How to use this menu
Six original tasks: two for criterion A (a short check and an end-of-unit test), one investigation for B, one communication task for C, and two real-life tasks for D that also assess C. Set at least one task for each criterion across the unit; nobody needs all six. The timings spread assessment through the unit: A1 after the discrete half, B1 before the vector work is finished, A2 and the D tasks at the end.
Every task fits one lesson unless its format says otherwise. The structure follows the unit: lessons 1 to 3 are networks (degree, Euler and Hamilton routes, shortest paths, spanning trees) and lessons 4 to 8 are vectors (column vectors, sums and multiples, position vectors, proof, navigation).
Descriptors: Year 11 Extended is assessed with the MYP Year 5 descriptors, paraphrased below each task. Check them against your school's current guide.
The menu at a glance
| Task | Criteria | When | Time | Format |
|---|---|---|---|---|
| A1: Quick check: degree, routes and cheapest networks | A | After lesson 3 | 20 minutes | Individual, written, in class. 17 marks. |
| A2: End-of-unit test | A | End of unit, after lesson 8 | 50 minutes | Individual, written, in class, one lesson. 26 marks. A scientific calculator is allowed. |
| B1: Grids of vertices | B | Mid-unit, after lesson 3 | 50 minutes | In class. Students may collect results in pairs, but each writes up their own findings. |
| C1: Explain it to someone who missed it | C | After lesson 6 | 35 minutes | Individual, written, in class. A page or two of explanation; diagrams welcome. |
| D1: Solar lamps for the island paths | D and C | End of unit, after lesson 8 | 50 minutes | Individual, in class, one lesson; question 7 may be finished at home. A written solution with explanations and a short reply to the planner. |
| D2: Crossing the channel to the island | D and C | End of unit, after lesson 8 | 50 minutes | Individual, in class, one lesson; question 6 may be finished at home. A written solution with explanations and a short reply to the captain; a scientific calculator is expected. |
D1 and D2 are alternatives. Both assess criteria D and C. D1 is a network problem (lessons 1 to 3) and D2 is a vector problem (lessons 4 to 8): set the one that matches the half of the unit you want to assess, not both.
The unit's inquiry
- Key concept. Relationships: how things are joined, and how a quantity in one place fixes a quantity in another.
- Related concepts. Representation, Generalization, Models
- Global context. Globalization and sustainability: connecting places efficiently, and moving between them with the least cost.
- Statement of inquiry. Representing joins as graphs and movements as vectors turns questions about routes and positions into arithmetic that can be checked and proved.
MYP Year 5 descriptors, paraphrased
Each task's clarification, on its student sheet below, says what these levels look like in that task.
| Level | Criterion A: Knowing and understanding |
|---|---|
| 1–2 | Selects suitable mathematics for simple problems in familiar situations and applies it with some success. |
| 3–4 | Selects suitable mathematics for more complex problems in familiar situations and applies it successfully. |
| 5–6 | Selects suitable mathematics for challenging problems in familiar situations, applies it successfully and generally solves these problems correctly in a range of contexts. |
| 7–8 | Selects suitable mathematics for challenging problems in both familiar and unfamiliar situations, applies it successfully and generally solves these problems correctly in a range of contexts. |
| Level | Criterion B: Investigating patterns |
|---|---|
| 1–2 | With support, applies problem-solving techniques to find simple patterns and states predictions that fit them. |
| 3–4 | Applies problem-solving techniques to find simple patterns and suggests general rules that fit the findings. |
| 5–6 | Applies problem-solving techniques to find complex patterns, describes them as general rules that fit the findings and verifies that the rules hold. |
| 7–8 | Selects and applies problem-solving techniques to find complex patterns, describes them as general rules that fit correct findings, and proves, or verifies and justifies, those rules. |
| Level | Criterion C: Communicating |
|---|---|
| 1–2 | Uses limited mathematical language and limited forms of representation; the reasoning is hard to follow. |
| 3–4 | Uses some appropriate mathematical language and forms of representation; the reasoning can be followed and the work is adequately organized. |
| 5–6 | Usually uses appropriate language, moves between forms of representation with some success, and the reasoning is complete and coherent in a mostly logical structure. |
| 7–8 | Consistently uses appropriate language, moves effectively between forms of representation, and the reasoning is complete, coherent and concise in a logical structure. |
| Level | Criterion D: Applying mathematics in real-life contexts |
|---|---|
| 1–2 | Identifies some relevant elements of the real-life situation and selects, with some success, strategies to model it. |
| 3–4 | Identifies the relevant elements, selects adequate strategies to model the situation, applies them to reach a solution and discusses whether it makes sense. |
| 5–6 | Identifies the relevant elements, selects adequate modelling strategies and applies them to reach a valid solution, explains the degree of accuracy and explains whether the solution makes sense. |
| 7–8 | Identifies the relevant elements, selects appropriate modelling strategies and applies them successfully to reach a correct solution, justifies the degree of accuracy and justifies whether the solution makes sense. |
Task A1: Quick check: degree, routes and cheapest networks
Why set it. A short check of lessons 1 to 3 while there is still time to reteach: degree and the handshake result, Euler routes, shortest paths and minimum spanning trees. Questions 7 and 8 are unfamiliar, so the top band is open to everyone.
When: After lesson 3. Time: 20 minutes. Format: Individual, written, in class. 17 marks.
The student sheet, with its task-specific clarification
When: After lesson 3. Time: 20 minutes. Calculator: no calculator. Criterion: A, knowing and understanding.
Answer every question you can and show your working: a correct method earns credit even if the final answer slips. Questions 7 and 8 are meant to make you think, so try something rather than leaving them blank.
