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

Glencoe Math Course 2 Volume 1 Chapter 1 Practice Exercise 1.2 Solutions Page 21   Exercise 12   Problem 1

Given that, the percent is 13\(\frac{1}{3}\) %

We need to write the given percent as a fraction and simplify it.

Given percent is 13\(\frac{1}{3}\) %

Converting the given mixed fraction into an improper fraction, we get

13\(\frac{1}{3}\) = \(\frac{40}{3}\) %

Thus, the percentage becomes

\(\frac{40}{3}\) percent = \(\frac{40}{3}\)÷ \(\frac{100}{1}\)

Multiply it by the reciprocal of \(\frac{100}{1}\) , we get

\(\frac{40}{3}\) percent = \(\frac{40}{3}\)÷ \(\frac{100}{1}\)

= \(\frac{40}{3}\)× \(\frac{1}{100}\)

= \(\frac{4}{3}\)× \(\frac{1}{10}\)

= \(\frac{2}{3}\)× \(\frac{1}{5}\)

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

The given percent 13\(\frac{1}{3}\) % as a fraction in the simplest form is \(\frac{2}{15}\)

Read and Learn More Glencoe Math Course 2 Volume 1 Common Core Student Edition Solutions

Glencoe Math Course 2 Volume 1 Page 22   Exercise 15   Problem 2

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

For example: 

\(\frac{\frac{5}{10}}{2}, \frac{11}{\frac{6}{5}}, \frac{\frac{13}{12}}{\frac{9}{8}}\). These are all complex fractions.

Solving a complex fraction:

Let us consider a complex fraction that involves ratios

Simplifying it, we get

​\(\frac{\frac{5}{2}}{10}\)

= \(\frac{5}{2}\) × \(\frac{1}{10}\)

= \(\frac{1}{2}\) × \(\frac{1}{2}\)

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

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.

Glencoe Math Course 2 Volume 1 Page 22   Exercise 16   Problem 3

We need to write three different complex fractions that simplify to \(\frac{1}{4}\)

So, we will use the definition

The three different complex fractions and their simplification is given below

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

Simplifying it, we get

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

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

2)\(\frac{\frac{5}{4}}{5}\)

Simplifying it, we get

\(\frac{5}{4}\)× \(\frac{1}{5}\)

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

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

Simplifying it, we get

\(\frac{\frac{1}{3}}{\frac{4}{3}}\) = \(\frac{1}{3}\)×\(\frac{3}{4}\)

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

The three different complex fractions that simplify to \(\frac{1}{4}\) are \(\frac{\frac{1}{2}}{2}\), \(\frac{\frac{5}{4}}{5}\), \(\frac{\frac{1}{3}}{\frac{4}{3}}\)

Glencoe Math Course 2 Volume 1 Page 22   Exercise 17   Problem 4

We need to determine the value of ,\(\frac{15}{124} \cdot \frac{230}{30} \div \frac{230}{124} \)

Simplifying it, we get

\(\frac{15}{124} \cdot \frac{230}{30} \div \frac{230}{124} \) = \(\frac{15}{124} \times \frac{230}{30} \times \frac{124}{230}\)

= \(\frac{15}{1}\) × \(\frac{1}{30}\) × \(\frac{1}{1}\)

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

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

The value of \(\frac{15}{124} \cdot \frac{230}{30} \div \frac{230}{124} \) = \(\frac{1}{2}\)

Common Core Chapter 1 Ratios and Proportional Reasoning Practice Exercise 1.2 answers Page 22   Exercise 18   Problem 5

We need to determine which statement explains how to use the model to simplify the complex fraction.

Count the twelfths that fit within  \({2}{3}\) of the above equation

There are 12 number of  \(\frac{1}{12}\)‘s are there.

Thus , \(\frac{2}{3}\) rd of the are

\(\frac{2}{3}\) ×12 =  2 × 4 = 8

The result of the given complex fraction is

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

= \(\frac{2}{1}\)×\(\frac{4}{1}\)

= 8

Thus, both are the same. Hence, this statement is correct.

Count the twelfths that fit within \(\frac{2}{3}\). This statement explains how to use the model to simplify the complex fraction.

