Marksparks

Glencoe Math Course 2 Volume 1 Common Core Chapter 1 Ratios and Proportional Reasoning

 

Page 33   Exercise 1   Problem 1

Let the two ratios be a:b and c:d

If both the ratios are equal a:b=c:d

Then the given ratio is said to be in proportion.

Furthermore, if the sets fluctuate in the same proportion, the ratio is indeed in proportion.

In terms of objects, if two objects have the same form but different sizes, they are perhaps the same shape but different sizes.

For instance, suppose the two objects are spheres, but one is smaller and the other is larger.

Despite the fact that the sizes differ, the corresponding angles would be the same.

As a result, their ratios are proportional.

As a result, the two objects are proportional.

When two objects have the same form, and same angles, but various sizes, they all seem to be in proportion.

As a result, their ratios are proportional.

 

Page 36   Exercise 1   Problem 2

The Vista Marina rents boats for $25 per hour.

In addition to the rental fee, there is a $12 charge for fuel.

We need to determine whether the number of hours you can rent the boat is proportional to the total cost.

As per the given information

 

For each hour, the relationship between the cost and the rental time as a ratio in its simplest form will be,
​
\(\frac{\text { cost }}{\text { Time }}=\frac{37}{1}\) = 37

\(\frac{62}{2}\)  =  31

\(\frac{87}{3}\)  =  29

The ratios of the two quantities are not the same.

Therefore, the number of hours you can rent the boat is not proportional to the total cost

The number of hours you can rent the boat is not proportional to the total cost.

 

Page 36   Exercise 2   Problem 3

We need to find out which situation represents a proportional relationship between the hours worked and the amount earned for Matt and Jane.

The given is

 

The given is

For each hour, the relationship between the time and Matt’s earnings as a ratio in its simplest form will be

\(\frac{\text {Earning}}{\text { Time }}=\frac{12}{1}\) = 12

\(\frac{20}{2}\)  = 10

\(\frac{31}{3}\)  =  10.33

The values of the ratios are different. Thus, they are not proportional.

Similarly
For each hour, the relationship between the time and Jane’s earnings as a ratio in its simplest form will be

\(\frac{12}{1}\)  =  12

\(\frac{24}{2}\)  =  12

\(\frac{36}{3}\)  =  12

The ratios of the two quantities are the same. Therefore, they are proportional.

Jane’s situation represents a proportional relationship between the hours worked and a mount earned.

 

Page 37   Exercise 1   Problem 4

Given that, an adult elephant drinks about 225 liters of water each day.

We need to determine whether the number of days the water supply lasts is proportional to the number of liters of water the elephant drinks or not.

For each day, the elephant drinks 225 L of water.

As per the given information, completing the table as below

For each day, the relationship between the number of days and the number of liters of water the elephant drinks as a ratio in its simplest form will be

\(\frac{225}{1}\) = 225

\(\frac{450}{2}\) = 225

\(\frac{675}{3}\) = 225

\(\frac{900}{4}\) = 225

The ratios of the two quantities are the same.

Therefore, the number of days the water supply lasts is proportional to the number of liters of water the elephant drinks.

The number of days the water supply lasts is proportional to the number of liters of water the elephant drinks.

 

Page 37   Exercise 2   Problem 5

Given that an elevator ascends, or goes up, at a rate of 750 feet per minute.

We need to determine whether the height to which the elevator ascends is proportional or not to the number of minutes it takes to get there.

For each minute, the elevator rises 750 feet.

As per the given information, completing the table as below

For each minute, the relationship between the height and the time as a ratio in its simplest form will be

\(\frac{750}{1}\)   =  750

\(\frac{1500}{2}\) =  750

\(\frac{2250}{3}\) = 750

\(\frac{3000}{4}\) = 750

The ratios of the two quantities are the same.

Therefore, the height to which the elevator ascends is proportional to the number of minutes it takes to get there.

The height to which the elevator ascends is proportional to the number of minutes it takes to get there.

 

Page 37   Exercise 3   Problem 6

We need to determine which among the given situation represents a proportional relationship between the number of laps run by each student and their time

 

The given table is

For each second, the relationship between the laps and Desmond’s time as a ratio in its simplest form will be,

\(\frac{146}{2}\) = 73

\(\frac{292}{4}\) = 73

\(\frac{584}{8}\) = 73

The ratios of the two quantities are the same.

Therefore, the number of laps is proportional to Desmond’s time.

 

For each second, the relationship between the laps and Maria’s time as a ratio in its simplest form will be,

\(\frac{150}{2}\)= 75

\(\frac{320}{4}\) = 80

\(\frac{580}{6}\) = 96.67

The ratios of the two quantities are not the same.

Therefore, the number of laps is not proportional to Maria’s time.

Desmond’s situation represents a proportional relationship between the number of laps run by him and their time.

 

Page 37   Exercise 4   Problem 7

Given that, Plant A is 18 inches tall after one week, 36 inches tall after two weeks, and 56 inches tall after three weeks.

Plant B is 18 inches tall after one week, 36 inches tall after two weeks, and 54 inches tall after three weeks.

We need to determine which situation represents a proportional relationship between the plants’ height and the number of weeks.

From the given information, forming a table with those values

 

Also

 

The ratio in its simplest form for Plant A is

\(\frac{18}{1}\) = 18

\(\frac{36}{2}\) = 18

\(\frac{56}{3}\) = 18.67

The ratios of the two quantities are not the same.

Therefore, the plant’s height is not proportional to the number of weeks.

Similarly, The ratio in its simplest form for Plant B is

\(\frac{18}{1}\) = 18

\(\frac{36}{2}\) = 18

\(\frac{54}{3}\) = 18

The ratios of the two quantities are the same.

Therefore, the plant B’s height is proportional to the number of weeks.

Plant B’s situation represents a proportional relationship between the plants’ height and number of weeks.

 

Page 38   Exercise 7   Problem 8

Given that, Blake ran laps around the gym. His times are shown in the table below.

Blake is trying to decide whether the number of laps is proportional to the time. We need to find his mistake and correct it.

 

As per the given information

We need to determine whether the number of laps is proportional to the time.

The relationship between the number of laps and the time as a ratio in its simplest form will b

\(\frac{4}{1}\) =  4

\(\frac{6}{2}\)  =  3

\(\frac{8}{3}\)  =  2.67

\(\frac{10}{4}\)  =  2.5

The ratios of the two quantities are not the same.

Therefore, the number of laps is not proportional to the time.

The number of laps is not proportional to the time.

 

Page 38   Exercise 8   Problem 9

We need to determine whether the cost for ordering multiple items that will be delivered is sometimes, always, or never proportional.

Also, we need to explain it.

Consider the price of an item, here we have considered a pen drive which costs $10.

If we ordered 10 pen drives, the cost will be

10 dollars × 10 = 100 dollars

If we ordered 20

20 × 10 = 200 dollars

For 30 pen drives

30 × 10 = 300 dollars

The relationship between the number of units and the cost as a ratio in its simplest form will be,

\(\frac{100}{10}\) =  10

\(\frac{200}{20}\) = 10

\(\frac{300}{30}\) = 10

Thus, the ratios are the same.

Thus, the cost for ordering multiple items that will be delivered is always proportional.

The cost for ordering multiple items that will be delivered is always proportional.

 

Page 38   Exercise 9   Problem 10

We need to determine which of the given relationship has a unit rate of 60 miles per hour.

Calculating the unit rate one by one

1) 300 miles in 6 hours

​Unit rate \( = \frac{\text { Number of miles }}{\text { Number of hours }}\)

=  \(\frac{300}{6}\)

=  50 miles/ hour

​Unit rate=  50 miles/hour
​

2) 300 miles in 5 hours

Unite rate = \(\frac{300}{5}\)

​Unit rate = 60 miles/hour

 

3) 240 miles in 6 hours

Unite rate = \(\frac{240}{6}\)

​Unit rate = 40 miles/hour
​

4)  240 miles in 5 hours

​Unit rate = \(\frac{240}{5}\)

​Unit rate = 48 miles/hour
​
300 Miles in 5 hours have a unit rate of 60 miles per hour.

