5. Using Ratios and Solving Proportions
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Chapter 1
5.
Using Ratios and Solving Proportions
This lesson delves into the concept of ratios and proportions, focusing on their applications in various real-world scenarios. For instance, it discusses how to solve problems related to cooking recipes, where you need to figure out the amount of ingredients based on a given ratio. It also covers how to determine rates of work, like how long it will take to complete a task at a certain pace. The Cross Products Property is highlighted as a key method for solving proportions. The lesson is rich with examples, including those that involve variables, making it a valuable lesson for anyone looking to understand the practical applications of ratios and proportions.
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| | 14 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Using Ratios and Solving Proportions
Slide of 14
In this lesson, using real-world examples, the comparison of two quantities will be practiced using a particular concept. Additionally, it will be shown how to use the concept to form and solve equations.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
Predicting the Time to Run an Ultra-Marathon
Jordan trained all year to be able to run in the esteemed Summer Ultra-Marathon, which took place in her hometown Tallahassee, Florida. She comfortably ran the first 6 miles of the ultra-marathon in 54 minutes. 
If she is able to maintain the same pace, how long will it take her to finish the remaining 29 miles? Give the exact answer in hours.
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Explore
Comparing the Number of Movie Theaters Across Countries
Jordan's best friend Emily really loves watching movies! As they walked home from the cinema, Emily wondered aloud, "Which country's population, as a whole, has the easiest access to movie theaters?" After an intense weekend of internet research, she created the following table.
| Number of Theaters | |
|---|---|
| China | 54164 |
| U.S. | 40246 |
| India | 11000 |
| France | 5741 |
| Canada | 3114 |
| South Africa | 800 |
Well, China has the greatest amount of theaters, so people in China must have the easiest access — so it seems. But wait! Countries have different populations. What if the number of theaters is largest simply because the population is largest? Emily decides it is better to calculate the number of movie theaters in the respective country per 10000 people.
| Number of Theaters | Population | Number of Theaters per 10000 People | |
|---|---|---|---|
| China | 54164 | 1433783686 | 0.37 |
| U.S. | 40246 | 329064917 | 1.22 |
| India | 11000 |
1366417754 | 0.08 |
| France | 5741 | 65129728 | 0.88 |
| Canada | 3114 | 37411047 | 0.83 |
| South Africa | 800 | 58558270 | 0.14 |
Discussion
Defining a Ratio
A relation in which two values, such as the number of theaters and the number of people, are compared is called a ratio.
Concept
Ratio
A ratio is a comparison of two quantities that describes how much of one thing there is compared to another. Ratios are commonly represented using colon notation or as fractions. They are read as the ratio of a to b,
where b is a non-zero number.
FractionbaColon Notationa:b
The ratio a:b means that for every a units of one quantity, there are b units of another quantity. Ratios can be part-to-part or part-to-whole.
| Part-To-Part | Part-To-Whole | |
|---|---|---|
| Explanation | Describes how two different groups are related | Describes the relationship between a specific group to a whole |
| Example 1 | The number of sophomores to freshmen on the basketball team is 7:15. | The number of sophomores to all basketball team members is 7:22. |
| Example 2 | The number of mangoes to jackfruits the vendor has is 10:20. | The number of mangoes to all fruits the vendor has is 10:42. |
Explore
Comparing Two Ratios
Consider different pairs of ratios. Compare their values and determine whether they are equal or not.
If the ratios are equal, how can this fact be written algebraically using only mathematics symbols?
Discussion
The Concept of Proportion
When two ratios are equal in value, an equals sign can be written between them, which results in creating a proportion.
Concept
Proportion
A proportion is an equation showing the equivalence of two ratios, or fractions, with different numerators and denominators.
In this case, one-third of a pizza is the same amount of pizza as two-sixths or four-twelfths. If the simplified forms of two fractions are equal, then they are said to be proportional. For example, one-third is proportional to two-sixths and four-twelfths. 
ba=dcora:b=c:d
The first and last numbers in the proportion are called the extremes, while the other two numbers are called the means.
ad↓means:↑b=↑c:extremesd↓
As an example of proportionality, consider slices of pizza. Depending on the number of times it has been sliced, the same amount of pizza could be cut into 1, 2, or 4 pieces.
