Potwb-16-combined5-6.pdf

 

 

 

The problems in this booklet are organized into strands. A problem often appears in multiple strands. The problems are suitable for most students in Grade 5 or higher.

 

Problem of the Week Problem B A Probability Puzzle Jacob has the 600 pieces of a rectangular puzzle in a box. He knows that • the finished puzzle will have a perimeter of 200 cm, with the width being of the length, and

2 3

• the straight edge on each edge piece is 2 cm long, including the two edges of each corner piece. a) If Jacob puts his hand in the puzzle box without looking, and pulls out one piece, what is the probability that it will be a corner piece? b) What is the probability that it will be an edge piece, but not a corner piece?

Strand

Data Management and Probability

Problem of the Week Problem B A Probability Puzzle

Problem Jacob has the 600 pieces of a rectangular puzzle in a box. He knows that • the finished puzzle will have a perimeter of 200 cm, with the width being and

2 3

of the length,

• the straight edge on each edge piece is 2 cm long, including the two edges of each corner piece. a) If Jacob puts his hand in the puzzle box without looking, and pulls out one piece, what is the probability that it will be a corner piece? b) What is the probability that it will be an edge piece, but not a corner piece?

Solution a) Since only 4 of the 600 puzzle pieces are corner pieces, the theoretical probability that Jacob will pull out a corner piece is 4 1 Number of desired outcomes = = . Total number of possible outcomes 600 150 b) Since the perimeter of the puzzle is 200 cm, the length + width = 100 cm. Thus the width must be 40 cm and the length must be 60 cm, in order that the width be 32 of the length. On each 60 cm length, there will be thirty 2 cm edges, while on each 40 cm width, there will be twenty 2 cm edges. But the corner pieces each have two such edges. Thus there are 28 non-corner edge pieces on each 60 cm length, and 18 non-corner pieces on each 20 cm width, giving a total of 28 × 2 + 18 × 2 = 92 non-corner edge pieces. So the probability that Jacob will draw a non-corner edge piece is 92 46 23 = = . 600 300 150

Problem of the Week Problem B What Are the Chances? Many myths persist about how prior events affect the chances of future events. a) Philip Tails flips a coin 19 times and gets tails every time. He thinks that on his 20th flip he will have a better chance of getting heads because of what happened on the first 19 flips. Is Philip correct? Explain why or why not. b) For a certain lottery, a player selects 6 different numbers from 1 to 49. The lottery randomly picks six different numbers from 1 to 49 (the order of the selection does not matter). You win if your numbers match the lottery selections. Last week, Lou Key won by choosing the numbers 1, 2, 3, 4, 5, 6. He decides he has a better chance of winning again next week if he picks different numbers. Is he correct? Explain why or why not. c) A bag of marbles contains 10 red marbles, 8 blue marbles, and 6 yellow marbles. (Express your answers in (i) and (ii) reduced to lowest terms.) (i) If you stick your hand in the bag and draw out one marble, what is the probability that it will be red? (ii) If you draw a red marble, and then a blue marble, and remove them from the bag, what are the chances of drawing a red marble on your third draw? (iii) How is the outcome in part (ii) different from that in parts a) and b)?

Strand

Data Management and Probability

Problem of the Week Problem B What Are the Chances? Problem Many myths persist about how prior events affect the chances of future events. a) Philip Tails flips a coin 19 times and gets tails every time. He thinks that on his 20th flip he will have a better chance of getting heads because of what happened on the first 19 flips. Is Philip correct? Explain why or why not. b) For a certain lottery, a player selects 6 different numbers from 1 to 49. The lottery randomly picks six different numbers from 1 to 49 (the order of the selection does not matter). You win if your numbers match the lottery selections. Last week, Lou Key won by choosing the numbers 1, 2, 3, 4, 5, 6. He decides he has a better chance of winning again next week if he picks different numbers. Is he correct? Explain why or why not. c) A bag of marbles contains 10 red marbles, 8 blue marbles, and 6 yellow marbles. (Express your answers in (i) and (ii) reduced to lowest terms.) (i) If you stick your hand in the bag and draw out one marble, what is the probability that it will be red? (ii) If you draw a red marble, and then a blue marble, and remove them from the bag, what are the chances of drawing a red marble on your third draw? (iii) How is the outcome in part (ii) different from that in parts a) and b)?

Solution a) Phillip is not correct. Each flip of the coin is unaffected by any previous event. This is called an independent event because the outcome of the next event is NOT dependent on the outcome of any previous events. b) No, picking different numbers will not give Lou a better chance of winning the lottery, as long as there is an equal chance of any of the numbers 1 to 49 being selected for each of the six numbers drawn. c) For the bag of marbles containing 10 red marbles, 8 blue marbles, and 6 yellow marbles: (i) the theoretical probability of drawing a red marble is 10 10 5 the number of desired outcomes = = = . the total number of possible outcomes 10 + 8 + 6 24 12

(ii) If a red marble and a blue marble are removed, there will now be 9 red, 7 blue, and 6 yellow marbles in the bag. Thus the theoretical probability of drawing a red marble is now 9 9 = . 9 + 7 + 6 22 (iii) In parts a) and b), none of the possible outcomes changed. Here, marbles are removed from the bag, and so the number of outcomes, and hence the probabilities, both change.

Problem of the Week Problem B “Draft/Schmaft" In a hockey draft lottery, teams are given a certain percent chance of winning the # 1 pick, based on how they placed the year before. Their chances are represented by coloured ping pong balls, with each team having a percentage of balls based on their standings in the previous year. Suppose a possible set of results from last year gives the following chances: • Columbus Red Coats 20% (20 chances out of 100 to win the number one pick); • Toronto Maple Stumps 13.5% (13.5 chances out of 100 to win the number one pick); • Arizona Toy Poodles 11.5%; • Carolina Gentle Breezes 10%; • Buffalo Lightsabres 8%; • the remaining teams in the league have a combined 37% of the wins. a) Why can’t a total of only 100 ping pong balls be used for the draw? b) What is the fewest number of balls needed to run the lottery? c) How many ping pong balls would Columbus get in the draw, assuming the fewest number of balls were used? d) How many balls would Toronto get? Extension: If Columbus wins the first pick and all their ping pong balls are removed from the draw, what is the probability that Buffalo will get the second pick? Give your answer as a fraction in lowest terms. Strands

Number Sense and Numeration, Data Management and Probability

Problem of the Week Problem B “Draft/Schmaft" Problem In a hockey draft lottery, teams are given a certain percent chance of winning the # 1 pick, based on how they placed the year before. Their chances are represented by coloured ping pong balls, with each team having a percentage of balls based on their standings in the previous year. Suppose a possible set of results from last year gives the following chances: • • • • • •

Columbus Red Coats 20% (20 chances out of 100 to win the number one pick); Toronto Maple Stumps 13.5% (13.5 chances out of 100 to win the number one pick); Arizona Toy Poodles 11.5%; Carolina Gentle Breezes 10%; Buffalo Lightsabres 8%; the remaining teams in the league have a combined 37% of the wins.

a) Why can’t a total of only 100 ping pong balls be used for the draw? b) What is the fewest number of balls needed to run the lottery? c) How many ping pong balls would Columbus get in the draw, assuming the fewest number of balls were used? d) How many balls would Toronto get? Extension: If Columbus wins the first pick and all their ping pong balls are removed from the draw, what is the probability that Buffalo will get the second pick? Give your answer as a fraction in lowest terms.

Solution a) A total of 100 ping pong balls is not enough, since it doesn’t give any way of representing 0.5%. b) Since no fraction of a percent occurs other than 0.5% in the known data, the lottery can be run with a minimum of 200 balls. That is, one ball for each 0.5% in 100%. c) If 200 balls are used, Columbus would get 20% of the total, 20 balls for each hundred or a total of 2 × 20 = 40 balls.. We can get the same result multiplying 200 by 0.2. d) If 200 balls are used, Toronto would get 13.5% of 200, or 27 balls. Extension: If Columbus wins the first pick, and their 40 balls are removed, there will now be 200 − 40 = 160 balls remaining. Buffalo’s has 8% of 200 balls, or 16 balls of the remaining 160 balls. Thus the probability of Buffalo getting the second pick is

16 160

=

1 . 10

Problem of the Week Problem B Cookie Monsters! POTW cookies have two chocolate chip cookies on the outside which hold together a delicious creamy center. While sharing your POTW cookies with your friends, you notice that there are many different ways that people eat them. You decide to conduct a survey to determine the favourite ways these cookies are eaten. a) Discuss with your classmates different ways the cookies can be eaten. Decide on the four most popular methods. b) Survey your class, friends, and family, and make a tally chart showing which of the four methods they prefer. Remember: the more people from whom you get data, the better your survey. c) Record the data from your tally chart in the table. d) Display your data on an appropriate graph, using software if desired. e) Based on your research, what conclusions can you draw about how the people you surveyed like to eat POTW cookies? Method Method Method Method 1 2 3 4

Strand

Data Management and Probability

Problem of the Week Problem B Cookie Monsters! Problem POTW cookies have two chocolate chip cookies on the outside which hold together a delicious creamy center. While sharing your POTW cookies with your friends, you notice that there are many different ways that people eat them. You decide to conduct a survey to determine the favourite ways these cookies are eaten. a) Discuss with your classmates different ways the cookies can be eaten. Decide on the four most popular methods. b) Survey your class, friends, and family, and make a tally chart showing which of the four methods they prefer. Remember: the more people from whom you get data, the better your survey. c) Record the data from your tally chart in the table. d) Display your data on an appropriate graph, using software if desired. e) Based on your research, what conclusions can you draw about how people like to eat POTW cookies?

Solution Here are the results of one class survey. Four different methods of eating the cookies were discovered: 1. Straight bites 2. Lick the creme 3. Eat the creme inside first 4. Eat the non-creme outsides first Method Method Method Method 1 2 3 4 ///

//// /// ///

3

13

8

4

Based on the results of this survey, most people in this class liked to eat the creme first, either by licking it off or eating the whole creme side of the cookie. The least popular way is by eating straight bites.

Problem of the Week Problem B These Teachers are Golden Three teachers have classes practicing their gymnastics routines for their physical education open house. • Ms. Korbut has four groups comprised of 2, 3, 4, and 5 students. • Mrs. Miller has groups of 4, 5, 6, and 7. • Mrs. Comaneci has groups of 6, 7, 8, and 9. a) If each teacher wants to have the same total number of students to supervise, then which single group should be moved to another teacher? b) What other arrangements of these groups would achieve this balance?

Strand

Data Management and Probability

Problem of the Week Problem B These Teachers are Golden Problem Three teachers have classes practicing their gymnastics routines for their physical education open house. • Ms. Korbut has four groups comprised of 2, 3, 4, and 5 students. • Mrs. Miller has groups of 4, 5, 6, and 7. • Mrs. Comaneci has groups of 6, 7, 8, and 9. a) If each teacher wants to have the same total number of students to supervise, then which single group should be moved to another teacher? b) What other arrangements of these groups would achieve this balance?

Solution a) Since Ms. Korbut has a total of 14 students, Mrs. Miller has 22 students, and Mrs. Comaneci has 30 students, there are 14 + 22 + 30 = 66 students altogether. Thus each of the three teachers must have 66 ÷ 3 = 22 students to supervise in order for them all to be equal. So moving the group of 8 students from Mrs. Comaneci’s class to Ms. Korbut’s class will achieve the desired equal-sized classes. b) This balance can also be achieved by any rearrangement of the groups that gives a total of 22 students in each class. There are groups of {2}, {3}, {4}, {4}, {5}, {5}, {6}, {6}, {7}, {7}, {8}, and {9}. One possibility would be classes of 3 + 6 + 6 + 7 = 22, 4 + 5 + 5 + 8 = 22, and 2 + 4 + 7 + 9 = 22. Another possibility is classes of 6 + 7 + 9 = 22, 3 + 4 + 7 + 8 = 22, and 2 + 4 + 5 + 5 + 6 = 22. Did you find other possibilities?

 

Problem of the Week Problem B Do Please STOP It!

On the left below is the word STOP. On the right are six objects which may or may not be the result of a transformation of the whole word STOP, i.e., one or more translations (slides), reflections (flips), or rotations (turns). Cross out the objects on the right which are NOT possible transformations of the whole word STOP.

Discuss your conclusions with one of your classmates, explaining your reasoning.

Strand

Geometry and Spatial Sense

Problem of the Week Problem B Do Please STOP It! Problem On the left below is the word STOP. On the right are six objects which may or may not be the result of a transformation of the whole word STOP, i.e., one or more translations (slides), reflections (flips), or rotations (turns). Cross out the objects on the right which are NOT possible transformations of the whole word STOP.

Discuss your conclusions with one of your classmates, explaining your reasoning.

Solution The solution and explanations are shown in the diagrams below.

NOTE: Other answers are possible. For example, object 4 could be regarded as the result of a slide to the right of one width of the word, followed by a flip about a vertical line at the left end of the word.

Problem of the Week Problem B A Fitting Solution The figure below can be ’tiled’ (that is, it will be completely covered with no spaces) by a variety of arrangements of pattern blocks of different shapes. a) Using exactly 10 pattern blocks, how many of each shape are required to completely cover this figure? b) What fraction of the total area of the figure is covered by hexagons? by trapezoids? by rhombuses? by triangles? Express each of your answers reduced to lowest terms. c) In what other ways can you cover this figure with the four shapes from the pattern blocks (but not necessarily 10 blocks)?

To Think About Each of the hexagon, trapezoid and rhombus can be covered completely by a whole number of triangles. How many triangles are needed for each of the shapes?

Strands

Geometry and Spatial Sense, Number Sense and Numeration

Problem of the Week Problem B A Fitting Solution Problem The figure below can be ’tiled’ (that is, it will be completely covered with no spaces) by a variety of arrangements of pattern blocks of different shapes. a) Using exactly 10 pattern blocks, how many of each shape are required to completely cover this figure? b) What fraction of the total area of the figure is covered by hexagons? by trapezoids? by rhombuses? by triangles? Express each of your answers reduced to lowest terms. c) In what other ways can you cover this figure with the four shapes from the pattern blocks (but not necessarily 10 blocks)? Solution a) There is just one combination of shapes that can be used to cover this figure using exactly 10 pattern blocks. A possible solution is shown at the right. There is one hexagon, one trapezoid, one rhombus and seven triangles. 1 of the figure, the b) Since each green triangle covers 18 fractions of the total area of the figure covered by each colour are as follows: 6 1 yellow hexagon = six triangles = 18 = 13 3 = 16 1 red trapezoid = three triangles = 18 2 1 blue rhombus = two triangles = 18 = 19 7 7 green triangles = 18

c) There are seven other ways to cover this figure with four pattern blocks, as follows: 2 yellow hexagons, 1 red trapezoid, 1 blue rhombus, 1 green triangle (5 blocks); 1 yellow hexagon, 3 red trapezoids, 1 blue rhombus, 1 green triangle (6 blocks); 1 yellow hexagon, 2 red trapezoids, 2 blue rhombi, 2 green triangles (7 blocks); 1 yellow hexagon, 2 red trapezoids, 1 blue rhombus, 4 green triangles (8 blocks); 1 yellow hexagon, 1 red trapezoid, 4 blue rhombi, 1 green triangle (7 blocks); 1 yellow hexagon, 1 red trapezoid, 3 blue rhombi, 3 green triangles (8 blocks); 1 yellow hexagon, 1 red trapezoid, 2 blue rhombi, 5 green triangles (9 blocks). Convince yourself that these are correct by drawing diagrams for each. Think about why these are the only covers that use only four colours of pattern blocks.

Problem of the Week Problem B Look At Rick Shaw Go! Rick Shaw transports tourists through a park on his human-powered cart. He takes passengers around on several different routes (The Basic, The Classic, The Plus and The Ultimate), depending on how much they are willing to pay. On each route, he travels along each path (line segment) exactly once. He may pass through a node (vertex) more than once to complete a route. He does not have to start and end at the same place.

