S OLUTIONS M ANUAL D ANIEL S. M ILLER Niagara County Community College T HINKING M ATHEMATICALLY S EVENTH E DITION Robert Blitzer Miami Dade College TABLE OF CONTENTS T HINKING M ATHEMATICALLY , 7 E Chapter 1: Problem Solving and Critical Thinking 1 Chapter 2: Set Theory 25 Chapter 3: Logic 51 Chapter 4: Number Representation and Calculation 125 Chapter 5: Number Theory and the Real Number System 159 Chapter 6: Algebra: Equations and Inequalities 227 Chapter 7: Algebra: Graphs, Functions, and Linear Systems 285 Chapter 8: Personal Finance 353 Chapter 9: Measurement 411 Chapter 10: Geometry 429 Chapter 11: Counting Methods and Probability Theory 469 Chapter 12: Statistics 527 Chapter 13: Voting and Apportionment 587 Chapter 14: Graph Theory 643 Chapter 1 Problem Solving and Critical Thinking 1 Check Points 1.1 1. Counterexamples will vary. Example: 40 40 1600   2. a. Add 6 each time. b. Multiply by 5 each time. 3 + 6 = 9 2  5 = 10 9 + 6 = 15 10  5 = 50 15 + 6 = 21 50  5 = 250 21 + 6 = 27 250  5 = 1250 27 + 6 = 33 2, 10, 50, 250, 1250 3, 9, 15, 21, 27, 33 c. Cycle multiplying by 2, 3, 4. d. Cycle adding 8, adding 8, subtracting 14. 3  2 = 6 1 + 8 = 9 6  3 = 18 9 + 8 = 17 18  4 = 72 17 – 14 = 3 72  2 = 144 3 + 8 = 11 144  3 = 432 11 + 8 = 19 432  4 = 1728 19 – 14 = 5 1728  2 = 3456 5 + 8 = 13 6, 18, 72, 144, 432, 1728, 3456 13 + 8 = 21 21 – 14 = 7 9, 17, 3, 11, 19, 5, 13, 21, 7 3 . a. Starting with the third number, each number is the sum of the previous two numbers, 29 + 47 = 76 b. Starting with the second number, each number one less than twice the previous number, 2(129) 1 257   4. The shapes alternate between rectangle and triangle. The number of little legs cycles from 1 to 2 to 3 and then back to 1. Therefore the next figure will be a rectangle with 2 little legs. 5. a. Conjecture based on results: The original number is doubled. Select a number. 4 10 0 3 Multiply the number by 4. 4 4 16   10 4 40   0 4 0   3 4 12   Add 6 to the product. 16 6 22   40 6 46   0 6 6   12 6 18   Divide this sum by 2. 22 2 11   46 2 23   6 2 3   18 2 9   Subtract 3 from the quotient. 11 3 8   23 3 20   3 3 0   9 3 6   Summary of results: 4  8 10  20 0  0 3  6 b. Select a number: n Multiply the number by 4: 4 n Add 6 to the product: 4 6 n  Divide this sum by 2: 4 6 4 6 2 3 2 2 2 n n n      Subtract 3 from the quotient: 2 3 3 2 n n    Chapter 1 Problem Solving and Critical Thinking 2 Concept and Vocabulary Check 1.1 1. counterexample 2. deductive 3. inductive 4. true Exercise Set 1.1 1. Counterexamples will vary. Example: President Obama was younger than 65 at the time of his inauguration. 2. Counterexamples will vary. Example: Beyoncé Knowles is a singer who appears in movies. 3. Counterexamples will vary. Example: 3 multiplied by itself is 9, which is not even. 4. Counterexamples will vary. Example: 100 is a three-digit number and 100 100 200,   which is not a four-digit number. 5. Counterexamples will vary. Example: Adding 1 to the numerator and denominator of 1 2 results in 2 3 which is not equal to 1 2 . 6. Counterexamples will vary. Example: 8 3 5,   which is odd, but 8 and 3 are not both odd. 7. Counterexamples will vary. Example: When 1  is added to itself, the result is 2,  which is less than 1.  8. Counterexamples will vary. Example: When 1 is divided by 2,  the result is 1 2 ,  which is greater than 2.  