Download Comparing Bits and Pieces

At a Glance | Problem 1.1
Be sure to discuss to discuss Chung’s statement, which is incorrect. Students may
need to play out some examples to notice that it should be “For every +1 the
eighth grade plans to raise, the seventh grade plans to raise +3.” The numbers are
reversed, which is a common error in writing ratio statements.
C. The teachers’ goal is +360. One way to know
this is that the teachers’ goal is more than
each of the two grades (+210 more than
one grade, and for every +60 the teachers
plan to raise, one grade is raising only +50).
That means the teachers’ goal is less than
the seventh-grade goal; in fact it is 4
5 of the
seventh-grade goal. Four-fifths of +450 is
+360.
Assignment Guide for Problem 1.1
Applications:1–2 | Connections:35–40
Answers to Problem 1.1
A. All the claims are true except Chung’s.
Reasoning will vary. Check students’ work.
B. Answers will vary. Possible answers: The
eighth-grade goal is 1
3 of the seventh-grade
goal. For every +10 the 3 eighth graders
plan to raise, the seventh graders plan to
raise +30. The seventh-grade goal is +300
more than the eighth-grade goal.
Comparing Bits and Pieces
2
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Investigation1
At a Glance | Problem 1.2
The language of comparison statements is challenging. Help students to
understand that in the statement, Theratioofthesixth-gradegoaltothe
seventh-gradegoalis60to90, it is important that the first number matches
the first described quantity and the second number matches the second.
2. Answers will vary. Possible answers: For
every +30 the sixth graders plan to raise,
the eighth graders plan to raise +15. For
every +90 the sixth graders plan to raise,
the teachers plan to raise +108. For every
+30 the eighth graders plan to raise, the
seventh graders plan to raise +90.
Assignment Guide for Problem 1.2
Applications:3–4 | Connections:41–43
Extensions:65–70
Answers to Problem 1.2
C. 1. Each number represents a dollar amount.
For every 60 dollars in the sixth-grade
goal, there are 90 dollars in the seventhgrade goal.
A.
Goal
$300
Goal
$450
Goal
$150
Goal
$360
$270
$405
$135
$324
$240
$360
$120
$288
$210
$315
$105
$252
$180
$270
$90
$216
$150
$225
$75
$180
$120
$180
$60
$144
$90
$135
$45
$108
$60
$90
$30
$72
$30
$45
$15
$36
2. Answers will depend on the answers to
Question B, part 2. Possible answers:
The ratio of the sixth-grade goal to the
eighth-grade goal is 30 to 15. The ratio
of the sixth-grade goal to the teachers’
goal is 90 to 108. The ratio of the eighthgrade goal to the seventh-grade goal is
30 to 90.
D. 1. The sum of two even numbers is even.
$0
Sixth
grade
$0
Seventh
grade
$0
Eighth
grade
(Two rectangles with height 2 can be put
together to form a larger rectangle with
height 2.)
2. Students may notice that if you double (or
halve) both numbers in the ratio, you get
an equivalent ratio. Some may notice the
same pattern based on tripling (or cutting
in three), etc.
$0
Teachers
B. 1. Ben’s second claim is also true. The sixth
graders plan to raise 10 sets of +30. The
seventh graders plan to raise 10 sets of
+45. So for each set of +30 in the sixthgraders’ goal, there is a set of +45 in the
seventh-graders’ goal.
Comparing Bits and Pieces
3. Some students may notice that
when ratios are equivalent, the same
multiplication relationship has to hold
between the two numbers in the ratio.
For example, the ratio 60 to 180 is
equivalent to the ratio 10 to 30 because
the second number in each ratio is equal
to three times the first number.
4. Here are more examples of equivalent
ratios: 60 to 30 and 30 to 15; 45 to 54
and 90 to 108; 60 to 180 and 30 to 90.
2
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Investigation1
At a Glance | Problem 1.3
7 is 7 .
D. 1. a. The distance between 0 and 10
10
3
7
1.
b. The distance between 5 and 10 is 10
7 and 1 is 3 .
c. The distance between 10
10
3
2
d. The distance between 5 and 1 is 5 .
1
2. a. The distance between 0 and 1
3 is 3 .
1 1
b. The distance between 1
3 and 2 is 6 .
2 1
c. The distance between 1
3 and 3 is 3 .
2 1
d. The distance between 1
2 and 3 is 6 .
1
e. The distance between 1
2 and 1 is 2 .
1
f. The distance between 2
3 and 1 is 3 .
