Unit 1 – Number Sense and Algebraic Thinking Lesson 1 & 2 – Whole Number Problem Solving

Black – Whole Number Problem Solving

1. Mr. Wayne was three times as old as Colleen 5 years ago. Their total age now is 42 years. How old is Colleen now?

2. Sam bought a racket, a pair of shoes and a bag for $242 altogether. The pair of shoes cost $71 more than the racket while the racket cost $27 less than the bag. How much did the racket cost?

3. 5 books and 9 magazines cost $156 altogether. 9 books and 5 magazines cost $180 altogether. How much more does each book cost than each magazine?

4. Ryan has three times as many stamps as Hal and twice as many stamps as Jimmy. If they have 220 stamps altogether, how many more stamps does Jimmy have than Hal?

5. Lucy and Jean have 240 stickers altogether. If Lucy gives 10 of her stickers to Jean, she will have three times as many stickers as Jean. How many stickers does Lucy have?

6. Peter and Mike have 180 stamps altogether. After Mike gave 20 of his stamps to Peter, he had twice as many stamps as Peter. How many more stamps did Mike have than Peter at first?

7. If Amy gives 10 of her stamps to Robin, she will have four times as many stamps as Robin. If she gives 20 of her stamps to Robin, she will have three times as many stamps as Robin. How many stamps do they have in all?

8. 8 clocks and 5 watches cost $695 altogether. 3 clocks and 3 watches cost $300 altogether. How much more does each clock cost than each watch?

9. There were 476 paper-clips in. box A and 52 paper-clips in box B. After an equal number of paper-clips was added into each box, box A had five times as many paper clips as box B. How many paper-clips were added into box B?

10. An overnight package delivery service charged $16 for every, package safely delivered but would pay a penalty of $48 for' every package delivered damaged. If it received $9152 for a delivery of 600 packages, how many packages were safely delivered?

11. Peter, Tom and Vincent have 240 cards altogether. Peter gave some of his cards to Tom and the number of Tom's cards was doubled. Then Tom gave some of his cards to Vincent and the number of Vincent's cards was doubled. If the three boys had the same number of cards in the end, how many cards did Peter have at first?

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12. Multiplication Mystery This week we're going to use our multiplication skills... but not in the usual fashion! Instead, take a look at the problem below:

Each letter represents a digit. Your job is to discover what each is. Your short Answer this week should state the final product. Your Explanation must include the correct digits for each letter above as well as the logic you used to determine each one.

13. Victoria in Bloom Victoria, the capital of British Columbia, Canada, is proud of its reputation as the City of Gardens. Every year the residents spend one week in February, while most of Canada is still under ice and snow, counting the flowers in bloom in their city! In 2005 during that week 4,773,559,314 blossoms were counted. The record, counted in 2002, was in 8,521,514,876 flowers in bloom! http://www.tourismvictoria.com/flowercount/ Last year Rupert Prince and his friends counted the blossoms in the park near their house in Victoria. Well, they didn't actually count each one individually. To account for all the flowering trees and shrubs, the city asks people to use these approximations:  Small tree full of blossoms: 250,000 blossoms  Medium tree full of blossoms: 500,000  Large tree full of blossoms: 750,000  Small Heather bush: 500  Medium Heather bush: 1,000  Large Heather bush: 2,000 Using these guidelines Rupert and his friends found that the park had 2,005,500 flowers.

Here's one combination they might have found: Type How many? Flowers on each Total Small tree 2 250,000 500,000 Medium tree 3 500,000 1,500,000 Small bush 1 500 500 Medium bush 1 1,000 1,000 Large bush 2 2,000 4,000

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Total 9 2,005,500 Find two more combinations of trees and/or shrubs that they might have found in the park.

Tell how you solved the problem and show how you know you are correct.

Extra: Find the greatest number of trees and/or shrubs that would produce that total, and the smallest number that could be possible. Explain.

14. A Million in Time When T.J. doesn't want to do something, like clean up his toys, he'll exaggerate the time he thinks it will take to do the job. "That would take too long, Mom. It would take a million minutes to clean that up." If he's really feeling overwhelmed, he'll change that to a million hours, days, weeks...or "to infinity and beyond!" This made me wonder which of these claims is even possible, considering the life span of human beings. Do we live a million seconds? How about minutes? hours? days? weeks? To earn credit this week, you'll need to determine the longest amount of time we can reasonably expect to live. In your explanation, be sure to include the calculations for the options mentioned in the previous paragraph that are necessary to support your claim.

15. Darlene's Dart Board Darlene challenged her friends to a game of darts on her new dart board. Players take turns throwing darts. A player wins by scoring exactly 100 points. 1. List one combination of scores that will win the game. Explain how you found it. 2. Is it possible to score 100 points with more darts or fewer darts than you listed? Explain how you know. Extra: Find at least two other combinations that will win this game. List them and describe your strategy for finding them.