 

Page 39   Exercise 10    Problem 11

Given that a vine grows 7.5 feet every 5 days.

We need to determine whether the length of the vine on the last day is proportional to the number of days of growth.

The given table is

We need to determine whether the length of the vine on the last day is proportional to the number of days of growth.

The relationship between the time and the length as a ratio in its simplest form will be,

\(\frac{7.5}{5}\)  =  1.5

\(\frac{15}{10}\)  =  1.5

\(\frac{22.5}{15}\)  =  1.5

\(\frac{30}{20}\)  =  1.5

The ratios of the two quantities are the same.

Therefore, the length of the vine on the last day is proportional to the number of days of growth.

The length of the vine on the last day is proportional to the number of days of growth

 

Page 39   Exercise 11   Problem 12

Given that, To convert a temperature in degrees Celsius to degrees Fahrenheit, multiply the Celsius temperature by \(\frac{9}{5}\)and then add 32 degrees

We need to determine whether the temperature in degrees Celsius is proportional to its equivalent temperature in degrees Fahrenheit.

As per the given information, the completed table will be

We need to determine whether the temperature in degrees Celsius is proportional to its equivalent temperature in degrees Fahrenheit.

The relationship between the Celsius and the Fahrenheit as a ratio in its simplest form will be

\(\frac{0}{32}\)  =  0

\(\frac{10}{50}\) =  \(\frac{1}{5}\) =  0.2

\(\frac{20}{68}\)  =  0.294

\(\frac{30}{86}\)  =  0.3488

The ratios of the two quantities are not the same.

Therefore, the temperature in degrees Celsius is not proportional to its equivalent temperature in degrees Fahrenheit.

The temperature in degrees Celsius is not proportional to its equivalent temperature in degrees Fahrenheit.

 

Page 40    Exercise 14   Problem 13

Given that, Mr. Martinez is comparing the price of oranges from several different markets.

We need to determine which market’s pricing guide is based on a constant unit price

Determine whether the total cost and the number of oranges are proportional or not.

 

 

1)  \(\frac{3.50}{5}\) =  0.7

\(\frac{6}{10}\)  =  0.6

\(\frac{8.50}{15}\)  =  0.5667

\(\frac{11}{20}\) =  0.55

The ratio values are different.

 

2)  \(\frac{3.50}{5}\)  =  0.7

\(\frac{6.50}{10}\)  =  0.65

\(\frac{9.50}{15}\)  =  0.633

\(\frac{12.50}{20}\)  =  0.467

The ratio values are different.

 

3) \(\frac{3}{5}\) =  0.6

\(\frac{5}{10}\) =  0.5

\(\frac{7}{15}\) =  0.467

\(\frac{9}{20}\) =  0.45

The ratio values are different.

 

4) \(\frac{3}{5}\)  =  0.6

\(\frac{6}{10}\) =  0.6

\(\frac{9}{15}\)  =  0.6

\(\frac{12}{20}\) =  0.6

The ratio values are the same.

Hence, Option  (4)  is the market’s pricing guide is based on a constant unit price.

Option  (4)  is the market’s pricing guide is based on a constant unit price.

 

Page 40    Exercise 15   Problem 14

Given that, the middle school is planning a family movie night where popcorn will be served.

The constant relationship between the number of people n and the number of cups of popcorn p is shown in the table.

We need to determine how many people can be served with 519 cups of popcorn.

 

 

Given that the relationship between the number of people and the number of cups of popcorn is constant.

Using the proportionality relationship

Let x be the number of people who can be served with 519 cups of popcorn.

\(\frac{30}{90}\) = \(\frac{x}{519}\)

\(\frac{1}{3}\)= \(\frac{x}{519}\)

x =  \(\frac{519}{3}\)

x =  173

173 people can be served with 519 cups of popcorn.

 

Page 40   Exercise 16   Problem 15

Given that, x = 12

We need to find the value of 3x

The given expression is 3x

Also x = 12

Thus, the value becomes

​3x = 3 (12)

3x = 36

The value of 3x = 36

 

Page 40   Exercise 17   Problem 16

Given that, x = 12

We need to find the value of 2x − 4

The given expression is 2x − 4

Also x = 12

Thus, the value becomes

​2x − 4 = 2(12) − 4

2x − 4 = 24 − 4

2x − 4 = 20
​
The value of 2x − 4 = 20

 

Page 40  Exercise 18   Problem 17

Given that, x = 12

We need to find the value of 5x + 30

The given expression is 5x + 30

Also x = 12

Thus, the value becomes

​5x + 30 = 5(12) + 30​

5x + 30 = 60 + 30

5x + 30 = 90

The value of 5x + 30 = 90

 

Page 40   Exercise 19   Problem 18

Given that, x = 12

We need to find the value of 3x − 2x

The given expression is 3x − 2x

Also x = 12

Thus, the value becomes

​3x − 2x = 3(12) − 2(12)

3x − 2x = 36 − 24

3x − 2x = 12
​
The value of 3x − 2x = 12

 

Page 40   Exercise 22   Problem 19

Given that, Brianna downloads 9 songs each month onto her MP3 player.

We need to show the total number of songs downloaded after 1,2,3, and 4 months.

For the first month, the number of songs downloaded is 9.

For the second month

9 × 2 = 18 songs

For the third month

9 × 3 = 27 songs

For the fourth month

9 × 4 = 36 songs

Thus, the table becomes

 

The total number of songs downloaded after 1,2,3, and 4 months is shown below

 

Page 44   Exercise 1  Problem 20

We need to define complex fractions and also give two examples of complex fractions.

A complex fraction is nothing but a fraction that has fractions in its denominator or in the numerator or in both.

For example

\(\frac{5}{\frac{12}{3}}, \frac{\frac{10}{15}}{25}, \frac{\frac{14}{8}}{\frac{8}{5}}\) These are all complex fractions.

Solving a complex fraction:

Let us consider a complex fraction \(\frac{5}{\frac{12}{3}}\)

Simplifying it, we get

\(\frac{5}{\frac{12}{3}}\)= \(\frac{5}{2}\)× \(\frac{1}{10}\)

= \(\frac{1}{2}\)×\(\frac{1}{2}\)

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

A complex fraction is a fraction that has more than one fraction.

That is, fractions will be in their denominator or in the numerator, or in both.

The two examples of complex fractions are \(\frac{11}{\frac{9}{7}} \text { and } \frac{\frac{22}{8}}{\frac{6}{7}}\)

 

Page 44   Exercise 2   Problem 21

A rate is a fraction of a ratio of two different quantities.

A ratio differs from a rate as the ratio is the relationship between two quantities of the same or different units.

The rate only deals with different units.

When we simplify a rate to make their denominator as 1, then it is said to be the unit rate.

Unit rate is nothing but the rate per unit of a quantity.

For example, if we buy 8 flowers for $10.

The unit rate is the cost of 1 flower.

When a rate is simplified so that it has a denominator of 1 unit, it is called a unit rate.

 

Page 44   Exercise 3  Problem 22

Given that, 750 yards in 25 minutes.

We need to determine the unit rate of the given and round the unit rate obtained to the nearest hundredth if needed.

Number of yards  = 750

Number of minutes  = 25

The unit rate is given by

Unite rate \(=\frac{\text { Number of yards }}{\text { Number of minutes }}\)

=  \(\frac{750}{25}\)

=  30 yards per minute

The unit rate of the given is 30 yards per minute.

 

Page 44   Exercise 4   Problem 24

Given that, $420 for 15 tickets.

