Chapter 4 Answers. Chapter 4 Problems of the Week, p For example, I added each player s total points as follows:

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1 Chapter 4 Problems of the Week, p For example, I added each player s total points as follows: Chamberlain Jordan Abdul-Jabbar Since Chamberlain earned the most points, he was the best player. 2. For example, the Japanese man lived from June 29, 1865 through June 29, 1985, which is = 120 full years, after which he continued to live until February 21, Since the dates June 29 and February 21 are both close to the end of their months I counted the months between July and March as full months and got 8 months. I then converted 120 years to months by multiplying by 12: = 1440 months and added the 8 months to get 1448 months. So the Japanese man lived about 1448 months. I then calculated the months the U.S. woman lived in the same way. From November 18, 1874 to November 18, 1990 is = 116 years. Since November 18 and February 14 are both near the middle of their months I rounded each date to the next full month and counted the months in between to get 3 full months. Then I converted 116 years to months by multiplying by 12 to get 1392 and added the 3 months to get 1395 months. So the Japanese man lived about = 56 months longer than the U.S. woman. 3. For example, if the centipede can travel 1.8 km in an hour and the giant tortoise only 0.27 km in the same time, the centipede will travel much further than the tortoise over 3 hours. To find out how much, I can add an extra zero at the end of 1.8 km and then subtract 0.27 km from it. This will tell me the difference in the distance each animal can travel per hour. Then I can multiply the answer by 3 to find out how much farther the centipede will travel in 3 hours km 0.27 km 1.53 km 3 = 4.59 km more in 3 hours Chapter 4 Mental Math (Master) pp a) For example, = 1256 b) For example, = 148 c) For example, = 800; = 750 d) For example, = 10; = 90; = 1290 e) For example, = 1054 f) For example, = = a) For example, = 1000; = 2000 b) For example, = 2000; = 1500; = 2500 c) For example, = 1000; = ; = d) For example, = 1000; = Chapter 4: Addition and Subtraction Copyright 2005 by Thomson Nelson

2 3. Trip Driving Distance (km) Rounded Distance (km) Vancouver Calgary Calgary Winnipeg Winnipeg Toronto Toronto Halifax a) = 2000 b) = 4000 c) = a) Sean s school raised $500 + $1500 = $2000; Natasha s school raised $ $ $200 = $2200 $1 = $2199. Natasha s school raised $2199 $2000 = $199 more. b) 1750 kg kg = 4000 kg; 1500 kg kg = 4000 kg; 4000 kg kg = 8000 kg c) m 2500 m = 7500 m; 7500 m 7250 m = 250 m higher 6. a) = 1.0; = 2.0 b) = 1.0; = 13 c) = 1.00; = 4.00 d) = 1.00; = 3.00 e) = 2.00; = 10.00; = f) 2.00 = 1.50 = 3.50; = a) For example, = = 5.1 b) For example, = = 3.01 c) For example, = = 5.00 d) For example, = = a) For example, $ $25.00 = $30.00; $ $10.00 = $40.00; $ = $ b) For example, $ $0.01 = $25.00; $ $10.00 = $35.00 c) For example, $ $0.75 = $1.00; $ $27.00 = $28.00 d) For example, $ $50.00 = $69.00; $69.00 $0.01 = $68.99 e) For example, $ $3.00 = $11.00; $11.00 $0.01 $0.01 = $ a) For example, $ $30.98 = $ $0.88 = $50.88 b) For example, $ $29.99 = $ $0.01 $0.01 = $50.00 c) For example, $ $0.25 = $20.00; $30.25 $0.25 = $30.00; so $ $30.25 = $50.00 d) For example, $ $29.99 = $50.00 $0.01 $0.01 = $49.98 e) For example, $ $29.98 = $50.00 $0.02 $0.02 = $ a) For example, $ $0.02 = $ The change is $0.02. b) For example, $ $0.50 = $18.00; $ $2.00 = $ The change is $2.50. c) For example, $20.00 $15.00 = $5.00; $5.00 $0.50 = $4.50. The change is $4.50. d) For example, $ $0.75 = $12.00; $ $3.00 = $15.00; $ $5.00 = $ The change is $8.75. e) For example, $ $ $13.00; $20.00 $13.00 = $7.00; $7.00 $0.25 = $6.75. The change is $ a) For example, $ $15.00 = $40.00, so $40.00 $15.00 = $25.00 b) For example, $ $15.00 = $40.00, so $ $14.99 = $40.00 $0.01 = $39.99 c) For example, $ $15.00 = $40.00, so $ $15.00 = $40.00 $0.01 = $39.99 d) For example, $40.00 $25.00 = $15.00, so $40.00 $24.99 = $ $0.01 = $15.01 Copyright 2005 by Thomson Nelson 81