4 =
E. 1. Answers will vary. Possible answers: 12
1=2=3= 5 = 6
3 6 9 15 18
2
2. Answers will vary. Possible answers: 1
4, 5,
3 3 4 3
5
8 , 10 , 10 , 12 and 12
3. Answers will vary. Possible answers: With
Assignment Guide for Problem 1.3
Applications:5–18
Connections:44–46,52–53,55–64,71–80
Answers to Problem 1.3
A. 1. Check students’ strips.
2. Answers will vary. Possible answers: For
thirds: Make an S with a strip and press
it together, keeping all three pieces the
same length. For thirds: Fold a strip
around itself until all three parts are
equal length and then press it together,
keeping all three sections of equal length.
For sixths: Fold thirds, then fold in half.
For fifths: Fold a strip around two fingers
two-and-a-half times. Take the strip off
your fingers—still rolled—and flatten the
roll, making the five sections as close as
possible to the same length.
fraction strips, you line up the ends and
look for the marks that match up exactly.
With number lines, you can cut lengths
into smaller pieces (such as cutting each
fourth into two pieces to get eighths).
With numbers, you can multiply (or
divide) the numerator and denominator
by the same number to get equivalent
fractions.
B. 1. Fold halves in half, then in half again.
2. One part of a halves strip is four times the
size of one part of an eighths strip.
3. You can fold a thirds strip in half to make
sixths. You can fold a thirds strip in thirds
to make ninths. You can fold a thirds strip
in fourths to make twelfths, etc.
4. He is correct. You can fold a strip in thirds
4. Halves, thirds, fourths, sixths, eighths,
and then use this to mark the number
line from 0 to 1. This locates the point 1
3
on the line. You can also use the fraction
strip to measure a distance from 2
3 to 1,
5
1
or from 2 to 6 . In both these cases the
distance is 1
3.
ninths and tenths. The fifths strip may
have marks that are close to marks on
the twelfths strip, but these marks will
not line up exactly (unless the folds are
6 1
4
inaccurate). This suggests 1
2 = 12 , 3 = 12 ,
1 = 3 , 1 = 2 , 3 = 4 , 5 = 6 , etc.
4 12 6 12 9 12 10 12
5. This is a good idea. You would have to
C. 1. Additional fractions will vary, but include
10 12
14
15 , 18 , and 21 .
2. Answers will vary. Possible answers: If you
make sure that the whole strip matched
the whole thermometer. Then you could
fold the fraction strip in a way such that
one piece matched the shaded portion
of the thermometer, or some multiple
of folded pieces matched the shaded
portion.
double (or triple, or...) both the numerator
and denominator, you get a fraction
equivalent to the one you started with. If
you add the numerator to itself and the
denominator to itself, you get a fraction
equivalent to the one you started with.
Comparing Bits and Pieces
2
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Investigation1
At a Glance | Problem 1.4
C. 1. These dollar amounts are correct. +75
is 1
4 of +300 because 300 = 4 * 75. The
other dollar amounts on the thermometer
are multiples of +75.
2. (See Figure 1, next page.)
D. 1. Using only fractions for which students
2 3
have made fraction strips: Day 4: 1
3, 6, 9,
3 6 9
5 10
4
12 . Day 8: 4 , 8 , 12 . Day 10: 6 , 12 .
2. One segment on the fourths strip is equal
to two segments on the eighths strip.
E. 1. 5
4 or equivalent
2. (See Figure 2, next page.)
Comparing Bits and Pieces
2
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Investigation1
At a Glance | Problem 1.4
Figure 1
Figure 2
4
4
Goal
$300
Goal
$300
3
3
Goal
$300
9
9
~$266.67
8
9
Goal
$300
~$233.33
2
3
$200
6
9
~$166.67
5
9
2
4
$150
1
3
$100
4
9
$100
3
9
1
4
$75
2
9
Comparing Bits and Pieces
5
6
4
5
$200
4
6
$150
3
6
$100
2
6
$50
1
6
1
9
~$33.33
Day 4
$250
Goal
$300
1
4
$75
~$66.67
Day 2
2
4
$150
~$133.33
6
6
3
4
$225
$200
Goal
$300
$240
7
9
3
4
$225
4
4
Day 6
Day 8
3
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Day 10
Day 9
Investigation1
At a Glance | Problem 1.5
4. Thinking in units of +50, Kate would write
of 150 make 450. Margarita’s drawing
suggests that she is thinking about
folding the strip into 450 pieces. 150 of
these small pieces are in each third. 300
450
represents +300 raised out of a goal
of +450.
5 fifty-dollar bills : 6 fifty-dollar bills, or
5 : 6. Twenty is not a factor of both 250
and 300, but she might think of +250 as
1
12 1
2 twenty dollar bills and write 12 2 : 15.