Above is a picture of Rick and his four possible routes. a) Determine a way to travel each of the routes. Record the results in the following chart. For The Basic, a successful route could start at A and end at C as follows: A→B→C→A→D→C (This is one possible route; there are others.) b) At each node in any route there are a number of intersecting paths. For each route, record the total number of intersecting paths, the number of intersecting paths at the start point and the number of intersecting paths at the end point. For The

Basic, there are 3 intersecting paths at node A, 2 intersecting paths at node B, 3 intersecting paths at node C, and 2 intersecting paths at node D. There is a total of 3 + 2 + 3 + 2 = 10 intersecting paths, 3 intersecting paths at the starting point A and 3 intersecting paths at the endpoint C. No. of Intersecting Paths Route Total Start End Basic: A → B → C → A → D → C

10

3

3

Classic: Plus: Ultimate:

c) What trends do you notice about the number of intersecting paths at the start and end points of each successful route? d) What trend do you see in the total number of intersecting paths? e) Try to draw two routes, one that has a successful path and one that does not. Switch your new routes with a partner. Predict which of your partner’s routes will work and which will not. Then confirm your predictions. Strands Geometry and Spatial Sense, Patterning and Algebra

Problem of the Week Problem B Look At Rick Shaw Go! Problem Rick Shaw transports tourists through a park on his human-powered cart. He takes passengers around on several different routes (The Basic, The Classic, The Plus and The Ultimate), depending on how much they are willing to pay. On each route, he travels along each path (line segment) exactly once. He may pass through a node (vertex) more than once to complete a route. He does not have to start and end at the same place.

Above is a picture of Rick and his four possible routes. a) Determine a way to travel each of the routes. Record the results in the following chart. (There may be more than one successful route.) b) At each node in any route there are a number of intersecting paths. For each route, record the total number of intersecting paths, the number of intersecting paths at the start point and the number of intersecting paths at the end point. c) What trends do you notice about the number of intersecting paths at the start and end points of each successful route? d) What trend do you see in the total number of intersecting paths? e) Try to draw two routes, one that has a successful path and one that does not. Switch your new routes with a partner. Predict which of your partner’s routes will work and which will not. Then confirm your predictions.

Solution a) b) The completed table is shown below. There are other possible routes. Route

No. of Intersecting Paths Total Start End

Basic: A → B → C → A → D → C

10

3

3

Classic: C → D → A → E → B → C → A → B

14

3

3

Plus: D → F → A → B → E → A → D → C → F → B → C

20

3

3

Ultimate: H → K → C → H → G → C → B → F → A → B → E → A → D → C

26

3

5

c) All successful routes start and end at a point with an odd number of intersecting paths. d) All the total numbers of intersecting paths are even. See the next page for further comments on parts c), d) and e), and on routes in general.

In each of the given diagrams (graphs) in the question, there were two nodes (vertices) with an odd number of intersecting paths. All other nodes had an even number of intersecting paths. In general you would be able to find a successful route if exactly two of the nodes in your diagram had an odd number of intersecting paths and all of the other nodes had an even number of intersecting paths. An Euler, pronounced “oiler”, path is a path that uses every edge of the graph exactly once, but starts and ends at a different node. An Euler circuit is a path that uses every edge of the graph exactly once, but starts and ends at the same node. In our problem, each route was an Euler path. If a graph has no nodes with an odd number of intersecting paths, you can find an Euler circuit. If a graph has exactly two nodes with an odd number of intersecting paths, you can find an Euler path by starting at either of those nodes. If a graph has three or more nodes with an odd number of intersecting paths, you cannot complete an Euler path or circuit. The following graphs are fairly similar. Exactly one of the graphs contains an Euler path, exactly one contains an Euler circuit and exactly one contains neither an Euler path or an Euler circuit. Can you correctly identify which type of path is contained in each graph.

Problem of the Week Problem B Who Was Pythagoras? Part of the side length of the large square shown in Figure 1 is 5 units. The remainder of the side length is 12 units. It follows that the total side length of the square in Figure 1 is 5 + 12 = 17 units. Within the large square in Figure 1, there are four right-angled triangles, which we shall call A1, A2, A3 and A4, and two squares S1 and S2. a) How do you know that all four triangles are congruent to one another? b) What are the areas of S1 and S2?

Figure 1

c) Figure 2 contains the same large square of side length 17 as in Figure 1. Figure 2 contains five geometric shapes. What must be the shape of the interior figure S? Explain your answer, stating the length of the sides of S, and the area of S. d) Now compare Figure 1 to Figure 2. Which areas are the same in both Figure 1 and Figure 2? What areas of Figure 1 must sum to give the area of S? e) Write your result from d) as an equation. This will tell you how the side lengths 5, 12, 13 of a right-angled triangle are related. Strand

Geometry and Spatial Sense

Figure 2

Problem of the Week Problem B Who Was Pythagoras? Problem Part of the side length of the large square shown in Figure 1 is 5 units. The remainder of the side length is 12 units. It follows that the total side length of the square in Figure 1 is 5 + 12 = 17 units. Within the large square in Figure 1, there are four right-angled triangles, which we shall call A1, A2, A3 and A4, and two squares S1 and S2. a) How do you know that all four triangles are congruent to one another? b) What are the areas of S1 and S2? c) Figure 2 contains the same large square of side length 17 as in Figure 1. Figure 2 contains five geometric shapes. What must be the shape of the interior figure S? Explain your answer, stating the length of the sides of S, and the area of S. d) Now compare Figure 1 to Figure 2. Which areas are the same in both Figure 1 and Figure 2? What areas of Figure 1 must sum to give the area of S?

Figure 1

e) Write your result from d) as an equation. This will tell you how the side lengths 5, 12, 13 of a right-angled triangle are related.

Solution

Figure 2

a) The four triangles are congruent to one another because each one is exactly half of a rectangle with sides 5 and 12, created by a diagonal of length 13. (Alternatively, we know they are congruent because they all have three sides of the same lengths.)

b) Since S1 and S2 are squares of sides 5 and 12 respectively, they have areas 5 × 5 = 25 and 12 × 12 = 144 (sometimes 5 × 5 is written as 52 and 12 × 12 is written as 122 ). c) Since the four corner triangles are identical, their third side must be the same, with length 13 in each case (from Figure 1). Further, since all four corners of S are right angles, S is a square with side length 13 and area 13 × 13 = 169 (sometimes 13 × 13 is written as 132 ). d) The four corner triangles of Figure 2 have the same areas as triangles A1, A2, A3, A4 in Figure 1. Since the large outer square in both figures has side length 17, the total area for both must be the same. Thus the area of S in Figure 2 must equal the sum of the areas of S1 and S2 in Figure 1. e) Expressed as an equation, the result of part d) states that Area S1 + Area S2 = Area S 25

+

144

=

169

5 × 5 + 12 × 12 = 13 × 13 This can be written

52

+

122

=

132

This result is a specific example illustrating the famous Pythagorean Theorem. In a right-angled triangle the side opposite the right angle is called the hypotenuse. If the length of the hypotenuse is c and the lengths of the two shorter sides are a and b, then the Pythagorean Theorem relates the three sides with a2 + b2 = c2 . If you do this problem again using a instead of 5, b instead of 12, and c instead of 13, then through your discovery you will actually prove the Pythagorean Theorem! This generalization is shown on the next page.

a) How do you know that all four triangles are congruent to one another? The four triangles are congruent to one another because each one is exactly half of a rectangle with sides a and b, created by a diagonal of length c. (Alternatively, we know they are congruent because they all have three sides of the same lengths.) b) What are the areas of S1 and S2? Since S1 and S2 are squares of sides a and b respectively, they have areas a × a and b × b (sometimes written as a2 and b2 ).

Figure 1

c) Figure 2 contains the same large square of side length c as in Figure 1. Figure 2 contains five geometric shapes. What must be the shape of the interior figure S? Explain your answer, stating the length of the sides of S, and the area of S. Since the four corner triangles are identical, their third side must be the same, with length c in each case (from Figure 1). Further, the symmetry of the diagram implies that all four of the corner angles of S must be right angles. Thus S is a square with side length c and area c × c, or c2 .

Figure 2

d) Now compare Figure 1 to Figure 2. Which areas are the same in both Figure 1 and Figure 2? What areas of Figure 1 must sum to give the area of S? The four corner triangles of Figure 2 have the same areas as triangles A1, A2, A3, A4 in Figure 1. Since the large outer square in both figures has side a + b, the total area for both must be the same. Thus the area of S in Figure 2 must equal the sum of the areas of S1 and S2 in Figure 1. e) Write your result from d) as an equation. This will tell you how the side lengths a, b, c of a right-angled triangle are related. Expressed as an equation, the result of part d) states that Area S1 + Area S2 = Area S

This can be written

a×a

+

a2

+

b×b = c×c b2

=

c2

This result illustrates the famous Pythagorean Theorem which relates the side lengths a and b of a right-angled triangle to the hypotenuse, its longest side, length c. And through this discovery, the result has actually been proven!

Problem of the Week Problem B Looks Can Be Deceiving Using grid paper, you may wish to construct each of the diagrams shown below. This may be helpful as you work through the problem. In each of the three diagrams below, there are two squares which overlap in various ways. The bottom square is fixed and has side length 12. The overlapping square has one vertex located at the centre C of the bottom square. The darker shaded regions are common to the two squares and are labeled A1, A2 and A3, respectively.

a) Determine the areas of A1, A2 and A3, the overlapping areas in each of the above diagrams. How is each of these areas related to the total area of the square of side length 12? b) Below is a diagram in which the overlapping area, A4, is in an arbitrary position. Show that the blank square of side length 12 can be completely tiled by four copies of A4 and use this to determine the overlapping area, A4.

Extension: The above example shows us that the overlapping area of the two squares is 36 units2 , one-quarter of the total area. Would it make any difference if the overlapping figure were a rectangle? A right-angled triangle? Another polygon? Explain your thinking.

Strands

Geometry and Spatial Sense, Measurement

Problem of the Week Problem B Looks Can Be Deceiving Problem Using grid paper, you may wish to construct each of the diagrams shown below. This may be helpful as you work through the problem. In each of the three diagrams below, there are two squares which overlap in various ways. The bottom square is fixed and has side length 12. The overlapping square has one vertex located at the centre C of the bottom square. The darker shaded regions are common to the two squares and are labeled A1, A2 and A3, respectively.

a) Determine the areas of A1, A2 and A3, the overlapping areas in each of the above diagrams. How is each of these areas related to the total area of the square of side length 12? b) Below is a diagram in which the overlapping area, A4, is in an arbitrary position. Show that the blank square of side length 12 can be completely tiled by four copies of A4 and use this to determine the overlapping area, A4.

Extension: The above example shows us that the overlapping area of the two squares is 36 units2 , one-quarter of the total area. Would it make any difference if the overlapping figure were a rectangle? A right-angled triangle? Another polygon? Explain your thinking. Solution The solution is on the next page.

a) A1 is a triangle of height 6 and base 12, and thus has area

1 2

× 12 × 6 = 36 square units.

A2 is a square of side 6, with area 6 × 6 = 36 square units. A3 consists of two right-angled triangles of base length 3 and height 6, plus a rectangle with side lengths 3 and 6. The area of each right-angled triangle is 12 × 3 × 6 = 9 square units. The area of the rectangle is 3 × 6 = 18 square units. The total area of A3 is 9 + 9 + 18 = 36 square units. So all three overlapping areas A1, A2, and A3 have exactly the same area!

B) The way in which A4 tiles the square of side 12 is illustrated at the right. Since this shows that the overlapping area is 36 square units (i.e., 41 of the area of the bottom square) for any position in which the overlapping figure is an irregular quadrilateral, and part a) shows that this also holds for the two special cases, we see that the overlapping area is always the same, 36 square units or 41 of the area of the square with side length 12. Extension: As long as the overlapping polygon has one vertex which is a right angle pinned at the centre, C, of the base square and the two sides forming that vertex are long enough that they intersect the sides of the bottom square, the shape of the overlapping area would always be one of type A1, A2, or A4, with area equal to 41 of the area of the bottom square.

 

Problem of the Week Problem B The Time of Your Life! As you answer these questions, recall that there are 365 days in a year (except for a leap year), 24 hours in a day, 60 minutes in each hour, and 60 seconds in each minute. a) The oldest human ever to have lived, according to the Guiness Book of Records, was 122 years and 164 days old at the time of her death. Jeanne Louise Calment (France) was born on February 21, 1875, and died at a nursing home in Arles, in southern France, on August 4, 1997. How many minutes was this person alive (assuming she was born and died at midnight on the given dates)? b) If you were born on January 1, 2005, how many years old would you be when you celebrate the billionth second that you have been alive?

Jeanne Louise Calment 1996 Source: Wikipedia

Strands

Number Sense and Numeration, Measurement

Problem of the Week Problem B The Time of Your Life! Problem As you answer these questions, recall that there are 365 days in a year (except for a leap year), 24 hours in a day, 60 minutes in each hour, and 60 seconds in each minute. a) The oldest human ever to have lived, according to the Guiness Book of Records, was 122 years and 164 days old at the time of her death. Jeanne Louise Calment (France) was born on February 21, 1875, and died at a nursing home in Arles, in southern France, on August 4, 1997. How many minutes was this person alive (assuming she was born and died at midnight on the given dates)? b) If you were born on January 1, 2005, how many years old would you be when you celebrate the billionth second that you have been alive?

Jeanne Louise Calment 1996 Source: Wikipedia

Solution a) To determine how many minutes Jeanne Louise Calment was alive, we must convert her age of 122 years and 164 days to minutes. Since leap years occur every four years, we shall assume there are 365.25 days in a year, on average. Thus the total number of days is 164 days + 122 years = 164 + (122 × 365.25) = 44 724.5 days.

Since there are 60 × 24 = 1 440 minutes in each 24 hour day, we see that she was alive for a total of 44 724.5 × 1 440 = 64 403 280 minutes. Note: If, instead of using the average number of days in a year, you count the number of leap years (31) and non-leap years (91) that Jeanne Louise was alive, you will get a very slightly different answer, namely 64 404 000 minutes. It is an interesting class discussion to think about why this answer is slightly larger. b) We need to figure out how many years there are in one billion seconds. Working upwards in the size of the time unit, we have: 1 000 000 000 seconds equals 1 000 000 000 ÷ 60 = 16 666 666 32 minutes; 16 666 666 23 minutes equals 16 666 666 23 ÷ 60 ≈ 277 777.778 hours; 277 277.778 hours equals 277 777.778 ÷ 24 ≈ 11 574.074 days; 11 574.074 days equals 11 574.074 ÷ 365.25 ≈ 31.7 years. Thus you will be more than 31 years old when you celebrate your billionth second of life. Extension: How long have YOU been alive? • How many whole number years? • How many whole number days (don’t forget leap years)? • How many whole number hours (assuming you were born at midnight on your birthday)? • How many minutes? • How many seconds?

Problem of the Week Problem B “B” There or “B” Square! A and B are two squares such that the perimeter of Square A is a multiple of the perimeter of Square B by a whole number. We wish to explore how the areas of the two squares are related.

a) If the perimeter of Square A is 3 times that of Square B, what is the ratio of the area of Square B to the area of Square A? Illustrate your answer with a diagram on graph paper. Express your answer as both a ratio and a fraction. b) What if the perimeter of Square A is 4 times that of B? 5 times that of B? c) Explain how you would figure out the area ratio if the perimeter of Square A is 10 times that of Square B.

Strands

Measurement, Number Sense and Numeration

Problem of the Week Problem B “B” There or “B” Square! Problem A and B are two squares such that the perimeter of Square A is a multiple of the perimeter of Square B by a whole number. We wish to explore how the areas of the two squares are related. a) If the perimeter of Square A is 3 times that of Square B, what is the ratio of the area of Square B to the area of Square A? Illustrate your answer with a diagram on graph paper. Express your answer as both a ratio and a fraction. b) What if the perimeter of Square A is 4 times that of B? 5 times that of B? c) Explain how you would figure out the area ratio if the perimeter of Square A is 10 times that of Square B.