3 12. For example, a) = 1.0; = 0.5 b) = 3.00; = 6.50 c) = 1.00; = 0.75; = 0.50 d) = 6.00; = 2.00; = 4.00 e) = f) = 2.00; = 3.00; = a) For example, = 1.50, so = = 1.51 b) For example, must be 0.01 more than = 2.75, or 2.76 c) For example, 1.95 is 0.05 less than 2.00, so = is 1.05 d) For example, 4.49 is 0.01 less than 5.00, so = = 4.99 Scaffolding for Getting Started Activity (Master) pp Counting on: 20, 40, 50, 60, 70, 75, 80, 85, 90; you have $90.00 worth of certificates. A. For example, adults cost $12.95, which is about $13, and children cost $7.50, which is between $7.00 and $8.00. To be safe, I will round to $8.00. There should be at least 2 adults, so that will cost $13 + $13 = $26. I would also like to take 6 friends, which means 7 children including me. So 7 $8 = $56, and $26 + $56 = $82. I have $90 in certificates. So $90 82 = $8.00. It looks like I could take one more friend if I wanted, for a total estimate of $90 for 2 adults and 8 children. B. For example, 2 adults cost $13 + $13 = $ This is 10 cents too much, so the actual cost is $ Children cost about $8.00 each, so $ is $ However, that is 8 $0.50, or $4.00 too much. So $64.00 $4.00 = $ $25.90 for the adults equals $85.90 in total. C. For example, I used mental math for the estimate because rounded numbers are easy to add. However, to find the actual cost I used a calculator, because it was faster to multiply $ that way than with pencil and paper. I could have also used mental math by multiplying $ = $15.00 and then multiplying $15.00 by 4. D. For example, my estimate is $90; my answer is $ My answer is reasonable because it is fairly close to my estimate. E. For example, I will use all the certificates, the two $20, three $10, and four $5, to pay for 2 adults, 7 friends, and me. This makes $ F. For example, my change will be $90.00 $85.90 = $4.10. G. For example, my new problem is that I want to take 2 adults and 2 children, including me, to 2 movies. Do I have enough certificates to pay? First I estimate: 2 adults will cost about $25.90 for 1 movie, so the cost will be about $50.00 for 2 movies. 2 children will cost 2 $7.50 = $15 for one movie, so the cost will be $30.00 for two movies. Altogether that is about $80.00, so I think I will have enough. My solution is $ $25.90 = $50 + $1.80 = $51.80 $ $15.00 = $30.00 $ $30.00 = $81.80 Scaffolding for Do You Remember? (Master) p For example, there are about 800 children. Therefore, there are about = 5200 adults. I estimated by rounding 767 to the nearest 100 and then subtracted that from For example, Method 1: = 300; then subtract = 225. Method 2: is 200. I took away 25 too little, so I need to add 25 to 200; = Chapter 4: Addition and Subtraction Copyright 2005 by Thomson Nelson

