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 rather than leaving it all to the end.
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: simplify, combine, find the missing value | A | After lesson 3 | 30 minutes | Individual, written, in class. 23 marks. |
| A2: End-of-unit test | A | End of unit, after lesson 7 | 50 minutes | Individual, written, in class, one lesson. 30 marks in three sections. |
| B1: Adding a list of fractions | B | Mid-unit, after lesson 4 | 50 minutes | In class. Students may collect results in pairs, but each writes up their own findings. |
| C1: Put Tomas's working right and explain it | C | After lesson 6 | 40 minutes | Individual, written, in class or as homework. One page of words, working and a labelled sketch. |
| D1: Printing cloth bags for a market stall | D and C | End of unit, after lesson 7 | 50 minutes | Individual, in class. A written solution with explanations and a short note to the group. |
| D2: A courier's mean speed | D and C | End of unit, after lesson 7 | 50 minutes | Individual, in class, one lesson. A written solution with explanations and a short reply to the courier. |
D1 and D2 are alternatives. Both assess criteria D and C, in two different real-life contexts, one built on average cost and one on mean speed: set one, not both.
The unit's inquiry
- Key concept. Relationships: how one quantity depends on another when the dependence is a fraction with a letter in the bottom.
- Related concepts. Equivalence, Representation, Models
- Global context. Globalization and sustainability: how the cost and speed of making or moving things behave as the amount grows, and where a model stops being sensible.
- Statement of inquiry. Writing a rational expression in its simplest equivalent form, and reading it as a graph with asymptotes, shows what a cost or a speed does as quantities grow and which values the model can never take.
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: simplify, combine, find the missing value
Why set it. A short check of lessons 1 to 3 while there is still time to reteach: factorizing and cancelling, multiplying and dividing, and adding and subtracting algebraic fractions. Questions 7 and 8 are unfamiliar, so the top band is open to everyone.
When: After lesson 3. Time: 30 minutes. Format: Individual, written, in class. 23 marks.
The student sheet, with its task-specific clarification
When: After lesson 3. Time: 30 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. Simplifying
- Simplify (x² − 9)/(x² + 5x + 6). (3 marks)
- For which values of x is the original fraction in question 1 not defined? (2 marks)
Part 2. Products and quotients
- Simplify (2x + 6)/(x² − 4) × (x − 2)/(x + 3). (3 marks)
- Simplify (x² − x − 12)/(x + 1) ÷ (x − 4)/(3x + 3). (3 marks)
Part 3. Sums and differences
- Write 2/(x − 1) + 3/(x + 2) as a single fraction. (3 marks)
- Write 5/(x² − 1) − 2/(x + 1) as a single fraction. (3 marks)
Part 4. Think it through
- Dalia writes: (x + 4)/(x + 6) = 4/6 = 2/3, “because the x cancels”. Choose a value of x that shows she is wrong, and say what she should have noticed about the top and the bottom. (2 marks)
- The fraction (x² + kx + 6)/(x² − 4) can be simplified for two different whole-number values of k. Find both values and simplify the fraction each time. (4 marks)
How your work is judged
| Level | Criterion A in task A1: what it looks like |
|---|---|
| 1–2 | You answer some of questions 1 to 6 correctly, for example one factorization or one common denominator. |
| 3–4 | You answer most of questions 1 to 6 correctly, with the factorizing shown, and name an excluded value in question 2. |
| 5–6 | You answer questions 1 to 6 correctly with clear working, give every excluded value, and make a good start on question 7 or 8. |
| 7–8 | You answer questions 1 to 6 correctly and solve questions 7 and 8, with reasoning that shows why each answer works. |
Markscheme
| Task A1 question | Answer and marking |
|---|---|
| 1 | (x − 3)/(x + 2). 1 mark for (x − 3)(x + 3), 1 for (x + 2)(x + 3), 1 for cancelling to the answer. Cancelling x² to leave −9/(5x + 6) is the “cancel a term” slip. |
| 2 | x = −2 and x = −3. 1 mark each. The (x + 3) cancels, but x = −3 still makes the original bottom zero. An answer of only −2 scores 1. |
| 3 | 2/(x + 2). 2(x + 3)/((x − 2)(x + 2)) × (x − 2)/(x + 3). 1 mark for factorizing, 1 for cancelling both factors, 1 for the answer. |