9. Pattern: Add 4 24 + 4 = 28 8, 12, 16, 20, 24, 28 10. Pattern: Add 5 39 + 5 = 44 19, 24, 29, 34, 39, 44 11. Pattern: Subtract 5 17 – 5 = 12 37, 32, 27, 22, 17, 12 12. Pattern: Subtract 4 17 – 4 = 13 33, 29, 25, 21, 17, 13 13. Pattern: Multiply by 3 243  3 = 729 3, 9, 27, 81, 243, 729 14. Pattern: Multiply by 4 512  4 = 2048 2, 8, 32, 128, 512, 2048 15. Pattern: Multiply by 2 16  2 = 32 1, 2, 4, 8, 16, 32 16. Pattern: Multiply by 5 125  5 = 625 1, 5, 25, 125, 625 17. Pattern: 1 alternates with numbers that are multiplied by 2 16  2 = 32 1, 4, 1, 8, 1, 16, 1, 32 18. Pattern: 1 alternates with numbers that are increased by 3 10 + 3 = 13 1, 4, 1, 7, 1, 10, 1, 13 19. Pattern: Subtract 2 4  – 2 = 6  4, 2, 0, 2  , 4  , 6  20. Pattern: Subtract 3 6  – 3 = 9  6, 3, 0, 3  , 6  , 9  21. Pattern: Add 4 to the denominator 1 1 18 4 22   1 2 , 1 6 , 1 10 , 1 14 , 1 18 , 1 22 22. Pattern: Add 1 to the denominator 1 1 5 1 6   1, 1 2 , 1 3 , 1 4 , 1 5 , 1 6 Section 1.1 Inductive and Deductive Reasoning 3 23. Pattern: Multiply the denominator by 3 1 1 27 3 81   1, 1 3 , 1 9 , 1 27 , 1 81 24. Pattern: Multiply the denominator by 2 1 1 8 2 16   1, 1 2 , 1 4 , 1 8 , 1 16 25. Pattern: The second number is obtained by adding 4 to the first number. The third number is obtained by adding 5 to the second number. The number being added to the previous number increases by 1 each time. 33 9 42   26. Pattern: The second number is obtained by adding 3 to the first number. The third number is obtained by adding 4 to the second number. The number being added to the previous number increases by 1 each time. 27 8 35   27. Pattern: The second number is obtained by adding 3 to the first number. The third number is obtained by adding 5 to the second number. The number being added to the previous number increases by 2 each time. 38 13 51   28. Pattern: The second number is obtained by adding 3 to the first number. The third number is obtained by adding 5 to the second number. The number being added to the previous number increases by 2 each time. 37 13 50   29. Pattern: Starting with the third number, each number is the sum of the previous two numbers. 27 44 71   30. Pattern: Starting with the third number, each number is the sum of the previous two numbers. 19 31 50   31. Pattern: Cycle by adding 5, adding 5, then subtracting 7. 13 5 18   32. Pattern: Cycle by adding 6, adding 6, then subtracting 10. 13 6 19   33. Pattern: The second number is obtained by multiplying the first number by 2. The third number is obtained by subtracting 1 from the second number. Then multiply by 2 and then subtract 1, repeatedly. 34 1 33   34. Pattern: The second number is obtained by multiplying the first number by 3. The third number is obtained by subtracting 1 from the second number. Then multiply by 3 and then subtract 1, repeatedly. 123 1 122   35. Pattern: Divide by 4  1 1 ( 4) 4     64, 16  , 4, 1  , 1 4 36. Pattern: Divide by 5  1 1 ( 5) 5     125, 25  , 5, 1  , 1 5 37. Pattern: The second value of each pair is 4 less than the first. 3 – 4 = 1  1 1 2 2 (6, 2), (0, 4), (7 , 3 ), (2, 2), (3, 1)    38. Pattern: The second value of each pair is the square of the first.   2 16 4 7 49           5 25 16 2 4 1 1 4 3 9 5 25 6 36 7 49 , , , , (7, 49), , , ,   39. The figure cycles from square to triangle to circle and then repeats. So the next figure is 40. The figure rotates 90  counterclockwise. So the next figure is Chapter 1 Problem Solving and Critical Thinking 4 41. The pattern is to add one more letter to the previous figure and use the next consecutive letter in the alphabet. The next figure is shown at right. 42. The figure alternates from triangle to square and gains one line on the bottom. The next figure is . 43. a. Conjecture based on results: The original number is doubled. Select a number. 4 10 0 3 Multiply the number by 4. 4 4 16   10 4 40   0 4 0   3 4 12   Add 8 to the product. 16 8 24   40 8 48   0 8 8   12 8 20   Divide this sum by 2. 24 2 12   48 2 24   8 2 4   20 2 10   Subtract 4 from the quotient. 12 4 8   24 4 20   4 4 0   10 4 6   Summary of results: 4  8 10  20 0  0 3  6 b. 4 n 4 8 4 8 4 8 2 4 2 2 2 2 4 4 2 n n n n n n          44. a. Conjecture based on results: The result is always 2. Select a number. 