ACTIVITY 1
To use a table to arrange information
[LO 2.1, 2.2]
1. Alice is making necklaces by using triangular motifs that she places on a wire necklet (as shown in Diagram 1 below). Each motif is made from three black beads, six white beads and six coloured beads. It is important that she has the right number of beads of each colour. She wants to make 50 necklaces with black, white and red beads, 40 with black, white and yellow beads, 40 with black, white and blue beads and 30 with black, white and green beads.
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Question: Calculate how many beads of each colour she has to buy.
2. She also combines three of these motifs in a bigger necklace (as shown in Diagram 2 above). The two top motifs in each necklace have the same design as before, but the bottom motif is made of 15 coloured beads only. She makes only half as many of these necklaces in each colour as she makes of the smaller necklaces referred to in Question 1.
Question: How many beads of each colour will she need for these necklaces?
3. She finds that the bigger necklaces are very popular, so she decides to use bigger motifs, and to combine more motifs in one necklace. The next two diagrams show her new plans.
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Alice wants to use four colours in these motifs. Use the diagram above to make your own four-colour design.
Question: Repeat the calculation exercises above for these necklaces.
4. These triangular motifs can be made larger and larger, of course. Not all of them are suitable for necklaces though!
Exercise: Below you see equilateral triangular Motifs 1 to 4. Draw Motifs 6 and 7 in the same sequence.
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The Motif 3 above has 1 black bead, 3 white beads and six red beads. You will need to refer to it in problem 6 below.
5. Data gathering: The beads in the motifs have a diameter of 1 cm each. Thus, the surrounding triangle in Motif 1 in the sequence has a side length of 2 cm. Complete the table below by referring to the triangles above as well as the ones you have drawn.
| Side length of triangle in centimetres | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Number of beads per triangle | 3 | 6 | 10 | ||||
| Perimeter of triangle | 6 | 9 |
6. Investigation: Alice makes her necklaces by combining triangular motifs as in diagrams 2 and 4. She uses the 10-bead motifs (Motif 3). The smallest necklace has one motif (size 1) and the next necklace has three motifs (size 2) with a triangular gap in the middle. Visualise the next sizes (3, 4, etc) of the necklaces (or draw them) and complete the table below. Try to complete the last column as well.
| Size of necklace | 1 | 2 | 3 | 4 | 5 | X |
| Number of triangular motifs | 6 | 10 | ||||
| Number of triangular spaces | ||||||
| Number of beads on each side of triangular motif | ||||||
| Total number of beads in necklace | ||||||
| Number of black beads | ||||||
| Total perimeter of pendant with 1 cm-diameter beads | 9 |
7. Investigation: Do the same for the next table, if Alice now uses the 15-bead motif that we saw in the very first diagram.
| Size of necklace | 1 | 2 | 3 | 4 | 5 | x |
| Number of triangular motifs | 1 | 3 | ||||
| Number of triangular spaces | 0 | 1 | ||||
| Number of beads on each side of triangular motif | 5 | 10 | ||||
| Total number of beads in necklace | 15 | 45 | ||||
| Number of black beads | 3 | 9 | ||||
| Total perimeter of pendant with 1 cm-diameter beads | 12 | 27 |
Putting information in a table makes it much easier to notice the patterns that the numbers form. That is why tables are very often used to arrange information. You will see many cases where tables are used in mathematics. Use a table whenever you think that it will be helpful when you are working on a problem.
ACTIVITY 2
To investigate relationships between variables
[LO 2.1, 2.6]
Mr and Mrs Peters want to hire a caravan for their holiday. They made enquiries about the cost from three caravan hire firms. Now they need to decide which caravan to hire, and for how long. We will help them decide. Here are the verbal descriptions of the details they obtained from the firms:
1. Arranging data: Complete the following table from the descriptions above.
| Number of days: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| Away-van: | R750 | ||||||||||
| Best Caravans: | R1560 | R1920 | |||||||||
| Car-a-holiday: | R1490 | R2030 |
2. Can you tell from the table which option will be the best depending on the length of the holiday?
Write a short summary of your conclusions.
