Session 4 Proportional Reasoning - Learner · PDF filePatterns, Functions, and Algebra 91 Session 4 Session 4 Proportional Reasoning ... In Mixture A, there is one more blue beaker

Patterns, Functions, and Algebra 91 Session 4

Session 4

Proportional Reasoning

Key Terms for This SessionNew in This Session

• absolute comparison

• relative comparison

• proportional relationship

• direct variation

• origin

Introduction and ReviewIn Session 3, we looked at functions and found them to be a relationship between inputs and outputs where thereis exactly one output for each input. In this session, we’ll look at a special kind of functional relationship: the pro-portional relationship. We will develop proportional reasoning skills by comparing quantities, looking at the rela-tive ways numbers change, and thinking about proportional relationships in linear functions. [SEE NOTE 1]

Learning ObjectivesIn this session, we’ll explore direct variation and proportional reasoning. We will:

• Differentiate between relative and absolute meanings of “more” and determine which of these is a proportional relationship

• Compare ratios without using common denominator algorithms

• Differentiate between additive and multiplicative processes and their effects on scale and proportionality

• Interpret graphs that represent proportional relationships or direct variation

NOTE 1. Many people have trouble reasoning proportionally.Typically, when people first begin to think about proportions, theythink in absolute, rather than in relative, terms.These different ways of thinking correspond to using additive (how much moreis 12 than 10?) rather than multiplicative (what is the ratio of 10 to 20 compared to 12 to 25?) reasoning.

This session introduces the idea that there are different meanings of “more” and distinguishes between relative and absolutecomparisons. To familiarize ourselves with the idea of equivalent ratios, we will use both additive and multiplicative methodsto explore different ways of making similar figures. We will look at mixture problems and explore ratios without using algo-rithms to convert them to common denominators. Finally, we will examine characteristics of equations and graphs that repre-sent direct variation.

Materials Needed: Graph paper, rulers, handouts of Quadperson, blank overheads

NOTE 1 cont’d. next page

Session 4 92 Patterns, Functions, and Algebra

Problem A1. Suppose there are two classes in a school, one with 20 students and one with 25. If the first class has10 girls, and the second class has 12, which class has more girls? [SEE NOTE 2]

What different definitions of “more” are used here? What is your definition of “more”?

If you said that the second class has more girls, you’re making an absolute comparison.You probably thought that12 is 2 more than 10, so there are more girls in the class with 12.

If you said that the first class has more girls, you’re making a relative comparison. You probably thought 10 is halfof 20, and 12 is less then half of 25, so there are more girls in the class with 10.

Clearly these are two different interpretations of “more.” Although both interpretations are correct, in some casesit is more appropriate to look at relative rather than absolute comparisons. For example, compare an all-girl classof 20 students with a class of 25 students, 22 of whom are girls. In a sense, there are “more” girls in the class with20.

Part A: Two Different Meanings of“More” (15 MINUTES)

VIDEO SEGMENT (approximate times: 3:08-4:31): You can find this segment onthe session video approximately 3 minutes and 8 seconds after the Annen-berg/CPB logo. Zero the counter on your VCR clock when you see theAnnenberg/CPB logo.

In this video segment, participants share their answers to Problem A1 withProfessor Cossey.Watch the segment after completing Problem A1 and com-pare your answer with those of the onscreen participants. If you get stuck onthe problem, you can watch the video segment to help you.

NOTE 1, CONT’D.

ReviewGroups: Discuss any questions about the homework. If time allows, take a few minutes to try out the number games with apartner. Pairs should show their networks to one another. One partner can choose a (secret) input, run it through the network,and reveal only the output. Then the other partner can use the “undoing” network to find the original number.

Groups: Take a minute and discuss iteration, along with Problems H4 and H5 from Session 3. It is likely to be a new idea.

NOTE 2. Groups: Begin Part A by discussing the meaning of “more.” Read through the exercise and share ideas about whichclass has “more” girls. Consider the difference between absolute and relative comparisons. Answer Problems A1-A5 in pairs orin small groups.


