Download 1:00 Activity: Grade Level Exploration of the Standards

June 10, 2008
page 1
Facilitator Notes
Professional Development: K-8 Mathematics Standards
Grades 6-8: Proportional Reasoning
1. OSPI is pleased to provide materials to use in teacher professional development sessions about the K-8 Mathematics Standards that were
approved by the State Board of Education on April 28, 2008. These materials provide a structure for two full days focused on helping
6-8 teachers understand the critical content embedded in these Standards. We hope that these materials will be used by local
schools and school districts, education service districts, and university teacher educators to help inservice and preservice teachers deepen
their personal understanding of key mathematics ideas. Feedback about the effectiveness of the materials and ways to improve them can
be sent to OSPI so improvements can be made.
2. The goal of these professional development sessions is to help participants deepen their personal understanding of some of the
mathematics embedded in the K-8 Mathematics Standards. With deeper understanding, teachers will be better able to (a) understand
students’ mathematics thinking, (b) ask targeted clarifying and probing questions, and (c) choose or modify mathematics tasks in order
to help students learn more.
3. The problem sets are presented to participants on separate “activity pages.” Sometimes you will want participants to work on all the
problems in a set, and then you will lead a debriefing of each of the problems. Sometimes you will want participants to work on
problems sequentially so that you can debrief each problem in order. The choice will depend on the particular problem set, your
preferences about facilitating discussions, and the needs of the particular group of participants you are working with.
4. Some of the problems may be appropriate for students to complete, but other problems are intended ONLY as work for the participants as
adult learners. After participants have solved problems, you might want to discuss which ones would be appropriate for students. There
are reflections after each problem set that can help participants begin to think about how knowledge of the underlying mathematics ideas
can help them plan more effective instruction for students.
5. Encourage participants to discuss their thinking with partners. This will help participants develop fluent language about the relevant
mathematics ideas. Sometimes you may choose to ask participants to work independently before talking with partners, but sometimes
you may choose to ask participants to work immediately with partners. Always allow sufficient time for participants to work on a
problem set (or on individual problems) before you begin the debriefing. Participants are more likely to contribute to the discussion if
they are confident about their answers and about their solution strategies. This will also model that it is important to give students ample
time to work on a problem before discussing answers to that problem.
6. Although there is no explicit attention to instructional practice in these content professional development sessions, discussing implications
for teaching will help deepen participants’ own understanding. You are encouraged to tailor those discussions to the needs of each
group of participants. For example, if participants are all using a common set of curriculum materials, you may want to lead some
June 10, 2008
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discussions related to those materials. Be careful, however, not to lose the emphasis on deepening participants’ knowledge of
mathematics.
7. Approximate times are given for each problem set, but you will need to create an agenda that responds to the specific parameters of how
you are working with participants. For example, you might schedule these sessions on two consecutive days or you might schedule
them across four half-days. Extra time will be needed for the Reflection at the end of the Problem Sets. There may be too many
problems for participants to complete comfortably in a two-day session, so you need to think carefully about which problems you ask
participants to solve in each Problem Set. Alternately, you may choose to omit some Problem Sets completely.
Logistics
These professional development materials were designed with the following assumptions about logistics for the meetings.
1. Participants will primarily be classroom teachers of mathematics from grades 6-8. (There are different sets of problems for teachers
from grades K-2 and 3-5.) Modifications may need to be made if there are significant numbers of ELL teachers or special education
teachers.
2. Participants should be seated at tables of 3-6 people each. Discussions among participants are strongly encouraged.
3. The problem sets need to be copied prior to the start of the sessions. You will also need a computer and projector for display of the
slides, chart paper and markers, graph paper, rulers, adding machine tape, and enough space for small groups of participants to work
comfortably. A document camera might also be useful.
The materials were developed by a team of Washington educators:
Kathryn Absten, ESD 114
George Bright, OSPI
Jewel Brumley, Yakima School District
Boo Drury, OSPI
Andrea English, Arlington School District
Karrin Lewis, OSPI
Rosalyn O’Donnell, Ellensburg School District
David Thielk, Central Kitsap School District
Numerous other people from Washington and from across the nation, provided comments about various drafts of these materials. We
greatly appreciate all of their help.
