(Solution Manual) A Problem Solving Approach to Mathematics for Elementary School Teachers

Billstein, Libeskind, Lott, Boschmans, 13

Math & Statistics

Solution Manual

13th Edition

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Author(s): Billstein, Libeskind, Lott, Boschmans

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SOLUTIONS MANUAL (All Chapters) For A PROBLEM SOLVI NG APPROACH TO MATHEMATICS FOR ELEMENTAR Y SCHOOL TEACHERS THIRTEENTH EDITION Rick Billstein Barbara Boschmans Shlomo Libeskind Johnny W. Lott iii Contents Chapter 1 An Introduction to Problem Solving 1 Chapter 2 Introduction to Logic and Sets 21 Chapter 3 Num eration Systems and Whole Number Operations 45 Chapter 4 Number Theory 87 Chapter 5 Integers 105 Chapter 6 Rational Numbers and Proportional Reasoning 127 Chapter 7 Decimals, Percen ts, and Real Numbers 155 Chapter 8 Algebraic Thinking 189 Chapter 9 Probability 217 Chapter 10 Data Analysis/St atistics: An Introduction 245 Chapter 11 Introductory Geometry 273 Chapter 12 Congruence and Similarity with Constructions 297 Chapter 13 Area, Pythagorean Theorem, and Volume 323 Chapter 14 Transformations 361 Copyright © 2020 Pearson Education, Inc. 1 CHAPTER 1 AN INTRODUCTION TO PROBLEM SOLVING Assessment 1-1A: Mathematics and Problem So lving 1. ( a ) List the numb ers: 12 9 8 9 9 99 98 2 1 100 100 100 100 ++ ++++ ++++ ++ There are 99 sums of 100. Thus the total can be found by computing 99 100 2.⋅=4950 (Another way of looking at this problem is to realize there are 99 249.5= pairs of sums, each of 100; thus 49.5  100 = 4950.) (b )The number of terms in any seque nce of numbers may be found by subtracting the first term from the last, dividing the res ult by the common difference between terms , and then adding 1 (because both ends mu st be acc ounted for). Thus 1001 1 215 0 1-+= terms. List the numbers: 1 3 999 1001 1001 999 3 1 1002 1002 1002 1002  ++ ++ ++ ++ ++ ++ There are 501 sums of 1002. Thus the total can be found by computing 5011002 2.⋅=251,001 (c)The number of terms in any sequence o f numbers may be found by subtracting the f irst term from the last, dividing the result by the common difference between terms, and th en adding 1 (because both ends must be accounted for). Thus 300 3 311 0 0-+= terms. List the numbers: 3 6 297 300 300 297 6 3 303 303 303 303  ++ ++ ++ ++ ++ ++ There are 100 sums of 303. Thus the total can be found by computing 100 303 2. 15,150⋅= (d )The number of terms in any sequence ofnumbers may be found by subtracting the first term from the last, dividing the res ult by the common difference between terms , and then adding 1 (because both ends mu st be acc ounted for). Thus 400 4 41 100-+= terms. List the numbers: 4 8 396 400 400 396 8 4 404 404 404 404  ++ ++ ++ ++ ++ ++ There are 100 sums of 404. Thus the total can be found by computing 100 404 2. 20, 200⋅= 2. ( a ) (b ) 2 Chapter 1: An Introduction to Problem Solving Copyright © 2020 Pearson Education, Inc. When the stack in (a) and a stack of the same size is placed differently next to the original stack in (a), a rectangle containing 100 (101) blocks is created. Since each block is represented twice, the desired sum is 100(101) 2 . =5050 While the above represents a specific example, the same thi nking can be used for any natural number n to arrive at a formula (1 ) 2 .nn 3. There are 147 36 111 1 2-+= terms. List the numbers: 36 37 146 147 147 146 37 36 183 183 183 183  ++ ++ ++ ++ ++ ++ There are 112 sums of 183. Thus the total can be found by computing 112 183 2.⋅=10, 248 4 . ( a ) Make a table as follows; there are 9 rows so there are 9 different ways . 6-cookie packages 2-cookie packages single-cookie packages 1 2 0 1 1 2 1 0 4 0 5 0 0 4 2 0 3 4 0 2 6 0 1 8 0 0 10 (b) Make a table as follo ws; there are 12 rows so there are 12 different ways . 6-cookie 2-cookie single-cookie packages packages packages 20 0 13 0 12 211 410 6 06 0 05 2 04 403 602 801 1 000 1 2 5 . If each layer of boxes has 7 more than the previous layer we ca n add powers of 7: 70 = 1 (red box) 71 = 7 (blue boxes) 72 = 49 (black boxes) 73 = 343 (yellow boxes) 74 = 2401 (gold boxes) 1 + 7 + 49 + 343 + 2401 = 2801 boxes altogether. 6 . Using strategies from Po yla’s problem solving list identify subg oals (solve simpler problems) and make diagrams to solve the original problem. 1 triangle; name this the “unit” triangle. This triangle is made of 4 unit tr iangles. Counting the la rge triangle there are 5 triangles Assessment 1-1A: Mathematics and Problem Solving 3 Copyright © 2020 Pearson Education, Inc. Unit triangles 4 unit triangles 9 unit triangles 93 1 13 total triangles Unit triangles 4 unit triangles 9 unit triangles 16 unit triangles 16 7 3 1 There are 27 triangles in the original figure . 7 . Observe that E = (1 + 1) + (3 + 1) +  + (97 + 1) = O + 49. Thus, E is 49 more than O. Alternative strategy: 123456 9 79 8 98(99)49(99)2 49(50)2(1 2 3 4 49) 2 49(50)2 49(99) 49(50) 49(49).OE E OOEE + =++++++ + + == æö÷ç =+ + + + += = ÷ç÷÷çèø =+-= - = So O is 49 less than E . 8 . Bubba is last; Cory mus t be between Alababa and Dandy; Dandy is faster than Cory. Listing from fastest to slowest, the finishing order is then Dandy, Cory, Alababa, and Bubba . 9 . Make a table. $20 bills $10 b ills $5 bills 2 1 0 2 0 2 1 3 0 1 2 2 1 1 4 1 0 6 0 5 0 0 4 2 0 3 4 0 2 6 0 1 8 0 0 10 There are twelve rows so there are twelve different ways . 1 0 . The diagonal from the left, top corner to the right, bottom co rner sums to 17 22 27 66.++= The first row sums to 17 7 24 . aa++= + So 66 24 . a=-= 42 The last column sums to 72 7 3 4 . bb++ = + So 66 34 . b=-= 32 The first column sums to 17 12 29 . cc++ =+ So 66 29 . c=-= 37 The second column sums to 42 22 64 . dd++ =+ So 66 64 . d=-= 2 1 1. Debbie and Amy began reading on the same day, since 72 pages for Debbie  9 pages per day = 8 days. Thus Amy is on 6 pages per day 8 days . page4 8 ´= 1 2 . The last three digits m ust sum to 20, so the second to last digit must be 20 (7 4) 9.-+= Since the sum of the 11th , 12th, and 13th digits is also 20, the 11th digit is 20 (7 9) 4.-+= We can continue in this fashion until we find that A is 9, or we can observe the repeating pattern from back to front, 4, 9, 7, 4, 9, 7, … and discover that A is 9 . 1 3 . Choose the box labeled Oranges and Apples (Box B). Retrieve a fruit from Box B. Since Box B is mislabeled, Box B should be labeled as having the f ruit you retrieved. For example, if you retrieved an apple, then Box B should be labeled Apples. Since Box A is mislabeled, the Oranges and Apples label should be placed on Box A. These leave only one possibility for Box C; it should be labeled Oranges. If an orange was retrieved from Box B, then Box C would be labeled Oranges and Apples and Box A should be labeled Apples. 1 4 . The electrician made $1 315 for 4 days at $50 per hour. She spent $15 per day on gasoline so 4 • $15 = $60 on gasoline. The total is then $1315 + $60 = $ 1375. At $50 per hour, she worked 1375 5027.5 hours . 15. Working backward: Top  6 rungs  7 rungs + 5 rungs  3 rungs = top  11 rungs, which is located at the middle. From the middle rung travel up 11 to the top or down 11 to the bottom. Along with the starting rung, then, there are 11 + 11 + 1 = 23 rungs. 