Solution Manual for Mathematical Reasoning for Elementary Teachers, 7th Edition

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Guide for Activities1Chapter 1Thinking CriticallyActivity 1Box 1: UNDERSTAND THE PROBLEM1.Answers will vary.2.Sample Responses: Shapes are congruent if they have exactly the same shape and size. Shapes arecongruent if they can be made to match exactly by placing one on top of the other.3.a.b. 3 ways4.Answers will vary.5.Answers will vary.Box 2: DEVISE A PLAN1.Squares could be moved around to try to construct the staircases or they could be sketched on graphpaper.2.Simplify the problem, look for a pattern, and make a table3.Answers will vary.Box 4: LOOK BACK3.a.3 waysb.Since the two staircases must fit together to form a rectangle with area 120, find the dimensionsof all rectangles with integral dimensions and an area of 120. The possible dimensions are1×120, 2×60, 3×40, 4×30, 5×24, 6×20, 8×15, 10×12.To form a staircase, both dimensions must be greater than 1, and one dimension must be evenand the other odd. This leaves 3 possibilities, a 3×40 rectangle, a 5×24 rectangle, and an8×15 rectangle.The 3×40 rectangle can be divided into two 3-step staircases. The bottom step is 21 squareslong, the middle step is 20 steps long, and the top step is 19 squares long.The 5×24 rectangle can be divided into two 5-step staircases. The bottom step is 14 squareslong, the next step is 13 squares long, the middle step is 12 steps long, the fourth step is11 squares long and the top step is 10 squares long.The 8×15 rectangle can be divided into two 8-step staircases. The steps are 11, 10, 9, 8, 7, 6,5, and 4 steps long.4.To determine whethernsquares can be used to construct two congruent staircases, find a pair offactors ofn. There is a pair of staircases for each factor pair in which both factors are greater than 1and one factor is odd and the other even.

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2Chapter 1 • Thinking CriticallyActivity 2Box 11.2.3.4.6,3,4,5,6,75.14,17,20,23,26,296.25,18,11,4,–3,–107.15,21,28,36,45,558.28,39,52,67,84,1039.32,64,128,256,512,102410.9,3,1,13,19,12711.34,55,89,144,233,37712.Each term of the sequence in Exercise 5 is the number of squares in the corresponding term of thesequence in Exercise 1. Similarly, each term of the sequence in Exercise 7 is the number oftriangles in the corresponding term of the sequence in Exercise 3.13.a.The number of triangles in each term of the sequence in Exercise 3 is half the number ofsquares in the corresponding term of the sequence in Exercise 2.b.Sample Response:(Continued on next page.),

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Guide for Activities3Activity 2 ContinuedBox 21.7,12,17,22,27,32,37,42,472.2,5,8,11,14,17,20,23,26Each term of the sequence in Exercise 1 on page 5 is a rectangular array 3 squares tall andnsquareslong (wherenis the term number) with the top right-hand square removed. Thus the number ofsquares used to form each term is 1 less than 3 timesn.3.44,41,38,35,32,29,26,23,204.7,13,23,37,55,77,103,133,1675.2,6,12,20,30,42,56,72,90Each term of the sequence in Exercise 2 on page 5 is a rectangular array of squares. The height ofthe rectangle is equal to the term number and its length is 1 more than the term number. Since thearea of the rectangle is the length times the width, the number of squares in the array isthe termnumbertimestheterm number plus 1. That is,the term numbertimesthe next term number.6.In Exercises 1–3, the difference between successive terms is a constant. In Exercise 1 the differenceis always 5, in Exercise 2 it is always 3, and in Exercise 3 it is –3. In each case, the difference is theconstant by which the term number was multiplied.The differences between successive terms in Exercises 4 and 5 are not constant. Note that, in thesecases, the rule for generating the sequence does not involve multiplying the term number by aconstant.7.Answers will vary. One possibility is to look for a pattern in the differences between successiveterms and to continue that pattern to find the next five terms in the sequence. For example, inExercise 5 the differences are the sequence of even numbers 4, 6, 8, 10, 12, 14, 16, 18… Thus, thenext five terms would be 90 +20= 110, 110 +22= 132, 132 +24= 156, 156 +26= 182, and182 +28= 210.8.2, 4, 8, 16, 30,52,84,128, 186,2609.The first four terms of the sequences are the same, but from there on they are different. So nomatter how many of the initial terms of a sequence you may know, there may be more than one wayto extend the sequence.