Glencoe Math Course 2 Volume 1 Page 23   Exercise 20   Problem 6

The objective is to Simplify the value of \(\frac{12}{\frac{3}{5}}\)

We will use the definition.

Given is \(\frac{12}{\frac{3}{5}}\)

Simplifying it, we get

\(\frac{12}{\frac{3}{5}}\) = 12  ÷  \(\frac{3}{5}\)

=  12  ×  \(\frac{5}{3}\)

=  4  ×  5

=  20

The value of \(\frac{12}{\frac{3}{5}}\)  =  20

Step-By-Step Guide For Glencoe Math Practice Exercise 1.2 Chapter 1 Problems Page 23 Exercise 21 Problem 7

We are given \( \frac{\frac{9}{10}}{9}\)

To Find: The objective is to simplify the given fraction \( \frac{\frac{9}{10}}{9}\)

Given is \( \frac{\frac{9}{10}}{9}\)

Simplifying it, we get

\( \frac{\frac{9}{10}}{9}\) = \(\frac{9}{10}\)÷ 9

= \(\frac{9}{10}\) × \(\frac{1}{9}\)

= \(\frac{1}{10}\)

The value of \( \frac{\frac{9}{10}}{9}\)= \(\frac{1}{10}\)

Exercise 1.2 practice solutions for Glencoe Math Course 2 Chapter 1 Ratios And Proportional Reasoning Page 23   Exercise 22   Problem 8

Simplify the value of \(\frac{\frac{1}{2}}{\frac{1}{4}}\)

Given is \(\frac{\frac{1}{2}}{\frac{1}{4}}\)

Simplifying it, we get

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

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

= 2

The value of \(\frac{\frac{1}{2}}{\frac{1}{4}}\)= 2

Glencoe Math Course 2 Volume 1 Page 23 Exercise 23 Problem 9

The objective is to Simplify the value of \(\frac{\frac{1}{12}}{\frac{5}{6}}\)

We will use the Definition.

Given \(\frac{\frac{1}{12}}{\frac{5}{6}}\)

Simplifying it, we get

\(\frac{\frac{1}{12}}{\frac{5}{6}}\)= \(\frac{1}{12}\) × \(\frac{6}{5}\)

= \(\frac{1}{2}\) × \(\frac{1}{5}\)

= \(\frac{1}{10}\)

Hence, The value of \(\frac{\frac{1}{12}}{\frac{5}{6}}\) = \(\frac{1}{10}\) 

Common Core Practice Exercise 1.2 Ratios and Proportional Reasoning Glencoe Math Course 2 Page 23  Exercise 24  Problem 10

Simplify the value of \(\frac{\frac{5}{6}}{\frac{5}{9}}\)

Given is \(\frac{\frac{5}{6}}{\frac{5}{9}}\)

Simplifying it, we get

\(\frac{\frac{5}{6}}{\frac{5}{9}}\) = \(\frac{5}{6}\)× \(\frac{9}{5}\)

= \(\frac{9}{6}\)

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

The value of \(\frac{\frac{5}{6}}{\frac{5}{9}}\) = \(\frac{3}{2}\)

Examples Of Problems From Practice Exercise 1.2 Ratios And Proportional Reasoning Chapter 1 Glencoe Math Page 23   Exercise 25   Problem 11

Given that, Mrs. Frasier is making costumes for the school play. Each costume requires 0.75 yards of fabric. She bought 6 yards of fabric.

We need to determine how many costumes Mrs. Frasier can make.

The number of yards needed to make one costume =  0.75

Total yards of fabric bought  =  6

The ratio is, \(\frac{6}{0.75}\)

=  8

She can make eight costumes out of it.

Hence, Mrs. Frasier can make 8 costumes.

Glencoe Math Course 2 Volume 1 Page 23 Exercise 26 Problem 12

Given that, A lawn company advertises that they can spread 7,500 square feet of grass seed in 2\(\frac{1}{2}\) hours.

We need to determine the number of square feet of grass seed that can be spread per hour.

Given that
Square feet of grass that they can spread = 7500

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

The unit rate is

Unite rate \( = \frac{7500}{2 \frac{1}{2}} \)

=  \(\frac{7500}{\frac{5}{2}}\)

=  7500  ×  \(\frac{2}{5}\)

= 1500 × 2

=  3000

Hence, The number of square feet of grass seed that can be spread per hour is 3000 square feet.

