 5 pitchers so we need to represent the quotient $\frac{3}{4}$ ÷ 5 using the rectangle.

The rectangle represents 1 whole gallon so first we need to divide it into 4 equal columns and shade 3 of them to represent the $\frac{3}{4}$ gallons.

Next, we need to divide the rectangle into 5 rows to represent the 5 pitchers.

The whole rectangle is divided into 20 parts in all. Each row has 3 parts that are shaded so the quotient is $\frac{3}{20}$. The division equation is then $\frac{3}{4} \div 5=\frac{3}{20}$ and each pitcher has $\frac{3}{20}$ gallon of juice.

Result

$\frac{3}{20}$ gallon

$\frac{3}{4} \div 5=\frac{3}{20}$

Page 36 Exercise 45 Answer

Five equation are give and we have to select all those which are true.

To divide a number with a fraction is the same as multiply the number with the reciprocal of that fraction. Thus, the following equations are true.

$14 \div \frac{7}{10}=14 \times \frac{10}{7}$ $10 \div \frac{3}{5}=10 \times \frac{5}{3}$ $20 \div 4=20 \times \frac{1}{4}$

Division is not commutative, which means it is not the same, for example, to divide 15 by 3, and to divide 3 by 15.

$16 \div \frac{4}{5} \stackrel{?}{=} \frac{1}{16} \times \frac{4}{5}$ $16 \div \frac{4}{5}=\frac{16}{1} \times \frac{5}{4}=\frac{16 \times 5}{1 \times 4}=\frac{80}{4}=20$ $\frac{1}{16} \times \frac{4}{5}=\frac{1 \times 4}{16 \times 5}=\frac{4}{80}=\frac{1}{20}$ $20 \neq \frac{1}{20}$ $16 \div \frac{4}{5} \neq \frac{1}{16} \times \frac{4}{5}$

The same logic is used to show that $12 \div \frac{2}{3} \neq \frac{1}{12} \times \frac{2}{3}$

Result

$14 \div \frac{7}{10}=14 \times \frac{10}{7}, 10 \div \frac{3}{5}=10 \times \frac{5}{3} \text {, and } 20 \div 4=20 \times \frac{1}{4}$

Page 36 Exercise 46 Answer

Five equation are give and we have to select all those which are true.

To divide a number with a fraction is the same as multiply the number with the reciprocal of that fraction. Thus, the following equations are true.

$\frac{1}{3} \div 3=\frac{1}{3} \div \frac{3}{1}=\frac{1}{3} \times \frac{1}{3}$ $\frac{4}{5} \div 5=\frac{4}{5} \div \frac{5}{1}=\frac{4}{5} \times \frac{1}{5}$ $\frac{2}{3} \div 6=\frac{2}{3} \div \frac{6}{1}=\frac{2}{3} \times \frac{1}{6}$

Every whole number can be written as a fraction with that number as the numerator and one as the denominator.

8 written as a fraction is $\frac{8}{1}$, not $\frac{1}{8}$.

Thus, the equation $\frac{7}{8} \div 8=\frac{7}{8} \div \frac{1}{8}=\frac{7}{8} \times \frac{8}{1}$ is not true.

The same logic is used to show that the equation $\frac{4}{9} \div 4=\frac{4}{9} \div \frac{1}{4}=\frac{4}{9} \times \frac{4}{1}$ is also not true.

Result

enVisionmath 2.0: Grade 6 Volume 1 Chapter 1 Use Positive Rational Numbers Mid Topic

Schindler — Tue, 29 Apr 2025 05:27:22 +0000

Chapter 1 Use Positive Rational Numbers

Mid Topic

Page 25 Exercise 1 Answer

Find the closest whole number to the given decimal number which can be divided by given whole number. That is the required compatible number. Dividing that compatible number by the given whole number gives us the estimate.

Result

The required compatible number is the closest whole number to the given decimal number. Divide that number by the given whole number to get an estimate.

Page 25 Exercise 2 Answer

Keaton is building a rectangular tabletop and wnats to put a metal border around the edge. The length of the tabletop is 1.83 meters and the width is 0.74 meter. Use the formula P = 2l + 2w to find the perimeter of the tabletop.

P = 2l + 2w = 2 x 1.83 + 2 x 0.74

P = 2 x 1.83 + 2 x 0.74 = 1.48 + 3.66

P = 1.48 + 3.66 = 5.14

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Result

The perimeter of the tabletop is 5.14 meters.

Page 25 Exercise 3 Answer

Norbert’s Nursery is having a sale. Flats of flowers are priced as follows: petunias 5.25 per flat, daisy 7.65 per flat, and begonia 8.40 per flat. Jake buys 2 flats of petunias, 3 flats of daisies, and 1 flat of begonias. If he pays with a 50 bill, how much change should Jake receive?

First, we must calculate the exact price of the flowers.