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4Chapter 1 • Thinking CriticallyBeginningof monthNumber ofnewbornpairsNumber of1-month-old pairsNumber ofpairs 2 monthsold or olderTotalnumber ofpairs110012010131012411135212563238753513Activity 3Box 11.2.a.The number of 1-month old pairs in monthN(N≥2) is equal to the number of newborn pairs inmonthN– 1.b.The number of pairs 2 months old or older in monthN(N≥2) is equal to the sum of thenumber of 1-month old pairs and the number of pairs 2-months old or older in monthN– 1.c.From the second month on, the number of newborn pairs each month is equal to the number ofpairs 2 months old or older.Box 2: CREATING A SPREADSHEET2.a.1 was added to the 1 in cell A2, so a 2 appears in cell A3.b.The formula “= A3 + 1” is copied into cell A4, the formula “= A4 + 1” is copied into cell A5,and so on. Thus for each cell A4 through A14, 1 is added to the number in the preceding cell.The result is that the numbers 3 through 13 appear in cells A4 through A14 respectively.c.To find the number of pairs after one year, you must extend the table to the beginning of the13th month.3.a.The content of cell B2 was copied into cell C3, so a 1 appears in cell C3.7.233 pairs8.From the third term on, each term equals the sum of the two preceding terms.MonthNewborn1 Month Old2 Months OldTotal Pairs11001201013101241113521256323875351388582191381334102113215511342134891255345514413895589233(Continued on next page.)

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Guide for Activities5Activity 3 ContinuedBox 3: Looking for PatternsFibonacci #sSumsFib. # – 1SquaresSum of SquaresProductsQuotients110111112012222414661.5372915151.66666666751242540401.68207641041041.6251333121692732731.6153846152154204417147141.6190476193488331156187018701.61764705955143543025489548951.6181818188923288792112816128161.6179775281443761432073633552335521.6180555562336092325428987841878411.6180257513779863761421292299702299701.61803713561015966093721006020706020701.6180327879872583986974169157623915762391.6180344481597418015962550409412664841266481.618033813258467642583667705610803704108037041.61803405641811094541801748076128284465282844651.61803396367651771067644576522574049690001.c.The first sum is 1 less than the third Fibonacci number, the second sum is 1 less than the fourthFibonacci number, and so on—thenth sum is 1 less than Fibonacci numbern+ 2.2.c.The numbers in the two columns are the same.d.The sum of the squares of the firstnFibonacci numbers equals the product of Fibonacci numbernand Fibonacci numbern+ 1.3.The quotients approach 1.61803.

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6Chapter 1 • Thinking CriticallyActivity 4Box 1Term Number12345678Term411182532394653Difference7777777Constant difference = 7TermConstantWhatToNumberDifferenceWas Done?Get1´7→7___–3___=4First Term2´7→__14____–3___=11Second Term3´__7__→__21____–3___=__18_Third Term10´__7__→__70____–3___=__67_Tenth Term50´__7__→_350____–3___=_347_Fiftieth TermEach term is 3 less than 7 times the term number,Rule:nth term = 7n– 3Box 2Rule25th100thTermTerm1.25,29,33,374T+ 51054052.30,37,44,517T– 51706953.5,7,9,112T– 5451954.90,88,86,84–2T+ 10050–1005.65,62,59,56–3T+ 805–220

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Guide for Activities7Activity 52.Number ofPeople2468102NNumber ofPairs12345NMin Numberof Moves38152435N2+ 2NSequence ofMovesRLRRLLRRLLRRLLRRRLLLRRRLLRRLLRRRLLLLRRRRLLLLRRRLLRRLLRRRLLLLRRRRRLLLLLRRRRRLLLLRRRLLR1R, 2L, 3R, ... , (N– 1)L,NR,NL,NR, (N– 1)L, ... ,3R, 2L, 1R ifNis odd1R, 2L, 3R, ... , (N– 1)R,NL,NR,NL, (N– 1)R, ... ,3R, 2L, 1R ifNis even4.12 people5.255 moves6.Number ofPeople13579NMin Numberof Moves15111929(N2+ 4N– 1)/4Sequence ofMovesRRLRRLRLLRRRLLRRLRLLRRRLLLRRRRLLLRRLRLLRRRLLLLRRRRRLLLLRRRRLLLRRLThe Legend of the Tower of Brahma1.Number of Disks12345nNumber of Moves13715312n– 12.a.210– 1 = 17 min 3 secb.230– 1≈34 yearsc.250– 1≈36 million yearsd.264– 1≈585 billion yearsActivity 6Box 11.The results for Steps 1-5 will vary depending on the number selected by each person.2.Sample Response: 10 because that was the answer I got for Persons 1 through 5.3.Step 1:nStep 2: 3nStep 3: 3n+ 30Step 4: (3n+ 30) ÷ 3 =n+ 10Step 5:n+ 10 –nThe result is 10.(Continued on next page.)