16. Lawn-Mowing Chore At my house, summer vacation means added chores for my three sons. Every Saturday, the lawn must be mowed. Mark starts the mower and completes 1/3 of the lawn. Sam takes over and mows exactly 1/4 of the grass. Josh then finishes off the last 700 square feet of the yard. What is the area of my yard? Bonus: The boys devised the plan for dividing the work and consider it to be fair based on their ages. My youngest son is nine. Can you match the names and ages of my three sons? Don't forget to explain how you found your answer, and say whether you think the

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plan is fair.

17. Imogene the Imp One hot summer night Imogene went to the Fun Park. The line at the Ferris Wheel was long. Imogene stepped into the line behind 39 other people. Imogene is not a patient child! She decided to sneak toward the front. Each time one person from the front of the line got on the ride, and the line inched forward, Imogene passed two people in front of her. How many people got onto the Ferris Wheel before Imogene was at the front of the line? Be sure to explain how you solved the problem.

18. Building Barns The workers at Amy's Barn Company usually work in teams of 5. It takes a crew of 5 workers 9 days to build a barn. A crew lost two of its workers. How many days should it take the remaining 3 workers to build a barn? Extra: If a customer needs to have a barn built in 7 days, what is the minimum number of workers that Amy would need to put on the job?

19. Sharing Smelt Larry spent Saturday morning ice fishing for smelt. He caught a bucketful! On the way home he stopped for lunch at his friend Daryl's house. He gave Daryl half of his catch plus 3 more fish. When he saw his mother at the post office, he gave her half of his remaining fish plus 3 more. Larry then went to his brother Calvin's house to watch a game on Calvin's large screen TV. They fried up half of the remaining smelt plus 3 more for a snack. When Larry got home, he looked in his bucket and found only 10 smelt left, which he saved for breakfast. How many fish did Larry catch? Explain how you found your answer. Show how you know it is correct. Extra: What if Larry found 20 smelt left in his bucket? Or 30? What can you say about the starting number of fish whenever the final remainder is a multiple of 10?

20. Catching Up! Marika and her father were both born on July 13. He was born in 1961 and she was born in 1991. Every year on their birthday, Marika's father smiles at her and tells her that she's catching up to his age. Marika knows he's saying something that he 4 Unit 1 – Number Sense and Algebraic Thinking Lesson 1 & 2 – Whole Number Problem Solving

thinks is funny, but she doesn't really understand what he means. He gave her five questions to help her think about it. Please help Marika answer these questions so that she will understand what he means. 1. What fraction of her father's age was Marika when she was 1 month old? 2. What fraction of his age was she when she was 1 year old? 3. What fraction of his age was she when she was 1 decade old? 4. On what date will she be 1/2 his age? 5. How do the answers to these questions show that Marika is "catching up" to her father? Extra: Marika will never actually "catch up" to her father's age. If he lives a long, healthy life, what do you think is the largest fraction of his age that she might reach?

21. More Mowing Madness Matt does such a good job with his mowing business that word-of-mouth has brought him more work than he can complete by himself. His friend Georgette has joined the business, and they have agreed to split the work evenly on each yard they mow. Since they will use only Matt's mower, which cuts a path 2 feet wide, one will mow while the other works on edging, shrub clipping, and any other yard care requested. Georgette is also a big math fan, so she would like to continue to use the "Matt- mowing" technique, starting on the outside portion of the lawn and mowing a path completely around the outside edge, then continuing with a second path around the remaining portion of the lawn, and so on.

Where should Matt and Georgette change places if each is to mow exactly half of a 72-ft. by 96-ft. yard? In other words, will Matt and Georgette switch places at the end of a complete "Matt-mowed" rectangle? If so, after which one? If not, how many more feet into the next rectangle should the first person mow before the switch?

22. Olympic Rings Repair Jellybean and I have been watching the Sydney 2000 Olympics, wishing we were in Australia! We both hope to attend an Olympics someday. Jellybean suggested we have a party to celebrate the accomplishments of all the athletes and participate in the international spirit of the Olympics. Jellybean created an Olympic rings decoration and was about to put it up when I ran into her. Jellybean was left holding the blue ring, but the yellow, black, green, and red rings had come apart and rolled to various spots in the room. I quickly apologized for my lapse in attention and helped her put the rings back together. We were careful to put the colors in the proper order. After we had completed the repairs and had hung the decoration on the wall, Jellybean asked me how many different ways there are to arrange the colors while keeping the same formation for the Olympic rings. I'm hoping you'll help us find the answer to her question. Bonus: Generalize your results above for other combinations (if you noticed a

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pattern, how would it extend to the 6th, 7th, 10th, and 100th rings?).

23. Stained Glass Window I have fallen in love with a particular shape I want made into a stained glass window for my front door. The square grid in the image shown is a six inch grid - that is, every grid line is six inches away from the next line.