Part 1. Degree and routes
- A graph has vertices P, Q, R, S and T, and edges PQ, PR, QR, QS, RS and ST. Write down the degree of each vertex. (2 marks)
- How many edges does that graph have? (1 mark)
- Does the graph have an Euler route? Give a reason. (2 marks)
- Is there a graph on 5 vertices in which every vertex has degree 3? Explain how you know. (2 marks)
Part 2. Weighted graphs
- The edge lengths are AB 5, AC 3, BC 1, BD 6, CD 4, CE 8 and DE 2. Find the shortest route from A to E and its length. (2 marks)
- For the same graph, find a minimum spanning tree. List its edges and give its total weight. (3 marks)
Part 3. Think it through
- Zoe says: “Join the shortest routes from A to every other vertex and you always get the minimum spanning tree.” Three vertices A, B and C have edges AB 4, AC 4 and BC 1. Does this triangle show Zoe is wrong? Show why. (2 marks)
- A tree has 10 vertices. Find its number of edges, and the sum of all its degrees. Could every vertex have degree 3? (3 marks)
How your work is judged
| Level | Criterion A in task A1: what it looks like |
|---|---|
| 1–2 | You find some degrees and answer some of questions 1 to 3 correctly, for example the count of edges. |
| 3–4 | You answer most of questions 1 to 6 correctly: degrees, an Euler route decision, a shortest route and a spanning tree with a few slips. |
| 5–6 | You answer questions 1 to 6 correctly with working that shows your method, and make a good start on question 7 or 8. |
| 7–8 | You answer questions 1 to 6 correctly and settle question 7 or 8 with reasoning that shows why your answer must hold. |
Markscheme
| Task A1 question | Answer and marking |
|---|---|
| 1 | P 2, Q 3, R 3, S 3, T 1. 2 marks; 1 mark if one degree is wrong. Counting S as 2 (forgetting ST) is the usual slip. |
| 2 | 6 edges. The degrees add to 2 + 3 + 3 + 3 + 1 = 12, so 12 ÷ 2 = 6. 1 mark. |
| 3 | No. Q, R, S and T are odd, which is four odd vertices; a route needs 0 or 2. 1 mark for no, 1 mark for the count of odd vertices. |
| 4 | No. Five vertices of degree 3 would make the degrees add to 15, but the sum of degrees is twice the number of edges, so it must be even. 1 mark for no, 1 mark for the sum of 15 and the parity reason. |
| 5 | A, C, D, E with length 3 + 4 + 2 = 9. Other routes: A, C, E is 11; A, C, B, D, E and A, B, C, D, E are 12; A, B, D, E is 13. 1 mark for the route, 1 mark for 9. |
| 6 | BC 1, DE 2, AC 3, CD 4; weight 10. AB 5 is skipped because A, B and C are already joined. 1 mark for the method (cheapest edges first, none that closes a cycle), 1 mark for the four edges, 1 mark for 10. A tree on 5 vertices has 4 edges, which is a check. |
| 7 | Yes. From A the shortest routes are AB (4) and AC (4), total 8, but the tree BC 1 and AB 4 (or AC 4) has weight 5. 1 mark for showing the two totals, 1 mark for stating that the claim fails in this graph. A comparison without numbers earns 1 mark at most. |
| 8 | 9 edges; degrees add to 18; no. A tree on 10 vertices has 10 − 1 = 9 edges, the degrees add to 2 × 9 = 18, and ten vertices of degree 3 would add to 30. 1 mark each. |
| Marks to a level | Total 17. 1 to 5 marks: 1 to 2. 6 to 10: 3 to 4. 11 to 14: 5 to 6. 15 to 17, with question 7 or 8 fully correct: 7 to 8. Use the marks as a guide, then check the student's profile against the clarification before settling the level. |
Exemplar answers
| Task A1 response | What it looks like |
|---|---|
| Level 3 to 4 profile | Questions 2, 5 and 8 right (1 + 2 + 3); question 1 gives S as 2 (1 mark); question 3 says no with the reason “it has odd vertices” (1 mark); question 6 lists BC, DE, AC, AB (weight 11, 1 mark for the method); nothing correct on 4 or 7. 9 marks, and the profile matches 3 to 4: most of questions 1 to 6 with slips, nothing unfamiliar. |
| Level 7 to 8 answer to question 7 | “Shortest routes from A: A to B is 4 directly (via C it is 4 + 1 = 5), and A to C is 4. Joined together that is AB and AC, weight 8. But BC is only 1, so AB and BC joins all three for 4 + 1 = 5. So the shortest routes together are not the cheapest tree: the tree only has to connect everything, so it can use the cheap edge BC.” |
| Level 7 to 8 answer to question 4 | “Each edge is counted at both of its ends, so the degrees add to twice the number of edges, an even number. Five vertices of degree 3 add to 15, which is odd. So no such graph exists.” |
Differentiation
A way in for anyone stuck
- Give the six edges of question 1 on a drawing with the vertices spaced out, so the degrees can be counted on the picture.
- For question 6, let students write the edges in order of weight first, then cross out any that close a cycle.
- Read questions 7 and 8 aloud on request. Reading is not what is being assessed.
Stretch
- Question 5 again, but with edge DE changed to 9. What is the shortest route to E now, and does the minimum spanning tree change? (Route A, C, E length 11; the tree swaps DE 9 for the next cheapest edge that reconnects E, which is CE 8, giving weight 16.)
- How many edges does a connected graph on 6 vertices need at least? Which graphs on 6 vertices have exactly that many? (5 edges; the trees.)
Students learning in English as an additional language
| Word in task A1 | Plain meaning |
|---|---|
| degree | the number of edges that meet at a vertex |
| edge | a line that joins two vertices |
| spanning tree | a set of edges that joins every vertex and has no cycle |
| cycle | a closed path that returns to its start without repeating a vertex |
| weight | the number written on an edge, for example its length or cost |
Sentence starters for task A1:
- The degree of … is … because …
- I took the edges in order of … and skipped … because …
- Zoe is / is not right because …
Task A2: End-of-unit test
Why set it. A full-unit test across networks and vectors. Sections 1 and 2 are familiar skills; Section 3 is unfamiliar, so the top band is open to students who can apply the ideas in a new setting.
When: End of unit, after lesson 8. Time: 50 minutes. Format: Individual, written, in class, one lesson. 26 marks. A scientific calculator is allowed.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 8. Time: 50 minutes. Calculator: scientific calculator. Criterion: A, knowing and understanding.
Show your working. Give angles and bearings to 1 decimal place. Section 3 contains problems you have not seen before: choose a method and try it, because a correct method earns credit.