16. There are several differ ent ways to solve this. One is to use a variable. So, let a be equal to the number of apple pies that are baked. This means the number of cherry p ies that are baked is 4 Chapter 1: An Introduction to Problem Solving Copyright © 2020 Pearson Education, Inc. represented by 4 – a. So, using 9 slices for an apple pie and 7 slices for a cherry pie, we get: 97 ( 4) 3 4 92 8 73 4 26 3aa aa a a     So the number of apple pies is 3 . Therefore, the number of cherry pies is 1 . Since the number of pies is small, another solution strategy is to make a table of all possible cases: Apple Pies Cherry Pies Apple slices Cherry slices Total slices 0 4 0 28 28 1 3 9 21 30 2 2 18 14 32 3 1 27 7 34 4 0 36 0 36 From the table, we can see there are 34 slices when there are three apple pies and one cherry pie. 17. Al made $50 . Examine it step b y step. First, Al spent $100 on the CD player. At this point, he is down $100. Once he sold it for $125, he is now $25 ahead. When he later bought it back for $150, he was down $125 ; but after selling it again for $175, he is now ahead $50. 1 8 . Since the bat is $49 more than the ball, and the total spent is $50, we c an use the guess and check method to solve. The table below represents possible guesses: Cost of ba t Cost of ball su m $49.00 $0.00 $49.00 $49.25 $0.25 $49.50 $49.50 $0.50 $50.00 Another method would be to use a variable. Let the cost of the ball (in dollars) be the variable b. The cost of a bat in dollars, therefore, would be b + 49. Together the two costs must add up to 50, so: b + (b + 49) = 50 2 b + 49 = 50 2 b = 1 b = ½. In terms of money, b = $0.50, 50 cents . Since the ball co sts 50 cents, the bat must cost $49.50 , and the sum of the price of bat and ball does equal $50. Assessment 1-1B 1 . ( a ) List the numbers: 12 4 84 9 49 48 2 1 50 50 50 50  ++++ ++++ ++++ There are 49 sums of 50. Thus the total can be found by computing 49 50 2.⋅=1225 (Another way of looking at this problem is to realize there are 49 224.5= pairs of sums, each of 50; thus 24.5 50 1225.)⋅= ( b ) The number of terms in any sequence of numbers may be found by subtracting the first term from th e last, dividing the result by the common difference between terms, and then adding 1 (because both ends must be accounted for). Thus 2009 1 21-+= 1005 terms. List the numbers: 1 3 2007 2009 2009 2007 3 1 2010 2010 2010 2010 ++ ++ ++ ++++ ++ There are 1005 sums of 2010. Thus the total can be found by computing 1005 2010 2. 1,010,025⋅= ( c ) The number of terms in any sequence of numbers may be found by subtracting the first term from th e last, dividing the result by the common difference between terms, and then adding 1 (because both ends must be accounted for). Thus 600 6 61 100 terms.-+= List the numbers: 6 12 594 600 600 594 12 6 606 606 606 606 ++ ++ ++ ++++ ++ There are 100 sums of 606. Thus the total can be found by computing 606 100 2. 30, 300⋅= ( d ) The number of terms in any sequence of numbers may be found by subtracting the first term from the last (or the last from the first if the first is greater than the last), dividing the result by the common Assessment 1-1B 5 Copyright © 2020 Pearson Education, Inc. difference between terms, and then adding 1 (because both ends must be accounted for). Thus 1000 5 512 0 0 t e r m s .-+= List the numbers: 1000 995 10 5 5 10 995 1000 1005 1005 1005 1005  ++ ++ ++ ++ ++ ++ There are 200 sums of 1005. Thus the total can be found by computing 1005 200 2. 100,500⋅= 2. (a) The diagram illustrates how the numbers can be paired to form 50 sums of 101. The sum of the first 100 natural numbers is 50(101) .=5050 ( b ) A diagram similar to the one in 2a would illustrate how the numbe rs can be paired to form 100 sums of 202 . Because there are an odd number of terms, the middle term, 101, is left unpaired. So, the sum of the first 201 natural numbers is 100 • 202 + 101 = 20,301 . 3 . There are 203 58 111 4 6 t e r m s .-+= terms List the numbers: 58 59 202 203 203 202 59 58 261 261 261 261  ++ ++ ++ ++ ++ ++ There are 146 sums of 261. Thus the total can be found by computing 146 261 2. 19,053⋅= 4 . There are many answers to this problem. A systematic list is a goo d approach. Using only two numbers and add ition, 5 rows give 5 different ways. One number Eleven minus the number 1 10 2 9 3 8 4 7 5 6 We can view 6 + 5 as a different way than 5 + 6 and continue in this manner to find 10 different ways. Or, we can find 7 more ways using three numbers as follows. 1 1 9 2 1 8 3 1 7 4 1 6 5 1 5 1 2 8 5 . Since there are two different color socks in the drawer, drawing only two socks does not guarantee finding a matching pair; you could get one sock of each col or. However, once you draw a third sock, you a re guaranteed to have a matching pair, since the third sock must match one or the other of the previous two socks. 6 . There are 13 squares of one unit each; 4 squares of four units each; and on e square 9 units; for a total of 18 squares. 7 . 1357. . .9 9 5 7 ... 99 101 (5 ... 99 101) (1 3 5 ... 99) (101) (1 3) 97P Q QP=++++ + =+ + + + -=++ + - ++++ =- += Q is larger than P by 97 . 8 . A diagram will help. The next step is 120 miles + 40 miles = 160 miles from Missoula. 9 . ( a ) Marc must have five pennies to make an even $1.00. The mini mum number of coins would have as many quarters as possible, or three quarters. The remaining 20c/ must consist of at least on e dime and one nickel; the only possibility is one dime and two nickels. The minimum is 5 pennies, 2 nickels, 1 dime, and 3 quarters, or 11 coins . ( b ) The maximum number of coins is achieved by having as many pennies as possible. It is a requirement to have one quarter, one dime, and one nickel 40 ,c=/ so there may then be 60 pennies for a total of 63 coins . 6 Chapter 1: An Introduction to Problem Solving Copyright © 2020 Pearson Education, Inc. 10. Adding all the numbers gives 99. This means that each row, diagonal, and column must add to 99  3 = 33. Write 33 as a sum of the numbers in all possible ways: 19 11 3 19 9 517 13 317 11 51 79715 13 515 11 713 11 9++++ ++++++ ++++ ++ Summarizing the pattern: Number Nr. sums with number 3 2 5 3 7 2 9 3 11 4 13 3 15 2 17 3 19 2 Thus 11 must be in the center of the square and 5, 9, 13, and 17 must be in the corners. One solution would be: 17 7 9 3 11 19 13 15 5 1 1 . Answers may vary; two s olutions might be to: ( a ) Put four marbles on each tray of the balance scale. Take th e heavier four and weigh two on each tra y. Take the heavier two and weigh one on each tray; the heavier marble will be e vident on this third weighing. ( b ) This alternative show s the heavier marble can be found m ore efficiently, two steps rather than three. Put three marbles in each tray of the balance scale. ( i) If the two trays are the same weight, the heavier marble is one of the remaining two. Weigh them to find the heavier. (ii) If one side is heavie r, take two of the three marbles and we igh them. If they are the same weight, the remaining marble is the heavier. If not, the heavier will be evide nt on this second weighing. 12. (a) There are: 1 partridge 12 days 12 gifts; 2 doves 11 days 22 gifts; 3h e n s 1 0d a y s 3 0g i f t s ; 4 birds 9 days 36 gifts; 5r i n g s 8d a y s 4 0g i f t s ; 6 geese 7 days 42 gifts; 7s w a n s 6d a y s 4 2g i f t s ; 8 maids 5 days 40 gifts;´= ´= ´= ´= ´= ´= ´= ´= 9 ladies 4 days 36 gifts; 10 lords 3 days 30 gifts; 11 pipers 2 days 22 gifts; and 12 drummers 1 day 12 gifts.