We need to determine the unit rate of the given and round the unit rate obtained to the nearest hundredth if needed.

Amount in dollars  =  $420

Number of tickets = 15

The unit rate is

Unite rate \(=\frac{\text { Amount in dollars }}{\text { Number of tickets }}\)

= \(\frac{420}{15}\)

= 28 dollars per ticket

The unit rate of giving is, 28 dollars per ticket

 

Page 44   Exercise 5   Problem 25

Given, the complex fraction is\(\frac{9}{\frac{1}{3}}\)

We need to simplify the given complex fraction.

Rewrite the given fraction as a division.

The expression becomes

\(\frac{9}{\frac{1}{3}}\) = \(\frac{9}{1}\) ÷ \(\frac{1}{3}\)

Multiply it by the reciprocal of \(\frac{1}{3}\) , we get 

\(\frac{9}{\frac{1}{3}}\) = \(\frac{9}{1}\) ÷ \(\frac{1}{3}\)

\(\frac{9}{\frac{1}{3}}\) = \(\frac{9}{1}\) × \(\frac{1}{3}\)

Simplifying it further, we get

\(\frac{9}{\frac{1}{3}}\) = \(\frac{9}{1}\) ×  \(\frac{1}{3}\)

= \(\frac{27}{1}\)

=  27

The value of \(\frac{9}{\frac{1}{3}}\) is equal to 27.

 

Page 44   Exercise 7   Problem 26

Given, the complex fraction is \(\frac{\frac{1}{6}}{1 \frac{3}{8}}\)

We need to simplify the given complex fraction.

Rewrite the given fraction as a division.

The expression becomes

\(\frac{\frac{1}{6}}{1 \frac{3}{8}}=\frac{1}{6} \div 1 \frac{3}{8}\)

 

\(\frac{\frac{1}{6}}{1 \frac{3}{8}}=\frac{1}{6} \div \frac{11}{8}\)

Multiply it by the reciprocal of \(\frac{11}{8}\) , we get

\(\frac{\frac{1}{6}}{1 \frac{3}{8}}=\frac{1}{6} \div \frac{11}{8}\)

 

\(\frac{\frac{1}{6}}{1 \frac{3}{8}}\)= \(\frac{1}{6}\) × \(\frac{8}{11}\)

Simplifying it further, we get

\(\frac{\frac{1}{6}}{1 \frac{3}{8}}\)= \(\frac{1}{6}\) ×  \(\frac{8}{11}\)

=  \(\frac{1}{3}\)×\(\frac{4}{11}\)

=  \(\frac{4}{33}\)

The value of \(\frac{\frac{1}{6}}{1 \frac{3}{8}}\) is equal to \(\frac{4}{33}\)

 

Page 44   Exercise 9   Problem 27

We need to find out which among the given is the same as 2,088 feet per minute.
(1) 696 Meters per minute
(2) 696 Yards per minute
(3) 696 Feet per minute
(4) 696 Yards per second

Convert the given, 2,088 feet per minute to meters per minute.

Converting we get \(\frac{2088 \text { feet }}{1 \text { minute }}=\frac{2088 \text { feet }}{1 \text { minute }} \times \frac{0.3048 \text { meters }}{1 \text { feet }}\)

= \(\frac{2088 \times 0.3048 \text { meters }}{1 \text { minute }}\)

=  636.42 meters per minute

Convert the given to yards per minute \(\frac{2088 \text { feet }}{1 \text { minute }}=\frac{2088 \text { feet }}{1 \text { minute }} \times \frac{0.3333 \text { yards }}{1 \text { feet }}\)

= \(\frac{2088 \times 0.3333 \text { yards }}{1 \text { minute }}\)

=  696 yards per minute

Thus. Option (2) is correct.

696 yards per minute is the same as 2,088 feet per minute.

Glencoe Math Course 2 Volume 1 Common Core Chapter 1 Ratios and Proportional Reasoning

 

Page 3  Exercise 1  Problem 1

Let the two ratios be a:b and c:d

If both the ratios are equal a:b = c:d

Then the given ratio is said to be in proportion.

Furthermore, if the sets fluctuate in the same proportion, the ratio is indeed in proportion.

 

In terms of objects, if two objects have the same form but different sizes, they are perhaps the same shape but different sizes.

For instance, suppose the two objects are spheres, but one is smaller and the other is larger.

Despite the fact that the sizes differ, the corresponding angles would be the same.

As a result, their ratios are proportional.

As a result, the two objects are proportional.

When two objects have the same form, same angles, but various sizes, they all seem to be in proportion. As a result, their ratios are proportional.

 

Page 6   Exercise 1   Problem 2

Given: On a seventh-grade field trip.

The number of students who went on that trip is  180.

The number of adults who went on that trip is  24.

The number of buses in which they went is  4.

 

We need to write the ratio of adults: Students as a fraction and in its simplest form.

The number of adults  = 24

The number of students  = 180

The ratio of adults: Students in its simplest form is given by

​Adults: Students  = 24:180

\(\frac{\text { Adults }}{\text { Students }}=\frac{24}{180}\)

 

=\(\frac{2}{15}\)

The ratio of adults: Students as a fraction in the simplest form is \(\frac{2}{15}\)

 

Page 6   Exercise 2  Problem 3

Given:  In a seventh-grade field trip.

The number of students who went on that trip is 180.

The number of adults who went on that trip is 24.

The number of buses in which they went is 4.

 

We need to write the ratio of the students: Buses as a fraction and in its simplest form.

The number of buses  = 4

The number of students  = 180

The ratio of students: Buses in its simplest form is given by

​Students : Buses  = 180:4

\(\frac{\text {Students}}{\text { Buses}}=\frac{180}{4}\)

 

=\(\frac{45}{1}\)

The ratio of students: buses as a fraction in the simplest form is \(\frac{45}{1}\)

 

Page 6   Exercise 3   Problem 4

Given:  in a seventh-grade field trip.

The number of students who went on that trip is 180.

The number of adults who went on that trip is 24.

The number of buses in which they went is 4.

 

We need to write the ratio of the buses: People as a fraction and in its simplest form.

The number of adults  =  24

The number of students  = 180

The number of buses  = 4

The total number of people on the bus = Adults + Students

​= 24 + 180

= 204

The total number of people on the bus = 204

​The ratio of buses: People in its simplest form is given by

​Buses: People = 4:204

\(\frac{\text {Buses}}{\text { People}}=\frac{4}{204}\)

 

= \(\frac{1}{51}\)

The ratio of buses: People as a fraction in the simplest form is \(\frac{1}{51}\)

 

Page 6  Exercise 5  Problem 5

Given: 

12 out of 20 doctors agree

15 out of 30 doctors agree

We need to determine their ratios and to check whether the ratios are equivalent.

Writing the given ratios in their simplest form.

12 out of 20 doctors agree The ratio is

= \(\frac{12}{20}\)

= \(\frac{6}{10}\)

= \(\frac{3}{5}\)

15 out of 30 doctors agree The ratio is

= \(\frac{15}{30}\)

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

Both the ratios are simplified to different fractions.

Thus, both ratios are not equivalent.

The ratios are not equivalent.

 

Page 8   Exercise 1  Problem 6

Given:  The ratio of the number of boys to the number of girls on the swim team is 4:2.

The total number of people on the swim team is 24 athletes.

We need to determine how many more boys than girls are there on the team.

Also, use the given bar diagram to solve.

 

Given: The  ratio is 4:2

The total number of boys and girls is 24

The obtained bar diagram is

 

Let each part of the ratio be x

Thus, the equation becomes

4x + 2x = 24

Solving the equation, we get

​4x + 2x = 24

6x = 24

x = \(\frac{24}{6}\)

x = 4
​

The total number of Boys will be

​4x = 4 × 4

= 16
​
The total number of Girls will be

​2x  = 2 × 4

=  8
​
Determining how many more boys than girls are there on the team, we get

16 − 8 = 8

Thus, 8 boys are more on the swim team than the girls.