Publication date: June 10, 2008
June 10, 2008
Flow of Activities
Introductions
page 3
Slides
Notes
It is important that participants feel comfortable about participating in professional
development sessions that address their own personal mathematical understanding.
Some participants may be a bit nervous about the prospect of possibly making
mathematical mistakes in front of colleagues. Assure participants that everyone
makes mistakes occasionally, and in these sessions there are no consequences of
doing so.
Bookkeeping activities (e.g., sign in sheets) can be done now.
Be sure to change the slide so that the names of the facilitators are correct.
Introduce the facilitators.
Have participants introduce themselves. If the group is small enough, this can be
done with the whole group, but if the group is large, have participants introduce
themselves to the people “close by.”
Survey the group if you wish to know the approximate distribution of teachers by
grade or by school or by school district.
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Remind participants that since we are working on personal understanding of the
mathematics underlying the K-8 Mathematics Standards, some of the problems are
appropriate for adults but may not be appropriate for students. Teachers need to
know more mathematics, and at a deeper level, than students.
Students entering middle grades may, or may not, have a firm understanding of
fractions. Proportional reasoning is easier to learn, however, if students can think
flexibly about what fractions mean and how to represent fractions. Proportional
reasoning builds on an understanding of fractions. Proportional reasoning, in turn,
is a foundation for direct variation and linear relations (e.g., linear functions).
Flexible thinking about fractions supports students in using a variety of strategies
to solve proportional reasoning problems.
Problem Set 1
The goal is to review understanding of fraction.
about 75 minutes for
working on problems and
debriefing, with extra time
for the Reflection
In debriefing the problems in Set 1, emphasize that deep understanding of fractions
is important support for learning proportional reasoning. Being flexible with
fractions allows students to think in multiple ways about proportions.
June 10, 2008
Debriefing
page 5
This problem helps participants become more flexible at visualizing fractional
parts.
3/5 is the shaded part of the whole rectangle (though other answers are possible)
2/3 is the unshaded part of the whole rectangle
5/3 is the whole rectangle, if the shaded part is the unit
3/5 is the shaded part of the whole rectangle, and 2/3 of that is equivalent to the
unshaded part; that is, the unshaded part is 2/3 of the shaded part
2/3 of the shaded part is the unshaded part, and 3/2 of that is the shaded part
5/3 of the shaded part is the whole rectangle, and 3/5 of that is the shaded part
3/5 of the whole rectangle is the shaded part, and 5/3 of that is the whole rectangle
Be sure that participants see that for these problems, the final answer is the
unit for the initial fraction (that is, the fraction following “of”).
Ask participants which task seemed easier; most are likely to choose the first one.
Probe their responses to find out why they thought that one task was easier. Many
people find it easier to think about fractions with the same denominator, since the
number of pieces (or segments between 0 and 1 on a number line) is the same.
When the denominators are different, it is more difficult to coordinate the
representations of the fractions, since the number of pieces is different.
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One strategy is to find a great enough common denominator (e.g., 28 or 56) that
will make it easy to find three fractions equally spaced in between (e.g., 34/56,
36/56, 38/56).
Another strategy is to represent the fractions on a number line and use a visual
strategy to find three equally spaced fractions.
In the general case, changing the fractions to 4a/4b and 4(a+1)/4b allows you to
identify three fractions equally spaced: (4a+1)/4b, (4a+2)/4b, and (4a+3)/4b
4.5/7 is halfway in between 4/7 and 5/7, because the value of a fraction is
DIRECTLY proportional to the value of the numerator. By middle grades,
students should be able to deal with symbolism like 4.5/7, even though this is not
the typical symbolism for fractions.
Again, participants may find an appropriate denominator that allows them to find
three equally spaced fractions in between 5/7 and 5/6; or they might use a visual
strategy (e.g., number line).
In the general case, changing the fractions to ab/b(b+1) and a(b+1)/b(b+1) allows
you to identify three fractions equally spaced: (ab+a/4)/b(b+1), (ab+2a/4)/b(b+1),
(ab+3a/4)/b(b+1), though you would have to rewrite these in simpler form.
5/6.5 is not halfway in between 5/7 and 5/6, because the value of a fraction is
INDIRECTLY proportional to the value of the denominator. Showing the graphs,
y = x and y = 1/x might help participants see the difference in these two situations.
