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

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SOLUTIONSMANUALBRIANBEAUDRIEBARBARABOSCHMANSNorthern ArizonaNorthern ArizonaUniversityUniversityto accompanyAPROBLEMSOLVINGAPPROACHTOMATHEMATICSFORELEMENTARYSCHOOLTEACHERSTHIRTEENTHEDITIONRick BillsteinUniversity of MontanaBarbara BoschmansNorthern Arizona UniversityShlomo LibeskindUniversity of OregonJohnny W. LottUniversity of Montana

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iiiContentsChapter 1 An Introduction to Problem Solving1Chapter 2 Introduction to Logic and Sets21Chapter 3 Numeration Systems and Whole Number Operations45Chapter 4 Number Theory87Chapter 5 Integers105Chapter 6 Rational Numbers and Proportional Reasoning127Chapter 7 Decimals, Percents, and Real Numbers155Chapter 8 Algebraic Thinking189Chapter 9 Probability217Chapter 10 Data Analysis/Statistics: An Introduction245Chapter 11 Introductory Geometry273Chapter 12 Congruence and Similarity with Constructions297Chapter 13 Area, Pythagorean Theorem, and Volume323Chapter 14 Transformations361

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1CHAPTER 1AN INTRODUCTION TO PROBLEM SOLVINGAssessment 1-1A: Mathematics andProblem Solving1.(a)List the numbers:129899999821100100100100++++++++++++There are 99 sums of 100. Thus the total canbe found by computing99 1002.⋅=4950(Another way of looking at this problem isto realize there are99249.5=pairs ofsums, each of 100; thus 49.5100 = 4950.)(b)The number of terms in any sequence ofnumbers may be found by subtracting thefirst term from the last, dividing the resultby the common difference between terms,and then adding 1 (because both ends mustbe accounted for). Thus1001121501-+=terms.List the numbers:1399910011001999311002100210021002++++++++++++There are 501 sums of 1002. Thus the totalcan be found by computing501 10022.⋅=251, 001(c)The number of terms in any sequence ofnumbers may be found by subtracting the firstterm from the last, dividing the result by thecommon difference between terms, and thenadding 1 (because both ends must beaccounted for). Thus300331100-+=terms.List the numbers:3629730030029763303303303303++++++++++++There are 100 sums of 303. Thus thetotal can be found by computing100 3032.15,150⋅=(d)The number of terms in any sequence ofnumbers may be found by subtracting thefirst term from the last, dividing the resultby the common difference between terms,and then adding 1 (because both ends mustbe accounted for). Thus400441100-+=terms.List the numbers:4839640040039684404404404404++++++++++++There are 100 sums of 404. Thus thetotal can be found by computing100 4042.20, 200⋅=2.(a)(b)

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2Chapter 1: An Introduction to Problem SolvingWhen the stack in (a) and a stack of thesame size is placed differently next to theoriginal stack in (a), a rectangle containing100 (101) blocks is created. Since eachblock is represented twice, the desired sumis100 (101) 2.=5050While the above represents a specificexample, the same thinking can be used forany natural number n to arrive at a formula(1) 2.n n3.There are1473611112-+=terms.List the numbers:36371461471471463736183183183183++++++++++++There are 112 sums of 183. Thus the total canbe found by computing112 1832.⋅=10, 2484.(a)Make a table as follows; there are 9 rowsso there are9different ways.6-cookiepackages2-cookiepackagessingle-cookiepackages1201121040500420340260180010(b)Make a table as follows; there are 12 rowsso there are12 different ways.6-cookie2-cookiesingle-cookiepackagespackagespackages200130122114106060052044036028011000125.If each layer of boxes has 7 more than theprevious layer we can 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 boxesaltogether.6.Using strategies from Poyla’s problem solvinglist identify subgoals (solve simpler problems)and make diagrams to solve the originalproblem.1 triangle; name this the “unit” triangle.This triangle is made of 4 unit triangles.Counting the large triangle there are5 triangles