The stained glass is quite expensive, so I'd like to know exactly what its area is. Also, to insert the stained glass into the clear glass, I'll have a line of lead between the two colors of glass. I need to know how long this perimeter is. You probably don't know a formula for the area or perimeter of the shape I'm making out of stained glass, but the curve is made up of circular pieces and it all fits into one 24-inch by 24-inch square. Make sure you describe how you found the area and perimeter in your answer. Bonus: Give these answers in exact form. 24. Miracle Miranda and the Mascot Miracle Miranda is the leading scorer of the Hoops basketball team. After they won the WNBA Championship, the manager offered her a contract for one million dollars a year for the next 25 years. At the same time, the Hoops also hired a new mascot. She agreed to be paid $1 the first year, $2 the second year, $4 the third year, $8 the fourth year, and so on, for the next 25 years. 1. What will be Miranda's salary in year 15? What will be the mascot’s? 2. In which year will the mascot’s salary overtake Miranda's? Be sure to explain how you solved the problem. Tell what you did and why. Extra: After 25 years who will have received more money altogether? How much more? How do you know?

25. Tatianna and the Troll A little while ago, Idle Ivan visited a Russian bridge troll. Let me remind you of the story: In an old Russian fable Idle Ivan sighed to himself: "Everybody tells me to get a job or I should make a deal with the troll. But I don't expect even he could help me get rich." No sooner had he said this than the troll appeared from under a nearby bridge. "So, you want to make money, Ivan?" asked the troll. Ivan nodded, lazily. "Then, Ivan, all you have to do is cross the bridge. Every time you do, the money in your pocket will double." Ivan headed for the bridge but the troll stopped him. "Since I'm so generous," the troll said craftily," I think you ought to give me a little for my pains. Will you give me eight rubles every time you cross the bridge?" Ivan agreed and quickly crossed the bridge. He put his hand in his pocket. As if by

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magic he found his money had doubled. So he threw 8 rubles to the troll. He crossed a second time and did the same. On the third crossing, however, his money doubled but he found when he had paid the troll he had nothing left! The troll laughed and vanished. Now Tatianna decides to try her luck. After she crosses the bridge three times she states the following: "Gee! I still have the same amount of money that I had before I met you, Mr. Troll. I'm no richer nor poorer than I was before. Bah! So much for magic bridges and all that stuff!" How much does Tatianna have in her pocket?

26. Georg Tricks the Troll A little review for those of you who've missed the earlier problems in this trilogy: Idle Ivan visited the troll and lost all his rubles. Tatianna visited the troll and neither lost nor gained rubles. And now the rest of the story... Along comes Georg, who has been quietly observing all the activity. An idea has come to him about how to really make some money, unlike his more unlucky predecessors. So he boldly approaches the troll and says, "Though I don't have much money on me, I, too, want to get rich. If you will double my rubles as you did for my two friends, Ivan and Tatianna, I promise to pay you each time I cross the bridge. However, it is I who will set the amount this time. It will be a fair fee and I will pay the same amount each time I cross. You should accept my offer, because if you don't, I'll return home right now and give you nothing. And you must admit, Mr. Troll, something is better than nothing." The troll thought for a moment and realized Georg had a point. Since it was easy for him to create money magically for others, he agreed to the deal. "Go across then, young man. But don't try to double-cross me. Trolls don't like to be double-crossed while doubling the money of those who cross their bridges!" So Georg crossed three times, and paid the troll a fixed amount each time, as agreed, after seeing his money change. As he returned home, he counted his rubles. Great! He now had twice as much as when he started the adventure. Magic does work, he thought. How many rubles did he have before he crossed the first time, and how many rubles did he pay the troll each time, given that both numbers are whole numbers less than 10? Bonus: How much would Georg have and pay if the restriction above were removed?

Solutions

1. 13 years 2. $48 3. $6

4. 20 5. 190 6. 100

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7. 200 8. $30 9. 54

10. 593 11. 140

12. My answer is that the final product was 158530. Here are the steps I took to solve the problem. 1. E must be 0 because the units digit of the answer is 0. 2. G must be 0 because 3 + G in the tens digit of the answer is 3. 3. 2 x 1B must be 30 because the first two digits of the answer are 30 as I showed above. So, B = 5 since 2 x 15 = 30. 4. C x 1B = C x 15 as I showed above. Then C x 15 = 20 or 120 or 220.....and so on, since I have shown that the second partial product 3F2G = 3F20. Because C x 15 has to end in 20, C can only be 8. Notice that C x 15 = 120, so there is a carry of 1 into the fourth column (above F) of the partial products. 5. 3 x 1B = 3 x 15 = I5 in the third partial product. Therefore I = 4 since 3 x 15 = 45. 6. Since there is a carry of 1 into the column above F, 1 + F + I = 1 + F + 4 = 8. Therefore F = 3. 7. 3 + 2 = J. So J must equal 5. 8. Since there is no carry into the H column, H = 1. 9. 3 x A = 12, so A must be equal to 4. 10. 2 x A = 2 x 4 = D, so D = 8. 11. D + 2 + 5 = 8 + 2 + 5 = 1K (K with a carry of 1), so K = 5. Finally, 415 x 382 ------830 3120 1245 ------158530 13. Here’s two combinations they might have found: 1 large tree, 1 medium tree, 1 small tree, 1 large bush, 1 med bush, and 1,005 small bushes, or 2 medium trees, 4 small trees, 2 large bushes, 1 medium bush and 1 small bush. I started by making a table like you had with the type of tree, how many of the tree, flowers on each, and total. Then I decided to put in 1 for each of the first five kinds and see how many small bushes there would need to be to have enough flowers. This is what my table looked like. Type How many? Flowers on each Total Large tree 1 750,000 750,000 Medium tree 1 500,000 500,000 Small tree 1 250,000 250,000