4 3. a) For example, 4566 is about 4500 and 1837 is about 2000, so the sum is about = The calculated answer is b) For example, 2756 is about 3000 and 4248 is about 4000, so the sum is about = The calculated answer is c) For example, 3000 is 3000 and 865 is about 900, so the difference is about = The calculated answer is d) For example, 3299 is about 3300, and 6348 is about 6350, so the estimated difference is The calculated answer is a) For example, = = 101 b) For example, = = 85; 85 1 = 84 c) For example, 100, 90, 85 d) For example, 80 subtract 20 is 60 and then add 2 is 62. Scaffolding for Lesson 3 (Master) p a) For example, = 8000, or = The calculated answer is b) For example, = The calculated answer is c) For example, = The calculated answer is a) 11 2 [carried nos.] b) 11 [carried nos.] For example, the crates could have masses of 2500 kg, 2300 kg, 2400 kg, and 2400 kg, or of 1200 kg, 3400 kg, 300 kg, and 4700 kg, or any other combination that adds up to 9600 kg. Scaffolding for Lesson 4 (Master) pp For example, 7 years and 3 months, or 87 months. I found the total months by changing 7 years to 84 months and then adding the other 3 months. 5. There are 24 hours in a day. Dieter is therefore awake about 24 9 = 15 hours each day. a) 7 days per week 15 h = about 105 hours per week. b) 4 weeks per month 105 hours per week = about 450 hours per month. c) 12 months per year 450 hours per month = about 5400 hours per year. 6. Step 1: The year was 1891, because years = Step 2: The answer depends on the current year. For example, = 115 years ago. 7. For example, my problem is that I have to find the age difference in days between two people I know when I only know their ages in years. The question is, How much older in days is Person 1 than Person 2? I picked my Uncle Calum, who is 31, as Person 1, and my cousin Ryan, who is 8, as Person 2. Ignoring leap years, that makes Calum = days old, and Ryan = 2920 days. Now I can solve the problem by subtracting Ryan s age in days from Calum s: = 8395 days. Uncle Calum is days older than my cousin Ryan. 8. For example, I know that there are 50 bags, each with 124 candies in them, both red and black. I also know that 3173 of the total number of candies are red. So if I find out how many candies there are in total, I can subtract the number of red candies to find the number of black candies. My solution is = 6200 total candies red candies = 3027 black candies. There must be 3027 black candies in the 50 bags. Chapter 4 Test (Master) pp a) = = 763 b) = = 198 c) = a) 136 b) 43 each day 3. a) reasonable; for example, = 5800; I rounded to numbers that are easier to subtract. Copyright 2005 by Thomson Nelson 83

5 b) not reasonable; for example, = 8500; I rounded the numbers to the nearest hundreds. 4. Estimate of book mass total: 1100 g g g = 5400 g. The student can therefore safely carry another book only if its mass is about 6800 g 5400 g = 1400 g or less. Calculation: 1125 g g g = 5417 g; 6800 g 5417 g = 1383 g. 5. a) = 8510; = b) In step 1 I added the attendance for the 3 days; in step 2 I subtracted this total from a) Mental math, because the numbers are only 1 apart. b) A calculator, because 2 of the numbers have more than 4 digits. c) Mental math, because I know that = 70, so the rest of the answer is easy. 7. a) Estimate: 35 3 = 32 b) Estimate: = 10 Subtract: = Add: =10.03 c) Estimate: = 2.00 Subtract: = m m m m = 38.1 m 9. $ $ $5.95 = $81.44; $85.00 $81.44 = $3.56 change Chapter 4 Task (Master) pp. 64 Sample answers are provided for Part 1 A C. Part A. Saturday I saw that all but one (397) number had a zero in the ones column. I therefore rounded 397 to 400 to make the numbers easier to add. Then I combined groups of numbers: = 400; = 1300; = 1830; = 980; and = 730. I then added these numbers together ( ) Finally, I subtracted the 3 that I added when I rounded 397 to 400. So Jose ate = 5237 Calories on Saturday. My strategy was to add groups of numbers together because the numbers were round, which made them easy to add in my head. It was too confusing to add that many numbers on paper. I then used paper and pencil to add the resulting four numbers because there were many different numbers in each place value column and I couldn t keep track of them all in my head. Sunday When I looked at the ones column I noticed that most of the numbers were 5. I also saw that I could add 2 to 313 to make it 315 and take 2 away from 847 to make it 845. I then counted by 5 s: 5, 10, 15, 20, 25, 30 and then added the remaining 1 (from 361) to equal 31ones. Then I looked at the tens column and saw that there were a lot of 6 s. I subtracted 10 from (my now) 315 to make 305, and added it to 50 to make 60. Next I subtracted 20 from 75 to make 55 and added 20 to (my now) 845 to get 865. I then counted how many 6 s I had in the 10 s column. There were six, and I knew that 6 60 = 360. I then added the two 5 s, or together to get 100, and then the 360 (from the 6 s) to get 460. I now had just a 9 and an 8 left to add together: = 170. So = Chapter 4: Addition and Subtraction Copyright 2005 by Thomson Nelson