8 more boys than girls are there on the team.

 

Page 8  Exercise 3  Problem 7

We are given a bar diagram.

We must write a real-world problem that could be represented by the bar diagram and also to solve it.

The bar diagram is as follows

 

The given bar diagram is

The number of parts given is 6 and 5

Therefore, the ratio will be 6:5

The total amount given is 220.

Thus, the real-world problem for the given bar diagram will be.

The ratio of the expenses shared by the father and the son is 6:5.

The total amount shared by them will be 220.

We need to determine how much amount is shared by them both.

 

The bar diagram will be

Writing an equation using the obtained ratio and the total amount to solve this, we get

6x + 5x = 220

11x = 220

x = \(\frac{220}{11}\)

x =  20

Thus, each part represents the amount of 20.

The expense amount shared by the father is

6x=6×20=120

The expense amount shared by the son is

5x = 5×20 = 100

 

A real-world problem that could be represented by the given bar diagram is.

The ratio of the expenses shared by the father and the son is 6:5.

The total amount shared by them will be 220.

Determine how much expense amount is shared by them both.

The corresponding bar diagram will be.

On solving the problem, we get the expense shared by the father is 120 and by the son is 100

 

Page 8  Exercise 4  Problem 8

Consider a real-world problem.

Let the bar diagram given for that particular real-world problem be.

Count the number of parts in each case to determine the ratio corresponding to it.

Here, the number of parts is 6 and 5.

Therefore, the ratio given will be 6:5

The total amount corresponding to the given ratio is 220.

Thus, for solving the given ratio, we need to take the total number of parts given and the total amount represented in the bar diagram.

Thus, the equation will be

​6x + 5x = 220

11x = 220

x=220/11

x  = 20
​

We can use a bar diagram to obtain the ratios of each part and the total amount represented by the real-world problem to solve it.

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Percents

 

Page 95   Exercise 1   Problem 1

Given:

To explain how can percent help you understand situations involving money?

The percent help to understand situations involving money

The interest rates are written as a percent.

Also, find the interest earned on a savings account and the amount of interest charged on bank loans and credit cards.

The sales tax is also indicated in percent.

Hence explained.

 

Page 98   Exercise 1  Problem 2

Given:

Expression: 300 × 0.02 × 8 =

To find: Find each product

Determine the product of the first two factors, then the product of the result, and the last factor

(300 × 0.02) × 8

⇒  6 × 8 = 48

(300 × 0.02) × 8 = 48

Finally, The product of the factors are 48.

 

Page 98   Exercise 2   Problem 3

Given:

Expression: 85 × 0.25 × 3 =

To find: Find each product

Determine the product of the first two factors, then the product of the result, and the last factor

85 × 0.25 × 3

⇒  21.25 × 3 = 63.75

85 × 0.25 × 3 = 63.75

Finally, The product of the factors is 63.75.

 

Page 98   Exercise 3  Problem 4

Given:

Suppose Nicole saves $2.50 every day. How much money will she have in 4 weeks?

To find: Find each product

Because there are seven days in a week, there are 28 days in four weeks

⇒  28 × $2.50 = $70.00

Finally, The product of the factors is $70.00.

 

Page 98  Exercise 4   Problem 5

Given:  0.675 =

To find: Write each decimal as a present.

By multiplying the decimal by 100 and adding a percent sign, you can rewrite it as a percent

⇒ 0.675 = 67.5 %

Finally, The Decimals as a percent  is  67.5 %

 

Page 98  Exercise 5  Problem 6

Given: 0.725 =

To find: Write each decimal as a present.

By multiplying the decimal by 100 and adding a percent sign, you can rewrite it as a percent

⇒  0.725 = 72.5 %

Finally, The Decimals as a percent are  72.5.

 

Page 98   Exercise 6  Problem 7

Given: 0.95 =

To find: Write each decimal as a present.

By multiplying the decimal by 100 and adding a percent sign, you can rewrite it as a percent

⇒  0.95 = 95 %

Finally, The Decimals expressed as a percent are 95 %

 

Page 98   Exercise 7  Problem 8

Given: Approximately 0.92 of a watermelon is water. What percent represents this decimal?

To find: Write each decimal as a present.

By multiplying the decimal by 100 and adding a percent sign, you can rewrite it as a percent

⇒ 0.92% =  92 %

Finally, The Decimals expressed as a percent are  92 %

 

Page 102   Exercise 3  Problem 9

Given:

To find:  The answers in blank boxes

Total  =  150

Percent  =  40%

Rate per hundred =  40/100

Therefore

Part = \(\frac{40}{100}\) ×(150)

=  60

The solution of part is 60

 

Page 102   Exercise 4  Problem  10

Given:

To find: The answers in blank boxes

Total = 150

Percent  =  50%

Rate per hundred  =  50/100

Therefore

Part =  ​\(\frac{50}{100}\) ×(150)

=  75
​
The solution of part is 75

 

Page 102   Exercise 5  Problem  11

Given:

To find: The pattern

Total  = 150

Percent = 40%

Rate per hundred = 40/100

Therefore

Part = \(\frac{40}{100}\) × (150)

= 60

 

​Total  = 150

Percent  =  50%

Rate per hundred = 50/100

Therefore

Part =  ​\(\frac{50}{100}\) × (150)
​
=  75
​
By analyzing the pattern we found that the part has been increasing by every 15. The part has been increased by every  15.

 

Page 102   Exercise 6  Problem  12

Given: The table is showing percentages equivalent to real numbers.

We have to write a real-world problem based on the values of the table.

This is done by equating the percentage to the real numbers.

According to the table

10 times, 10% = 10 × 10 = 100%

10times, 25 = 25 × 10 = 250, or

If, ​ 10% = 251%

= 2.5

∴  100%  =  250

Hence 10%  =  25 is verified to write in the form of a percentage expression.

10% = 25 is written in the form of a percentage based on the table.

 

Page 102  Exercise 7  Problem  13

4times, 25% = 25 × 4 = 100%

4times, 15 = 15 × 4 = 60 , or

If,   ​25% = 15

1% = ​\(\frac{15}{25}\)(100%)

=  ​\(\frac{15}{25}\)×100

∴  100%  =  60
​
Hence 100%  =  60 is verified to write in the form of a percentage expression.

100% = 60 is written in the form of a percentage based on the table.

 

Page 102   Exercise 8   Problem  14

Given:

How percent  used to solve a real-world problem

Explanation:

Percent diagram helps to display the information, making it easier to solve for what it is missing

A diagram in a way to help your to brain process a lot of information at once.

It is the visual planning tool that takes some of the pressure off of remembering every single detail

Sometimes we have hard questions that require sifting through a lot of information to figure them out.

This type of chart or diagram gives a quick and easy way to see a whole is divided into its constituent parts.

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Percents

 

Page 103  Exercise 1  Problem 1

We need to explain how can percent help you understand situations involving money.

The percent help to understand situations involving money

The interest rates are written as a percent.

Also, find the interest earned on a savings account and the amount of interest charged on bank loans and credit cards.

The sales tax is also indicated in percent.

Hence explained.

 

Page 103  Exercise 2  Problem 2

Given:

To show that the 60%  of  2000 = 1200

To prove that the 60%  of  2000 = 1200

\(\frac{60}{100}\) (2000) = 1200

60 × 20 = 1200

1200 = 1200

​Thus the  60% of 2000 = 1200 is proved

 

Page 106   Exercise 1  Problem 3

Given:

To find – The number from 8% of 50

8% of  50
​
\(\frac{8}{100}\)(50)

\(\frac{8}{2}\)

= 4

​8% of 50 is 4

8% of 50 is 4

 

Page 106  Exercise 2  Problem 4

Given:

Find each number. Round to the nearest tenth if necessary.