Pop Quiz
Finding Equal Ratios
Consider the given ratio. Then, analyze the values of the ratios in each answer and choose which one forms a proportion with the given ratio.
Discussion
Means-Extremes Property of Proportion
Up to this point, each proportion has dealt with fractions. However, every proportion can be rewritten as an equation without any fractions. This is a property that is especially useful when solving proportions for an unknown variable.
Rule
Cross Products Property
In a proportion, the product of the extremes is equal to the product of the means.
ba=dc⇒ad=bc
This property is also known as cross-multiplication or Means-Extremes Property of Proportion.
Proof
Cross Products PropertyThe Cross Products Property can be proven by using the Properties of Equality.
ba=dc
MultEqn
LHS⋅b=RHS⋅b
a=dc⋅b
MoveRightFacToNum
ca⋅b=ca⋅b
a=dcb
MultEqn
LHS⋅d=RHS⋅d
ad=cb
CommutativePropMult
Commutative Property of Multiplication
ad=bc
Example
Learning to Apply Cross Products Property Correctly
Jordan and Emily, exhausted from watching so many movies, sat down to finally do their Math homework. In one exercise, they were asked to determine whether the two given ratios 1815 and 4.53.75 form a proportion. Both girls used the Cross Products Property, but their solutions were different. Jordan gasped in bewilderment.
a Is either of the girls correct? Find and correct any mistakes made.
b Using the same method, determine whether 72 and 216 form a proportion.
Answer
a No, both girls made mistakes in their solutions.
b Yes, the ratios form a proportion.
Hint
a Recall what the Cross Products Property states. Identify the extremes and means in the proportion set by the girls. Are the second steps in either one's solutions correct?
b Write the given ratios as a potential proportion. Then, confirm if it is true by using the Cross Products Property.
Solution
a First, each solution will be analyzed separately. Then, the exercise will be solved correctly.
Jordan's Solution
In the first step of Jordan's solution, she writes the given ratios as a potential proportion. To highlight the fact that it is not yet known whether the ratios form a proportion, she put a question mark above the equals sign.
Then, it is shown that Jordan applied the Cross Products Property to obtain the following result.
In order to determine whether that property was applied properly, recall what it states.
ba=dc ⇒ ad=bc
According to the property, in any proportion, the product of the extremes is equal to the product of the means. Identify the extremes and means in the exercise girls were solving.
15⋅1815⋅4.5=3.75⋅4.5 ×=18⋅3.75 ✓
Hence, Jordan's solution is not correct. Emily's Solution
Next, Emily's solution can be analyzed. After writing the ratios as a potential proportion, she also chose to apply the Cross Products Property.
15⋅3.7515⋅4.5=18⋅4.5 ×=18⋅3.75 ✓
Therefore, Emily's solution is also incorrect. Correct Solution
Now, it is time to solve the exercise properly. Since the cross products are equal, the ratios can be concluded to form a proportion. b In order to determine whether 72 and 216 form a proportion, set those ratios equal in the form of a potential proportion. Then, use the Cross Products Proportion to verify if the proportion is true.
Pop Quiz
Solving Proportions
Solve the given proportion for the unknown variable x.
Discussion
Rate and Unit Rate
In real-life situations, two quantities that are being compared often have different units. Such ratios have a specific name.
Concept
Rate
A rate is a ratio that compares two quantities measured in different units. For example, a certain species of bamboo grows 27 feet in height in 2 years. Then, 2 years27 ft is its rate of growth. Here are some other possible examples of rate.