5. Answers will vary. Samples: The ratio of
the amount of money raised by the sixth
grade to the amount raised by the eighth
grade is +250 : +112.50, or 20 : 9. For
every +250 the sixth graders raised, the
eighth graders raised +112.50. Or for
every +2 the sixth graders raised, the
eighth graders raised +1.
2. The sixth graders raised +250 by the end
50 250
of Day 10. 5
6 = 60 = 300
3. The teachers raised $450 by the end of
50 450
Day 10. 5
4 = 40 = 360
C. 1. Yes, this is a correct statement.
Explanations will vary.
Sample: 5 : 6 = 250 : 300. Check
students’ work.
D. 1. Answers will vary. Possible answers: The
sixth graders raised 5
6 of their goal. The
2. The ratio of the amount of money raised
ratio of the amount of money raised by
the sixth graders to the amount of money
raised by the seventh graders is 5 to 6.
by the 6th graders to the amount of
money raised by the 7th graders is 25 : 30.
3. These ratios are equivalent. Students
2. Fractions are useful when you want to talk
might reason from thermometer
drawings. There are 10 sets of 25 in
$250 and 10 sets of +30 in +300. On
thermometers showing +250 and +300
totals, +25 and +30 would both be
1 mark. Or they might reason
at the 10
numerically. If you multiply both parts
of the ratio 25 : 30 by 10, you make an
equivalent ratio.
Comparing Bits and Pieces
about a part of a whole. Ratios are useful
when you want to compare numbers.
2
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Investigation1
At a Glance | Problem 2.1
B. 1. Different partitioning strategies will
lead to different forms of this rate.
(See Figure 3, next page.)
Assignment Guide for Problem 2.1
Applications:1–6 | Connections:25–26
Extensions:31–33
2. Different partitioning strategies will
lead to different forms of this rate.
(See Figure 4, next page.)
Answers to Problem 2.1
C. 1. Several answers are possible. The picture
will lead many students to say that could
be 8 people in her group.
A. 1. Different partitioning strategies will lead
to different forms of this unit rate. Some
students may see this as an implied
division with six segments divided up
among the four people.
(See Figure 1.)
2. There are multiple answers. For example,
there could be 4 people in her group.
There could be 2 people in her group.
It would be unusual to share the chewy
worm between 2 people for that drawing.
2. Different partitioning strategies will lead
to different forms of this unit rate.
(See Figure 2.)
Figure 1
Each person gets 1 1
2 segments.
1
2
3
4
1
2
3
4
1
2
3
4
Each person gets 3
2 segments.
1
2
3
4
1
2
3
4
Each person gets 4
6 segments.
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
Figure 2
Each person gets 1 1
3 segments.
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
Each person gets 4
3 segments.
1
2
3
4
Comparing Bits and Pieces
5
6
1
2
3
4
5
6
1
2
3
4
5
6
2
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Investigation2
At a Glance | Problem 2.1
3. The answer depends on the number of
the chewy worm. If 2 people share, there
would be 3 segments per person. Each
1
person is then getting 3
6 or 2 of the chewy
worm. If 3 people share, there would be
2 segments per person. Each person is
1
then getting 2
6 or 3 of the chewy worm.
If 6 people share, there would be 1
segment per person. Each person is then
getting 1
6 of the chewy worm.
people in the group. If 8 people share,
3
there would be 6
8 or 4 of a segment per
person. If 4 people share, there would be
6
1
4 or 1 2 segments per person. If 2 people
share, there would be 3 segments per
person. If 3 people share, there would be
2 segments per person. If 6 people share,
there would be 1 segment per person.
4. There are multiple answers. Note: The
D. Answers will vary. If the worms are the same
unit rate becomes the number of pieces
each person gets (numerator), and
the number of segments becomes the
whole (denominator). If 8 people share,
3
there would be 6
8 or 4 of a segment per
person. Each person is then getting
size, sharing a 6-segment worm among
4 people gives a bigger share: 6
4 of a worm
12
instead of 8 of a worm. If the segments
are the same size, then the worms are
different sizes and the shares would be
equal segments per person.
3
4
E. Possible answer: Every time I found a
6
of the chewy worm. If 4 people share,
per-person amount, I found a unit rate. This
told me how many segments of the chewy
worm each person got. This happened in all
of Questions A–D.
there would be 1 1
2 segments per person.
6
11
4
2
Each person is then getting
6
or
6
of
Figure 3
1
Each person gets 1
2 + 6 of a segment.
1
2
3
4
5
6
7
8
9
10
11
12
1
2
3
4
5
6
7
8
9 10 11 12
9 10 11 12 1
2
3
4
5
6
7
8
9 10 11 12
Each person gets 2
3 of a segment.
1
2
3
4
5
6
7
8
Figure 4
1
Each person gets 1
2 + 10 of a segment.