Glencoe Math Student Edition Chapter 1 Practice Exercise 1.2 Answers Guide Page 23   Exercise 28   Problem 13

Given that, the percent is 7 \(\frac{3}{4}\)%

We need to write the given percent as a fraction and simplify it.

Given percent is 7 \(\frac{3}{4}\)%

Converting the given mixed fraction into an improper fraction, we get

7 \(\frac{3}{4}\) = \(\frac{31}{4}\)%

Thus, the percent becomes

\(\frac{31}{4}\) percent = \(\frac{31}{4}\) ÷ \(\frac{100}{1}\)

Multiply it by the reciprocal of \(\frac{100}{1}\) , we get

\(\frac{31}{4}\) percent = \(\frac{31}{4}\) ÷ \(\frac{100}{1}\)

\(\frac{31}{4}\) percent = \(\frac{31}{4}\)×\(\frac{100}{1}\) we, get

= \(\frac{31}{400}\)

The given percent 7\(\frac{3}{4}\)% as a fraction in the simplest form is \(\frac{31}{400}\)

Glencoe Math Course 2 Volume 1 Page 23   Exercise 30    Problem 14

The value of a certain stock increased by 1 \(\frac{1}{4}\)%

We need to explain how to write it as a fraction in the simplest form

Given percent is 1\(\frac{1}{4}\)%

Converting it into fractions, we get

1\(\frac{1}{4}\) percent = \(\frac{5}{4}\)percent

= \(\frac{\frac{5}{4}}{100}\)

= \(\frac{5}{4}\)×\(\frac{1}{100}\)

= \(\frac{1}{4}\) × \(\frac{1}{20}\)

= \(\frac{1}{80}\)

Hence, The value of \(\frac{1}{4}\) percent = \(\frac{1}{80}\)

Chapter 1 Practice Exercise 1.2 Glencoe Math Course 2 Step-By-Step Solutions Page 24   Exercise 31    Problem 15

Given that, Debra can run 20\(\frac{1}{2}\) miles in 2 \(\frac{1}{4}\)

We need to determine how many miles per hour she can run.

Given that = 20\(\frac{1}{2}\)

= \(\frac{41}{2}\) miles

Time taken ​= 2 \(\frac{1}{4}\)

= \(\frac{9}{4}\)

​The unit rate is

Unite rate = \(\frac{\text { Number of miles }}{\text { Time taken }}\)

=  \(\frac{\frac{41}{2}}{\frac{9}{4}}\)

=  \(\frac{41}{2}\) × \(\frac{4}{9}\)

=  41 × \(\frac{2}{9}\)

= \(\frac{82}{9}\)

= 9 \(\frac{1}{9}\) miles per hour

She can run 9 \(\frac{1}{9}\) miles per hour

Page 24   Exercise 32   Problem 16

Among the given options.

We need to determine which of the given complex fraction is equivalent to the value

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

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

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

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

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

= 1

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

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

= 1

4) \(\frac{\frac{1}{8}}{\frac{1}{2}}\)= \(\frac{1}{8}\) × \(\frac{2}{1}\)

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

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

Here, Option (1) is equal to \(\frac{1}{2}\)

Hence, Option (1) \(\frac{\frac{1}{4}}{\frac{1}{2}}\) is equivalent to \(\frac{1}{2}\)

Glencoe Math Course 2 Volume 1 Page 24   Exercise 35   Problem 17

We need to determine how many inches does two feet measure using the equivalent customary measurement.

We will use the definitions

We know that the equivalent customary measurement of foot and inches will be

1 foot  =  12 inches

We need to calculate how many inches are there for two feet.

Thus

​2×1 foot  = 2 × 12 inches

2 feet  =  24 inches
​
Hence, 2 feet  = 24 inches

Page 24   Exercise 37  Problem 18

We need to determine how many quarts does 8 gallons measure using the equivalent customary measurement.

We know that the equivalent customary measurement of gallons and quarts will be

1 gallon  = 4.8038 quarts

We need to calculate how many quarts are there for 8 gallons.

Thus

​8 × 1 gallon  = 8 × 4.8038 quarts

8 gallons = 38.4304 quarts
​
8 gallons = 38.4304 quarts

Page 24    Exercise 40   Problem 19

We need to determine how many grams does one-kilogram measures using the equivalent metric measurement.