2 x 5.25 + 3 x 7.65 + 1 x 8.40 = ?

2 x 5.25 + 3 x 7.65 + 1 x 8.40 = 10.50 + 22.95 + 8.40

10.50 + 22.95 + 8.40 = 41.85

The exact price Jake must pay is 41.85.

Jake paid with a 50 bill, and he received change which is the difference between 50 and 41.85.

Result

Jake received $8.15 of change.

Page 25 Exercise 4 Answer

It is given that $\frac{1}{4}$ page in the book costs $ 15.50 and we need to complete the table to find the cost for $\frac{2}{4}$ page, $\frac{3}{4}$ page, and $\frac{4}{4}$ page.

Since $\frac{2}{4}=2 \times \frac{1}{4}$, we can multiply the cost for $\frac{1}{4}$ page by 2 to get the cost for $\frac{2}{4}$ page:

Similarly, we can find the costs for $\frac{3}{4}$ page and $\frac{4}{4}$ page by multiplying the cost for $\frac{1}{4}$ page by 3 and 4:

The table is then completed as:

Now we need to determine which of the given expressions can be used to find the cost for $\frac{3}{4}$ page:

3 x $15.50 was teh expression we used to find the cost for the table, so the expression 3 x $15.50 is correct.

$1.50 + $15.50 + $15.50 is the repeated addition form of 3 x $15.50. The expression $1.50 + $15.50 + $15.50 is then also correct.

We know that $\frac{3}{4} \times \$ 15.50 \neq 3 \times \$ 15.50$ so the expression $\frac{3}{4}$ x $15.50 is not correct.

Using the Distributive Property gives:

(4 x $15.50) – $15.50 = (4 – 1) x $15.50 = 3 x $15.50

The expression (4 x $15.50) – $15.50 is then also correct.

3 + $15.50 is not correct since we know the total cost needs to be 3 times $15.50, not 3 plus $15.50.

Result

3 x $15.50, $15.50 + $15.50 + $15.50, and (4 x $15.50) – $15.50

Page 25 Exercise 5 Answer

Latisha hired 4 babysitters in the last month, paid a total of $170 in babysitting fees, each babysitter worked a total of 5 hours and is paid the same amount. We need to find how much each babysitter earns per hour.

Each babysitter is then paid $42.50 in total.

Each babysitter is then paid $8.50 per hour.

Result

Each babysitter is paid $8.50 per hour.

Page 25 Exercise 6 Answer

It is given that Lucia walks $2 \frac{3}{4}$ miles on Monday and walks $1 \frac{1}{2}$ times farther on Tuesday.

To find how far she walked on Tuesday, we must multiply $2 \frac{3}{4}$ and $1 \frac{1}{2}$.

To multiply two mixed numbers, rewrite each mixed number as an improper fraction. Then multiply the numerators and multiply the denominators. Reduce the fraction and rewrite as a mixed number if necessary:

$2 \frac{3}{4} \times 1 \frac{1}{2}=\frac{11}{4} \times \frac{3}{2}=\frac{11 \times 3}{4 \times 2}=\frac{33}{8}=4 \frac{1}{8}$

Lucia then walked $4 \frac{1}{8}$ miles on Tuesday.

Result

$4 \frac{1}{8}$ miles

Page 26 Exercise 1a Answer

Team members Eric and Natalia secure a grant for $75.00 and need to buy 3 beams and 6 channels. We need to determine if the grant will cover the cost, and if so, how much of the grant will remain.

A beam costs $5.95 and a channel costs $8.50.

The cost is $68.85 so the grant will cover the cost.

After they buy beams and channels, there will be left $6.15 of the grant.

Result

The grant will cover the cost and $6.15 will remain.

Page 26 Exercise 1b Answer

We know 3 teams members will equally share the cost of 2 motor controllers and 4 wheels.

From the table, the cost of 1 motor controller is $99.75 and the cost of 1 wheel is $18.90. The cost of 2 controllers and the cost of 4 wheels are then:

The total cost of the 2 controllers and 4 wheels is then:

Since they share the total cost evenly, we can divide the total cost by 3 to find how much each person pays:

Each person then contributed $91.70.

Result

$91.70

Page 26 Exercise 1c Answer

Nyan Robotics has a budget of $99 to buy sprockets, axles, and gears and they spend $\frac{2}{3}$ of the budget on sprockets. We need to find how much money from the budget remains to buy axles and gears.

If they spent $\frac{2}{3}$ on sprockets, it means they spent $\frac{2}{3}$ of $99 on sprockets, and the rest was left to be spent on axles and gears.

$\frac{2}{3} \times 99=\frac{2}{3} \times \frac{99}{1}=\frac{2 \times 99}{3 \times 1}=\frac{198}{3}=66$

$66 was spent on sprockets.

99 – 66 = 33

$33 remains to buy axles and gears.

Result

$33 remains to buy axles and gears.