1
2
3
4
5
1 2 3 4 5
Each person gets 3
5 of a segment.
1
2
3
Comparing Bits and Pieces
4
5
1
2
3
4
5
1
2
3
3
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4
5
Investigation2
At a Glance | Problem 2.2
3. Yes. In both cases, the relationship
Some possible answers include:
Crystal’s
Segments
Alexa’s
Segments
Total
Segments
2
1
3
4
2
6
6
3
9
8
4
12
10
5
15
12
6
18
14
7
21
between the boys’ shares is the same. For
every 2 segments Jared gets, Peter gets
3 segments. In the case of 10 to 15, Jared
gets 2 segments 5 times, and Peter gets
3 segments 5 times.
4. There are two unit rates with one of the
boys getting 1 segment: Jared gets 2
3
of a segment for every 1 segment for
Peter. Jared gets 1 segment for every
11
2 segments for Peter.
C. 1. Possible answers: Caleb might be 8 years
old and Isaiah 6 years old. They might be
4 and 3. They might be 12 and 9.
B. 1. They could share a 25-segment
2. a. Crystal gets 2
3 of the worm she shares
worm (Jared gets 10 segments, Peter
gets 15), or a 5-segment worm (Jared
gets 2 segments, Peter gets 3), or any
multiple of 5 segments.
with Alexa. Alexa gets 1
3 of the worm.
b. Jared gets 2
5 of the worm he shares
with Peter. Peter gets 3
5 of the worm.
c. Answers will vary. Possible answer:
Some possible answers include:
Jared’s
Segments
Peter’s
Segments
Total
Segments
2
3
5
4
6
10
6
9
15
8
12
20
10
15
25
12
18
30
14
21
35
If I write the fraction of the worm
each person gets using the same
denominator, the ratio of the
numerators is equivalent to the
ratio of the number of segments
each person gets.
8
For example, Caleb gets 14
(or 4
7 ) of the
6
3
worm and Isaiah gets
(or ) of the
14
2. The ratio of the number of parts Jared
gets to the number of parts Peter gets is
10 to 15, or 2 to 3.
Some possible answers include:
Jared’s
Segments
Peter’s
Segments
Total
Segments
Ratio for
Jared to Peter
2
3
5
2 to 3
4
6
10
4 to 6
6
9
15
6 to 9
8
12
20
8 to 12
10
15
25
10 to 12
12
18
30
12 to 18
14
21
35
14 to 21
Comparing Bits and Pieces
7
worm. The ratio of the segments is 8 : 6
(or 4 : 3). If you add the two numbers
in the ratio, you get a number that can
be the denominator of the fraction of a
worm each person gets. For example,
if the ratio of segments is 4 : 3, then
one person gets 4 out of every (4 + 3)
segments, or 4
7 of a worm, and the other
3
gets 7 .
2
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Investigation2
At a Glance | Problem 2.3
Suggested Questions
• WhataresomepatternsyounoticeintheratetableinQuestionA?Arethere
morewaysthanonetofindmissingentriesintheratetable?
• Onepersondoubledthecostof10chewyfruitwormstogetthecostof
20chewyfruitworms.Anotherpersonfoundthecostof1chewyfruitworm
andthenmultipliedby20.Whydobothstrategiesgivethesameanswer?
3. 10 ounces of popcorn kernels are needed
Assignment Guide for Problem 2.3
to make 40 cups of popcorn. 25 ounces
of popcorn kernels are needed to make
100 cups of popcorn.
Applications:16–24 | Connections:29–30
Extensions:36–37
4. 50 ounces of popcorn kernels are
needed to make 200 cups of popcorn or
100 servings.
Answers to Problem 2.3
5. 1
4 ounce of popcorn kernels are needed
to make 1 cup of popcorn.
A. 1. (See Figure 1.)
C. 1. Rate tables show that ratios can be
2. It will cost +.30 for 3 worms. It will cost
multiplied or divided to find equivalent
ratios. For example, if you know one
ratio, you can find another equivalent
ratio by doubling, tripling, or halving, etc.
+30 for 300 worms.
3. You can buy 500 worms with +50. You can
buy 100 worms with $10.
4. The unit price for one worm is +.10. The
2. Unit rates are easy to work with because
unit rate for one worm is +.10, i.e., +.10
per worm.
you multiply them by the quantity or
number of units to find an equivalent
ratio. For example, if you know that the
unit rate is 4 cups to 1 ounce, then for
3 ounces you will get 3(4) = 12, or
12 cups.
B. 1. (See Figure 2.)
2. 48 cups of popcorn can be made from
12 ounces of popcorn kernels. 120 cups
of popcorn can be made from 30 ounces
of popcorn kernels.
