Solutions Manual Mathematics for Elementary and Middle School Teachers with Activities 6th Edition B

Solutions

1.1 The Counting Numbers 1-1

Chapter 1 Numbers and the Base-Ten System 1.1

The Counting Numbers

1.

Answers will vary. For example, when connecting the counting numbers as a list view of numbers with the number of objects in a set view of numbers, a child must learn to associate each number in the list in a one to one correspondence with each object in the set, starting with one. Also, the child must be able to learn that the last number from the list, used to connect with the last object in the set, is the number of objects in the set.

2.

Yes, there is a better way to respond. For instance, you could group the beads into sets of 10 beads in each group. Then you would have 3 groups of 10 beads in each group and there would be 5 left over beads. This grouping would facilitate a discussion about place value and allow the conversation to focus on 3 tens.

3.

a.

You could group the beads into sets of 10 beads in each group. Then you would have 4 groups of 10 beads in each group and there would be 7 left over beads.

Using the place value system of representing numbers, 4 tens and 7 ones is 47.

Figure 1.1 shows a simple math drawing that could be drawn.

Figure 1.1: Representation of 47 b.

You could bag the toothpicks into sets of 10 toothpicks in each bag. Then when you get 10 bags of 10 toothpicks in each, you could bundle, with a rubber band, 10 bags of 10 toothpicks to make sets of 100 toothpicks in each bundle. Then you would have 3 bundles of 100 toothpicks in each bundle (or 3 hundreds) and you would have 2 bags of 10 toothpicks in each bag (or 2 tens) and there would be 8 left over toothpicks. Using the place value system of representing numbers, 3 hundreds, 2 tens, and 8 ones is 328. Figure 1.2 shows a simple math drawing that could be drawn.

1-2 Chapter 1: Numbers and the Base-Ten System

Figure 1.2: Representation of 328 c.

You could bag the toothpicks into sets of 10 toothpicks in each bag. Then when you get 10 bags of 10 toothpicks in each, you could bundle, with a rubber band, 10 bags of 10 toothpicks to make sets of 100 toothpicks in each bundle. Then when you have 10 bundles of 100 toothpicks, you could get a giant gallon sized plastic bag and put them into it and group these 10 sets of 100 toothpicks into 1 set of 1000 toothpicks. Using the place value system of representing numbers, 1 thousand is represented as 1000. Figure 1.3 shows a simple math drawing that could be drawn.

Figure 1.3: Representation of 1000

1.1 The Counting Numbers 1-3

4. a.

Let’s say you have a relatively unorganized collection of 62 toothpicks as shown below in Figure 1.4

Figure 1.4: Representation of 62 toothpicks You cou ld bag the toothpicks into sets of 10 toothpicks in each b ag See Figure 1.5.

Figure 1.5: Math drawing of 10 ones regrouped as 1 ten.

You co uld continue doing that with all 62 toothpicks, that is, you can continue collecti ng sets of ten toothpicks and regrouping them into tens until you can no longer do that. When you are done, you will be able to count the n umber of tens.

If you look at the number 62, the digit in the tens place is a 6 and that corresponds with thenumber of tens that you find. Similarly, the digit is ones p lace is 2 and we have 2 single toothpicks not in bundles of ten. See Figure 1.6.

Figure 1.6: 62 represented in base-ten bundles. b. The 6 in the number 62 could represent 6 tens, as shown in Figure 1.6. If we took the bundles apart, we would have 60 ones so 6 can stand for that as well. In summary the 6 stands for 6 tens or 60 ones.

5. a.

You could bag the toothpicks into sets of 10 toothpicks in each bag. Then when you get 10 bags of 10 toothpicks in each, you could bundle, with a rubber band, 10 bags of 10 toothpicks to make sets of 100 toothpicks in each bundle. Then you would have 3 bundles of 100 toothpicks in each bundle (or 3 hundreds) and you would have 2 bags of 10 toothpicks in each bag (or 2 tens) and there would be 8 left over toothpicks. Using the place value system of representing numbers, 3

1-4 Chapter 1: Numbers and the Base-Ten System

hundreds, 5 tens, and 8 ones is 328. Figure 1.7 shows a simple math drawing that could be drawn.

Figure 1.7: Representation of 358 b. The 5 in the number 358 could represent 5 tens, as shown in Figure 1.7. If we took the bundles apart, we would have 50 ones so 5 can stand for that as well. In summary the 5 stands for 5 tens or 50 ones. c. The 3 in the number 358 could represent 3 hundreds, as shown in Figure 1.7. If we took the bundles apart, we would have 30 tens so 3 can stand for that as well. If we took apart the ten bundles, we would have 300 single toothpicks, so 3 could stand for that as well. In summary, the 3 stands for 3 hundreds, 30 tens or 300 ones.

6.

For instance, if you we are using Popsicle sticks, you could get a large collection (say three-hundred fifty-six sticks) in an unorganized pile. Then you could ask the students that are learning about place value to start counting them. After a bit, someone (you or some of the students) will likely start to group the Popsicle sticks into piles of equal size.

You could then count the Popsicle sticks faster by bundling groups of 10 together.

Perhaps you might bundle these groups together physically by tying a twist tie around each set of 10 Popsicle sticks.

After a while of doing this, you will have lots of bundled sets of 10 Popsicle sticks. At this stage, you could take 10 sets of twist-tied sets of 10 Popsicle sticks and put them in to a gallon sized plastic bag to make groups of 100 Popsicle sticks (consisting of 10 sets of 10 bundled Popsicle sticks). As you continue making twist-tied bundles of 10 and baggies of 100, you eventually will use up all of the Popsicle sticks. When this is done, you would end up with 3 baggies (or 3 hundreds) and 5 twist-tied bundles (or 5 tens) and 6 left over Popsicle sticks. Now you could point out that your baggies, bundles, and individual Popsicle sticks correspond directly with the base-ten representation of three hundred fifty-six Popsicle sticks (or 356). Since the digit in each place value is representing a count of units that consist of 10 bundles of units that are represented in the digit to the right, we see that each place value represents a number that is 10 times greater than the place value to its immediate right. For example, when we count the farthest left place value in this number (365), we count groups of hundred (3 baggies). Since each baggie is made up of 10 bundles, we note that the place value immediately to right of the hundreds place is the place value where we are counting the bundles (or tens).

7.

Young children must learn several key ideas about place value and overcome some linguistic hurdles to learn how to count in the base-ten structure. They must understand

1.1 The Counting Numbers 1-5

the key role that 10 plays in our base-ten structure or in representing numbers. They must understand how the location affects the value of the number or the unit of the digit in that particular location. They must overcome linguistic difficulties inherent to how we say numbers in English, especially the anomalies like eleven or the inconsistent order of twenty-two (2 tens and two ones) and nineteen (one nine and one ten).

8.

The number 1001 looks like 100 and 1 put together. Calling it “one hundred one” makes sense w/o understanding the structure of our number system. See Figure 1.8. Each small block represents 1.

Figure 1.8: Base-Ten representation of 1001 9.

If we count what we’ve got in the math drawing, we see 17 individual toothpicks and 15 bags of 10 toothpicks in each bag. Naively, we might write this as 1517 toothpicks, which would be misleading since as written it would represent one-thousand five-hundred seventeen toothpicks.

Since our place value system can only represent up to 9 of any particular place value, we have to regroup when we have more than 9 of a particular place value (or base-ten unit).

In terms of toothpicks, this means that since 17 is greater than 9, we have enough individual toothpicks to regroup into one more baggie of 10. This gives us 7 left over toothpicks but now 16 bags. Similarly, we also have enough bags to regroup (or bundle) them with a rubber band into one group of 100 toothpicks. This gives us 1 bundle of 100 toothpicks (or 1 hundred), 6 bags of ten toothpicks (or 6 tens) and 7 individual toothpicks. See Figure 1.9 for what Figure 1.15 (in the regular text) would look like once you’ve regrouped the numbers in a way that corresponds to the base-ten structure.

Figure 1.9: Representation of 167

1-6

10.

Chapter 1: Numbers and the Base-Ten System

Answers will vary. In the base-ten structure, the digits represent different values of objects. The place value is integrally related to the value that any particular digit represents. The number ten plays a vital role in the system and is the basis of the value of each place. The base-ten structure is much easier to represent large numbers than more primitive ways of representing numbers. However, the base-ten structure is not as intuitive and is harder to learn than more primitive systems, such as a simple tally mark system.

11. a.

You cannot tell where to plot 5 on the given number line. Without another point for reference, you cannot determine how much space on the given number line that a unit represents or that some other quantity of units represent.

b.

You cannot plot 100 on the given number line. The reasoning is the same as the answer in part a above.

c.

You cannot plot N+1 on the given number line. Without knowing what N represents, you don’t know how many partitions to make between 0 and N. Those partition sizes would be each 1 unit, so you cannot determine where one more unit past N would be.

a.

See Figure 1.10. In the first number line I first used the larger tick marks to represent 400 each. Then I counted up to 800 and then 1200 using these sized tick marks. Realizing that 900 wasn’t going to fall perfectly on a 400 tick mark, I partitioned the 400 tick marks spaces into 4 smaller spaces and then each shorter tick mark represented 100. I then counted one more shorter tick mark past 800 to get to 900.

b.

See Figure 1.10. For the second number line, we partitioned the space between 0 and 300 into 3 equal spaces so between taller tick marks represents 100. Then we partitioned each of these spaces representing 100 into 2 small spaces representing 50. I counted up to 200 with large tick marks and then over one more smaller space to 250.

See Figure 1.10. For the last number line, I let the spaces between taller tick marks represent 2000. I went up to 6,000. Next, since it takes ten 200s to make 2,000 and the gap between 6,000 and 8,000 is 2,000, then I made 10 spaces between 6,000 and 8,000 and plotted 6,200 one of these spaces above 6,000.

12.

c.

1.1 The Counting Numbers 1-7

Figure 1.10: Number lines 13. a.

The next two units in base-two after 16 are 32 and 64.

b.

See Table 1.1

11

12

13

14

15

16

17

18

19

20 Table 1.1: 11 through 20 in base-two representation c.

d.

11 = 10112 12 = 11002 13 = 11012 14 = 11102 15 = 11112 16 = 100002 17 = 100012 18 = 100102 19 = 100112 20 = 101002 See Figure 1.11. 35 = 1000112.

Figure 1.11: 35 written in base-two.

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Chapter 1: Numbers and the Base-Ten System

e.

See Figure 1.12. 45 = 1011012.

Figure 1.12: 45 written in base-two.

14. a. b.

The next two units in base-three after 3 are 9 and 27.

See Table 1.2.

1 2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29 30 Table 1.2: Numbers 1 to 30 in base-three math drawings

1.1 The Counting Numbers 1-9

c.

See Table 1.3.

1 = 13 11 = 1023 2 = 23 12 = 1103 3 = 103 13 = 1113 4 = 113 14 = 1123 5 = 123 15 = 1203 6 = 203 16 = 1213 7 = 213 17 = 1223 8 = 223 18 = 2003 9 = 1003 19 = 2013 10 = 1013 20 = 2023 Table 1.3: Numbers 1 to 30 in base-three

d.

See Figure 1.13. 45 = 1203.

21 = 2103 22= 2113 23 = 2123 24 = 2203 25 = 2213 26 = 2223 27 = 10003 28 = 10013 29 = 10023 30 = 10103

Figure 1.13: 45 written in base-three.

15. a.

The next two units in base-eight after 8 are 64 and 256.

b.

See Table 1.4.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20 Table 1.4: Numbers 1 to 20 in base-eight math drawings

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Chapter 1: Numbers and the Base-Ten System

c.

See Table 1.5.

1 = 18 5 = 58 9 = 118 13 = 158 2 = 28 6 = 68 10 = 128 14 = 168 3 = 38 7 = 78 11 = 138 15 = 178 4 = 48 8 = 108 12 = 148 16 = 208 Table 1.5: Numbers 1 to 20 in base-eight

d.