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8Chapter 1 • Thinking CriticallyActivity 6 ContinuedBox 31.Answers will vary.2.The display on the calculator shows the month and day of a person’s birthday. For example, 1021for October 21.3.Step 1mStep 2:5mStep 3:5m+ 20Step 4:20m+ 80Step 5:20m+ 73Step 6:100m+ 365Step 7:100m+ 365 +dStep 8:100m+ dIn Step 8, multiplying the month,m, by 100 moves the month to the hundreds place in the finalanswer. Ifmis a single digit number, that digit will be in the hundreds place of the answer. Ifmis atwo digit number, its tens digit will be in the thousands place of the answer and its units digit willbe in the hundreds place. When the day,d, is added, the digits ofdwill appear in the tens and unitsplaces of the final answer.Activity 72.a.1Subtracting 1 leaves 4 on the display.If Player A subtracts 1, that leaves 3 on the display and B wins by subtracting 3.If A subtracts 2, that leaves 2 on the display and B wins by subtracting 2.If A subtracts 3, it leaves 1 on the display and B wins by subtracting 1.b.To be certain to win, you want to be the first player. At the end of your turn you want to leave16, 12, 8, or 4 on the display.3.a.In this game you want to be the second player. At the end of your turn you want to leave 20, 15,10, or 5 on the display.b.In this game you want to be the second player. At the end of your turn you want to leave 40, 30,20, or 10 on the display.Activity 8Box 11.15802.Sample Response: From clues b and d, I knew the number had to be a multiple of 10. Thesenumbers are easy to pick out since the must have a 0 in the units place.Box 21.153 (or 1 if it isn’t a multi-digit number)2.Sample Response: Since 73 = 343, from the last two clues I knew the digits in the number had to beless than 7. From the second clue, I knew the digits could only be 1, 3, or 5.Box 325 cards or 85 cards

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Guide for Activities9Activity 9Box 11.No. If there were green gumballs in the jar, it would be labeled correctly, but no jar has a correctlabel.2.Red3.No. If so, the jar labeled GREEN would have GREEN gumballs, but all labels are incorrect.4.The correct label for the jar labeled Red - Green is Red, the correct label for the jar labeled Red isGreen, and the correct label for the jar labeled Green is Red - Green.Box 2Greatest amount = $1.19Coins = 1 half dollar, 1 quarter, 4 dimes, 4 penniesBox 31.AlabamaAlaskaOklahomaMinnesotaCamelliaForget Me NotMistletoeLady's slipperYellowhammerWillow ptarmiganFlycatcherLoon2.Clues b, d, and c are the keys to solving the problem.From b, we know that Alaska and forget me nots go together.From d, we know that Alabama and yellowhammer go together.From c, either camellia or lady 's slipper go with Minnesota. If camellia and Minnesota go together,then the state bird can't be the willow ptarmigan (clue e) or the loon (clue c) so it must be theflycatcher. But this would contradict clue a, so Minnesota, lady's slipper, and loons must gotogether.Box 4Two answers are possible. Freddie or Susie can be in either First or Fourth place.FirstSecondThirdFourthYellowPurpleGreenRedFreddie or SusieLizJoeSusie or Freddie

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Guide for Activities49Chapter 10Measurement: Length, Area, and VolumeActivity 1Box 11.Answers will vary.2.Answers will vary.Box 21.Answers will vary.2.Answers will vary.3.a.Answers will vary.b.Answers will vary.c.On the geo-board construct a right triangle with a horizontal base of 3 and a vertical altitudeof 2. The area is 3 square units. Pick up the geoband at the intersection of the hypotenuse andthe altitude. Move the band one peg to the right. Continue stretching theband, always movingit one peg to the right so that the triangle has a base of 3 units and an altitude of 2 units. Theperimeter will increase without bound, but the area will always be 3 square units. Thus it ispossible to obtain a perimeter of approximately100, 1000, or any large number.Activity 2Box 11.a rectangle2.Sample Response: The angles are all right angles.3.They base of the parallelogram has the same length as the base of the rectangle and the altitude ofthe parallelogram equals the altitude of the rectangle.4.Answers will vary.5.The areas are equal.6.a.Count the number of square units in the rectangle.b.Use the formula,A= base×altitude.Box 21.Answers will vary.2.The areas are equal.3.A= base×altitude.