4 10 0 3 Multiply the number by 3. 4 3 12   10 3 30   0 3 0   3 3 9   Add 6 to the product. 12 6 18   30 6 36   0 6 6   9 6 15   Divide this sum by 3. 18 3 6   36 3 12   6 3 2   15 3 5   Subtract the original from the quotient. 6 4 2   12 10 2   2 0 2   5 3 2   Summary of results: 4  2 10  2 0  2 3  2 b. 3 n 3 6 3 6 3 6 2 3 3 3 2 2 n n n n n n          45. a. Conjecture based on results: The result is always 3. Select a number. 4 10 0 3 Add 5 to the number. 4 5 9   10 5 15   0 5 5   3 5 8   Double the result. 9 2 18   15 2 30   5 2 10   8 2 16   Subtract 4. 18 4 14   30 4 26   10 4 6   16 4 12   Divide the result by 2. 14 2 7   26 2 13   6 2 3   12 2 6   Subtract the original number. 7 4 3   13 10 3   3 0 3   6 3 3   Summary of results: 4  3 10  3 0  3 3  3 b. 5 n  2( 5) 2 10 2 10 4 2 6 2 6 2 6 3 2 2 2 3 3 n n n n n n n n n                Section 1.1 Inductive and Deductive Reasoning 5 46. a. Conjecture based on results: The result is always 5. Select a number. 4 10 0 3 Add 3 to the number. 4 3 7   10 3 13   0 3 3   3 3 6   Double the result. 7 2 14   13 2 26   3 2 6   6 2 12   Add 4. 14 4 18   26 4 30   6 4 10   12 4 16   Divide the result by 2. 18 2 9   30 2 15   10 2 5   16 2 8   Subtract the original number. 9 4 5   15 10 5   5 0 5   8 3 5   Summary of results: 4  5 10  5 0  5 3  5 b. 3 n  2( 3) 2 6 2 6 4 2 10 2 10 2 10 5 2 2 2 5 5 n n n n n n n n n                47. Using inductive reasoning we predict 6 7 1 2 3 4 5 6 2        . Arithmetic verifies this result: 21 = 21 48. Using inductive reasoning we predict 18 7 3 6 9 12 15 18 2        . Arithmetic verifies this result: 63 = 63 49. Using inductive reasoning we predict 1 3 5 7 9 11 6 6        . Arithmetic verifies this result: 36 = 36 50. Using inductive reasoning we predict 5 9 (5 9) 59     . Arithmetic verifies this result: 5 9 (5 9) 59 5 9 (14) 59 45 (14) 59 59 59           51. Using inductive reasoning we predict 98765 9 3 888,888    . Arithmetic verifies this result: 98765 9 3 888,888 888,885 3 888,888 888,888 888,888       52. Using inductive reasoning we predict 54321 9 1 488,888    . Arithmetic verifies this result: 54321 9 1 488,888 488,889 1 488,888 488,888 488,888       53. The first multiplier increases by 33. 132 + 33 = 165 The second multiplier is 3367. The product increases by 111,111. 165  3367 = 555,555 is correct. Chapter 1 Problem Solving and Critical Thinking 6 54. The pattern implies we should attach a 6 to the right of the first multiplier. The second multiplier is always 8. The pattern implies we should add 6 to that product to obtain 987,654. 123,456  8 + 6 = 987,654 is correct. 55. b; The resulting exponent is always the first exponent added to twice the second exponent. 56. c; The resulting exponent is always half the sum of the three exponents. 57. deductive; The specific value was based on a general formula. 58. inductive; The general conclusion for all HMO patients was based on specific observations. 59. inductive; The general conclusion for all full-time four-year colleges was based on specific observations. 60. deductive; The specific grade was based on a general course policy. 61. a. 1, 3, 6, 10, 15, and 21 are followed by 21 + 7 = 28 28 + 8 = 36 36 + 9 = 45 45 + 10 = 55 55 + 11 = 66 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, and 66. b. 4 – 1 = 3 9 – 4 = 5 16 – 9 = 7 25 – 16 = 9 The successive differences increase by 2. 25 + 11 = 36 36 + 13 = 49 49 + 15 = 64 64 + 17 = 81 81 + 19 = 100 c. The successive differences are 4, 7, and 10. Since these differences are increasing by 3 each time. The next five numbers will be found by using differences of 13, 16, 19, 22, and 25. 