See Figure 1.14. 100 = 1448.

17 = 218 18 = 228 19 = 238 20 = 248

Figure 1.14: 100 written in base-eight.

16.

Answers will vary. For example, they could use a photocopier and shrink the poster to a small size (but where the dots are still distinguishable hopefully), then they could make 999 copies of this small sized poster. This would work since each of the 1000 shrunk down posters contain 1 million dots and 1,000,000 x 1,000 is 1 billion. It would probably be doable; however, one might need to find a special photocopier (like one that photocopies large maps) if the poster is large enough. Or for a more green approach, one could try to have each student try to draw a portion of the dots themselves. For example, if there were 25 students in the class then each student would need to represent 40 million dots. This would likely be not very feasible because if you worked every day on this for 4 weeks (for a total of 5x4 or 20 days) each student would still need to draw 40  20 million or 2 million dots a day. If they worked for 50 minutes on it, that would be 2, 000, 000 50 or 40,000 dots every minute. Even if they could do this, the class would likely revolt after two or three days of drawing dots day after day.

1.2 Decimals and Negative Numbers 1.

a.

It represents one tenth. Since it takes ten of those to make one, then each long strip is one tenth of a unit or one tenth of one.

b.

It represents one hundredth. Since it takes one hundred of those to make one, then each small square is one hundredth of a unit or one hundredth of one.

c.

The figure represents 1.74. As described above and in the directions, each large square represents one. In the figure, there is one large square so there is 1 one.

Similarly, there are 7 tenths (or 7 long strips of small squares) and 4 hundredths (or 4 small squares).

d.

See Figure 1.15. As described above and in the directions, each large square represents one. In the figure, there are 3 large squares, so we have 3 ones.

Similarly, there are 2 tenths (or 2 long strips of small squares) and 5 hundredths (or 5 small squares).

1.2 Decimals and Negative Numbers 1-11

Figure 1.15: 3.25 represented with base-ten blocks e.

See Figure 1.16. As described above and in the directions, each large square represents one. In the figure, there is 1 large square so there is 1 one. Similarly, there are 8 tenths (or 8 long strips of small squares).

Figure 1.16: 1.8 represented with base-ten blocks f.

See Figure 1.17. As described above and in the directions, each large square represents one. In the figure, there is 1 large square so there is 1 one. Similarly, there are 8 hundredths (or 8 small squares).

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Chapter 1: Numbers and the Base-Ten System

Figure 1.17: 1.08 represented with base-ten blocks g.

See Figure 1.18. As described above and in the directions, each large square represents one. In the figure, there are 4 tenths (or 4 long strips of small squares and 3 hundredths (or 3 small squares).

Figure 1.18: 0.43 represented with base-ten blocks h.

2.

To represent a thousandth, we would need a very thin vertical strip such that ten of those strips would make up the small square that represents a hundredth.

Answers will vary. For example, to create tenths, you must first establish what shape represents one. Once you have done that, you can partition that shape into ten equal parts.

Each of these parts will represent one tenth. Similarly, you can partition one tenth into ten equal parts and each of these parts will represent one hundredth. Finally, you can partition one hundredth into ten equal parts, which will each represent one thousandth. If you want to represent one thousandth and one unit using the same model, then you must pick a one that can be split easily into 1000 pieces so that the thousandths are visible.

Base-ten blocks are often used for this representation.

1.2 Decimals and Negative Numbers 1-13

3.

a.

See Figure 1.19. The large square is one. The lightly shaded column represents one tenth. If you count the columns you can see that there are a total of 10 columns, indicating that each one is made up of 10 tenths.

Figure 1.19: Representation of ones, tenths, and hundredths. b.

See Figure 1.19. The large square is one. The darkly shaded small square represents one hundredth. The columns each represent a tenth. If you count the number of small squares per column, you can see the answer is that 10 hundredths make a tenth.

c.

See Figure 1.19. The large square is one. The darkly shaded small square represents one hundredth. The columns each represent a tenth. From part a, we know there are 10 tenths in a one. From part b, we know that each of these tenths have 10 hundredths. To find how many hundredths make a one, we could just add 10 hundredths, 10 times or we could multiply and see that 10 × 10 = 100 total hundredths.

a.

See Figure 1.20. If each paperclip is a thousandth, then the darkly shaded red column in the upper left-hand corner represents a hundredth. This is true because we know that 10 thousandths make a hundredth and there are ten paperclips in that column.

4.

Figure 1.20: Base-Ten representations with paperclips

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Chapter 1: Numbers and the Base-Ten System

b.

See Figure 1.19. If each paperclip is a thousandth, then the squares represent a tenth. This is true because we know that there are 10 thousandths make a hundredth (or column). Also, we know that 10 hundredths make a tenth. We can see that there are ten columns in each square, so the squares are tenths.

c.

See Figure 1.19. We know that ten tenths (squares) make a one. Since we can see ten squares in the entire collection, then the entire collection is one.

5.

a. i.

See Figure 1.21. If each paperclip is a thousandth, then each red column represents a hundredth because there are 10 paperclips in each column and each black squares of toothpicks represents a tenth because there are 10 columns in each black square. In the figure, there are 3 black squares or 3 tenths, and 2 red columns or 2 hundredths and finally 9 additional paperclips or thousandths in blue. So, 0.329 has been represented.

Figure 1.21: 0.329 representation ii.

See Figure 1.22. In the figure, there are 2 red columns or 2 hundredths and 7 additional paperclips or thousandths in blue. So, 0.027 has been represented.

Figure 1.22: 0.027 representation

1.2 Decimals and Negative Numbers 1-15

iii.

See Figure 1.23. In the figure, there is 1 black square or 1 tenth and 4 red columns or 4 hundredths. So, 0.14 has been represented.

Figure 1.23: 0.14 representation

6.

b.

The 2 means 2 hundredths. However, since there are 10 thousandths in each hundredth, you could think of it as 20 thousandths. You can see this in the Figure 1.21 above as 20 toothpicks in the 2 red columns.

c.

The 3 means 3 tenths. However, since there are 10 hundredths in each tenth, then you could think of it as 30 hundredths. You can see this in Figure 1.21 above as 30 columns in the 3 black squares. Moreover, because each hundredth has 10 thousandths in it, the 3 tenths could also be thought of as 300 thousandths, which would be the total number of paperclips in the 3 black squares.

If one toothpick represents one thousandth. then the tubes of ten toothpicks would represent one hundredth because 10 thousandths make up 1 hundredth and there are 10 toothpicks in each tube. And the bundles of tubes would represent one tenth because 10 hundredths make up 1 tenth and there are 10 tubes in each bundle. In that case, the figure would represent the number 0.346 because there are 3 tenths (bundles) shown and 4 hundredths(tubes) shown, and 6 thousandths (individual toothpicks) shown.

However, if one toothpick represents one tenth. then the tubes of ten toothpicks would represent one because 10 tenths make up 1 one and there are 10 toothpicks in each tube.

And the bundles of tubes would represent one ten because 10 ones make up 1 one and there are 10 tubes in each bundle. In that case, the figure would represent the number 34.6 because there are 3 tens (bundles) shown and 4 ones(tubes) shown and 6 tenths (individual toothpicks) shown.

Finally, if one toothpick represents one ten. then the tubes of ten toothpicks would represent one hundred because 10 tens make up 1 hundred and there are 10 toothpicks in each tube. And the bundles of tubes would represent one thousand because 10 hundreds make up 1 thousand and there are 10 tubes in each bundle. In that case, the figure would represent the number 3,460 because there are 3 thousands (bundles) shown and 4 hundreds(tubes) shown and 6 tens (individual toothpicks) shown.

1-16

Chapter 1: Numbers and the Base-Ten System

As in Practice Exercise 3, one toothpick could represent any number of quantities. If one toothpick represents 1, then the given collection represents 346. Other possibilities are shown in the table. See Table 1.6.

If 1 toothpick represents: 100 10 1 1 10 1 100 1 1000

then the collection represents: 34,600 3,460 346 34.6 3.46 0.346

Table 1.6: Base-Ten representations of a math drawing 7.

Base-ten block sketches will be shown here. Similar sketches can be used for the bundled units model. a.

Figure 1.24 represents 0.26 if we consider each block to represent a hundredth.

Then by counting we see that we have 6 blocks (or 6 hundredths) and 2 bundles of 10 blocks (or 2 tenths). If we decided to have each block represent a tenth instead, then Figure 1.24 would represent 2.6; whereas if each block represented a hundred then Figure 1.24 would represent 2,600.

Figure 1.24: A representation of 0.26 b.

Figure 1.25 represents 13.4 if we consider each block to represent a tenth. Then by counting we see that we have 4 blocks (or 4 tenths) and 3 bundles of 10 blocks (or 3 ones) and 1 bundle of 100 blocks (or 1 ten). If we decided to have each block represent a hundredth instead, then Figure 1.25 would represent 0.134; whereas if each block stood for 1 unit then Figure 1.24 would represent 134.

Figure 1.25: A representation of 13.4

1.2 Decimals and Negative Numbers 1-17

c.

Figure 1.26 represents 1.28 if we consider each block to represent a tenth. Then by counting we see that we have 8 blocks (or 8 hundredths) and 2 bundles of 10 blocks (or 2 tenths) and 1 bundle of 100 blocks (1 one). If we decided to have each block represent a hundred thousandth of a unit instead, then Figure 1.26 would represent 0.00128; whereas if each block stood for 10 then Figure 1.26 would represent 1,280.

Figure 1.26: A representation of 1.28 d.

Figure 1.27 represents 0.000032 if we consider each block to represent a millionth. Then by counting we see that we have 2 blocks (or 2 millionths) and 3 bundles of 10 blocks (or 3 hundred-thousandths). If we decided to have each block represent a tenth instead, then Figure 1.27 would represent 3.2; whereas if each block stood for a thousandth, then Figure 1.27 would represent 0.032.

Figure 1.27: A representation of 0.000032 8.

See Figure 1.28. In the figure below the darkened strip next to the phrase “each 0.001”is meant to be 1 hundredth strip partitioned into 10 smaller strips. To represent 1.438 as a length, make a long strip by laying a 1 unit long strip next to 4 tenth unit long strips, 3 hundredth unit long strips and 8 thousandth unit long strips. To represent 0.804, lay 8 tenth unit long strips next to 4 thousandth unit long strips.

Figure 1.28: Representing base-ten numbers as lengths

1-18

Chapter 1: Numbers and the Base-Ten System

9.

a.

See Figure 1.29. If each paper clip is worth a hundredth, then each column must represent a tenth because 10 hundredths make up 1 tenth and there are 10 paper clips in each column. Similarly, the entire square represents one because 10 tenths make up 1 one and there are 10 columns in the square. The entire square has 10 columns with 10 paper clips in each column so there are a total of 10 × 10 = 100 paper clips. So, 100 hundredths make up a one.

Figure 1.29: Base-ten representation with paperclips b.

See Figure 1.29. If each paper clip is worth a tenth, then each column must represent one because 10 tenths make up 1 one and there are 10 paper clips in each column. Similarly, the entire square represents ten because 10 ones make up 1 ten and there are 10 columns in the square. The entire square has 10 columns with 10 paper clips in each column so there are a total of 10 × 10 = 100 paper clips. So, 100 tenths make up a ten.

c.

See Figure 1.29. If each paper clip is worth a thousandth, then each column must represent a hundredth because 10 thousandths make up 1 hundredth and there are 10 paper clips in each column. Similarly, the entire square represents a tenth because 10 hundredths make up 1 tenth and there are 10 columns in the square.

The entire square has 10 columns with 10 paper clips in each column so there are a total of 10 × 10 = 100 paper clips. So, 100 thousandths make up a tenth.

d.

See Figure 1.29. If each paper clip is worth a ten, then each column must represent a hundred because 10 tens make up 1 hundred and there are 10 paper clips in each column. Similarly, the entire square represents a thousand because 10 hundreds make up 1 thousand and there are 10 columns in the square. The entire square has 10 columns with 10 paper clips in each column so there are a total of 10 × 10 = 100 paper clips. So, 100 tens make up a thousand.

e.