Solution a) The diagram below reveals that when the perimeter of Square A is 3 times that of Square B, then the ratio of the area of Square B to the area of Square A is 1:9, or, in fraction form, 19 . This is due to the fact that each side of Square A is 3 times as long as that of Square B, in order to make the perimeter 3 times as long. Thus, if the side of Square B is taken as 1 unit, the side of Square A will be 3, and hence its area is 3 × 3 = 9 square units. (3 × 3 can also be written using exponents as 32 . We would say “3 squared”.)

b) The diagram also illustrates that when the perimeter of Square A is 4 times that of Square B, then the ratio of the area of Square B to the area of Square A is 1 : 16 = 1 : 42 , 1 or, in fraction form, 16 . Similarly, we would expect that when the side of Square A is 5 times that of Square B, then the areas will be in the ratio 1 : 25 = 1 : 52 , or, in fraction 1 form, 25 . c) Following the reasoning of parts a) and b), when the perimeter of Square A is 10 times that of Square B, then each side will be 10 times as long, and the areas will be in the 1 ratio 1 : 100 = 1 : 102 , or, in fraction form, 100 .

Problem of the Week Problem B A Dimension of Irregularity You are given the following distances between the vertices of the irregular polygon in the diagram below. AH = BF = CE = 4 cm, and AB = HF = CD = 5 cm In addition, you are told that the sides BC and F E are equal in length, and the points H, F , C, and D lie on the same straight line.

a) Describe how you would find the area of this irregular polygon. Include a description of what other measurements you would need in order to find the area. b) Make a geometric pattern with four copies of this polygon, using rotations and/or reflections. Colour your pattern in a way that pleases you. Did You Know? To find the area of a triangle, multiply the length of the base (b) times the height (h) and divide by 2. The base can be any side of the triangle but the height is a line segment perpendicular to the particular base from the opposite vertex.

Strand

Measurement

Problem of the Week Problem B A Dimension of Irregularity Problem You are given the following distances between the vertices of the irregular polygon in the diagram below. AH = BF = CE = 4 cm, and AB = HF = CD = 5 cm In addition, you are told that the sides BC and F E are equal in length, and the points H, F , C, and D lie on the same straight line.

a) Describe how you would find the area of this irregular polygon. Include a description of what other measurements you would need in order to find the area. b) Make a geometric pattern with four copies of this polygon, using rotations and/or reflections. Colour your pattern in a way that pleases you. Solution a) Using the given information shown on the diagram above, we can find the area of this polygon by adding the areas of its parts. Clearly we have enough information to find the areas of the rectangle ABF H (5 × 4 = 20 cm2 ), and the triangle CDE (5 × 4 ÷ 2 = 10 cm2 ). Further, we know that triangle BF C is congruent to triangle F CE, since they have three sides with the same lengths. Also, since HF C is a straight line, the angles BF C and F CE are right angles, so triangles BF C and CDE are right-angled triangles. To complete the calculation of the total area, we would need to know the length F C in order to find the area of triangles BF C and F CE, and the height of the triangle HF G in order to find the area of triangle HF G.

b) Here are two sample patterns which each use four copies of the given polygon.

Problem of the Week Problem B Block Out! Aputi (‘Snow on the Ground’), Siku (‘Ice’), and Qinu (‘Slushy Ice by the Sea’) are using rectangular blocks of snow to build a snow fort. a) They want to build one wall of their fort, but they can’t decide on the dimensions. If they have 24 blocks of snow, what are the possible different sizes of walls they could build, using whole blocks? Which of these would be sensible for a snow fort wall? b) The snow fort is to be rectangular, with four walls. Each wall is to be at least two, but no more than four blocks high. (i) If the front wall uses 24 blocks, and they can make up to 60 snow blocks in total, what possible dimensions could the other walls be? (Assume all four walls are the same height.) (ii) What is the total number of blocks they would need in each case? (iii) Which dimensions give the fort with greatest interior floor area? Which give the best protection? Organize your answers in the chart below. Note that b and h stand for base and height, respectively; all measurements are in blocks. The first line is for the sample shown below. Dimensions of the fort front×side×height

4×3×2

Strands

Dimensions of front and back walls b = 4, h = 2

No. of Blocks 16

Blocks for Side Walls (no corners) b = 1, h = 2

Measurement, Number Sense and Numeration

No. of Blocks 4

Total Blocks 20

Problem of the Week Problem B Block Out! Problem Aputi (‘Snow on the Ground’), Siku (‘Ice’), and Qinu (‘Slushy Ice by the Sea’) are using rectangular blocks of snow to build a snow fort. a) They want to build one wall of their fort, but they can’t decide on the dimensions. If they have 24 blocks of snow, what are the possible different sizes of walls they could build, using whole blocks? Which of these would be sensible for a snow fort wall? b) The snow fort is to be rectangular, with four walls. Each wall is to be at least two, but no more than four blocks high. (i) If the front wall uses 24 blocks, and they can make up to 60 snow blocks in total, what possible dimensions could the other walls be? (Assume all four walls are the same height.) (ii) What is the total number of blocks they would need in each case? (iii) Which dimensions give the fort with greatest interior floor area? Which give the best protection? Organize your answers in the chart below. Note that b and h stand for base and height, respectively; all measurements are in blocks. Dimensions of the fort 6×3×4

Dimensions of front and back walls b = 6, h = 4

8×3×3

front×side×height

No. of Blocks

Blocks for Side Walls (no corners)

No. of Blocks

Total Blocks

48

b = 1, h = 4

8

56

b = 8, h = 3

48

b = 1, h = 3

6

54

8×4×3

b = 8, h = 3

48

b = 2, h = 3

12

60

12 × 3 × 2

b = 12, h = 2

48

b = 1, h = 2

4

52

12 × 4 × 2

b = 12, h = 2

48

b = 2, h = 2

8

56

12 × 5 × 2

b = 12, h = 2

48

b = 3, h = 2

12

60

Solution a) Possible dimensions for the wall (length × height, in blocks) include 1 × 24, 2 × 12, 3 × 8, 4 × 6, 6 × 4, 8 × 3, 12 × 2, 24 × 1. Of these, only 6 × 4, 8 × 3, and 12 × 2 seem reasonable. b) The answers to parts (i) and (ii) are shown in the completed table above. (iii) The greatest interior floor area is in the 12 × 5 × 2 fort (with a 10 × 3 block floor area), while the best protection is the 6 × 3 × 4 fort (with the highest walls).

Problem of the Week Problem B Time IS Money Sky had a job at a local camp during the summer holidays. a) One week, she worked the following hours: Monday 8:45 - 11:30, and 12:15 - 17:00 Tuesday 10:20 - 11:30, and 12:15 - 17:50 Wednesday 12:30 - 20:00 Thursday 9:00 - 12:00, and 12:45 - 17:00 Friday 7:15 - 11:00 How many hours did Sky work in total that week? b) The following week, Sky worked from Monday to Friday, 9 a.m. to 4 p.m. each day, with a 30-minute lunch break each day. She is not paid during her lunch break. Did she work more or less hours this week than the previous week? c) If Sky earned $10.40 per hour during the first week of work, how much money did she earn in total that week? (You will need to convert measures in hours and minutes to decimal numbers of hours.) d) In between the first and second weeks, Sky’s hourly rate of pay increased slightly. If she earned exactly the same amount of money in both weeks, what was her hourly rate of pay in the second week?

Strands

Measurement, Number Sense and Numeration

Problem of the Week Problem B Time IS Money

Problem Sky had a job at a local camp during the summer holidays. a) One week, she worked the following hours: Monday Tuesday Wednesday Thursday Friday

8:45 - 11:30, and 12:15 - 17:00 10:20 - 11:30, and 12:15 - 17:50 12:30 - 20:00 9:00 - 12:00, and 12:45 - 17:00 7:15 - 11:00

How many hours did Sky work in total that week? b) The following week, Sky worked from Monday to Friday, 9 a.m. to 4 p.m. each day, with a 30-minute lunch break each day. She is not paid during her lunch break. Did she work more or less hours this week than the previous week? c) If Sky earned $10.40 per hour during the first week of work, how much money did she earn in total that week? (You will need to convert measures in hours and minutes to decimal numbers of hours.) d) In between the first and second weeks, Sky’s hourly rate of pay increased slightly. If she earned exactly the same amount of money in both weeks, what was her hourly rate of pay in the second week?

Solution a) The total hours Sky worked each day are: Monday 8:45 - 11:30, and 12:15 - 17:00, giving 2hr 45min + 4hr 45min = 7hr 30min; Tuesday 10:20 - 11:30, and 12:15 - 17:50, giving 1hr 10min+ 5hr 35min = 6hr 45min; Wednesday 12:30 - 20:00, giving 7hr 30min; Thursday 9:00 - 12:00, and 12:45 - 17:00, giving 3hr + 4hr 15min = 7hr 15min; Friday 7:15 - 11:00, giving 3hr 45min. Thus Sky worked a total of 7hr 30min + 6hr 45min + 7hr 30min+ 7hr 15min + 3hr 45min = 32hr 45min. b) The following week, Sky worked 7 hr minus 30 min = 6hr 30min each day. Thus she worked a total of 6hr 30min×5 = 32hr 30min, which is 15 min less than the previous week.

c) Since 32hr 45min = 32.75 hr, Sky earned a total of $10.40 ×32.75 = $340.60 during the first week. d) Since she earned the same during the second week for only 32hr 30min hr, or 32.5 hr, her new hourly rate was $340.60 ÷ 32.5 = $10.48 per hour.

Problem of the Week Problem B Leo’s Dog was Framed! Leo built a picture frame using the plan illustrated below. The picture inside the frame is 25 cm wide and 20 cm high. The horizontal and vertical distance from the edge of the picture to the outer edge of the frame is 5 cm. This information is marked on the plan.

What is the area of the border of the frame? There are many ways to solve this problem. Try to solve the problem in more than one way. Extension: The area of the border of the frame is larger than the area of the picture inside the frame. The width of the current border is 5 cm. What would the width of the border need to be so that the area of the border equals the area of the picture? Determine your answer correctly rounded to one decimal place.

Strand

Measurement

Problem of the Week Problem B Leo’s Dog was Framed! Problem Leo built a picture frame using the plan illustrated to the right. The picture inside the frame is 25 cm wide and 20 cm high. The horizontal and vertical distance from the edge of the picture to the outer edge of the frame is 5 cm. This information is marked on the plan. What is the area of the border of the frame? There are many ways to solve this problem. Try to solve the problem in more than one way. Extension: The area of the border of the frame is larger than the area of the picture inside the frame. The width of the current border is 5 cm. What would the width of the border need to be so that the area of the border equals the area of the picture? Determine your answer correctly rounded to one decimal place.

Solution Solution 1 Draw vertical and horizontal lines along the edge of the picture extending to the outside of the frame. This creates four 5 cm by 5 cm squares (in the corners), two 25 cm by 5 cm rectangles (along the top and bottom), and two 5 cm by 20 cm rectangles (along the left and right side). Area of Border = Area of four squares + Area of Two Rectangles + Area of Two Rectangles = 4 × 5 × 5 + 2 × 25 × 5 + 2 × 5 × 20 = 100 + 250 + 200 = 550 cm2 The area of the border of the frame is 550 cm2 .

Solution 2 Draw horizontal lines along the top and bottom edges of the picture extending to the outside of the frame. This creates two identical rectangles (top and bottom), and two other identical rectangles (left and right side of the frame). The length of the rectangles across the top and bottom is 5 + 25 + 5 = 35 cm and the width is 5 cm. The two side rectangles have length 20 cm and width 5 cm. Area of Border = Area of top and bottom rectangles + Area of the two side rectangles = 2 × 35 × 5 + 2 × 20 × 5 = 350 + 200 = 550 cm2 The area of the border of the frame is 550 cm2 .

Solution 3 A trapezoid is a quadrilateral with two parallel sides. To find the area of a trapezoid, we could break it into two triangles and a rectangle. There is a formula for the area of a trapezoid. To find the area, find the sum of the lengths of the two parallel sides, a and b. Multiply this sum by the distance between the two parallel sides, h. Divide the product by 2. This can be written A = h × (a + b) ÷ 2.

Draw diagonal lines from the four corners of the picture to the closest corner of the frame. This creates two identical trapezoids (top and bottom), and two other identical trapezoids (along the sides of the frame). The length of the parallel sides of the trapezoids across the top and bottom is 5 + 25 + 5 = 35 cm and 25 cm. The length of the parallel sides of the trapezoids along the two sides is 5 + 20 + 5 = 30 cm and 20 cm. The height of all four trapezoids is 5 cm.

Area of Border = Area of top and bottom trapezoids + Area of the two side trapezoids = 2 × 5 × (35 + 25) ÷ 2 + 2 × 5 × (30 + 20) ÷ 2 = 300 + 250 = 550 cm2 The area of the border of the frame is 550 cm2 .

Solution 4 In this solution, we find the area of the larger outside rectangle and then subtract the area of the small interior rectangle. This difference is the area of the border. The outside rectangle (the entire frame) has length 5 + 25 + 5 = 35 cm and width 5 + 20 + 5 = 30 cm. The area of the outer rectangle is 35 × 30 = 1050 cm2 . The inside rectangle (the picture) has length 25 cm and width 20 cm. The area of the outer rectangle is 25 × 20 = 500 cm2 . By subtracting the area of the inside rectangle from area of the large rectangle we determine the border area. Therefore, the area of the border of the frame is 1050 − 500 = 550 cm2 .

Solution to the extension For the extension we will use the idea from solution 4. We know that the area of the border is 550 cm2 when the width of the border is 5 cm. The area of the picture is 500 cm2 . We know that the width of the border must be less than 5 cm. We will basically try educated guesses to arrive at the width of the border correct to one decimal. Try a border width of 4.5 cm. Width of Border (in cm) 4.5

Length of Outer Rectangle (in cm)

Width of Outer Rectangle (in cm)

Area of Outer Rectangle) (in cm)2 )

4.5 + 25 + 4.5

4.5 + 20 + 4.5

34 × 29

= 34

= 29

= 986

Area of Inner Rectangle (in cm2 )

Area of Border (in cm2 ) 986 − 500

500

= 486

Since the area of the border is under 500 cm2 , the width of the border is too small. Try a border width of 4.7 cm. Width of Border (in cm) 4.7

Length of Outer Rectangle (in cm)

Width of Outer Rectangle (in cm)

Area of Outer Rectangle) (in cm)2 )

Area of Inner Rectangle (in cm2 )

4.7 + 25 + 4.7

4.7 + 20 + 4.7

34.4 × 29.4

500

= 34.4

= 29.4

= 1011.36

Area of Border (in cm2 ) 1011.36 − 500 = 511.36

Since the area of the border is over 500 cm2 , the width of the border is too large. The only possible one decimal border width left to check is 4.6 cm. Width of Border (in cm) 4.6

Length of Outer Rectangle (in cm)

Width of Outer Rectangle (in cm)

Area of Outer Rectangle) (in cm)2 )

Area of Inner Rectangle (in cm2 )

4.6 + 25 + 4.6

4.6 + 20 + 4.6

34.2 × 29.2

500

= 34.2

= 29.2

= 998.64

Area of Border (in cm2 ) 998.64 − 500 = 498.64

Since the area of the border is under 500 cm2 , the width of the border is too small. However, we want the width correct to one decimal so that the area of the border is as close to 500 cm2 as possible. When the width is 4.6 cm, the area of the border is 498.64 cm2 , 1.36 cm2 under the required area. When the width is 4.7 cm, the area of the border is 511.36 cm2 , 11.36 cm2 over the required area. Therefore, when the width of the border is 4.6 cm, the area of the border and the area of the picture are as close to being equal as possible under the condition of one decimal accuracy. This method is not very efficient. There is good news. You will learn an algebraic approach to this problem in later math studies that will get you a value for the width much more efficiently. You will have to wait.

Problem of the Week Problem B Speedy or Not, You May Be Caught! Various animals can travel at the following top speeds for short distances: • cheetahs race along at 120 km/h (that is, they travel 120 km every hour); • cows can gallop at 12.8 km/h; • snails can slither along at 0.013 m/s; A human’s top speed is 10.4 m/s. a) How many seconds would it take each animal to run a 100 m race? How long would it take a human to complete the 100 m? b) While hunting, cheetahs can only maintain their amazing speed for about 15 s, after which they generally give up the chase. Gazelles can run 60 km/h when evading predators. Suppose a cheetah spots a gazelle 300 m away. If both animals start at the same time and run in a straight line, will the gazelle escape? For further consideration: If the top speeds of the animals could be maintained over a long distance (physically impossible), approximately how long would it take the animals to run a 42.2 km marathon? How long would it take a human to complete the marathon?