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Large bush 1 2,000 2,000 Medium bush 1 1,000 1,000 Small bushes ? 500 ? Then I added up all the numbers to see how many flowers there were. 750,000 + 500,000 + 250,000 + 2,000 + 1,000 = 1,503,000 flowers. Then I subtracted that from the total number to find out how many flowers I still needed to get. 2,005,500 - 1,503,000 = 502,500 flowers. Since each small bush had 500 flowers, I divided by 500 to see how many small bushes there would be. 502,500 / 500 = 1,005 small bushes. Then I updated my table.

Type How many? Flowers on each Total Large tree 1 750,000 750,000 Medium tree 1 500,000 500,000 Small tree 1 250,000 250,000 Large bush 1 2,000 2,000 Medium bush 1 1,000 1,000 Small bushes 1,005 500 1,503,000

Total 1,010 2,005,500 Then I did it again and picked other numbers to fill out my table. These are the numbers I picked. Type How many? Flowers on each Total Large tree 0 750,000 0 Medium tree 2 500,000 1,000,000 Small tree 4 250,000 1,000,000 Large bush 2 2,000 4,000 Medium bush 1 1,000 1,000 Small bushes 1 500 500 Totals 10 2,005,500 Extra: To figure out the extra, I first figured the greatest number of shrubs. Since the small bushes had the least number of flowers, then having all small bushes would give you the most number of shrubs. I figured it out by dividing the total number of flowers by the number of flowers on a small bush.

2,005,500 / 500 = 4,011 small bushes. To figure out the least number of trees and shrubs, I figured you wanted to have as many of each large tree as you could, since they had the most flowers. So I divided the number of flowers by the number of flowers on a large tree.

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2,005,500 / 750,000 = 2.674 large trees. I knew that I would have 2 large trees and then some flowers left. To see how many flowers were on 2 large trees, I multiplied. 750,000 x 2 = 1,500,000 flowers in 2 large trees. 2,005,500 - 1,500,000 = 505,500 flowers left. Then I just figured the least trees for the most flowers. I kept subtracting the number of flowers and then figuring how many I still needed. I kept going until I got to the end. I figured 1 medium tree with 500,000 flowers. 505,500 - 500,000 = 5,500 Then 2 large heather bushes for 2 x 2,000 = 4,000 flowers. 5,500 - 4,000 = 1,500 flowers Then 1 large heather bush for 1,000 flowers. 1,500 - 1,000 = 500 flowers. Then 1 small heather bush for 500 flowers. 500 - 500 = 0 To add up the total number of trees and bushes, I went back to see how many of each I had. 2 + 1 + 2 + 1 + 1 = 7 trees and bushes for the smallest number. I looked to see if there was any combination that wouldn’t have at least one small bush. I found that it was not possible because that was the only thing that had 500 flowers. Everything else had “0” in the hundreds place, tens place and the ones place. So you couldn’t use them to add up to get the 500 you needed in 2,005,500.

14. In an average life span of 80 years, you would expect to live for 2,524,608,000 seconds. In minutes it would be 42,076,800. In hours it would be 701,280. In days it would be 29,220. In weeks it would be 4,160. So "T.J." would only be reasonable if he said "something" would take him a million seconds or minutes. As there are 60 seconds in one minute you have to multiply 60 by 60 to find how many seconds in an hour. 60 x 60 = 3,600. Then I determined a full day by multiplying this by 24. 3,600 x 24 = 86,400 seconds per day. To get how many seconds per year I multiplied 86,400 by 365 and got 31,536,000. As there is a Leap year every four years, I had to divide 80 by 4. This showed me that there are 20 Leap years. I then subtracted 20 from 80 to get 60. I then needed to fugrue out how many seconds in 60 years. This I did by multipliying 31,536,000 (full year in sec.) by 60 and got 1,892,160,000.