The arguments were essentially identical. When you are looking at base-ten place values that are two digits apart, 100 of the smaller units make up 1 of the larger unit.

1.2 Decimals and Negative Numbers 1-19

10.

We extend the base-ten structure from whole numbers to decimals by allowing the pattern established with whole numbers to continue. Specifically, as I move to the left from one place value to another, it takes 10 base-ten units from the right-hand spot to make 1 base-ten unit in the left-hand spot. For example, in the number 4,563, if we consider the digits 4 and 5, we note that the base-ten unit for the 5 digit is a hundred (5 is in the hundreds spot) and that the base-ten unit for the 4 digit is a thousand (4 is in the thousands spot). This works because as state above, it takes 10 hundreds (right hand spot) to make 1 thousand (left hand spot). This applies to decimals as well. If I consider the number 0.4563 and the same digits. Now the base-ten unit for 5 is hundredths and the base-ten unit for 4 is tenths. The reason those are the base-ten values is because it is true that 10 hundredths will make 1 tenth so the value of the digits follows the same pattern as before.

11.

Jerome is trying to follow the pattern shown on the number line by thinking that the number seven and ten tenths comes after seven and nine tenths. In fact, that thinking is correct. What is incorrect is to write seven and ten tenths as 7.10. Pointing out to Jerome that 7.10 is already on the number line (7.1) might help him understand his error. Baseten blocks could also help by allowing him to see that seven units and ten tenths of a unit is properly rearranged as eight units.

12.

Answers will vary. Examples in which zeros can be dropped include such numbers as 01, 1.0, and 00.20100. Examples in which zeros cannot be dropped include such numbers as 1.01, 2.00301, and 0.00002. The issue of whether one can drop the zero directly in front of the decimal point (e.g., .1 vs. 0.1) is not a mathematical issue but one of style.

Sometimes zeros which could be dropped mathematically are kept because of tolerance levels. This is discussed in the “What Is the Significance of Rounding When Working with Numbers That Represent Actual Quantities”, Section 1.4.

13.

Answers will vary. For example, see Figure 1.30.

Figure 1.30: Zooming in on a number line.

1-20

14.

Chapter 1: Numbers and the Base-Ten System

See Figure 1.31.

Figure 1.31: Zooming in on 7.0028 15.

See Figure 1.32.

Figure 1.32: Locating numbers on various scales 16.

Yes, Cierral may label the tick mark that way. Starting from the left, the tick marks should then be labeled 7.0001, 7.0002, 7.0003, …, 7.0009.

17.

Yes, Juan may plot 9.999 where he did. Starting from the left, the intervening tick marks should then be labeled 9.9991, 9.9992, …, 9.9999.

1.2 Decimals and Negative Numbers 1-21

18.

a.

This number line can be labeled in many different ways. Three examples are shown in Figure 1.33.

Figure 1.33: Three ways to label the tick marks b.

This number line can be labeled in many different ways. Three examples are shown in Figure 1.34.

Figure 1.34: Three ways to label the tick marks c.

This number line can be labeled in many different ways. Three examples are shown in Figure 1.35.

Figure 1.35: Three ways to label the tick marks d.

This number line can be labeled in many different ways. Three examples are shown in Figure 1.36.

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Chapter 1: Numbers and the Base-Ten System

Figure 1.36: Three ways to label the tick marks 19.

20.

a.

First, note that there are ten tick marks between A and B indicating that each of the spaces between tick marks represent a base-ten unit. The red dot is between the third and fourth tick mark. If we consider some options for A and B that work for the dot falling between the 3rd and 4th tick mark, we see that A could be 5 and B could be 6. This would make the tick marks tenths and the tick mark immediately to the left of the red mark would be 5.3 and the one immediately to the right of the red mark would be 5.4. Alternatively, A could be 5.34 and B could be 5.35. This would make the tick marks thousandths and the tick mark immediately to the left of the red dot would be 5.343 and the one immediately to the right of the red dot would be 5.344. The red dot is very close to the tick mark on the right. So, it must be a number that is very close to either 5.344 or 5.4.

5.3438 is only 2 ten thousandths away from 5.344 so that is the correct choice.

Therefore A = 5.34 and B = 5.35.

b.

Unlike part a, the red dot is slightly less than halfway between the two tick marks.

If we consider our options for the A and B, we can see that 5.34 is about halfway between 5.3 and 5.4. So that is our choice and A = 5 and B = 6.

c.

The red dot is two tick marks away from A. This could work if A was 1.2 and each tick mark was a hundredth (0.01). Or this could work if A was 1 and each tick mark was a tenth (0.1). If each tick mark were 0.01, then the 7 thousandths would indicate that the red dot should be closer to the third tick mark than the second (as shown in the number line shown in the text for part d below). So that leaves the other option as the right option. A = 1. Each tick mark is a tenth so two tick marks over is 1.2 and then you are closer to the second tick mark than the third tick mark so 1.227 makes sense. Ten tenths = 1 so B is 1 more so B = 2.

d.

A is 1.2 as indicated above in part c. The tick marks are hundredths so the 2nd tick mark represents 1.22 and the third tick mark 1.23. Since 1.227 is closer to 1.23 than 1.22 the tick mark is in the correct location. Ten hundredths = 1 more tenth so B = 1.3.

The distance between 0 and 1 is the unit. We place -1 one unit to the left of the 0. We place -2 two units to the left of the 0. For -1.68, we start at 0 and move 1 unit to the left.

1.2 Decimals and Negative Numbers 1-23

Then from this spot, we move 6 tenths of a unit to the left from this spot. Finally, we move 8 hundredths of a unit to the left from this spot to arrive at -1.68.

21.

Figure 1.37 shows the decimal numbers: -1, -0.92, -0.3, -0.03, 0, 0.07, 0.1, 0.3, 0.9, and 1.

These choices allow students to distinguish between numbers such as 0.07 and 0.1, which allows students to consider place value when plotting each number. These choices also allow a comparison of the decimals -0.03, -0.3, and 0.3, which helps students consider the place value meaning of -0.03 compared to -0.3 and also to consider that, like the numbers on the positive part of the number line, smaller decimals (-0.3 for instance) are farther to the left from larger decimals like -0.03.

Figure 1.37: Comparing decimals on a number line 22.

a.

See Figure 1.38.

Figure 1.38: -4.3 on a number line b.

See Figure 1.39.

Figure 1.39: -0.28 on a number line c.

See Figure 1.40.

Figure 1.40: -0.28 on a number line, zoomed in d.

See Figure 1.41.

Figure 1.41: -6.193 on a number line e.

See Figure 1.42.

Figure 1.42: -6.193 on a number line, zoomed in

1-24

23.

Chapter 1: Numbers and the Base-Ten System

Thinking of –N as the opposite of N, then if N were a negative number, then –N would be the opposite of a negative number. In other words, -N would be a positive number. N would also be a positive number. So –N and N would be the same number.

1.3 Reasoning to Compare Numbers in Base Ten 1.

2.

a.

3 hundreds and 19 tens and 2 ones is greater (even though 3 hundreds are fewer than 4 hundreds). This is because 10 tens make 1 hundred so 19 tens make 1 hundred and 9 tens so we can describe 3 hundreds and 19 tens and 2 ones as 4 hundreds and 9 tens and 2 ones and that is clearly bigger than 4 hundreds and 8 tens and 7 ones.

b.

When you write a number in base-ten notation like 198 and 213, you automatically have regrouped any number of lower place value units into their higher place value equivalents. For example, how would you even write 3 hundreds 19 tens and 2 ones? You couldn’t write 3192 because that makes it look like 3 thousand, one hundred and ninety-two. Instead, you have to regroup the 19 tens into 1 additional hundred and you are left with 9 tens. Then the number is written in a more traditional form, 492. It just so happens that 492 is the same as 3 hundreds 19 tens and 2 ones. Because 492 and 487 are now both in a form where everything that could be made into a higher value base-ten unit has been so changed, you can go ahead and just compare the hundreds and then the tens and if necessary then the ones to see which number is bigger

a.

4 ones is greater. This is because 10 tenths make 1 one so 37 tenths make 3 ones. and 9 tenths. 4 ones is greater than 3 ones and you don’t have enough tenths to make another one so 4 ones is greater than 37 tenths.

b.

61 tenths is greater. This is because 10 tenths make 1 one so 61 tenths make 6 ones and 1 tenths. 6 ones is greater than 4 so 61 tenths is greater than 4 ones.

3.

Since 10 hundredths make up each tenth, then 2 tenths can be broken down into 20 hundredths. This shows us that 2 tenths is indeed bigger than 13 hundredths, but that is because 20 hundredths is bigger than 13 hundredths. 2 is also less than 24 but 2 tenths is less than 24 hundredths because 20 hundredths is less than 24 hundredths. 2 tenths is always less than anything more than 20 hundredths, but is always greater than anything less than 20 hundredths because 2 tenths is exactly 20 hundredths.

4.

We compare numbers in the base-ten structure in the way that we do because base-ten places of larger value count more than the largest combined value made with lower places. For instance, when comparing 234 to 219, we see that both numbers have the same value in the hundreds place but that the 234 has a greater value in the tens place.

Since the tens place counts more than the largest amount you could have in the ones place, we can essentially ignore the values in the ones place.

1.3 Reasoning to Compare Numbers in Base-Ten 1-25

5.

See Figure 1.43.

Figure 1.43: Using bundled objects to compare decimals 6. a. b.

See the solution to practice exercise #4.

See Figure 1.44 Since 1.1 is farther to the right of 0.999 so 1.1 is greater than 0.999.

Figure 1.44: Comparing 1.1 and 0.999 with a number line. c.

See Figures 1.45 and 1.46. 1.1 is greater because it has more overall toothpicks.

Key = 1 toothpick = one hundredth

, Figure 1.45: Math drawing for 1.1

1-26

Chapter 1: Numbers and the Base-Ten System

Figure 1.46: Math drawing for 0.999.

7.

See Figure 1.47. a.

0 is to the left of 0.6 in the number line so 0 < 0.6. b.

0.00 = 0, which is to the left of 0.7, so 0.00 < 0.7. c.

3.00, 3.0, and 3 are all the same point. d.

3.7777 is to the right of 3.77 so 3.7777 > 3.77.

Figure 1.47: A number line to compare 0 and 0.6.

8.

a.

Mary has labeled the tick mark after 4.99 incorrectly as 4.100 instead of 5.00. She might be making this mistake by thinking of the decimal parts of the numbers like one might read them “four point ninety seven, four point ninety eight, four point ninety nine, and now four point one hundred”, since 100 comes after 99.

b.

You could help Mary see that her answer is wrong by helping her read the answer properly as four and one hundred thousandths. Also, you could show her that, using bundled objects, this number is the same as 4.10 (and 4.1) which she will probably recognize as less than 4.99. Showing this 4.1 on the number line would help reinforce this idea. To help her correct the error, it might be good to use bundled objects showing that each time you go up a tick mark on the number line that you are adding one of the objects representing one hundredth. But then when you have nine objects that each represent tenths and nine additional objects that

1.3 Reasoning to Compare Numbers in Base-Ten 1-27

each represent a hundredth, you get to regroup when you add the next hundredths.

Specifically, 10 hundredths regroup into one more tenth, giving us 10 tenths, which regroup into one more unit or one. This helps you see that the next tick mark would have 4+1 or 5 units, zero tenths, and zero hundredths so it would be 5.00 or just 5.

9.

Mark may think this because 178 > 25 when you ignore the decimal or perhaps because when you read 0.178 you use the word “hundred” and with 0.25 you only use the words “twenty” and “five” or even thinking 0.178 is longer than 0.25. Using money to model 8 decimals, 0.178 would represent 17¢ + 10 ¢, but 0.25 represents 25¢. Since 25¢ is greater than 17¢ and part of a cent, 0.25 must be the bigger of the two numbers. Using a number line, you could help Mark see that 0.178 lies to the left of 0.25 and so it is the smaller of the two numbers.