All photos source: Wikipedia

Strands

Number Sense and Numeration, Measurement

Problem of the Week Problem B Speedy or Not, You May Be Caught! Problem Various animals can travel at the following top speeds for short distances: • cheetahs race along at 120 km/h(that is, they travel 120 km every hour); • cows can gallop at 12.8 km/h; • snails can slither along at 0.0125 m/s; A human’s top speed is 10.4 m/s. a) How many seconds would it take each animal to run a 100 m race? How long would it take a human to complete the 100 m? b) While hunting, cheetahs can only maintain their amazing speed for about 15 s, after which they generally give up the chase. Gazelles can run 60 km/h when evading predators. Suppose a cheetah spots a gazelle 300 m away. If both animals start at the same time and run in a straight line, will the gazelle escape? For further consideration: If the top speeds of the animals could be maintained over a long distance (physically impossible), approximately how long would it take for the animals to run a 42.2 km marathon? How long would it take a human to complete the marathon? Solution a) Since we want our answer in seconds, it is helpful to convert hours to seconds: 1 h = 60 minutes = 60 × 60 = 3600 seconds For the cheetah:

Also, remember that 1 km = 1000 m. 120 km in 1 h 120 000 m in 3600 s ÷1200 100 m in 3600 ÷ 1200 = 3 s

For the cow: ÷128 For the snail: ÷0.0125 ×100 For the human: ÷10.4 ×100

12.8 km in 1 h 12 800 m in 3600 s 100 m in 3600 ÷ 128 ≈ 28.1 s 0.0125 m in 1 s 1 m in 1 ÷ 0.0125 = 80 s 100 m in 80 × 100 = 8000 s 10.4 m in 1 s 1 m in 1 ÷ 10.4 s 100 m in 1 ÷ 10.4 × 100 ≈ 9.6 s

To run 100 m, the cheetah would take 3 s, the cow would take approximately 28.1 s, the snail would take 8000 s, and the human would take approximately 9.6 s.

b) Since the cheetah will give up the chase after 15 s, we need only see where both animals are at that time. The gazelle runs 60 km in 60 minutes. This is the same as 1 km in 1 minute. This is the same as 1000 m in 60 seconds. Dividing by 4, we determine that the gazelle can run 250 m in 15 seconds. The cheetah runs 120 km in 60 minutes. This is twice as fast as the gazelle. Therefore, the cheetah can run 2 × 250 = 500 m in 15 seconds. Since the gazelle had a 300 m headstart, it will be 550 m from where the cheetah started. This is 50 m ahead of the cheetah and the gazelle will escape to live another day. ≈ 16.67 m and the cheetah Let’s look at this using a chart. In 1 second, the gazelle runs 1000 60 will run ≈ 33.33 m. Therefore, every second the cheetah will get ≈ 16.67 m closer to the gazelle. Every 3 seconds the cheetah will gain 50 m on the gazelle. The cheetah starts at 0 m and the gazelle starts at 250 m. The chart illustrates that the cheetah is getting closer but never reaches the gazelle in 15 seconds. Time (in s) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Location of cheetah (in m) 0 33.33 66.66 100 133.33 166.66 200 233.33 266.66 300 333.33 366.66 400 433.33 466.66 500

Location of Gazelle 300 316.67 333.33 350 366.67 383.33 400 416.67 433.33 450 466.67 483.33 500 516.67 533.33 500

Distance Apart (in m) 300 283.33 266.67 250 233.33 216.67 200 183.33 166.67 150 133.33 116.67 100 83.33 66.67 50

For further consideration: A marathon is 42.2 km or 42 200 m. We can obtain the times for a marathon by just multiplying our answers in a) by 422. For the cheetah, it will take 3 × 422 = 1266 s. If we divide this by 3600, we convert our answer to hours. It would take 0.35 h which is approximately 21 minutes. For the cow, it will take 28.1 × 422 = 11 858.2 s. If we divide this by 3600, we convert our answer to hours. It would take 3.3 h which is approximately 3 h 18 minutes. For the snail, it will take 8000 × 422 = 3 376 000 s. If we divide this by 3600, we convert our answer to hours. It would take 937.8 h which is approximately 937 h 47 minutes. This is just over 39 days! For the human, it will take 9.6 × 422 = 4051.2 s. If we divide this by 3600, we convert our answer to hours. It would take 1.125 h which is approximately 1 h 8 minutes.

Problem of the Week Problem B Bytes Rule! Measurements depend critically on the units used. Here are some sample measures of length, and of computer memory. a) When measuring length, the metric system uses different prefixes to go with the base unit, metres (m). Complete the table using the following metric prefixes: centi (c), deca (da), deci (d), hecto (h), milli (m) (Two rows of the chart are done for you.) term kilometre

prefix shortform conversion from m kilo

km

×1000 ×100 ×10

metre

m ÷10 ÷100 ÷1000

b) Computer memory also uses metric prefixes, but on a larger scale. The base unit for memory is a byte. Fill in the table using the prefixes. mega (M), tera (T), giga (G). term

prefix shortform

conversion ×1 000 000 000 000 ×1 000 000 000 ×1 000 000

kilobyte

kilo

kB

byte

×1000

B

c) What will happen as computer memory continues to expand? Do some research and fill in the table for the future of amounts of memory. term prefix shortform

conversion ×1 000 000 000 000 000 000 000 ×1 000 000 000 000 000 000 ×1 000 000 000 000 000

byte

b

Strand

Measurement

Problem of the Week Problem B Bytes Rule! Problem Measurements depend critically on the units used. Here are some sample measures of length, and of computer memory. a) When measuring length, the metric system uses different prefixes to go with the base unit, metres (m). Complete the table using the following metric prefixes: centi (c), deca (da), deci (d), hecto (h), milli (m)

term

prefix

shortform

conversion from m

kilometre

kilo

km

×1000

hectometre

hecto

hm

×100

decametre

deca

dam

×10

metre

m

decimetre

deci

dm

÷10

centimetre

centi

cm

÷100

millimetre

milli

mm

÷1000

b) Computer memory also uses metric prefixes, but on a larger scale. The base unit for memory is a byte. Fill in the table using the prefixes. mega (M), tera (T), giga (G). term

prefix

shortform

conversion

terabyte

tera

TB

×1 000 000 000 000

gigabyte

giga

GB

×1 000 000 000

megabyte

mega

MB

×1 000 000

kilobyte

kilo

kB

×1000

byte

B

c) What will happen as computer memory continues to expand? Do some research and fill in the table for the future of amounts of memory. term

prefix

shortform

conversion

zettabyte

zetta

ZB

×1 000 000 000 000 000 000 000

exabyte

exa

EB

×1 000 000 000 000 000 000

petabyte

peta

PB

×1 000 000 000 000 000

byte

B

Solution The answers are given in the completed tables within the problem statement above.

Problem of the Week Problem B Looks Can Be Deceiving Using grid paper, you may wish to construct each of the diagrams shown below. This may be helpful as you work through the problem. In each of the three diagrams below, there are two squares which overlap in various ways. The bottom square is fixed and has side length 12. The overlapping square has one vertex located at the centre C of the bottom square. The darker shaded regions are common to the two squares and are labeled A1, A2 and A3, respectively.

a) Determine the areas of A1, A2 and A3, the overlapping areas in each of the above diagrams. How is each of these areas related to the total area of the square of side length 12? b) Below is a diagram in which the overlapping area, A4, is in an arbitrary position. Show that the blank square of side length 12 can be completely tiled by four copies of A4 and use this to determine the overlapping area, A4.

Extension: The above example shows us that the overlapping area of the two squares is 36 units2 , one-quarter of the total area. Would it make any difference if the overlapping figure were a rectangle? A right-angled triangle? Another polygon? Explain your thinking.

Strands

Geometry and Spatial Sense, Measurement

Problem of the Week Problem B Looks Can Be Deceiving Problem Using grid paper, you may wish to construct each of the diagrams shown below. This may be helpful as you work through the problem. In each of the three diagrams below, there are two squares which overlap in various ways. The bottom square is fixed and has side length 12. The overlapping square has one vertex located at the centre C of the bottom square. The darker shaded regions are common to the two squares and are labeled A1, A2 and A3, respectively.

a) Determine the areas of A1, A2 and A3, the overlapping areas in each of the above diagrams. How is each of these areas related to the total area of the square of side length 12? b) Below is a diagram in which the overlapping area, A4, is in an arbitrary position. Show that the blank square of side length 12 can be completely tiled by four copies of A4 and use this to determine the overlapping area, A4.

Extension: The above example shows us that the overlapping area of the two squares is 36 units2 , one-quarter of the total area. Would it make any difference if the overlapping figure were a rectangle? A right-angled triangle? Another polygon? Explain your thinking. Solution The solution is on the next page.

a) A1 is a triangle of height 6 and base 12, and thus has area

1 2

× 12 × 6 = 36 square units.

A2 is a square of side 6, with area 6 × 6 = 36 square units. A3 consists of two right-angled triangles of base length 3 and height 6, plus a rectangle with side lengths 3 and 6. The area of each right-angled triangle is 12 × 3 × 6 = 9 square units. The area of the rectangle is 3 × 6 = 18 square units. The total area of A3 is 9 + 9 + 18 = 36 square units. So all three overlapping areas A1, A2, and A3 have exactly the same area!

B) The way in which A4 tiles the square of side 12 is illustrated at the right. Since this shows that the overlapping area is 36 square units (i.e., 41 of the area of the bottom square) for any position in which the overlapping figure is an irregular quadrilateral, and part a) shows that this also holds for the two special cases, we see that the overlapping area is always the same, 36 square units or 41 of the area of the square with side length 12. Extension: As long as the overlapping polygon has one vertex which is a right angle pinned at the centre, C, of the base square and the two sides forming that vertex are long enough that they intersect the sides of the bottom square, the shape of the overlapping area would always be one of type A1, A2, or A4, with area equal to 41 of the area of the bottom square.

Problem of the Week Problem B Fenced In! Cassie is building a cedar split-rail fence around her horse paddock, such as the one illustrated in the photo. The three rails are laid horizontally, with the top ones supported in the notch of each fourpost bundle, and the other two suspended by wires above them. Where they meet, each pair of rails overlap somewhat to provide stability for the fence. a) Suppose the 3 rails in Cassie’s fence have a 6 inch ( 21 ft) overlap, as shown in the diagram below. If she needs a fence 105 ft long, and each of her cedar rails is 10 ft long, how many rails will she need in total for the horizontal spans of her fence? b) Cassie also needs to construct the four-post bundles for her fence. If the posts are to be 5 ft high in each bundle, how many cedar rails will she need for the bundles? c) Suppose instead that Cassie can get cedar rails 13 ft long. How many of these rails would she need in total to construct her 105 ft fence?

Strands

Measurement, Number Sense and Numeration

Problem of the Week Problem B Fenced In! Problem Cassie is building a cedar split-rail fence around her horse paddock. The three rails are laid horizontally, with the top ones supported in the notch of each four-post bundle, and the other two suspended by wires above them. Where they meet, each pair of rails overlap somewhat to provide stability for the fence. a) Suppose the 3 rails in Cassie’s fence have a 6 inch ( 12 ft) overlap, as shown in the diagram below. If she needs a fence 105 ft long, and each of her cedar rails is 10 ft long, how many rails will she need in total for the horizontal spans of her fence? b) Cassie also needs to construct the four-post bundles for her fence. If the posts are to be 5 ft high in each bundle, how many cedar rails will she need for the bundles? c) Suppose instead that Cassie can get cedar rails 13 ft long. How many of these rails would she need in total to construct her 105 ft fence?

Solution a) First we need to think about how many spans of 3 rails Cassie needs. Since each rail is 10 ft long, and the fence is to be 105 ft long, she will need at least 11 spans. But we must also take into account the loss of 12 ft for each internal overlap. Here is a diagram to illustrate what one level of horizontal rails would look like.

Thus, if we look at having 11 spans, there would be 2 end rails of 9 21 ft, and 9 internal spans of 9 ft, plus 10 overlaps of 12 ft each. So 11 spans would give a total length of 2×9

1 1 + 9 × 9 + 10 × = 19 + 81 + 5 = 105 ft, 2 2

exactly the length of fence Cassie needs. Thus she would need 11 × 3 = 33 cedar rails in total for the horizontal spans of her fence.

b) The 11 horizontal spans will require 12 four-post bundles, since she needs them at both ends of the fence. Thus she will need 12 × 4 = 48 posts, each 5 ft high. Since each cedar rail is 10 ft long, she can get 2 posts from each rail, so she will need 24 cedar rails for the bundles. In total, Cassie would need 33 + 24 = 57 of the 10 ft long cedar rails. c) Since 105 ÷ 13 ≈ 8.1, we will need more than 8 spans of the 13 ft horizontal rails. Let’s try using 9 spans; replacing 9 by 12 in our calculation from part a), we see that 9 spans would give a total length of 2 × 12

1 1 + 7 × 12 + 8 × = 25 + 84 + 4 = 113 ft. 2 2

Thus 9 spans are sufficient. Since we exceeded the 105 ft by 8 ft the final span would only need to be 5 ft long. Cassie would only need two of the 13 ft rails to get the three horizontal 5 ft pieces for that end. So for the 9 horizontal spans, she would need 8 × 3 + 2 = 26 of the 13 ft cedar rails. In addition, she would need 10 four-post bundles, each of which would require 2 cedar rails, since she could only cut 2 posts 5 ft high from each 13 ft cedar rail. Thus she would need 20 cedar rails for the bundles, giving a total of 46 rails for the entire fence.

 

Problem of the Week Problem B The Time of Your Life! As you answer these questions, recall that there are 365 days in a year (except for a leap year), 24 hours in a day, 60 minutes in each hour, and 60 seconds in each minute. a) The oldest human ever to have lived, according to the Guiness Book of Records, was 122 years and 164 days old at the time of her death. Jeanne Louise Calment (France) was born on February 21, 1875, and died at a nursing home in Arles, in southern France, on August 4, 1997. How many minutes was this person alive (assuming she was born and died at midnight on the given dates)? b) If you were born on January 1, 2005, how many years old would you be when you celebrate the billionth second that you have been alive?

Jeanne Louise Calment 1996 Source: Wikipedia

Strands

Number Sense and Numeration, Measurement

Problem of the Week Problem B The Time of Your Life! Problem As you answer these questions, recall that there are 365 days in a year (except for a leap year), 24 hours in a day, 60 minutes in each hour, and 60 seconds in each minute. a) The oldest human ever to have lived, according to the Guiness Book of Records, was 122 years and 164 days old at the time of her death. Jeanne Louise Calment (France) was born on February 21, 1875, and died at a nursing home in Arles, in southern France, on August 4, 1997. How many minutes was this person alive (assuming she was born and died at midnight on the given dates)? b) If you were born on January 1, 2005, how many years old would you be when you celebrate the billionth second that you have been alive?

Jeanne Louise Calment 1996 Source: Wikipedia

Solution a) To determine how many minutes Jeanne Louise Calment was alive, we must convert her age of 122 years and 164 days to minutes. Since leap years occur every four years, we shall assume there are 365.25 days in a year, on average. Thus the total number of days is 164 days + 122 years = 164 + (122 × 365.25) = 44 724.5 days.

Since there are 60 × 24 = 1 440 minutes in each 24 hour day, we see that she was alive for a total of 44 724.5 × 1 440 = 64 403 280 minutes. Note: If, instead of using the average number of days in a year, you count the number of leap years (31) and non-leap years (91) that Jeanne Louise was alive, you will get a very slightly different answer, namely 64 404 000 minutes. It is an interesting class discussion to think about why this answer is slightly larger. b) We need to figure out how many years there are in one billion seconds. Working upwards in the size of the time unit, we have: 1 000 000 000 seconds equals 1 000 000 000 ÷ 60 = 16 666 666 32 minutes; 16 666 666 23 minutes equals 16 666 666 23 ÷ 60 ≈ 277 777.778 hours; 277 277.778 hours equals 277 777.778 ÷ 24 ≈ 11 574.074 days; 11 574.074 days equals 11 574.074 ÷ 365.25 ≈ 31.7 years. Thus you will be more than 31 years old when you celebrate your billionth second of life. Extension: How long have YOU been alive? • How many whole number years? • How many whole number days (don’t forget leap years)? • How many whole number hours (assuming you were born at midnight on your birthday)? • How many minutes? • How many seconds?