Problem A2. Describe the numerical calculations you would do with two numbers to make an absolute compar-ison.

Problem A3. Describe the numerical calculations you would do with two numbers to make a relative comparison.

Problem A4. Give some examples of situations in which it is more useful to make an absolute comparison. Givesome examples of situations in which it is more useful to make a relative comparison.

Problem A5. Suppose the average height of eighth-grade students is greater than the average height of seventh-grade students. Is this an absolute or relative comparison? Why?

Part A, cont’d.


Comparing MixturesThe scientists at the research lab for Whodunit Jeans are trying to decide on just the right shade of blue for a newline of jeans. Being scientists, not mathematicians, the researchers decide to choose a color by mixing pure blueliquid and clear water together until they get just the right shade. [SEE NOTE 3]

The scientists have several beakers of liquid, some with blue liquid and some with clear water. They plan to mixthese together in big bowls. Before they mix the liquids, they guess how blue the mixture will be.

In Problems B1-B7, there are two sets (A and B) of blue-clear combinations to mix. Predict which set will be bluer,and explain your reasoning. Assume you do not know how to compare fractions with unlike denominators. (Con-verting to a common denominator and comparing would make these problems trivial.) [SEE NOTE 4]

Problem B1.

[SEE TIP B1, PAGE 106]

Part B: The Mixture Blues (45 MINUTES)

NOTE 3. We will now look at ways to compare ratios in the context of mixing colors. In these problems, it is important to under-stand that we’ll be thinking about these mixtures without using the common algorithms for comparing fractions. That is, weneed to assume we don’t have the skills for finding a common denominator to compare—for example, 4/7 and 3/8.This forcesus to compare ratios by relying on a more basic understanding of what a ratio is.

NOTE 4. Groups: Work in pairs on Problems B1-B13. Discuss solutions to Problem B5 in particular.

Consider the following solution:

In Mixture A, there is one more blue beaker than clear, and in mixture B, there is one more blue beaker than clear. Therefore,the mixtures are the same color.

Try to find the fault in this explanation. This is the same reasoning, by the way, that students use when they claim that 2/3 isthe same as 3/4, since the numerator in each fraction is one less than the denominator. They are, in fact, using additive strate-gies.

A lovely way of solving mixture problems like this is by comparing the effect these beakers have on the entire mixture. Forexample, if mixture A had 2 clear beakers and 3 blue, and mixture B had 99 clear beakers and 100 blue, the extra blue in A wouldhave a greater effect on the entire mixture than the extra blue in B, which is spread over 199 total beakers. Therefore, A wouldbe bluer (3/5 compared to 100/199).

Some people may use a canceling technique to compare these ratios. They will form blue/clear beaker pairs and cross themout. For example, in Problem B6, they will cancel all the beakers in A after forming 2 blue-clear pairs, and do the same in B with1 clear left over, claiming that B is less blue. This technique will not work in Problem B5, however, since after using the “cancel-ing” technique, each mixture has 1 blue beaker left, suggesting they’re the same shade of blue. Once again, though, the effectof the extra blue in A is greater than the effect in B because there are fewer beakers in A.

Another approach to these problems is to equalize the number of beakers in each situation by creating multiples of mixtures.For example, in Problem B7, if you double mixture B, you get the same number of beakers as in mixture A, making them easyto compare.


Problem B2.

Problem B3.

Problem B4.

Problem B5.

Problem B6.

Part B, cont’d.


Problem B7.

Why would an absolute comparison not be useful in solving Problems B1-B7?

Mixing BluesSome of the scientists in the lab decided to see what would happen if they took two different mixtures and mixedthem together.They called this the “union”of the two mixtures. For example, in Problem B4, they took the two mix-tures A and B and formed the union of the mixtures. Standard mathematical notation for the union of two things(usually sets) is ∪, so they named their new mixture A ∪ B.

Problem B8. In this example, which is bluer: A, B, or A ∪ B?

Part B, cont’d.