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Assessment 1-1A: Mathematics and Problem Solving3Unit triangles4 unit triangles9 unit triangles93113 total trianglesUnit triangles 4 unit triangles 9 unit triangles 16 unit triangles16731There are 27 triangles in the original figure.7.Observe thatE=(1 + 1) + (3 + 1) ++(97 + 1)=O+ 49. Thus,Eis 49 more thanO.Alternative strategy:123456979898(99)49(99)249(50)2(123449)249(50)249(99)49(50)49(49).OEEOOEE+=++++++++==æö÷ç=+++++==÷ç÷÷çèø=+-=-=So O is 49 less than E.8.Bubba is last; Cory must be between Alababaand Dandy; Dandy is faster than Cory. Listingfrom fastest to slowest, the finishing order isthenDandy, Cory, Alababa, and Bubba.9.Make a table.$20 bills$10 bills$5 bills2102021301221141060500420340260180010There are twelve rows so there aretwelvedifferent ways.10.The diagonal from the left, top corner to theright, bottom corner sums to17222766.++=The first row sums to17724.aa++=+So6624.a=-=42The last column sums to72734.bb++=+So6634.b=-=32The first column sums to171229.cc++=+So6629.c=-=37The second column sumsto422264.dd++=+So6664.d=-=211.Debbie and Amy began reading on the sameday, since 72 pages for Debbie9 pages perday = 8 days. Thus Amy is on6 pages per day8 days.page 48´=12.The last three digits must sum to 20, so thesecond to last digit must be20(74)9.-+=Since the sum of the 11th, 12th, and 13thdigits isalso 20, the 11thdigit is20(79)4.-+=We can continue in this fashion until we findthat A is 9, or we can observe the repeatingpattern from back to front, 4, 9, 7, 4, 9,7, … and discover thatA is 9.13.Choose the box labeled Oranges and Apples(Box B). Retrieve a fruit from Box B. SinceBox B is mislabeled, Box B should be labeledas having the fruit you retrieved. For example,if you retrieved an apple, then Box B should belabeled Apples. Since Box A is mislabeled, theOranges and Apples label should be placed onBox A. These leave only one possibility for BoxC; it should be labeled Oranges. If an orangewas retrieved from Box B, then Box C would belabeled Oranges and Apples and Box A shouldbe labeled Apples.14.The electrician made $1315 for 4 days at $50per 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 worked13755027.5 hours.15.Working backward: Top6 rungs7 rungs + 5rungs3 rungs = top11 rungs, which islocated at the middle. From the middle rungtravel up 11 to the top or down 11 to thebottom. Along with the starting rung, then, thereare 11 + 11 + 1 =23 rungs.16.There are several different ways to solve this.One is to use a variable. So, letabe equal to thenumber of apple pies that are baked. This meansthe number of cherry pies that are baked is

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4Chapter 1: An Introduction to Problem Solvingrepresented by 4 –a. So, using 9 slices for anapple pie and 7 slices for a cherry pie, we get:97(4)34928734263aaaaaaSo the number ofapple pies is 3. Therefore, thenumber ofcherry pies is 1.Since the number of pies is small, anothersolution strategy is to make a table of allpossible cases:ApplePiesCherryPiesAppleslicesCherryslicesTotalslices040282813921302218143231277344036036From the table, we can see there are 34 sliceswhen there are three apple pies and one cherrypie.17.Al made $50. Examine it step by step. First, Alspent $100 on the CD player. At this point, he isdown $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 itagain for $175, he is now ahead $50.18.Since the bat is $49 more than the ball, and thetotal spent is $50, we can use the guess andcheck method to solve. The table belowrepresents possible guesses:Cost of batCost of ballsum$49.00$0.00$49.00$49.25$0.25$49.50$49.50$0.50$50.00Another method would be to use a variable. Letthe cost of the ball (in dollars) be the variableb.The cost of a bat in dollars, therefore, would beb+ 49. Together the two costs must add up to50, so:b+ (b+ 49) = 502b+ 49 = 502b= 1b= 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 balldoes equal $50.Assessment 1-1B1.(a)List the numbers:12484949482150505050++++++++++++There are 49 sums of 50. Thus the total canbe found by computing49 502.⋅=1225(Another way of looking at this problem isto realize there are49224.5=pairs ofsums, each of 50; thus24.5501225.)⋅=(b)The number of terms in any sequenceof numbers may be found by subtractingthe first term from the last, dividing theresult by the common difference betweenterms, and then adding 1 (because bothends must be accounted for). Thus2009121-+=1005 terms.List the numbers:132007200920092007312010201020102010++++++++++++There are 1005 sums of 2010. Thus thetotal can be found by computing1005 20102.1, 010, 025⋅=(c)The number of terms in any sequenceof numbers may be found by subtractingthe first term from the last, dividing theresult by the common difference betweenterms, and then adding 1 (because bothends must be accounted for). Thus600661100 terms.-+=List the numbers:612594600600594126606606606606++++++++++++There are 100 sums of 606. Thus the totalcan be found by computing606 1002.30, 300⋅=(d)The number of terms in any sequenceof numbers may be found by subtractingthe first term from the last (or the last fromthe first if the first is greater than the last),dividing the result by the common