We need to calculate how many grams are there in one kilogram.

We know that kilo refers to the number 1000

According to the equivalent metric measurement of kilograms and grams

1 kilogram = 1000 grams

1 kilogram = 1000 grams

Glencoe Math Course 2 Volume 1 Page 24   Exercise 40   Problem 20

We need to determine how many grams does one-kilogram measures using the equivalent metric measurement.

We need to calculate how many grams are there in one kilogram.

We know that kilo refers to the number 1000

According to the equivalent metric measurement of kilograms and grams

1 kilogram = 1000 grams

1 kilogram = 1000 grams

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

Glencoe Math Course 2 Volume 1 Chapter 1 Exercise 1.3 Solutions Page 25   Exercise 2   Problem 1

We need to determine the given calculations using the equivalent metric measurement.

1 minute = _____ seconds

1 hour = ______ seconds

We need to calculate how many seconds are there in one minute.

According to the equivalent metric measurement of minutes and seconds, we get

1 minute = 60 seconds

Thus, for one hour

We know that

​1 hour  =  60 minutes

=  60 × 1 minute

=  60 × 60 seconds

=  3600 seconds
​
1 minute = 60 seconds

1 hour = 3600 seconds

Page 25  Exercise 3  Problem 2

Given:

We need to determine the number of feet per second a squirrel can run.

Read and Learn More Glencoe Math Course 2 Volume 1 Common Core Student Edition Solutions

From the given table, the squirrel can run 10 mph Thus

Speed  =  \(\frac{10 \text { miles }}{1 \text { hour }}\)

We know that

​1 mile = 5280 feet

10 miles = 52800 feet
​
Similarly

1 hour = 60 minutes

=  60 × 60 seconds

=  3600 seconds
​
The speed of the squirrel be

Speed  \(=\frac{52800 \text { feet }}{3600 \text { seconds }}\)

=  14. 6667 feet per second

≈  14.67 feet per second

A squirrel can run 14.67 feet per second.

Glencoe Math Course 2 Volume 1 Common Core Chapter 1 Page 25   Exercise 4   Problem 3

We need to determine the number of feet per second does it measure for 10 miles per hour

Given that, 10 miles per hour.

We know that

​1 mile = 5280 feet

10 miles = 52800 feet

Also, by converting hour to second, we get

​1 hour = 60 minutes

= 60×60 seconds

= 3600 seconds

Therefore

= 14.667 feet per second

≈ 14.67 feet per second.
​
10 , miles per hour = 14.67 feet per second.

The number of feet per second  does it measure for 10 miles per hour is 14.67 feet per second.

Common Core Student Edition Ratios And Proportional Reasoning Exercise 1.3 Answers Page 28   Exercise 2   Problem

Given that, A skydiver is falling at about 176 feet per second.

We need to determine how many feet per minute is he falling.

Given falling speed is 176 feet per second.

We know that 1 minute  =  60 seconds

Converting the given, we get

\(\frac{176 \text { feet }}{1 \text { second }}=\frac{176 \text { feet }}{1 \text { second }} \times \frac{60 \text { second }}{1 \text { minutes }}\)

=  \(\frac{176 \times 60 \text { feet }}{1 \text { minutes }}\)

=  10560 feet per minute

He is falling at 10560 feet per minute.

Glencoe Math Course 2 Volume 1 Common Core Chapter 1 Page 28   Exercise 4 Problem 4

Given that, the ratio of\(\frac{3 \text { feet }}{1 \text { yard }}\) has a value of one.

We need to determine how it is equal to one.

Given that the ratio is having two different units.

Converting the two units into one.

We know that 1 yard = 3 feet

Thus, the given ratio becomes

\(\frac{3 \text { feet }}{1 \text { yard }}=\frac{1 \text { yard }}{1 \text { yard }}\) = 1

Thus, the value of the ratio is one.

Hence, The ratio \(\frac{3 \text { feet }}{1 \text { yard }}\) = 1 since the value of 3 feet = 1 yard

Chapter 1 Exercise 1.3 Glencoe Math Course 2 Volume 1 Workbook Solutions Page 29   Exercise 1   Problem 5

Given that, A go-kart’s top speed is 607,200 feet per hour.