10.

Answers will vary. For example, 3.2405, 3.2407, 3.24032 are all between 3.24 and 3.241.

11.

There are infinitely many such numbers. For example, 8.46, 8.456, 8.4502, and so on.

12.

a.

Many answers are possible. For example, 3.82 is between 3.8 and 3.9. The number line can be marked in many ways. One way is shown in Figure 1.48.

Figure 1.48: A number line for problem 12

13.

b.

3.8 is like $3.80, and 3.9 is like $3.90. To find a number between 3.8 and 3.9, think of adding some cents (fewer than ten) to $3.80.

a.

The number line can be marked in many ways. One way is shown in Figure 1.49.

Figure 1.49: A number line for problem 13a. b.

The number line can be marked in many ways. One way is shown in Figure 1.50.

Figure 1.50: A number line for problem 13b. c.

The number line can be marked in many ways. One way is shown in Figure 1.51.

1-28

Chapter 1: Numbers and the Base-Ten System

Figure 1.51: A number line for problem 13c.

14.

On a number line, –8 is three units to the left of –5. Thinking of debt, to owe someone $8 is to have less net value than if you owed $5.

15.

On a number line, –3 is to the left of –1, so –3.25 is to the left of –1.4. Thinking of debt, to owe someone $3.25 is to have less net value overall than if you owed $1.40.

16.

a.

See Figure 1.52.

Figure 1.52: Comparing decimals on a number line

17.

b.

–0.7 < –0.06 < 0 < 0.6 < 1.

a.

Answers will vary. -1.52 is between -1.6 and -1.5. The number line can be marked in many ways. One way is shown in Figure 1.53.

Figure 1.53: A number line for problem 17a. b.

Answers will vary. -34.98 is between -34.9835 and -34.9714. The number line can be marked in many ways. One way is shown in Figure 1.54

Figure 1.54: A number line for problem 17b. c.

Answers will vary. -7.835 is between -7.834 and -7.83561. The number line can be marked in many ways. One way is shown in Figure 1.55.

Figure 1.55: A number line for problem 17c

1.4 Reasoning to Round Numbers 1-29

18.

One list that fits the requirements of the problem follows. Note that the numbers are getting smaller: 3.5141, 3.51401, 3.514001, 3.5140001, 3.51400001 …

19.

No. This problem is essentially the same as Practice Exercise 11. We can continue with a similar list of decimal numbers getting closer and closer to 2 without ever reaching 2.

This would be 2.1, 2.01, 2.001, 2.0001, 2.00001, Another solution involving averages (or midpoints) can easily show a list of numbers getting smaller, but with all the numbers that are bigger than 2.

2.1 , 2.05, 2.025, 2.0125, 2.00625,… This list is generated by simply dividing the decimal part of the number by 2. In essence, if you considered a number line, you start at 2.1 and then find the midpoint between 2 and 2.1, then the midpoint between 2.05 and 2, then between 2.025 and 2, and so on. Since you will always be able to find a new midpoint, you’ll always get closer to 2, but still bigger than 2.

20.

Since A < B then the 0 point would be closer to A. As the values of the digits increase, negative numbers get increasingly smaller and farther away from 0. For instance, -8 is farther away from 0 than -6 is. Since A<B, then if we were to plot –B, it would be farther from 0 and thus left of -A. Since 0 must be exactly halfway between B and –B, then the 0 would be closer to –A.

21.

Since A>B, then the 0 point would be closer to B. As the values of the digits increase, negative numbers get increasingly smaller and farther away from 0. With A>B, then if we were to plot –B, it would be closer to 0 and thus right of -A. Since 0 must be exactly halfway between B and –B, then the 0 would be closer to B.

1.4 Reasoning to Round Numbers 1.

2.1349 would round to 2.13. This is because of the 4 in the thousandths place, 2.1349 is less than 2.135. Thus, 2.1349 is closer to 2.13 than to 2.14. Figure 1.56 shows the relationship on a number line.

Figure 1.56: Rounding to the nearest hundredth 2.

27,003 would round to 27,000. 27,003 is much less than 27,050, so it is closer to 27,000 than to 27,100. Figure 1.57 shows the relationship on a number line.

Figure 1.57: Rounding to the nearest hundred

1-30

3.

Chapter 1: Numbers and the Base-Ten System

9,995.2 would round to 10,000. The 5 in the ones place puts 9995 exactly halfway between 9990 and 10,000. The conventional rounding rule is to round up in this case. But we do not have rely on just convention, because the 2 in the tenths place actually puts the number over halfway to 10,000. Figure 1.58 shows the relationship on a number line.

Figure 1.58: Rounding to the nearest ten 4.

Adam’s rule is not valid. For example, apply his algorithm to round 1.492 to the nearest unit: 1.492 → 1.49 → 1.5 → 2 .

But 1.492 is actually closer to 1 than to 2 because it is less than 1.5.

5.

If the number has been rounded to the nearest whole number then it is possible that a serving of the snack food might have any amount less than 0.5 grams of trans fat, since these would be rounded to 0. For example, if they food testers found that the product had 0.2512 grams of trans fat, then if they were supposed to round the answer to the nearest whole number, they would round it to 0.

6.

No, it is possible and even likely that the weight has been rounded to the nearest thousand pounds. If so, the weight could possibly be as low as 11,500 pounds and up to (but not including) 12,500 pounds.

7.

No, the population may have been rounded to the nearest hundred. The exact population is at least 2,650 but less than (but not including) 2,750.

Chapter 1 Review 1.

The child needs to be able to make a one-to-one correspondence between the number words 1, 2, 3, 4, 5 and the beads. For example, the child might point twice to one of the beads and count two numbers for that bear, or the child might skip over a bead while counting. The child must also understand that the last number word that is said while counting the beads tells how many beads there are in the collection.

2.

The 2 really means 2 tens. For instance, you could group the sticks into sets of 10 sticks in each group. Then you would have 2 groups of 10 sticks in each group and there would be 4 sticks left over. However, if you recognize also that there are 20 single sticks in the two groups, then you could say that the 2 stands for 20 ones as well.

3.

a.

Bundle every set of ten toothpicks by putting a rubber band about them. These will represent tens. When you do this, you will get 24 tens and have 7 individual

Chapter 1 Review 1-31

toothpicks left over. Now put every set of 10 bundles (or tens) into a hundred by putting them into piles of 10 ten bundles or 10 tens. These piles will represent a hundred. You will get 2 piles and you will have 4 bundles left and you will still have those 7 individual toothpicks as well. That will be representing 2 hundreds, 4 tens and 7 ones.

4.

b.

The 4 represents 4 bundles or 4 tens.

c.

The 2 represents 2 piles or 2 hundreds.

a.

See Figure 1.59. If each paperclip is a thousandth, then each red column represents a hundredth because there are 10 paperclips in each column and each black square of toothpicks represents a tenth because there are 10 columns in each black square. In the figure, there is 3 black square or 1 tenth, and 4 red columns or 4 hundredths and finally 6 additional paperclips or thousandths in blue. So, 0.146 has been represented.

Figure 1.59: Representation of 0.146 b.

5.

The 4 stands for 4 hundredths. However, since 10 thousandths make up 1 hundredth, 4 also stands for 40 thousandths. You can see this in the picture by counting the number of paper clips (thousandths) that are found in the 4 red columns.

See Figure 1.60.

1-32

Chapter 1: Numbers and the Base-Ten System

Figure 1.60: 4.3752 on number lines with various levels of precision 6.

7.

a.

8 ones is greater than 73 tenths. Each one can be broken down into 10 tenths.

Taking 8 ones we get 80 tenths. Now both quantities are represented with the same base-ten unit, tenths and 80>73 so 80 tenths (or 8 ones) is greater than 73 tenths.

b.

8 ones is smaller than 93 tenths. Each one can be broken down into 10 tenths.

Taking 8 ones we get 80 tenths. Now both quantities are represented with the same base-ten unit, tenths and 80<93 so 80 tenths (or 8 ones) is smaller than 93 tenths.

103>80 and 1.03>0.8. If the single bead is a base-ten unit, then as shown in the picture, ten beads can be put together with a string and since 10 base-ten units make 1 base-ten unit of the next place value, then the stringed column represents the adjacent higher place value base-ten unit. And then 100 beads made the square and since 100 base-ten units make 1 base-ten unit of two place values higher, then the square represents the next higher place value base-ten unit after the one represented by the stringed column. In other words, whatever the 3 beads represent, the 1 square must represent a place value two spots higher (or to the left) of the digit 3. Only 103, 1.03, and 0.103 fit that requirement.

And the 8 stringed columns must represent a place value one spot higher that than what the 3 beads represent. Of the choices left, only 103>80 and 1.03>0.8 fit.

Chapter 1 Review 1-33

8.

6.8 is between 6.777 and 6.888. See Figure 1.61.

Figure 1.61: Finding a number between 6.777 and 6.888 9.

See Figure 1.62.

Figure 1.62: Plotting 0,1,-1,-1.7, 1.58 on a number line 10.

On a number line, –6 is one unit to the left of –5. The farther left you go on a number line, the smaller the numbers get. Thinking of debt, to owe someone $6 is to have less net value than if you owed $5.

11.

a.

274.48 would round to 274.5 if rounded to the nearest tenth. This is because of the 8 in the hundredths place, 274.48 is more than 274.45. Thus, 274.48 is closer to 274.5 than to 274.4. Figure 1.63 shows the relationship on a number line.

Figure 1.63: Rounding 274.48 using a number line to the nearest tenth b.

274.48 would round to 270 if rounded to the nearest ten. This is because of the 4 in the ones place, 274 is smaller than 275. Thus, 274.48 is closer to 270 than to 280. Figure 1.64 shows the relationship on a number line.

Figure 1.64: Rounding 274.48 using a number line to the nearest ten

2-2

Chapter 2: Fractions and Problem Solving

Chapter 2 Fractions and Problem Solving 2.1 Defining and Reasoning About Fractions 1.

Answers will vary. Michael needs to learn that the whole or unit amount can be comprised of a collection of objects. The number of dark marbles is indeed fewer than the entire number of marbles, and in that sense, they can represent a number that is fewer than 1. As Michael is able to recognize that since there are three rows, we can group them together and call each row 1 of the total number of marbles.

3

2.

The 3 means that the whole or the unit amount has been partitioned into three parts that are equal in some way. By itself, 2 is implied to mean 2 of a unit or 1. Saying 3 is the 3

3

whole can be misleading because one may assume then that you are taking 2 of 3. This 3

could be especially confusing in situations where the whole is defined instead of the implied unit. For example, if we were taking 2 of 8 pounds of apples. Here 1 unit is the 3

group of 8 pounds of apples, so saying “3 is the whole” confuses the situation greatly. Or if you ate 2 of one apple, you might think that you are taking about 2 of 3 apples, which 3

3

could be confusing as well.

3.

a.

4 and 40 are similar. For example, 4 can be thought of as 4 things. Those things

7

7

are each 1 of a unit. Similarly, 40 can be thought of as 4 things using the base-ten 7

structure. Those things are each 10 units. For example, I could have 4 of a pound 7

of sugar. I would have 4 equal amounts that are each a seventh of a pound. Or I could have 40 pounds of sugar. I would have 4 equal amounts each amount is ten pounds of sugar. b.

5 and 500 are similar because both represent 5 things. For 5, the things are thirds

3

of a unit and for 500 they are groups of 100 units. c.

2 and 0.2 are similar because both represent 2 things. For 2, the things are ninths

9

of a unit and for 0.2 the things are tenths of a unit. d.

3

9

3 and 0.03 are similar because both represent 3 things. For 3 , the things are

19

19

nineteenths of a unit and for 0.03 the things are hundredths of a unit.

4.