VIDEO SEGMENT (approximate times: 18:48-20:05): You can find this segmenton the session video approximately 18 minutes and 48 seconds after theAnnenberg/CPB logo. Zero the counter on your VCR clock when you see theAnnenberg/CPB logo.

In this video segment, Frederick uses relative comparison to solve ProblemB4. Watch this segment after you have completed Problems B1-B7 and com-pare your reasoning with Frederick’s.


Problem B9. What can you say about the relative “blueness” of A and B if A ∪ B is bluer than A? If it is bluer thanB? [SEE TIP B9, PAGE 106]

Problem B10. Can A ∪ B ever be bluer than both A and B? If not, why? If so, when? [SEE NOTE 5] [SEE TIP B10, PAGE 106]

Problem B11. Can A ∪ B ever be just as blue as either A or B? If not, why? If so, when?

Someone invented a term called the “blueness quotient” (BQ) of a mixture. In the mixture in Problem B4, the BQof A is 3/5, and the BQ of B is 2/4.

Problem B12. Come up with a rule for computing the BQ of A ∪ B if you know the BQs of A and B.

Take It FurtherPROBLEM B13. Mixture A has a BQ of 1/3, and the lab has decided that it would like A ∪ B to have a BQ of 1/2.Whatcould be the BQ of mixture B? Is there more than one answer? Why?

Part B, cont’d.

NOTE 5. Problem B10 is an interesting one and may require spending some time thinking about how to prove this assertion.In fact, A ∪ B can never be bluer than both A and B. You can use some algebra to prove this. Find a relationship between a, b,c, and d such that:

(a + c) / (b + d) is greater than a / b and (a + c) / (b + d) is greater than c / d.

These two inequalities become:

c / d is greater than a / c and a / b is greater than c / d, a contradiction.

Therefore, A ∪ B can never be bluer than both A and B.


Playing With QuadpersonIn Part B, you made relative comparisons involving mixtures. Scaling is another type of relative comparison prob-lem. Scaling is used in graphic design, cartography, construction, and many other areas. In fact, if you have everdoubled a recipe or built a model airplane, you have dealt with problems of scale. [SEE NOTE 6]

This activity compares the effects of absolute and relative comparisons on a picture.Think about what would hap-pen to a drawing if every line were made 1/2 as long. Would it look like the same picture at all? What would hap-pen if you made every line a 1/2-inch shorter? In this activity, you will see the effects of these changes onQuadperson, a drawing of a face made of quadrilaterals.

On a piece of graph paper, draw a picture of Quadperson so that each line in your drawing is 1/2 as long as thecorresponding line in the picture below.Then, make another picture of Quadperson so that each line in your draw-ing is 1/2-inch less in length than the corresponding line in the original Quadperson.

Part C: Quadperson (45 MINUTES)

NOTE 6. Groups: Work in pairs on the Quadperson activity. One person should do Problem C1 and one should try Problem C2.

After completing the activity, consider which of the new Quadpersons looks more like the original Quadperson, and why. Dis-cuss this in pairs and as a whole group. Explore which line segment drawn in Problem C2 was most dramatically affected bycutting half an inch of the length. In fact, it is the rectangle “nose” that becomes a line segment. Relative to the length of thissegment, cutting off half an inch had a dramatic effect.

What would have happened if the original rectangle nose had been a mile long? What effect would cutting off half an inchhave had then? Which line segment was most dramatically affected in Problem C1 by taking half its length?

Try It Online!

This problem can be explored online asan Interactive Activity. Go to thePatterns, Functions, and Algebra Website at www.learner.org/learningmathand find Session 4, Part C, Playing WithQuadperson.


Problem C1. Compare the “before” and “after” Quadperson when you multiplied each length by 1/2.

Problem C2. Compare the “before” and “after” Quadperson when you subtracted 1/2 inch from each length.

Problem C3. Of Problems C1 and C2, which is a relative comparison? Which is an absolute comparison? Explainhow you know.