Section 1. Vectors
- Let p = (6, −8) and q = (−2, 5). Find (a) the magnitude of p; (b) p + q and its magnitude; (c) 2p − 3q. (5 marks)
- Relative to an origin O, A is (1, 2) and B is (7, −2). Find (a) the vector AB; (b) the position vector of the midpoint of AB; (c) the point C so that ABCD is a parallelogram with D at (−1, 5). (5 marks)
- A vector is (5, 2). Find its angle with the horizontal, and its bearing (the angle clockwise from north). (2 marks)
Section 2. Networks
- A graph has 6 vertices with degrees 3, 3, 2, 2, 2 and 2. How many edges does it have? Does it have an Euler route, and if so where must the route start and end? (3 marks)
- A weighted graph on vertices A to F has edges AB 7, AC 9, BC 5, BD 8, CD 3, CE 6, DE 4, DF 10 and EF 2. Find (a) a minimum spanning tree, listing its edges and total weight; (b) the shortest route from A to F and its length. (5 marks)
Section 3. Unfamiliar problems
- OACB is a parallelogram with its vertices in that order, OA = a and OB = b. P is the midpoint of AC and Q is the midpoint of BC. Prove that PQ is parallel to AB, and find the ratio of their lengths. (3 marks)
- A boat steers on bearing 120 at 10 km/h through the water, in a current that flows due north at 4 km/h. Find (a) the boat's speed over the ground, to 1 decimal place; (b) the bearing of its actual path; (c) its position relative to the start after 3 hours. (3 marks)
How your work is judged
| Level | Criterion A in task A2: what it looks like |
|---|---|
| 1–2 | You do some of the routine steps in sections 1 and 2 correctly, for example a magnitude or a list of edges. |
| 3–4 | You complete most of sections 1 and 2 correctly, using a method that is the right one even where a slip appears. |
| 5–6 | You complete sections 1 and 2 correctly with your method shown, and make a good start on question 6 or 7. |
| 7–8 | You complete sections 1 and 2 correctly and carry out question 6 or 7 to a correct conclusion, choosing your own method. |
Markscheme
| Task A2 question | Answer and marking |
|---|---|
| 1(a) | 10. √(36 + 64) = √100. 1 mark. |
| 1(b) | p + q = (4, −3); magnitude 5. 1 mark for the sum, 1 mark for √(16 + 9) = 5. |
| 1(c) | (18, −31). 2p = (12, −16) and 3q = (−6, 15), so 2p − 3q = (12 + 6, −16 − 15). 1 mark for 2p and 3q, 1 mark for the answer. (6, −1) comes from adding. |
| 2(a) | (6, −4). End minus start: (7 − 1, −2 − 2). 1 mark. (−6, 4) is BA. |
| 2(b) | (4, 0). A + half of AB = (1, 2) + (3, −2), or the mean of the coordinates. 1 mark for the method, 1 mark for the answer. |
| 2(c) | C = (5, 1). In a parallelogram ABCD, AB = DC, so C = D + AB = (−1 + 6, 5 − 4). 1 mark for AB = DC, 1 mark for the point. |
| 3 | Angle 21.8°; bearing 068.2°. tan−1(2 ÷ 5) = 21.80°; the bearing is 90 − 21.8 = 68.2°, or tan−1(5 ÷ 2). 1 mark each. 21.8 given as the bearing is the usual error. |
| 4 | 7 edges; yes; start and end at the two vertices of degree 3. The degrees add to 14, so 7 edges. Exactly two vertices are odd. 1 mark for 7, 1 mark for yes with the two odd vertices, 1 mark for start and end at those two. |
| 5(a) | EF 2, CD 3, DE 4, BC 5, AB 7; weight 21. CE 6 is skipped (C, D, E are already joined); AC 9 would close a cycle. 1 mark for the method, 1 mark for the edges, 1 mark for 21. A tree on 6 vertices has 5 edges. |
| 5(b) | A, C, E, F with length 9 + 6 + 2 = 17. Other routes: A, B, C, E, F is 20; A, C, D, E, F is 18; A, B, D, F is 25. 1 mark for the route, 1 mark for 17. The path through the minimum spanning tree (A, B, C, D, E, F, length 21) is not the shortest. |
| 6 | PQ = (1/2)AB, so PQ is parallel to AB and half as long. AC is equal to OB = b and BC is equal to OA = a, so OP = a + (1/2)b and OQ = b + (1/2)a. Then PQ = OQ − OP = (1/2)b − (1/2)a = (1/2)(b − a) = (1/2)AB. 1 mark for OP and OQ in terms of a and b, 1 mark for PQ = (1/2)(b − a), 1 mark for the conclusion (scalar multiple, so parallel; ratio 1 : 2). |
| 7 | (a) 8.7 km/h. (b) 096.6°. (c) 26.0 km east and 3.0 km south. The boat's own velocity is east 10 sin 120 = 8.66 and north 10 cos 120 = −5; the current adds 4 north, so the resultant is east 8.66 and north −1. Speed √(75 + 1) = √76 = 8.72; the path is 1 km/h south for every 8.66 east, so the bearing is 180 − tan−1(8.66 ÷ 1) = 96.6°; after 3 hours the position is 3 × (8.66, −1) = (26.0, −3.0). 1 mark each. Taking 10 cos 120 as +5 (north 9, speed 12.5 km/h) is the usual error. |
| Marks to a level | Total 26. 1 to 5 marks: 1 to 2. 6 to 12: 3 to 4. 13 to 19: 5 to 6. 20 to 26, with at least 4 of the 6 marks in section 3: 7 to 8. Check the profile against the clarification before settling the level. |
Exemplar answers
| Task A2 response | What it looks like |
|---|---|
| Level 3 to 4 profile | Section 1 right except that 2p − 3q is given as (6, −1) (1 mark of 2) and the bearing as 21.8° (1 mark of 2); question 4 gives 7 edges and yes but not where the route starts (2 of 3); question 5(a) lists a tree of weight 23 that uses AC 9 in place of AB 7 (method mark only); 5(b) gives A, B, C, D, E, F (no marks); questions 6 and 7 attempted but incomplete (1 mark). The count is in the 5 to 6 range, but the profile has method slips in section 1 and no success in section 3, so 3 to 4 is the better fit: the clarification decides. |
| Level 7 to 8 answer to question 6 | “In the parallelogram AC = OB = b and BC = OA = a. So OP = OA + (1/2)AC = a + (1/2)b, and OQ = OB + (1/2)BC = b + (1/2)a. PQ = OQ − OP = (1/2)b − (1/2)a = (1/2)(b − a), and AB = b − a. So PQ = (1/2)AB: a scalar multiple, so PQ is parallel to AB and half as long, a ratio of 1 : 2.” |
| Level 7 to 8 answer to question 7 | “Boat: east 10 sin 120 = 8.66, north 10 cos 120 = −5. Current: north 4. Total: east 8.66, north −1. Speed = √(8.662 + 12) = √76 = 8.7 km/h. The path is just south of east: the angle below east is tan−1(1/8.66) = 6.6°, so the bearing is 090 + 6.6 = 096.6°. In 3 hours: 26.0 km east and 3.0 km south.” |
Differentiation
A way in for anyone stuck
- Allow a sheet of the vector rules (column notation, magnitude, position vectors) and the two spanning tree and shortest path methods in words, but not worked examples.