Push participants on whether it makes sense to think of 7/3 as 7 parts out of 3. If
they say, “No, because there are only 3 parts and you can’t have 7 parts if there are
only 3 parts to start with.” then ask them to create an alternate way of thinking
about 7/3.
Thinking of fractions as ratios or as a “multiplier” might be a better mental image
for fractions greater than 1 (or even fractions less than 0).
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No; multiplying 8 and 12 by counting numbers will “miss” lots of fractions
equivalent to 8/12, for example, 2/3, 4/6, 6/9, 10/15, 12/18, etc.
Yes, since we begin the process with the fraction in “most reduced form,”
multiplying 2 and 3 by each of the counting numbers will generate all of the
fractions equivalent to 2/3.
This difficult question raises some interesting mathematical issues. Depending in
the time available, you may choose to skip this problem.
Each of these sets of fractions, {8n/12n} and {2n/3n} is countably infinite (that is,
in a 1-to-1 correspondence with the set of whole numbers), so the sets are the same
size. Yet, {8n/12n} is a proper subset of {2n/3n}. You could do an entire
workshop on infinite sets, but that is not the point here. Many participants will say
that {8n/12} is 1/4 the size of {2n/3n}, since there are three elements of {2n/3n} in
between each element of {8/n/12n}. Actually, these two sets are the same size;
that is, they can be put in 1-to-1 correspondence with each other, so asking about
the percentage is not really meaningful. This question goes far beyond what
students in middle grades are expected to know, but teachers should understand
some of the mathematics behind this issue.
When teachers have a deep understanding of fractions it is easier for them to
understand students’ thinking. These problems provide a slightly unusual “twist”
on fraction understanding.
You may want to ask participants to decide which Core Content areas these
problems illustrate. Most likely, participants will select from among 6.1, 6.3
7.1, 7.2, 8.1, and 8.5.
June 10, 2008
Problem Set 2
about 75 minutes for
working on problems and
debriefing, with extra time
for the Reflection
page 8
The goal is to explore multiplicative reasoning and connect multiplication and
division. Deep understanding of the relationship between multiplication and
division will support deeper understanding of proportional reasoning.
Introduce these problems by noting that multiplying is an action, and
multiplication is an idea. Multiplicative reasoning involves use of the meaning of
multiplication and division to make sense of problems.
This is a “warm up” to get participants thinking about multiplicative reasoning.
Help them realize that when they answer part c, they may be using the distributive
property; that is, 150% of 40 can be thought of as 100% of 40 plus 50% of 40.
Some participants may say that in order to compute the volume, you have to have
three numbers, so in this case you cannot compute the volume. Help participants
see that they can think of volume of a rectangular prism in two ways: V = LWH as
three separate and distinct values or V = (LW)H, with two of the values already
multiplied. This is an application of the associative property of multiplication. Of
course, the volume of the box is 17x8 cm3 = 136 cm3.
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This problem helps participants see that the area of a rectangle is a very flexible
way to think about multiplication of two quantities: (a) whole numbers, (b)
fractions or decimals, and even (c) binomials.
Making 2 copies of 4 1/5 is easy, but making 2/3 of a copy of 4 1/5 may challenge
some participants. One strategy is to think of 2/3 of 3 (easy) and then 2/3 of 1 1/5.
Making 2/3 of 1 is easy, but 2/3 of 1/5 may take a bit more thinking. This strategy
yields:
2/3 of 4 1/5 = 2/3 of 3 + 2/3 of 1 + 2/3 of 1/5 = 2 + 2/3 + 2/15
Challenge participants to write the answer as a single mixed number and explain
how they found that number.
Many participants will begin by converting one of the measurements into the units
of the other measurement. This is not incorrect, but challenge participants to think
about how to solve this problem without doing any conversions.
Of course, any rectangular area could easily be the unit of measure, so a rectangle
that is 1 inch by 1 centimeter could be a unit of measure. If so, this rectangle has
an area of 15 in-cm units. Many teachers are uncomfortable with a unit of
measure like an in-cm, but there is no mathematical reason for avoiding such units.