´= ´= ´= ´= S o t h e gifts given the most by your true love was 42 geese and 42 swans . ( b ) 12 22 22 12++++= 364  gifts total . 13. (a) There must be 1 or 3 quarters for an amount ending in 5. Then dimes can add to $1.15 plus 4 pennies to realize $1.19. Thus: Quarters Dimes Pennies Total 3 4 4 $1.19 1 9 4 $1.19 and in neither case can change for $1.00 be made. ( b ) Two or zero quarter s would allow an amount ending in 0. Then more combinations of dimes or pennies could add to $1.00. 14. If the price of 15 sandwiches equals the price of 20 salads, each sandwich will buy 20 4 15 3= salads. Thus 3 sandwiches ()4 33.== 4s a l a d s 15. Use a variable and a table 12 AM 5 AM 9 AM 12 PM T T 15 2(T 15) 2(T 15) 10-- - + Assessment 1-2A: Explor ations with Patterns 7 Copyright © 2020 Pearson Education, Inc. So, 2(T 15) 10 32 2T 30 10 32 2T 20 32 2T 52 T. 26 de grees-+= -+= -= == 16. One way to solve this problem is to write an equation using a variab le. For example, let p equal the number of puzzles Seth bought. So the number of trucks Seth bought would be 5 – p. Since each puzzle cost $9 and each truck cost $5, we get 95 ( 5) 3 3 92 5 53 3 48 2pppp p p   So Seth bought 2 puzzles . Since he bought 5 gifts all together, he also bought 3 trucks. Another way to solve it is to guess and check, making a table to keep track of results Puzzles Trucks Cost of puzzles Cost of Trucks Total cost 0 5 $0 $25 $25 1 4 $9 $20 $29 2 3 $18 $15 $33 3 2 $27 $10 $37 The table shows that by buying 2 puzzles and 3 trucks , Seth will spend $33. 17. The first line tells us that hamburgers equal $10. Using that information, we can use the second line to figure out 4 hot dogs cost $8, so each hot dog costs $2. Going to the third line, since two hot dogs cost $4, each drink must cost $1. So on the fourth line, one drink and one hamburger cost $11. The value of the question mark is $11. 18. This is difficult to visua lize, so the strategy of examining a simpler cas e and looking for a pattern might be the best method to solve this problem. It would also help if you are able to make a model of the shapes and act out the situation. Imagine a 333´´ large cube. It would be made up of a total of 27 small cubes. Taking off one layer all around would involve taking off the top layer, the botto m layer, the front, the back, the left side, and the right side. Doing this one step at a time: taking off the top layer takes away 9 cubes. The same if you take off the bottom layer, so you’ve removed 18 cubes so far. Taking off the fron t and the ba ck would remove 3 more cubes each (the other 6 cubes on those faces were removed when the top and bottom were removed). Finally, taking off the left and right sides removes one more cube each. In total, 26 cubes are removed, leaving only 1 cube. Now, imagine a 444´´ cube. There would be a total of 64 small cubes making it. Removing the top and bottom layers takes a total of 32 cubes away; removing the front and the back would take away 16 more cubes; finally, removing the left and right sides takes away 8 more cubes, leaving only 8 cubes. So, when starting with a 333´´ large cube, we are left with 1 small cube (1 1 1 1);´´= when starting with a 444´´ large cube, we are left with 8 small cubes ( 222 8 ) .´´= Therefore, the pattern seems to be, when given large cube of dimension , nnn´´ to find the number of small cubes that make it after removing one layer of small cubes all around the larger cube, you would take (2 ) (2 ) (2 ) .nnn-´-´- So for a 10 10 10´´ large cube, the number of small cubes left would be 888´´= 512 small cubes. Assessment 1-2A: Explorations with Patterns 1 . ( a ) Each figure in the sequence adds one box each to the top and bottom rows. The next would be: ( b ) Each figure in the sequence adds one upright and one inverted triangle. The next would be: ( c ) Each figure in the sequence adds one box to the base and one row to the overall triangle. The next would be: 8 Chapter 1: An Introduction to Problem Solving Copyright © 2020 Pearson Education, Inc. 2 . ( a ) Terms that continue a pattern are 17, 21, 25, , … . This is an arithmetic sequence because each successive term is obtained from the previous term by addition of 4. (b) Terms that continue a pattern are 220, 270, 320, … . This is arithmetic because each successive term is obtained from the previous term by addition of 50. ( c ) Terms that continue a pattern are 27, 81, 243, … . This is geometric because each successive term is obtained from the previous term by m ultiplying by 3. ( d ) Terms that continue a pattern are 91 11 310 ,10 ,10 , … . This is geometric because each successive term is obtained from the previ ous term by mu ltiplying by 102. ( e ) Terms that continue a pattern are ,30 30 30193 + 10× 2 ,193 + 11× 2 ,193 + 12× 2 … . This is arithmetic because each successive term is obtained from the previous term by addition of 302. 3 . In these problems, let na represent the nth term in a sequence, 1a represent the first term, d represent the comm on difference between terms in an arithmetic sequence, and r represent the common ratio between terms in a geometric sequence. In an arithmetic sequence, 1(1 ) ;naand=+- in a geometric sequence 1 1 .n naa r-= Thus: ( a ) Arithmetic sequence: 1 1 a= and 4: d= ( i ) 100 1( 1 0 01 ) 4 19 9 4 .a 397=+ -⋅=+ ⋅= ( ii ) 1( 1 ) 4 14 4 .nan n 43=+ -⋅=+ -= - n (b) Arithmetic sequence: 1 70 and 50: ad== ( i ) 100 70 (100 1) 50 70 99 50 .a 5020=+ - ⋅ =+⋅= ( ii ) 70 ( 1) 5070 50 50or . nan n 50n + 20=+ - ⋅=+ - ( c ) Geometric sequence: 1 1a n d 3 : ar== ( i ) 100 1 100 13 . a993-=⋅ = (ii ) 113 .n nan-13-=⋅ = ( d ) Geometric sequence: 2 1 10 and 10 : ar== ( i ) 2( 1 0 0 1 ) 29 9 100 19810 (10 ) 10 (10 ) 10 10 .a 19910-=⋅ =⋅ =⋅ = ( ii ) 2( 1 ) (2 2)10 (10 ) 10 10 .n n na 2n-110- -=⋅=⋅ = ( e ) Arithmetic sequence: 30 30 1 193 7 2 and 2 : ad=+ ⋅ = ( i ) 30 30 100 30 30193 7 2 (100 1) 2 193 7 2 99 2 .a 30193 + 106 × 2=+ ⋅+ - ⋅=+ ⋅+ ⋅ = ( ii ) 30 30193 7 2 ( 1) 2 .nan 30193 + (n + 6) × 2=+ ⋅+ - ⋅ = 4 . 2, 7, 12 , … . Each term is the 5th number on a clock face (clockwise) from, the preceding term. 5 . ( a ) Make a table. Number of term Term 1 111 1⋅⋅= 2 222 8⋅⋅= 3 333 2 7⋅⋅= 4 444 6 4⋅⋅= 5 555 1 2 5⋅⋅= 6 666 2 1 6⋅⋅= 7 777 3 4 3⋅⋅= 8 888 5 1 2⋅⋅= 9 999 7 2 9⋅⋅= 10 10 10 10 1000⋅⋅= 11 11 11 11 1331⋅⋅= T h e 1 1th term 1331 is the least 4-digit number greater than 1000. (b) The 9th term 729 is the greatest 3-digit number in this pattern. ( c ) 1 04 = 10,000; The great est number less than 104 is 21 21 21 . 9261 (d) The cell A14 corresponds to the 14th term, which is 14 14 14 . 2744 ⋅⋅= 6 . ( a ) The number of matchstick squares in each windmill form an arithmetic sequence with 1 5 a= and 4. d= The number of matchstick squares required to build the Assessment 1-2A: Explorations with Patterns 9 Copyright © 2020 Pearson Education, Inc. 10th windmill is thus 5( 1 0 1 ) 4+- ⋅ = 59 4 . 41 squares +⋅= (b) The nth windmill would require 5( 1 ) 4 54 4 nn+- ⋅ = +- = 41n+ squares . ( c ) There are 16 matchsticks in the original windmill. Each additional windmill adds 12 matchsticks. This is an arithmetic sequence with 1 16 a= and 12, d= so 16na=+ (1 ) 1 2n-⋅ = 12 4n+ matchsticks . 