We need to determine its speed in miles per hour.

The speed of the go-kart = \(\frac{607,200 \text { feet }}{1 \text { hour }}\)

We must convert the speed in feet to miles.

We know that 1 feet  = 0.000189394 miles

Thus, by converting, we get

= \(\frac{607200 \text { feet }}{1 \text { hour }} \times \frac{0.000189394 \text { miles }}{1 \text { feet }}\)

=\(\frac{607200 \times 0.000189394 \text { miles }}{1 \text { hour }}\)

=  115 miles per hour

Hence, The speed is 115 miles per hour.

Glencoe Math Course 2 Volume 1 Common Core Chapter 1 Page 29   Exercise 2   problem 6

Given that, the fastest a human has ever run is 27 miles per hour.

We need to determine how many miles per minute the human ran.

Given:

The fastest speed of the human = \(\frac{27 \text { miles }}{1 \text { hour }}\)

Converting the hour into minutes, we get

1 hour = 60 minutes

Thus, the speed becomes

=  \(\frac{27 \text { miles }}{60 \text { minutes }}\)

=  \(\frac{0.45 \text { miles }}{1 \text { minute }}\)

The human ran 0.45 miles per minute.

Glencoe Math Chapter 1 Ratios And Proportional Reasoning Exercise 1.3 Explanation Page 29   Exercise 3   Problem 7

Given that, A peregrine falcon can fly 322 kilometers per hour.

We need to determine how many meters per hour the falcon can fly.

Given:

Speed of the falcon = \(\frac{322 \text { Kilometers }}{1 \text { hour }}\)

Converting kilometers to meters, we get

=  \(\frac{322 \text { kilometers }}{1 \text { hour }}=\frac{322 \text { kilometers }}{1 \text { hour }} \times \frac{1000 \text { meters }}{1 \text { kilometer }}\)

=  \(\frac{322 \times 1000 \text { meters }}{1 \text { hour }}\)

=  \(\frac{322 \text { meters }}{1 \text { hour }}\)

The falcon can fly 322000 meters per hour.

Glencoe Math Course 2 Volume 1 Common Core Chapter 1 Page 29    Exercise 4   Problem 8

Given that, A pipe is leaking at 1.5 cups per day.

We need to determine how many gallons per week the pipe is leaking.

Also, given 1 gallon is equivalent to 16 cups.

Determine one cup is equivalent to how many gallons.

​1 gallon = 16 cups

\(\frac{1}{16}\) gallon = 1 cup
​

Determine one day is equivalent to how many weeks

​1 week = 7 days

\(\frac{1}{7}\) week = 1 day
​
Converting the given cups to gallons, we get
​
\(\frac{1.5 \text { cups }}{1 \text { day }}=\frac{1.5 \text { cups }}{1 \text { day }} \times \frac{\frac{1}{16} \text { gallon }}{1 \text { cup }} \times \frac{1 \text { day }}{\frac{1}{7} \text { week }}\)

=  \(\frac{1.5 \times \frac{1}{16} \text { gallons }}{\frac{1}{7} \text { week }}\)

=  \(\frac{0.09375 \text { gallons }}{\frac{1}{7} \text { week }}\)

= 0.65625 gallons per week
​
The pipe is leaking at 0.65625 gallons per week.

Student Edition Exercise 1.3 Ratios And Proportional Reasoning Glencoe Math Course 2 Page 30    Exercise 7    Problem 9 

Given that, the speed at which a certain computer can access the Internet is 2 megabytes per second.

We need to determine how fast is this in megabytes per hour.

Given:

Speed of the computer = \(\frac{2 \text { megabytes }}{1 \text { second }}\)

We know that

​1 hour = 3600 seconds

\(\frac{1}{3600}\) hour = 1 second
​
Converting we get

\(\frac{2 \text { megabytes }}{1 \text { second }}\) = \(\frac{2 \text { megabytes }}{1 \text { second }} \times \frac{1 \text { second }}{\frac{1}{3600} \text { hour }}\)

= \(\frac{2 \text { megabytes }}{\frac{1}{3600} \text { hour }}\)

= 7200 megabytes per hour

The computer can access the internet in 7200 megabytes per hour.