3. This is a flow diagram for the Car-a-holiday prices. Calculate the missing values for the blocks:
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4. Make flow diagrams for the other two options.
5. Bradley is holidaying in America. He would like to hire a cell phone. He has investigated three different offers. Below are the following: A description in words of offer 1, A table of values for offer 2 and A flow diagram for offer 3.
There are spaces for you to complete the other two offers in each case. After you have completed the six, do the next exercise.
Verbal descriptions:
Offer 1: “ADVANCED MOBILE! Lowest call cost! Popular handset! $20 when you sign, plus 60 cents per call!”
Offer 2: “GENIE RENTALS
Offer 3: “HI–PRO
Tables:
| Number of calls: | 10 | 20 | 30 | 40 | 50 | 60 |
| Advanced mobile: | ||||||
| Genie rentals: | $24 | $38 | $52 | $66 | $80 | $940 |
| Hi–Pro: |
Flow diagrams:
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7. Now write down which one of the three representations was the most useful to you when you answered question 6. Give complete reasons for your statements.
ACTIVITY 3
To create a model to explain relationships
[LO 2.2, 2.3]
Olga likes chocolate-covered raisins. She has been making a list of the contents of the packets that she has been buying. She always buys from the same shop, but sometimes she buys small packets (50g), sometimes medium (100g) and sometimes large (200g) packets.
She has made a table of the facts she has discovered.
| Packet size | 50g | 100g | 200g |
| Average number of raisins | 78 | 153 | 304 |
| Cost per packet | R3,80 | R7,40 | R14,50 |
Olga contacted the manufacturers and found out from them that part of the price is for the packaging and the other part is for the contents. The packet cost is very similar for the three sizes, and the greatest part of the price is for the raisins. The unit cost of the contents stays the same regardless of the size of the packet.
Another interesting fact she found is that the factory controls very carefully that the packets do not contain too few raisins. They aim to have at least 75 in the 50g packet, 150 in the 100g packet and 300 in the 200g packet. To be sure that this happens, they are careful to make the packet of raisins with the same final mass. They also put a few extra raisins in most packets. Olga was very happy to tell them that her figures agreed with their standards.
From the given information, find out how much the raisins (excluding the packaging) cost. Give your answer in rand per kilogram.
| LO 2 |
| Patterns, Functions and AlgebraThe learner will be able to recognise, describe and represent patterns and relationships, as well as to solve problems using algebraic language and skills. |
| We know this when the learner: |
| 2.1 investigates, in different ways, a variety of numeric and geometric patterns and relationships by representing and generalising them, and by explaining and justifying the rules that generate them (including patterns found in nature and cultural forms and patterns of the learner’s own creation; |
| 2.2 represents and uses relationships between variables in order to determine input and/or output values in a variety of ways using: |
| 2.2.1 verbal descriptions; |
| 2.2.2 flow diagrams; |
| 2.2.3 tables; |
| 2.2.4 formulae and equations; |
| 2.3 constructs mathematical models that represent, describe and provide solutions to problem situations, showing responsibility toward the environment and health of others (including problems within human rights, social, economic, cultural and environmental contexts); |
| 2.4 solves equations by inspection, trial-and-improvement or algebraic processes (additive and multiplicative inverses, and factorisation), checking the solution by substitution; |
| 2.5 draws graphs on the Cartesian plane for given equations (in two variables), or determines equations or formulae from given graphs using tables where necessary; |
| 2.6 determines, analyses and interprets the equivalence of different descriptions of the same relationship or rule presented: |
| 2.6.1 verbally; |
| 2.6.2 in flow diagrams; |
| 2.6.3 in tables; |
| 2.6.4 by equations or expressions; |
| 2.6.5 by graphs on the Cartesian plane in order to select the most useful representation for a given situation; |
| 2.8 uses the laws of exponents to simplify expressions and solve equations; |
| 2.9 uses factorisation to simplify algebraic expressions and solve equations. |
Discussion
Answers:
1 480 black; 960 white; 300 red; 240 yellow; 240 blue and 180 green
2 480 black; 960 white; 675 red; 540 yellow; 540 blue and 405 green
5.