22 + 13 = 35 35 + 16 = 51 51 + 19 = 70 70 + 22 = 92 92 + 25 = 117 d. If a triangular number is multiplied by 8 and then 1 is added to the product, a square number is obtained. 62. Each row begins and ends with 1. Other numbers are the sum of the two values that are diagonally above. 66. does not make sense; Explanations will vary. Sample explanation: Such conclusions would be certain. 67. makes sense Section 1.1 Inductive and Deductive Reasoning 7 68. does not make sense; Explanations will vary. Sample explanation: Though this sample was 51%, it is not certain that this exact percentage will hold for the entire population. 69. makes sense 70. The pattern suggests that the compatible expression is the square of the first number minus twice the product of the two numbers, plus the square of the second number. 2 (11 7) 121 154 49     71. a. The sums are all 30: 16 3 11 5 10 15 9 17 4 b. The sums are all 36: 17 5 14 9 12 15 10 19 7 c. For any values of a , b , and c , the sums of all rows, all columns, and both diagonals are the same. d. The sums of the expressions in each row, each column, and each diagonal is 3 a . e. Finding each sum verifies the conjecture that they are all 3 a . First row: ( ) ( ) ( ) 3 a b a b c a c a        Second row: ( ) ( ) ( ) 3 a b c a a b c a        Third row: ( ) ( ) ( ) 3 a c a b c a b a        First Column: ( ) ( ) ( ) 3 a b a b c a c a        Second Column: ( ) ( ) ( ) 3 a b c a a b c a        Third Column: ( ) ( ) ( ) 3 a c a b c a b a        First Diagonal: ( ) ( ) ( ) 3 a b a a b a      Second Diagonal: ( ) ( ) ( ) 3 a c a a c a      72. Answers will vary. Possible answer: 5, 10, 15 or 5, 10, 20. 5  1 = 5 0 5 2 5   5  2 = 10 1 5 2 10   5  3 = 15 2 5 2 20   73. a. The result is a three- or four- digit number in which the thousands and hundreds places represent the month of the birthday and the tens and ones places represent the day of the birthday. b. 5[4(5 6) 9] 165 5[20 24 9] 165 5[20 33] 165 100 165 165 100 M D M D M D M D M D                    Chapter 1 Problem Solving and Critical Thinking 8 74. a. 6  6 = 36 66  66 = 4356 666  666 = 443,556 6666  6666 = 44,435,556 b. An additional digit of 6 is attached to the numbers being multiplied. An additional digit of 4 is attached to the left of the result and an additional digit of 5 is placed between the 3 and the 6. c. 66666  66666 = 4,444,355,556 666,666  666,666 = 444,443,555,556 d. Inductive reasoning; it uses an observed pattern and draws conclusions from that pattern. 75. a. 3367  3 = 10101 3367  6 = 20202 3367  9 = 30303 3367  12 = 40404 b. The first multiplier is always 3367. The second multipliers are successive multiples of 3. The product increases by 10101. c. 3367  15 = 50505 3367  18 = 60606 d. Inductive reasoning; it uses an observed pattern and draws conclusions from that pattern. Check Points 1.2 1. a. The digit to the right of the billions digit is less than 5. Thus, replace all the digits to the right with zeroes. 7,476,242,056 rounded to the nearest billion is 7,000,000,000. b. The digit to the right of the ten millions digit is 5 or greater. Thus, add 1 to the ten millions digit and replace all the digits to the right with zeroes. 7,48,242,056 rounded to the nearest hundred millions is 7,480,000,000. 2. a. The digit to the right of the tenths digit is less than 5. Thus, 3.141593 rounded to the nearest tenth is 3.1. b. The digit to the right of the ten-thousandths digit is greater than 5. Thus, 3.141593 rounded to the nearest ten-thousandth is 3.1416. 