Glencoe Math Course 2 Volume 1 Common Core Chapter 1 Page 30   Exercise 8   Problem 10

Given that, the approximate metric measurement of length is given for the U.S.

a customary unit of length. We need to use our estimation skills to complete the graphic organizer below.

Fill in each blank with the foot, yard, inch, or mile.

We know that the metric to customary measurements are
​1 inch    =    2.54 centimeters
1 feet    =    0.30 meter
1 yard   =    0.91 meter
1 miles  =   1.61 kilometer
​

Thus

Hence, The Complete Figure is given below:

Ratios And Proportional Reasoning Glencoe Math Course 2 Chapter 1 Exercise 1.3 Guide Page 30  Exercise 9  Problem 11

The unit rate is nothing but the rate per one unit of the material or a thing.

For example, if we come across a grocery store in the market and if we want to buy five kilograms of tomato, the shopkeeper tells us that the price of the tomatoes per kilogram will be $5

In this way, we can easily calculate how much the price will be if we want to buy more kilograms of tomatoes.

In this case, we need to buy 5 kilograms.

Thus, the price will be

​Unit price × Number of Kilograms of tomatoes we need to buy

=  5  ×  5

=  25  dollars
​
Like this, we can easily calculate the price of the thing for any quantity if we know the unit price.

The rate or price of anything for any varying quantity can be easily deduced if we know the unit rate or unit price.

Glencoe Math Course 2 Volume 1 Common Core Chapter 1 Page 30   Exercise 10   Problem 12

Given that, we need to determine and explain if we convert 100.

feet per second to inches per second, will there be more or less than 100 inches.

Given that

1 feet = 12 inches

Converting the given, we get

= \(\frac{100 \text { feet }}{1 \text { second }}=\frac{100 \text { feet }}{1 \text { second }} \times \frac{12 \text { inches }}{1 \text { feet }}\)

=  \(\frac{100 \times 12 \text { inches }}{1 \text { second }}\)

=  \(\frac{1200 \text { inches }}{1 \text { second }}\)

When you convert 100 feet per second to inches per second, there be more than 100 inches.

Since 100 feet per second = 1200 inches per second

Glencoe Math Chapter 1 Ratios And Proportional Reasoning Exercise 1.3 Explanation Page 30   Exercise 11 Problem 13

We need to convert 7 meters per minute to yards per hour.

We know that

​1 meters = 1.09361 yards

1 hour =  60 minutes

\(\frac{1}{60}\)hour = 1 minute

Converting the given, we get

=  7.65529 × 60

=  459.3174 yards per hour
​
Thus, 7 meters per minute will be 459.3174 yards per hour

Glencoe Math Course 2 Volume 1 Common Core Chapter 1 Page 30  Exercise  12   Problem 14

Given that, A salt truck drops 39 kilograms of salt per minute.

We need to determine how many grams of salt the truck drops per second From the given options.
(1)  600
(2)  625
(3)  650
(4)  6,000
​
Converting the given, we get

=  650 grams per second

Hence, Option (3) 650 grams of salt the truck drops per second.

Page 31   Exercise 13   Problem 15

We need to convert 20 mi/h to feet per minute.

Converting the given we get

=  \(\frac{20 \times 5280 \mathrm{feet}}{60 \text { minutes }}\)

=  1760 feet per minute

20 mi/h = 1760 feet/ minute

Glencoe Math Course 2 Volume 1 Common Core Chapter 1 Page 31   Exercise 14   Problem 16

Converting the given we get

=  \(\frac{16 \times \frac{1}{100} \mathrm{~m}}{\frac{1}{60} \text { hour }}\)

=  \(\frac{16}{100} \times 60 \text { meter / hour }\)

=  \(\frac{16 \times 6}{10} \text { meter / hour }\)

=  9.6 meter/hr

16 cm/min =  9.6 meter/hr

Page 31   Exercise 16   Problem 17

We need to convert 26 cm/s to m/min.

Converting the given we get

=  15.6 meter/ minute

26cm/s  = 15.6 m/minute

Page 31   Exercise 17   Problem 18

We need to convert 24 mi/h to feet per second.