When we write numbers in base ten, the digits in each place tell us how many base-ten units there are for that place’s value. For example, 542 has numbers like 5 which is in the hundreds place and that tells us that we should have 5 base-ten units representing 100 each while the 4 tells us we will have 4 base-ten units worth 10 each. Similarly, with

2.1 Defining and Reasoning about Fractions 2-3

fractions the numerator tells us how many units we will have. For example, 3 tells us 11

that we have 3 elevenths of a unit. But the denominator, rather than place value, tells us the size of those 3 things (in this case elevenths of a unit).

5.

This problem is similar to Practice Exercise 1. The given array of x’s represents four parts, each part a column of x’s. Since each part is 1 of the original rectangle, then we 5 want to show five parts or columns, so we add 1 more column of 5 x’s as shown below. x x x x x

6.

x x x x x

x x x x x

x x x x x

This problem is very similar to Practice Exercise 2. The given array of x’s represents six parts, each part being a row of x’s. Since each part or row is 1 of the original rectangle, 5 then we want to only show five parts or rows, so we remove 1 row of 5 x’s, as shown. x x x x x

7.

x x x x x

x x x x x

x x x x x

x x x x x

x x x x x

Because the strip represents 4 of kilometer, then according to our definition of fractions, 3

the strip represents 4 parts that are each 1 of a kilometer. If we partitioned the strip into 4 3

parts, each part would then be 1 of a kilometer and only 3 of them would be needed for 1 3

kilometer. See Figure 2.1 below. The shaded region represents 1 kilogram.

Figure 2.1: 4 of a kilogram 3

8.

Because the strip represents 2 of mile, then according to our definition of fractions, the 3

strip represents 2 parts that are each 1 of a mile. Looking at the strip, we don’t necessarily 3

immediately see 2 parts. We see 6 parts. However, if we group those parts together, we can find 2 larger parts that make up the strip. See Figure 2.2. The red shaded region is 1 of these larger parts and the blue shaded region is the other larger part. These larger parts

2-4

Chapter 2: Fractions and Problem Solving

consist of 3 of the original smaller parts. To draw a full mile, we would need a strip consisting of 3 of these larger pieces. See Figure 2.3.

Figure 2.2: 2 of a mile 3

Figure 2.3: 1 full mile.

9.

a.

You could use different size units to describe a 1-foot ruler. For example, if your unit was 1 inch, then a ruler would be described using the number 12 as in “The ruler is 12 inches long”.

b.

If your unit was a yard, then a 1-foot ruler would be described using the number 1

3

as in “The ruler is 1

of a yard long.” You could use other measurements. The famous American author Samuel Clements, otherwise known by his pen name Mark Twain, picked his pen name based on a measurement that steamboats would use to make sure they didn’t run aground in the shallow parts of the river. The measurement “Mark” was sometimes used as another name for a fathom, which is 6 feet. So, rulers could also be “1 of a fathom long” or “1 of a mark long”. Any 6 6 unit bigger than a foot would mean a 1-foot ruler could be described with a fraction less than 1 unit.

3

10.

a.

Cameron got 1 of the candy bar. See Figure 2.4. Arianna got 1 of the candy bar.

6

2

Arianna then partitioned that portion into three equal pieces. So, Cameron got 1 of one half of the candy bar. Partitioning the other half of the candy bar into three equal pieces shows that each piece is 1 of the candy bar.

6

Figure 2.4: Kaitlyn’s candy bar

3

2.1 Defining and Reasoning about Fractions 2-5

b.

The entire candy bar is one of the unit amounts (as in Arianna got 1 of the candy 2

bar or the portion given to Cameron is 1 of the candy bar). Another unit amount is 6

1 of the candy bar, because Cameron got 1 of one half of the candy bar.

2

c.

3

Depending on which unit amount you use, Cameron’s piece could be viewed as 1

3

or 1.

6

11.

a.

1 of a recipe. See Figure 2.5. Partitioning a cup into 3 parts, we see that two of

2

these parts represents the amount (or portion) needed for a full recipe. Then we look at 1 of a cup of cocoa (or the amount we actually have on hand) and compare 3

this to amount needed to make a full recipe. We see that 1 of the amount needed to 2

make a full recipe is the amount we actually have on hand. So, we can only make 1 of the recipe.

2

Figure 2.5: Fractions of a cup of cocoa b.

An entire cup of cocoa is one reference unit (as in you need 2 of a cup of cocoa 3

for one batch and you have 1 of a cup of cocoa). The amount needed for the full 3

recipe (or 2 of a cup of cocoa) is another reference unit as in 1 of the amount 3

2

needed for the full recipe. c.

The amount that you have on hand is 1 of the amount needed for the full recipe 2

and also it is 1 of a cup of cocoa.

3

12.

a.

3 of the butter she should have used. See Figure 2.6. The recipe calls for 5 of a

5

4

cup, and Susan uses 3 of a cup. As the math drawing shows, Susan used three 4

parts out of the five called for by the recipe. So, she used 3 of the amount called for by the recipe.

5

2-6

Chapter 2: Fractions and Problem Solving

Figure 2.6: Fractions of a cup of butter b.

A cup of butter is one unit amount (5 of a cup). The amount of butter called for by 4

the recipe is another unit amount (3 of the amount called for by the recipe). The 5

amount that Susan used could be thought of as 3 of the amount called for by the 5

recipe or as 3 of a cup of butter.

4

13.

4 of the daily value. See Figure 2.7. The daily value of calcium is what is found in 3 of a

3

4

cup of snack food. So 1 of a cup of snack food contains 1 of the daily value. One cup of 4

3

snack food is really 4 of a cup of snack food, which gives 4 of the full daily value.

4

3

Figure 2.7: Daily value of calcium 14.

Answers will vary widely. For example: 2 of the sixth-grade class was on a field trip to 3

the county courtroom. The students were divided into two equal groups for separate tours. What fraction of the class was in each group? (1) What fraction of the students on 3

the trip was in each group? (1) 2

15.

a.

Ben has the correct answer, but incorrect reasoning. If Marla ate 1 of each cake, 12

then she ate 1

12

of the cakes combined (see part c). While it is true that 1 = 2 , the 12

denominator in the phrase “ 2

24

24

of the cake”, does not represent the 24 pieces in the

math drawing but rather 2 of the amount of cake. If the cakes were of the same 24

size, then there would be 24 equal-sized pieces of cake and Ben’s reasoning would be correct. b.

Seyong seems to believe that the equation 1 + 1 = 2 applies in this situation.

12

12

12

But each twelfth is part of a different whole or unit amount, so this is not a meaningful computation. However, if the pieces were of cakes of the same size, say one pound, then Seyong would be right in a sense because the fractions would

2.1 Defining and Reasoning about Fractions 2-7

be parts of the same size whole. Seyong could say that Marla ate 2 of a pound of 12

cake. However, Seyong is not correct to say that Marla ate 2

of the cake in either case. Unless the total amount of cake was split up into 12 pieces, then the piece sizes are not one-twelfth of the cake. If the cakes were exactly the same sizes, then the piece sizes would be 1 of the cake since there are 24 equal pieces that 24 make up the entire amount of cake.

12

c.

The cakes can be shared equally among 12 people by giving each person a serving like Marla’s. Since there are 12 servings for the total, each serving is 1 of the 12

cake. See Practice Exercise 5.

16.

Yes, Harry’s method is valid. See the solution to Practice Exercise 6.

17.

See Figure 2.8. As in Practice Exercise 5, if you shade 3 of each individual pie, then the 8

shaded parts make 3 of the 3 pies combined. Imagine preparing to serve 8 people each 8

getting one piece from each pie. Person 1 gets all the pieces marked with 1, Person 2 gets all the pieces marked with 2, etc. Altogether, each person’s serving size is 1 of the 3 pies 8

combined. If you have only served the shaded portions (or persons 1,2 and 3), then you have only served 3 of the 3 pies combined.

8

Figure 2.8: Shading three-eighths 18.

We need at least three units to show 7, but that can get confusing because the unit amount 3

could be viewed as one object, the entire collection of units, or individual pieces of an object. A number line can be a useful way to show improper fractions.

19.

In Figure 2.9, the 1-unit has been partitioned into 3 equal-length parts, so the distance between tick marks is 1 of the unit. Then 1 is the first tick mark to the right of zero and 2 3

3

is the second tick mark to the right of zero.

3

2-8

Chapter 2: Fractions and Problem Solving

Figure 2.9: Number line with 1 and 2 marked.

3

20.

3

In Figure 2.10, each unit has been partitioned into 3 equal-length parts, so the distance between tick marks is 1 of the unit. Then 5 is the fifth tick mark to the right of zero.

3

3

Figure 2.10: Number line with 5 marked.

3

21.

See Figure 2.11.

Figure 2.11: Plotting 0, 1, 4 and 11 3

22.

The shaded region could be described as 7

8

3

of the total region or as 21 rectangles. An 3

unambiguous question would be: If the two rectangles combined is considered to be the unit amount, what fraction names the shaded region? This question is unambiguous because it clearly states what the unit amount will be.

23.

9

is correct if the entire figure (three rectangles) is meant to represent the referent unit amount.

9 is correct if one rectangle (partitioned into four pieces) is meant to represent the referent 12

4

unit amount.

An unambiguous question might be: If the three rectangles together make the unit amount, what fraction of this unit amount is shaded? Any unambiguous question about the shaded region must clearly state what the unit amount is.

24.

a.

No. The hexagons take up 2 of the area. We do not know how many shapes were 3

in NanHe’s design, but based on the fractions in the problem, we can know that the design was made using 11 shapes or a number of shapes that is a multiple of eleven. Thus, the shapes in each design can be organized into one or more groups of the same eleven shapes: 4 hexagons, 5 rhombi, and 2 triangles. In each group of 11 shapes, if you were to use triangles to make the hexagons, you would need 24 triangles. To do the same with the rhombi would require 10 triangles. So, each group in the design could be recreated using 36 triangles, which has the effect of breaking any of these groups of 11 shapes up into equal parts. Thus, in each group

2.2 Reasoning About Equivalent Fractions 2-9

of 11 shapes, the hexagons use 24 of the triangles and 24 = 2 . If the area taken up 36

3

by the hexagons is 2 of each group of 11 shapes, then 2 is also the fraction of the 3

3

entire design taken up by hexagons b.

If we view the hexagons, triangles, and rhombi as all just shapes, then they are equal, and this allows us to view each as 1 of the total number of shapes in each group and 11 overall. But if we are talking about area, then the hexagon no longer equal to a triangle. Instead, we see that the hexagon is the same as 6 triangles and we think of each shape in terms of how many triangles make up the shape, giving us equal parts again.

25.

a.

If School A has 600 students, then 800 = 8 of the students attend the after1500 15 school program. If School A has 900 students, then 700 = 7 of the students 1500 15 attend the after school program. In the first situation, 1 of 600 students is 200 3

students, and 2 of 900 students is 600 students. So, 800 out of the 1500 students 3

are in the after school program. b.

Answers will vary. For example: if School A has 300 students and School B has 1200 students, then the fraction of students in the after school program is 100+800 900 = 3. = 1500

1500

5

Or: If each school has 750 students, then the fraction of students in the after school 250+500 750 = 1. program is = 1500

c.

1500

2

Jamie is correct only when the same number of students attend each school. Jamie is wrong for any other situation because the unit amounts will be different.

2.2 Reasoning About Equivalent Fractions 1.

This problem is very similar to Practice Exercise 1. If you have a rod that is partitioned into four equal parts and 3 are shown shaded, and then you partition each of the 4 parts into 3 smaller parts, the shaded portion will then consist of 3 × 3 parts and the unit amount will consist of 4 × 3 parts. So, the shaded portion of the unit amount can be described both as 3 of the rod and as 3×3 = 9 of the rod. See Figure 2.12.

4

4×3

12

Figure 2.12: Equivalent fractions 3 = 9 4

12

2-10

2.