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To find out how many seconds in a Leap year I multiplied 86,400( a full day) by 366 (days in a Leap year). This is 31,622,400 seconds. Now I had to multiply this by the 20 Leap years and got 632,448,000 seconds. I then added this to my total seconds for the 60 years and got my answer of 2,524,608,000 seconds in an average 80 year life span. Going through this same process again I figured minutes. Multiplying 60 by 24 gave me the total of minutes in a day, 1,440. 1,440 x 365 (days per year) = 525,600 525,600 x 60 years = 31,536,000 For the 20 Leap years, I did 1,440 x 366 = 527,040. 527,040 x 20 (leap years) = 10, 540,800. The 20 Leap years added to the other 60 years, 31,536,000 + 10,540,800= 42,076,800 minutes in an average 80 years life span. For Hours the process became shorter. 24(hours in a day) x 365=8,760. 8,760 x 60 years = 525,600. 24 hours per day x 366 (leap year) = 8,784. 8,784 x 20 leap years = 175,680. 175,680 (20 leap years) + 525,600 (60 years) = 701,280 hours in an average 80 year life span. Figuring days was very quick. 365 x 60 = 21,900. 366 x 20 = 7,320. So, 21,900 + 7,320 = 29,220 days in an average 80 year life span. Weeks: 52 weeks per year x 80 = 4,160.

15. 1. One combination is 16,16,16,16,18,18. 2. It is not possible. To find my first combination I used the strategy of guess and check. I tried multiplying 16 by 4 and got the answer of 64. Then I added 18 to it twice because I knew that 64 plus 18 plus 18 is 100. For the second question I figured out that it is only possible with a combination of 6 numbers. The biggest number that you could use and that times 5 is less than 100. I used the number five because it is one less number than six. If 19 times 5 is to small then 6 is the smallest number. 15 is the biggest number that you can use and 15 times 7 is more than 100. I multiplied 15 by 7 to find out if there could be a combination of 7 numbers. There could not. Extra: 2 other combinations that I found are 19,19,15,15,16,16 and 17,17,18,18,15,15. My strategy for finding these combinations is that I picked three numbers that equaled 50 and put the numbers down twice because 50 times 2 is 100.

16. Area of the yard 1680 sq. ft. Bonus: Sam is 9, Mark is 12, Josh is 15. Since 1/3 + ¼ = 4/12 + 3/12 = 7/12, Josh was left to mow 5/12 of the total yard which is given to be 700 s.f. Now if 700 is 5/12 of the total, then the total is equal 700 x 12/5 = 1680 sq.ft.

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Bonus: I assumed that the youngest had the least to mow and the oldest had the biggest share. Since the proportions of areas mowed were 3/12, 4/12, 5/12 the ages should maintain the same proportions that is 3 : 4 : 5, so if the youngest is 9, the second should be 12=(9 x 4/3)and the oldest 15(12 x 5/4). What is left now is to identify the names: the youngest child mowed 1/4 of the yard the 2nd 1/3 of the yard and the oldest 5/12 of the yard. So Sam is 9, Mark is 12, Josh is 15. Since I'm the oldest kid in my family I would say that this is unfair, but without asking, I'm sure that my brother feels differently about the fairness of this.

17. Thirteen people get on to the ferris wheel before Imogene. Extra: For 100 people 34 people would get on before Imogene. 1. I made 39 tick mark to represent each person in line. I then put an "x" through one person on the end who got on the ferris wheel and moved Imogene ahead two space on the opposite end of the line. I continued this process until Imogene was moved ahead and was next to the person getting on the ferris wheel. This operation occured thirteen times before she was next to get on. I checked my answer by adding the number of people getting on the ride and the number of spaces Imogene moved ahead. 2 + 1 = 3 I then divided the number of people in line be the sum above to confirm the number of times people got on and Imogene moved forward. 39/3 = 13 Extra: I tried 30, 50 and 90 people in line. Each time the number of people in line was divided by 3 to determine the number of people who would get on before Imogene. When the line length is evenly divisible by three the quotient is the exact number of people getting on ahead of Imogene. When the length is not evenly divisible by three, the quotient must be rounded up to the next whole number. The rule is; the sum of the # of people getting on + the # of people she passes, divided into the line length equals how many people will get on the ride before Imogene.

18. It will take 3 workers 15 days to build a barn

Extra: Amy will need 7 workers, but one of them can be assigned to other projects 57% of the time. List of Known: • It takes 5 workers 9 days to build a barn. • There are 3 workers • It takes 45 worker days to build a barn The question is: How many days will it take with 3 workers. I need to know the total amount of time the workers put into the project. And that equals 5 workers times 9 days is 45 worker days (5• 9). 5 workers • 9 days = 45 worker days.

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(45 worker days) ------= 15 days. ( 3 workers ) It will take 3 workers 15 days to build a barn. Extra: • It still takes 45 worker days, to build a barn. there are only 7 days to build it. (45 worker days) ------= 6.43 ( 7 days) Amy will need 7 workers, but one of them can be assigned to other projects 57% of the time.