Converting the rates, we get

\(\frac{24 \text { miles }}{1 \text { hour }}=\frac{24 \text { miles }}{1 \text { hour }} \times \frac{5280 \text { feet }}{1 \text { mile }} \times \frac{1 \text { hour }}{3600 \text { seconds }}\)

= \(\frac{24 \times 5280 \text { feet }}{3600 \text { seconds }}\)

=  35.2 ft/sec

24 mi/h = 35.2 ft/s

Glencoe Math Course 2 Volume 1 Common Core Chapter 1 Page 31   Exercise 18   Problem 19

We need to convert 105.6 L/h to L/min.

Converting the given we get

\(\frac{105.6 \mathrm{~L}}{1 \text { hour }}\)=\(\frac{105.6 \mathrm{~L}}{1 \text { hour }} \times \frac{1 \text { hour }}{60 \text { minutes }}\)

=  \(\frac{105.6 \mathrm{~L}}{60 \text { minutes }}\)

=  1.76 L/min

1056 L/h  =  1.76L/min

Page 32    Exercise 20   Problem  20

We need to convert Thirty-five miles per hour to feet per minute.

Converting the rates, we get

=  3080 feet per minute
​
Thirty-five miles per hour is the same rate as 3,080 feet per minute.

Page 32   Exercise 21   Problem  21

Given that, a boat is traveling at an average speed of 15 meters per second.

We need to determine how many kilometers per second the boat is traveling.

Converting the rates, we get

= 0.015 Kilometer/ second

A boat is traveling at an average speed of 15 meters per second.

The boat is traveling at 0.015 kilometers per second.

Page 32  Exercise 22    Problem  22

Given that, an oil tanker empties at 3.5 gallons per minute.

We need to convert this rate to cups per second.

Converting the rates, we get

=  \(\frac{3.5 \times 16 \mathrm{cups}}{60 \text { seconds }}\)

= 0.93 cups/second

An oil tanker empties at 3.5 gallons per minute.

This rate in cups per second is 0.93 cups /second

Glencoe Math Course 2 Volume 1 Common Core Chapter 1 Page 32   Exercise 23   Problem  23

Given that

$36 for 4 basketball hats; $56 for 7 basketball hats.

We need to show that they are equivalent are not.

Find the values of each ratio:

For the first case

For the second case

Both the values of the ratios are different.

Thus, they are not equivalent.

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

Page 32    Exercise 24    Problem  24

Given that 12 posters for 36 students; 21 posters for 63 students

We need to show that they are equivalent are not.

Find the values of each ratio:

For the first case:

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

For the second case:

\(\frac{21}{63}\) = \(\frac{1}{3}\)

Both the values of the ratios are the same. Thus, they are equivalent.

The values of the ratios are the same. Thus, they are equivalent.

Glencoe Math Course 2 Volume 1 Common Core Chapter 1 Page 32   Exercise 25    Problem  25

Given that, an employer pays $22 for 2 hours.

We need to use the ratio table to determine how much she charges for 5 hours.

The unit rate per hour is

\(\frac {22}{2}\)=\(\frac{11 \text { dollars }}{1 \text { hour }}\)

Thus, the employer pays 11 dollars per hour. Thus, for 5 hours

5 hours \(\times \frac{11 \text { dollars }}{1 \text { hour }}=5 \times 11 \text { dollars }\)

=  55 dollars

For 5 hours, she charges $55 26 

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

Glencoe Math Course 2 Volume 1 Chapter 1 Exercise 1.4 Solutions 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

Read and Learn More Glencoe Math Course 2 Volume 1 Common Core Student Edition Solutions

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.

Common Core Chapter 1 Ratios And Proportional Reasoning Exercise 1.4 Answers 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.

Step-By-Step Guide For Glencoe Math Chapter 1 Exercise 1.4 Problems 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.

Exercise 1.4 Glencoe Math Course 2 Ratios And Proportional Reasoning Solutions Explained 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.

Examples Of Problems From Exercise 1.4 Ratios And Proportional Reasoning Chapter 1 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.

Student Edition Glencoe Math Chapter 1 Exercise 1.4 Answers Guide 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.

Chapter 1 Exercise 1.4 Glencoe Math Course 2 Step-By-Step Solutions 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

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

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

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

Glencoe Math Course 2 Chapter 1 Ratios And Proportional Reasoning Exercises With Solutions 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.