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50Chapter 10 • Measurement: Length, Area, and VolumeActivity 3Box 11.a parallelogram2.Sample Response: SidesMIandEKare parallel and congruent.3.A= 124.125.A= 6Box 21.The area of the triangle is one-half the area of the parallelogram.2.The base and altitude of the triangle are the same as the base and altitude of the parallelogram.3.A=b×a4.A=12b×aActivity 4Box 11.a parallelogram2.MLis parallel toJK.3.Answers will vary.4.The area of the trapezoid is one-half the area of the parallelogram.5.Answers will vary.Box 21.The area of the trapezoid is one-half the area of the parallelogram.2.The measure of the base of the parallelogram is equal to the sum of the measures of the bases of thetrapezoid.3.A(parallelogram) =bxa, soA(trapezoid) =12(b1+b2)a.

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Guide for Activities51Activity 5Box 11.m∠X= 56°,m∠Y= 90°,m∠Z= 34°,2.right3.Area of SquareA+ Area of SquareB= Area of SquareC.Box 21.Area of SquareA+ Area of SquareB= Area of SquareC.2.rightActivity 6Box 11.The area of the square is 64.2.The area of the rectangle is 65.3.The area of the other figure is 63.4.AB+BC>AC.This shows thatACis not a straight line, and the parts do nit fit together as shownin the drawing.Box 2The next four Fibonacci numbers are 13, 21, 34, and 55.Box 31.The area of the square is 1692.The area of the rectangle is 168.3.The area of the other figure is 170.(Continued on next page.)

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52Chapter 10 • Measurement: Length, Area, and VolumeActivity 6 ContinuedBox 41.212.Divide two sides of the square into segments of 8 and 133.The area of the rectangle is 442. The area of the other figure is 440.EXTENSIONSFor any Fibonacci numbern, the area of the square isn2. The area of the related rectangle formed bypartitioning the square and rearranging the pieces is the product of the preceding Fibonacci number andthe next Fibonacci number. If the area of the rectangle is one more than the area of the square, the area ofthe other figure is one less. If the area of the rectangle is one less than the area of the square, the area ofthe other figure is one more.Activity 7Box 1Area of theLargest SquareArea of theSmallest SquareArea of the ThirdSquareSum of the Areasof the TwoSmaller SquaresIs the TriangleAcute, Right, orObtuse?6493645obtuse1003664100right36162541acuteThe remaining entries in the table will vary.Box 2Entries in the tables will vary.a.acuteb.rightc.obtuseActivity 8Box 11.a.9, 6, 3, 2, 121b.6, 6, 5, 4, 3, 3, 2, 2, 1, 133c.12, 12, 10, 7, 4, 2472.a.24b.27c.253.a.45b.60c.72(Continued on next page.)

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Guide for Activities53Activity 8 ContinuedBox 2LengthWidthHeightNumber ofCubes352304651204446478316865.5103301436.527315956758.7135565.51.V=lwh2.Number of cubes = area of the base×height3.a.A1= 2×5 = 10A2= 5×8 = 40A3= 2×8 = 16b.V= 10×8 = 80V= 40×2 = 80V= 16×5 = 804.Volume of cylinder = area of the base×heightActivity 9Box 11.They are equal.2.They are equal.3.Answers will vary.4.35.136.Volume of a pyramid =13×area of the base×altitudeBox 21.132.Volume of a cone =13×area of the base×altitude or13(πr2)h

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54Chapter 10 • Measurement: Length, Area, and VolumeActivity 10Box 11.trapezoidal prism2.square pyramid3.rectangular prism4.triangular prismBox 2Since measures are not accurate and printing of the nets may reduce or enlarge the figures, all answers areapproximate. Measures of surface area are in cm2and measures of volume are in cm3.1.S.A.≈87V≈512.S.A.≈50V≈183.S.A.≈117V≈834.S.A.≈133V≈72EXTENSIONVolume of the prism is 54.Volume of each pyramid is 18.

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