Figure 1
Chewy Fruit Worm Pricing
Number of Worms
Reduced Price
1
5
10
15
30
90
150
180
$.10
$.50
$1
$1.50
$3
$9
$15
$18
Figure 2
Popcorn Table
Number of Cups of Popcorn
4
8
12
16
20
24
28
32
36
40
44
48
Number of Ounces of
Popcorn Kernels
1
2
3
4
5
6
7
8
9
10
11
12
Comparing Bits and Pieces
2
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Investigation2
At a Glance | Problem 3.1
E. 1. a. This is possible. In the image on the
Assignment Guide for Problem 3.1
left, the temperature was 10° C; in the
photo on the right, the temperature
was -10° C. This means that each
day’s temperature was 10º from
freezing—one day was above freezing;
the other day was below. The two
temperatures are 20 degrees apart.
Applications:1–15 | Connections:20–22,24
Extensions:89–91
Answers to Problem 3.1
b. Yes. If the bird and the fish are
A. 1. (See Figure 1.)
6 7 8 9 5
2. 5
4 , 4 , 4 , 4 , 4 , - 4 . All those that
equidistant from sea level, the height
of the bird’s position would be the
positive value above sea level and the
depth of fish’s position would be the
negative value below sea level.
have numerators greater than the
denominators.
B. 1. (See Figure 2.)
2
1
1
2
2. 11
3 , 23 , 3, 33 , -13 , and -13 . All those
2. a. Aaron is correct. If he gets the answer
right, he will have 300 points. If he
gets the answer wrong, he will have
-300 points. The absolute value
of each of these numbers of points
is 300.
that are whole or mixed numbers.
C. 1. - 1
2
2. 1
2
3. 0
()
b. The point values of the questions
1
4. - 1
2 =2
D. 1. 1 and -1
could be any pair of opposites.
5
2. Two numbers: 5
4 and - 4
3. One number, 0
Figure 1
−
5
4
3
2
1
− − − −
4
4
4
4
4
−2
−1
0
4
1
4
2
4
3
4
0
4
4
5
4
6
4
7
4
1
8
4
9
4
2
3
Figure 2
−1 23 −1 13
−2
−
−1
Comparing Bits and Pieces
1
3
1
3
0
1 13
2 23
1
2
3 13
3
2
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4
Investigation3
At a Glance | Problem 3.2
3 3
1
3. 2
3 7 9; 9 is equivalent to 3 , which is less
than 2
3.
6
4. 13
12 6 5 ; When I rewrite the improper
1
fractions as mixed numbers, I have 112
13
1
1
1
and 15 . 12 is less than 5 so 12 is less
than 6
5.
3 2
5. - 4 6 5 ; Negative numbers are always less
4. Yes; Every positive number is greater than
its opposite, but this isn’t true for every
number. Zero is equal to its opposite.
Negative numbers are less than their
opposites.
0 0
0 0
6
2
2
5. Blake is correct. - 6
5 = 5 and - 3 = 3 .
6 2
6
5 7 3 so 5 is further to the right of zero
(making it greater) and - 6
5 is further to
the left of zero (making it lesser).
than positive numbers.
1
6. -11
5 7 -13 ; Both mixed numbers have
1 has the whole part so, I look to the
fraction part. - 1
5 is closer to zero than
1 is greater than -11 .
-1
so
-1
5
3
3
6. Blake’s strategy will always work for
comparing two negative numbers; the
number with the greater absolute value
is farther to the left, so it is the lesser.
When comparing two positive numbers,
the number with the greater absolute
value is still farther away, but to the
right of zero, making it greater. When
comparing numbers with different signs,
finding the distance from zero is not
helpful.
D. 1. The account activities on October 9th and
October 23rd have the same absolute
value. This information tells Brian that on
October 29th he spent the same amount
of money he deposited on October 23rd.
2. Yes. The absolute value of the account
6
C. 1. 5
8 6 8 ; Since they have the same
withdrawal on October 27th is less
than the absolute value of the account
withdrawal on October 21st.
denominator, I look to the numerator.
6
Five is less than six so 5
8 is less than 8 .
5
2. 5
6 7 8 ; They have the same numerator
so I compare the denominators and find
that one sixth is greater than one eighth.
5
Therefore, 5
6 is greater than 8 .
