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Solution Manual for A Problem Solving Approach to Mathematics for Elementary School Teachers, 13th Edition

Solution Manual for A Problem Solving Approach to Mathematics for Elementary School Teachers, 13th Edition gives you access to textbook solutions that make learning easier and more efficient.

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    S OLUTIONS M ANUAL B RIAN B EAUDRIE B ARBARA B OSCHMANS Northern Arizona Northern Arizona University University to accompany A P ROBLEM S OLVING A PPROACH TO M ATHEMATICS F OR E LEMENTARY S CHOOL T EACHERS T HIRTEENTH E DITION Rick Billstein University of Montana Barbara Boschmans Northern Arizona University Shlomo Libeskind University of Oregon Johnny W. Lott University of Montana

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    iii Contents Chapter 1 An Introduction to Problem Solving 1 Chapter 2 Introduction to Logic and Sets 21 Chapter 3 Numeration 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, Percents, and Real Numbers 155 Chapter 8 Algebraic Thinking 189 Chapter 9 Probability 217 Chapter 10 Data Analysis/Statistics: 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

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    1 CHAPTER 1 AN INTRODUCTION TO PROBLEM SOLVING Assessment 1-1A: Mathematics and Problem Solving 1. (a) List the numbers: 1 2 98 99 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 2 49.5 = pairs of sums, each of 100; thus 49.5  100 = 4950.) (b) The number of terms in any sequence of numbers may be found by subtracting the first term from the last, dividing the result by the common difference between terms, and then adding 1 (because both ends must be accounted for). Thus 1001 1 2 1 501 - + = 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 501 1002 2 . ⋅ = 251, 001 (c) The number of terms in any sequence of numbers may be found by subtracting the first term from the last, dividing the result by the common difference between terms, and then adding 1 (because both ends must be accounted for). Thus 300 3 3 1 100 - + = 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 of numbers may be found by subtracting the first term from the last, dividing the result by the common difference between terms, and then adding 1 (because both ends must be accounted for). Thus 400 4 4 1 100 - + = 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)

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    2 Chapter 1: An Introduction to Problem Solving 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 thinking can be used for any natural number n to arrive at a formula ( 1) 2. n n  3. There are 147 36 1 1 112 - + = 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 follows; there are 12 rows so there are 12 different ways . 6-cookie 2-cookie single-cookie packages packages packages 2 0 0 1 3 0 1 2 2 1 1 4 1 0 6 0 6 0 0 5 2 0 4 4 0 3 6 0 2 8 0 1 10 0 0 12 5. If each layer of boxes has 7 more than the previous layer we can add powers of 7: 7 0 = 1 (red box) 7 1 = 7 (blue boxes) 7 2 = 49 (black boxes) 7 3 = 343 (yellow boxes) 7 4 = 2401 (gold boxes) 1 + 7 + 49 + 343 + 2401 = 2801 boxes altogether. 6. Using strategies from Poyla’s problem solving list identify subgoals (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 triangles. Counting the large triangle there are 5 triangles

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    Assessment 1-1A: Mathematics and Problem Solving 3 Unit triangles 4 unit triangles 9 unit triangles 9 3 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: 1 2 3 4 5 6 97 98 98(99) 49(99) 2 49(50) 2(1 2 3 4 49) 2 49(50) 2 49(99) 49(50) 49(49). O E E O O E E     + = + + + + + + + + = = æ ö ÷ ç = + + + + + = = ÷ ç ÷ ÷ ç è ø = + - = - = So O is 49 less than E . 8. Bubba is last; Cory must 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 bills $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 . 10. The diagonal from the left, top corner to the right, bottom corner sums to 17 22 27 66. + + = The first row sums to 17 7 24 . a a + + = + So 66 24 . a = - = 42 The last column sums to 7 27 34 . b b + + = + So 66 34 . b = - = 32 The first column sums to 17 12 29 . c c + + = + So 66 29 . c = - = 37 The second column sums to 42 22 64 . d d + + = + So 66 64 . d = - = 2 11 . 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 . page 48 ´ = 12. The last three digits must sum to 20, so the second to last digit must be 20 (7 4) 9. - + = Since the sum of the 11 th , 12 th , and 13 th digits is also 20, the 11 th 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 . 13. 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 fruit 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. 14. The electrician made $1315 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 different 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 pies that are baked is