- For question 2(c), let students sketch the parallelogram first and mark which sides are equal.
- Let students tick off each edge as they take or skip it in question 5(a); a drawing of the graph on a separate sheet is allowed.
Stretch
- In question 5, a new edge BF is added. How cheap must it be before the minimum spanning tree changes? (Under 5: BF closes the cycle B, C, D, E, F, and the largest tree edge on it is BC 5, so BF replaces BC if it costs less than 5.)
- In question 7 the captain wants the path over the ground to run due east. Which bearing must the boat steer? (113.6°: the north component 10 cos t + 4 must be 0, so cos t = −0.4 and t = 113.6°.)
Students learning in English as an additional language
| Word in task A2 | Plain meaning |
|---|---|
| magnitude | the length of a vector |
| bearing | an angle measured clockwise from north |
| parallelogram | a four-sided shape with two pairs of parallel sides |
| Euler route | a route that uses every edge exactly once |
| resultant | the single vector that has the same effect as two or more vectors together |
Sentence starters for task A2:
- The magnitude is … because …
- I took the edges in order: …, and I skipped … because …
- PQ is parallel to AB because …
Task B1: Grids of vertices
Why set it. A pattern hunt built on lessons 1 and 2: counting edges and odd vertices of rectangular grids. The sheet never states the pattern. Students organize results, describe rules, test them on a case that breaks one of them, and explain.
When: Mid-unit, after lesson 3. Time: 50 minutes. Format: In class. Students may collect results in pairs, but each writes up their own findings.
The student sheet, with its task-specific clarification
In MYP Year 5, support you give (a prompt, a table outline) is allowed at levels 1 to 2 only: note it on the student's work.
When: Mid-unit, after lesson 3. Time: 50 minutes. Calculator: no calculator. Criterion: B, investigating patterns.
Picture a rectangle of dots, with a rows of b dots in each row. Every dot is joined to the dot next to it in its row and to the dot below it in its column, and to nothing else (no diagonals). Call this an a by b grid. A 2 by 3 grid has six dots and one edge between each neighbouring pair. This task asks what you can say about grids of any size, without drawing the large ones.
Part 1. Collect results
- Draw the grids 2 by 2, 2 by 3, 2 by 4, 3 by 3, 3 by 4 and 4 by 4. For each, record the number of edges and the number of vertices whose degree is odd. Design the table yourself.
- Add the 4 by 5 grid to your table, using its drawing or a method you have found.
Part 2. Describe what you find
- Describe, in words and then with letters, a rule that gives the number of edges of an a by b grid.
- Describe a rule that gives the number of vertices of odd degree. Say for which grids it works.
Part 3. Use and test your rules
- Predict the number of edges and the number of odd vertices in a 5 by 6 grid, and in a 10 by 10 grid. Show how your rules give each answer. Does the 10 by 10 grid have an Euler route?
- Test your rules on a single row of 5 dots (a 1 by 5 grid). Say what is the same and what is different, and what that tells you about the limits of your rules.
Part 4. Explain
- Explain why your rule for the edges must be right, without counting edges one by one.
- Find all the grids with at least two rows and two columns that have an Euler route, and explain how you know there are no others.