To find – The number of 95 % of 40

Given:

95 % of 40

95 % = \(\frac{95}{100}\)

Then \(\frac{95}{100}\) of 40

=  \(\frac{95}{100}\) × 40

⇒   \(\frac{95}{5}\) × 2

⇒  19 × 2 = 38

So, 95 % of 40 is 38.

The solution is 38.

 

Page 106  Exercise 3  Problem 5

Given:

Find each number. Round to the nearest tenth if necessary.

To find – The number of 110 % of 70

Given:
​
​110 % of 70

110 % = \(\frac{110}{100}\)

Then \(\frac{110}{100}\) of 70

=  \(\frac{110}{100}\) × 70

⇒  11 × 7=  77

So, 110 % of 70 is 77.

The solution is 77.

 

Page 106  Exercise 4  Problem 6

Given:

Mackenzie wants to buy a backpack that costs $50.If the tax rate is 6.5%, how much tax will she pay?

To find – The amount of tax.

Cost of backpack = $50

Tax rate =  6.5%

Amount of tax =  6.5 %  of  50

6.5 % =  \(\frac{6.5}{100}\)

Then ​\(\frac{6.5}{100}\) of  50

=  \(\frac{6.5}{100}\)  ×  50

⇒  \(\frac{6.5}{2}\)

= 3.25

So the amount of tax = $3.25

The solution is $3.25

 

Page 106   Exercise 5  Problem 7

Given:

Building on the essential question give an example of a real-world situation in which you would find the percent of a number.

To find – An example.

Take an example of 17% of 150 oranges are bad.

To find the number of oranges are bad.

17 % of 150

​17 % = \(\frac{17}{100}\)

\(\frac{17}{100}\) of  150

=  \(\frac{17}{100}\) × 150

⇒  \(\frac{17}{2}\) × 3 = 26

So, 26 oranges are bad.

Hence the example has been found.

 

Page 107  Exercise 2   Problem 8

Given:

Find each number. Round to the nearest tenth if necessary.

45% of $432

To find – The number.

Given:
​
​45 % of 432

45 % \(\frac{45}{100}\)

Then,\(\frac{45}{100}\) ​ of  432

= \(\frac{45}{100}\)  × 432

& ⇒ \(\frac{9}{5}\) × 108  =  194.4

So,45% of $432 is $194.4

The solution is 194.4

 

Page 107   Exercise 3  Problem 9

Given:

Find each number. Round to the nearest tenth if necessary.

3.23% of $640

To find –The number.

Given:

23 % of 640

23 % = \(\frac{23}{100}\)

Then \(\frac{23}{100}\)  of  640

= \(\frac{23}{100}\) × 640

=  \(\frac{23}{5}\) ×  32

=  147.2

So, 23 % of $640 is $147.2

The solution is 147.2

 

Page 107  Exercise 4  Problem 10

Given:

Find each number. Round to the nearest tenth if necessary.

4.130% of 20

To find –  The number.

Given:
​
​130 % of 20

130 % = \(\frac{130}{100}\)

Then \(\frac{130}{100}\)  of  20

=  \(\frac{130}{100}\) × 20

⇒ 13 × 2 = 26

So,130 % of 20 is 26

The solution is 26

 

Page 107  Exercise 7  Problem 11

Given:

Find each number. Round to the nearest tenth if necessary.

7.32% of 4

To find  –  The number.

Given:
​
​32 %of 4

32 = \(\frac{32}{100}\)

Then, \(\frac{32}{100}\)  of  4

=  \(\frac{32}{100}\) × 4

⇒  \(\frac{32}{25}\)

= 1.28

So, 32 % of 4 is 1.28.

The solution is 1.28

 

Page 107   Exercise 9  Problem 12

Given:

Find each number. Round to the nearest tenth if necessary.

9.23.5%of 128

To find –  The number.

Given:

⇒ 23.5

Then, \(\frac{23.5}{100}\)  of  128

= \(\frac{23.5}{100}\)  × 128

=\(\frac{23.5}{25}\) × 32

=  30.08
​
So,23.5 % of 128 is 30.08

The solution is 30.08

 

Page 107  Exercise 10  Problem 13

Given:

Suppose there are 20 questions on a multiple-choice test.

If 25% of the answers are choice B, how many of the answers are not choice B?

To find the number of answers are not choice B.

Total number of questions = 20

25% of the answers are choice B

Therefore, the number of answers are choice B = 25% of 20
​
​\(\frac{32}{100}\)of 20 = \(\frac{32}{100}\) × 20

⇒ \(\frac{1}{4}\) × 20

= 5

The number of answers are choice B = 5

So, the number of answers are not choice B = 20 − 5

= 15

The number of answers are not choice B = 15

 

Page 107   Exercise 11  Problem 14

Given:

 

 

To find – The dollar amount of the group discount each student would receive at each park.

At pirate bay 20% discount of $35.95

\(\frac{20}{100}\) of 35.95 = \(\frac{20}{100}\) ×35.95

⇒  \(\frac{1}{5}\)  × 35.95

=  7.19

​At funtopia 15% discount of $29.75
​
​\(\frac{15}{100}\)of 29.75 = ​\(\frac{15}{100}\) ×29.75

⇒ ​\(\frac{3}{20}\) × 29.75 = 4.46

​At zoomland 25% discount of $38.49
​
\(\frac{25}{100}\) of 38.49 = \(\frac{25}{100}\)  × 38.49

⇒ \(\frac{1}{4}\) × 38.49

= 9.62
​
$7.19 at pirate bay,$4.46 at funtopia, and $9.62 at zoomland.

The dollar amount of the group discount each student would receive is $7.19 at pirate bay,$4.46 at funtopia, and $9.62 at zoomland.

 

Page 108   Exercise 14   Problem 15

Given:

Find each number. Round to the nearest hundredth.

5\(\frac{1}{2}\)% of 60

To find – The number.

Given:

5\(\frac{1}{2}\)% of 60

5\(\frac{1}{4}\) = \(\frac{11}{2}\)%

=  \(\frac{11/2}{100}\)

=  \(\frac{11}{200}\)

5\(\frac{1}{2}\)% of 60 = \(\frac{11}{200}\)× 60

\(\frac{11}{10}\) × 3 = 3.3

So 5\(\frac{1}{2}\)% of 60 is 3.3.

The solution is 3.3

 

Page 108  Exercise  15 Problem 16

Given:

Find each number. Round to the nearest hundredth.

20\(\frac{1}{4}\)% of 3

To find –  The number.

Given:

​20\(\frac{1}{4}\)% of 3

20\(\frac{1}{4}\) =  \(\frac{81}{4}\)%

\(\frac{81/4}{100}\) = \(\frac{81}{400}\)

​20\(\frac{1}{4}\)% of 3 = \(\frac{81}{400}\) ×  3

⇒  0.6075 ≃ 0.608

So ,  ​20\(\frac{1}{4}\)% of 3 is 0.608

The solution is 0.608

 

Page 108   Exercise 16  Problem 17

Given:

Find each number. Round to the nearest hundredth.

1,000 % of 99

To find – The number.