Problem C4. Did the Quadperson look the same after a change made by an absolute comparison? Did the Quad-person look the same after a change made by a relative comparison? Why?

Would every absolute comparison change the shape of Quadperson? Would every relative comparison maintainthe shape of Quadperson?

Can you explain why the two-dimensional area would change in the manner described in this segment?

Part C, cont’d.


In this video segment, Andrea and Deanna compare the two versions ofQuadperson and discuss why one Quadperson is more in proportion withthe original than the other.Watch it after you’ve completed Problems C1-C4,and reflect on the onscreen participants’ reasoning about their results.


In this video segment, the onscreen participants discover the proportionalchange in area determined by the relative change in Quadperson.


Analyzing the ResultsLet’s try to generalize our results from the two ways of creating a new Quadperson. [SEE NOTE 7]

Problem C5. Let y1 be a function that takes an input x (the length of a line segment) and outputs the segment’slength as described by Problem C1. Write a formula for y1. [SEE TIP C5, PAGE 106]

Problem C6. Let y2 be a function that takes an input x (the length of a line segment) and outputs the segment’slength as described by Problem C2. Write a formula for y2.

Problem C7. Graph the functions you created in Problems C5 and C6. Describe any similarities or differences inthe graphs. [SEE TIP C7, PAGE 106]

A relationship is proportional if the ratio of inputs and outputs is the same for all inputs and outputs. This kind ofrelationship is also sometimes referred to as direct variation.

Problem C8. Look at the graphs you created in Problems C5 and C6. Use the graphs to decide whether or noteither function represents a proportional relationship.

Part C, cont’d.

NOTE 7. Groups: Work on Problems C5-C8 in pairs or small groups. In these problems, create line segments that illustrate theeffect of the two techniques. Notice that the graph for y2 = (1/2)x represents a proportional relationship; that is, when x doubles, y doubles. This is also known as direct variation. The ratio of any two pairs of coordinates is always the same.

Notice that although this line passes through the origin (as does any graph of a proportional relationship), the graph for situ-ation B, in which y1 = x + 1/2, does not.


Would the graphs of two different proportional relationships intersect? Where?

Graphs of Proportional RelationshipsProblem C9. Draw a third picture of Quadperson, so that each line in this new drawing is double the length of theoriginal corresponding line.

Problem C10. Describe this relationship with a table and an equation.

Problem C11. How do you know that the graph you made in Problem C10 represents a proportional relationship?What does this graph have in common with others that you have drawn?

Problem C12. What conclusions can you draw about the graphs of proportional relationships? Explain how youcould determine whether or not a line graph represents a proportional relationship.

Part C, cont’d.

VIDEO SEGMENT (approximate times: 14:34-16:13): You can find this segmenton the session video approximately 14 minutes and 34 seconds after theAnnenberg/CPB logo. Zero the counter on your VCR clock when you see theAnnenberg/CPB logo.

In this video segment, the onscreen participants compare the graphs inProblem C7 and use the graphs to explain their observations about Quad-person. Watch it after you have completed Problems C5-C8 and compareyour observations about the graphs with those of the onscreen participants.


In Parts B and C we examined the proportionality of mixtures and scale models. In this activity, we will take a closelook at graphs to learn more about the relationships between distance and time and between steepness andspeed. [SEE NOTE 8]

Suppose seven cars are all near an intersection.The graph below shows the distances between cars and the inter-section as time passes. Study this graph carefully and then try to answer the following questions.

PROBLEM D1. In what direction is each car moving in relation to the intersection?

Part D: Speeds, Rates, Steepness, andLines (45 MINUTES)

NOTE 8. The graph in this section is designed to uncover a variety of common misunderstandings. First of all, some may inter-pret the graph as a picture of the trips the cars took, a flawed interpretation. If the graph has a segment that is horizontal, theremay be more than one interpretation of the car’s trip.The car could be stopped, or it could be going in a circle around the inter-section. This is often an interesting insight for people.