6 Lastly I added up the hundreds: = = = = 1700, and then all my totals together (from the hundreds column) 630 (from the tens column) + 31 (from the ones column) Calories for Sunday My strategy was to add each column separately because the numbers were not round numbers. I added and subtracted the ones to form as many numbers ending in 5 as I could because it is easy to count by 5 s. In the tens column there were already four 6 s. By adding and subtracting tens I was able to form six 6 s in this column. They were easy to multiply because I know my 6 times table. I added two of the remaining numbers together to form 100 because it was easy to add on 100 more, and then added the last 2 digits together before adding them to my tens column total. Because there weren t many hundreds it was easy to add them together. I also looked for doubles to make it easier. It was then easy to add each place value separately before adding the totals together because I had no numbers to carry. Part B. Saturday I rounded 5237 to Then I subtracted: = I used 2200, because it is the lower end of the Calorie range. Next I subtracted: = I used 300, because it is the difference between 2500, the upper end of the range, and So I estimated that Jose s Saturday meal was between 2700 and 3000 Calories over the recommended daily allowance. Then I calculated: = 3037, and = Since my calculations are only 37 more than my estimates, I think my estimates are reasonable. Sunday Jose s total of 2361 for Sunday is greater than 2200 and less than 2500, so it falls within the recommended range of Calories. Saturday I first rounded $2.89 to $3, $1.99 to $2, $7.99 to $8, $3.99 to $4, and $1.99 to $2. Then = = $19, after which I added $19 + $ 4 + $3 ($ $1.50) = $26. From this total I next subtracted the 15 that I had added when rounding up: $ = $ Next, I rounded the costs for lunch: $4.29 to $4, $3.29 to $3, and $2.19 to $2. I then added these numbers: = $9, and I added this total to my previous sum: $ $9 = $ Lastly, I summed up the amounts left off when I rounded the lunch costs down: $ $0.29 = $ $0.20 = $0.78 $0.01 = $0.77, and I added this amount to get the final total. 11 $ $35.62 My strategy was to round costs that ended in 99 and 89 cents to the nearest dollar because it was easy to add whole numbers. I then had to subtract the number of cents I added when rounding the numbers up. When adding the cents I added the doubles first. I also rounded the lunch costs to the nearest dollar, then added the cents I subtracted when rounding down. I rounded 19 to 20 as it was easier to add a round number and then just subtract 1 from the total. Copyright 2005 by Thomson Nelson 85