7 . ( a ) Each cube adds four squares to the preceding figure; or 6, 10, 14, … . This is an arithmetic sequence with 1 6 a= and 4. d= Thus 15 6( 1 5 1 ) 4 a=+ -⋅= 62 squares to be painted in the 10th figure. ( b ) This is an arithmetic sequence with 1 6 a= and 6. d= The nth term is thus: 6( 1 ) 4 .nan=+ -⋅= 42n+ 8 . Since the first year beg ins with 700 students, after the first year there would be 760, after the second there would be 820, … , and after the twelfth year the number of students would be the 13 th term in the sequence. This then is an arithmetic sequence with 1 700 a= and 60, d= so the 13th term (current enrollment twelve more years) + is: 700 (13 1) 60+- ⋅= 1420 students. 9 . Using the general expression for the nth term of an arithmetic sequence with 1 24,000 a= and 9 31,680 yields: a= 31680 24000 (9 1) 31680 24000 8 960,d dd=+ -=+  = the amount by which Juan’s income increased each year. To find the year in which his income was $45120: 45120 24000 ( 1) 960 45120 23040 960 23.n n n=+ - ⋅=+ = Juan’s income was $45,120 in his 23rd year . 1 0 . The number that fits int o the last triangle is 8. The numbers inside the triangle are found by multiplying the number at the top of the triangle by the number at the bottom left of the triangle; then subtracting from that the number at the bottom right of the triangle. So, 25 2 . 8 ´-= 11. (a) To build an up-down-up staircase with 3 steps up and 3 steps down, each current step will have a block placed on it; plus a block will be added to each end. In total, there will be five more blocks, for a total of 9 blocks used. To build an up-down-up staircase with 4 steps up and 4 steps down, each step from the previous iteration will have a block placed on it (5 blocks) plus a block at each end (2 blocks), adding seven total blocks, bringing the total to 9 + 7 = 16 blocks. The table below illustrates the pattern: Steps up/down Blocks added Total blocks 1 0 1 2 3 4 3 5 9 4 7 16 Therefore, to build an up-down-up staircase with 5 steps up and 5 steps down, 9 blocks need to be added to the previous 16 blocks to arrive at a total of 25 blocks . (b) Based on the table and answer above, the total blocks are always the square of the number of steps up and down. So, if the number of steps up and down are n, then the total number of blocks will be n2. 12. (a) Using the general expression for the nth term of an arithmetic sequence with 1 51, 251,n aa== and 1 d= yields: 251 51 ( 1) 1 251 50 201.n nn=+- ⋅ =+  = There are 201 terms in the sequence. ( b ) Using the general expression for the nth term of a geometric sequence with 1 1, a= 602,na= and 2 r= yields: 60 1 121 ( 2 ) 2 16 0 6 1 .nn nn--== -= = There are 61 terms in the sequence. ( c ) Using the general expression for the nth term of an arithmetic sequence with 1 10, a= 2000,na= and 10 d= yields: 2000 10 ( 1) 102000 10 200. n nn=+- ⋅= = There are 200 terms in the sequence. 10 Chapter 1: An Introduction to Problem Solving Copyright © 2020 Pearson Education, Inc. ( d ) Using the general expression for the nth term of a geometric sequence with 1 1, a= 1024,na= and 2 r= yields: 1 10 11024 1(2) 22 1 1 0 11.n nn n- -= = - = = There are 11 terms in the sequence. 13. (a) 2 2 2 2 2First term: (1) 2 ; Second term: (2) 2 ; Third term: (3) 2 ; Fourth term: (4) 2 ; and Fifth term: (5) 2 .3 6 11 18 27+= += += += += (b) First term: 5(1) 1 ; Second term: 5(2) 1 ; Third term: 5(3) 1 ; Fourth term: 5(4) 1 ; and Fifth term: 5(5) 1 .6 11 16 21 26+= += += += += (c) (1) (2) (3) (4) (5)First term: 10 1 ; Second term: 10 1 ; Third term: 10 1 ; Fourth term: 10 1 ; and Fifth term: 10 1 .9 99 999 9999 99999-= -= -= -= -= (d) First term: 3(1) 2 Second term: 3(2) 2 Third term: 3(3) 2 Fourth term: 3(4) 2 andFifth term: 3(5) 21; 4; 7; 10; 13.-= -= -= -= -= 14. Answers may vary; examples are: (a) If 5, n= then 55 525 16 .+=¹+ = (b) If 2,n= then 22(2 4) 6 36+= = does not equal 222 4 20.+= 15. (a) There are 1, 5, 11, 19, 29 tiles in the five figures. Each figure a dds 2n tiles to the preceding figure, thus a6 the 6th term has 29 12 .+= 41 tiles ( b ) 21, 4, 9, 16, 25, n= … . Adding (1 )n- to n2 yields 1, 5, 11, 19, 29, … , which is the proper sequence. Thus the n th term has .2+1- nn (c) If 2(1 ) 1 2 5 9 ; nn+-= Then 21260 0. nn+- = This implies ( 35)( 36) 0, so 35.nn n-+ = = There are 1259 tiles in the 35th figure . 16. The nth term of the arithmetic sequence is 200 (200). n+ The sequence can also be generated by adding 200 to the previous term. The nth term of the geometric sequence is 2n. The sequence can als o be generated by multiplying the previous term by 2. Make a table. Number of the term Arithmetic term Geometric term 7 1600 128 8 1800 256 9 2000 512 10 2200 1024 11 2400 2048 12 2600 4096 W i t h t h e 12th term , the geometric sequence is greater. 17. (a) Start with one piece of paper. Cutting it into five pieces gives us 5. Taking each of the pieces and cutting it into five pieces again gives 55 2 5⋅= pieces. Continuing this process gives a ge ometric sequence: 1, 5, 25, 125, … . After the 5th cut there are 55 = 3125 pieces of paper. ( b ) The number of pieces after the nth cut would be .n5 18. (a) For an arithmetic sequence there is a common difference between the terms. Between 39 and 69 there are three differences so we can find the common difference by subtracting 39 from 69 and dividing the answer by three: 69 39 30and 30 3 10.   The common difference is 10 and we can find the missing terms: 39 – 10 = 29 and 39 + 10 = 49 and 49 + 10 = 59. (b) For an arithmetic sequence there is a common difference between the terms. Between 200 and 800 there are three differences so we can find the common difference by subtracting 200 from 800 and dividing the answer by three: 800 200 600and 600 3 200.   The Assessment 1-2B 11 Copyright © 2020 Pearson Education, Inc. common difference is 2 00 and we can find the missing terms: 200 – 200 = 0 and 200 + 200 = 400 and 400 + 200 = 600. (c) For a geometric sequence there is a common ration between the terms. Between 54 and 510 there are three common ratios used so we can find the common ratio by dividing 510 by 54 and then taking the cube root: 1 3 10 4 6 6 255 5 a n d ( 5 ) 5 .  The common ratio is 52 and we can find the missing terms: 42 4 2 6 255 , 5 5 , 5 5 .  268555 19. (a) Let’s call the missing terms a, b, c, d, e and f, then the sequence becomes: , ,1,1, , , , 11 0 10 11 11 2 11 2 3 23 5 35 8 .ab cde f bb ab a a cc cd d d cd e e e de f f        The missing terms are 1, 0, 2, 3, 5 , and 8. (b) Let’s call the missing terms a, b, c, and d, then the sequence becomes: ,,, 1 0 , 1 3 , , 3 6 , 5 9 10 13 3 10 3 10 7 73 4 10 13 23abc d cc bc b b ab c a dd        The missing terms are -4, 7, 3 , and 23. (c) If a Fibonacci-type sequence is a sequence in which the first two terms are arbitrary and in which every ter m starting from the third is the sum of th e previous two terms, then we can add 0 and 2 to get the third term and continue the pattern: 02 2 22 4 246461 0 61 0 1 6 10 16 26    The missing terms are 2, 4, 6, 10, 16 , and 26. 2 0 . (a) Year 1 80 .05(80) 84Year 2 84 .05(84) 88.2Year 3 88.2 .05(88.2) 92.61Year 4 92.61 .05(92.61) 97.2405Year 5 97.2405 .05(97.2405) 102.102525 .     $102.10 (b) This is a geome tric sequence with 180a and r = 1.05, so the price after n years is 80 • 1.05n. Assessment 1-2B 1 . ( a ) In a clockwise direction, the shaded area moves to a new position separated from the original by one open space, then two open spaces, then by three, e tc. The separation in each successive step increases by one unit; next would be: ( b ) Each figure in the sequence adds one row of boxes to the base. Next would be: ( c ) Each figure in the sequence adds one box to the top and each leg of the figure. Next would be: 2 . ( a ) Terms that continue a pattern are 18, 22, 26, … . This is an arithmetic sequence because each successive term is obtained from the previous term by addition of 4. ( b ) Terms that continue a pattern are 39, 52, 65, … . This is an arithmetic sequence because each successive term is obtained from the previous term by addition of 13. ( c ) Terms that continue a pattern are 44, 45, 46, … . This is a geometric sequence because each successive term is obtained from the previ ous term by mu ltiplying by 4. 12 Chapter 1: An Introduction to Problem Solving Copyright © 2020 Pearson Education, Inc. ( d ) Terms that continue a pattern are 214, 218, 222, … . This is a geometric sequence because each successive term is obtained from the previ ous term by mu ltiplying by 24. ( e ) Terms that continue a pattern are ,....50 50 50100 + 10× 2 ,100 + 12× 2 ,100 + 14× 2 This is an arithmetic sequence because each successive term is obtained from the previous term by adding by 2•2 50. 3 . In these problems, a n represents the nth term in a sequence, a 1 represents the first term, d represent the comm on difference between terms in an arithmetic sequence, and r represents the common ratio between terms in a geometric sequence. In an arithmetic sequence, 1(1 ) ;naand=+- in a geometric sequence, 1 1 .n naa r-= Thus: (a) Arithmetic sequence:1 2 a= and 4. d= (i) 100 2( 1 0 0 1 ) 4 . a 398 =+ -⋅= (ii) 2( 1 ) 4 24 4 .nan n 42=+-⋅ =+ -= - n ( b ) Arithmetic sequence: 1 0 a= and 13. d= ( i) 100 0( 1 0 0 1 ) 1 3 . a 1287 =+ -⋅ = ( ii) 0( 1 ) 1 3 .nan 13n 13=+-⋅=- ( c ) Geometric sequence: 1 4 a= and 4. r= (i) 99 100 44 . a1004 =⋅ = ( ii) 144 .n na 4-=⋅ =n ( d ) Geometric sequence: 2 1 2 a= and 42. r= (i) 2 4 99 2 396 100 2( 2 ) 22 . a3982 =⋅ =⋅ = ( ii) 24 ( 1 ) 24 42( 2 ) 22 .n n na 4n 22- --=⋅ =⋅ = ( e ) Arithmetic sequence: 50 50 51 1 100 4 2 and 2 2 2 : ad=+ ⋅ = ⋅= ( i) 50 51 100 51 51 51 50100 4 2 (100 1) 2 100 2 2 99 2 100 101 2100 101 2 2 .a 50100 + 202 × 2=+ ⋅+- ⋅ =+ ⋅+ ⋅ =+⋅=+⋅ ⋅= ( ii) 50 51 51 51 51 50100 4 2 ( 1) 2 100 2 2 ( 1) 2 100 ( 1) 2 100 ( 1) 2 2 .nan n n n 50100 + 2(n + 1)2=+ ⋅+ - ⋅ =+ ⋅+ - ⋅ =+ + ⋅ =+ + ⋅ ⋅ = 4 . The hands must move 8 hours to move from 1 to 9 on the clock face. To move from 9 to 5, the hand must move 8 hours also. To move from 5 to 1, the hand must move another 8 hours. If we add 8 hours to 1 o’clock, we will land on the 9. This pattern will con tinue, so the next three terms are 9, 5, 1 . 5 . ( a ) Answers may vary: t wo possible answers are: (i) 2 e. ., 1 3 5 7 4 .ns g2The sum of the first odd number is ;+++= n ( ii) ()217 2 2e. ., 1357 4 .gSquare the average of the first and last terms; ++++= = (b) There are 35 1 211 8-+= terms in this sequence. (i) 21357 3 51 8 .324 ++++ + = = (ii) ()213 5 2 218 . 324+== 6 . ( a ) Note that 5 toothpick s are added to form each succeeding hexagon. This is an arithmetic sequence 1 6 a= and 5, d= Assessment 1-2B 13 Copyright © 2020 Pearson Education, Inc. so 10 6( 1 0 1 ) 5 a=+ -⋅= 69 5+⋅= 51 toothpicks . ( b ) n hexagons would require 6 ( 1) 5 6 5 5 nn+- ⋅ = +- = 51n+ toothpicks . 7 . ( a ) Looking at the second figure, there are 3 + 1 = 4 triangles. In the third figure, there are 5319++= triangles. The fourth figure would then have 753 11 6+++= triangles. An alternative to simply adding 7, 5, 3, and 1 together is to note that 718+= and 538 .+= There are 4 22= of these sums, and 28 1 6 .⋅= Then the 100th figure would have 100 99 199+= triangles in the base, 99 98 197+= triangles in the second row, and so on until the 100th row where there would be 1 triangle. 199 1 200;+= 197 3 200;+= etc. and so the sum of each pair is 200 and there are 100 250= of these pairs. 50 200 10,000,⋅= or 10,000 triangles in the 100th figure. (b) The number of triangles in the nth figure is 2(number of triangles in base 1).n+ The number of triangles in the base is (1 ) , nn+- or 21 .n- (2 1)n-+ 12 . n= Then 2 2(2 )nnn= , or n2 triangles in the nth figure. 8 . This is a geometric sequence with 115,360 2a= and 1 2r= . The nth term of a geometric sequence is 1 1 ;n naa r-= thus the 10th term would be ()101 215360 . =15 liters Note the progression of terms in the following table: After Day Amount of Water Remaining 1 1 215,360 7680 liters⋅= 2 1 27680 3840 liters⋅=   9 1 260 30 liters⋅= 10 1 230 15 liters⋅= 9 . This is an arithmetic sequence with 11 68 a= (i.e., 8 a.m. plus 10 minutes, or 10 60 of an hour) and 5 6d= (or 50 60 of an hour). Thus 5 18 668( 8 1 ) 1 4 , o r a 2:00 p.m. =+ - ⋅ = (14 is 2:00 p.m. on a 24-hour clock.) 10. Answers will be a rota tion of the following figure: 11. (a) In the first drawing, there are 6 toothpicks; in the second drawing, t here are 10, and the third drawing has 14 t oothpicks. This it is an arithmetic sequence with 1 6, 4. ad== So, for the tenth figure, we would have 10 10 106( 1 0 1 ) 4 69 4a a a 42.=+ -⋅ =+⋅ = ( b ) The nth term for this arithmetic sequence is 6( 1 ) 4 64 4n n nan an a 4n + 2.=+-⋅ =+ - = (c) For a total of 102 toothpicks, to find the figure, we have 102 4 2 100 4 .n n n 25=+ = = 12. (a) T h e nth term for this geometric sequence is 13.n- Thus 99 133 .n-= S o 99 1, n=- and 100. n= There are 100 terms in the sequence. (b) The nth term for this arithmetic sequence is 9( 1 ) 4 . n+- ⋅ Thus 353 9=+ (1 ) 4 .n-⋅ Solving for n, 87. n= There are 87 terms in the sequence. (c) The nth term for this arithmetic sequence is 38 ( 1) 1. n+- ⋅ Thus 238 38=+ (1 ) 1 .n-⋅ Solving for n, 201. n= There are 201 terms in the sequence. 14 Chapter 1: An Introduction to Problem Solving Copyright © 2020 Pearson Education, Inc. 13. (a) First term: 5(1) 1 Second term: 5(2) 1 Third term: 5(3) 1 Fourth term: 5(4) 1 Fifth term: 5(5) 14 9 141924-= -= -=-= -= ( b ) First term: 6(1) 2 Second term: 6(2) 2 Third term: 6(3) 2 Fourth term: 6(4) 2Fifth term: 6(5) 2-= -= -= -= -=4 10 16 22 28 ( c ) First term: 5 1 1 Second term: 5 2 1 Third term: 5 3 1 6 Fourth term: 5 4 1 Fifth term: 5 5 1 6 11 1 21 26⋅+= ⋅+= ⋅+= ⋅+= ⋅+= (d) 2 2 2 2 2First term: 1 1 Second term: 2 1 Third term: 3 1 Fourth term: 4 1 Fifth term: 5 1 0 3 8 15 24-= -= -= -= -= 14. Answers may vary; examples are: ( a ) If 6, n= then 36 336 .+=¹ ( b ) If 4, n= then 22 2(4 2) 4 4 2 12.-= ¹ - = 15. (a) The first figure has 2 tiles, the second has 5 tiles, the third has 8 tiles, … . This is an arithmetic sequence where the nth term is 2( 1 ) 3 . n+- ⋅ T h u s t h e 7th term has 2( 7 1 ) 3+- ⋅ = 20 tiles. ( b ) The nth term is 2( 1 ) 32 n+- ⋅ =+ 33 .n-= 31n- (c) The question can be written as: Is there an n such that 3 1 449.n-= Since 3 1 449 3 450 150,nn n-=  =  = the answer is yes, the 150th figure . 16. T h e nth term of the arithmetic sequence is 100 (300). n-+ The sequence can also be generated by adding 300 to the previous term. The nth term of the geometric sequence is 13n-. The sequence can als o be generated by multiplying the previous term by 3. Make a table Number of the term Arithmetic term Geometric term 7 2000 729 8 2300 2187 9 2600 6561 W i t h t h e 9th term , the geometric sequence is greater. 