Some may assume two cars that are the same distance from the intersection are side by side. Others may think that the twocars in this situation are on opposite sides of the intersection, facing each other, or that they could be approaching the inter-section along intersecting streets.

A negative slope indicates that a car is moving toward the intersection. A positive slope indicates that the car is moving awayfrom the intersection, since the distance is increasing over time. Some may wonder where the intersection is. In fact, the graphis not a picture of the car’s path, but a representation of the relationship between distance and time.

Keep in mind that speed is always a positive quantity, while velocity can be positive or negative, corresponding to positive andnegative slopes.

Groups: End the session by discussing Problem D5. Each group can share the “story” of their car’s trip.


PROBLEM D2. Compare the cars’ speeds. How do their speeds relate to the steepness of the lines? [SEE TIP D2, PAGE 106]

PROBLEM D3. Does the distance vary directly with time for any of the cars? That is, is the relationship between dis-tance and time proportional for any of the cars? How do you know?

PROBLEM D4. Do any of the cars stop during their trips? If so, which cars?

PROBLEM D5. Choose one of the seven cars, and describe your trip in this car. Use your imagination! Think of yourobservations as a passenger or driver in this car, and give the highlights of your trip for these 15 seconds. Includewhere and when you started the trip and what you saw going on around you—in front of the car, to the sides, andthrough the rearview mirror.

Part D, cont’d.

The car problem is taken from IMPACT Mathematics Course 2, developed by Education Development Center, Inc. (New York:Glencoe/McGraw-Hill, 2000), pp. 328-329.


Problem H1. Sandy made some iced tea from a mix, using 12 tablespoons of mix and 20 cups of water. Chris andPat thought it tasted great, but they needed 30 cups of tea for their party. Lee arrived, and they found they dis-agreed about how to make 30 cups that tasted just the same:

Chris: It’s easy: Just add 10 tablespoons of tea and 10 cups of water. Increase everything by 10.

Pat: Wait a minute. Thirty is just 1 1/2 times 20, so since you add 1/2 as much water, add 1/2 the tea: add 10cups of water and 6 tablespoons of tea.

Sandy: I think about it this way: We used 12 tablespoons for 20 cups, so 12/20 = 3/5 tablespoons for 1 cup, so for30 cups we should use 30 × 3/5 = 18 tablespoons.

Lee: Wait: 20 - 12 = 8, so you want to keep the difference between water and tea at 8. Since there are 30 cupsof water, we should use 30 - 8 = 22 tablespoons of tea. That will keep everything the same.

Critique each of these methods. Which methods are the same? Which methods will really produce tea that tastesthe same? [SEE TIP H1, PAGE 106]

Problem H2.

a. Draw a right triangle with legs 3 and 4 cm and hypotenuse 5 cm.

b. Draw a triangle whose legs are double those in H2(a).

c. Draw a triangle whose side lengths are each 2 cm more than those of the first triangle. That is, thelengths are 5 cm, 6 cm, and 7 cm.

d. Which of the new triangles looks similar to the original triangle?

Homework


Problem H3. Five brothers ran a race. The twins began at the starting line. Their older brother began behind thestarting line, and their two younger brothers began at different distances ahead of the starting line. Each boy ranat a fairly uniform speed. Here are the rules for the relationship between distance (d meters) from the starting lineand time (t seconds) for each boy.

Adam: d = 6t Brett: d = 4t + 7 Caleb: d = 5t + 4

David: d = 5t Eric: d = 7t - 5

a. Which brothers are the twins? How do you know?

b. Which brother is the oldest? How do you know?

c. For each brother, describe how far from the starting line he began the race and how fast he ran. [SEE TIP

H3(C), PAGE 106]

d. Which line below represents which brother? What events match the intersection points of the lines?

1.

2.

3.

4.

5.

e. What is the order of the brothers 2 seconds after the race began?

f. Which two brothers stay the same distance apart throughout the race? Howdo you know, based on their graphs? Howdo you know, based on their equations?

g. If the finish line was 30 meters from thestarting line, who won?

h. Which brothers’ relationships betweendistance from the starting line and timeare proportional? How do you know?