Problem of the Week Problem B Making Your Mark A group of five students decided to mark up a 100s chart in different ways. • Zaffar shaded the numbers divisible by 2; • Yolanda marked the numbers divisible by 3 with a check mark in the upper right corner; • Xerxes circled the numbers divisible by 5; • William marked the numbers divisible by 7 with an X in the upper left corner; and • Violet underlined the prime numbers. a) Did at least one symbol appear on all 100 numbers on the chart? b) Which numbers had the most symbols? c) How many symbols appeared on the numbers in part b)? d) If the chart went beyond 100, what would be the least number that has the symbols of Zaffar, Yolanda, Xerxes, and William? e) Is there a number greater than 100 which would satisfy the conditions of all five students? Explain.

Strands

Number Sense and Numeration, Patterning and Algebra

Problem of the Week Problem B Making Your Mark Problem A group of five students decided to mark up a 100s chart in different ways. • Zaffar shaded the numbers divisible by 2; • Yolanda marked the numbers divisible by 3 with a check mark in the upper right corner; • Xerxes circled the numbers divisible by 5; • William marked the numbers divisible by 7 with an X in the upper left corner; and • Violet underlined the prime numbers. a) Did at least one symbol appear on all 100 numbers on the chart? b) Which numbers had the most symbols? c) How many symbols appeared on the numbers in part b)? d) If the chart went beyond 100, what would be the least number that has the symbols of Zaffar, Yolanda, Xerxes, and William? e) Is there a number greater than 100 which would satisfy the conditions of all five students? Explain.

Solution The completed chart is shown following the answers below. a) At least one symbol appeared on all numbers except 1. b) The numbers which had the greatest number of symbols were 30, 42, 60, 70, 84, and 90. c) Each of the numbers in part b) had 3 symbols. The numbers 30, 60, and 90 are divisible by 2, 3, and 5; the numbers 42 and 84 are divisible by 2, 3, and 7; and the number 70 is divisible by 2, 5, and 7. d) A number which satisfies the conditions of Zaffar, Yolanda, Xerxes, and William would have to be a multiple of each of 2, 3, 5, and 7. Thus the least such number would be 2 × 3 × 5 × 7 = 210.

e) To satisfy the conditions of all five students, a number would have to be both prime AND a multiple of 210. This is a contradiction, since a prime is only divisible by 1 and by itself. Thus no such number exists.

Problem of the Week Problem B Simply Interesting! A boy named Richard has $1000 to invest in one of two longterm plans. He intends to withdraw only the interest at the end of each year, but leave his $1000 invested. The two plans are as follows: Plan 1: This plan pays $50 interest each year as long as the original investment is untouched. This $50 is called 5% simple interest. Plan 2: This plan pays $500 interest at the end of the first year. (This $500 interest is 50% of the original investment.) At the end of each of the following years, the $1000 investment earns half as much interest as the year before. (So in the second year it would only earn $250 in interest.) a) After 5 years, what is the total interest that Richard would make from each plan? After 10 years? b) Which option should he choose to maximize the total interest if he invests for 15 years? c) If Richard left his $1000 invested for a very long time, which plan do you think would provide the greatest total interest? Plan 1

Plan 2

Year

Investment

Interest

Total Interest

Investment

Interest

Total Interest

1

$1000

$50

$ 50

$1000

$500

$500

2

$1000

$50

$100

$1000

$250

$750

3

$1000

$1000

4

$1000

$1000

5

$1000

$1000

6

$1000

$1000

7

$1000

$1000

8

$1000

$1000

9

$1000

$1000

10

$1000

$1000

11

$1000

$1000

12

$1000

$1000

13

$1000

$1000

14

$1000

$1000

15

$1000

$1000

Strand

Number Sense and Numeration

Problem of the Week Problem B Simply Interesting! Problem A boy named Richard has $1000 to invest in one of two longterm plans. He intends to withdraw only the interest at the end of each year, but leave his $1000 invested. The two plans are as follows: Plan 1: This plan pays $50 interest each year as long as the original investment is untouched. This $50 is called 5% simple interest. Plan 2: This plan pays $500 interest at the end of the first year. (This $500 interest is 50% of the original investment.) At the end of each of the following years, the $1000 investment earns half as much interest as the year before. (So in the second year it would only earn $250 in interest.) a) After 5 years, what is the total interest that Richard would make from each plan? After 10 years? b) Which option should he choose to maximize the total interest if he invests for 15 years? c) If Richard left his $1000 invested for a very long time, which plan do you think would provide the greatest total interest? Solution

The completed table on the next page reveals the answers to parts a) and b). There are two sets of answers for Plan 2: the first carries 4 decimals in the calculations and is the more accurate; the bracketed numbers use rounding to the nearest cent, and are inaccurate, especially after year 6, due to round-off error. Answers will vary depending on how many decimals are used. Clearly Plan 2 will earn Richard more in 15 years. However, if he leaves the $1000 invested indefinitely, Plan 2 total interest appears to reach a maximum of at $1000, whereas Plan 1 total interest continues to grow at $50 each year, and will surpass $1000 by year 21.

Plan 1

Plan 2

Year

Investment

Interest

Total Interest

Investment

Interest

Total Interest

1

$1000

$50

$ 50

$1000

$500.00

$500.00

2

$1000

$50

$100

$1000

$250.00

$750.00

3

$1000

$50

$150

$1000

$125.00

$875.00

4

$1000

$50

$200

$1000

$ 62.50

$937.50

5

$1000

$50

$250

$1000

$ 31.25

$968.75

6

$1000

$50

$300

$1000

$15.625 (15.63)

$984.375 (984.38)

7

$1000

$50

$350

$1000

$7.8125 (7.81)

$992.1875 (992.19)

8

$1000

$50

$400

$1000

$3.9063 (3.91)

$996.0938 (996.10)

9

$1000

$50

$450

$1000

$1.9531 (1.95)

$998.0469 (998.05)

10

$1000

$50

$500

$1000

$0.9766 (0.98)

$999.0234 (999.03)

11

$1000

$50

$550

$1000

$0.4883 (0.49)

$999.5117 (999.52)

12

$1000

$50

$600

$1000

$0.2441 (0.24)

$999.7559 (999.76)

13

$1000

$50

$650

$1000

$0.1221 (0.12)

$999.8779 (999.88)

14

$1000

$50

$700

$1000

$0.0610 (0.06)

$999.9390 (999.94)

15

$1000

$50

$750

$1000

$0.0305 (0.03)

$999.9695 (999.97)

The completed table reveals the answers to parts a) and b). Total interest from Plan 1 after 5 years is $250 whereas the total interest after 5 years from Plan 2 is $968.75. Total interest from Plan 1 after 10 years is $500 whereas the total interest after 10 years from Plan 2 is $999.02. There are two sets of answers for Plan 2: the first carries 4 decimals in the calculations and is the more accurate; the bracketed numbers use rounding to the nearest cent, and are inaccurate, especially after year 6, due to round-off error. Answers will vary depending on how many decimals are used. For c), clearly Plan 2 will earn Richard more in 15 years. However, if he leaves the $1000 invested indefinitely, Plan 2 total interest appears to reach a maximum of $1000, whereas Plan 1 total interest continues to grow at $50 each year, and will surpass $1000 by year 21.

Problem of the Week Problem B “B” There or “B” Square! A and B are two squares such that the perimeter of Square A is a multiple of the perimeter of Square B by a whole number. We wish to explore how the areas of the two squares are related.

a) If the perimeter of Square A is 3 times that of Square B, what is the ratio of the area of Square B to the area of Square A? Illustrate your answer with a diagram on graph paper. Express your answer as both a ratio and a fraction. b) What if the perimeter of Square A is 4 times that of B? 5 times that of B? c) Explain how you would figure out the area ratio if the perimeter of Square A is 10 times that of Square B.

Strands

Measurement, Number Sense and Numeration

Problem of the Week Problem B “B” There or “B” Square! Problem A and B are two squares such that the perimeter of Square A is a multiple of the perimeter of Square B by a whole number. We wish to explore how the areas of the two squares are related. a) If the perimeter of Square A is 3 times that of Square B, what is the ratio of the area of Square B to the area of Square A? Illustrate your answer with a diagram on graph paper. Express your answer as both a ratio and a fraction. b) What if the perimeter of Square A is 4 times that of B? 5 times that of B? c) Explain how you would figure out the area ratio if the perimeter of Square A is 10 times that of Square B.

Solution a) The diagram below reveals that when the perimeter of Square A is 3 times that of Square B, then the ratio of the area of Square B to the area of Square A is 1:9, or, in fraction form, 19 . This is due to the fact that each side of Square A is 3 times as long as that of Square B, in order to make the perimeter 3 times as long. Thus, if the side of Square B is taken as 1 unit, the side of Square A will be 3, and hence its area is 3 × 3 = 9 square units. (3 × 3 can also be written using exponents as 32 . We would say “3 squared”.)

b) The diagram also illustrates that when the perimeter of Square A is 4 times that of Square B, then the ratio of the area of Square B to the area of Square A is 1 : 16 = 1 : 42 , 1 or, in fraction form, 16 . Similarly, we would expect that when the side of Square A is 5 times that of Square B, then the areas will be in the ratio 1 : 25 = 1 : 52 , or, in fraction 1 form, 25 . c) Following the reasoning of parts a) and b), when the perimeter of Square A is 10 times that of Square B, then each side will be 10 times as long, and the areas will be in the 1 ratio 1 : 100 = 1 : 102 , or, in fraction form, 100 .

Problem of the Week Problem B Confusion Rules a) Tian and Mary disagree about the value of the expression 6 × 4 − 8 ÷ 2 + 5. Tian thinks it should be 13, but Mary says it should be 25. Try to explain how each of them arrived at their answer. b) The expression 6 × 4 − 8 ÷ 2 + 5 mixes the operations of addition and subtraction with that of multiplication and division. To avoid confusion in evaluating expressions like these, a special rule is followed: “Perform all multiplication and division first. Then perform all the addition and subtraction.” To help keep things straight, we can put brackets around the terms involving multiplication or division. The above expression could be written (6 × 4) − (8 ÷ 2) + 5. If the special rule is followed in the expression 6 × 4 − 8 ÷ 2 + 5, which student, Tian or Mary, has the correct answer? c) Evaluate each of the following expressions using the special rule. (i) 6 ÷ 3 + 4 − 2 × 2 + 5 (ii) 2 × 3 × 4 − 11 − 6 × 2 + 20 ÷ 5 (iii) 0.5 × 24 + 15 ÷ 5 − 13 Extension: Find the values of A and B so that each of the following expressions equals zero. (i) 22 ÷ 11 + 3 × A − 7 × 2 (ii) 24 ÷ B − 2 × 3 + 7 − 3 × 3

Strand

Number Sense and Numeration

Problem of the Week Problem B and Solution Confusion Rules Problem a) Tian and Mary disagree about the value of the expression 6 × 4 − 8 ÷ 2 + 5. Tian thinks it should be 13, but Mary says it should be 25. Try to explain how each of them arrived at their answer. b) The expression 6 × 4 − 8 ÷ 2 + 5 mixes the operations of addition and subtraction with that of multiplication and division. To avoid confusion in evaluating expressions like these, a spacial rule is followed: “Perform all multiplication and division first. Then perform all the addition and subtraction.” To help keep things straight, we can put brackets around the terms involving multiplication or division. The above expression could be written (6 × 4) − (8 ÷ 2) + 5. If the special rule is followed in the expression 6 × 4 − 8 ÷ 2 + 5, which student, Tian or Mary, has the correct answer? c) Evaluate each of the following expressions using the special rule. (i) 6 ÷ 3 + 4 − 2 × 2 + 5 (ii) 2 × 3 × 4 − 11 − 6 × 2 + 20 ÷ 5 (iii) 0.5 × 24 + 15 ÷ 5 − 13 Extension: Find the values of A and B so that each of the following expressions equals zero. 22 ÷ 11 + 3 × A − 7 × 2 24 ÷ B − 2 × 3 + 7 − 3 × 3

(i) (ii)

Solution a) Tian performed his operations moving left to right. 6 × 4 − 8 ÷ 2 + 5 → 24 − 8 ÷ 2 + 5 → 16 ÷ 2 + 5 → 8 + 5 → 13 Mary performed her operations by doing multiplication and division first followed by subtraction and addition last. 6 × 4 − 8 ÷ 2 + 5 → 24 − 8 ÷ 2 + 5 → 24 − 4 + 5 → 20 + 5 → 25 b) Using the special rule, Mary has the correct answer. 6 × 4 − 8 ÷ 2 + 5 = 24 − 8 ÷ 2 + 5 = 24 − 4 + 5 = 20 + 5 = 25

c) Evaluating each of the expressions using the special rule, we obtain: = = = = =

(6 ÷ 3) + 4 − (2 × 2) + 5 2 + 4 − (2 × 2) + 5 2+4−4+5 6−4+5 2+5 7

= = = =

(0.5 × 24) + (15 ÷ 5) − 13 12 + (15 ÷ 5) − 13 12 + 3 − 13 15 − 13 2

(i)

(iii)

(ii) = = = = = = =

(2 × 3 × 4) − 11 − (6 × 2) + (20 ÷ 5) (6 × 4) − 11 − (6 × 2) + (20 ÷ 5) 24 − 11 − (6 × 2) + (20 ÷ 5) 24 − 11 − 12 + (20 ÷ 5) 24 − 11 − 12 + 4 13 − 12 + 4 1+4 5

Extension: This part may have been challenging. (i)

(22 ÷ 11) + (3 × A) − (7 × 2) = 2 + (3 × A) − 14

We want 2 plus something to equal 14 so that when we subtract 14, the result is 0. The something would have to be 12. But 12 is 3 × A so A = 4. We can verify that we have the correct value for A by letting A = 4 in the expression. (22 ÷ 11) + (3 × 4) − (7 × 2) = 2 + (3 × 4) − (7 × 2) = 2 + 12 − (7 × 2) = 2 + 12 − 14 = 0 (ii) We will work out the multiplication parts and then determine the value of B. (24 ÷ B) − (2 × 3) + 7 − (3 × 3) = (24 ÷ B) − 6 + 7 − (3 × 3) = (24 ÷ B) − 6 + 7 − 9 Here you can play with possible values of B. If B = 3, 24 ÷ 3 = 8, 8 − 6 = 2, 2 + 7 = 9 and 9 − 9 = 0. B = 3 is the required value.

Problem of the Week Problem B Make A Deal! At this time, the Canadian dollar ($ CA) is worth less than the American dollar ($ US). In fact, $100 CA is only worth $75 US. For example, if you wanted to buy a notebook for $4 CA in Canada you would pay $3 US for the same notebook in the United States. Katie saw a shirt at Wieners in Canada for $23 CA. She can purchase the same shirt at Tarjay in the U.S. for $18 US. a) Where should Katie make her purchase if she wishes to pay the lowest price? To put it another way, is $18 US worth more, less or the same as $23 CA? b) What would $100 CA have to be worth in $ US in order to make $23 CA equivalent to $18 US?

Strand

Number Sense and Numeration

Problem of the Week Problem B Make A Deal! Problem At this time, the Canadian dollar ($ CA) is worth less than the American dollar ($ US). In fact, $100 CA is only worth $75 US. For example, if you wanted to buy a notebook for $4 CA in Canada you would pay $3 US for the same notebook in the United States. Katie saw a shirt at Wieners in Canada for $23 CA. She can purchase the same shirt at Tarjay in the U.S. for $18 US. a) Where should Katie make her purchase if she wishes to pay the lowest price? To put it another way, is $18 US worth more, less or the same as $23 CA? b) What would $100 CA have to be worth in $ US in order to make $23 CA equivalent to $18 US?