3. a. $3.40 $2.25 $5.60 $5.40 $3.40 $2.85 3.95 $3 $2 $6 $5 $3 $3 4 $26               b. The given bill is not reasonable. It is too high. 4. a. Round $52 per hour to $50 per hour and assume 40 hours per week. 40 hours $50 $2000 week hour week   The architect’s salary is $2000  per week. b. Round 52 weeks per year to 50 weeks per year. $2000 50 weeks $100,000 week year year   The architect’s salary is $100, 000  per year. 5. a. 0.48 2148.72  b. 0.5 2100 1050   Your family spent approximately $1050 on heating and cooling last year. 6. a. The yearly increase in life expectancy can be approximated by dividing the change in life expectancy by the change in time from 1950 to 2020. 81.9 71.1 10.8 0.15 yr 2020 1950 70     for each subsequent birth year. b.   number of years life expectancy yearly from 1950 to 2050 in 1950 increase 71.1 0.15 ( 2050 1950 ) 71.1 0.15(100) 71.1 15 86.1 yr         7. a. about 22% b. The greatest rate of increase in the percentage of college students who smoked cigarettes can be found by identifying the portion of the graph with the largest upward slope. This occurs between 1994 and 1998. c. Approximately 24% of college students smoked cigarettes in 1982 and 1994. d. 2014 was the year in which the least percentage of college students smoked cigarettes. The percentage that year was 13%. Section 1.2 Estimation, Graphs, and Mathematical Models 9 8. a. The yearly increase in tuition and fees can be approximated by dividing the change in tuition and fees by the change in time from 2000 to 2016. $33, 480 $15,518 $17, 962 $1123 2016 2000 16     b.  Cost in yearly 2000 increase 15,518 1123 T x     c. 2020 is 20 years after 2000. Thus, 15,518 1123 15,518 1123(20) $37,978 T x      Concept and Vocabulary Check 1.2 1. estimation 2. circle graph 3. mathematical model 4. true 5. true 6. false Exercise Set 1.2 1. a. 39,144,200 b. 39,145,000 c. 39,140,000 d. 39,100,000 e. 39,000,000 f. 40,000,000 2. a-f. Answers will vary depending upon state chosen. 3. 2.718 4. 2.7183 5. 2.71828 6. 2.718282 7. 2.718281828 8. 2.7182818285 9. 350 600 950   Actual answer of 955 compares reasonably well 10. 250 800 1050   Actual answer of 1045 compares reasonably well 11. 9 1 19 29    Actual answer of 29.23 compares quite well 12. 8 3 24 35    Actual answer of 35.34 compares quite well 13. 32 11 21   Actual answer of 20.911 compares quite well 14. 46 15 31   Actual answer of 30.893 compares quite well 15. 40 6 240   Actual answer of 218.185 compares not so well 16. 80 7 560   Actual answer of 512.98 compares not so well 17. 0.8 400 320   Actual answer of 327.06 compares reasonably well 18. 0.7 200 140   Actual answer of 141.37 compares quite well 19. 48 3 16   Actual answer of 16.49 compares quite well 20. 55 5 11   Actual answer of 11.62 compares quite well 21. 30% of 200,000 is 60,000 Actual answer of 59,920.96 compares quite well 22. 40% of 300,000 is 120,000 Actual answer of 122,432.52 compares reasonably well 23. $3.47 $5.89 $19.98 $2.03 $11.85 $0.23      $3 $6 $20 $2 $12 $0 $43        24. $4.23 $7.79 $28.97 $4.06 $13.03 $0.74      $4 $8 $29 $4 $13 $1 $59        Chapter 1 Problem Solving and Critical Thinking 10 25. Round $19.50 to $20 per hour. 40 hours per week (40  $20) per week = $800/week Round 52 weeks to 50 weeks per year. 50 weeks per year (50  $800) per year = $40,000 $19.50 per hour $40,000 per year  26. Round $29.85 to $30 per hour. 40 hours per week (40  $30) per week = $1200/week Round 52 weeks to 50 weeks per year. 50 weeks per year (50  $1200) per year = $60,000 $29.85 per hour $60,000 per year  27. Round the $605 monthly payment to $600. 3 years is 36 months. Round the 36 months to 40 months. $600  40 months = $24,000 total cost. $605 monthly payment for 3 years  $24,000 total cost. 28. Round the $415 monthly payment to $400. 4 years is 48 months. Round the 48 months to 50 months. $400  50 months = $20,000 total cost. $415 monthly payment for 3 years  $20,000 total cost. 29. Round the raise of $310,000 to $300,000. Round the 294 professors to 300. $300,000 ÷ 300 professors = $1000 per professor. $310,000 raise  $1000 per professor. 