17. Use a table of Fibonacci numbers to find the pattern, nF is the nth Fib onacci number: Generation Male Female Number in Generation Total 1 1 0 1 1 2 0 1 1 2 3 1 1 2 4 4 1 2 3 7 5 2 3 5 12 6 3 5 8 20      n F n–2 Fn–1 Fn F n+2–1 The sum of the first n Fibonacci numbers is 21.nF+- 12 144, F= so there are 143 bees in all 10 generations. 18. (a) For an arithmetic sequence there is a common difference between the terms. Between 49 and 64 there are three differences so we can find the common difference by subtracting 49 from 64 and dividing the answer by three: 64 49 15and15 3 5.   The common difference is 5 and we can find the missing terms: 49 – 5 = 44 and 49 + 5 = 54 and 54 + 5 = 59. (b) For a geometric sequence there is a common ratio between the terms. Between 1 and 625 there are four common ratios used so we can find the common ratio by dividing 625 by 1 and then taking the fourth root: 1 4625 1 625and 625 5.  The common ratio is 5 and we can find the missing terms: 15 , 55 , 2 55 .  5 25 125 Mathematical Connections 1-2: Review Problems 15 Copyright © 2020 Pearson Education, Inc. (c) For a geometric sequence there is a common ratio between the terms. Between 310 and 319 there are three common ratios used so we can find the common ratio by dividing 319 by 310 and then taking the cube root: 1 3 19 10 9 9 333 3 a n d ( 3 ) 3 .  The common ratio is 33 and we can find the missing terms: 10 3 10 3 13 333 , 3 3 , 3 3 .  71 31 633 3 ( d ) For an arithmetic sequence there is a common difference between the terms. Between a and 5 a there are four differences so we can find the common difference by subtracting a from 5 a and dividing the answer by four: 54 a n d 4 4 .aa a a a  The common difference is a and we can find the missing terms: a + a = 2a, 2a + a = 3a, 3a + a = 4a. 19. (a) Let’s call the missing terms x and y, then the sequence becomes 1, x, y, 7, 11 and if it is a Fibonacci-typ e sequence then: 1 7 71 1 1 1 74 and 7 4 7 3.xy xy yy xy x x      The missing terms are 3 and 4 . (b) Let’s call the missing terms x, y. and z, then the sequence becomes x, 2, y, 4, z and if it is a Fibonacci-typ e sequence then: 2 24 2 42 4 6 and 2 2 0.xy yy yz z z xx    The missing terms are 0, 2, and 6 . (c) Let’s call the missing terms x, y. and z, then the sequence becomes x, y, 3, 4, z and if it is a Fibonacci-typ e sequence then: 3 34 4 31 34 7 and 3 1 3 2.xy yy xy x x    The missing terms are 2, 1, and 7 . Mathematical Connections 1-2: Review Problems 1 5 . Order the teams from 1 to 10, and consider a simpler problem of counting how many games are played if each team p lays each other once. The first team plays nine teams. The second team also plays nine teams, but one of these games has already been counted. The third team also plays 9 teams, but two of these games were counted in the previous two summands. Continuing in this manner, the total is 10 + 9 + 8 + … + 3 + 2 + 1 = 9(10) / 2 = 45 games. Double this amount to obtain 90 games must be played for each team to play each other twice. 16. 7 ways . Make a table: Quarters Dimes Nickels 1 1 1 1 0 3 0 4 0 0 3 2 0 2 4 0 1 6 0 0 8 1 7 . If the problem is interpr eted to stated that at least one 12-person tent is used, then there are 10 ways . This can be seen b y the table below, which illustrates the ways 2-,3-,5-, and 6-person tents can be combined accommodate 14 people. 6-Person 5-Person 3-Person 2-Person 2 0 0 1 1 1 1 0 1 0 2 1 1 0 0 4 0 2 0 2 0 1 3 0 0 1 1 3 0 0 4 1 0 0 2 4 0 0 0 7 16 Chapter 1: An Introduction to Problem Solving Copyright © 2020 Pearson Education, Inc. Chapter 1 Review 1 . Make a plan. Every 7 day s (every week) the day will change from Sunday to Sunday. 365 days per year 7¸ days per week 52» weeks per year 1 7+ weeks per year. Thus the day of the week will change from Sunday to Sunday 52 times and then change from Sunday to Monday. July 4 will be a Monday . 2 . $5.90 2 $2.95 more on one of the items. That is $20 $2.95 $22.95 for the more expensive item and $20 $2.95 $17.05 for the less expensive item. Check that both items add up to $40: $22.95 $17.05 $40. 3. (a) 15, 21, 28. Neither . The successive differences of terms in creases by on e; e.g., 10 + 5, 15 + 6, … . ( b ) 32, 27, 22. Arithmetic Subtract 5 from each term to obtain the subsequent term. (c) 400, 200, 100. Geometric Each term is half the previous term. ( d ) 21, 34, 55. Neither Each term is the sum of the previous two terms—this is the Fibonacci sequence. ( e ) 17, 20, 23. Arithmetic Add 3 to each term to obtain the subsequent term. ( f ) 256, 1024, 4096. Geometric Multiply each term by 4 to obtai n the subsequent term. ( g ) 16, 20, 24. Arithmetic Add 4 to each term to obtain the subsequent term. (h) 125, 216, 343. Neither Each term is the 3rd power of the counting numbers =13, 23, 33, … . 4 . ( a ) The successive differences are 3. Each term is 3 more than the previous term. This suggests that it is an arithmetic sequence of the form 3? .n+ Since the first term is 5, 3(1) ? 5.+= The nth term would be . 32n+ ( b ) Each term given is 3 times the previous term. This suggests that the sequence is geometric. nth term will be 3n. ( c ) The only number that changes in successive terms is th e exponent. For the first term, the exponent is 2; for the second term, the exponent is 3; for the third term, the exponent is 4; and so on. So, for the n th term, the exponent will be n + 1. Therefore, the nth term will be 2n+1 – 1. 5. (a) 3(1) 2 ; 3(2) 2 ;3(3) 2 ;3(4) 2 ; and3(5) 2 .1 4 7 10 13-= -= -= -= -= ( b ) 2 2 2 2 211 ; 22; 33 ; 44 ; a n d 55 .2 6 12 20 30+= += += += += (c) 4(1) 1 ; 4(2) 1 ; 4(3) 1 ; 4(4) 1 ; and 4(5) 1 .3 7 11 1519-=-=-=-=-= 6. (a) 1 2, 2, 200.n ad a== = S o 200 2 ( 1) 2 100. nn=+-⋅= S u m i s 100(2 200) 2. 10,100+= (b) 1 51, a= 1, 151.n da== S o 151 51 ( 1) 1 101. nn =+- ⋅  = S u m i s 101 (51 151) 2. 10, 201⋅+= 7 . (a) Answers will vary; for example 5 and 3 are odd numbers, but 5 + 3 = 8, which is not odd. (b) 15 is odd; and it does not end in a 1 or a 3. ( c ) The sum of any two even numbers is always even. An even number is one divisible by 2, so any even number can be represented by 222 .+++  Regardless of how many twos are added, the result is always a multiple of 2, or an even number. 8. All rows, columns, and diagonals must add to 34; i.e., the sum of the digits in row 1. Complete rows or columns with one number missing, then two, etc. to work through the square: 16 3 2 13 10 97 1 2 41 451 1 8 6 15 1 Chapter 1 Review 17 Copyright © 2020 Pearson Education, Inc. 9. The ten middle tables will hold two each and the two end tables will ho ld three each, totaling 26 people. 10. (a) 60 61 61 60 60 60 6022 22 2(2 ) 2 2( 2 1 )    602 ( b ) 2 2625 625  25 11. 100 5 20¸= plus 1=21 posts . 1 must be added because both end posts must be counted. 12. 1 mile 5280 feet.= 5280 6 feet 880 turns per mile.¸= 880 50000 miles .´= 44,000,000 turns 13. There are 9 students b etween 7 and 17 (8 through 16). There must be 9 between them in both directions, since they are direct opposites. 992 .++= 20 students 14. Let l be a large box, m be a medium box, and s be a small box: 3 (3 2 each) [(3 2) 5 each] 36 3 0 .ll m m s lm s+´ +´ ´ ++= 39 total boxes 15. Extend the pattern of doubling the number of ants each day. This is a geometric sequence with 1 1500, a= 100,000,na= and 2. r= 1 1 2 3100,000 1500 2 66 2 .n n- -=⋅ = Since 71 81 22 332 66 and 2 66 , the ant farm will fill sometime between the 7th and 8th day . 