How your work is judged
| Level | Criterion B in task B1: what it looks like |
|---|---|
| 1–2 | With help, you draw some grids, count edges and odd vertices and say something you notice. |
| 3–4 | You complete the table and suggest a rule for the number of edges or for the number of odd vertices. |
| 5–6 | You organize your results clearly, describe a rule for each count that fits all your findings, and check both on the 5 by 6 grid and the 1 by 5 row. |
| 7–8 | You find rules for both counts, use them correctly for large grids, test them on the single row, and prove or justify why they hold, including where the odd vertex rule stops working. |
Markscheme
| Task B1 question | Answer and marking |
|---|---|
| The pattern (do not share it before the task) | For a by b grids with a, b at least 2: edges = a(b − 1) + b(a − 1) = 2ab − a − b; odd vertices = 2(a − 2) + 2(b − 2) = 2(a + b − 4). Degrees: corners 2, other vertices on the outer edge 3, inside vertices 4. A 1 by n row has n − 1 edges (the edge rule still works) but exactly 2 odd vertices (the odd-vertex rule gives 2n − 6, which is wrong). |
| 1 | Table. (rows by columns: edges, odd vertices) 2 by 2: 4, 0. 2 by 3: 7, 2. 2 by 4: 10, 4. 3 by 3: 12, 4. 3 by 4: 17, 6. 4 by 4: 24, 8. Credit each correct row; a table that is organized so the pattern shows is evidence for the upper bands. |
| 2 | 4 by 5: 31 edges, 10 odd vertices. 4 × 4 + 5 × 3 = 31; 2(2) + 2(3) = 10. |
| 3 | Any correct description of the edge count: “across edges plus down edges”, a(b − 1) + b(a − 1), or 2ab − a − b. A rule for the pattern of the counts only (for example “adds 3, then 3, then 7”) is evidence for levels 3 to 4. |
| 4 | 2(a + b − 4), or “two on each side for each dot beyond the two corners”, valid for a and b at least 2. Credit a student who says it fails for a single row as soon as they test it (see question 6). |
| 5 | 5 by 6: 49 edges, 14 odd vertices. 10 by 10: 180 edges, 32 odd vertices; no Euler route. 5 × 5 + 6 × 4 = 49; 2(5 + 6 − 4) = 14; 2 × 10 × 10 − 20 = 180; 2(16) = 32. A route needs 0 or 2 odd vertices. |
| 6 | 1 by 5 has 4 edges and 2 odd vertices. The edge rule 2 × 5 − 1 − 5 = 4 holds; the odd-vertex rule gives 2(1 + 5 − 4) = 4, which is wrong. The ends have degree 1 and the three in the middle have degree 2. This is why the odd vertex rule needs two rows and two columns. |
| 7 | Edges. Each of the a rows has b − 1 edges across and each of the b columns has a − 1 edges down. Or: the degrees add to 4 × 2 + 3 × (2(a − 2) + 2(b − 2)) + 4(a − 2)(b − 2) and the edges are half of that. 7 to 8 expects a reason of this kind. |
| 8 | Only 2 by 2 and 2 by 3 (and 3 by 2). An Euler route needs at most 2 odd vertices, so 2(a + b − 4) is 0 or 2, which means a + b is 4 or 5. With a, b at least 2 this gives 2 by 2 and 2 by 3. A single row is excluded by the condition, but also has an Euler route (it is a path). |
Exemplar answers
| Task B1 response | What it looks like |
|---|---|
| Level 3 to 4 | “Edges: 4, 7, 10, 12, 17, 24. Odd vertices: 0, 2, 4, 4, 6, 8. For 2 by n the edges go up by 3 each time. I think the edges are a times b, minus something. 5 by 6 is 30 − 11 = 19.” The table is right and a rule is suggested, but the edge rule is wrong (49), the odd vertex rule is missing and no test is made beyond 4 by 5. |
| Level 7 to 8 | “Each row has b − 1 edges and there are a rows; each column has a − 1 edges and there are b columns, so edges = a(b − 1) + b(a − 1) = 2ab − a − b. For odd vertices: a corner has 2 neighbours, a vertex on a side has 3, an inside vertex has 4. So only the vertices on the sides, not the corners, are odd: each side has (its length − 2) of them, so 2(a − 2) + 2(b − 2). 5 by 6: edges 49, odd vertices 14. For 1 by 5 the edge rule works (4) but the odd vertex rule gives 4 and the real answer is 2, because every vertex is on the side.” |
Differentiation
A way in for anyone stuck
- Give a two-column table outline for Part 1 and let the student write the headings.
- Suggest counting the degree of every vertex on a 3 by 4 grid and writing it inside each dot before counting the odd ones.
- Prompt with a question, never the rule: “Which dots have an odd number of neighbours?”
Stretch
- Add one diagonal in every square of a 3 by 3 grid. How many edges does it have now? (16.) Is there an Euler route? Find out.
- For a by b grids, when is a Hamilton cycle possible? (When a × b is even and both are at least 2; an odd number of vertices cannot be alternately coloured.)
Students learning in English as an additional language
| Word in task B1 | Plain meaning |
|---|---|
| grid | dots arranged in rows and columns |
| row | a line of dots going across |
| column | a line of dots going down |
| predict | say what will happen before you check |
| rule | a sentence or formula that tells you what happens every time |
Sentence starters for task B1:
- I noticed that the number of …
- My rule is: …
- I tested my rule on … and it worked / did not work because …
Task C1: Explain it to someone who missed it
Why set it. A communication task on lesson 6: position vectors and points on a line. The students are not marked on getting the numbers right (that is criterion A) but on how clearly and completely they explain, with notation, diagrams and order.
When: After lesson 6. Time: 35 minutes. Format: Individual, written, in class. A page or two of explanation; diagrams welcome.
The student sheet, with its task-specific clarification
When: After lesson 6. Time: 35 minutes. Calculator: no calculator. Criterion: C, communicating.
Dara was away for lesson 6 and tried the question below from the lesson notes. Dara's working is shown. Your job is to write a clear explanation Dara can follow: what went wrong, why, and how to do it properly.
The question. P is the point (2, −3) and Q is the point (8, 9). Find (a) the midpoint of PQ and (b) the point R on PQ that is one third of the way from P to Q.
Dara's working. “PQ = (8 − 2, 9 − (−3)) = (6, 12). Half of PQ is (3, 6), so the midpoint is (3, 6). One third of PQ is (2, 4), so R is (2, 4).”
Your explanation
- Say which of Dara's steps is correct and what each number in them stands for (a movement or a point).
- Explain why (3, 6) and (2, 4) are not the answers, using a diagram.
- Show Dara a correct method, using position vectors, for the midpoint and for R, with each step labelled.
- Finish by giving Dara one short sentence to remember that stops the same mistake happening again.