30. Round the raise of $310,000 to $300,000. Round the 196 professors to 200. $300,000 ÷ 200 professors = $1500 per professor. $310,000 raise  $1500 per professor. 31. Round $61,500 to $60,000 per year. Round 52 weeks per year to 50 weeks per year. 50 weeks  40 hours per week = 2000 hours $60,000 ÷ 2000 hours = $30 per hour $61,500 per year  $30 per hour 32. Round $38,950 to $40,000 per year. Round 52 weeks per year to 50 weeks per year. 50 weeks  40 hours per week = 2000 hours $40,000 ÷ 2000 hours = $20 per hour $38,950 per year  $20 per hour 33. 80 365 24 700,800 hr    34. 40 365 24 350, 400 hr    35. 0.2 100 20 40 0.5 0.5    Actual answer of 42.03 compares quite reasonable. 36. 0.5 90 45 180 0.25 0.25    Actual answer of 169.62 compares somewhat reasonable. 37. The given information suggests $30 would be a good estimate per calculator. $30 10 $300   which is closest to choice b. 38. The given information suggests $7 would be a good estimate per calculator. $7 10 $70   which is closest to choice c. 39. The given information suggests 65 mph would be a good rate estimate and 3.5 would be a good time estimate. 65 3.5 227.5   which is closest to choice c. 40. The given information suggests 45 mph would be a good rate estimate and 3.5 would be a good time estimate. 45 3.5 157.5   which is closest to choice b. 41. The given information suggests you can count 1 number per second. 10000 2.77 or 3 hours 60 60   42. The given information suggests you can count 1 number per second. 1,000, 000 11.57 or 12 days 60 60 24    43. 0.10 16,000,000 1,600, 000    10% of 16,000,000 is 1,600,000 high school teenagers. 44. 0.20 16,000,000 3, 200, 000    20% of 16,000,000 is 3,200,000 high school teenagers. 45. a. about 85 people per 100 b. (85 23) 87 5400    46. a. about 66 people per 100 b. (66 27) 72 2800    Section 1.3 Problem Solving 11 47. a. 26.3 9.7 16.6 0.5 2012 1980 32     The annual increase is about 0.5%. b. 9.7 0.5(2020 1980) 9.7 0.5(40) 29.7      In 2020 the percentage will be approximately 29.7%. 48. a. 21.7 6.7 15.9 0.5 2012 1980 32     The annual increase is about 0.5%. b. 6.7 0.5(2020 1980) 6.7 0.5(40) 26.7      In 2020 the percentage will be approximately 26.7%. 49. a. The percent body fat in 45-year-old women is about 37%. b. The percent body fat in women reaches a maximum at age 55 of about 38% c. Women have 34% body fat at age 25. 50. a. The percent body fat in 25-year-old men is about 23%. b. The percent body fat in men reaches a maximum at age 65 of about 26% c. Men have 24% body fat at age 35. 51. a. 401 310 91 1.4 ppm per year 2015 1950 65     b. 310 1.4 C x   c. 2050 is 100 years after 1950. 310 1.4(100) 450 ppm C    52. a. 58.44 56.98 1.46 0.02 2015 1950 65 F      b. 56.98 0.02 T x   c. 2050 is 100 years after 1950. 56.98 0.02(100) 58.98 T F     66. makes sense 67. does not make sense; Explanations will vary. Sample explanation: Very large numbers and very small numbers often must be estimated when using a calculator. 68. makes sense 69. does not make sense; Explanations will vary. Sample explanation: Some mathematical models can break down over time. 70. Since there are infinitely many digits, the digits can not be reversed. 71. a 72. d 73. b 74. c 75. 20 16 50 16, 000 hours    . 16, 000 667 days 24  667 1.8 yr 365  76. Round days in a year to 400. $1, 000, 000,000 1, 000,000 days $1000 / day 1,000,000 days 400 days/year 2500 years    Check Points 1.3 1. The amount of money given to the cashier is unknown. 