16. The best strategy would be one of guessing and checking: (i) Ten 3’s + two 5’s 40= close but too low. (ii) Nine 3’s + three 5’s 42= still too low. (iii) Eight 3’s + Four 5’s 44.= They must have answered four 5-point questions . 17. Let length of the longest piece,= length of the middle-sized piece, and lenth of the shortest piece.m s= = Then 3 and 10. So 90 3( 1 0 ) 9 0 5 100.ms m ms mm m m== - ++=  ++ - =  =  Thus ; 3; a n d 10 .m m sm20cm 60cm 10 cm= ===-= 18. Make a diagram that dem onstrates all the ways four-digit numbers can b e formed from left (thousands place) to right (ones place). 12 four- digit numbers can be formed. 19. Answer may vary. Fill the 4-cup container with water and pour the water into the 7-cup container. Fill the 4-cup container again and pour water into the 7-cup container until it is full. Four minus three ( 1) cups of water will remain in the 4-cup container. Empty the 7-cup container and pour the contents of the 4-cup container into the 7- cup container. The 7-cup container now holds 1 cup of water. Refill the 4-cup container and pour it into the 7-cup container. The 7-cup container now contains exactly 5 cups of water. 18 Chapter 1: An Introduction to Problem Solving Copyright © 2020 Pearson Education, Inc. 20. A possible pattern is to increase each rectangle by one row of dots and one column of dots to obtain the next term in the sequence. Make a table. Number of the term Row of dots Column of dots Term (row × column) 1 1 2 2 2 2 3 6 3 3 4 12 4 4 5 20 5 5 6 30 6 6 7 42 7 7 8 56  100 100 101 10100  n n 1n+ (1 )nn+ We also observe that the number of the term corresponds to the number of rows in the arrays and that the number of columns in the array is the number of the term p lus one. Thus, the next three terms are 30, 42, and 56 . The 100th term is 10,100 and the nth term is . (+ 1 )nn 21. A possible pattern is that each successive figure is constructed by adjoining another pentagon to the previous figure (a) ( b ) Observe that the perimeter of the first figure is 5 and that when a new pentagon is adjoined 4 new sides are added and one side (where the new pentagon is adjoined) is lost. Make a table. Number of terms 1-unit sides (perimeter) 1 5 2 514 8 -+ = 3 814 11 -+ = 4 11 1 4 14 -+ = ( c ) and (d) Looking at the terms in the sequence and noting that the differe nce of terms is three, we suspect that the sequence is arithmetic and conjecture that the nth term is 3 n+2. However, we need to be sure. Looking at the 4th term (part a) we observe that the pentagons on the end Contribute 4 sides to the perimeter and the “middle” pentagons contribute 3 sides. Thus, in the nth figure there will be 2 “end” pentagons that contribute 4 1-unit sides and 2 n- “middle” pentagons that contribute 3( 2) 1n- unit sides. The total will be 3( 2) 2(4)n-+ = . 3n + 2 units Thus the 100th term is 3(100) 2 . 302+= 22. (a) The circled terms will constitute an arithmetic sequence because the common difference will be twi ce the difference in the original series. (b) The new sequence will be a geometric sequence because the ratio will be the square of the ratio of the original series. 23. When n = 1, then n2 – n = 12 – 1 = 0, so the first term, a1 = 0. When n = 2, then n2 – n = 22 – 2 or 2. Thus a1 + a2 = 2; hence a2 = 2. For n = 3, n2 – n = 32 – 3 = 6. Substituting for a1 and a2, we get a3 = 6 – 2 =4. For n = 4, 42 – 4 = 12. Substituting for a1, a2, and a3, we get 0 + 2 + 4 + a4 = 12. Hence a4 = 6. 24. (a) Let’s call the missing terms a, and b then the sequence becomes: 13, , ,27 13 13 27 13 27 24 0 20 13 7ab ab ab ab b b b b ab a        S o 7 and 20 are the missing terms. ( b ) Let’s call the missing terms a, and b then the sequence becomes: 137, , ,163 137 137 163 137 163 23 0 0 150 137 13ab ab ab ab b b b b ab a        S o 13 and 150 are the missing terms. Chapter 1 Review 19 Copyright © 2020 Pearson Education, Inc. (c) Let’s call the missing terms x, and y, then the sequence becomes: ,,, 2 2 2 22 2 2bxya bx y x yb xya y bya ya b aby ab ab bxb x abx         So the missing terms are  +and22ab ab. 25. The first line tells us that since three cylinders =15, each cylinder must be worth 5. Using that information, we can use the second line to figure out that a circle must be worth 4. That information can be used on the third line to determine that the cup-lik e shape must be equal to 1. So, putting it all together in the fourth line, we have 4 + 5 + 1 = 10. 26. The pattern involves taking two squares that are diagonal, summing their numbers, and putting that sum in the square that is below one of the diagonal squares and to the left of the other diagonal square. For example, the top square is 4; the square diagonal to it (to the right) is 2; their sum is 6, which is the number that appears in the square below the 4 and to the left of the 2. You can also see it where 9 + 10 = 19, and 1 + 6 =7. So using this pattern, the question mark must be equal to 3. 27. Since 1, 2, and 3 are alre ady on the triangle, the numbers 4-9 will be used to fill in the six question marks. On the left side, 1+2 = 3, so the two question marks on that side must sum to 14. Similarly on the bottom, the two question marks must sum to 13, and on the right side, the two question marks must sum to 12. So on the right side with the available numbers, only two possibilities, 5+7, or 4+8, sum to 12. If the two numbers on the right side are 5 and 7, the bottom two numbers (which must sum to 13) are 4 and 9, and the two numbers on the left side must be 6 and 8. If the two numbers on the right side are 4 and 8, the two bottom numbers must be 6 and 7, leaving 5 and 9 to be the numbers on the right. Below are the two solutions presented on the triangle: Copyright © 2020 Pearson Education, Inc. 21 CHAPTER 2 INTRODUCTION TO LOGIC AND SETS Assessment 2-1A: Reasoning and Logic: An Introduction 1 . (a) False statement . A statement is a sentence that is either true or false, but not both. (b) False statement. Los Angeles is a city, not a state. (c) Not a statement. Questions are not statements. (d) True statement. 2 . (a) There exists at least one natural number n such that 81 1 . n+= (b) There exists at least one natural number n such that 24. n= (c) F o r all natural numbers n, 33 . nn+=+ (d) F o r all natural numbers n, 549 .nnn+= 3 . (a) F o r all natural numbers n, 81 1 . n+= (b) F o r all natural numbers n, 24. n= (c) T h e r e i s no natural number x such that 33 .x x +=+ (d) T h e r e i s no natural number x such that 54 9 .x xx+= 4 . (a) The book does not have 500 pages. (b) 3 5 15.⋅¹ (c) Some dogs do not have four legs. (d) No rectangles are squares. (e) All rectangles are squares. (f) Some dogs have fleas. 5 . (a) If 4, n= or 5, n= then 6 n< and 3, n> so the statement is true, since it can be shown to work for some natural numbers . n (b) All natural numbers ar e greater than zero; so, since the condition is that 0 n> or 5, n< the statement is true. 6 . (a) ~~ ( ~ ) T Fpp p FT TF (b) ~~~ T Fp p pp pp FT F TT F (c) Yes. The truth table entries are the same. (d) No. The truth table entri es are not the same. 7 . ( a ) ~~ p q p qpp q TT T F T TF F F F FT T T T FF T T Τ ( b ) ~ . pq p qº  Answers will vary. Here are two possibl e examples:1. Let p be “the Bobcats win” and q be “the Bobcats make the playoffs”. Then Column 3 would read “If the Bobcats win, then the Bobcats make the playoffs”. Column 5 would read “The Bobcats lose or the Bobcats make the playoffs.”2. Let p be “it is summer vacation” and q be “I am at home”. Then Column 3 would read “If it is summer vacation, then I am at home”. Column 5 would read “It is not summer vacation or I am at home.” ( c ) In this problem, p is the statement “2 + 3 = 5” and q is the statement “4 + 6 = 10”. That would make the statement in this problem be in the form of ~pq. So, it will be logically equivalent to a statement in the form of pq or “if 2 + 3 = 5, then 4 + 6 = 10.” 