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Assessment 1-1B5difference between terms, and then adding1 (because both ends must be accountedfor). Thus1000551200 terms.-+=List the numbers:100099510551099510001005100510051005++++++++++++There are 200 sums of 1005. Thus thetotal can be found by computing1005 2002.100, 500⋅=2.(a)The diagram illustrates how the numberscan be paired to form 50 sums of 101. Thesum of the first 100 natural numbers is50(101).=5050(b)A diagram similar to the one in 2a wouldillustrate how the numbers can be paired toform 100 sums of 202. Because there arean odd number of terms, the middle term,101, is left unpaired. So, the sum of thefirst 201 natural numbers is 100 • 202 +101 =20,301.3.There are2035811146 terms.-+=termsList the numbers:58592022032032025958261261261261++++++++++++There are 146 sums of 261. Thus the total canbe found by computing146 2612.19, 053⋅=4.There are many answers to this problem. Asystematic list is a good approach. Using onlytwo numbers and addition, 5 rows give 5different ways.One numberEleven minus the number11029384756We can view 6 + 5 as a different way than 5 + 6and continue in this manner to find 10 differentways. Or, we can find 7 more waysusing three numbers as follows.1192183174165151285.Since there are two different color socks in thedrawer, drawing only two socks does notguarantee finding a matching pair; you couldget one sock of each color. However, once youdraw a third sock, you are guaranteed to have amatching pair, since the third sock must matchone or the other of the previous two socks.6.There are 13 squares of one unit each; 4 squaresof four units each; and one square 9 units; for atotal of 18 squares.7.1357...9957...99101(5...99101)(135...99)(101)(13)97PQQP=+++++=++++-=+++-++++=-+=Q is larger than P by 97.8.A diagram will help.The next step is 120 miles + 40 miles =160milesfrom Missoula.9.(a)Marc must have five pennies to make aneven $1.00. The minimum number of coinswould have as many quarters as possible,or three quarters. The remaining20c/mustconsist of at least one dime and one nickel;the only possibility is one dime and twonickels. The minimum is 5 pennies, 2nickels, 1 dime, and 3 quarters, or11 coins.(b)The maximum number of coins is achievedby having as many pennies as possible. It isa requirement to have one quarter, onedime, and one nickel40 ,c=/so there maythen be 60 pennies for a total of63 coins.