2. Step 1: Understand the problem. Bottles: 128 ounces costs $5.39 Boxes: a 9-pack of 6.75 ounce boxes costs $3.15 We must determine whether bottles or boxes are the better value. Step 2: Devise a plan. Dividing the cost by the number of ounces will give us the cost per ounce. We will need to multiply 9 by 6.75 to determine the total number of ounces the boxes contain. The lower cost per ounce is the best value. Step 3: Carry out the plan and solve the problem. Unit price for the bottles: $5.39 $0.042 per ounce 128 ounces  Unit price for the boxes: $3.15 $3.15 $0.052 per ounce 9 6.75 ounces 60.75 ounces    Bottles have a lower price per ounce and are the better value. Step 4: Look back and check the answer. This answer satisfies the conditions of the problem. Chapter 1 Problem Solving and Critical Thinking 12 3. Step 1: Understand the problem. We are given the cost of the computer, the amount of cash paid up front, and the amount paid each month. We must determine the number of months it will take to finish paying for the computer. Step 2: Devise a plan. Subtract the amount paid in cash from the cost of the computer. This results in the amount still to be paid. Because the monthly payments are $45, divide the amount still to be paid by 45. This will give the number of months required to pay for the computer. Step 3: Carry out the plan and solve the problem. The balance is $980 $350 $630.   Now divide the $630 balance by $45, the monthly payment. $45 month 630 months $630 $630 14 months. month $45 45      Step 4: Look back and check the answer. This answer satisfies the conditions of the problem. 14 monthly payments at $45 each gives 14 $45 $630.   Adding in the up front cash payment of $350 gives us $630 $350 $980.   $980 is the cost of the computer. 4. Step 1: Understand the problem. The total change must always be 30 cents. One possible coin combination is six nickels. Another is three dimes. We need to count all such combinations. Step 2: Devise a plan. Make a list of all possible coin combinations. Begin with the coins of larger value and work toward the coins of smaller value. Step 3: Carry out the plan and solve the problem. Quarters Dimes Nickels 1 0 1 0 3 0 0 2 2 0 1 4 0 0 6 There are 5 combinations. Step 4: Look back and check the answer. Check to see that no combinations are omitted, and that those given total 30 cents. Also double-check the count. 5. Step 1: Understand the problem. We must determine the number of jeans/T-shirt combinations that we can make. For example, one such combination would be to wear the blue jeans with the beige shirt. Step 2: Devise a plan. Each pair of jeans could be matched with any of the three shirts. We will make a tree diagram to show all combinations. Step 3: Carry out the plan and solve the problem. There are 6 different outfits possible. Step 4: Look back and check the answer. Check to see that no combinations are omitted, and double-check the count. 6. Step 1: Understand the problem. There are many possible ways to visit each city once and then return home. We must find a route that costs less than $1460. Step 2: Devise a plan. From city A fly to the city with the cheapest available flight. Repeat this until all cities have been visited and then fly home. If this cost is above $1460 then use trial and error to find other alternative routes. Step 3: Carry out the plan and solve the problem. A to D costs $185, D to E costs $302, E to C costs $165, C to B costs $305, B back to A costs $500 $185 + $302 + $165 + $305 + $500 = $1457 The route A , D , E , C , B , A costs less than $1460 Step 4: Look back and check the answer. This answer satisfies the conditions of the problem. Trick Questions 1.3 1. The farmer has 12 sheep left since all but 12 sheep died. 2. All 12 months have [at least] 28 days. 3. The doctor and brother are brother and sister. 4. You should light the match first. Concept and Vocabulary Check 1.3 1. understand 2. devise a plan 3. false 4. false