Rates might be most useful when finding how much of something is per 1 unit of something else. Such a comparison is called a unit rate. If the given rates are not already unit rates, they can be determined by some calculations. Dive deeper into two of the previous examples.
| Scenario | Rate | Unit Rate |
|---|---|---|
| Kriz finds 20 Pokémon every 10 days. | 20 Pokémon per 10 days, 10 Pokémon per 5 days |
2 Pokémon per 1 day, 730 Pokémon per 1 year |
| At a party, 42 candies were eaten by 6 kids. | 42 candies per 6 kids, 21 candies per 3 kids |
7 candies per 1 kid |
Example
Using Rate and Unit Rate in Cooking
After finishing their homework, Jordan and Emily decided to bake some cookies together. The recipe they found calls for 2 cups of water for every 5 cups of flour. However, the girls wanted to cook more cookies to be able to share with their classmates. They used 12 cups of flour and now they need to figure out how many cups of water they need.
Help Jordan and Emily solve this problem so that they can share the delicious cookies!
a How many cups of water do the girls need to follow the recipe precisely?
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Hint
a What is the rate of water to flour given in the recipe? Set a proportion and solve it by using the Means-Extremes Property of Proportion.
b Think of how many cups of flour are needed per 1 cup of water.
Solution
a In the recipe, it is given that for every 2 cups of water 5 cups of flour are needed. Using this information, the following ratio representing the rate of water to flour can be formed.
5 cups of flour2 cups of water
It is also known that the girls used 12 cups of flour. Let x represent the number of cups of water they need for that amount of flour. Using these two values, a new ratio can be formed. 12 cups of flourx cups of water
Since Jordan and Emily want to follow the recipe precisely, this ratio should be equivalent to the rate of water to flour given in the recipe. Therefore, the ratios can be set equal to form a proportion.
5 cups of flour2 cups of water=12 cups of flourx cups of water
By solving this proportion for x, the amount of needed cups of water can be calculated. The equation will be solved by applying the Means-Extremes Property of Proportion.
52=12x
CrossMult
Cross multiply
2⋅12=5x
Multiply
Multiply
24=5x
DivEqn
LHS/5=RHS/5
524=x
CalcQuot
Calculate quotient
4.8=x
RearrangeEqn
Rearrange equation
x=4.8
b Start by recalling that a unit rate is rate that tells the amount of one item in comparison to 1 of another item. Therefore, finding the unit rate of flour to water means finding the number of cups of flour needed per 1 cup of water.
1 cup of water? cups of flour
It is given that the recipe calls for 2 cups of water for every 5 cups of flour. This is equivalent to saying that for every 5 cups of flour 2 cups of water are required.
2 cups of water5 cups of flour
Now, to find how many cups of flour needed for just 1 cup of water, both the numerator and denominator of the fraction should be divided by 2.
2/2 cups of water5/2 cups of flour⇕1 cups of water2.5 cups of flour
Therefore, the unit rate of flour to water is 2.5 cups of flour to 1 cup of water.
Example
Calculating the Rates of Making Invitations
Emily, eager to watch more movies, planned to host a movie night. To invite her friends, she started making colorful invitations. She made the first batch in just 21 minutes, but then her mother asked her to help in the kitchen. She had to take a break from making invitations.
When she returned to making invitations, her younger sister offered her help in creating 5 invitations. After 7 minutes, Emily realized that together they already have made twice the number of invitations she made in the first sitting.
a How many invitations did Emily make in the first sitting?
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b How long does it take Emily to make 1 invitation?
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c Who, Emily or her sister, makes invitations faster?
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Hint
a Write the expressions of Emily's rate of work in the first and seconds sittings. Then set and solve a proportion.
b Calculate the quotient of the time Emily worked on the first batch of invitations and the number of invitations she made.
c Note that Emily's sister finished all 5 invitations in 7 minutes. Compare Emily's rate of work and her sister's rate.
Solution
a Let x be the number of invitations that Emily made in the first sitting. It is given that she worked on them for 21 minutes, so the quotient of x and 21 represents the rate of Emily's work.
21x
Together with her sister, Emily made 2x invitations in 7 minutes. By dividing 2x by 7, the rate of their work together can be found. However, in this case determining only Emily's rate would be more useful. To find it, subtract 5 invitations that Emily's sister made and then divide by 7.
72x−5
Since Emily worked with the same rate the whole time, the following proportion can be formed.