How your work is judged
| Level | Criterion C in task C1: what it looks like |
|---|---|
| 1–2 | You use a few correct words, such as vector or point, and your explanation is hard for Dara to follow in places. |
| 3–4 | You use some correct words and notation, and your steps are in a sensible order, so Dara could follow most of it even if some parts are unclear. |
| 5–6 | You usually use correct notation, move between a diagram, symbols and words, and your explanation is complete and in a logical order. |
| 7–8 | You consistently use correct notation, move smoothly between a diagram, symbols and words, and your explanation is complete, brief and organized so that each step leads to the next. |
Markscheme
| Task C1 question | Answer and marking |
|---|---|
| What a complete explanation contains | (1) Dara's PQ = (6, 12) is correct and is a movement (a vector). (2) (3, 6) and (2, 4) are half and one third of that movement, still movements from P, not points. (3) A position vector fixes a point from the origin, so the point is P plus the movement: midpoint = (2, −3) + (3, 6) = (5, 3); R = (2, −3) + (2, 4) = (4, 1). (4) A diagram showing P, Q, the movement PQ and the two points, with a sensible scale. (5) A memory sentence, for example “a fraction of PQ is a movement; add it to P to land on the point”. |
| Check (for the marker) | Midpoint by averages: ((2 + 8) / 2, (−3 + 9) / 2) = (5, 3). Third of the way: (2 + 6 / 3, −3 + 12 / 3) = (4, 1). The correct points are P plus Dara's answers. Reading these does not decide the level; the explanation does. |
| How to level | Place the work by how well it communicates, using the clarification: language and notation (vector, position vector, column form, labelled points), the forms used (words, symbols, diagram) and the order and completeness of the reasoning. Mathematical slips count only where they break the line of reasoning Dara has to follow; accuracy itself is criterion A. |
Exemplar answers
| Task C1 response | What it looks like |
|---|---|
| Level 3 to 4 | “Dara found PQ right. But half of PQ is only half the journey, so (3, 6) is how far to go, not where you are. You need to add it on to P. Then it is (5, 3). For R it is (4, 1).” The words are mostly right and the order is sensible, but there is no diagram, no notation (OP, p, q) and the sentence at the end is missing. |
| Level 7 to 8 | “PQ = q − p = (6, 12) is a movement from P to Q. Half of it, (3, 6), is the movement from P to the midpoint M; it is a vector, not a point. A point is a position vector from O: OM = OP + PM = (2, −3) + (3, 6) = (5, 3). Diagram: P at (2, −3), an arrow of (6, 12) to Q at (8, 9), M halfway. Same for R: OR = OP + one third of PQ = (2, −3) + (2, 4) = (4, 1). Remember: a fraction of PQ is a journey, so add it to where you started.” |
Differentiation
A way in for anyone stuck
- Give a sentence frame for each part: “Dara's number (6, 12) stands for …, because …”.
- Provide an empty coordinate grid with P and Q already plotted, so the diagram is about the explanation, not the plotting.
- Allow a labelled sketch with a short written caption in place of the third bullet's full prose.
Stretch
- Write the same explanation for the point S that is two thirds of the way from P to Q, and say how the position vector formula changes. (S = (6, 5) = (1 − 2/3)P + (2/3)Q.)
- Explain why the midpoint formula is the average of the coordinates, using position vectors. (OM = OP + (1/2)(OQ − OP) = (1/2)(OP + OQ).)
Students learning in English as an additional language
| Word in task C1 | Plain meaning |
|---|---|
| position vector | a vector from the origin O to a point; it tells you where the point is |
| movement | a change in position: how far across and how far up or down |
| midpoint | the point exactly halfway along a line |
| one third of the way | start at the first point and go a third of the distance to the second |
Sentence starters for task C1:
- Dara's number (6, 12) is the vector …, which means …
- This is not a point because …
- To land on the point you must add … to …
Task D1: Solar lamps for the island paths
Why set it. A real-life network problem in an international setting: a minimum spanning tree for the cost of cable, a shortest route for a maintenance trip, and a judgement about robustness. Students decide what to model, how accurate the answer is, and whether it makes sense.
When: End of unit, after lesson 8. Time: 50 minutes. Format: Individual, in class, one lesson; question 7 may be finished at home. A written solution with explanations and a short reply to the planner.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 8. Time: 50 minutes. Calculator: no calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.
Six hamlets, A to F, on a small island in the Andaman Islands are to be linked by one cable that powers the solar lamps along the paths. Each possible path between two hamlets has a cost to lay the cable, in thousands of rupees. The costs are estimates, correct to about 10%. The planner has listed them: AB 12, AC 18, BC 9, BD 15, CD 11, CE 20, DE 7, DF 16 and EF 10. The planner asks you for the cheapest cable layout that reaches every hamlet, and for advice on whether it is a good plan.
Your task
- Say what the vertices, the edges and the weights are in this model, and one thing the model leaves out.
- Find the cheapest layout that reaches every hamlet. Say which method you used, list the paths chosen and give the total cost.
- Check that your layout has the right number of paths for six hamlets and explain what that tells you.
- A technician must walk from hamlet A to hamlet F along the paths in the planner’s list. Find the shortest route and its length. How much longer is the route that follows only your cheapest layout?
- The costs are only correct to about 10%. Give the total cost with a range, and say whether the planner should quote it as exact.
- The planner would like every hamlet to stay linked if any single path is damaged by a storm. Show whether adding the cheapest extra path achieves that, and say what you recommend.
- Write a short reply to the planner: your layout, its cost, the one thing you are least sure of and your recommendation.