Chapter 2: Fractions and Problem Solving

If you have a unit amount that is partitioned into three parts and two are shaded, and then you partition each of the three parts into four smaller parts, the shaded amount will then consist of 2 × 4 parts and the unit amount will consist of 3 × 4 parts. So, the shaded portion of the unit amount can be described both as 2 of the unit amount and as 2×4 = 8 3

of a unit amount. Alternatively, consider multiplying 2 by one as shown here:

3×4

12

3

2 2 4 2×4 8 ×1= × = = 3 3 4 3×4 12

3.

To give two fractions of the same reference unit common denominators, we use a common multiple of the two numbers. For each fraction, we multiply both the numerator and denominator by a factor that will change the denominator to this common multiple.

For the fractions given, this common multiple would be 12. Specifically, we multiply as shown below.

2 2×4 8 3 3×3 = 9 . = = and = 3

3×4

12

4

4×3

12

In giving the fractions common denominators, we are partitioning the parts of each fraction into equivalent sized smaller parts. The total number of pieces and the size of the pieces change, while the amount of the unit amount that would be represented by the shaded pieces remains the same.

4.

See Figure 2.13. We join together small pieces of the rod to make larger rod pieces. Here we join 2 smaller pieces to make a larger piece because both 6 and 8 can by divided by 2 (thus 8 pieces or 6 pieces could both be put into groups of 2). When we group the 6 shaded smaller pieces into groups of 2, this yields a total of 6 ÷ 2 larger shaded rod pieces. Similarly, when we group the 8 total smaller pieces into groups of 2, this yields a total of 8 ÷ 2 rod pieces.

Figure 2.13: Equivalent fractions 6 = 3 8

4

Division is typically thought of as getting smaller. Here the pieces are getting larger. But the number of pieces is getting smaller since you had 8 total pieces (and 6 shaded) and now via division you have 4 total rod pieces with 3 shaded rod pieces. We are seeing 3 groups of 2 shaded (6) out of 4 groups of 2 in the unit amount. Arithmetically, we have

2.2 Reasoning About Equivalent Fractions 2-11

5.

To simplify a fraction, you look for common factors of the numerator and denominator.

This makes sense because we are trying to regroup the smaller pieces of the fraction into larger pieces so that the unit amount is partitioned into fewer parts. Specifically, the six shaded pieces representing the numerator in 6 could be grouped into two pieces, with 9

three smaller pieces in each group. Likewise, the nine pieces representing the denominator can be grouped into three pieces, again with three smaller pieces in each group. See Figure 2.14.

Figure 2.14 Equivalent fractions 6 = 2 9

3

We then end up with two shaded pieces out of three pieces overall, that is, 2.

3

Arithmetically, we have

The total number of pieces and the size of the pieces change, while the amount of the unit amount that would be represented by the shaded pieces remains the same.

6.

See Figure 2.15. Eighths and thirds of a unit are different sized parts. To get parts of the same size in relation to the whole, get a common denominator. Since 3 = 9 and 1 = 8 , 8

24

3

24

let each tick mark represent 1

24 of the unit. Put a dot at the 0 location, at 8 marks (8

spaces) over to represent 1 = 8 , and 9 marks over to represent 3 = 9 .

3

24

8

24

Figure 2.15: Plotting 0, 1 and 3 3

7.

8

See Figure 2.16. Halves, fifths, and eighths of a unit are different sized parts. To get parts of the same size in relation to the whole, get a common denominator. Since 1 = 20, 3 = 24, and 5 = 25, let each tick mark represent 1

40

8

40

40

2

40 5

of the unit amount. If you consider a

2-12

Chapter 2: Fractions and Problem Solving

random spot on the number line to be the 1 = 20 mark, then count over 4 spaces to the 4th 2

40

mark to the right of 1 represent 3 = 24 and one more mark for 5 = 25.

2

5

40

8

40

Figure 2.16: Plotting 1, 3, and 5 2 5

8.

8

See Figure 2.17. Note that 0.3 = 3 . Tenths and fifths of a unit are different sized parts.

10

To get parts of the same size in relation to the whole, get a common denominator.

3 = 6 so put the dots at 3 marks over from 0 to represent 0.3 = 3 and 6 marks over to 5

10

10

represent 3 = 6 .

5

10

Figure 2.17: Plotting 0, 0.3 and 3

5

9.

See Figure 2.18. Sixths and sevenths of a unit are different sized parts. To get parts of the 8= same size in relation to the whole, get a common denominator. Since 5 = 35 and 6

42

7

48, let each tick mark represent 1 . If you consider the farthest left spot on the number line

42

42

to be the 5 = 35 mark, then count over 7 marks to represent 1 = 42 and then 6 more 6

42

42

marks to represent 8 = 48.

7

42

Figure 2.18: Plotting 1, 5, and 8 6

10.

7

See Figure 2.19. To have sevenths and thirds both falling on tick marks, we must partition the number line into twenty-firsts. I accomplished this by first dividing the interval from 0 to 2 in half to locate 1. I then partitioned the interval from 0 to 1 into 7

7

thirds to locate 1 , 2 , 3 = 1. So, 1 is the same as 7 .

21 21 21

7

3

21

Figure 2.19: Plotting 1

3

7

2.2 Reasoning About Equivalent Fractions 2-13

11.

See Figure 2.20. To place fifths and halves on the same number line, we must use a common denominator of tenths.

Figure 2.20: Plotting 1

2

12.

See Figure 2.21. To place eighths and twelfths on the same number line, we must use a common denominator of twenty-fourths.

Figure 2.21: Plotting 5

12

13.

See Figure 2.22. 47 = 7 5 .

6

6

Figure 2.22: Plotting 47 6

14.

Erin is wrong; the tick mark should be labeled 2 2 = 2 1. Since the interval from 2 to 3 is 8

4

partitioned into 8 pieces, each tick mark represents 1 rather than 1 .

8

15.

a.

10

5

12 of a cup. See Figure 2.23. Representing the two fractions as twelfths allows us

to remove the 4 (or 1) of a cup from the 9 (or 3) of a cup and see that there are 5 12

3

12

4

parts left, each part being 1 of a cup of water.

12

Figure 2.23: Removing some of the water

2-14

Chapter 2: Fractions and Problem Solving

b.

3 appears as 9 . 1 appears as 4 .

4

16.

a.

12 3

12

3 of a cup. See Figure 2.24. Expressing 3 of a cup of water as 6 of a cup of water

8

4

8

allows us to remove half, or 3 of a cup of water, leaving 3 of a cup in the bowl.

8

8

Figure 2.24: Removing half of the water b.

3 appears as 6.

4

17.

a.

8

3 of a recipe. See Figure 2.25. Expressing 2 of a cup of flour and 1 of a cup of

4

3

2

flour in terms of sixths of a cup of flour allows us to see that of four parts (1ths of 6

a cup of flour) needed for a full recipe, we have three of these parts, so we can make 3 of a recipe.

4

Figure 2.25: What fraction of the recipe? b.

2 and 1 appear respectively as 4 and 3.

3

18.

a.

2

6

6

8 of his order. See Figure 2.26. Expressing the fractions as twelfths of a ton of

9

gravel shows that Ken got eight out of nine parts ordered.

Figure 2.26: Received 8 out of 9 parts

2.2 Reasoning About Equivalent Fractions 2-15

b.

3 and 2 also appear as 9

4

19.

a.

3

and 8 .

12

12

2 of a cup. See Figure 2.27. Representing 2 of a cup as 6 of a cup shows that 1 of

9

3

9

3

this amount is taking two parts out of the six, which represent 2 of a cup. (Note 9

that since measuring cups rarely come in ninths, Here, a person could argue that 1

4

of a cup would be a more practical answer.)

Figure 2.27: One-third of two-thirds b.

2 also appears as 6.

3

20.

9

See Figure 2.28. Since 1 of a mile corresponds to 4 of Benny’s total running distance then 2

5

thinking of Benny’s distance ran so far as being made up of 4 parts, I split the number line region from 0 to 1 into 4 parts, each corresponding to one of the 4 parts Benny has 2

run so far. The bold dark arrow shows how far he has run so far. Extending out the next is 1 of a mile to make the full unit mile, we see that these parts we made are is 1 of a mile, 2

8

while at the same time being 1 of Benny’s total running distance. So, Benny needs to go 5

one more 1 of a mile to finish for a total of 5 parts run, each of size 1 of a mile. So, Benny 8

8

will have run a total of 5 of a mile. This is show in the dotted arrow below in Figure 2.28.

8

Figure 2.28: Four-fifths of the running distance is one-half of a mile 21.

See Figure 2.29. since 3 of a mile is equivalent to 6 of mile, I can split how far has been 2

4

run into 2 equal parts, each being 3 of a mile while at the same time each part is also 1 of 4

5

the total running distance. Then she needs to run 5 of those parts. And so, she will have run 3 + 3 + 3 + 3 + 3 = 15 of a mile.

4

4

4

4

4

4

2-16

Chapter 2: Fractions and Problem Solving

Figure 2.29: Two-fifths of the total running distance is three-halves of a mile 22.

a.

It would take 31 copies of strip A to make strip B exactly. Since strip B and strip 2

A are made of equal sized parts, we can use these parts to solve this problem.

Each little part is 1 of strip A. You will need 7 of these parts to make strip B. So, 2

you have 7 parts, each part is 1 of strip A. So, if you make 7 of strip A you will get 2

2

strip B. Since 2 parts would make an entire copy of A, then you can also say you have 31 copies of strip A. 2

When you consider the parts in relation to strip B, you can see that each part is 1

7

of strip B. Since strip A only uses 2 of these parts, then you can make strip A with only 2 of strip B. 7

b.

It would take 11 copies of strip C to make strip D exactly. Since strip D and strip 4

C are made of equal sized parts, we can use these parts to solve this problem.

Each little part is 1 of strip C. You will need 5 of these parts to make strip D. So, 4

you have 5 parts, each part is 1 of strip C. So, if you make 5 of strip C you will get 4

4

strip D. Since 4 parts would make an entire copy of C, then you can also say you have 11 copies of strip C. On the other hand, you could also think of the candies 4

as the way you are comparing the strips. Each candy strip is made up of the same size and type of candy pieces. For strip C you’ve got 24 pieces of candy, so each piece of candy represents 1 of strip C. Strip D needs 30 of the candy pieces. If 24 you made 30 of strip C you would get strip D. 5 and 30 are equivalent fractions.

24

4

24

When you consider the parts in relation to strip D, you can see that each part is 1

5

of strip D. Since strip C only uses 4 of these parts, then you can make strip C with only 4 of strip D. In terms of candy pieces, you see that each candy piece is 1 of 5

30

strip D but you only need 24 to make strip C, so if you made 24 of strip D you would get strip C. 4 and 24 are equivalent fractions.

5

30

30

2.2 Reasoning About Equivalent Fractions 2-17

c.

It would take 11 copies of strip E to make strip F exactly. Since strip F and strip E 4

are made of equal sized parts, we can use these parts to solve this problem. Each little part is 1 of strip E. You will need 5 of these parts to make strip F. So, you 4

have 5 parts, each part is 1 of strip E. So, you if you make 5 of strip E you will get 4

4

strip F. Since 4 parts would make an entire copy of E, then you can also say you have 11 copies of strip E. On the other hand, you could also think of the candies 4

as the way you are comparing the strips. Each candy strip is made up of the same size and type of candy pieces. For strip E you’ve got 20 pieces of candy, so each piece of candy represents 1 of strip E. Strip F needs 25 of the candy pieces. If 20 you made 25 of strip E, you would get strip F. 5 and 25 are equivalent fractions.

20

4

20

When you consider the parts in relation to strip F, you can see that each part is 1

5

of strip F. Since strip E only uses 4 of these parts, then you can make strip E with only 4 of strip F. In terms of candy pieces, you see that each candy piece is 1 of 5

25

strip F but you only need 20 to make strip E, so if you made 20 of strip F you 25

would get strip E. 4 and 20 are equivalent fractions.