Homework, cont’d.

Problem H3 is taken from IMPACT Mathematics Course 2, developed by Education Development Center, Inc. (New York:Glencoe/McGraw-Hill, 2000), pp. 331-332.

Session 4: Tips 106 Patterns, Functions, and Algebra

Part B. The Mixture BluesTIP B1. Consider whether you are making absolute or relative comparisons in Problems B1-B7.

TIP B9. Your results from Problems B1-B7 may be helpful here.

TIP B10. Consider this problem from a practical standpoint.

Part C. QuadpersonTIP C5. If x is the input, describing a rule to determine the output will help you find the formula.

TIP C7. One way to make these graphs is to create a table of values for each function, then plot and connect thepoints on each graph.

Part D. Speeds, Rates, Steepness, and LinesTIP D2. How could you approximate the cars’ speeds using the labels on the axes?

HomeworkTIP H1. Which are absolute comparisons, and which are relative ones? What type of comparison is more usefulhere?

TIP H3(C). Which number in each equation is related to that brother’s speed? A table may help you to see therelationship.

Tips

Patterns, Functions, and Algebra 107 Session 4: Solutions

Part A: Two Different Meanings of “More”Problem A1. Either answer is defendable. The first class has a higher percentage of girls—half its students aregirls—while girls make up less than half the second class. The second class, however, has a larger number of girls.Essentially, the meaning of “more” is the real issue in this problem.

Problem A2. An absolute comparison is done by counting: 22 is more than 20, because you count to 20 beforecounting to 22. Or it can be done by subtraction: 22 is more than 20 because 22 - 20 = 2, a positive number.

Problem A3. A relative comparison is done by finding percentages or fractions or by finding a rate. On a quiz, 22out of 25 is worse than 18 out of 20, even though 22 is larger in absolute terms.

Problem A4. You might use absolute comparisons when looking at annual salaries, but a relative comparisonwhen looking at per-hour wages. An absolute comparison might tell you that extended cable TV is more expen-sive than basic cable, but a relative comparison might tell you that you get more channels per dollar on extendedcable. A truck may be able to travel further than a compact car before needing to be refueled (an absolute com-parison), but the compact car may travel more miles per gallon (a relative comparison).

Problem A5. This is a relative comparison, because it compares heights “per student” by using the average. Anabsolute comparison might compare the total height of all eighth-graders to the total height of all seventh-graders.

Part B: The Mixture BluesProblem B1. A will be bluer, because B’s mix is the same as A’s, but with two additional clear beakers thrown in.

Problem B2. B will be bluer. Its mix is the same as A’s, but with an extra blue beaker thrown in.

Problem B3. B will be bluer, for the same reason as in Problem B2. Or, consider what would happen if you tried tomake equal doses of A and B. (Common denominators!)

Problem B4. A will be bluer. One explanation is that B is half blue, while A is more than half blue.

Problem B5. A will be bluer. A and B each have one more blue than clear beaker, but in A, each beaker makes moreof a difference. In other words, A has 1/5 more blue than clear, and B has 1/7 more blue than clear, and since 1/5 islarger than 1/7, A is more blue. Another explanation is that to get B from A, you would need to add 1 blue and 1clear beaker, which (as a mix) is lighter than A. Therefore, B will end up being lighter than A.

Problem B6. A will be bluer. Its mix is the same as B’s, with one less beaker of clear water.

Problem B7. They are identical in color. One explanation is that each mix has twice as much blue as clear. A sec-ond explanation is that mixture A can be made by doubling the ingredients of mixture B.

Problem B8. The bluest is A. A is already bluer than B, so adding B’s mix into A won’t make it a deeper blue (thinkof what would happen with paint). This means that A ∪ B cannot be bluer than A, so A is the bluest of the three.Another explanation is that A has one more blue than clear, and so does A ∪ B, but in A ∪ B the extra blue doesnot contribute as much to the mixture (since there are more beakers of liquid in it).