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    4 Chapter 1: An Introduction to Problem Solving represented by 4 – a . So, using 9 slices for an apple pie and 7 slices for a cherry pie, we get: 9 7(4 ) 34 9 28 7 34 2 6 3 a a a a 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 by 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. 18. Since the bat is $49 more than the ball, and the total spent is $50, we can use the guess and check method to solve. The table below represents possible guesses: Cost of bat Cost of ball sum $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 costs 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: 1 2 48 49 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 2 24.5 = 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 the last, dividing the result by the common difference between terms, and then adding 1 (because both ends must be accounted for). Thus 2009 1 2 1 - + = 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 the last, dividing the result by the common difference between terms, and then adding 1 (because both ends must be accounted for). Thus 600 6 6 1 100 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

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    Assessment 1-1B 5 difference between terms, and then adding 1 (because both ends must be accounted for). Thus 1000 5 5 1 200 terms. - + = 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 numbers 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 1 1 146 terms. - + = 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 good approach. Using only two numbers and addition, 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 color. However, once you draw a third sock, you are 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 one square 9 units; for a total of 18 squares. 7. 1 3 5 7 ... 99 5 7 ... 99 101 (5 ... 99 101) (1 3 5 ... 99) (101) (1 3) 97 P Q Q P = + + + + + = + + + + - = + + + - + + + + = - + = 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 minimum number of coins would have as many quarters as possible, or three quarters. The remaining 20 c / must consist of at least one 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 .

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    6 Chapter 1: An Introduction to Problem Solving 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 5 17 13 3 17 11 5 17 9 7 15 13 5 15 11 7 13 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 11. Answers may vary; two solutions might be to: (a) Put four marbles on each tray of the balance scale. Take the heavier four and weigh two on each tray. Take the heavier two and weigh one on each tray; the heavier marble will be evident on this third weighing. (b) This alternative shows the heavier marble can be found more 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 heavier, take two of the three marbles and weigh them. If they are the same weight, the remaining marble is the heavier. If not, the heavier will be evident on this second weighing. 12. (a) There are: 1 partridge 12 days 12 gifts; 2 doves 11 days 22 gifts; 3 hens 10 days 30 gifts; 4 birds 9 days 36 gifts; 5 rings 8 days 40 gifts; 6 geese 7 days 42 gifts; 7 swans 6 days 42 gifts; 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. ´ = ´ = ´ = ´ = So the 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 quarters 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 3 3 . = = 4 salads 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 - - - +

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    Assessment 1-2A: Explorations with Patterns 7 So, 2(T 15) 10 32 2T 30 10 32 2T 20 32 2T 52 T . 26 degrees - + = - + = - = = = 16. One way to solve this problem is to write an equation using a variable. 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 9 5(5 ) 33 9 25 5 33 4 8 2 p p p p 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 visualize, so the strategy of examining a simpler case 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 3 3 3 ´ ´ 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 bottom 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 front and the back 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 4 4 4 ´ ´ 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 3 3 3 ´ ´ large cube, we are left with 1 small cube (1 1 1 1); ´ ´ = when starting with a 4 4 4 ´ ´ large cube, we are left with 8 small cubes (2 2 2 8). ´ ´ = Therefore, the pattern seems to be, when given large cube of dimension , n n n ´ ´ 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). n n n - ´ - ´ - So for a 10 10 10 ´ ´ large cube, the number of small cubes left would be 8 8 8 ´ ´ = 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:

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    8 Chapter 1: An Introduction to Problem Solving 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 multiplying by 3. (d) Terms that continue a pattern are 9 11 13 10 ,10 ,10 , … . This is geometric because each successive term is obtained from the previous term by multiplying by 10 2 . (e) Terms that continue a pattern are , 30 30 30 193 + 10× 2 ,193 + 11× 2 ,193 + 12× 2 … . This is arithmetic because each successive term is obtained from the previous term by addition of 30 2 . 3. In these problems, let n a represent the nth term in a sequence, 1 a represent the first term, d represent the common 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) ; n a a n d = + - in a geometric sequence 1 1 . n n a a r - = Thus: (a) Arithmetic sequence: 1 1 a = and 4: d = ( i ) 100 1 (100 1) 4 1 99 4 . a 397 = + - ⋅ = + ⋅ = ( ii ) 1 ( 1) 4 1 4 4 . n a n n 4 3 = + - ⋅ = + - = - n (b) Arithmetic sequence: 1 70 and 50: a d = = ( i ) 100 70 (100 1) 50 70 99 50 . a 5020 = + - ⋅ = + ⋅ = ( ii ) 70 ( 1) 50 70 50 50 or . n a n n 50n + 20 = + - ⋅ = + - (c) Geometric sequence: 1 1 and 3: a r = = ( i ) 100 1 100 1 3 . a 99 3 - = ⋅ = ( ii ) 1 1 3 . n n a n-1 3 - = ⋅ = (d) Geometric sequence: 2 1 10 and 10 : a r = = ( i ) 2 (100 1) 2 99 100 198 10 (10 ) 10 (10 ) 10 10 . a 199 10 - = ⋅ = ⋅ = ⋅ = ( ii ) 2 ( 1) (2 2) 10 (10 ) 10 10 . n n n a 2n-1 10 - - = ⋅ = ⋅ = (e) Arithmetic sequence: 30 30 1 193 7 2 and 2 : a d = + ⋅ = ( i ) 30 30 100 30 30 193 7 2 (100 1) 2 193 7 2 99 2 . a 30 193 + 106 × 2 = + ⋅ + - ⋅ = + ⋅ + ⋅ = ( ii ) 30 30 193 7 2 ( 1) 2 . n a n 30 193 + (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 1 1 1 1 ⋅ ⋅ = 2 2 2 2 8 ⋅ ⋅ = 3 3 3 3 27 ⋅ ⋅ = 4 4 4 4 64 ⋅ ⋅ = 5 5 5 5 125 ⋅ ⋅ = 6 6 6 6 216 ⋅ ⋅ = 7 7 7 7 343 ⋅ ⋅ = 8 8 8 8 512 ⋅ ⋅ = 9 9 9 9 729 ⋅ ⋅ = 10 10 10 10 1000 ⋅ ⋅ = 11 11 11 11 1331 ⋅ ⋅ = The 11 th term 1331 is the least 4-digit number greater than 1000. (b) The 9 th term 729 is the greatest 3-digit number in this pattern. (c) 10 4 = 10,000; The greatest number less than 10 4 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

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    Assessment 1-2A: Explorations with Patterns 9 10th windmill is thus 5 (10 1) 4 + - ⋅ = 5 9 4 . 41 squares + ⋅ = (b) The n th windmill would require 5 ( 1) 4 5 4 4 n n + - ⋅ = + - = 4 1 n + 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 16 n a = + ( 1) 12 n - ⋅ = 12 4 n + 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 (15 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 n th term is thus: 6 ( 1) 4 . n a n = + - ⋅ = 4 2 n + 8. Since the first year begins 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 13 th term (current enrollment twelve more years) + is: 700 (13 1) 60 + - ⋅ = 1420 students. 9. Using the general expression for the n th 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 d d = + - = +  = 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 . 10. The number that fits into 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, 2 5 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 n 2 . 12. (a) Using the general expression for the nth term of an arithmetic sequence with 1 51, 251, n a a = = and 1 d = yields: 251 51 ( 1) 1 251 50 201. n n n = + - ⋅  = +  = There are 201 terms in the sequence. (b) Using the general expression for the n th term of a geometric sequence with 1 1, a = 60 2 , n a = and 2 r = yields: 60 1 1 2 1(2) 2 1 60 61. n n n n - - = =  - =  = There are 61 terms in the sequence. (c) Using the general expression for the n th term of an arithmetic sequence with 1 10, a = 2000, n a = and 10 d = yields: 2000 10 ( 1) 10 2000 10 200. n n n = + - ⋅  =  = There are 200 terms in the sequence.