Figure 1
−1 12 to −1
−1 13 , − 54
Closer
to
−1 12
−1 to − 12
− 12 to 0
0 to 12
1
to
2
1 to 112
− 23 , − 34 , − 56 ,− 67
− 13 , − 38
1 1 3
,
,
,
5 10 12
4 6 7 7
,
, , ,
5 10 8 9
3 3 1
, ,
8 7 3
3 8 4 2
,
, ,
4 10 7 3
1
5 9
112
,8
Half- Closer Closer Half- Closer Closer Half- Closer Closer Half- Closer Closer Half- Closer Closer Half- Closer
way to−1 to−1 way to
to
way to 0
to 0
way
way to 1
to 1
way to
to 1 to 1
1
1
1
−1 13 − 54
−2 −2
− 56
− 34 − 23 − 13
− 67
− 38
Comparing Bits and Pieces
2
1
5
1
10
3
12
3
8
3
7
1
3
12
2
6
10
4
7
2
3
2
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3
4
4
5
7
8
7
9
8
10
5
112
9
8
Investigation3
At a Glance | Problem 3.3
Be sure that the strategy of writing equivalent fractions with a denominator of 100
comes out in the discussion. Ask students to rewrite each decimal as a fraction. This
may help them see how the denominator relates to the place value of a decimal.
C. 1. Other nice numbers include 2, 5, 10, 20,
Assignment Guide for Problem 3.3
25 and 50. Any factor of 100 will give a
number of people with whom Ann can
share without cutting into smaller pieces.
Applications:53–69 | Connections:93
Extensions:None
2. 3, 7 or any number that is not a factor of
100 will need the lasagna cut into 100
servings, in addition to fractional servings.
Answers to Problem 3.3
1 of a pan
D. Sonam is wrong. 0.1 represents 10
10
of lasagna, which is 10 servings (or 100
of
9
a pan), and 0.09 represents 100 of the pan,
1,
A. 1. Each co-worker gets 10 servings, or 10
or 0.1 of a pan.
2. Answers may vary. If students think
which 9 servings. Therefore, 0.1 7 0.09.
in terms of a share as a fraction, each
1 of 400 square inches.
co-worker gets 10
If they mark a grid, they may describe the
share in terms of length and width; each
co-worker may get a 2-inch-by-20-inch
piece or a 4-inch-by-10-inch piece.
25
B. 1. Each teacher gets 25 servings, or 1
4 or 100
or 0.25 of a pan.
2. Each sixth-grader gets one half of a
1 or 1 or 5 or 0.005
serving, or 200
1000
2
of a pan.
100
1
3. Each neighbor gets 12 1
2 servings, or 8 or
121
2
100
125
or 1000
or 0.125 of a pan.
Comparing Bits and Pieces
2
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Investigation3
At a Glance | Problem 3.4
1
3. 1
8 is half of 4 . The absolute value of the
decimal equivalent for - 1
8 is half of the
2. Fractions that are easy to write as decimals,
1 and 3 , have a power of ten in
such as 10
5
the denominator or can be easily renamed
with a power of 10 in the denominator.
Some fractions, such as 1
8 , are difficult but
still possible to rename with a power of
ten in the denominator. Other fractions,
such as 1
7 , cannot be renamed with a
denominator that is a power of 10 because
the denominator does not share a factor
with any power of 10.
absolute value of the decimal equivalent
for - 1
4.
1
4. 1
6 is half of 3 . The decimal equivalent for
1 is half of the decimal equivalent for 1 .
6
3
1
1
5. a. 6
b. - 2
1
c. - 1
d. 10
4
3. a. 0.333
F.
b. No.
3
4
C. 1. - 2
5 = -0.4; - 5 = -0.6; - 5 = -0.8;
-6
5 = -1.2
5
2
4
2. 8 = 0.25; 3
8 = 0.375; 8 = 0.50; 8 = 0.625;
2. 0.3 6 0.33
3. 0.25 = 0.250
4. 0.12 6 0.125
5. -0.1 6 0.1
6 = 0.75; 7 = 0.875
8
8
3
1
3. 3 = 0.33333…; 2
3 = 0.6666…; 3 = 1;
4 = 1.33333…
6. -0.3 6 -0.27
7. Answer may vary. Sample answer:
One the number line, 0.3 is to the left
of 0.33.
3
4. Answers will vary. Students might use
8. Answer may vary. Sample answer:
the unit fractions for which they know
decimal equivalents and find multiples.
For example, 1
5 = 0.2
One the number line, 0.1 is to the right
of - 0.1.
9. One the number line, 0.25 and 0.250
so 2
5 = 2 * 0.2 = 0.4. They also
might rename each fraction with a
denominator that is a power of ten. So
3
375
8 = 1000 = 0.375. Students who prefer to
work with models may use their fraction
strips with the related tenths strips,
hundredths strips, or number lines.
D. 1. 0.85
1. 0.1 6 0.9
share the same point.
