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