| Side length of triangle in centimetres | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Number of beads per triangle | 3 | 6 | 10 | 15 | 21 | 28 | 36 |
| Perimeter of triangle | 6 | 9 | 12 | 15 | 18 | 21 | 24 |
6.
| Size of necklace | 1 | 2 | 3 | 4 | 5 | x |
| Number of triangular motifs | 1 | 3 | 6 | 10 | 15 | x+(x–1)+(x–2)+ … +1 |
| Number of triangular spaces | 0 | 1 | 3 | 6 | 10 | (x–1)+(x–2)+ … +1 |
| Number of beads on each side of triangular motif | 4 | 8 | 12 | 16 | 20 | 4x |
| Total number of beads in necklace | 10 | 30 | 60 | 100 | 150 | 10{x+(x–1)+(x–2)+ … +1} |
| Number of black beads | 1 | 3 | 6 | 10 | 15 | (x–1)+(x–2)+ … +1 |
| Total perimeter of pendant with 1cm-diameter beads | 9 | 21 | 33 | 45 | 57 | 3(4x–1) |
7.
| Size of necklace | 1 | 2 | 3 | 4 | 5 | x |
| Number of triangular motifs | 1 | 3 | 6 | 10 | 15 | x+(x–1)+(x–2)+ … +1 |
| Number of triangular spaces | 0 | 1 | 3 | 6 | 10 | (x–1)+(x–2)+ … +1 |
| Number of beads on each side of triangular motif | 5 | 10 | 15 | 20 | 25 | 5x |
| Total number of beads in necklace | 15 | 45 | 90 | 150 | 225 | 15{x+(x–1)+(x–2)+ … +1} |
| Number of black beads | 3 | 9 | 18 | 30 | 45 | 3{x+(x–1)+(x–2)+ … +1} |
| Total perimeter of pendant with 1cm-diameter beads | 12 | 27 | 42 | 57 | 72 | 3(5x–1) |
ACTIVITY 2
1.
| Number of days: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| Away-van: | R750 | 1500 | 2250 | 3000 | 3750 | 4500 | 5250 | 6000 | 6750 | 7500 | 8250 |
| Best Caravans: | R1560 | R1920 | 2280 | 2640 | 3000 | 3360 | 3720 | 4080 | 4440 | 4800 | 5160 |
| Car-a-holiday: | R1490 | R2030 | 2570 | 3110 | 3650 | 4190 | 4730 | 5270 | 5810 | 6350 | 6890 |
If they want to go for only three days then Away-van is the cheapest. Best Caravans is the cheapest for holidays of from 4 to 11 days. Car-a-holiday is never the cheapest option, even if the holiday is longer than11 days.
3. Input = 9; output = 540 × 5 + 950 = 3 650
4.
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5. Bradley and his phones:
Tables:
| Number of calls: | 10 | 20 | 30 | 40 | 50 | 60 |
| Advanced mobile: | $26 | $32 | $38 | $44 | $50 | $56 |
| Genie rentals: | $24 | $38 | $52 | $66 | $80 | $94 |
| Hi-Pro: | $40 | $50 | $60 | $70 | $80 | $90 |
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6. Genie Rentals is the cheapest as long as he won’t want to make more than about 10 calls. Hi-Pro is never the cheapest. He is likely to get the best deal from Advanced Mobile if he wants to stay for a while.
7. It is easier to compare costs from the table. A graph would be easier still.
ACTIVITY 3
TEST