Solution a) Since $100 CA is equivalent to $75 US, dividing each number by 100 reveals that $1 CA is equivalent to $0.75 US (or 75 cents US). Thus, $23 CA must be equivalent to 23 × $0.75 = $17.25 US. This means that the $23 CA purchase price is equivalent to a $17.25 US purchase price. Since the Tarjay price is $18 US, it is cheaper for Katie to buy the shirt in Canada. b) We will work backwards to find the appropriate value. If $23 CA were equivalent to $18 US, then dividing each number by 23 reveals that $1 CA would have to be equivalent to $0.7826 US (to 4 decimals). Then $100 CA would be equivalent to 100 × $0.7826 = $78.26 US. Thus we can conclude that the two shirts would be of equivalent value at both stores if $100 CA were equivalent to $78.26 US.

Problem of the Week Problem B A Fitting Solution The figure below can be ’tiled’ (that is, it will be completely covered with no spaces) by a variety of arrangements of pattern blocks of different shapes. a) Using exactly 10 pattern blocks, how many of each shape are required to completely cover this figure? b) What fraction of the total area of the figure is covered by hexagons? by trapezoids? by rhombuses? by triangles? Express each of your answers reduced to lowest terms. c) In what other ways can you cover this figure with the four shapes from the pattern blocks (but not necessarily 10 blocks)?

To Think About Each of the hexagon, trapezoid and rhombus can be covered completely by a whole number of triangles. How many triangles are needed for each of the shapes?

Strands

Geometry and Spatial Sense, Number Sense and Numeration

Problem of the Week Problem B A Fitting Solution Problem The figure below can be ’tiled’ (that is, it will be completely covered with no spaces) by a variety of arrangements of pattern blocks of different shapes. a) Using exactly 10 pattern blocks, how many of each shape are required to completely cover this figure? b) What fraction of the total area of the figure is covered by hexagons? by trapezoids? by rhombuses? by triangles? Express each of your answers reduced to lowest terms. c) In what other ways can you cover this figure with the four shapes from the pattern blocks (but not necessarily 10 blocks)? Solution a) There is just one combination of shapes that can be used to cover this figure using exactly 10 pattern blocks. A possible solution is shown at the right. There is one hexagon, one trapezoid, one rhombus and seven triangles. 1 of the figure, the b) Since each green triangle covers 18 fractions of the total area of the figure covered by each colour are as follows: 6 1 yellow hexagon = six triangles = 18 = 13 3 = 16 1 red trapezoid = three triangles = 18 2 1 blue rhombus = two triangles = 18 = 19 7 7 green triangles = 18

c) There are seven other ways to cover this figure with four pattern blocks, as follows: 2 yellow hexagons, 1 red trapezoid, 1 blue rhombus, 1 green triangle (5 blocks); 1 yellow hexagon, 3 red trapezoids, 1 blue rhombus, 1 green triangle (6 blocks); 1 yellow hexagon, 2 red trapezoids, 2 blue rhombi, 2 green triangles (7 blocks); 1 yellow hexagon, 2 red trapezoids, 1 blue rhombus, 4 green triangles (8 blocks); 1 yellow hexagon, 1 red trapezoid, 4 blue rhombi, 1 green triangle (7 blocks); 1 yellow hexagon, 1 red trapezoid, 3 blue rhombi, 3 green triangles (8 blocks); 1 yellow hexagon, 1 red trapezoid, 2 blue rhombi, 5 green triangles (9 blocks). Convince yourself that these are correct by drawing diagrams for each. Think about why these are the only covers that use only four colours of pattern blocks.

Problem of the Week Problem B Block Out! Aputi (‘Snow on the Ground’), Siku (‘Ice’), and Qinu (‘Slushy Ice by the Sea’) are using rectangular blocks of snow to build a snow fort. a) They want to build one wall of their fort, but they can’t decide on the dimensions. If they have 24 blocks of snow, what are the possible different sizes of walls they could build, using whole blocks? Which of these would be sensible for a snow fort wall? b) The snow fort is to be rectangular, with four walls. Each wall is to be at least two, but no more than four blocks high. (i) If the front wall uses 24 blocks, and they can make up to 60 snow blocks in total, what possible dimensions could the other walls be? (Assume all four walls are the same height.) (ii) What is the total number of blocks they would need in each case? (iii) Which dimensions give the fort with greatest interior floor area? Which give the best protection? Organize your answers in the chart below. Note that b and h stand for base and height, respectively; all measurements are in blocks. The first line is for the sample shown below. Dimensions of the fort front×side×height

4×3×2

Strands

Dimensions of front and back walls b = 4, h = 2

No. of Blocks 16

Blocks for Side Walls (no corners) b = 1, h = 2

Measurement, Number Sense and Numeration

No. of Blocks 4

Total Blocks 20

Problem of the Week Problem B Block Out! Problem Aputi (‘Snow on the Ground’), Siku (‘Ice’), and Qinu (‘Slushy Ice by the Sea’) are using rectangular blocks of snow to build a snow fort. a) They want to build one wall of their fort, but they can’t decide on the dimensions. If they have 24 blocks of snow, what are the possible different sizes of walls they could build, using whole blocks? Which of these would be sensible for a snow fort wall? b) The snow fort is to be rectangular, with four walls. Each wall is to be at least two, but no more than four blocks high. (i) If the front wall uses 24 blocks, and they can make up to 60 snow blocks in total, what possible dimensions could the other walls be? (Assume all four walls are the same height.) (ii) What is the total number of blocks they would need in each case? (iii) Which dimensions give the fort with greatest interior floor area? Which give the best protection? Organize your answers in the chart below. Note that b and h stand for base and height, respectively; all measurements are in blocks. Dimensions of the fort 6×3×4

Dimensions of front and back walls b = 6, h = 4

8×3×3

front×side×height

No. of Blocks

Blocks for Side Walls (no corners)

No. of Blocks

Total Blocks

48

b = 1, h = 4

8

56

b = 8, h = 3

48

b = 1, h = 3

6

54

8×4×3

b = 8, h = 3

48

b = 2, h = 3

12

60

12 × 3 × 2

b = 12, h = 2

48

b = 1, h = 2

4

52

12 × 4 × 2

b = 12, h = 2

48

b = 2, h = 2

8

56

12 × 5 × 2

b = 12, h = 2

48

b = 3, h = 2

12

60

Solution a) Possible dimensions for the wall (length × height, in blocks) include 1 × 24, 2 × 12, 3 × 8, 4 × 6, 6 × 4, 8 × 3, 12 × 2, 24 × 1. Of these, only 6 × 4, 8 × 3, and 12 × 2 seem reasonable. b) The answers to parts (i) and (ii) are shown in the completed table above. (iii) The greatest interior floor area is in the 12 × 5 × 2 fort (with a 10 × 3 block floor area), while the best protection is the 6 × 3 × 4 fort (with the highest walls).

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Problem of the Week Problem B My How You've Grown, eh?

In 1970, the population of Canada was estimated at 21.32 million people. In 2010, the population was estimated to be 34.01 million. a) If Canada continues to grow at the same rate as this data predicts, what will be its population in 2030? b) If this rate of growth persists, when will Canada reach a population of 50 million?

Strands

Number Sense and Numeration,

Patterning and Algebra

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Problem of the Week Problem B My How You've Grown, eh?

Problem In 1970, the population of Canada was estimated at 21.32 million people. In 2010, the population was estimated to be 34.01 million.

a) If Canada continues to grow at the same rate as this data predicts, what will be its population in 2030? b) If this rate of growth persists, when will Canada reach a population of 50 million?

Solution a) The given data states that, over the four decades from 1970 to 2010, Canada's population increased by

12 690 000 (34 010 000 − 21 320 000)

people, or

12 690 000 ÷ 4 = 3 172 500

people per decade.

Since 2030 is another two decades after 2010, we would expect an addition to the population of size

3 172 500 × 2 = 6 345 000

people. Thus the total

population predicted for 2030 would be

34 010 000 + 6 345 000 = 40 355 000

people.

b) The additional population in the three decades from 2030 to 2060 would be

3 172 500 × 3 = 9 517 500, giving total population 40 355 000 + 9 517 500 = 49 872 500 The predicted annual growth is

by 2060.

3 172 500 ÷ 10 = 317 250

people per year.

Thus, by 2061, the total population would be

49 872 500 + 317 250 = 50 189 750. So the 50 million mark would be surpassed some time in 2061.

Problem of the Week Problem B “Draft/Schmaft" In a hockey draft lottery, teams are given a certain percent chance of winning the # 1 pick, based on how they placed the year before. Their chances are represented by coloured ping pong balls, with each team having a percentage of balls based on their standings in the previous year. Suppose a possible set of results from last year gives the following chances: • Columbus Red Coats 20% (20 chances out of 100 to win the number one pick); • Toronto Maple Stumps 13.5% (13.5 chances out of 100 to win the number one pick); • Arizona Toy Poodles 11.5%; • Carolina Gentle Breezes 10%; • Buffalo Lightsabres 8%; • the remaining teams in the league have a combined 37% of the wins. a) Why can’t a total of only 100 ping pong balls be used for the draw? b) What is the fewest number of balls needed to run the lottery? c) How many ping pong balls would Columbus get in the draw, assuming the fewest number of balls were used? d) How many balls would Toronto get? Extension: If Columbus wins the first pick and all their ping pong balls are removed from the draw, what is the probability that Buffalo will get the second pick? Give your answer as a fraction in lowest terms. Strands

Number Sense and Numeration, Data Management and Probability

Problem of the Week Problem B “Draft/Schmaft" Problem In a hockey draft lottery, teams are given a certain percent chance of winning the # 1 pick, based on how they placed the year before. Their chances are represented by coloured ping pong balls, with each team having a percentage of balls based on their standings in the previous year. Suppose a possible set of results from last year gives the following chances: • • • • • •

Columbus Red Coats 20% (20 chances out of 100 to win the number one pick); Toronto Maple Stumps 13.5% (13.5 chances out of 100 to win the number one pick); Arizona Toy Poodles 11.5%; Carolina Gentle Breezes 10%; Buffalo Lightsabres 8%; the remaining teams in the league have a combined 37% of the wins.

a) Why can’t a total of only 100 ping pong balls be used for the draw? b) What is the fewest number of balls needed to run the lottery? c) How many ping pong balls would Columbus get in the draw, assuming the fewest number of balls were used? d) How many balls would Toronto get? Extension: If Columbus wins the first pick and all their ping pong balls are removed from the draw, what is the probability that Buffalo will get the second pick? Give your answer as a fraction in lowest terms.

Solution a) A total of 100 ping pong balls is not enough, since it doesn’t give any way of representing 0.5%. b) Since no fraction of a percent occurs other than 0.5% in the known data, the lottery can be run with a minimum of 200 balls. That is, one ball for each 0.5% in 100%. c) If 200 balls are used, Columbus would get 20% of the total, 20 balls for each hundred or a total of 2 × 20 = 40 balls.. We can get the same result multiplying 200 by 0.2. d) If 200 balls are used, Toronto would get 13.5% of 200, or 27 balls. Extension: If Columbus wins the first pick, and their 40 balls are removed, there will now be 200 − 40 = 160 balls remaining. Buffalo’s has 8% of 200 balls, or 16 balls of the remaining 160 balls. Thus the probability of Buffalo getting the second pick is

16 160

=

1 . 10

Problem of the Week Problem B Time IS Money Sky had a job at a local camp during the summer holidays. a) One week, she worked the following hours: Monday 8:45 - 11:30, and 12:15 - 17:00 Tuesday 10:20 - 11:30, and 12:15 - 17:50 Wednesday 12:30 - 20:00 Thursday 9:00 - 12:00, and 12:45 - 17:00 Friday 7:15 - 11:00 How many hours did Sky work in total that week? b) The following week, Sky worked from Monday to Friday, 9 a.m. to 4 p.m. each day, with a 30-minute lunch break each day. She is not paid during her lunch break. Did she work more or less hours this week than the previous week? c) If Sky earned $10.40 per hour during the first week of work, how much money did she earn in total that week? (You will need to convert measures in hours and minutes to decimal numbers of hours.) d) In between the first and second weeks, Sky’s hourly rate of pay increased slightly. If she earned exactly the same amount of money in both weeks, what was her hourly rate of pay in the second week?

Strands

Measurement, Number Sense and Numeration

Problem of the Week Problem B Time IS Money

Problem Sky had a job at a local camp during the summer holidays. a) One week, she worked the following hours: Monday Tuesday Wednesday Thursday Friday

8:45 - 11:30, and 12:15 - 17:00 10:20 - 11:30, and 12:15 - 17:50 12:30 - 20:00 9:00 - 12:00, and 12:45 - 17:00 7:15 - 11:00

How many hours did Sky work in total that week? b) The following week, Sky worked from Monday to Friday, 9 a.m. to 4 p.m. each day, with a 30-minute lunch break each day. She is not paid during her lunch break. Did she work more or less hours this week than the previous week? c) If Sky earned $10.40 per hour during the first week of work, how much money did she earn in total that week? (You will need to convert measures in hours and minutes to decimal numbers of hours.) d) In between the first and second weeks, Sky’s hourly rate of pay increased slightly. If she earned exactly the same amount of money in both weeks, what was her hourly rate of pay in the second week?

Solution a) The total hours Sky worked each day are: Monday 8:45 - 11:30, and 12:15 - 17:00, giving 2hr 45min + 4hr 45min = 7hr 30min; Tuesday 10:20 - 11:30, and 12:15 - 17:50, giving 1hr 10min+ 5hr 35min = 6hr 45min; Wednesday 12:30 - 20:00, giving 7hr 30min; Thursday 9:00 - 12:00, and 12:45 - 17:00, giving 3hr + 4hr 15min = 7hr 15min; Friday 7:15 - 11:00, giving 3hr 45min. Thus Sky worked a total of 7hr 30min + 6hr 45min + 7hr 30min+ 7hr 15min + 3hr 45min = 32hr 45min. b) The following week, Sky worked 7 hr minus 30 min = 6hr 30min each day. Thus she worked a total of 6hr 30min×5 = 32hr 30min, which is 15 min less than the previous week.

c) Since 32hr 45min = 32.75 hr, Sky earned a total of $10.40 ×32.75 = $340.60 during the first week. d) Since she earned the same during the second week for only 32hr 30min hr, or 32.5 hr, her new hourly rate was $340.60 ÷ 32.5 = $10.48 per hour.

Problem of the Week Problem B Digital Reflections A prime number is a whole number greater than 1 that has only two factors, itself and 1. For example, 13 is a prime number since the only factors of 13 are 13 and 1. M and N are two different single digit numbers from the set 2, 3, 4, 5, 6, 7, 8, 9, so that • One of M or N is a prime number, and the other is not a prime number, • One of M or N is an even number, and the other is an odd number, and • If you add the two digit numbers MN and NM, their sum is KK, where K is another prime number. a) Which of the single digit numbers 2, 3, 4, 5, 6, 7, 8, 9 are prime, and which are not prime? b) Find all solutions for M and N?

Strand

Number Sense and Numeration

Problem of the Week Problem B Digital Reflections Problem A prime number is a whole number greater than 1 that has only two factors, itself and 1. For example, 13 is a prime number since the only factors of 13 are 13 and 1. M and N are two different single digit numbers from the set 2, 3, 4, 5, 6, 7, 8, 9, so that • One of M or N is a prime number, and the other is not a prime number, • One of M or N is an even number, and the other is an odd number, and • If you add the two digit numbers MN and NM, their sum is KK, where K is another prime number. a) Which of the single digit numbers 2, 3, 4, 5, 6, 7, 8, 9 are prime, and which are not prime? b) Find all solutions for M and N?

Solution a) The single digit numbers 2, 3, 5, and 7 are prime, while 4, 6, 8, and 9 are not prime. b) Since M and N must be different, one prime and one not prime, one even and one odd, the only possible combinations of M and N are 2 and 9, 3 and 4, 3 and 6, 3 and 8, 5 and 4, 5 and 6, 5 and 8, 7 and 4, 7 and 6, or 7 and 8. Of these pairs, three have two-digit sums MN + NM = KK, namely 34 + 43 = 77, 45 + 54 = 99 and 36 + 63 = 99. However, 9 is not a prime number. Thus the only solutions are M= 3, N= 4, or M= 4, N= 3.