Problem B9. If A ∪ B is bluer than A, then B must be bluer than A. This is true because A ∪ B is formed by addingB to A; if we add something to A and it becomes a deeper blue, whatever we added (B) must be bluer than whatwe started with (A). The reverse is also true: If A ∪ B is bluer than B, then A is bluer than B.

Problem B10. No such magical mixing device exists. Take the bluer of A and B; A ∪ B can never be bluer than it.You could also solve this problem using algebra and common denominators.

Solutions

Session 4: Solutions 108 Patterns, Functions, and Algebra

Problem B11. Yes, but only if A and B are equally blue. If A is bluer than B, adding B dilutes A, and the result (A ∪ B) will not be as blue as A. If A and B are equally blue, A ∪ B will be identical to both.

Problem B12. Use the total number of blue beakers, divided by the total number of beakers used. If the BQ of Ais M / N and the BQ of B is P / Q, then the BQ of A ∪ B is:

(M + P) / (N + Q)

Problem B13. There are several possible BQs for mixture B. Mixture B could contain only one beaker of blue (BQ = 1/1). A ∪ B would then have BQ 2/4 = 1/2. If we wanted A ∪ B to have BQ 3/6 (an equivalent fraction to 1/2),then mixture B would have to have BQ 2/3. Mixture B can have BQ 1/1, 2/3, 3/5, 4/7, 5/9, or any fraction of the formN / (2N - 1). Mixture B’s only requirement is that it must have a BQ larger than 1/2 (see Problem B10 for the reasoning).

Part C. QuadpersonProblem C1. The “after” Quadperson has the same scale to its facial features; the nose is still 4 times as wide as itis tall, and so forth.

Problem C2.This “after”Quadperson does not have the same shape as the original. In particular, the nose becomesa flat line, but other features are scaled differently from “before.”

Problem C3. Problem C1 is the relative comparison. Think about the angles and measurements in the body; weexpect, for example, the head to be a certain fraction of the size of the torso, and so forth.

Problem C4.The change in Problem C1, a relative comparison, keeps these measurements in proportion, while thechange in Problem C2, an absolute comparison, does not. In Problem C2, short lengths are made way too short(the nose, for example) by giving an absolute change in length, rather than a proportional change in length.

Problem C5. y1 = x/2.

Problem C6. y2 = x - 1/2.

Problem C7. Both graphs are straight lines. The graph of y1 goes through the origin (0, 0), while the graph of y2

does not. Additionally, the graph of y2 becomes negative if x < 1/2, not a good thing when measuring lengths.

Solutions, cont’d.

a. b.


Problem C8. Yes, the graph of y1 is proportional, since the input is always twice the output. Or, the output is halfthe input. (Compare that to the formula y1 = x / 2.) The graph of y2 is not proportional; try finding the outputs fortwo different values of x, then determine if they are proportional. This produces the different shape of Quadper-son in Problem C2.

Problem C9.

Problem C10. The equation for this table is y3 = 2x.

Problem C11. It is a proportional relationship because every output is twice the input, and if we multiply theinput by any number, we multiply the output by the same number. This graph, like the last proportional graph,passes through the origin (0, 0).

Problem C12. All proportional relationships have the equation y = kx, where k is some constant number. A linegraph represents a proportional relationship only when the line goes through the origin (0, 0).



Part D. Speeds, Rates, Steepness, and LinesProblem D1. First, the graph only allows us to decide whether a car is moving toward or away from the intersec-tion, and does not tell us any specific direction. In relation to the intersection, the movement of each car is as fol-lows:

• Car 1 moves to the intersection, then stops there.

• Car 2 does not move at all.

• Car 3 is moving away from the intersection at all times.

• Car 4 starts at the intersection and moves away from it at all times.

• Car 5 moves away from the intersection for 12 seconds, then stops.

• Car 6 stays at the intersection for 5 seconds, then moves away from it for 7 seconds, then stops.

• Car 7 moves toward the intersection, then stops a distance away from it.