5

d.

25

See the answers to part b and c. In both answers our reasons are similar to the reasoning given in part a and we get the same answers. See the first explanation given in the answers to parts b and c. In those explanations, we focused on the fact that Strip C (in part b) and Strip E (in part c) were made of 4 equal size parts, while the other strip (Strip D in part b and strip F in part c) were made of 5 of those same size parts. So, in both problems, we determined that 11 copies of the 4

shorter strip (Strip C and Strip E) were needed to make the longer strip (Strip D and Strip F), and 4 of the longer strip was needed to make the shorter strip.”

5

23.

a.

Rachel has read 3 as many books as Leah has. Because the parts of the strip 4

diagrams are of equal size, we can see that Leah has read four portions of books while Rachel has read 3 portions. Since we are comparing to Leah’s amount, then Leah’s amount is the whole or unit amount, which tells us that each portion is 1 of 4

the amount that Leah has read. Rachel has 3 of these 1 portions so she has read 3 as many books as Leah has. b.

4

4

Leah has read 4 as many books as Rachel has. Since we are comparing to Rachel’s 3

amount, then Rachel’s amount is the whole or unit amount, which tells us that each portion is 1 of the amount that Rachel has read. Leah has 4 of these 3

1 portions so she has read 4 as many books as Rachel has.

3

3

2-18

24.

Chapter 2: Fractions and Problem Solving

a.

See Figure 2.30. Starting with Joseph’s strip, dividing it into six equal pieces, then you make Adam’s strip equal in length to five of those pieces.

Figure 2.30: Adam and Joseph’s books b.

25.

Joseph has read 6 as many books as Adam this year.

5

Becky should pay 1 of the July/August bill, and her two friends should pay 5 of the bill 6

12

each. See Figure 2.31. Dividing the first month’s cost between the two friends and the second month’s cost among the three girls yields pieces of different sizes. Expressing everything as twelfths of the total July/August usage makes it much easier to express each person’s share.

Figure 2.31: Sharing the electricity bill The cases in parts a. and b. result in the same division of cost. In part a. the bulk of the electricity is shared, so the cost should be shared equally. In part b. the bulk of the electricity is used individually, but the individuals are doing similar things, and thus using roughly equal amounts of electricity, so sharing the cost equally still makes sense.

2.3 Reasoning to Compare Fractions 1.

Since the whole or unit is not the same, comparing areas of the two shaded regions is not a valid way to compare the sizes of these fractions. When you compare two fractions like 3 and 5 you are assuming that the unit amount is the same.

4

8

2.

See the text.

3.

See the text.

2.3 Reasoning to Compare Fractions 2-19

4.

A smaller denominator means the same sized unit it partitioned into a smaller number of pieces. This means each piece must be larger. If numerators are the same, the fraction involving larger pieces is larger overall.

5.

36 104 is greater. First, notice that the numerator of 36 is greater, so it has more pieces than 104

the other fraction, 35 . Also notice that partitioning a unit into 104 pieces yields larger 109

pieces than partitioning it into 109 pieces. So 36 not only has more pieces than 35 , but it 104

109

also has a larger number of larger pieces, so 36 is greater.

104

6.

23 is greater. Both fractions are very close to one half. In fact, they are both less than one

47

half. To make one half when a whole is partitioned into 31 pieces, we would need 15 1 pieces. So 15 is 1 of a piece less than one half of the unit amount, where the pieces are 31

2

2

1 of the unit amount in size. To make one half when a unit amount is partitioned into 47

31

pieces, we would need 23 1 pieces. So 23 is 1 of a piece less than one half of the unit 2

47

amount, where the pieces are 1

47

smaller than the 1

31

2

of the unit amount in size. Since these 1 pieces are 47

pieces, then half of each piece size remains similar, half of a 1 piece 47

is smaller than half of a 1 piece. So 23 is closer to one half of the unit amount.

31

7.

47

27 is larger because it is only 1 away from 1 whereas the other fraction is 1

28

28

20

away, and

since partitioning a unit into 28 pieces yields smaller pieces than partitioning it into 20 pieces, 1 is the smaller amount, so 27 has less to go to reach 1 than 19 so 27 is larger.

28

8.

28

20

28

The order is -1.1,-1, -0.98, 11 , 9 , 1.2. One problematic explanation might be explaining 10 8

why 11 is smaller than 9. Since one-tenth is smaller than one-eighth, 11 is closer to 1 10

8

10

(farther to the left) than 9. Another problematic explanation might be explaining why 9 is 8

smaller than 1.2. Writing 9 as 1 1 and 1.2 as 1 2 = 1 1 helps explain this ordering.

8

9.

8

10

5

23 ≈ 0.2738 and 29 ≈ 0.2959. The number picked will vary. For example: 7

84

See Figure 2.32.

8

98

25

Figure 2.32: A number line

= 0.28.

2-20

10.

Chapter 2: Fractions and Problem Solving 78

134

≈ 0.58209 and 124 ≈ 0.58216. The number picked will vary. For example: 5821 = 213

10000

0.5821. See Figure 2.33.

Figure 2.33: A number line 11.

Many answers are possible. Since 2 = 4 = 6 = ⋯ and 3 = 6 = 9 = ⋯ then we can 5

10

15

5

10

15

find as many fractions between 2 and as 3 we want to. For instance, both 7 5

12.

a.

a.

5 61

3×15

15

45

45

45

so choose, for example 0.61 and 0.63 which become

3

and 63 .

100

49

7

49

49

49

5 ≈ 0.7143 and 6 ≈ 0.8571 so choose, for example 0.75 and 0.80 which become

7 75

100

14.

and 8 work.

5 = 35 and 6 = 42, so choose, for example, 37 and 41.

7

b.

45

3 = 0.6 and 2 = 0.66666 …

100

13.

15

3×9 = 27 and 2×15 = 30 so choose, for example, 28 and 29.

5×9

b.

5

7

3

8

= 4.

4

10

5

= and

See Figure 2.34. As shown in this figure, 1 is not halfway between 1 and 1. Instead, 5 is 5

4

6

24

halfway between 1 and 1. This fact is most easily illustrated by finding a common 4

6

denominator between the two fractions: 1 = 4

6 and 1 =

24

6

4 . Just as 5 is halfway between

24

4 and 6, 5 is halfway between 4 and 6 . But 1 = 5 , which is slightly less than 5 .

24

24

24

5

25

24

Figure 2.34: Showing 1 is closer to 1.

5

15.

6

Sam is not accounting for the numerators. While a fraction with a larger denominator may have smaller pieces, if it has enough of those smaller pieces (the numerator) then it will represent a larger amount. For example, 9 is bigger than 1 even though the halves 10

2

are bigger than the tenths. But having 9 of these tenths makes up for the fact that they are small and is bigger than a single half.

16.

Answers will vary. Here are two examples:

2.3 Reasoning to Compare Fractions 2-21

a.

Which fraction is greater: 1 or 2? Since 1 × 3 = 3, 1 is greater even though its 3

9

3

3

9 3

numerator and denominator are both smaller than in 2.

9

b.

Which fraction is greater: 2 or 5 ? Neither, because they both are equivalent to 1.

4

10

2

So even though the numerator and denominator of 5 are greater than in 2, the 10

fractions are equal.

17.

4

Malcom’s reasoning is not correct. The problem lies in the fact that although 8 does have 11

more pieces of the unit amount represented than 7 does, the pieces in 8 are smaller and 10

11

this complicates the situation. For example, 5 < 3 , but 5 > 3 and 11 > 5.

11

5

Many counterexamples exist, another is 8 < 7 .

11

18.

a.

10

Yes, it is valid: 30 = 3  10 = 3 and 20 = 2  10 = 2 . “Canceling” zeros comes 7  10

70

7

50

5  10

5

from canceling the factor of ten from the numerator and denominator.

19.

15

3 = 3 5 =  15 and 105 = 21 5 = 21  105 .

25 5  5 5 25 205 41 5 41 205

b.

No, it is not valid:

c.

Canceling happens when the same factor appears in the numerator and denominator when they are written as a product of factors. Cancelling is not valid by simply cancelling the same final digit.

a.

The numerators and denominators are both following the Fibonacci sequence. The next numerator (or denominator) is found by adding the last two numerators (or denominators).

1 2 3 5 8 13 21 34 55 89 , , , , , , , , , , 1 1 2 3 5 8 13 21 34 55

b.

The fractions are increasing in size. When comparing the first and third fractions, the first fraction is one-half less than the third. The third fraction is one-tenth less than the fifth fraction. The fifth fraction is one-sixty-fifth less than the seventh fraction, etc.

c.

An analogous pattern occurs in the even numbered fractions except that the fractions are decreasing. The fourth fraction is one-third less than the second. The sixth fraction is one-twenty-fourth less than the fourth, etc.

d.

1, 2, 1.5, 1.6, 1.6, 1.625, 1.6154,1.6190, 1.6176, 1.618. See Figure 2.35.

2-22

Chapter 2: Fractions and Problem Solving

e.

The numbers do appear to be getting closer and closer to a particular number. In fact, the sequence approaches the golden ratio, with a decimal approximation around 1.61803399.

Figure 2.35: Fractions on the number line 20.

2

3  , 3  4 , 1  2 , 3  4 , 23  24 .

3 4 5 6 2 3 7 8 30 31

a.

For example:

b.

Frank is correct, but his reasoning is flawed. For example, 3 has more parts than 2, 5

3

but 2 is larger. Looking only at the number of parts does not account for the sizes 3

of the parts. c.

The claim is that 𝐴+1 > 𝐴. This is only true when (𝐴+ 1) × 𝐵 > 𝐴 × (𝐵+ 1), 𝐵+1

𝐵

which can be simplified to 𝐴× 𝐵 + 𝐵 > 𝐴× 𝐵 + 𝐴 or 𝐵 > 𝐴. In other words, 𝐴+1 > 𝐴 is true if 𝐵 > 𝐴 . Since 𝐴 is a proper fraction then we know that 𝐵 > 𝐴 is

𝐵+1

𝐵

𝐵

true so 𝐴+1 > 𝐴 is also true. Another way to look at is that the fraction 𝐴 is 𝐵−𝐴 𝐵+1

𝐵

less than 1. 𝐴+1 is 𝐵+1

(𝐵+1)−(𝐴+1) 𝐵+1

𝐵

𝐵

= 𝐵−𝐴 less than 1. Since B is always less than B + 1 𝐵+1

in these situations, and since 𝐵 > 𝐴 in these situations, 𝐵−𝐴 > 𝐵−𝐴 which means 𝐴+1 is closer to 1 than 𝐴, which means 𝐴 < 𝐴+1.

𝐵+1

𝐵

𝐵

𝐵

𝐵+1

𝐵+1

2.4 Reasoning About Percent 1.

285 grams. See Figure 2.36. We can find the value by building up the value as follows: 95% = 50% + 25% + 5% + 5% + 5% + 5%; but once we know 5%, we can simply deduct that from the total:100% − 5% = 95%. So, 300 – 15 = 285 grams. Numerically, we have 95 ∙ 300 = 0.95 ∙ 300 = 285 grams.

100

2.4 Reasoning about Percentages 2-23

Figure 2.36: Daily value of carbohydrates We can also solve it with a percent table. See Table 2.1.

Table 2.1: A percent table used to calculate 95% of 300 is ?

2.

100% → 10% →

300

5% → 100% − 5% = 95% →

15

30 300 − 15 = 285

60%. See Figure 2.37. The bar representing 100% of the order has five pieces, with each piece representing one-half ton. Each piece also represents 20% of the order, since five of them make up the total. The three shaded pieces represent the delivered gravel. Since each piece is 20%, the crew has received 20 × 3 = 60% of the order.

Numerically, 𝑃× 5 = 3 ⟹ 𝑃 = 3 × 2 = 6 = 60 = 60%.

2

2

2

5

10

100

Figure 2.37: What percent of the gravel?

We can also solve it with a percent table. See Table 2.2.