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6Chapter 1: An Introduction to Problem Solving10.Adding all the numbers gives 99. This meansthat each row, diagonal, and column must add to993 = 33. Write 33 as a sum of the numbersin all possible ways:19113199517133171151797151351511713119++++++++++++++++Summarizing the pattern:NumberNr. sums with number32537293114133152173192Thus 11 must be in the center of the squareand 5, 9, 13, and 17 must be in the corners.One solution would be:1779311191315511.Answers may vary; two solutions might be to:(a)Put four marbles on each tray of thebalance scale. Take the heavier four andweigh two on each tray. Take the heaviertwo and weigh one on each tray; theheavier marble will be evident on this thirdweighing.(b)This alternative shows the heavier marblecan be found more efficiently, two stepsrather than three. Put three marbles in eachtray of the balance scale.(i)If the two trays are the same weight,the heavier marble is one of theremaining two. Weigh them to find theheavier.(ii)If one side is heavier, take two of thethree marbles and weigh them. If theyare the same weight, the remainingmarble is the heavier. If not, theheavier will be evident on this secondweighing.12.(a)There are:1 partridge12 days12 gifts;2 doves11 days22 gifts;3 hens10 days30 gifts;4 birds9 days36 gifts;5 rings8 days40 gifts;6 geese7 days42 gifts;7 swans6 days42 gifts;8 maids5 days40 gifts;´=´=´=´=´=´=´=´=9 ladies4 days36 gifts;10 lords3 days30 gifts;11 pipers2 days22 gifts; and12 drummers1 day12 gifts.´=´=´=´=So thegifts given the most by your truelove was 42 geese and 42 swans.(b)12222212++++=364giftstotal.13.(a)There must be 1 or 3 quarters for anamount ending in 5. Then dimes can add to$1.15 plus 4 pennies to realize $1.19. Thus:QuartersDimesPenniesTotal344$1.19194$1.19and in neither case can change for $1.00 bemade.(b)Two or zero quarters would allow anamount ending in 0. Then morecombinations of dimes or pennies couldadd to $1.00.14.If the price of 15 sandwiches equals the priceof 20 salads, each sandwich will buy204153=salads. Thus 3 sandwiches()433.==4 salads15.Use a variable and a table12 AM5 AM9 AM12 PMTT152(T15)2(T15)10---+

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Assessment 1-2A: Explorations with Patterns7So,2(T15)10322T3010322T20322T52T.26 degrees-+=-+=-===16.One way to solve this problem is to write anequation using a variable. For example, letpequal the number of puzzles Seth bought. So thenumber of trucks Seth bought would be 5 –p.Since each puzzle cost $9 and each truck cost$5, we get95(5)33925533482ppppppSo Seth bought2 puzzles. Since he bought 5gifts all together, he also bought3 trucks.Another way to solve it is to guess and check,making a table to keep track of resultsPuzzlesTrucksCost ofpuzzlesCost ofTrucksTotalcost05$0$25$2514$9$20$2923$18$15$3332$27$10$37The table shows that by buying2 puzzles and 3trucks, Seth will spend $33.17.The first line tells us that hamburgers equal $10.Using that information, we can use the secondline to figure out 4 hot dogs cost $8, so each hotdog costs $2. Going to the third line, since twohot dogs cost $4, each drink must cost $1. So onthe fourth line, one drink and one hamburgercost $11.The value of the question mark is$11.18.This is difficult to visualize, so the strategy ofexamining a simpler case and looking for apattern might be the best method to solve thisproblem. It would also help if you are able tomake a model of the shapes and act out thesituation.Imagine a333´´large cube. It would bemade up of a total of 27 small cubes. Taking offone layer all around would involve taking offthe top layer, the bottom layer, the front, theback, the left side, and the right side. Doing thisone step at a time: taking off the top layer takesaway 9 cubes. The same if you take off thebottom layer, so you’ve removed 18 cubes sofar. Taking off the front and the back wouldremove 3 more cubes each (the other 6 cubes onthose faces were removed when the top andbottom were removed). Finally, taking off theleft and right sides removes one more cubeeach. In total, 26 cubes are removed, leavingonly 1 cube.Now, imagine a444´´cube. There wouldbe a total of 64 small cubes making it.Removing the top and bottom layers takes atotal of 32 cubes away; removing the front andthe back would take away 16 more cubes;finally, removing the left and right sides takesaway 8 more cubes, leaving only 8 cubes.So, when starting with a333´´large cube,we are left with 1 small cube(1111);´´=when starting with a444´´large cube, weare left with 8 small cubes(2228).´´=Therefore, the pattern seemsto be, when given large cube ofdimension,nnn´´to find the number ofsmall cubes that make it after removing onelayer of small cubes all around the larger cube,you would take(2)(2)(2).nnn-´-´-So for a101010´´large cube, the number of smallcubes left would be888´´=512 smallcubes.Assessment 1-2A: Explorationswith Patterns1.(a)Each figure in the sequence adds onebox each to the top and bottom rows.The next would be:(b)Each figure in the sequence adds oneupright and one inverted triangle. The nextwould be:(c)Each figure in the sequence adds one boxto the base and one row to the overalltriangle. The next would be:

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8Chapter 1: An Introduction to Problem Solving2.(a)Terms that continue a pattern are17, 21,25, , … . This is anarithmeticsequencebecause each successive term is obtainedfrom the previous term by addition of 4.(b)Terms that continue a pattern are220, 270,320, … . This isarithmeticbecause eachsuccessive term is obtained from theprevious term by addition of 50.(c)Terms that continue a pattern are27, 81,243, … . This isgeometricbecause eachsuccessive term is obtained from theprevious term by multiplying by 3.(d)Terms that continue a pattern are9111310 ,10,10,… . This isgeometricbecause each successive term is obtainedfrom the previous term by multiplying by102.(e)Terms that continue a pattern are,303030193 + 10× 2,193 + 11× 2,193 + 12× 2… . This isarithmeticbecause eachsuccessive term is obtained from theprevious term by addition of302.3.In these problems, letnarepresent the nth termin a sequence,1arepresent the first term,d represent the common difference betweenterms in an arithmetic sequence, and r representthe common ratio between terms in a geometricsequence. In an arithmetic sequence,1(1) ;naand=+-in a geometric sequence11.nnaa r-=Thus:(a)Arithmetic sequence:11a=and4:d=(i)1001(1001)41994.a397=+-⋅=+⋅=(ii)1(1)4144.nann43=+-⋅=+-=-n(b)Arithmetic sequence:170 and50:ad==(i)10070(1001)50709950.a5020=+-⋅=+⋅=(ii)70(1)50705050 or.nann50n + 20=+-⋅=+-(c)Geometric sequence:11 and3:ar==(i)100110013.a993-=⋅=(ii)113.nnan-13-=⋅=(d)Geometric sequence:2110 and10 :ar==(i)2 (1001)2 9910019810(10 )10(10 )1010.a19910-=⋅=⋅=⋅=(ii)2 (1)(22)10(10 )1010.nnna2n-110--=⋅=⋅=(e)Arithmetic sequence:3030119372and2:ad=+⋅=(i)3030100303019372(1001)219372992.a30193 + 106 × 2=+⋅+-⋅=+⋅+⋅=(ii)303019372(1)2.nan30193 + (n + 6) × 2=+⋅+-⋅=4.2, 7, 12, … . Each term is the 5th number ona clock face (clockwise) from, the precedingterm.5.(a)Make a table.Number of termTerm11111⋅⋅=22228⋅⋅=333327⋅⋅=444464⋅⋅=5555125⋅⋅=6666216⋅⋅=7777343⋅⋅=8888512⋅⋅=9999729⋅⋅=101010101000⋅⋅=111111111331⋅⋅=The 11thterm1331is the least 4-digitnumber greater than 1000.(b)The 9thterm729is the greatest 3-digitnumber in this pattern.(c)104= 10,000; The greatest number lessthan 104is21 21 21.9261(d)The cell A14 corresponds to the 14th term,which is141414.2744⋅⋅=6.(a)The number of matchstick squares in eachwindmill form an arithmetic sequence with15a=and4.d=The number ofmatchstick squares required to build the