21x=72x−5
Finally, it can be solved for x by using cross-multiplication.
21x=72x−5
CrossMult
Cross multiply
7x=21(2x−5)
DivEqn
LHS/7=RHS/7
x=3(2x−5)
Distr
Distribute 3
x=6x−15
SubEqn
LHS−6x=RHS−6x
-5x=-15
DivEqn
LHS/(-5)=RHS/(-5)
x=3
b In Part A, it was calculated that Emily made 3 invitations in 21 minutes. By dividing 21 by 3, the amount of time she needs to make one invitation can be found.
321=7 min
c It is given that after working for 7 minutes the girls made twice the amount of invitations Emily made by herself before combining forces. This means that Emily's sister finished 5 invitations in 7 minutes.
Emily’s Sister’s Rate:7 minutes5 invitations
At the same time, it was determined in Part B that Emily makes 1 invitation in 7 minutes.
Emily’s Rate:7 minutes1 invitation
By comparing those rates, it can be concluded that Emily's sister is working much faster.
7 minutes5 invitations>7 minutes1 invitation
It is a good thing she offered to help!
Closure
Calculating the Time to Finish an Ultra-Marathon
Using the information learned in this lesson, the challenge presented at the beginning can now be solved. Recall that Jordan participated in the Summer Marathon in her hometown Tallahassee, Florida. She comfortably ran the first 6 miles of the marathon in 54 minutes. 
If she is able to maintain the same speed, how long will it take her to finish the remaining 29 miles? Give the exact answer in hours.
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Hint
Calculate Jordan's speed in miles per hour. Then, use it to set a proportion.
Solution
In order to determine the time Jordan needs to finish running the marathon, her speed should be calculated. It is given that Jordan ran 6 miles in 54 minutes.
Now, let y represent the number of hours Jordan needs to run 29 miles. Since it is said that Jordan maintains the same rate, the following proportion can be set.
If Jordan maintains the same rate, it will take her an impressive 4.35 hours or 4 hours and 21 minutes to finish the remaining 29 miles of the ultra-marathon.
54 min6 mi
By dividing both the numerator and denominator of the fraction by 54, Jordan's unit rate of miles per minute can be found.
5454 min546 mi⇔1 min546 mi
However, since the answer should be given in hours, the unit rate of miles per hour would be more useful. To find it, expand the fraction by 60 and then simplify.
1 min546 mi
ExpandFrac
ba=b⋅60a⋅60
1⋅60 min546⋅60 mi
▼
Simplify
OneMult
1⋅a=a
60 min546⋅60 mi
Rewrite
Rewrite 60 min as 1 h
1 h546⋅60 mi
MoveRightFacToNum
ca⋅b=ca⋅b
1 h546⋅60 mi
ReduceFrac
ba=b/6a/6
1 h96⋅10 mi
Multiply
Multiply
1 h960 mi
ReduceFrac
ba=b/3a/3
1 h320 mi
1 h320 mi=y h29 mi
Use the Means-Extremes Property to solve it and calculate the value of y.
1320=y29
CrossMult
Cross multiply
320y=1⋅29
OneMult
1⋅a=a
320y=29
MultEqn
LHS⋅3=RHS⋅3
20y=87
DivEqn
LHS/20=RHS/20
y=4.35
In a movie theater, there are 2 kids and 7 adults present 15 minutes before the start of a movie. More people come in and take their seats over the next 15 minutes. When the movie starts, the ratio between kids and adults is 43.
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We know that the numbers of children and adults were originally 2 and 7, respectively. We can use these values to write the ratio of children to adults. Children/Adults → 2/7 Since an equal number of adults and children entered the movie theater in the 15 minute length of time, let x represent the numbers of both the adults and kids who came in during this time. We can find the final number of kids and adults in the theater by adding x to the initial numbers. Children/Adults → 2+ x/7+ x This final ratio of children to adults in the theater is 3 4. Let's set these two ratios equal to each other to form a proportion. 2+x/7+x=3/4 Finally, we can solve this proportion by applying the Cross Products Property.