19. Larry caught 122 fish. Extra: If he found 20 fish left in his bucket, he must have caught 202 fish. If he found 30, he must have caught 282. I decided to try doing this problem backwards. The last person Larry shared his fish with was Calvin. So I started with Calvin. I pretended Larry got mean and took back the three extra fish that he gave to Calvin. Then Larry and Calvin would have one half each of the fish Larry had when he went to Calvin’s house. This way Larry would have 13 fish, and if 13 fish = 1/2, 1 whole equals 26!!! So I knew when Larry went to Calvin’s house he had 26 fish! With this new information I went on. Larry gave a half and 3 fish to Larry’s mom. Again he got mean and took back those 3 fish, adding them to his 26 fish. 26 + 3 = 29. Since 26 equals one half, one whole would be 26 + 26 = 58!!! So I knew when Larry met his mom he had 58 fish. Then for the last time (backwards), Larry gave away a half and 3 fish to Daryl. Again, Larry took away the extra 3 fish back. 58 + 3 = 61 (1 half). So one whole would be 61 + 61 = 122!!! So, believe it or not, Larry caught 122 fish. To Check: ½ x 122 = 61, and 61 – 3 = 58 ½ x 58 = 29, and 29 – 3 = 26 ½ x 26 = 13, and 13 – 3 = 10! Larry ended up with 10 smelt, so I was correct. For the Extra!!! If Larry found 20 smelt left in his bucket, he should have caught 202 smelt! Here’s how I found it out:

I solved this exactly like I did when Larry ended up with 10 fish. I went backwards. I imagined Larry giving a half and 3 more fish to Calvin. Then, taking back the 3 fish, he added them to his 20, making 23, which was a half. So 23 + 23 = 46 (1 whole). He had 46 fish when he came to Calvin’s house. So I went on like this in my head, until I got 202 fish. Then I checked; Here is how I did it.

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½ x 202 = 101, and 101 – 3 = 98 ½ x 98 = 49, and 49 – 3 = 46 ½ x 46 = 23, and 23 – 3 = 20! Then I went on to 30 fish at the end, and I thought just like before and got 282 fish. To check, I did: ½ x 282 = 141, and 141 – 3 = 138 ½ x 138 = 69, and 69 – 3 = 66 ½ x 66 = 33, and 33 – 3 = 30!!! To help me find out why the starting numbers always end with 2, I made a chart. If Larry ends up with… he should have caught… (final remainder) (starting number) 0 42 10 122 20 202 30 282 I noticed in the left column the number is always 10 more. In the right column, the number is always 80 more. I wondered why the difference was 80. I thought: Larry shared half of 80 with Calvin (40), and half of 40 with Mom (20), and lastly, half of 20 with Daryl (10)! Note: Why I didn’t put -3, +3, etc. is that it’s already done when I came up with 42. So I think that’s a good reason why the difference in the starting number is 80, and the difference in final remainder is 10. So you see, since 80 ends with 0 in the ones column, if Larry ended up with 0 fish and he caught 42 (see chart), all the other starting numbers have to end with 2 in the ones column.

20. #1: Marika at 1 month old is 1/361 of her father's age #2: Marika at 1 year old is 1/31 of her Father's age. #3: Marika at age 10 is 1/4 of her Father's age. #4: At age 30 Marika will 1/2 the age of her Father. #5: The Ratio's Prove that Marika is catching up to her father because she is becoming less ; less younger than him, ratio wise. #1: Marika at 1 month old is 1/361 of her father's age Process: 1991 Marika's Age -1961 Father's age ------30 Difference Between *12 Convert to months ------360 Difference in months + 1 Add one month for increase in father's age ------361 Age of Father in months 1 Age of Marika in months 361 Age of Father in months

14 Unit 1 – Number Sense and Algebraic Thinking Lesson 1 & 2 – Whole Number Problem Solving

1/361 Ratio of Marika to Father #2: Marika at 1 year old is 1/31 of her Father's age. Process: 30 Difference in ages (years) + 1 Add 1 year for increase in age ------31 Father's age 1 Age of Marika in years 31 Age of Father in years 1/31 Ratio of Marika to Father

#3: Marika at age 10 is 1/4 of her Father's age. Process: 30 Difference in ages (years) + 10 Add 10 year for increase in age ------40 Father's age 10 Age of Marika in years 40 Age of Father in years

1/4 Reduced Ratio of Marika to Father #4: At age 30 Marika will 1/2 the age of her Father. Process: 30 Difference in ages (years) * 2 Multiply by two for 1/2 ratio ------60 Father's age 30 Age of Marika in years 60 Age of Father in years

1/2 Reduced Ratio of Marika to Father #5: The Ratio's Prove that Marika is catching up to her father because she is becoming less and less younger than him, ratio wise. In actuality she will always be 30 years younger than him. Extra: If her father lived to be 100 years old, then the largest ratio Marika could ever be would be 7/10, However that is extreamly unlikely even if he did live a very long life.