2. 0.32
4. - 0.88
3. 0.82
5. 0.500
1
1
E. 1. - 1
2 = -0.5; 3 ≈ 0.33; - 4 = -0.25;
1 = 0.2; 1 ≈ 0.17; - 1 = -0.125; 1 = 0.1.
5
6
8
10
2.
− 12 = −0.5
1
6
− 14 = −0.25
≈ 0.17
1
3
≈ 0.33
1
− 18 = −0.125 0 1 = 0.1 5 = 0.2
10
Comparing Bits and Pieces
2
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Investigation3
At a Glance | Problem 3.5
8
B. 10
= 0.8 kilograms of cheddar cheese,
23
10 = 2.3 kilograms of peanut butter, and
Assignment Guide for Problem 3.5
4 apples for each box (with 5 leftover apples).
This is a case where it does not make sense
to find a part of a whole with the context of
apples.
Applications:85–87 | Connections:92
Extensions:100–105
Answers to Problem 3.5
7 = 0.5 kilograms of raisins, 13 ≈ 0.929
C. 14
14
13
A. 1. 3
6 = 0.5 kilograms of wheat crackers, 6
21
or 6 = 2.1666… kilograms of powdered
milk, and 24
6 = 4 oranges.
2. Mary and Meleck are saying the same
kilograms of saltine crackers, 77
14 = 5.5
kilograms of powdered milk (51
2 as a mixed
number), 13 oranges (with 13 leftover
oranges), 39
14 ≈ 2.786 kilograms of peanut
number in different forms, as an improper
fraction and as a mixed number. It’s
possible that Mary either thought of
dividing each kilogram into 6 parts or
she is saying 13 , 6. Meleck also likely
butter and
14
= 21
28 = 0.75 kilograms of
3
Swiss cheese (equivalent fraction of 21
28 = 4 ).
D. The sharing in this problem suggests that
we should divide the item we are sharing
by the number of shares. In this case, that
means dividing the number of kilograms of
food (which is the numerator) by the number
of boxes being packed (the denominator).
The result is an equivalent decimal, similar
to Funda’s strategy. Students may also use
ratios of kilograms to boxes instead of the
fraction of a kilogram being put in each box,
similar Scooter’s strategy.
thought 13 , 6 to get 21
6 and then used
1
his benchmark for 6 . Funda’s answer is
another version of Meleck’s benchmark.
All answers are correct.
3. Scooter is correct. He is using an
equivalent ratio to find the unit rate of
oranges per 1 box.
Comparing Bits and Pieces
101
2
2
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Investigation3
At a Glance | Problem 4.1
C. Will uses 25,, 50,, 75, and 100, as
2 3
benchmarks that correspond to 1
4 , 4 , 4 , and
4 . Alisha uses benchmarks that correspond to
4
D. 1. Using Will’s benchmarks, Angela and
Christina have very close free-throw
percentages. Both girls have higher
percentages than Emily. (See Figure 5.)
tenths: 10,, 20,, etc.
Figure 1
∼32
63
∼95
126
0%
25%
50%
75%
100%
0
∼64
∼129
∼193
257
25%
50%
75%
100%
∼32
63
∼95
126
0%
25%
50%
75%
100%
0
∼64
∼129
∼193
257
25%
50%
75%
100%
50
100
150
200
0%
25%
50%
75%
100%
0
50
100
150
200
25%
50%
75%
100%
0
Team 1
Team 2
0%
Figure 2
0
Team 1
Team 2
0%
Figure 3
0
Team 1
Team 2
0%
Figure 4
0
∼13
∼25
∼38
∼50
63
∼76
∼88
∼101
∼113
126
Team 1
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
0
∼26
∼51
∼77
∼103
∼129
∼154
∼180
∼206
∼231
257
20%
30%
40%
50%
60%
70%
80%
90%
100%
Team 2
0%
10%
Comparing Bits and Pieces
2
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Investigation4
At a Glance | Problem 4.1
Using Alisha’s benchmarks, Angela’s
free-throw percentage is 80,, while
Christina’s is slightly greater than 80,.
(See Figure 6.)
E. Using the example of free throws, percents
are like fractions in that they express part
of the whole. The part is the number of free
throws that were successful; the whole is the
number that were attempted. Percents are
like ratios in that they tell us how many free
throws were made for every 100 attempted.
Christina has the highest free-throw
percentage; Angela’s is next, and Emily’s
is the least.
Will and Alisha are both correct. If we
emphasize part of a whole, we are
thinking about fractions. If we emphasize a
comparison between the two quantities, we
are thinking about ratios.
2. Students may make percent bars, ratios,
fractions, or decimals. Using ratios to
predict, Angela will make 24 baskets
(12 : 15 = 24 : 30), Emily will make 22 or
23 (15 : 20 = 7.5 : 10 = 22.5 : 30), and
Christina will make 24. (This one is hard to
rename with 30 as the second part of the
ratio. Using a decimal approximation for
13
16 , 0.81, as a unit rate, we get 30 * 0.81
or about 24.)