Given:

​1,000 % of 99

1000 % \(\frac{1000}{100}\)  = 10

1,000 % of 99 = 10 × 99

⇒  990

So, 1,000 % of 99 is 990

The solution is 990

 

Page 108   Exercise 17  Problem 18

Given:

To convert a percentage to a number 520% of 100

520% of 100
​
\(\frac{520}{100}\)(100)

= 520
​
520% of 100  = 520

The answer is 520

 

Page 108  Exercise 19  Problem 19

Given:

To convert a percentage to a number 200% of 79

200% of 79
​
\(\frac{200}{100}\)(79)

= 2(79)

= 158

200% of 79 = 158

​The answer 200% of 79 is 158

 

Page 108   Exercise 21  Problem 20

Given:

To convert a percentage to a number 0.28% of 50

0.28% of 50
​
\(\frac{0.28}{100}\)(50)

=  \(\frac{0.28}{10}\)(5)

=  \(\frac{0.25}{2}\)

=  0.14

0.28% of 50 =  0.14
​
The answer 0.28% of 50 is 0.14

 

Page 108  Exercise 23  Problem 21  

Given:

To explain the percentage has been done easier with fractions or decimal

Solution :

Its easier to use a decimal

Percent means a part of 100  x /100 = x%

If you have a decimal, just move the decimal place to the left of 2 places

If you have a fraction, try to get the denominator equal to 100, or divide it out and move the

A decimal place to the left 2 places

Decimal is easier than a fraction

The answer 0 is a decimal is easier than a fraction

 

Page 108   Exercise 24  Problem 22

Given:

To choose the correct option in how much he left to spend

Solution :

Option c is correct

(1) He will spend 18% of the repair

\(\frac{18}{100}\)(300)

= (18)(3)

= 54
​
He spends $54 on repair

(2)  20% of savings
​
​\(\frac{20}{100}\) (300)

= 60

​
(C) 35% of the canvas
​
​\(\frac{35}{100}\)(300)

= 105
​
He will spend $54 on repair and $60in savings and $105 in canvas and leaving him $81

The answer $81 option c is correct

 

Page 109  Exercise 25  Problem 23

Given:

To convert the percentage into number 54% of 85

Solution :

54% of 85
​
​\(\frac{54}{100}\)(85)

= 54(0.85)

= 45.9

54% of 85 = 45.9
​
The answer for 54% of 85 is 45.9

 

Page 109  Exercise 26  Problem 24

Given:

To convert the percentage into number 12% of $230

Solution:

12% of $230
​
​\(\frac{12}{100}\)(230)

=  0.12(230)

=  27.6

12% of $230 =  27.6
​
The answer for 12% of $230 is $27.6

 

Page 109  Exercise 27  Problem 25

Given:

To convert the percentage into number 98% of 15

Solution :

98% of 15
​​
​\(\frac{98}{100}\)(15)

=  0.98(15)

=  14.7

98% of 15 =  14.7

​The answer for 98% of 15 = 14.7

 

Page 109   Exercise 28  Problem 26

Given:

To convert the percentage into number 250% of 25

Solution :

250% of 25
​
​\(\frac{250}{100}\)(25)

=  2.5(25)

=  62.5

250% of 25 =  62.5

​The answer for 250% of 25 is 62.5

 

Page 109  Exercise 31  Problem 27

Given:

To convert the percentage into number 0.5% of 60

Solution :

0.5% of 60
​
​\(\frac{0.5}{100}\)(60)

= ​\(\frac{0.5}{10}\) (6)

= 0.3

0.5% of 60 = 0.3
​
The answer for 0.5% of 60 is 0.3

 

Page 109  Exercise 32  Problem 28

Given:

To convert the percentage into number 2.4% of 20

Solution :

2.4% of 20
​
​\(\frac{2.4}{100}\)(20)

= ​ \(\frac{2.4}{10}\)(2)

= 0.48

2.4% of 20 = 0.48
​
The answer for 2.4% of 20 = 0.48

 

Page 109   Exercise 34   Problem 29

Given:

To find – How many households watched the finals

Solution :

17.7% of 110.2

Turn percent to decimals
​
​​\(\frac{17.7}{100}\)(110.2)

= 0.177(110.2)

= 19.51

17.7% of 110.2 = 19.51
​
19.51 million households watched the finals

The answer is 19.51 million households watched the finals

 

Page 109   Exercise 35   Problem 30

Given:

To find – What will be the cost for internet access after the increase

Solution :

The amount family paid = $19

The cost will increase by 5%

5% of 19
​
​\(\frac{5}{100}\)(19)

= 0.05(19)

=  19.95

5% of 19 =  19.95

​The cost would be $19.95

The answer is the cost would be $19.95

 

Page 110  Exercise 38   Problem 31

Given:

To choose the answer for how many costumers prefer horror movies

Solution :

Total no of costumer 200

The percentage of customers who prefers horror movies is 46%

46% of 200

​\(\frac{46}{100}\)(200)

=  0.46(200)

=  92

46% of 200 =  92
​
92 customer prefer horror movies

The answer is 92 customer prefer horror movies is correct

 

Page 110  Exercise 39   Problem 32

Given:

The bar graph shows Ramirez’s family budget.

Ramirez monthly income is 3000 dollars.

To find:

By satisfying the following condition and finding which statement is true.

Solution :

The monthly income of Ramirez is $3000

 

 

1. The family budget is $ 1000 for rent.

To find – It in percent
​
​​\(\frac{1000}{3000}\) × 100

= 33.33

But the bar diagram shows 50% has been spent for rent so it is not correct.

 

2. The family budget is $ 600 for food.
​
​\(\frac{600}{3000}\) × 100

= 20

The diagram shows 20% for their food Hence it is true

 

3. The family budget is $100 more for utilities than for other

\(\frac{100}{3000}\) × 100

= 3.33

But the diagram shows more than 10 for this, Hence this is false.

 

4. The family budget is $900 more for food than for rent.
​
​\(\frac{900}{3000}\)×100

= 90
​
The bar diagram shows 10-20 % for this, hence it is false.

Answer : All of this option (2)  is 20% for their food is true.

By checking all the conditions with the bar graph, The answer option g, 20% for their food is true.

 

Page 110  Exercise 40  Problem 33

Given:

To multiply the given number

Solution :

Multiply With digits

1.7 × 54  =  91.8

The answer is 91.8

 

Page 110   Exercise 41  Problem 34

Given:

Multiply:

41.1.5 × 3.65

To multiply.

Multiplying
​
​1.5 × 3.65

= 5.475

The solution is  5.475

 

Page 110   Exercise 42   Problem 35

Given:

Multiply:

42. 49.6×2.7

To multiply.

Multiplying
​
​49.6 × 2.7

= 133.92
​
The solution is 133.92

 

Page 110   Exercise 43   Problem 36

Given:

43 .trent spent 50 minutes at the neighbor’s house.

He spent 2/5 of the time swimming.

How many minutes did trent spend swimming?

To find the time trent spends swimming.

Total time trent spent in neighbor’s house = 50 minutes

Time he spent in swimming = \(\frac{2}{5}\) of 50 minutes

⇒ \(\frac{2}{5}\)  × 50  =  20

The time trent spent in swimming = 20 minutes.

The time trent spent in swimming = 20 minutes.

 

Page 110  Exercise 44   Problem 37

Given:

There are 240 seventh-graders at Yorktown middle school.

Two-thirds of the students participate in after-school activities.

To find –  The number of students who participate in after-school activities.

Number of seventh-graders at Yorktown middle school = 240

Students participate in after-school activities = \(\frac{2}{3}\) of 240

⇒ \(\frac{2}{3}\) × 240 = 160

The number of students participated in after-school activities = 160

The number of students participated in after-school activities = 160

Glencoe Math Course 2 Volume 1 Common Core  Chapter 3 Integers

 

Page 191  Exercise 1   Problem 1

Because the first three operations close the set of integers:
Add
Subtract
Multiply

These operations will return a set of integers as a result.

When you divide two numbers, though, the result is when the first and second integers are not multiples of each other (In other words, when the second integer is not a factor of the first integer), Then, rather than an integer, the outcome will be the Ratio between the two integers.

Hence the term “rational” for such numbers.

Finally, we concluded that when we add, subtract, and multiply will return a set of integers as a result. and when we divide integers the outcome will be the Ratio between the two integers. Hence the term “rational” for such numbers.