Problem of the Week Problem B Sweet Profit In addition to making great contests and awesome problems, the CEMC (Centre for Education in Mathematics and Computing) produces amazing POTW cookies. We sell cookies with POW. This year’s batch has finally arrived. Each bag contains 20 cookies. You purchase four bags of cookies for $20 all together.

a) While munching away on these delicious morsels, you wonder what an individual cookie is worth. Find the cost of a single cookie; show each step of your calculation. b) It only costs the CEMC 15 cents to make a single POTW cookie. (Any profits are used to support fun math activities.) How much profit was made from your total purchase? c) Alain, John and Rob work for the CEMC and are able to buy cookies at cost. They want to sell cookies to raise money for an overnight problem solving event. They need to raise $1440. How many customers (who buy the same amount of cookies as you) do they need in order to reach their goal?

Strand

Number Sense and Numeration

Problem of the Week Problem B Sweet Profit Problem In addition to making great contests and awesome problems, the CEMC (Centre for Education in Mathematics and Computing) produces amazing POTW cookies. We sell cookies with POW. This year’s batch has finally arrived. Each bag contains 20 cookies. You purchase four bags of cookies for $20 all together.

a) While munching away on these delicious morsels, you wonder what an individual cookie is worth. Find the cost of a single cookie; show each step of your calculation. b) It only costs the CEMC 15 cents to make a single POTW cookie. (Any profits are used to support fun math activities.) How much profit was made from your total purchase? c) Alain, John and Rob work for the CEMC and are able to buy cookies at cost. They want to sell cookies to raise money for an overnight problem solving event. They need to raise $1440. How many customers (who buy the same amount of cookies as you) do they need in order to reach their goal?

Solution a) Since four bags cost $20, one bag costs $20 ÷ 4 = $5 . Since each bag contains 20 cookies, each cookie costs $5 ÷ 20 = $0.25 , i.e., 25 cents. b) If the CEMC pays 15 cents per cookie, they make a profit of 25 − 15 = 10 cents per cookie, or 20 × 10 cents = 200 cents, or $2.00 per bag. Thus, if you buy 4 bags, they make a total of $8.00 profit from your purchase. c) Since each customer will bring a profit of $8.00, Alain, Rob and John will need a total of $1440 ÷ $8 = 180 customers to reach their goal. The cookies are so good that this should be no problem at all.

Problem of the Week Problem B Speedy or Not, You May Be Caught! Various animals can travel at the following top speeds for short distances: • cheetahs race along at 120 km/h (that is, they travel 120 km every hour); • cows can gallop at 12.8 km/h; • snails can slither along at 0.013 m/s; A human’s top speed is 10.4 m/s. a) How many seconds would it take each animal to run a 100 m race? How long would it take a human to complete the 100 m? b) While hunting, cheetahs can only maintain their amazing speed for about 15 s, after which they generally give up the chase. Gazelles can run 60 km/h when evading predators. Suppose a cheetah spots a gazelle 300 m away. If both animals start at the same time and run in a straight line, will the gazelle escape? For further consideration: If the top speeds of the animals could be maintained over a long distance (physically impossible), approximately how long would it take the animals to run a 42.2 km marathon? How long would it take a human to complete the marathon?

All photos source: Wikipedia

Strands

Number Sense and Numeration, Measurement

Problem of the Week Problem B Speedy or Not, You May Be Caught! Problem Various animals can travel at the following top speeds for short distances: • cheetahs race along at 120 km/h(that is, they travel 120 km every hour); • cows can gallop at 12.8 km/h; • snails can slither along at 0.0125 m/s; A human’s top speed is 10.4 m/s. a) How many seconds would it take each animal to run a 100 m race? How long would it take a human to complete the 100 m? b) While hunting, cheetahs can only maintain their amazing speed for about 15 s, after which they generally give up the chase. Gazelles can run 60 km/h when evading predators. Suppose a cheetah spots a gazelle 300 m away. If both animals start at the same time and run in a straight line, will the gazelle escape? For further consideration: If the top speeds of the animals could be maintained over a long distance (physically impossible), approximately how long would it take for the animals to run a 42.2 km marathon? How long would it take a human to complete the marathon? Solution a) Since we want our answer in seconds, it is helpful to convert hours to seconds: 1 h = 60 minutes = 60 × 60 = 3600 seconds For the cheetah:

Also, remember that 1 km = 1000 m. 120 km in 1 h 120 000 m in 3600 s ÷1200 100 m in 3600 ÷ 1200 = 3 s

For the cow: ÷128 For the snail: ÷0.0125 ×100 For the human: ÷10.4 ×100

12.8 km in 1 h 12 800 m in 3600 s 100 m in 3600 ÷ 128 ≈ 28.1 s 0.0125 m in 1 s 1 m in 1 ÷ 0.0125 = 80 s 100 m in 80 × 100 = 8000 s 10.4 m in 1 s 1 m in 1 ÷ 10.4 s 100 m in 1 ÷ 10.4 × 100 ≈ 9.6 s

To run 100 m, the cheetah would take 3 s, the cow would take approximately 28.1 s, the snail would take 8000 s, and the human would take approximately 9.6 s.

b) Since the cheetah will give up the chase after 15 s, we need only see where both animals are at that time. The gazelle runs 60 km in 60 minutes. This is the same as 1 km in 1 minute. This is the same as 1000 m in 60 seconds. Dividing by 4, we determine that the gazelle can run 250 m in 15 seconds. The cheetah runs 120 km in 60 minutes. This is twice as fast as the gazelle. Therefore, the cheetah can run 2 × 250 = 500 m in 15 seconds. Since the gazelle had a 300 m headstart, it will be 550 m from where the cheetah started. This is 50 m ahead of the cheetah and the gazelle will escape to live another day. ≈ 16.67 m and the cheetah Let’s look at this using a chart. In 1 second, the gazelle runs 1000 60 will run ≈ 33.33 m. Therefore, every second the cheetah will get ≈ 16.67 m closer to the gazelle. Every 3 seconds the cheetah will gain 50 m on the gazelle. The cheetah starts at 0 m and the gazelle starts at 250 m. The chart illustrates that the cheetah is getting closer but never reaches the gazelle in 15 seconds. Time (in s) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Location of cheetah (in m) 0 33.33 66.66 100 133.33 166.66 200 233.33 266.66 300 333.33 366.66 400 433.33 466.66 500

Location of Gazelle 300 316.67 333.33 350 366.67 383.33 400 416.67 433.33 450 466.67 483.33 500 516.67 533.33 500

Distance Apart (in m) 300 283.33 266.67 250 233.33 216.67 200 183.33 166.67 150 133.33 116.67 100 83.33 66.67 50

For further consideration: A marathon is 42.2 km or 42 200 m. We can obtain the times for a marathon by just multiplying our answers in a) by 422. For the cheetah, it will take 3 × 422 = 1266 s. If we divide this by 3600, we convert our answer to hours. It would take 0.35 h which is approximately 21 minutes. For the cow, it will take 28.1 × 422 = 11 858.2 s. If we divide this by 3600, we convert our answer to hours. It would take 3.3 h which is approximately 3 h 18 minutes. For the snail, it will take 8000 × 422 = 3 376 000 s. If we divide this by 3600, we convert our answer to hours. It would take 937.8 h which is approximately 937 h 47 minutes. This is just over 39 days! For the human, it will take 9.6 × 422 = 4051.2 s. If we divide this by 3600, we convert our answer to hours. It would take 1.125 h which is approximately 1 h 8 minutes.

Problem of the Week Problem B How Will Dennis Fare? To encourage Dennis to work harder at his math problems, his mother promised she would pay him 10 cents for each right answer, but subtract 5 cents for each wrong answer. a) If he earned 20 cents after doing 32 problems, how many problems did Dennis get correct? How many did he get wrong? b) Dennis answered another 32 problems. What is the least number of problems he would have to get correct in order to earn more than one dollar? Think about how to explain this mathematically in an efficient way.

Strands

Patterning and Algebra, Number Sense and Numeration

Problem of the Week Problem B How Will Dennis Fare? Problem To encourage Dennis to work harder at his math problems, his mother promised she would pay him 10 cents for each right answer, but subtract 5 cents for each wrong answer. a) If he earned 20 cents after doing 32 problems, how many problems did Dennis get correct? How many did he get wrong? b) Dennis answered another 32 problems. What is the least number of problems he would have to get correct in order to earn more than one dollar? Think about how to explain this mathematically in an efficient way. Solution Dennis earns 10 cents for each right answer, but loses 5 cents for each wrong answer. a) Since two wrong answers cancel the earnings of one right answer, Dennis would earn nothing if he had twice as many wrong answers as right. If he earned 20 cents on 32 problems, then he must have done better than that. Let’s try some possibilities, starting near a two-to-one split of wrong versus right. Trial 1: 22 wrong and 10 right gives 10 × 10 cents − 22 × 5 cents = a 10 cent loss. Trial 2: 21 wrong and 11 right gives 11 × 10 cents − 21 × 5 cents = +5 cents. Trial 3: 20 wrong and 12 right gives 12 × 10 cents − 20 × 5 cents = +20 cents. So Dennis got 12 problems right and 20 problems wrong. b) Using the trials in a), it appears that each increase of 1 in the number of correct answers (and a corresponding decrease of 1 in the number of wrong answers) gains Dennis 15 cents. To earn at least $1.00, he needs to earn at least 80 cents more than the 20 cents he earned in part a). Since each increase in correct answers gains him 15 cents, he will earn more than $1.00 if he has 6 more than in a), since he will gain 6 × 15 = 90 cents. Checking: 18 right and 14 wrong gives 18 × 10 cents − 14 × 5 cents = +110 cents, or $1.10. (However, 17 right and 15 wrong gives 17 × 10 cents − 15 × 5 cents = +95 cents.) So Dennis needs at least 18 right answers to earn more than $1.00.

Problem of the Week Problem B Don’t Spill the Beans! On a shelf in an old-fashioned candy store sits a jar of 120 jelly beans, some red, some white, and some green. a) If half of the jelly beans are red, and 20% are green, how many candies are there of each colour? b) Juming loves red jelly beans. She takes a scoopful of red jelly beans from the jar. After these are removed, only 40% of the remaining jelly beans in the jar are now red. What percentage of the remaining beans are green or white? (Keep in mind that the remaining beans in the jar are the new 100%.) c) How many red jelly beans did Juming scoop from the jar?

Strand

Number Sense and Numeration

Problem of the Week Problem B Don’t Spill the Beans!

Problem On a shelf in an old-fashioned candy store sits a jar of 120 jelly beans, some red, some white, and some green. a) If half of the jelly beans are red, and 20% are green, how many candies are there of each colour? b) Juming loves red jelly beans. She takes a scoopful of red jelly beans from the jar. After these are removed, only 40% of the remaining jelly beans in the jar are now red. What percentage of the remaining beans are green or white? (Keep in mind that the remaining beans in the jar are the new 100%.) c) How many red jelly beans did Juming scoop from the jar?

Solution a) There is a total of 120 jelly beans in the jar. • Half of 120 is • 20% of 120 is

120 2 120 5

= 60, so 60 jelly beans are red. = 24, so 24 jelly beans are green.

Thus the number of white jelly beans is 120 − 60 − 24 = 36 . b) Since the remaining jelly beans are the new 100%, and 40% of these are red, therefore 100% − 40% = 60% of the jelly beans is in the jar are now green or white. 2 c) Since 40% is 40 60 = 3 of 60%, and there are still 24 + 36 = 60 green or white jelly beans, we see that there are 23 × 60 = 40 red jelly beans left in the jar. Thus, of the original 60 red jelly beans, Juming scooped 60 − 40 = 20 red jelly beans from the jar.

Problem of the Week Problem B Fenced In! Cassie is building a cedar split-rail fence around her horse paddock, such as the one illustrated in the photo. The three rails are laid horizontally, with the top ones supported in the notch of each fourpost bundle, and the other two suspended by wires above them. Where they meet, each pair of rails overlap somewhat to provide stability for the fence. a) Suppose the 3 rails in Cassie’s fence have a 6 inch ( 21 ft) overlap, as shown in the diagram below. If she needs a fence 105 ft long, and each of her cedar rails is 10 ft long, how many rails will she need in total for the horizontal spans of her fence? b) Cassie also needs to construct the four-post bundles for her fence. If the posts are to be 5 ft high in each bundle, how many cedar rails will she need for the bundles? c) Suppose instead that Cassie can get cedar rails 13 ft long. How many of these rails would she need in total to construct her 105 ft fence?

Strands

Measurement, Number Sense and Numeration

Problem of the Week Problem B Fenced In! Problem Cassie is building a cedar split-rail fence around her horse paddock. The three rails are laid horizontally, with the top ones supported in the notch of each four-post bundle, and the other two suspended by wires above them. Where they meet, each pair of rails overlap somewhat to provide stability for the fence. a) Suppose the 3 rails in Cassie’s fence have a 6 inch ( 12 ft) overlap, as shown in the diagram below. If she needs a fence 105 ft long, and each of her cedar rails is 10 ft long, how many rails will she need in total for the horizontal spans of her fence? b) Cassie also needs to construct the four-post bundles for her fence. If the posts are to be 5 ft high in each bundle, how many cedar rails will she need for the bundles? c) Suppose instead that Cassie can get cedar rails 13 ft long. How many of these rails would she need in total to construct her 105 ft fence?

Solution a) First we need to think about how many spans of 3 rails Cassie needs. Since each rail is 10 ft long, and the fence is to be 105 ft long, she will need at least 11 spans. But we must also take into account the loss of 12 ft for each internal overlap. Here is a diagram to illustrate what one level of horizontal rails would look like.

Thus, if we look at having 11 spans, there would be 2 end rails of 9 21 ft, and 9 internal spans of 9 ft, plus 10 overlaps of 12 ft each. So 11 spans would give a total length of 2×9

1 1 + 9 × 9 + 10 × = 19 + 81 + 5 = 105 ft, 2 2

exactly the length of fence Cassie needs. Thus she would need 11 × 3 = 33 cedar rails in total for the horizontal spans of her fence.

b) The 11 horizontal spans will require 12 four-post bundles, since she needs them at both ends of the fence. Thus she will need 12 × 4 = 48 posts, each 5 ft high. Since each cedar rail is 10 ft long, she can get 2 posts from each rail, so she will need 24 cedar rails for the bundles. In total, Cassie would need 33 + 24 = 57 of the 10 ft long cedar rails. c) Since 105 ÷ 13 ≈ 8.1, we will need more than 8 spans of the 13 ft horizontal rails. Let’s try using 9 spans; replacing 9 by 12 in our calculation from part a), we see that 9 spans would give a total length of 2 × 12

1 1 + 7 × 12 + 8 × = 25 + 84 + 4 = 113 ft. 2 2

Thus 9 spans are sufficient. Since we exceeded the 105 ft by 8 ft the final span would only need to be 5 ft long. Cassie would only need two of the 13 ft rails to get the three horizontal 5 ft pieces for that end. So for the 9 horizontal spans, she would need 8 × 3 + 2 = 26 of the 13 ft cedar rails. In addition, she would need 10 four-post bundles, each of which would require 2 cedar rails, since she could only cut 2 posts 5 ft high from each 13 ft cedar rail. Thus she would need 20 cedar rails for the bundles, giving a total of 46 rails for the entire fence.

 

Problem of the Week Problem B Don’t Be Square...Give Me a High Five! a) These numbers belong in a group: 25, 40, 115, 55. These numbers do not belong in this group: 33, 71, 4, 106. (i) Which of these numbers belong in this group: 75, 205, 87, 43? (ii) What is the rule which describes numbers in this group? b) These numbers belong in another group: 42,18,108,462. These numbers do not belong in this group: 21, 46, 99, 104. (i) Which of these numbers belong in this group: 63, 84, 118, 456? (ii) What is the rule which describes numbers in this group? c) In simple codes, a number is assigned to each letter of the alphabet by a rule known to the decoder. For example, suppose the rule is to multiply the letter’s order in the alphabet by 5. Then the first three words of the title of this problem, encoded according to this rule, would give 207570100 1025 958510559025 for “Don’t Be Square”. Decode the following sentence, which was encoded according to the same rule: 1154510040 6551004095 12575105 15570 2075 57012510040457035 d) Make up your own coding rules, and use them to code the title of this problem. Trade rules with a classmate and decode each other’s result as a check.