Figure 5
∼4
7.5
∼11
15
0%
25%
50%
75%
100%
0
5
10
15
20
0%
25%
50%
75%
100%
0
4
8
12
16
25%
50%
75%
100%
0
Angela
Emily
Christina
0%
Figure 6
0
1.5
3
4.5
6
7.5
9
10.5
12
13.5
15
Angela
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
0
2
4
6
8
10
12
14
16
18
20
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
0
1.6
3.2
4.8
6.4
8
9.6
11.2
12.8
14.4
16
20%
30%
40%
50%
60%
70%
80%
90%
100%
Emily
Christina
0%
10%
Comparing Bits and Pieces
3
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Investigation4
At a Glance | Problem 4.2
C. A rate table has two labeled rows, and the
B. 1. There are 34 students in Marjorie’s class.
data is paired in columns, with each pair of
data related in the same way (or in the same
ratio). The percent bar has two rows also: the
top of the bar is labeled with the raw data
and the bottom of the bar is labeled with the
percents. These numbers are paired in the
same way; that is, the relationship between
12 people and 40, is the same as 30 people
and 100,.
2. Answers will vary. Students might say that
to get the number of students at 10, we
have to figure 34 , 10, because 10% is a
tenth of 100,. This will produce 34 , 10
or 3.4. To get the number of students at
40, they might multiply 3.4 by 4, or they
4 or 2 and find
might think that 40, is 10
5
a fifth of 34, i.e., 34 , 5 = 64
or
6.8, and
5
then double this.
3. 15 people in Marjorie’s class have dimples.
4. 44, of the people in Marjorie’s class
have dimples.
5. She used 10, benchmarks, then cut
the interval between 40, and 50, into
10 equal spaces, each worth 1%. At the
same time she cut the interval between
13.6 and 17 into 10 parts each worth
0.34. Four of these intervals of 0.34 is
approximately 15, 13.6 + 1.36.
6. Answers will vary depending on the class.
Figure 1
∼7
0
15
∼22
30
Attached Earlobes
0%
25%
50%
75%
100%
0
∼7
15
∼22
30
0%
25%
50%
75%
100%
0
∼7
15
∼22
30
0%
25%
50%
75%
100%
0
∼7
15
∼22
30
25%
50%
75%
100%
Dimples
Straight Hair
Widow’s Peak
0%
Comparing Bits and Pieces
2
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Investigation4
At a Glance | Problem 4.3
Percents Encourage students to use percent bars to express themselves. Point out
that their fraction work leads to the percent work.
Ratios Bring out the part-to-part comparison. Students may find the fraction of
artworks chosen by the public and then the comparative fraction of 200. Take the
opportunity to connect this correct strategy to the equivalent ratios strategy. In
a ratio strategy, express the ratio of the two kinds of artwork, and then scale that
ratio until the total of the two parts is 200.
D. 1. Approximately 67% of the works were
Assignment Guide for Problem 4.3
chosen by the public and approximately
33% were chosen by the curators.
Students might make percent bars,
equivalent ratios, or rate tables to find
these percentages. If 36 pictures are
100% of the total, then 3.6 pictures are
10,, so 24 pictures are about 67, of
the total.
Applications:21–25 | Extensions:41–44
Answers to Problem 4.3
A. Answers will vary. Many students will guess
a number close to 100. Students will want to
see the right-hand side of the exhibit. Some
will notice that the exhibit is cut off on the
left edge of the photograph, and will want to
know how many artworks are cropped out.
Some students will want to know whether
each frame represents a different work of art
because some of the artwork appears to go
together.
2. If there are 200 works of art, the public
chose approximately 133 and the curators
chose approximately 67. Students might
make percent bars, equivalent ratios, or
rate tables to find these percentages. If
200 pictures are 100, of the total, then
what number of pictures would match
67, chosen by the public.
E. Answers will vary. 67
33 would be a name that
B. This information should greatly reduce the
more closely approximates the ratio of
publicly chosen to curator chosen artworks.
estimated number of works of art in the
exhibit.
C. 1. (See Figure 1.)
2. Students will likely estimate 24 works of
art on the left-hand side. If this is 2
3 of the
total, then there should be a total of 36
works of art in this part of the exhibit. On
the right, there are 8 works of art (and
part of a seventh). If this is 2
3 of the total,
then there should be a total of 12 in this
part. 24 + 12 = 36.
Figure 1
48
?
2
3
Comparing Bits and Pieces
2
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3
3
Investigation4