The three parts of this problem help participants connect multiplication and
division. The three parts are different “views” of the same (or at least very
similar) mathematics ideas.
a. Many participants will compute 6 ÷ 2/3 by using the “invert and multiply” rule.
b. Some participants may struggle a bit at drawing a picture. The most common
one is subdividing 6 wholes into parts each of which is 2/3 of the whole. There
are 9 of these subdivisions.
c. Some participants will reason that if 6 is 2/3 of something, then 3 is 1/3 of that
thing, so 3/3 of that thing must be 9.
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If students understand how multiplication and division are inverse operations, they
will be more flexible in their thinking about proportional situations.
You may want to ask participants to decide which Core Content areas these
problems illustrate.
Problem Set 3
about 45 minutes for
working on problems and
debriefing, with extra time
for the Reflection
The goal is to explore solutions to problems that “obviously” require proportional
reasoning.
Some participants may view these problems as easier than some of the fraction
problems, and they are probably correct. Fraction problems were presented first,
however, because a strong base of fraction knowledge will support the learning of
proportional reasoning.
Brian’s view is supported by computing differences in populations between years.
Darlene’s view is supported by computed percentage of change in populations
between years. Ask participants which view they think is better, and then ask
them to explain their choices. [Darlene’s view aligns with the impact of the
increased population on the two towns.]
The numbers in this problem are “nice,” since 0.86 = 2 x 0.43, so there are many
ways to reason about this problem without setting up and solving a proportion
algebraically. For example, 0.01 pound costs $0.02, so 0.50 pound costs $1.00.
Or, $0.01 buys 0.43/86 = 0.005 pound, so $1.00 buys 0.5 pound.
This problem may elicit reasoning that is more aligned with number sense than
with proportional reasoning. If there are a variety of strategies used by
participants, be sure that several different strategies are presented during the
debriefing.
June 10, 2008
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This problem is computationally more challenging, since the numbers are not so
“nice.” However, the underlying mathematics is exactly the same. For
example, 0.01 pound costs (86/46) cents, so for $1.00 you can buy 1/(86/46) =
46/86 pounds. Or, $0.01 buys 0.46/86 pound, so $1.00 buys 46/86 pound. If
participants have calculators, they are likely to give the calculator output
(0.5348837, approximately) as “the” answer. This would be a chance to discuss
the difference between the exact answer (46/86 pound) versus the approximate
answer (0.5348837 pound). Mathematically, there is no reason NOT to leave the
answer as an indicated division (e.g., 46/86).
You may also want to probe participants’ reactions to Problems 3.2 and 3.3.
Which was more difficult? Why?
If we assume that the ratios are exact, then the number of students is some multiple
of 30. We want to have that same number of students also be a multiple of 25, so
the number of students must be some multiple of 150. So every 5 existing teachers
must become 6 teachers; that is, the school has to hire 20% more teachers.
While debriefing this problem, it is important to highlight the assumptions that are
being made (e.g., the number of students is a multiple of 30). You might also ask
participants if they would have approached the problem differently if the question
had been, “What percentage increase in teachers is needed in order to reduce the
ratio to 25:1?”
Proportional reasoning involves multiplicative thinking rather than additive
thinking. Additive thinking is characterized by computing differences between
numbers, while multiplicative reasoning is characterized by computing ratios of
numbers.
You may want to ask participants to decide which Core Content areas these
problems illustrate.
June 10, 2008
Problem Set 4
page 12
The goal is to explore more complex proportional reasoning problems.
about 60 minutes for
working on problems and
debriefing, with extra time
for the Reflection
The “Mr. Tall/Mr. Short” problem is a classic problem in educational research.
Students who are thinking additively give 8 paper clips as the answer (i.e., 6 + 2),
while students who are thinking multiplicatively give 9 paper clips as the answer
(i.e., 6 x (6/4)).
Problems 4.1 and 4.2 may be easy for participants, but they are very appropriate
for students. This may be an opportunity to lead a discussion about how
participants select tasks to present to students. Teachers should not rely ONLY on
the choices in their curriculum materials to choose problems that align with the
needs of students, but some participants may not have thought about this issue
explicitly.
This problem involves two slides.
The answers to these problems are:
1. Thermos A
2. Thermos B
June 10, 2008
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3. Thermos B
4. indeterminate: We don’t know how much stronger B is than A at the start. A
becomes stronger and B becomes weaker, but we can’t make a quantitative
comparison of the end result.