8 . (a) . qr Both q and r are true. (b) . r~ q r is true or q is not true. (c) . ~( q r ) q and r are not both true. (d) .~q q is not true. 9 . (a) False . The statement is a conjunction. In order for a conjunction to be true, both p and q must be true; otherwise, the conjunction is false. T he two parts of the conjunction could be stated as such: p is the statement 235+= and q is the 22 Chapter 2: Introduction to Logic and Sets Copyright © 2020 Pearson Education, Inc. statement 471 0+= . In this situation, p is true, but q is false. (b) False . The United States Supreme Court has nine justices when every seat is filled. (c) True . The only triangles that have three sides of the same length are equilateral triangles. In every case, an equilateral triangle will have two sides the same length as well. (d) False . Isosceles triangles have two sides equal in length, but the third side does not have to be equal to the other two. 1 0 . (a) By DeMorgan’s Laws, the negation of pq is ~~pq . Therefore, the answer is 2+3 5¹ or 4+7 1 0¹ . (b) With every seat f illed, the Supreme Court of the United St ates does not have 12 justices . (c) By problem #7 and DeMorgan’s Laws, the negation of pq is ~ .pq So, the statement would read: the triangle has three sides of the same length and the triangle does not have two sides of the same length . (d) The triangle has two sides of the same length and the triangle does not have three sides of the same length . In both (c) and (d) above , the negation of a conditional statement pq is ~pq . 11. (a) ~~~ ~~ ( ) TT F F F F TF F T T F FT T F T FFF T T T Tpq p q p q pq  Since the truth values for ~~pq are not the same as for ~( ),pq the statements are not logically equivalent . ( b ) ~~ ~ ( ) ~ ~ TT F F T F F TF F T F T F FT T F F T FFF T T F T Tp q p q pq pq p q  Since the truth values for ~( )pq are not the same as for ~~ ,pq the statements are not logically equivalent . 12. Alex is a male teacher w ho does not teach math and is 30 years old or younger. 13. If p=“it is raining” and q=“the grass is wet”: (a) .. (b) ~ pq pq (c) .~ pq (d) . pq The hypothesis is “it is raining;” the conclusion is “t he grass is wet.” (e) .. ~~ ( f )qp q p« 14. (a) Converse : If a triangle has no two sides of the same length, then the triangle is scalene. Inverse : If a triangle is not scalene, then the triangl e has (at least) two sides of the same length. Contrapositive : If a triangle does not have two sides of the same length, then th e triangle is not scalene. (b) Converse: If an angle is a right angle, then it is not acute. Inverse : If an angle is acute, then it is not a right angle. Contrapositive: If an angle is not a right angle, then it is acute. Note that the original statement and the contrapos itive are not tr ue, while the converse and inverse are true. (c) Converse : If Maria is no t a citizen of Cuba, then she is a U.S. citizen. Inverse : If Maria is not a U.S. citizen, then she is a citizen of Cuba. Contrapositive : If Maria is a citizen of Cuba, then she is not a U.S. citizen. Note that the o riginal statement and the contrapositive ar e true, while the converse and inverse are not true (d) Converse : If a number is not a natural number, then it is a whole number. Inverse : If a number is not a whole number, then it is a natural number. Contrapositive : If a number is a natural number, then it is not a whole number. 1 5 . The statements are negations of each other. ~~ ~ ( ~ ) ~ ( ) TT F F T T F TF T T F F T FT F F T T FFF T F T T Fpq qp q p q p q p q    Assessment 2-1B 23 Copyright © 2020 Pearson Education, Inc. 1 6 . The contrapositive is logically equivalent: “ If a number is not a multiple of 4 then it is not a multiple of 8 .” 1 7 . (a) Valid. This is valid by the transitivity property. “All squares are quadrilaterals” is ; pq “all quadrilaterals are polygons” isqr ; and “all squares are polygons” is pr . (b) Invalid. We do not know what will happen to students who are not freshman. There is no statement “sophomores, juniors, and seniors do not take mathematics.” 18. (a) Since all students in Integrated Mathematics I make A’s, and some of those students are in Beta Club, then some Beta Club students make A’s . (b) Let p= I study for the final, q= I pass the final, r= I pass the course, s= I look for a teaching job. Then , pq if I study for the final, then I will pass the final. , qr if I pass the fina l, then I will pass the course. , rs if I pass the course, I will look for a t eaching job. So , ps if I study for the f inal, then I will look for a teaching job . (c) The first statement could be rephrased as “If a triangle is equ ilateral, then it is isosceles.” Let p= equilateral triangle and q= isosceles triangl e. So the first statement is , pq . The second statement is simply p; then the conclusion should be q or there exist triangles that are isosceles . 1 9 . (a) If a figure is a square, then it is a rectangle. (b) If a number is an integer, then it is a rational number. (c) If a polygon has exactly three sides, then it is a triangle. 2 0 . (a) If ~ ~ ~ then .pq p q   32 6 1 1 3º  ⋅¹+ = or (b) If ~ ~ ~pq p q   º  then you cannot pay me now and yo u cannot pay me later . Assessment 2-1B 1 . (a) Not a statement . A statement is a sentence that is either true or false. (b) Not a statement . A statement must be either true of false; this could be either since “he” and “town” are not specified. ( c ) True statement . (d) False statement . 238 .+¹ ( e ) Not a statement . A statement is a sentence that is either true or false. 2 . (a) For all natural numbers n, 0. nn+= (b) There exists no natural number n such that 12 .nn+=+ (c) There exists at least one natural number n such that 3( 2) 12.n+= (d) There exists at least one natural number n such that 38. n= 3 . (a) There is no natural number n such that 0. nn+= (b) There exists at least one natural number n such that 12 .nn+=+ (c) For all natural numbers n, 3( 2 ) 1 2 .n⋅+¹ (d) For all natural numbers n, 38. n¹ 4 . (a) Six is greater than or equal to 8 . Another possible answer is 6 is not less than 8. (b) All cats have nine lives . Another way to express this would be to say that no cats do not have nine lives . (c) There exists a square that is not a rectangle . Another way to express this would be to say some squares are not rectangles . (d) All numbers are positive . (e) No people have blond hair. 5 . (a) If 10, n= then 5 n> and 2, n> so the statement is true. (b) n could equal 5, so the statement is false . 6 . (a) pq is false only if both p and q are false, so if p is true the statement is true regardless of the truth value of q. (b) An implication is false only when p is true and q is false, so if p is false then the statement is true regardless of the truth value of q.