21. I found that Matt mows exactly 6 paths of the yard to mow half. To solve this problem, I started out by drawing a rectangle with a length of 96 feet and a width of 72 feet. Next I calculated the area of the rectangle by multiplying 96 feet x 72 feet = 6912 square feet. One half of 6912 = 3456 square feet. Now I know that Matt should mow 3456 square feet. To figure out how many rectangles Matt has to mow, I first made a path just inside the border that went all the way around the rectangle and represented a width of 2 feet. On the long side of the rectangle, I wrote 96 for 96 feet. On the short side I

15 Unit 1 – Number Sense and Algebraic Thinking Lesson 1 & 2 – Whole Number Problem Solving

wrote 72 for 72 feet. The length from end point to endpoint on the 96-foot side, is 96 feet long, but the length from end point to endpoint on the 72 foot side is 68 feet long. This is because at each end of the 68-foot side there would be 2 feet that would already have been mowed when the long side was mowed. Therefore, the perimeter would be 68 + 68 + 96 + 96, which equals 328 feet x 2 feet = 656 square feet. I multiplied by 2 because the lawnmower cuts a 2 foot path. Next I subtract 656 from 3456 = 2800. Next I calculate the second rectangle, which will have a length of 92 feet and a width of 64 because I have to subtract out the area already mowed. Once again I compute the perimeter by adding 92 + 92 + 64 + 64 = 312. Then I multiply 312 by 2 = 624. I subtracted 624 from 2800 = 2176. After the 2nd rectangle, Matt would have 2176 more square feet to mow. Next I calculate the third rectangle, which will have a length of 88 and a width of 60. Once again I compute the perimeter by adding 88 + 88 + 60 + 60 = 296 square feet. Then I multiply 296 square feet by 2 feet = 592 square feet. I subtract 592 from 2176 = 1584 square feet left to go. At this point, I saw a pattern. I saw that every time I made a new path, the new path is 16 feet shorter. So for the fourth rectangle, I just subtract 16 from 296 and I get 280. I multiply 280 x 2 = 560 square feet, which I subtract from 1584 to get 1024. For the fifth, rectangle I subtract 16 from 280 and I get 264 which I multiply by 2 to get 528. I subtracted 528 from 1024 to get 496. For the sixth rectangle I subtracted 16 from 264 to get 248 and multiplied by 2 to get 496. When I subtract 496 from 496 I see that after mowing the sixth rectangle Matt is done with his turn.

22. There are 120 different ways to arrange the 5 rings. Bonus - There are 720 way to arrange 6 rings, 5040 ways to arrange 7 rings, and 3,628,800 ways to arrange 10 rings. First we found the possible combinations for two rings. Blue, Yellow and Yellow, Blue So for 2 rings there are 2 combinations. Then we did three rings. Blue, Yellow, Black - Blue, Black, Yellow - Black, Blue, Yellow Black, Yellow, Blue - Yellow, Blue, Black - Yellow, Black, Blue So for 3 rings there are 6 combinations of colors. We did the same method for four rings and found 24 combinations for four rings. Next we looked at our results - 2 rings = 2 combinations of colors 3 rings = 6 combinations of colors 4 rings = 24 combinations of colors We found two patterns The first pattern was you take the number of combinations and multiply it by the next number of rings. Like this - 2 rings have 2 combinations so you multiply the 2 combinations by the next number of rings so 2 x 3 = 6 combinations for 3 rings. Using this method we figured out that 5 rings would have 120 combinations because 4 rings

16 Unit 1 – Number Sense and Algebraic Thinking Lesson 1 & 2 – Whole Number Problem Solving

= 24 combinations and 24 x 5 (the next number of rings) = 120. Then we found a shorter way to do it. We noticed that you just had to multiply the number of rings together like this – 1 x 2 x 3 = 6 combinations for 3 rings (we used 1 because 1 ring would be 1 combination). 1 x 2 x 3 x 4 = 24 combinations for 4 rings so 1 x 2 x 3 x 4 x 5 =120 combinations for 5 rings. For the bonus we would use our short way of finding combinations - 6 rings would have – 1 x 2 x 3 x 4 x 5 x 6 = 720 combinations of colors 7 rings would have – 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040 combinations of colors 10 rings would have – 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = 3,628,800 combinations of colors We would do 100 rings the same way. We didn’t do it though because the number would be way too big for our calculators.

23. The area of the blue shape is 288 square inches and the perimeter of it is 24 times pi (or approximately 75.36) inches. I cut out the blue section of the stained glass window and then I cut it along the grid lines. This made four blue squares and eight parts of the blue squares. I put the eight blue parts together like a puzzle and I got four more complete blue squares. That made eight blue squares in all. Since each square had an area of 36 square inches (because they were cut from a six inch grid), then to find how much area 8 squares would make I just multiplied 36 by 8 and I got 288 square inches. To find the perimeter of the blue shape, I cut it apart again into the grid squares and then put the pieces together to form two circles, each with a diameter of 12 inches. To get the circumference of a circle you just have to multiply the diameter times pi or 3.14. If the circumference of 1 circle is 12 times pi, then the circumference of the 2 circles is 24 times pi (or 75.36) inches.