Since 2 quarts = 2 x 4 cups = 8 cups = 8 x 8 oz. = 64 oz., there are 64/6 = 10 2/3
glasses of lemonade in 2 quarts. Each glass requires 3 tablespoons of lemonade
mix, so 10 2/3 glasses requires 32 tablespoons of lemonade mix.
Participants might make a table, write a proportion, guess and check, draw a graph,
or model with a picture.
Some participants may question whether the lemonade mix “disappears” as part of
the mixture or whether the volume of a glass is 6 oz. plus 3 tablespoons.
Typically, we expect students to assume that a glass of lemonade is exactly 6 oz.
No. 3.5/8.5 ≠ 5/11, so the two shapes are not similar. Since enlargement creates
similar figures, you cannot enlarge the photo to the size of 8.5 x 11 inches.
June 10, 2008
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These problems are basic kinds of proportional reasoning problems. The often
appear (perhaps in different contexts) on standardized tests. Be sure that
participants are very comfortable with the solutions to these problems.
Proportional reasoning appears in many real-world applications of mathematics. If
students do not understand proportional reasoning, they are not likely to be
successful in high school Algebra.
You may want to ask participants to decide which Core Content areas these
problems illustrate.
Problem Set 5
about 90 minutes for
working on problems and
debriefing, with extra time
for the Reflection
The goal is to explore more complicated proportional reasoning problems in a
variety of contexts.
In general, these problems are likely to be challenging problems for many
participants. This set may take a considerable amount of time to complete and
debrief. You may want to ask participants to solve these problems one at a
time, rather than as a set. This will allow you a chance to debrief the
problems in order.
One of the assumptions in this problem is that the interest paid each month is the
same. Even though months do not all have the same number of days, this
simplifying assumption makes the problem accessible.
a. A monthly payment of $1,000 is equal to interest of $12,000 per year; this is
a 6% mortgage.
b. Her new payment is 165,000/200,000 of $1,000 or 165/200 x $1,000 or $825.
You may want to debrief participants’ understanding of this problem by asking
some of these questions:
• Is this a proportional reasoning problem? Why or why not?
• Do you need to compute the yearly interest on the new balance to find the
monthly interest?
• If Gertrude paid off half of her mortgage, what would her monthly payment
be? Why?
June 10, 2008
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This may be a rather difficult problem for participants, and it will probably take
significant time for participants to complete and then debrief. If time is short,
you may want to skip this problem.
The answer is 2/3 of his ORIGINAL daily ration. The 24 days of the man’s water
is changed to 18 days of water for two people. Thinking proportionally, every 4
days of the man’s water is changed to 3 days of water for two people (which is
equivalent to 6 days of water for one person). So 4 days of the man’s water is
changed to 6 days of the woman’s water; that is, she is getting 2/3 of his original
portion each day.
A common incorrect answer is 4/9, generated when participants use 27 days rather
than 24 days of water remaining. (They overlook the comment, “After 3 days”
which means that the man drank 3 days of water before the woman arrived on the
scene.)
An alternate correct solution is that after day 3, the man’s ration was 1/24 of the
total each day. When the woman arrived, she was to get 1/36 of the total each day.
The answer is to find a ratio of these two quantities: (1/36):(1/24) which is 2/3.
Debrief this problem with some of these questions:
• Would you give this problem to students? If so, at what grade or in what
course?
• What would you want students to learn from solving this problem?
• How would you support them in working on this problem?
Remember, however, that the discussion of the assumptions is not as important as
recognizing that we need to be aware that (a) the assumptions we are making and
(b) how those assumptions affect the solution to a problem. Problem solvers
sometimes are expected to make “reasonable” assumptions, but those assumptions
need to be explicitly acknowledged.
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This is becoming a “classic” problem. The key to creating a solution is
remembering that the number of married men and the number of married
women are equal.
One strategy is to find a “common numerator;” this number represents the number
of married couples. So, for men, 2/3 = 6/9 and for women, 3/5 = 6/10. For every
9 men, 6 are married, and for every 10 women, 6 are married. The smallest set of
people that “works” is 6 married men, 3 unmarried men, 6 married women, and 4
unmarried women. That is, 12 out of 19 people are married.