13 children and 13 adults entered the movie theater in the 15 minutes before the movie began, giving a total increase of 13+13=26 people.
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Paulina plays the saxophone in the school band. In her favorite songs playlist, there are saxophone covers of 12 classical songs and 48 pop songs.
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Let's start by calculating the total number of songs in Paulina's playlist. There are 12 classical songs and 48 pop songs in her playlist. 12+48 = 60songs This means that the ratio of classical songs to all songs is 1260. Classical/All songs → 12/60 Let's now add s classical songs to the playlist, which will also increase the total number of songs by s. As such, we need to add s to both the numerator and denominator of the obtained fraction. Classical/All songs → 12+ s/60+ s This ratio should equal 12. Using this information, we can set a proportion. 12+s/60+s = 1/2 Let's solve this equation for s.
Paulina should add 36 classical songs to make the number of classical songs be half the total number of songs in her playlist.
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Find ab given the following proportions.
b−2a+2=38andb+2a−2=74
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We are given the following two proportions. a+2/b-2=8/3 and a-2/b+2=4/7 To calculate the value of b a, we first need to find what values a and b satisfy both proportions simultaneously. Let's start by analyzing the first proportion. a+2/b-2=8/3 A proportion is an equality of two ratios. One possible scenario is that those equal ratios, or fractions, have the same numerators and denominators. If that is the case, let's find what values a and b will have. l a+2= 8 b-2= 3 ⇒ la=6 b=5 Now we can substitute these values into the second proportion and see if it remains true.
Since we obtained a true statement, the values we found for a and b satisfy both given proportions. Finally, we can calculate the value of ba. la= 6 b= 5 ⇒ b/a=5/6
There is another method of solving the given task. We still need to find the values of a and b that satisfy both given proportions.
Finding a Pair of Corresponding Values
We can set a value of one variable and calculate the corresponding value of the other variable. Let's say that a=1 and calculate the value of b.
Testing Values in the Second Proportion
We will now test if these values of a and b also satisfy the second proportion.
Since we ended up with a false statement, we know that this pair of numbers does not satisfy both proportions.
Repeat the Process
Let's repeat this process until we find a pair of values that satisfies both proportions.
| a | a+2/b-2=8/3 | b | a-2/b+2=4/7 | Simplified |
|---|---|---|---|---|
| 1 | 1+2/b-2=8/3 | 17/8 | 1-2/178+2? =4/7 | - 8/33≠ 4/7 |
| 2 | 2+2/b-2=8/3 | 7/2 | 2-2/72+2? =4/7 | 0/11≠ 4/7 |
| 3 | 3+2/b-2=8/3 | 31/8 | 3-2/318+2? =4/7 | 8/47≠ 4/7 |
| 4 | 4+2/b-2=8/3 | 17/4 | 4-2/174+2? =4/7 | 8/25≠ 4/7 |
| 5 | 5+2/b-2=8/3 | 37/8 | 5-2/378+2? =4/7 | 24/53≠ 4/7 |
| 6 | 6+2/b-2=8/3 | 5 | 6-2/5+2? =4/7 | 4/7= 4/7 |
We see that a= 6 and b= 5 satisfy both proportions.
Finding ba
Finally, we can evaluate ba for the values we found. la= 6 b= 5 ⇒ b/a=5/6 The value of ba is equal to 56.
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The age of some trees can be determined by counting how many concentric rings their trunks have. Each ring corresponds to one year of growth. An oak tree with a diameter of 15 inches has 40 rings.
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We know that an oak tree with a diameter of 15 inches has 40 rings. We can use these values to write the ratio of rings to diameter. Rings/Diameter → 40/15 Let r be the number of rings when the diameter is 22 inches. To find the value of r, we will need to form another ratio. Rings/Diameter → r/22 Since it is said that the tree continues to grow at the same rate, we can set these ratios equal to each other to form a proportion. 40/15 = r/22 Let's now solve this equation for r.
Therefore, there will be approximately 59 rings when the diameter of the tree is 22 inches.
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Using Ratios and Solving Proportions
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