 

Page 191  Exercise 1  Problem 2

The bottom of a snowboarding half-pipe is 5 meters below the top, Circle the integer you would you use to represent this position 5 (or)(−5)

In weather forecasting, negative numbers are used to show the temperature of a region.

On the Fahrenheit and Celsius scales, negative numbers are used to represent the temperature.

Finally, we concluded that the result to represent this position is (−5)

 

Page 194  Exercise 1  Problem 3

Given: $16

To write an integer.

A deposit increases the balance and thus it is best to take a positive integer

Therefore, the positive integer will be

⇒ 16

Finally, we concluded that the result is  ⇒  16

 

Page 194  Exercise 2   Problem 4

Given: A loss of 11 yards

To write an integer.

A loss of 11 yards is a decrease of the number of yards and thus it is best to take a negative integer.

Therefore, the negative integer will be

⇒  −11

A loss of 11 yards = −11

Finally, we concluded that the result is ⇒ −11

 

Page 194  Exercise 4  Problem 5

Given:−9=____

To evaluate the expression.

The absolute value of a number is the distance from the number to zero.

Therefore, the absolute value is

⇒ −9 = 9

Finally, we concluded that the result is ⇒ 9

 

Page 194  Exercise 5  Problem 6

Given: |18|−|−10| =____

To evaluate the expression.

The absolute value of a number is the distance from the number to zero.

Therefore, the absolute value is

​|18|−|−10| = 18 − 10

=8

​|18|−|−10| =8
​
The value of  |18|−|−10| is 8

 

Page 194  Exercise 7  Problem 7

Given:{11,−5,−8}

To Graph the set of integers on a number line.

The given set is {11,−5,−8}

Here,11 Lies in the middle between 10 and 12, while−5 lies in the middle between−4 and−6 and −8 lies in the middle between−7 and−9

The graph on a number line

 

 

The given set of integers {11,−5,−8} is graphed on a number line.

 

Page 194  Exercise 8  Problem 8

The absolute value of a number is positive because the distance of the number is always positive or zero if the number is zero.

Finally, we concluded that the absolute value of a number is positive since the distance is positive.

 

Page 195  Exercise 1  Problem 9

Given: A profit of $9

To write an integer.

A profit is an amount that was gained and thus it is best to use a positive integer.

Therefore, the positive integer will be

⇒ 9

Finally, we concluded that the result is  ⇒  9

 

Page 195  Exercise 2  Problem 10

Given: A bank withdrawal of $50

To write an integer.

A bank withdrawal decreases the balance of the bank account and thus it is best to use a negative integer.

Therefore, the negative integer will be

⇒ −50

Finally, we concluded that the result is ⇒ −50

 

Page 195   Exercise 3  Problem  11

Given: $53 below zero

To write an integer.

The temperature below zero is best described by a negative number.

Therefore, the negative integer will be

⇒ −53

Finally, we concluded that the result is ⇒−53

 

Page 195   Exercise 4  Problem 12

Given: 7 inches more than normal

To write an integer.

The number of inches more than normal is best described by a positive integer.

Therefore, the positive integer will be

⇒  7

Finally, we concluded that the result is ⇒ 7

 

Page 195  Exercise 7  Problem 13

Given: 10 =____

To evaluate the expression.

The absolute value of a number is the distance from the number to zero.

Therefore, the absolute value is

⇒  10 = 10

Finally, we concluded that the result is ⇒ 10

 

Page 195  Exercise 8  Problem 14

Given:−7−5 =____

To evaluate the expression.

The absolute value of a number is the distance from the number to zero.

Therefore, the absolute value is

​⇒ −7 − 5 = 7 − 5 = 2

⇒  2
​
Finally, we concluded that the result is ⇒ 2

 

Page 196  Exercise 13  Problem 15

Given: |−199.99|+|−39.99|+|−59.99|

To find the amount Mr. Chavez spent altogether.

The absolute value of a number is the distance of the number to zero.

Therefore, the absolute value is

​|−199.99| + |−39.99| + |−59.9|

= 199.99 + 39.99+59.99

= $299.97

|−199.99|+|−39.99|+|−59.99| = $299.97
​
Mr. Chavez spends ⇒ $299.97 altogether.

 

Page 196  Exercise 14   Problem 16

Given: x = 3

To find the value of x?

The absolute value of a number is the distance of the number to zero.

Hence, x has to be  3 or −3 (since 3 and −3 lie a distance of 3 from zero).

Therefore, the absolute value is

⇒  3  or −3

Finally we concluded that the value of x ⇒ 3 or − 3

 

Page 196  Exercise 15  Problem 17

The inequality is always true Because the absolute value of a number is positive.

Finally, we concluded that inequality is always true.

 

Page 196  Exercise 16  Problem 18

Given:

To find the expression that is not equal to the other three.

The absolute value of a number is the distance of the number to zero.

Evaluating each expression:

​15 − |−5| = 15 − 5

= 10

|−4| + 6 = 4 + 6

= 10

− |7 + 3|  = −10

|−10| = 10
​
Therefore, the expression does not equal the other three is

⇒ −|7 + 3| = −10

−|7+3| is the expression that does not equal the other three expressions.

 

Page 196  Exercise 17  Problem 19

Given:

1. − 11 °F

2. −10 °F

3. 10 °F

4. 11 °F

To find Which integer represents the temperature shown on the thermometer?

The integer−11 (thus below zero)is denoted on the thermometer and thus the temperature shown is −11 °F

Therefore, the integer represents the temperature shown on the thermometer A.(−11°F)

Finally, we concluded that the integer represents the temperature shown on the thermometer. (−11°F)

 

Page 197  Exercise 18  Problem 20

Given:

2 feet below flood level

To Write an integer for the situation?

The number of feet below flood level is represented by a negative integer

Therefore, the negative integer is −2

Finally, we concluded that the result is ⇒ −2

 

Page 197   Exercise 19  Problem 21

Given:

An elevator goes upto 12 floors

To Write an integer for the situation

The elevator goes up. Hence, it is represented by a positive integer.

Therefore, the positive integer is 12

Finally, we concluded that the result is  ⇒ 12

 

Page 197  Exercise 20  Problem 22

Given: {3,−7,6}

Graph the given set of integers on a number line.

The graph of the number is the point on this line that corresponds to each number. 3Lies in the middle between 2and4

While−7lies in the middle between−6 and−8,and 6lies in the middle between 5 and 7

 

 

The given set of integers {3,−7,6} is marked on a number line.

 

Page 197  Exercise 21   Problem 23

Given: (−2,−4,−6,−8)

To Graph the set of integers on a number line?

− 2 Lies in the middle between −1 and−3.While−4 lies in the middle between−3 and−5 , and −6 lies in the middle between−5 and−7,−8 Lies in the middle between −7 and -9

 

Finally, we Graph the set of integers on a number line.

 

Page 197  Exercise 22  Problem 24

Given: −12=____

To evaluate the expression.

The absolute value of a number is the distance from the number

The distance cannot be negative.

Therefore, the absolute value is
​
​⇒  12
​
Finally, we concluded that the result is  ⇒  12

 

Page 197  Exercise 23  Problem 25

Given: 7+4=____

To evaluate the expression

The absolute value of a number is the distance from the number to zero.

Therefore, the absolute value is

​⇒   7 + 4 = 7 + 4

⇒ 11
​
Finally, we concluded that the result is ⇒ 11

 

Page 197   Exercise 24   Problem 26

Given: |−9| + |−5| =____

Consider the given operations and evaluate the expression.

The absolute value function determines a number’s magnitude regardless of its sign.
​
​|−9| + |−5| = 9 + 5

= 14

​|−9| + |−5| = 14
​
The value of ∣−9∣+∣−5∣ is 14

 

Page 197   Exercise 25  Problem 27

Given: |−10| ÷ 2 × |5| = ____

Consider the signs and evaluate the expression.