A second strategy is to use an example. Suppose there are 60 women. Then 3/5 of
them (36) are married. So there are 36 married men; these are 2/3 of the men, so
there are 54 men. The total number of people is 60 + 54 = 114, and 72 of them are
married. The proportion of married people to people is 72/114 = 36/57 = 12/19.
12/19 of the people are married.
A third strategy is to “algebratize” this work by starting with X women and
repeating essentially the same reasoning with variables instead of numbers.
A fourth strategy is to draw a picture (or use counters or tiles) to represent the sets
of men and women (see the 2002 NCTM Yearbook).
Help participants make connections between the different representations. For
example, finding a common numerator can be demonstrated visually (e.g., by
using tiles). For some people, the visual representation will generate a solution
more quickly than using algebra.
Collectively, participants are likely to write two non-equivalent equations: Y = 3F
and F = 3Y. Since a foot is a smaller unit of measure than a yard, the NUMBER
of feet will always be greater than the NUMBER of yards.
One of the points of confusion here is that 1 yd. = 3 ft. is also a true equation.
However, F and ft. are not the same; F is the NUMBER of feet and ft. is the unit of
measure, foot. This confusion between the symbol used as a variable and the
symbol used as an abbreviation for the unit of measure is a confusion that students
often have. It is useful for teachers to explicitly recognize this kind of confusion
in students’ thinking.
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Speed vs. time and distance vs. time graphs are often difficult to distinguish. (This
might be analogous to the confusion between area and perimeter.) The slope of a
distance vs. time graph is speed, and the slope of a speed vs. time graph is
acceleration. (Yes, this is physics, but it is also important mathematics.)
The speed vs. time graph for Joe’s walk is composed of 4 horizontal line segments:
a. the first horizontal segment is slightly above zero
b. the second horizontal segment is zero; that is, it is on the horizontal axis.
This signifies that Joe was standing still.
c. the third horizontal segment is much more positive
d. the fourth horizontal segment is mathematically negative (sloping down,
indicating going in the reverse direction), but some participants will argue that
speed is always positive (and that is okay for this problem)
Participants may think that these problems are too difficult for their students.
However, there are some middle grades students can solve these problems.
Teachers need to have some “advanced” problems in their repertoire for those
advanced students.
You may want to ask participants to decide which Core Content areas these
problems illustrate.
Problem Set 6
The goal is to show an application of proportional reasoning.
about 75 minutes for
working on problems and
debriefing, with extra time
for the Reflection
This activity is taken from Navigating through Measurement in Grades 6-8
published by the National Council of Teachers of Mathematics.
June 10, 2008
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If you create a ruler marking in π units, you can use the relationship, C = π d, to
find the diameter of a round object by measuring the number of π units needed to
measure the circumference of that object. For example, if a tree is 10π inches
around, it has a diameter of 10 inches.
This activity is one that is specifically designed to use with students, so debriefing
this problem provides an opportunity to discuss ways of helping students
understand proportional reasoning. Remember, however, that the goal of this
professional development session is deepening participants’ understanding of
mathematics ideas.
This activity will engage students in a real-word application of one of the standard
formulas of middle grades mathematics: C = πd. Knowing how mathematics can
be applied will help develop understanding of that mathematics.
You may want to ask participants to decide which Core Content areas these
problems illustrate.
Problem Set 7
The goal is to reflect on personal learning about proportional reasoning.
about 60 minutes for
working on problems and
debriefing, with extra time
for the Reflection
These questions are intended to help teachers reflect on what they have learned
and how they might use that new knowledge in planning instruction for students.
One alternative is to change the order of Problem Set 7 and Problem Set 8.
June 10, 2008
page 19
Some terms that are likely to come up are numerator, denominator, ratio, part to
whole, part to part, rational number, equal parts, etc.
The “teacher” definition may be written in language that appears to be more
mathematical than the “student” definition. If so, be sure that the underlying
mathematics ideas in the two definitions are equivalent. We do not want the
“student” definition to be less accurate than the “teacher” definition.
Problems 7.1 and 7.2 should generate some interesting discussion among
participants, and it may be difficult to reach complete consensus on the answers. It
is more important for participants to hear different perspectives than it is to impose
particular answers for these questions.