How your work is judged
| Level | Criterion D in task D1: what it looks like |
|---|---|
| 1–2 | You pick out some of the hamlets and paths and find a layout, but it is not clear that it is the cheapest or that it reaches everyone. |
| 3–4 | You identify the hamlets, paths and costs, apply a spanning tree method to reach a total, and say whether the answer is sensible. |
| 5–6 | You choose a suitable method, reach a valid layout and total, find the shortest route, describe the accuracy of the cost and discuss whether the plan makes sense for the island. |
| 7–8 | You model the situation, reach a correct layout, route and total, justify the range on the cost, and give a justified recommendation that deals with storms and what the model leaves out. |
| Level | Criterion C in task D1: what it looks like |
|---|---|
| 1–2 | Your working is hard to follow and you use few correct terms. |
| 3–4 | Your steps can be followed and you use some correct terms such as vertex, edge and weight. |
| 5–6 | Your explanation is mostly logical, with a clear list of paths and a clearly set out reply. |
| 7–8 | Your explanation is complete and concise, with the method, the numbers and the recommendation all easy to find. |
Markscheme
| Task D1 question | Answer and marking |
|---|---|
| 1 | Vertices: the six hamlets; edges: the possible cable paths; weights: the cost of each. The model leaves out: the actual path length or terrain, a cost that varies by season, whether paths could share a trench, or that paths could be damaged. Credit any real omission. |
| 2 | DE 7, BC 9, EF 10, CD 11, AB 12; total 49 thousand rupees. Take edges in order of cost and skip any that would close a cycle: BD 15, DF 16, AC 18 and CE 20 are skipped. 49 < any other spanning tree. 1 mark for method, 1 for edges, 1 for 49. |
| 3 | 5 paths for 6 hamlets. A tree on n vertices has n − 1 edges. If a layout had 6 or more it would contain a cycle and be paying for a path nobody needs; with 4 or fewer some hamlet is cut off. |
| 4 | A, B, D, F with length 12 + 15 + 16 = 43; 6 longer. Along the layout the route is A, B, C, D, E, F = 12 + 9 + 11 + 7 + 10 = 49, so 49 − 43 = 6. Other routes: A, B, D, E, F is 44; A, C, D, F is 45. A technician can use any path that exists, but the cable can only run where it is laid, so the shorter route needs BD and DF to be built too. |
| 5 | About 49 thousand rupees; range 44.1 to 53.9. 10% of 49 is 4.9. The planner should quote 'about 44 to 54 thousand rupees', not '49' as if exact. Because the costs were estimates, more than 2 significant figures is false precision. |
| 6 | No single extra path is enough. The layout is the line A, B, C, D, E, F, so every path in it is a single point of failure. The cheapest extra path, BD at 15 (total 64), only protects BC and CD. A is reached only by AB and AC, and F only by DF and EF, so a storm-proof network must use all four; the cheapest one is AB, AC, BD, CD, DE, DF and EF, costing 89 thousand rupees, about 40 more than the cheapest layout. Credit any reasoned recommendation, such as building the 49 layout and storing spare cable, or paying for the 89 network if the lamps are a safety need. |
| 7 (criteria D and C) | A strong reply gives the layout, the cost to a sensible accuracy (about 44 to 54 thousand rupees), says what the model leaves out, notes that one storm can cut off A, E or F, and recommends (a) building the cheapest layout and (b) a survey for an A to F path. Credit any reasoned recommendation either way. |
Exemplar answers
| Task D1 response | What it looks like |
|---|---|
| Level 3 to 4 (D) | “I took the smallest costs: 7, 9, 10, 11, 12. That is 5 paths and 49. The shortest route A to F is 12 + 9 + 11 + 7 + 10 = 49.” Correct layout and total, but the route question uses the tree route rather than comparing, there is no range and no judgement about storms. |
| Level 7 to 8 (D and C) | “Layout DE, BC, EF, CD, AB, 49 thousand rupees; it is a line A to F. Shortest route 43 via B, D, F but that needs two paths not in the layout, 6 shorter than going along the cable. The costs are about 10%, so the total is 44 to 54 thousand rupees, and I would say 'about 49, between 44 and 54'. BD alone does not protect A, so I recommend the layout and a survey for a path joining A and F.” |
Differentiation
A way in for anyone stuck
- Give the nine paths as a table sorted by cost, so the method is about skipping cycles rather than sorting.
- For question 4, give a sketch of the graph on which students can trace routes.
- Let students estimate the range using a calculator-free method: 10% of 49 is about 5.
Stretch
- If AC had cost 8 instead of 18, what would the cheapest layout be? (AC 8, DE 7, BC 9, EF 10, CD 11: total 45; AB 12 is no longer needed.)
- How many different layouts reach every hamlet using 5 paths? Is the cheapest one unique? (The cheapest is unique because all nine costs are different.)
Students learning in English as an additional language
| Word in task D1 | Plain meaning |
|---|---|
| cable | a thick wire that carries electricity |
| hamlet | a very small village |
| layout | the way the paths are arranged |
| estimate | a careful guess; not exact |
| recommend | say what you think should be done |
Sentence starters for task D1:
- I took the cheapest path first, then …, and skipped … because …
- The total is about … because the costs are …
- I recommend … because …
Task D2: Crossing the channel to the island
Why set it. A real-life vector problem in an international setting: combining a boat's velocity with a current (lesson 8), finding the heading that cancels the drift, and judging how reliable the answer is when the speeds are estimates.
When: End of unit, after lesson 8. Time: 50 minutes. Format: Individual, in class, one lesson; question 6 may be finished at home. A written solution with explanations and a short reply to the captain; a scientific calculator is expected.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 8. Time: 50 minutes. Calculator: scientific calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.
A ferry crosses a channel to an island that is 30 km due north of its harbour. Through still water the ferry travels at 15 km/h. A current flows due east at 7 km/h. The ferry has fuel for 3 hours. The captain has asked you to plan the crossing and tell her what to steer. The speeds are measured to the nearest km/h.
Your task
- Write the ferry's velocity through still water, steering due north, and the current's velocity, as column vectors. Find the resultant velocity and the ferry's speed over the ground.
- Find the bearing of the ferry's actual path, to 1 decimal place, when it steers due north.
- If the captain steers due north the whole way, how long does it take to be 30 km north of the harbour, and how far east of the island is the ferry then?
- Find the bearing the captain should steer, to 1 decimal place, to travel directly to the island. How long does the crossing take? Is there enough fuel?
- The current is only correct to the nearest km/h, so it could be anything from 6.5 to 7.5 km/h. Find the range of bearings the captain should be ready to steer.
- Write a short reply to the captain: what to steer, how long, how far you trust the figures and one thing that could make the plan fail.