The absolute value function determines a number’s magnitude regardless of its sign.

​|−10| ÷ 2 × |5| = 10 ÷ 2 × 5

= 5 × 5

= 25

​|−10| ÷ 2 × |5| = 25
​
The value of |−10| ÷ 2 × |5| is 25

 

Page 197  Exercise 26  Problem 28

Given: 12 − |−8| + 7 =____

Consider the signs and evaluate the expression.

The absolute value function determines a number’s magnitude regardless of its sign.

​​12−|−8| + 7 = 12 − 8 + 7
​
=  4 + 7

= 11
​
The value of 12−|−8| + 7 is 11

 

Page 197  Exercise 27  Problem 29

Given: 27 ÷ 3−4 = ____

To evaluate the expression.

The absolute value of a number is the distance from the number to zero.

Therefore, the absolute value is
​
​⇒  27÷3−4 = 27 ÷ 3 − 4 = 9 − 4 =5

⇒ 5

​Finally, we concluded that the result is ⇒ 5

 

Page 197   Exercise 28  Problem  30

Given: Jasmine’s pet guinea pig gained 8 ounces in one month

To Write an integer to describe the amount of weight her pet gained.

Since the weight increased, it is described the amount of weight gained by a positive integer

Therefore, the positive integer is

⇒ 8

Finally, we concluded that the result is ⇒ 8

 

Page 198   Exercise 30  Problem  31

1. A $100 check deposited in a bank can be represented by  + 100.

2. loss of 15 yards in a football game can be represented by − 15

3. A temperature of 20 below zero can be represented by −20.

4. A submarine diving 300 feet underwater can be represented by + 300

To find Which of the following statements about these real-world situations is not true?

A situation below or under a level is best represented by a negative integer, while a situation above or over a level is represented by a positive Integer.

Therefore 4. A submarine diving 300 feet underwater can be represented by + 300 is not true.

Therefore, the statement 4 Option is not true.

Finally, we concluded that the result is  ⇒ 4 Option

 

Page 198  Exercise 31  Problem  32

Given:

 

To find which day was the low temperature the farthest from 0°F?

The temperature farthest from 0°Fis the temperature of −8°F on Wednesday

Therefore, the lowest temperature was on Wednesday

Finally, we concluded that the low temperature the farthestfrom0∘Fis on Wednesday at −8°F

 

Page 198  Exercise 32  Problem  33

Given:

To Write the ordered pair corresponding to each point graphed.

The first number is the x-coordinate, while the second number is the y-coordinate.

Therefore J(−2,4)

Finally, we concluded that the ordered pair corresponding to each point was graphed. ⇒ J(−2,4)

 

Page 198  Exercise 33  Problem  34

Given:

To Write the ordered pair corresponding to each point graphed.

The first number is the x-coordinate, while the second number is the y-coordinate.

Therefore K(0,2)

Finally, we concluded that the ordered pair corresponding to each point was graphed.  ⇒ K(0,2)

 

Page 198  Exercise 34   Problem  35

Given:

To Write the ordered pair corresponding to each point graphed.

The first number is the x-coordinate, while the second number is the y-coordinate.

Therefore L(−3,−1)

Finally, we concluded that the ordered pair corresponding to each point was graphed. ⇒ L(−3,−1)

 

Page 198  Exercise 35  Problem 36

Given:

To Write the ordered pair corresponding to each point graphed.

The first number is the x-coordinate, while the second number is the y-coordinate.

Therefore M(1,1)

Finally, we concluded that the ordered pair corresponding to each point was graphed. ⇒ M(1,1)

 

Page 198  Exercise 36  Problem 37

Given: A(2,4)

To Graph and label each point on the coordinate plane.

The first number is the x-coordinate, while the second number is the y-coordinate.

The required graph is:

 

 

Finally, we Graph and label each point on the coordinate plane

 

Page 198  Exercise 37  Problem 38

Given: B(−3,1)

To Graph and label each point on the coordinate plane.

The first number is the x-coordinate, while the second number is the y-coordinate.

 

 

Finally, we Graph and label each point on the coordinate plane

 

Page 198  Exercise 38  Problem 39

Given: C(2,0)

To Graph and label each point on the coordinate plane.

The first number is the x-coordinate, while the second number is the y-coordinate

The required graph is:

 

 

Finally, we plotted Graph and labeled each point on the coordinate plane

 

Page 198   Exercise 39  Problem 40

Given:D(−3,−3)

To Graph and label each point on the coordinate plane.

The first number is the x-coordinate, while the second number is the y-coordinate.

The required graph is:

 

 

Finally, we Graph and label each point on the coordinate plane

 

Page 201  Exercise 1  Problem 41

Given:

5 + 6 =_____

To Find each sum. Show your work using drawings.

Given equation is
​
​5 + 6 = 11

⇒ 11
​

Drawings:

Finally, we find the sum ⇒ 11

 

Page 201  Exercise 3  Problem 42

Given:

−5 + (−4) =_____

To Find each sum. Show your work using drawings.

Given equation is
​
​−5 + (−4) = −9

⇒ −9

Drawings:

Finally, we find the sum  ⇒ −9

 

Page 201  Exercise 4  Problem 43

Given:

7 + 3 = _____

To Find each sum. Show your work using drawings.

Given equation is

​7 + 3 = 10

⇒ 10

​Drawings:

 

Finally, we find the sum  ⇒ 10

 

Page 201  Exercise 6  Problem  44

Given:

−2 + 7 =_____

To Find each sum. Show your work using drawings.

−2 + 7 =  5

⇒  5

Drawings:

Finally, we find the sum  ⇒ 5

 

Page 201  Exercise 7  Problem  45

Given:

8 + (−3) =_____

To Find each sum. Show your work using drawings.

Given equation is

​8 + (−3) = 5

⇒ 5

​Drawings:

Finally, we find the sum ⇒ 5

 

Page 201  Exercise 8  Problem 46 

Given:

3 + (−6)=_____

To Find each sum. Show your work using drawings.

Given equation is
​
​3 + (−6) =−3

⇒ −3
​

Drawings:

Finally, we find the sum  ⇒−3

 

Page 202  Exercise 10  Problem  47

Given:

To complete the table.

The given equation is

​7 + (−12) = − 5

⇒ −5
​

Sign of addend with greatest absolute value: Negative

Sign of sum: Negative

 

Finally, we complete the table.

 

Page 202  Exercise 11  Problem 48

Given: 

To complete the table.

Given equation is

​−4 + 9 = 5

⇒  5
​

Sign of addend with greatest absolute value: Positive

Sign of sum: Positive

 

Finally, we complete the table.

 

Page 202  Exercise 12  Problem 49 

Given:

To complete the table.

-12 + 20 =  8

⇒  8

Sign of addend with greatest absolute value: Positive

Sign of sum: Positive

 

Finally, we complete the table.

 

Page 202  Exercise 13  Problem  50

Given:

To complete the table

The given expression is

​15 + (−18) = −3

⇒ −3
​
Sign of addend with greatest absolute value: Negative

Sign of sum: Negative

 

Finally, we complete the table.

 

Page 202  Exercise 17  problem 51

Given:

First Round = −100

Second Round = −250

Third Round = 500

To Find the contestant’s total number of points. Explain your reasoning

The total number of points is the sum of all points.

The equation is

​⇒ −100 −250 + 500

⇒  (−350) + 500

⇒ 150
​
Finally, we concluded that the contestant’s total number of points  ⇒ 150 points

 

Page 202  Exercise 18  Problem  52

The sum of two integers is Negative, If

Both integers are negative

One integer is negative and the other is positive, while the absolute value of the negative integer is more than the absolute value of the positive integer.

Finally, we concluded that the sum of two integers is Negative. If Both integers are negative or One integer is negative and the other is positive, while the absolute value of the negative. An integer is more than the absolute value of the positive integer.