A fraction is a number, but it is “made up of” two numbers (numerator and
denominator). Students (and teachers) need to understand that fractions are
numbers.
There is not total agreement about definitions for ratio and rate. A ratio is often
considered to be a comparison of two quantities that have the same label; for
example, 6 ft.:3 ft. A rate is often considered to be a comparison of two quantities
that may not have the same label; for example, 500 miles:10 hours. In a ratio, the
labels “cancel,” so a ratio can be considered as a number. In a rate, the labels do
not “cancel,” so a rate often has a number part (e.g., 500:10) and a measurement
part (e.g., miles:hour or miles per hour).
June 10, 2008
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Participants should have already spent two days in professional development on
the K-8 Mathematics Standards, so Problem 7.4 will help connect the proportional
reasoning problems with the Standards that teachers are responsible for teaching.
This problem could take quite a bit of time to complete, so you may want to ask
different groups of participants to take different Problem Sets and classify the
problems as addressing one or multiple Performance Expectations.
You may want to ask participants to decide which Core Content areas these
problems illustrate.
Problem Set 8
about 90 minutes for
working on problems and
debriefing, with extra time
for the Reflection
The goal is to explore extensions of some of the proportional reasoning problems
already solved. Again, some participants will view these problems as easier than
some other problems already completed. Comments about the relative ease of
problems can become an opening to discuss ways of sequencing problems for
students to help them generate deep understanding of proportional reasoning.
There are many correct answers, but the data must show a strictly proportional
relationship. For example,
x:
y:
2
1
4
2
6
3
8
4
10
5
12
6
14
7
16
8
June 10, 2008
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There are many correct answers, but the data must show a linear relationship for
each of the segments, with a “more positive” relationship for the second segment.
For example,
x:
y:
2
1
4
2
6
3
8
4
10
7
12
10
14
13
16
16
There are many correct answers, but the data must show a linear relationship for
each of the segments, with a “less positive” relationship for the second segment.
For example,
x:
y:
2
2
4
4
6
6
8
8
10
9
12
10
14
11
16
12
Possible follow-up questions for these three graphs are given below:
• What is the difference between a proportional relationship and a linear
relationship?
• How did you choose values before and after the point at which the slope changed
abruptly?
• How do you know by looking at someone else’s table of values where the point
of changing slopes occurs?
The boy can bike 8 miles in 40 minutes and walk 8 miles in 160 minutes, so he
saves 120 minutes or 2 hours.
June 10, 2008
page 22
If B is the actual time biking, then the distance is 10B. The walking distance is the
same, so the walking time is 10B/4 or (10/4)B or 2.5B; that is, the walking time is
2.5 as long as the biking time.
The average speed is the distance (i.e., twice the biking distance, or 2x10B = 20B)
divided by the total time (i.e., the biking time plus the walking time, or
B + 2.5B = 3.5 B).
average speed - (20B)/(3.5B) = 20/3.5 = 40/7 mph.
The average speed does not depend on the exact biking time.
a. 1 ft./sec. = 1 ft./sec. x 1 mile/5280 ft. x 60 sec./1 min. x 60 min./1 hr.
= 3600/5280 mile/hr. = 15/22 mile/hr., which is about 0.68 mph
b. 50 ft./20 sec. = (50/20)(15/22) mph = 75/44 mph, which is about 1.70 mph
c. The solution for (b) could repeat the steps from part (a) or you can use the
fraction, 15/22, to help, as shown above.
Changing units is something that happens a lot in science, but less often in
mathematics.
a. compute:
(31 mL/53 sec.)(60 sec./1 min.) = (31x60)/53 mL/min., which is about 35
milliliters per minutes
b. compute:
(18 mL/97 sec.)(60 sec./1 min.)(60 min./1 hr.) = (18x60x60/97) mL/hr.,
which is about 668 mL/hr.
These problems provide another opportunity to distinguish between exact answers
(in indicated multiplication/division form) and approximate answers (such as
might be generated on a calculator).
If there is time, you may want to ask participants to decide which Core
Content areas these problems illustrate.
June 10, 2008
Closing
page 23
Be sure to thank participants for their engagement in the activities.
You may need to distribute clock hours forms or complete other paperwork.