Read and Learn More Glencoe Math Course 2 Volume 1 Common Core Student Edition Solutions

Chapter 1 Ratios And Proportional Reasoning Glencoe Math Course 2 Exercise Answers 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

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

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

Glencoe Math Course 2 Volume 1 Common Core 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

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

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

Solutions For Glencoe Math Course 2 Volume 1 Chapter 1 Exercises 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

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

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

Glencoe Math Course 2 Volume 1 Common Core 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.

Step-By-Step Guide To Ratios And Proportional Reasoning Chapter 1 Exercises 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.

Glencoe Math Course 2 Volume 1 Common Core 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

Glencoe Math Course 2 Volume 1 Common Core 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

Glencoe Math Course 2 Volume 1 Chapter 2 Percents Exercise Solutions 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.

Common Core Chapter 2 Percents Exercise Answers Glencoe Math Course 2 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.

Step-By-Step Guide For Chapter 2 Percent Exercises In Glencoe Math Course 2 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.

Read and Learn More Glencoe Math Course 2 Volume 1 Common Core Student Edition Solutions

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 %

Exercise Solutions For Chapter 2 Percents Glencoe Math Course 2 Volume 1 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 %

Common Core Percents Chapter 2 Exercises With Solutions Glencoe Math Course 2 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

Examples of problems from Chapter 2 Percent Exercises In Glencoe Math Course 2 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 15.

Page 102   Exercise 6  Problem  12

Given: The table shows 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.

Student Edition Glencoe Math Course 2 Chapter 2 Percents Exercise Guide 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.

Step-By-Step Answers For Chapter 2 Percents In Glencoe Math Course 2 Volume 1 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

Glencoe Math Course 2 Volume 1 Chapter 2 Exercise 2.1 Solutions 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.

Glencoe Math Course 2 Volume 1 Chapter 2 Exercise 2.1 Solutions 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

Read and Learn More Glencoe Math Course 2 Volume 1 Common Core Student Edition Solutions

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 106   Exercise 1  Problem 3

Given:

To find – The number from 8% of 50

8% of  50
​
\(\frac{8}{100}\)(50)

= 4

​8% of 50 is 4

8% of 50 is 4

Common Core Chapter 2 Percents Exercise 2.1 Answers Glencoe Math Course 2 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.

Step-by-step guide for Exercise 2.1 Chapter 2 Percents in Glencoe Math Course 2 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.

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 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

Exercise 2.1 Solutions For Chapter 2 Percents Glencoe Math Course 2 Volume 1 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.

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 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

Common Core Percents Exercise 2.1 Chapter 2 Solutions Glencoe Math Course 2 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

Glencoe Math Course 2 Volume 1 Common Core Chapter 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

Examples Of Problems From Exercise 2.1 Chapter 2 Percents In Glencoe Math Course 2 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

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 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

Student Edition Glencoe Math Course 2 Chapter 2 Percents Exercise 2.1 Guide 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

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 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

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 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

Step-By-Step Answers For Exercise 2.1 Chapter 2 Percents In Glencoe Math Course 2 Volume 1 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

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 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

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 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

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 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

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 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

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 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

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 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

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 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

Glencoe Math Course 2 Volume 1 Chapter 3 Integers Exercise Solutions 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.

Common Core Chapter 3 Integers Exercise Answers Glencoe Math Course 2 Page 191  Exercise 1  Problem 2

The bottom of a snowboarding half-pipe is 5 meters below the top, Circle the integer you would 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)

Read and Learn More Glencoe Math Course 2 Volume 1 Common Core Student Edition Solutions

Step-By-Step Guide For Chapter 3 Integers Exercises In Glencoe Math Course 2 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

Exercise Solutions For Chapter 3 Integers Glencoe Math Course 2 Volume 1 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

Common Core Integers Chapter 3 Exercises With Solutions Glencoe Math Course 2 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.

Examples of problems from Chapter 3 Integers exercises in Glencoe Math Course 2 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

Detailed Explanation Of Chapter 3 Integers Exercises Glencoe Math Course 2 Solutions 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

Student Edition Glencoe Math Course 2 Chapter 3 Integers Exercise Guide 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

Step-By-Step Answers For Chapter 3 Integers In Glencoe Math Course 2 Volume 1 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.