24. 1. Miranda's salary in year 15 is $1,000,000. Mascot's salary in year 15 is $16,384. 2. Mascot's salary overtakes Miranda's salary in year 21. Extra: After 25 years Mascot has received more money altogether. The Mascot has received $8,554,431 more 1. Every year Miranda makes 1 million dollars since she gets a fixed amount each year. So in year 15 her salary will still be 1 million dollars. The mascot, on the other hand, gets her salary doubled every year starting with 1 dollar. I made the following table to see what will be the mascot's salary in year 15: Salary: 1 2 x 1 = 2 2 x 2 = 4 4 x 2 = 8 8 x 2 = 16 ...... Year: 1 2 3 4 5 ...... It took me one whole page to do this!

I did the doubling by hand and later checked my answers using a calculator (I had to fix a couple of mistakes that I made in my "by hand" calculations). In the end, I got $16,384 for year 15. 2. I continued with my table until year 26 (you will see why in my answer to the Extra). Then I looked at my table and the first time the mascot's salary passed

17 Unit 1 – Number Sense and Algebraic Thinking Lesson 1 & 2 – Whole Number Problem Solving

Miranda (which is 1 million dollars) is in year 21. In that year the mascot makes $1,048,576. Extra. After 25 years Miranda has 25 million dollars, because she makes 1 million every year and 1 x 25 = 25. The Mascot's altogether income in 25 years is $33,554,431. So the Mascot has more. I know this because I found a pattern in my table. The pattern looked like this: Year: 1 2 3 4 5 6 ...... Salary in current year: 1 2 4 8 16 32 ...... Total salary up to now: 1 3 7 15 31 63...... As you notice, for example in year 5, the total salary is 31 which is one less than the salary in year 6. This pattern continues. Here is the reason why we see such a pattern. Look at year 10 for example. Because salary is doubled every year, Salary in year 10 - salary in year 9 = salary in year 9.

Similarly Salary in year 10 - (salary in year 9 + year 8) = salary in year 8 Continuing this way Salary in year 10 - (salary in year 9 + year 8 + ...year 1) = salary in year 1; which is 1 dollar. This shows salary in year 10 is one dollar more than the total salary for the first nine years. So total salary up to year 25 will be the salary in year 26 minus 1 which equals (from my earlier table) $33,554,431. To see how much more the mascot made, I took away Miranda's total money out of the mascot's total money. I did this: 33,554,431 - 25,000,000 ------8, 554, 431 ------So the Mascot has $8,554,431 more.

25. Tatianna started with 8 rubbles and ended with 8 rubbles. To find out how many rubbles Tatianna started with I used the guess and check method. First I started with the number seven and followed the same procedure that Ivan followed and wrote down my results. When I used seven it looked like this: (7 x 2 – 8 = 6) then (6 x 2 – 8 = 4) and finally (4 x 2 – 8 = 0) and I did not come out

18 Unit 1 – Number Sense and Algebraic Thinking Lesson 1 & 2 – Whole Number Problem Solving

with the number that I started with so I moved on to the next number. The next number was 8. After I followed the same procedure I came out with the same number that I started with so that was my answer. I later realized that I could have just have multiplied the eight rubbles that she gave to the troll by 2 and I could have followed that procedure and gotten the same answer.

26. He had 7 rubles in his pocket to start with. I used a trial and error method. I broke it down to what he paid to the troll and then tried all the numbers less than 10 for the rubles in his pocket at the beginning. If he paid 2 then using all the numbers less than 10 it does not work. If he paid 3 then it's not becuase it came out to not be 2X because it is odd. If he paid 4 it did not work. If he paid 5 it's not but getting closer I think. If he paid 6, then 7 works.. I got it!!! First I eliminated some numbers and guessed at the number of rubles he had in his pocket and what he had to pay the troll for crossing the bridge. Then when I got to figure out that he had 7 rubles in his pocket in the beginning and every time he crossed he paid the troll 6 rubles.

7 + 7 = 14 double rubles 14 - 6 = 8 pay toll of 6 rubles for first crossing 8 + 8 = 16 double rubles 16 - 6 = 10 pay toll of 6 rubles for second crossing 10 + 10 = 20 double rubles 20 - 6 = 14 pay toll of 6 rubles for third crossing Because 7 + 7 = 14 and he had 7 rubles in his pocket to start and it is doubled. Therefore, he started out with 7 rubles in his pocket.

19 Unit 1 – Number Sense and Algebraic Thinking Lesson 1 & 2 – Whole Number Problem Solving Bibliography Information

Teachers attempted to cite the sources for the problems included in this problem set. In some cases, sources may not have been known.

Problems Bibliography Information

Lee, Joseph D. Primary Mathematics 5: All examples and Challenging Word Problems. Singapore: EPB problems 1 – 4 Pan Pacific, 2006.

Lee, Joseph D. Primary Mathematics 6: 5 – 11 Challenging Word Problems. Singapore: EPB Pan Pacific, 2006.

The Math Forum @ Drexel 12 – 26 (http://mathforum.org/)

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