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    10 Chapter 1: An Introduction to Problem Solving (d) Using the general expression for the n th term of a geometric sequence with 1 1, a = 1024, n a = and 2 r = yields: 1 10 1 1024 1(2) 2 2 1 10 11. n n n n - - =  =  - =  = There are 11 terms in the sequence. 13. (a) 2 2 2 2 2 First 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 and Fifth term: 3(5) 2 1; 4; 7; 10; 13. - = - = - = - = - = 14. Answers may vary; examples are: (a) If 5, n = then 5 5 5 2 5 1 6. + = ¹ + = (b) If 2, n = then 2 2 (2 4) 6 36 + = = does not equal 2 2 2 4 20. + = 15. (a) There are 1, 5, 11, 19, 29 tiles in the five figures. Each figure adds 2n tiles to the preceding figure, thus a 6 the 6th term has 29 12 . + = 41 tiles (b) 2 1, 4,9, 16, 25, n = … . Adding ( 1) n - to n 2 yields 1, 5, 11, 19, 29, … , which is the proper sequence. Thus the n th term has . 2 + 1 - n n (c) If 2 ( 1) 1259; n n + - = Then 2 1260 0. n n + - = This implies ( 35)( 36) 0, so 35. n n 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 n th term of the geometric sequence is 2 n . The sequence can also 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 With the 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 5 5 25 ⋅ = pieces. Continuing this process gives a geometric sequence: 1, 5, 25, 125, … . After the 5 th cut there are 5 5 = 3125 pieces of paper. (b) The number of pieces after the nth cut would be . n 5 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 30 and 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: 800 200 600 and 600 3 200.     The

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    Assessment 1-2B 11 common difference is 200 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 5 4 and 5 10 there are three common ratios used so we can find the common ratio by dividing 5 10 by 5 4 and then taking the cube root:   1 3 10 4 6 6 2 5 5 5 and (5 ) 5 .    The common ratio is 52 and we can find the missing terms: 4 2 4 2 6 2 5 5 ,5 5 ,5 5 .       2 6 8 5 5 5 19. (a) Let’s call the missing terms a, b, c, d, e and f, then the sequence becomes: , ,1,1, , , , 1 1 0 1 0 1 1 1 1 2 1 1 2 3 2 3 5 3 5 8. a b c d e f b b a b a a c c c d d d c d e e e d e 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: , , ,10,13, ,36,59 10 13 3 10 3 10 7 7 3 4 10 13 23 a b c d c c b c b b a b c a d d                       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 term starting from the third is the sum of the previous two terms, then we can add 0 and 2 to get the third term and continue the pattern: 0 2 2 2 2 4 2 4 6 4 6 10 6 10 16 10 16 26             The missing terms are 2, 4, 6, 10, 16 , and 26 . 20. (a) Year 1 80 .05(80) 84 Year 2 84 .05(84) 88.2 Year 3 88.2 .05(88.2) 92.61 Year 4 92.61 .05(92.61) 97.2405 Year 5 97.2405 .05(97.2405) 102.102525 .            $102.10 (b) This is a geometric sequence with 1 80 a  and r = 1.05, so the price after n years is 80 • 1.05 n . 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, etc. 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 4 4 , 4 5 , 4 6 , … . This is a geometric sequence because each successive term is obtained from the previous term by multiplying by 4.