Strand

Patterning and Algebra

Problem of the Week Problem B Don’t Be Square...Give Me a High Five! Problem a) These numbers belong in a group: 25, 40, 115, 55. These numbers do not belong in this group: 33, 71, 4, 106. (i) Which of these numbers belong in this group: 75, 205, 87, 43? (ii) What is the rule which describes numbers in this group? b) These numbers belong in another group: 42,18,108,462. These numbers do not belong in this group: 21, 46, 99, 104. (i) Which of these numbers belong in this group: 63, 84, 118, 456? (ii) What is the rule which describes numbers in this group? c) In simple codes, a number is assigned to each letter of the alphabet by a rule known to the decoder. For example, suppose the rule is to multiply the letter’s order in the alphabet by 5. Then the first three words of the title of this problem, encoded according to this rule, would give 207570100 1025 958510559025 for “Don’t Be Square”. Decode the following sentence, which was encoded according to the same rule: 1154510040 6551004095 12575105 15570 2075 57012510040457035 d) Make up your own coding rules, and use them to code the title of this problem. Trade rules with a classmate and decode each other’s result as a check.

Solution a) The rule appears to be that the numbers in this group are each multiples of 5. Thus 75 = 15 × 5 and 205 = 41 × 5 belong to this group, while 87 and 43 do not. b) The rule in this part is not as obvious. However, it appears to be that the numbers in this group are each multiples of 6. Thus 84 = 14 × 6 and 462 = 77 × 6 belong to this group, while 63 and 118 do not. (Some may have thought that the numbers in this group were all even. However, if you look at the given numbers not in the group, you will see even numbers,)

c) The decoded sentence is “With maths you can do anything”. A table similar to the following may have been helpful in your decoding process. Letter Position Value 5×(Position Value) Letter Position Value 5×(Position Value)

A 1 5 N 14 70

B 2 10 O 15 75

C 3 15 P 16 80

D 4 20 Q 17 85

E 5 25 R 18 90

F 6 30 S 19 95

G 7 35 T 20 100

H 8 40 U 21 105

I 9 45 V 22 110

J 10 50 W 23 115

K 11 55 X 24 120

L 12 60 Y 25 125

M 13 65 Z 26 130

Then you can use the table to help split the digits to match the appropriate values in the table. 115 45 100 40

65 5 100 40 95

125 75 105

W I

MA T H S

Y O U

T H

15 5 70

20 75

5 70 125 100 40 45 70 35

CAN

DO

AN Y

T H I N G

La solution pour le problème en français est : Tu as bien réussi d) Answers will vary depending on the coding rule chosen by each student.

Problem of the Week Problem B Making Your Mark A group of five students decided to mark up a 100s chart in different ways. • Zaffar shaded the numbers divisible by 2; • Yolanda marked the numbers divisible by 3 with a check mark in the upper right corner; • Xerxes circled the numbers divisible by 5; • William marked the numbers divisible by 7 with an X in the upper left corner; and • Violet underlined the prime numbers. a) Did at least one symbol appear on all 100 numbers on the chart? b) Which numbers had the most symbols? c) How many symbols appeared on the numbers in part b)? d) If the chart went beyond 100, what would be the least number that has the symbols of Zaffar, Yolanda, Xerxes, and William? e) Is there a number greater than 100 which would satisfy the conditions of all five students? Explain.

Strands

Number Sense and Numeration, Patterning and Algebra

Problem of the Week Problem B Making Your Mark Problem A group of five students decided to mark up a 100s chart in different ways. • Zaffar shaded the numbers divisible by 2; • Yolanda marked the numbers divisible by 3 with a check mark in the upper right corner; • Xerxes circled the numbers divisible by 5; • William marked the numbers divisible by 7 with an X in the upper left corner; and • Violet underlined the prime numbers. a) Did at least one symbol appear on all 100 numbers on the chart? b) Which numbers had the most symbols? c) How many symbols appeared on the numbers in part b)? d) If the chart went beyond 100, what would be the least number that has the symbols of Zaffar, Yolanda, Xerxes, and William? e) Is there a number greater than 100 which would satisfy the conditions of all five students? Explain.

Solution The completed chart is shown following the answers below. a) At least one symbol appeared on all numbers except 1. b) The numbers which had the greatest number of symbols were 30, 42, 60, 70, 84, and 90. c) Each of the numbers in part b) had 3 symbols. The numbers 30, 60, and 90 are divisible by 2, 3, and 5; the numbers 42 and 84 are divisible by 2, 3, and 7; and the number 70 is divisible by 2, 5, and 7. d) A number which satisfies the conditions of Zaffar, Yolanda, Xerxes, and William would have to be a multiple of each of 2, 3, 5, and 7. Thus the least such number would be 2 × 3 × 5 × 7 = 210.

e) To satisfy the conditions of all five students, a number would have to be both prime AND a multiple of 210. This is a contradiction, since a prime is only divisible by 1 and by itself. Thus no such number exists.

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Problem of the Week Problem B My How You've Grown, eh?

In 1970, the population of Canada was estimated at 21.32 million people. In 2010, the population was estimated to be 34.01 million. a) If Canada continues to grow at the same rate as this data predicts, what will be its population in 2030? b) If this rate of growth persists, when will Canada reach a population of 50 million?

Strands

Number Sense and Numeration,

Patterning and Algebra

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Problem of the Week Problem B My How You've Grown, eh?

Problem In 1970, the population of Canada was estimated at 21.32 million people. In 2010, the population was estimated to be 34.01 million.

a) If Canada continues to grow at the same rate as this data predicts, what will be its population in 2030? b) If this rate of growth persists, when will Canada reach a population of 50 million?

Solution a) The given data states that, over the four decades from 1970 to 2010, Canada's population increased by

12 690 000 (34 010 000 − 21 320 000)

people, or

12 690 000 ÷ 4 = 3 172 500

people per decade.

Since 2030 is another two decades after 2010, we would expect an addition to the population of size

3 172 500 × 2 = 6 345 000

people. Thus the total

population predicted for 2030 would be

34 010 000 + 6 345 000 = 40 355 000

people.

b) The additional population in the three decades from 2030 to 2060 would be

3 172 500 × 3 = 9 517 500, giving total population 40 355 000 + 9 517 500 = 49 872 500 The predicted annual growth is

by 2060.

3 172 500 ÷ 10 = 317 250

people per year.

Thus, by 2061, the total population would be

49 872 500 + 317 250 = 50 189 750. So the 50 million mark would be surpassed some time in 2061.

Problem of the Week Problem B Even Dots are Odd The numbers 1, 3, 6, 10, 15, 21, and so on, form a sequence. That is, there is a rule which tells you how to find each number, or term in the sequence, from the previous term. a) These numbers are sometimes called Triangular Numbers because they can be represented by dots arranged in triangles. Here are the first three terms:

Term

Number

1

1

2

3

3

6

4

10

5

15

6

21

7 Draw the dot-triangles that would represent the next three terms, 10, 15, and 21.

8

b) Write a pattern rule for this sequence.

9

c) Enter the next five numbers in the sequence in the table.

10 11

Extension: What is the pattern of odd and even numbers in this sequence? Why does this happen?

Strand

Patterning and Algebra

Problem of the Week Problem B Even Dots are Odd Problem The numbers 1, 3, 6, 10, 15, 21, and so on, form a sequence. That is, there is a rule which tells you how to find each number, or term in the sequence, from the previous term. a) These numbers are sometimes called Triangular Numbers because they can be represented by dots arranged in triangles. Here are the first three terms:

Draw the dot-triangles that would represent the next three terms, 10, 15, and 21. b) Write a pattern rule for this sequence. c) Enter the next five numbers in the sequence in the table.

Term

Number

1

1

2

3

3

6

4

10

5

15

6

21

7

28

8

36

9

45

10

55

11

66

Extension: What is the pattern of odd and even numbers in this sequence? Why does this happen?

Solution a) Here are the dot-triangles for terms 4, 5, and 6.

b) Each number in the sequence is obtained by adding the number (label) of the new term to the value of the previous term. For example, Term 7 = Term 6 +7 = 21 + 7 = 28, Term 8 = Term 7 +8 = 28 + 8 = 36, and so on. Thus the pattern rule is Term(N+1) = Term(N) + (N+1). In the dot-triangles, this corresponds to adding a row to the bottom of the previous dot triangle, with one more dot in the new row. c) The next five numbers are shown in the completed table above. Extension: The pattern of odd and even numbers is odd, odd, even, even, odd, odd, even, even, i.e., 2 odd, then 2 even, repeatedly. This happens because the sum of two odd numbers is even, but the sum of an even number and an odd number is odd. Suppose we designate the parity of the term number in upper case (EVEN or ODD), and the parity of the value of the term in lower case italics (even or odd ). If we start with an EVEN term number which has an even value, then the next term number will be ODD, and its value will thus be an ODD number plus an even number, which is odd (e.g., Term 4 has value 10, and next comes Term 5, which is 10 + 5 = 15). So as the term numbers alternate between EVEN and ODD, the term values will follow this pattern: ODD term number + even value = odd value , EVEN term number + odd value = odd value , ODD term number + odd value = even value , EVEN term number + even value = even value , ODD term number + even value = odd value , with the last statement restarting the pattern, which repeats indefinitely.

Problem of the Week Problem B Look At Rick Shaw Go! Rick Shaw transports tourists through a park on his human-powered cart. He takes passengers around on several different routes (The Basic, The Classic, The Plus and The Ultimate), depending on how much they are willing to pay. On each route, he travels along each path (line segment) exactly once. He may pass through a node (vertex) more than once to complete a route. He does not have to start and end at the same place.

Above is a picture of Rick and his four possible routes. a) Determine a way to travel each of the routes. Record the results in the following chart. For The Basic, a successful route could start at A and end at C as follows: A→B→C→A→D→C (This is one possible route; there are others.) b) At each node in any route there are a number of intersecting paths. For each route, record the total number of intersecting paths, the number of intersecting paths at the start point and the number of intersecting paths at the end point. For The

Basic, there are 3 intersecting paths at node A, 2 intersecting paths at node B, 3 intersecting paths at node C, and 2 intersecting paths at node D. There is a total of 3 + 2 + 3 + 2 = 10 intersecting paths, 3 intersecting paths at the starting point A and 3 intersecting paths at the endpoint C. No. of Intersecting Paths Route Total Start End Basic: A → B → C → A → D → C

10

3

3

Classic: Plus: Ultimate:

c) What trends do you notice about the number of intersecting paths at the start and end points of each successful route? d) What trend do you see in the total number of intersecting paths? e) Try to draw two routes, one that has a successful path and one that does not. Switch your new routes with a partner. Predict which of your partner’s routes will work and which will not. Then confirm your predictions. Strands Geometry and Spatial Sense, Patterning and Algebra

Problem of the Week Problem B Look At Rick Shaw Go! Problem Rick Shaw transports tourists through a park on his human-powered cart. He takes passengers around on several different routes (The Basic, The Classic, The Plus and The Ultimate), depending on how much they are willing to pay. On each route, he travels along each path (line segment) exactly once. He may pass through a node (vertex) more than once to complete a route. He does not have to start and end at the same place.

Above is a picture of Rick and his four possible routes. a) Determine a way to travel each of the routes. Record the results in the following chart. (There may be more than one successful route.) b) At each node in any route there are a number of intersecting paths. For each route, record the total number of intersecting paths, the number of intersecting paths at the start point and the number of intersecting paths at the end point. c) What trends do you notice about the number of intersecting paths at the start and end points of each successful route? d) What trend do you see in the total number of intersecting paths? e) Try to draw two routes, one that has a successful path and one that does not. Switch your new routes with a partner. Predict which of your partner’s routes will work and which will not. Then confirm your predictions.

Solution a) b) The completed table is shown below. There are other possible routes. Route

No. of Intersecting Paths Total Start End

Basic: A → B → C → A → D → C

10

3

3

Classic: C → D → A → E → B → C → A → B

14

3

3

Plus: D → F → A → B → E → A → D → C → F → B → C

20

3

3

Ultimate: H → K → C → H → G → C → B → F → A → B → E → A → D → C

26

3

5

c) All successful routes start and end at a point with an odd number of intersecting paths. d) All the total numbers of intersecting paths are even. See the next page for further comments on parts c), d) and e), and on routes in general.

In each of the given diagrams (graphs) in the question, there were two nodes (vertices) with an odd number of intersecting paths. All other nodes had an even number of intersecting paths. In general you would be able to find a successful route if exactly two of the nodes in your diagram had an odd number of intersecting paths and all of the other nodes had an even number of intersecting paths. An Euler, pronounced “oiler”, path is a path that uses every edge of the graph exactly once, but starts and ends at a different node. An Euler circuit is a path that uses every edge of the graph exactly once, but starts and ends at the same node. In our problem, each route was an Euler path. If a graph has no nodes with an odd number of intersecting paths, you can find an Euler circuit. If a graph has exactly two nodes with an odd number of intersecting paths, you can find an Euler path by starting at either of those nodes. If a graph has three or more nodes with an odd number of intersecting paths, you cannot complete an Euler path or circuit. The following graphs are fairly similar. Exactly one of the graphs contains an Euler path, exactly one contains an Euler circuit and exactly one contains neither an Euler path or an Euler circuit. Can you correctly identify which type of path is contained in each graph.

Problem of the Week Problem B How Will Dennis Fare? To encourage Dennis to work harder at his math problems, his mother promised she would pay him 10 cents for each right answer, but subtract 5 cents for each wrong answer. a) If he earned 20 cents after doing 32 problems, how many problems did Dennis get correct? How many did he get wrong? b) Dennis answered another 32 problems. What is the least number of problems he would have to get correct in order to earn more than one dollar? Think about how to explain this mathematically in an efficient way.

Strands

Patterning and Algebra, Number Sense and Numeration

Problem of the Week Problem B How Will Dennis Fare? Problem To encourage Dennis to work harder at his math problems, his mother promised she would pay him 10 cents for each right answer, but subtract 5 cents for each wrong answer. a) If he earned 20 cents after doing 32 problems, how many problems did Dennis get correct? How many did he get wrong? b) Dennis answered another 32 problems. What is the least number of problems he would have to get correct in order to earn more than one dollar? Think about how to explain this mathematically in an efficient way. Solution Dennis earns 10 cents for each right answer, but loses 5 cents for each wrong answer. a) Since two wrong answers cancel the earnings of one right answer, Dennis would earn nothing if he had twice as many wrong answers as right. If he earned 20 cents on 32 problems, then he must have done better than that. Let’s try some possibilities, starting near a two-to-one split of wrong versus right. Trial 1: 22 wrong and 10 right gives 10 × 10 cents − 22 × 5 cents = a 10 cent loss. Trial 2: 21 wrong and 11 right gives 11 × 10 cents − 21 × 5 cents = +5 cents. Trial 3: 20 wrong and 12 right gives 12 × 10 cents − 20 × 5 cents = +20 cents. So Dennis got 12 problems right and 20 problems wrong. b) Using the trials in a), it appears that each increase of 1 in the number of correct answers (and a corresponding decrease of 1 in the number of wrong answers) gains Dennis 15 cents. To earn at least $1.00, he needs to earn at least 80 cents more than the 20 cents he earned in part a). Since each increase in correct answers gains him 15 cents, he will earn more than $1.00 if he has 6 more than in a), since he will gain 6 × 15 = 90 cents. Checking: 18 right and 14 wrong gives 18 × 10 cents − 14 × 5 cents = +110 cents, or $1.10. (However, 17 right and 15 wrong gives 17 × 10 cents − 15 × 5 cents = +95 cents.) So Dennis needs at least 18 right answers to earn more than $1.00.