How your work is judged
| Level | Criterion D in task D2: what it looks like |
|---|---|
| 1–2 | You pick out some of the speeds and directions and find one correct quantity, such as the resultant, with little support for why it makes sense. |
| 3–4 | You identify the velocities, add them correctly, find the drift or the heading, and say whether the answer is sensible for the crossing. |
| 5–6 | You choose suitable vector methods, find a valid heading and time, describe the accuracy using the range of the current and discuss whether the plan is practical. |
| 7–8 | You model the crossing, find a correct heading and time, justify the range of bearings from the uncertainty in the current and give a justified recommendation to the captain. |
| Level | Criterion C in task D2: what it looks like |
|---|---|
| 1–2 | Your working is hard to follow; you use few correct terms such as velocity or bearing. |
| 3–4 | Your steps can be followed, with some correct terms and a sketch of the vectors. |
| 5–6 | Your explanation is mostly logical with a clear vector diagram and a clear reply to the captain. |
| 7–8 | Your explanation is complete and concise, with a labelled vector diagram, each result in context and a reply the captain could act on. |
Markscheme
| Task D2 question | Answer and marking |
|---|---|
| 1 | (0, 15) and (7, 0); resultant (7, 15); speed 16.6 km/h. √(49 + 225) = √274 = 16.55. 1 mark for the vectors, 1 mark for the resultant, 1 mark for 16.6. East is x and north is y. |
| 2 | 025.0°. tan−1(7 ÷ 15) = 25.02° clockwise from north, so 025.0°. 25.0 measured from the east is wrong (65.0°). |
| 3 | 2 hours; 14 km east of the island. The northward speed is 15 km/h, so 30 ÷ 15 = 2 h. The drift is 7 × 2 = 14 km. Using the speed over the ground (16.6 km/h) gives the wrong time because the 30 km is the northward distance only. |
| 4 | 332.2°; about 2 h 16 min; yes. The ferry must steer west of north so that the westward component of its velocity cancels the current: 15 sin θ = 7, so θ = sin−1(7 ÷ 15) = 27.8°, and the bearing is 360 − 27.8 = 332.2°. The northward speed is √(152 − 72) = √176 = 13.3 km/h, so 30 ÷ 13.27 = 2.26 h = 136 min = 2 h 16 min, within 3 hours of fuel with about 44 minutes spare. |
| 5 | About 330.0° to 334.3°. At 6.5 km/h: sin−1(6.5 ÷ 15) = 25.7°, bearing 334.3°. At 7.5 km/h: sin−1(7.5 ÷ 15) = 30.0°, bearing 330.0°. A bearing to the nearest degree is the most the figures support: 330° to 334°. |
| 6 (criteria D and C) | A strong reply gives the heading to a sensible accuracy (about 332°, within 330° to 334°), the time (about 2 h 16 min, so plan for 2½ hours) with the 44 minutes spare as the margin, and a real risk: the current is not constant, the ferry may be slowed by waves, or a crew that steers north and corrects late arrives to the east of the island. Credit any reasoned recommendation. |
Exemplar answers
| Task D2 response | What it looks like |
|---|---|
| Level 3 to 4 (D) | “Resultant (7, 15), speed 16.6. If it steers north, 30 ÷ 16.6 = 1.8 hours and it drifts about 13 km. To go straight it should steer a bit west, about 30°.” The resultant is right but the time uses the wrong speed, the heading is estimated without a method and there is no accuracy or fuel check. |
| Level 7 to 8 (D and C) | “Steering north: resultant (7, 15), speed 16.6 km/h, bearing 025.0°. North speed 15, so 30/15 = 2 h and drift 14 km east. To cancel the current: 15 sin θ = 7, θ = 27.8° west of north, bearing 332.2°; north speed √176 = 13.3, time 2 h 16 min, 44 min of fuel to spare. With the current from 6.5 to 7.5 km/h the heading is 330.0° to 334.3°, so I would tell the captain to steer about 332° and check the position often, because a current stronger than 7.5 km/h pushes the ferry east of the island.” |
Differentiation
A way in for anyone stuck
- Give a labelled vector triangle with the 15 km/h, 7 km/h and the resultant marked, so the work is the calculation.
- For question 4, suggest the picture first: the ferry's own velocity must have a westward part of 7 km/h to cancel the current.
- Allow a table: speed, bearing, time, with one row per question.
Stretch
- The current becomes 16 km/h. Can the ferry reach the island by steering any bearing? Explain. (No: its own speed is 15 km/h, less than the current, so it cannot cancel the drift; sin−1(16/15) does not exist.)
- At what current speed would the crossing need all 3 hours of fuel? (30 ÷ √(225 − c2) = 3 gives c = √(225 − 100) = 11.2 km/h.)
Students learning in English as an additional language
| Word in task D2 | Plain meaning |
|---|---|
| current | water moving in one direction |
| drift | being pushed sideways by a current or wind |
| fuel | what an engine burns to run |
Sentence starters for task D2:
- The ferry's velocity is … and the current's is …, so the resultant is …
- To cancel the drift I must steer … because …
- The plan could fail if …
Practicalities
Every task is original and written for this unit; none is copied or adapted from a published book.
The student sheets, with their task-specific clarifications, are public on the unit page so students know what each level looks like before they start. Markschemes and exemplars are only on this page.
A1, B1 and D1 need no calculator. A2 allows a scientific calculator (questions 3 and 7 need trigonometry; the rest is mental or written arithmetic); D2 needs one throughout. C1 needs none.
All contexts and distances are invented. No task uses data from a real school, class or student.
The sheets are public, so a student may have seen A2. For an unseen A2, use this parallel version (same marks, same markscheme notes). Q1 p = (−5, 12), q = (5, 0): |p| = 13; p + q = (0, 12), magnitude 12; 3p − 2q = (−25, 36). Q2 A(−2, 3), B(4, −5): AB = (6, −8); midpoint (1, −1); D(0, 2) gives C = (6, −6) for parallelogram ABCD. Q3 vector (5, 12): angle with the horizontal 67.4 degrees, bearing 022.6 degrees. Q4 a graph on 7 vertices with degrees 4, 4, 3, 3, 2, 2, 2: 10 edges, and yes there is an Euler route because exactly two vertices are odd. Q5 edges AB 4, AC 6, BC 3, BD 9, CD 7, CE 5, DE 2, DF 8, EF 6: minimum spanning tree DE, BC, AB, CE, EF of weight 20; shortest A to F is A, C, E, F of length 17. Q6 OACB is a parallelogram with OA = a and OB = b; P is on AC with AP one third of AC and Q is on BC with BQ one third of BC: OP = a + (1/3)b, OQ = b + (1/3)a, so PQ = (2/3)(b − a) = (2/3)AB, ratio 2 : 3. Q7 boat on bearing 060 at 10 km/h with a current of 3 km/h due south: east 8.66, north 5 − 3 = 2, speed 8.9 km/h, bearing 077.0 degrees, position 26.0 km east and 6.0 km north after 3 hours.
The Year 5 descriptor bands on this page are band summaries written for this site; check them against the current IB MYP Mathematics guide before relying on them for reporting.