Problem D2. The steepness of the line gives the speed of each car, in meters per second. For example, Car 4 startsat the intersection, then is 120 meters away after 4 seconds. Its rate of speed is then 120 / 4 = 30 meters per sec-ond. Using the same technique, and any two points on the line, we can find the speed for each car.

• Car 1 moves at 20 meters per second, then stops.

• Car 2 does not move at all.

• Car 3 moves at 20 meters per second.

• Car 4 moves at 30 meters per second.

• Car 5 moves at approximately 12 meters per second, then stops.

• Car 6 starts at rest, then moves at approximately 23 meters per second, then stops.

• Car 7 moves at approximately 13 meters per second, then stops.

Problem D3. Yes, Car 4’s distance varies directly with time. Its graph is a straight line passing through the origin(0, 0).

Problem D4. Yes, any horizontal line represents a stopped car (because the distance from the intersection is notchanging). Cars 1, 2, 5, 6, and 7 are stopped at some time.

You could argue that Car 2, or any car that stays the same positive distance away from the intersection, is movingin a circle around the intersection! It may be true that Cars 5, 6, and 7 all entered a traffic circle at the same time;as mentioned in Problem D1, the graph does not give us enough information to tell us in what direction the carsare moving, only their distance.

Problem D5. Answers will vary. Here’s one scenario: Car 6 started out smoothly from a red light, and continuedmoving away from the intersection at 20 meters per second. Then, suddenly, Car 6 was stopped short by a colli-sion just ahead between Cars 7 and 5.



HomeworkProblem H1. Chris’s method will produce tea that has too much mix.This method would work only if the originalrecipe called for the same amount of tea and water, which it doesn’t. In terms of fractions, 22/30 is not the sameproportion as 12/20.

Pat’s method is correct: 6/10 is the same proportion as 12/20, so the new mixture will have the right proportion ofmix.

Sandy’s method is also correct: 18/30 is the same proportion as 12/20.

Lee’s method is identical to Chris’s method. If you keep the difference between tea and water the same, there willbe too much tea mix added at 30 cups.

Chris and Lee’s suggestions are incorrect because they are absolute comparisons, when a relative comparison isneeded to keep the proportion of tea mix the same.

Problem H2.

d. The triangle with side lengths 6, 8, and 10 is similar looking. A good comparison is the measures of thethree angles of the triangle.

Problem H3.

a. The twins are Adam and David. We know they start at the starting line, so they must be the ones withoutany constants in their equations.

b. The oldest must be Eric, since he is the one whose equation includes the instruction “- 5”, which meanshe begins 5 meters behind the starting line.

c. Adam started at the starting line, and ran at 6 meters per second.

Brett started 7 meters ahead, and ran at 4 meters per second.

Caleb started 4 meters ahead, and ran at 5 meters per second.

David started at the starting line, and ran at 5 meters per second.

Eric started 5 meters behind the starting line, and ran at 7 meters per second.



The points of intersection on the graph represent when one brother passes another during the race. Theirtimes (on the horizontal axis) and their distance from the start (on the vertical axis) are the same.

e. Use t = 2 in all five equations, or refer to the graph. The order is:

Brett (15 m), Caleb (14), Adam (12), David (10), and Eric (9).

f. Caleb and David, who run at the same speed, stay the same distance apart. We know this because theirgraphs form parallel lines, which always stay the same distance apart.

g. Find the finishing times for the five by solving the equations for d = 30, or referring to the graph (draw a horizontal line at distance d = 30). Solving the equations shows that Adam and Eric tie for first (5 seconds), then Caleb (5.2), Brett (5.75), and David (6).

h. Adam and David’s relationships are proportional. Their graphs pass through the origin (0, 0), and theirequations are in the form y = kx.


1. Brett

2. Caleb

3. Adam

4. David

5. Eric

d.

Session 4 Proportional Reasoning - Learner · PDF filePatterns, Functions, and Algebra 91 Session 4 Session 4 Proportional Reasoning ... In Mixture A, there is one more blue beaker

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