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Assessment 1-2A: Explorations with Patterns910th windmill is thus5(101)4+-⋅=594.41 squares+⋅=(b)Thenth windmill would require5(1)4544nn+-⋅=+-=41n+squares.(c)There are 16 matchsticks in the originalwindmill. Each additional windmill adds12 matchsticks.This is an arithmetic sequence with116a=and12,d=so16na=+(1)12n-⋅=124n+matchsticks.7.(a)Each cube adds four squares to thepreceding figure; or 6, 10, 14, … . This isan arithmetic sequence with16a=and4.d=Thus156(151)4a=+-⋅=62 squaresto be painted in the 10th figure.(b)This is an arithmetic sequence with16a=and6.d=Thenth term is thus:6(1)4.nan=+-⋅=42n+8.Since the first year begins with 700 students,after the first year there would be 760, afterthe second there would be 820, … , and after thetwelfth year the number of students wouldbe the 13thterm in the sequence.This then is an arithmetic sequence with1700a=and60,d=so the 13thterm(current enrollmenttwelve more years)+is:700(131)60+-⋅=1420 students.9.Using the general expression for thenth term ofan arithmetic sequence with124, 000a=and931, 680 yields:a=3168024000(91)31680240008960,ddd=+-=+=the amount by which Juan’s income increasedeach year.To find the year in which his income was$45120:4512024000(1)960451202304096023.nnn=+-⋅=+=Juan’s income was $45,120 in his23rd year.10.The number that fits into the last triangle is8.The numbers inside the triangle are found bymultiplying the number at the top of the triangleby the number at the bottom left of the triangle;then subtracting from that the number at thebottom right of the triangle. So,252.8´-=11.(a)To build an up-down-up staircase with 3 stepsup and 3 steps down, each current step willhave a block placed on it; plus a block will beadded to each end. In total, there will be fivemore blocks, for a total of 9 blocks used. Tobuild an up-down-up staircase with 4 steps upand 4 steps down, each step from the previousiteration will have a block placed on it (5blocks) plus a block at each end (2 blocks),adding seven total blocks, bringing the totalto 9 + 7 = 16 blocks. The table belowillustrates the pattern:Stepsup/downBlocksaddedTotalblocks1012343594716Therefore, to build an up-down-up staircasewith 5 steps up and 5 steps down, 9 blocksneed to be added to the previous 16 blocksto arrive at a total of25 blocks.(b)Based on the table and answer above, thetotal blocks are always the square of thenumber of steps up and down. So, if thenumber of steps up and down aren, thenthe total number of blocks will ben2.12.(a)Using the general expression for the nthterm of an arithmetic sequence with151,251,naa==and1d=yields:25151(1)125150201.nnn=+-⋅=+=There are201 termsin the sequence.(b)Using the general expression for thenthterm of a geometric sequence with11,a=602,na=and2r=yields:601121(2)216061.nnnn--==-==There are61 termsin the sequence.(c)Using the general expression for thenth termof an arithmetic sequence with110,a=2000,na=and10d=yields:200010(1)10200010200.nnn=+-⋅==There are200 termsin the sequence.

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10Chapter 1: An Introduction to Problem Solving(d)Using the general expression for thenthterm of a geometric sequence with11,a=1024,na=and2r=yields:110110241(2)2211011.nnnn--==-==There are11 termsin the sequence.13.(a)22222First term:(1)2;Second term:(2)2;Third term:(3)2;Fourth term:(4)2; andFifth term:(5)2.36111827+=+=+=+=+=(b)First term:5(1)1;Second term:5(2)1;Third term:5(3)1;Fourth term:5(4)1; andFifth term:5(5)1.611162126+=+=+=+=+=(c)(1)(2)(3)(4)(5)First term:101;Second term:101;Third term:101;Fourth term:101; andFifth term:101.999999999999999-=-=-=-=-=(d)First term:3(1)2Second term:3(2)2Third term:3(3)2Fourth term:3(4)2andFifth term:3(5)21;4;7;10;13.-=-=-=-=-=14.Answers may vary; examples are:(a)If5,n=then5552516.+=¹+=(b)If2,n=then22(24)636+==does not equal222420.+=15.(a)There are 1, 5, 11, 19, 29 tiles in the fivefigures. Each figure adds 2n tiles to thepreceding figure, thusa6the 6th term has2912.+=41 tiles(b)21, 4,9, 16, 25,n=… . Adding(1)n-ton2yields1, 5, 11, 19, 29,… , which is theproper sequence. Thus the nthterm has.2+1-nn(c)If2(1)1259;nn+-=Then212600.nn+-=This implies(35)(36)0, so35.nnn-+==There are 1259 tiles in the35th figure.16.The nth term of the arithmetic sequence is200(200).n+The sequence can also begenerated by adding 200 to the previous term.Thenthterm of the geometric sequence is 2n.The sequence can also be generated bymultiplying the previous term by 2. Make atable.Number ofthe termArithmetictermGeometricterm716001288180025692000512102200102411240020481226004096With the12th term, the geometric sequence isgreater.17.(a)Start with one piece of paper. Cutting itinto five pieces gives us 5. Taking each ofthe pieces and cutting it into five piecesagain gives5525⋅=pieces. Continuingthis process gives a geometric sequence: 1,5, 25, 125, … . After the 5thcut there are55= 3125pieces of paper.(b)The number of pieces after the nth cutwould be.n518.(a)For an arithmetic sequence there is acommon difference between the terms.Between 39 and 69 there are threedifferences so we can find the commondifference by subtracting 39 from 69 anddividing the answer by three:693930 and 30310.The commondifference is 10 and we can find themissing terms: 39 – 10 =29and 39 + 10 =49and 49 + 10 =59.(b)For an arithmetic sequence there is acommon difference between the terms.Between 200 and 800 there are threedifferences so we can find the commondifference by subtracting 200 from 800 anddividing the answer by three:800200600 and 6003200.The