7 Sunday Using much the same strategy, I rounded these amounts to the nearest dollar and kept track of how much I increased each number, as shown. $0.76 rounded to 1 ($0.24 increase) $0.88 rounded to 1 ($0.12 increase) $0.85 rounded to 1 ($0.15 increase) $2.86 rounded to 3 ($0.14 increase) $0.99 rounded to 1 ($0.01 increase) $0.98 rounded to 1 ($0.02 increase) $5.94 rounded to 6 ($0.06 increase) $0.99 rounded to 1 ($0.01 increase) I then added the dollar amounts: = = = = $15. Then I added up the amount of cents that I added to each number when rounding = 25; = = 30; = 20; = = 75. I then subtracted this amount from the dollar total: $ = $ Now I added the remaining amounts: = = = = I then added this amount to my previous total: = = $ ( ) = $ So Jose s food costs for Sunday were $ To find the difference between the two days costs of meals, I first estimated by rounding Saturday s cost of $35.62 to $35, and Sunday s costs of $15.09 to $15. I then subtracted $15 from $35 to get $20. So my estimate told me that Saturday s costs were about $20 more than Sunday s. Then I found the actual difference with a calculator: $35.62 $15.09 = $ Since this was only 53 more than my estimate, I knew that my calculations were correct. Part C. Student choices, regardless of source used, should fall within the Calorie range. Student work should indicate that the Task Checklist has been followed. Lesson 8 Answers (continued from p. 47) 5. a) For example: watch and book, total cost = $ $ $29.98 $73.73 b) For example, $ $20.00 = $93.73; $ = $93.75; $ = $94.00; $ $6.00 = $100.00; $ $6.00 = $ cost = $17.75; change = $2.25. If 25 more than his purchase is $18.00, he must have made a purchase of $ = $ He was given 25 and then $2.00 more for total change of $2.25.) Lesson 9 Answers (continued from p. 51) 7. a) For example, I estimate 2 1 = b) For example, I estimate = c) For example, I estimate, 4 2 = d) For example, I estimate 4 3 = 1. I count up from 2.85: = e) For example, I estimate = 2. I count up from 0.75 to 2: = f) For example, I estimate 12 5 = 7. I count up from 4.75: = Chapter 4: Addition and Subtraction Copyright 2005 by Thomson Nelson

8 8. a) For example, any whole number larger than 4, because it had to be equal to 4.75 plus a decimal number. It could have been 5, 6, 7, or 10, for instance. b) For example, 0.25 from 5 to get 4.75 or 5.25 from 10 to get Chapter Review Answers (continued from p. 57) 5. a) Estimate: = = My answer is close to my estimate, so it is reasonable b) Estimate: = = My answer is very close to my estimate, so it is reasonable c) Estimate: = = My answer is close to my estimate, so it is reasonable No. The total mass is 8057 kg, which is greater than the safe limit of 8000 kg. 7. a) The total attendance for first 3 days was The fourth day s attendance was = b) I used addition to calculate the total attendance for the first 3 days; then I used subtraction to calculate the attendance on the fourth day. c) For example, I used pencil and paper because I know how to add three 4-digit numbers and subtract a 4-digit number from another 4-digit number. 8. a) For example, I would use mental math to add to get Then I would use paper and pencil to add 3 numbers instead of 4. (Answer = ). b) For example, I would use paper and pencil because the numbers are too hard to subtract mentally and I prefer paper and pencil over a calculator for subtraction. (Answer = 2573). c) For example, I would use a calculator because is a 5-digit number. (Answer = ). d) For example, I would use mental math because I noticed that = So = = , plus add another 25 to get a) Estimate: = My answer is close to my estimate, so it is reasonable b) Estimate: = My answer is close to my estimate, so it is reasonable c) Estimate: = My answer is close to my estimate, so it is reasonable d) Estimate: = My answer is close to my estimate, so it is reasonable Copyright 2005 by Thomson Nelson 87

9 m. 11. a) cost = $34.23; change = $5.77 b) cost = $43.97; change = $16.03 c) cost = $95.99; change = $ a) Estimate: = My answer is close to my estimate, so it is reasonable b) Estimate: = My answer is close to my estimate, so it is reasonable c) Estimate: = My answer is close to my estimate, so it is reasonable d) Estimate: = My answer is close to my estimate, so it is reasonable a) The two perimeters are 5.67 m and m. b) The difference in the perimeters is 6.53 m. 88 Chapter 4: Addition and Subtraction Copyright 2005 by Thomson Nelson