Table 2.2: A percent table used to calculate ?% of 5 is 3 2

100%

→

100% ÷ 5

→

20%

→

20% ∙ 3

→

60%

→

5 ÷5 2

1∙3 2

2

2-24

3.

Chapter 2: Fractions and Problem Solving

20 grams. See Figure 2.38. Looking for 80% is like looking for 20% since 80% + 20% = 100%. Also, 20% ∙ 5 = 100% so we can partition the unit amount into five pieces 80 • 25 = (0.8)(25) = 20 representing 20% each. Numerically, 100 grams.

Figure 2.38: 80% of your daily fiber We can also solve the problem with a percent table. See Table 2.3.

Table 2.3: A percent table used to calculate 80% of 25 is ?

25

100%  5

→ →

20%

→

5

20%•4

→

5∙4

80%

→

20

100%

4.

25 ÷ 5

240,000. See Figure 2.39. The math drawing shows building up 100% by putting together 6 groups of 15% each and then 2 of 15%. So 3

100% = 36,000•6+ 23 •36,000 = 216,000+ 24,000 = 240,000. Numerically, 15% •T = 36, 000  T =

36, 000 0.15

= 240, 000 .

Figure 2.39: Building up 100% 5.

3600 mg. See the Table 2.4 below. 5% ∙ 20 = 100%, so 180 ∙ 20 = 3,600 mg.

180 = 3600 mg.

Numerically, 0.05•T =180T =0.05 Table 2.4: A percent table used to calculate 5% of ? = 180 5% 5% ∙ 20

6.

→ →

180 180 ∙ 20

100% → 3600 70%. See Figure 2.40. Each part represents 1-inch of rain. The whole or unit amount 2

represents the average July rainfall. The shaded portion represents this year’s July

2.4 Reasoning about Percentages 2-25

rainfall. Seven parts out of ten are shaded, which is 70%. Numerically, 70 P•5 = 3.5 P = 3.5 = 35 = 100 = 70%.

5 50

Figure 2.40: July rainfall 7.

66 2 %. See Figure 2.41. Half a cup of cereal can be seen as two parts out of three, or 3

2

=

3

2

66

3. Numerically, 𝑃∙

100

3 4

1

1 4

2

2

2 3

3

= ⟹ 𝑃= ∙ = .

Figure 2.41: Part of a serving of cereal 8.

a.

See Table 2.5.

Table 2.5: A percent table used to calculate ?% of 250 = 225 250 →

100%

25 →

10%

250 − 25 = 225 →

100% - 10% = 90%

25 is 10% of 250. 225 is 25 less than 250, so it is 10% less than 100%. So, 225 is 90% of 250. b. c.

P = 225 = 225 25 = 9 = 90 . So, P = 90%.

100 250 250  25 10 100 50% of 1600 is 800. 10% of 1600 is 160. Since 960 = 800 + 160, then 960 is 60% of 1600. See Table 2.6.

Table 2.6: A percent table used to calculate ?% of 1600 = 960 1600

→

100%

800 → 160 →

50% 10%

800 + 160 = 960

→

10% + 50% = 60%

2-26

9.

Chapter 2: Fractions and Problem Solving

d.

P 960 960 16 60 . So, P = 60%. = = = 100 1600 1600 16 100

a.

3000. See Table 2.7.

Table 2.7: A percent table used to calculate 690 is 23% of ?

23%

b.

690

1%

→ →

100%

→

30 ∙ 100 = 3,000

690 ÷ 23 = 30

7.5%. See Table 2.8.

Table 2.8: A percent table used to calculate ?% of 40 is 3

c.

40

→

100%

1

→

100% ÷ 40 = 2.5%

1∙3

→

2.5% ∙ 3 = 7.5%

7000. See Table 2.9.

Table 2.9: A percent table used to calculate 630 is9% of ?

9%

→

630

9% ÷ 9 = 1%

→

630 ÷ 9 = 70

→ 70 ∙ 100 = 7,000

1% ∙ 100 = 100% d.

4000. See Table 2.10.

Table 2.10: A percent table used to calculate 12% of ? is 480 12%

10.

480

12% ÷ 12 = 1%

→ →

1% ∙ 100 = 100%

→

40 ∙ 100 = 4,000

480 ÷ 12 = 40

75%. See Figure 2.42. The bar represents 100% of your full daily value of folic acid, or 2

3

cup, as 4 cup. Then 1 cup is three parts out of four, or 75% of your full daily value of folic 6

2

acid. Numerically, 𝑃× 2 = 1 ⟹ 𝑃 = 1 × 3 = 3 = 75 = 75%.

3

2

2

2

4

100

2.4 Reasoning about Percentages 2-27

Figure 2.42: Part of a serving of cereal 11.

4,370. This problem is very similar to Problem 1. See Table 2.11. 5% of 100% is 201 , so 5% of 4,600 is 4,600 ÷ 20 = 230. So Biggo Corporation hopes all but 230 employees, or 4,600 – 230 = 4,370 employees will participate. Numerically, 95 100 •4,600 = 0.95•4,600 = 4,370 .

Table 2.11: A percent table used to calculate 95% of 4600 is ?

100% 100%  20 = 5% 100% − 5% = 95%

12.

→ → →

4600 4600  20 = 230 4600 − 230 = 4370

85%. See Table 2.12.

Table 2.12: A percent table used to calculate ?% of 4 is 3.4 3 0.4

→ →

75% 10%

3+0.4=3.4

→

75% + 10% = 85%

Split the 3.4 acres and consider 3 acres + 0.4 acres. 3 acres is 75% of 4 acres (think 3), 4

and 0.4 acres is 10% of 4 acres. So, with 3.4 acres, the company has 75% + 10%. Or solve the equation: 4 ∙ 𝑃 = 3.4.

13.

$4.8 million. See Table 2.13.

Table 2.13: A percent table used to calculate 75% of ? is 3.6 million 75% →

3,600,000

𝟕𝟓% ÷ 𝟑 = 𝟐𝟓% →

3,600,000 ÷ 3 = 1,200,000

→

1,200,000 ∙ 4 = 4,800,000

25% ∙ 4 = 100%

Recognizing that 75% and 100% are both multiples of 25%, I first divided 75% by 3 and 3,600,00 by 3 to show that 1,200,000 matched with 25%. Then to get to 100%, I multiplied 25% by 4 and also 1,200,000 by 4. Or solve the equation: 0.75 ∙ 𝑁 = 3.6.

14.

50,400. See Table 2.14

2-28

Chapter 2: Fractions and Problem Solving

Table 2.14: A percent table used to calculate 60% of 84,000 is ?

100%

→

84 , 000

50%

→

42 , 000

10%

→ →

8 , 400

50% + 10% = 60%

42 , 000 + 8 , 400 = 50 , 400

The problem asks, “what is 60% of 84,000?” 50% is 42,000, and 10% is 8,400. So, 60% is 42,000 + 8,400 = 50,400. Or solve the equation: 0.60 ∙ 84,000 = N. 15.

125%. See Figure 2.43. Each square represents 20% of Denise’s distance but 25% of Frank’s distance. The math drawing shows that Frank ran 80% = 80 = 4 of Denise’s 100

5

distance. It also shows that Denise ran 5 of Frank’s distance. 5 = 1 1 = 125%.

4

4

4

Figure 2.43: Denise’s distance as a percentage of Frank’s 16.

87%. 115% of BigMart’s soda sales is the same as GrandMart’s soda sales. So 1.15• B = G  B =1.15G =1.151 •G  0.87G , which means BigMart sells about 87% as much soda as GrandMart.

17.

a.

No. Benton spent 133 13 % of Connie’s ticket price. Figure 2.44 shows that Benton spent 4 of Connie’s ticket price. Each square is 25% of Benton’s ticket price, but 1

3

333 % of Connie’s ticket price.

Figure 2.44: Connie spent 75% as much as Benton b.

No. Connie spent 80% of Benton’s ticket price. Figure 2.45 shows that Connie spent 4 of Benton’s ticket price. Each square is 20% of Benton’s ticket price and 5 25% of Connie’s ticket price.

2.4 Reasoning about Percentages 2-29

Figure 2.45: Benton spent 125% as much as Connie 18.

No. 25% of the items are not newspapers. If these items cost much more than the newspapers or much less than the newspapers, then the income from these items would not equal 25%. The income from newspapers would be 75% of the total only if all the items are the same price.

19.

200. 30% of 500 is 150 style A handbags. 70% of 500 is 350 style B handbags. To bring the balance to 50% style A handbags, we need to increase the number of style A bags sold from 150 to 350.

20

a.

We actually cannot determine the percent of students in the county that speak Spanish at home because we don’t know the populations of the two schools.

Furthermore, by Tom’s logic we would have to say that 150% of the students do not speak Spanish at home because 70% do not at one school and 80% do not at the other. This cannot be true.

b.

i. No, the correct percent is 28%. In this scenario, 30% of School A is 120 students, and 20% of School B is 20 students. So, 140 students in the county speak Spanish at home out of a total population is 500 students. So, the = 28 percent is 140 500 100 . ii. No, the correct percent is 22%. In this scenario, 30% of School A is 30 students, and 20% of School B is 80 students. So, 110 students in the county speak Spanish at home out of 500 students altogether. So, the percent is 110 22 500 = 100 . iii. If the two schools have equal populations, then the correct percent is the average of the two percents. For example, if both schools have 100 students, then 30 students in School A and 20 students in School B speak Spanish at home. This gives us 50 students out of 200, which is 25%.

21.

a.

Think of the can of soup as being 20 parts solid and 80 parts water. Mixing a can of water adds another 100 parts of water. So, the mixture has 20 parts solid and 180 parts water, to make 200 parts total. With 180 parts water out of 200 parts total is equivalent to 90 parts water out of 100 parts total, which is 90%.

2-30

Chapter 2: Fractions and Problem Solving

6632 %. Instead of 20 parts solid and 80 parts water, the concentrated soup is 20

b.

parts solid and 40 parts water. So, it is 40 parts water out of 60 parts total, which is equivalent to 2 which is 66 32 %.

3

c.

22.

75%. Continuing with the line of the previous two parts of the solution we assume the 20 parts solid remains constant. To make this 50% of the contents of the can, we need solids to be 20 parts out of 40. So, the concentrated soup must be 20 parts water. The soup started with 80 parts water. Removing 60 parts of the water means removing 60 out of 80, which is equivalent to 68 = 34 = 75%.

60 cans. 60% of 300 cans is 180 cans of brand A beans. 40% of 300 cans is 120 cans of brand B beans. After removing some cans of brand B, the 180 cans of brand A now represent 75% of the total. This means that the 180 cans are 3 of the total because there 4

are 3 parts each being 1 of the total. Since 180 ÷ 3 = 60, then we know that 60 cans are 1 4

4

of the total. So, we will have 180 cans of brand A and 60 cans of brand B, which means 60 cans of brand B were removed.

23.

100 pounds. The fresh cucumbers are 99% water, and 1% plant material. This means the plant material weighs 0.01 200 = 2 pounds. Later, the cucumbers are 98% water. The weight of the actual plant material does not change, only the weight from water. So now, the 2 pounds of plant 100 material constitute 2% of the weight. That is: 2 •W = 2 W = 2• =100 pounds.

100

2

Chapter 2 Review 1.

For 6 we would have 6 equal parts that are 1 of the unit. 1 or 5 of the unit would be only 5 5

5

5

of those equal parts. In Figure 2.46, each unit has been partitioned into 5 equal-length parts, so the distance between tick marks is 1 of the unit. 6 is indicated at 6 ticks marks 5

5

from zero and 1 or 5 at 5 of those tick marks.

5

Figure 2.46: Plotting 1 on a number line 2. a.

See Figure 2.47. The tick marks initially are seen as representing 1 of a mile.

8

Then, three tick marks over represents 3 of a mile, while five tick marks over 8

represents 5 of a mile. These points are marked with dots. However, if we view 8