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Assessment 1-2B11common difference is 200 and we can findthe missing terms: 200 – 200 =0and 200 +200 =400and 400 + 200 =600.(c)For a geometric sequence there is acommon ration between the terms.Between 54and 510there are three commonratios used so we can find the commonratio by dividing 510by 54and then takingthe cube root:13104662555and (5 )5 .Thecommon ratio is 52 and we can find themissing terms:42426255,55,55.26855519.(a)Let’s call the missing terms a, b, c, d, e andf, then the sequence becomes:, ,1,1, ,, ,11010111121123235358.a bc d e fbbabaacccdddcdeeedeffThe missing terms are1, 0, 2, 3, 5, and8.(b)Let’s call the missing terms a, b, c, and d,then the sequence becomes:, , ,10,13,,36,5910133103107734101323a b cdccbcbbabcaddThe missing terms are-4, 7, 3, and23.(c)If a Fibonacci-type sequence is a sequencein which the first two terms are arbitraryand in which every term starting from thethird is the sum of the previous two terms,then we can add 0 and 2 to get the thirdterm and continue the pattern:022224246461061016101626The missing terms are2, 4, 6, 10, 16,and26.20.(a)Year 180.05(80)84Year 284.05(84)88.2Year 388.2.05(88.2)92.61Year 492.61.05(92.61)97.2405Year 597.2405.05(97.2405)102.102525.$102.10(b)This is a geometric sequence with180aandr= 1.05, so the price afternyears is80 • 1.05n.Assessment 1-2B1.(a)In a clockwise direction, the shaded areamoves to a new position separated from theoriginal by one open space, then two openspaces, then by three, etc. The separation ineach successive step increases by one unit;next would be:(b)Each figure in the sequence adds one rowof boxes to the base. Next would be:(c)Each figure in the sequence adds one boxto the top and each leg of the figure. Nextwould be:2.(a)Terms that continue a pattern are18, 22,26, … . This is anarithmeticsequencebecause each successive term is obtainedfrom the previous term by addition of 4.(b)Terms that continue a pattern are39, 52,65, … . This is anarithmeticsequencebecause each successive term is obtainedfrom the previous term by addition of 13.(c)Terms that continue a pattern are44, 45,46, … . This is ageometricsequencebecause each successive term is obtainedfrom the previous term by multiplying by4.

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Solution Manual for A Problem Solving Approach to Mathematics for Elementary School Teachers, 13th Edition - Page 15 preview image