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    12 Chapter 1: An Introduction to Problem Solving (d) Terms that continue a pattern are 2 14 , 2 18 , 2 22 , … . This is a geometric sequence because each successive term is obtained from the previous term by multiplying by 24. (e) Terms that continue a pattern are ,.... 50 50 50 100 + 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 common 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) ; n a a n d = + - in a geometric sequence, 1 1 . n n a a r - = Thus: (a) Arithmetic sequence: 1 2 a = and 4. d = ( i ) 100 2 (100 1) 4 . a 398 = + - ⋅ = ( ii ) 2 ( 1) 4 2 4 4 . n a n n 4 2 = + - ⋅ = + - = - n (b) Arithmetic sequence: 1 0 a = and 13. d = ( i ) 100 0 (100 1) 13 . a 1287 = + - ⋅ = ( ii ) 0 ( 1) 13 . n a n 13n 13 = + - ⋅ = - (c) Geometric sequence: 1 4 a = and 4. r = ( i ) 99 100 4 4 . a 100 4 = ⋅ = ( ii ) 1 4 4 . n n a 4 - = ⋅ = n (d) Geometric sequence: 2 1 2 a = and 4 2 . r = ( i ) 2 4 99 2 396 100 2 (2 ) 2 2 . a 398 2 = ⋅ = ⋅ = ( ii ) 2 4 ( 1) 2 4 4 2 (2 ) 2 2 . n n n a 4n 2 2 - - - = ⋅ = ⋅ = (e) Arithmetic sequence: 50 50 51 1 100 4 2 and 2 2 2 : a d = + ⋅ = ⋅ = ( i ) 50 51 100 51 51 51 50 100 4 2 (100 1) 2 100 2 2 99 2 100 101 2 100 101 2 2 . a 50 100 + 202 × 2 = + ⋅ + - ⋅ = + ⋅ + ⋅ = + ⋅ = + ⋅ ⋅ = ( ii ) 50 51 51 51 51 50 100 4 2 ( 1) 2 100 2 2 ( 1) 2 100 ( 1) 2 100 ( 1) 2 2 . n a n n n n 50 100 + 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 continue, so the next three terms are 9, 5, 1 . 5. (a) Answers may vary: two possible answers are: ( i ) 2 e. ., 1 3 5 7 4 . n s g 2 The sum of the first odd number is ; + + + = n ( ii ) ( ) 2 1 7 2 2 e. ., 1 3 5 7 4 . g Square the average of the first and last terms; + + + + = = (b) There are 35 1 2 1 18 - + = terms in this sequence. ( i ) 2 1 3 5 7 35 18 . 324  + + + + + = = ( ii ) ( ) 2 1 35 2 2 18 . 324 + = = 6. (a) Note that 5 toothpicks are added to form each succeeding hexagon. This is an arithmetic sequence 1 6 a = and 5, d =

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    Assessment 1-2B 13 so 10 6 (10 1) 5 a = + - ⋅ = 6 9 5 + ⋅ = 51 toothpicks . (b) n hexagons would require 6 ( 1) 5 6 5 5 n n + - ⋅ = + - = 5 1 n + toothpicks . 7. (a) Looking at the second figure, there are 3 + 1 = 4 triangles. In the third figure, there are 5 3 1 9 + + = triangles. The fourth figure would then have 7 5 3 1 16 + + + = triangles. An alternative to simply adding 7, 5, 3, and 1 together is to note that 7 1 8 + = and 5 3 8. + = There are 4 2 2 = of these sums, and 2 8 16. ⋅ = 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 2 50 = of these pairs. 50 200 10, 000, ⋅ = or 10,000 triangles in the 100th figure. (b) The number of triangles in the n th figure is 2 (number of triangles in base 1). n + The number of triangles in the base is ( 1), n n + - or 2 1. n - (2 1) n - + 1 2 . n = Then 2 2 (2 ) n n n = , or n 2 triangles in the n th figure. 8. This is a geometric sequence with 1 15,360 2 a = and 1 2 r = . The n th term of a geometric sequence is 1 1 ; n n a a r - = thus the 10th term would be ( ) 10 1 2 15360 . = 15 liters Note the progression of terms in the following table: After Day Amount of Water Remaining 1 1 2 15,360 7680 liters ⋅ = 2 1 2 7680 3840 liters ⋅ =   9 1 2 60 30 liters ⋅ = 10 1 2 30 15 liters ⋅ = 9. This is an arithmetic sequence with 1 1 6 8 a = (i.e., 8 a.m. plus 10 minutes, or 10 60 of an hour) and 5 6 d = (or 50 60 of an hour). Thus 5 1 8 6 6 8 (8 1) 14, or a 2:00 p.m. = + - ⋅ = (14 is 2:00 p.m. on a 24-hour clock.) 10. Answers will be a rotation of the following figure: 11. (a) In the first drawing, there are 6 toothpicks; in the second drawing, there are 10, and the third drawing has 14 toothpicks. This it is an arithmetic sequence with 1 6, 4. a d = = So, for the tenth figure, we would have 10 10 10 6 (10 1) 4 6 9 4 a a a 42. = + - ⋅ = + ⋅ = (b) The n th term for this arithmetic sequence is 6 ( 1) 4 6 4 4 n n n a n a n 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) The n th term for this geometric sequence is 1 3 . n - Thus 99 1 3 3 . n - = So 99 1, n = - and 100. n = There are 100 terms in the sequence. (b) The n th 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 n th 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.

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