12Chapter 1: An Introduction to Problem Solving(d)Terms that continue a pattern are214, 218,222, … . This is ageometricsequencebecause each successive term is obtainedfrom the previous term by multiplying by24.(e)Terms that continue a pattern are,....505050100 + 10× 2,100 + 12× 2,100 + 14× 2This is anarithmeticsequence becauseeach successive term is obtained from theprevious term by adding by 2•250.3.In these problems, anrepresents the nthterm in a sequence, a1represents the first term,d represent the common difference betweenterms in an arithmetic sequence, and rrepresents the common ratio between terms in ageometric sequence.In an arithmetic sequence,1(1);naand=+-in a geometric sequence,11.nnaa r-=Thus:(a)Arithmetic sequence:12a=and4.d=(i)1002(1001)4.a398=+-⋅=(ii)2(1)4244.nann42=+-⋅=+-=-n(b)Arithmetic sequence:10a=and13.d=(i)1000(1001)13.a1287=+-⋅=(ii)0(1)13.nan13n13=+-⋅=-(c)Geometric sequence:14a=and4.r=(i)9910044.a1004=⋅=(ii)144.nna4-=⋅=n(d)Geometric sequence:212a=and42 .r=(i)24 9923961002(2 )22.a3982=⋅=⋅=(ii)24 (1)2442(2 )22.nnna4n22---=⋅=⋅=(e)Arithmetic sequence:505051110042and222:ad=+⋅=⋅=(i)50511005151515010042(1001)210022992100101210010122.a50100 + 202 × 2=+⋅+-⋅=+⋅+⋅=+⋅=+⋅⋅=(ii)50515151515010042(1)210022(1)2100(1)2100(1)22.nannnn50100 + 2(n + 1)2=+⋅+-⋅=+⋅+-⋅=++⋅=++⋅⋅=4.The hands must move 8 hours to move from 1to 9 on the clock face. To move from 9 to 5, thehand must move 8 hours also. To move from 5to 1, the hand must move another 8 hours. If weadd 8 hours to 1 o’clock, we will land on the 9.This pattern will continue, so the next threeterms are9, 5, 1.5.(a)Answers may vary: two possible answersare:(i)2e. ., 13574 .nsg2The sum of the firstodd numberis;+++=n(ii)()21722e. .,13574 .gSquare the average of the first andlast terms;++++==(b)There are3512118-+=terms in thissequence.(i)213573518.324+++++==(ii)()21352218.324+==6.(a)Note that 5 toothpicks are added to formeach succeeding hexagon. This is anarithmetic sequence16a=and5,d=

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Solution Manual for A Problem Solving Approach to Mathematics for Elementary School Teachers, 13th Edition - Page 16 preview image

Assessment 1-2B13so106(101)5a=+-⋅=695+⋅=51toothpicks.(b)nhexagons would require6(1)5655nn+-⋅=+-=51n+toothpicks.7.(a)Looking at the second figure, there are 3 +1 = 4 triangles. In the third figure, there are5319++=triangles. The fourthfigure would then have753116+++=triangles. Analternative to simply adding7, 5, 3,and 1together is to note that718+=and538.+=There are422=of thesesums, and2816.⋅=Then the 100thfigure would have10099199+=triangles in the base,9998197+=triangles in the second row, and so on untilthe 100th row where there would be 1triangle. 1991200;+=1973200;+=etc. and so the sum ofeach pair is 200 and there are100250=ofthese pairs.5020010, 000,⋅=or10,000trianglesin the 100th figure.(b)The number of triangles in thenth figure is2(number of triangles in base1).n+Thenumber of triangles in the base is(1),nn+-or21.n-(21)n-+12 .n=Then22(2 )nnn=, orn2trianglesin thenth figure.8.This is a geometric sequence with115,3602a=and12r=. Thenth term of ageometric sequence is11;nnaa r-=thus the10th term would be()101215360.=15 litersNote the progression of terms in the followingtable:AfterDayAmount of Water Remaining11215,3607680 liters⋅=21276803840 liters⋅=9126030 liters⋅=10123015 liters⋅=9.This is an arithmetic sequence with1168a=(i.e., 8 a.m. plus 10 minutes, or1060of an hour)and56d=(or5060of an hour). Thus518668(81)14, ora2:00 p.m.=+-⋅=(14 is 2:00 p.m. on a 24-hour clock.)10.Answers will be a rotation of the followingfigure:11.(a)In the first drawing, there are 6 toothpicks;in the second drawing, there are 10, and thethird drawing has 14 toothpicks. This it isan arithmetic sequence with16,4.ad==So, for the tenth figure, wewould have1010106(101)4694aaa42.=+-⋅=+⋅=(b)Thenthterm for this arithmetic sequence is6(1)4644nnnanana4n + 2.=+-⋅=+-=(c)For a total of 102 toothpicks, to find thefigure, we have102421004.nnn25=+==12.(a)Thenthterm for this geometric sequence is13.n-Thus99133.n-=So991,n=-and100.n=There are100 termsin the sequence.(b)Thenthterm for this arithmetic sequenceis9(1)4.n+-⋅Thus3539=+(1)4.n-⋅Solving forn,87.n=Thereare87 termsin the sequence.(c)Thenthterm for this arithmetic sequence is38(1)1.n+-⋅Thus23838=+(1)1.n-⋅Solving forn,201.n=There are201 termsin the sequence.

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