| 4 | 3(x + 3), or 3x + 9. Invert and multiply 1 mark; (x − 4)(x + 3) and 3(x + 1) 1 mark; answer 1 mark. Multiplying by the second fraction without inverting it is the usual error. |
| 5 | (5x + 1)/((x − 1)(x + 2)). 1 mark for the common denominator, 1 for 2(x + 2) + 3(x − 1), 1 for the answer. Adding tops and bottoms (5/(2x + 1)) scores 0. |
| 6 | (7 − 2x)/((x − 1)(x + 1)). The common denominator is (x − 1)(x + 1): 5 − 2(x − 1) = 5 − 2x + 2. 1 mark for the denominator, 1 for the numerator, 1 for the answer. 3 − 2x comes from 5 − 2x − 2: the minus sign did not reach the −1 inside the bracket. |
| 7 | Any x that makes the sides differ, such as x = 2: 6/8 = 3/4, not 2/3; 1 mark for a value and the two numbers. 1 mark for saying that only a factor can be cancelled, and x + 4 and x + 6 share no factor. x = 0 does not work: both sides are then 2/3. |
| 8 | k = −5 gives (x − 3)/(x + 2); k = 5 gives (x + 3)/(x − 2). The bottom is (x − 2)(x + 2), so the top needs a factor x − 2 (zero at x = 2: 4 + 2k + 6 = 0) or x + 2 (zero at x = −2: 4 − 2k + 6 = 0). 1 mark for each k, 1 for each simplification. |
| Marks to a level | Total 23. 1 to 5 marks: 1–2. 6 to 11: 3–4. 12 to 18: 5–6. 19 to 23, with at least 3 of question 7 and 8's 6 marks: 7–8. 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 3–4 profile | Question 1 right, question 2 gives only x = −2 (1 mark), question 3 right; question 4 factorized but not inverted (1 mark); question 5 adds 2 + 3 over (x − 1)(x + 2) (0); question 6 finds the common denominator (1 mark); question 7 gives a value but no reason (1 mark); nothing on question 8. 3 + 1 + 3 + 1 + 0 + 1 + 1 + 0 = 10 marks, and the profile matches 3–4. |
| Level 7–8 answer to question 8 | “The bottom is (x − 2)(x + 2). The top can only cancel if it has one of those as a factor, so it must be zero at x = 2 or at x = −2. At x = 2: 4 + 2k + 6 = 0, so k = −5, and x² − 5x + 6 = (x − 2)(x − 3), giving (x − 3)/(x + 2). At x = −2: 4 − 2k + 6 = 0, so k = 5, and x² + 5x + 6 = (x + 2)(x + 3), giving (x + 3)/(x − 2).” |
| Level 7–8 answer to question 7 | “Put x = 2: the left side is 6/8 = 3/4 and her answer is 2/3, so they are different. x + 4 and x + 6 have no common factor; the x is only part of each, not a factor of the whole top or bottom, so nothing cancels.” |
Differentiation
A way in for anyone stuck
- Give the factorizations of x² − 9 and x² + 5x + 6 as a reference line, so question 1 tests cancelling and not factorizing.
- Offer a common-denominator frame for questions 5 and 6: three boxes for the new top of each fraction and the shared bottom.
- Read question 8 aloud on request and suggest asking “what must be true of the top at x = 2?” Reading is not what is being assessed.
Stretch
- Question 8 again with (x² + kx + 12)/(x² − 9). How many whole numbers k now work, and what are the fractions? (Two: k = −7 gives (x − 4)/(x + 3), k = 7 gives (x + 4)/(x − 3).)
- Find a fraction with top and bottom both quadratic that simplifies to 1/(x + 5) for every x except two values. State the two values you must exclude.
Students learning in English as an additional language
| Word in task A1 | Plain meaning |
|---|---|
| factorize | write an expression as a product of brackets, such as x² − 9 = (x − 3)(x + 3) |
| cancel | divide the top and the bottom by the same factor |
| not defined | has no value, because the bottom would be zero |
| common denominator | one bottom that both fractions can be rewritten with |
| whole number | a number with no fraction or decimal part, such as −5, 0 or 3 |
Sentence starters for task A1:
- The original is not defined when … because …
- I factorized the top as … and the bottom as …, so … cancels.
- Dalia is wrong because …
Task A2: End-of-unit test
Why set it. The whole unit at three levels of demand. Section 3 is unfamiliar on purpose: it is where the evidence for 7–8 comes from.
When: End of unit, after lesson 7. Time: 50 minutes. Format: Individual, written, in class, one lesson. 30 marks in three sections.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 7. Time: 50 minutes. Calculator: no calculator. Criterion: A, knowing and understanding.
Three sections, from familiar questions to ones you have not met in this form. Show your working everywhere: it earns credit even when an answer slips. Check every answer to an equation in the original equation.
Section 1. Familiar questions
- Simplify (3x² + 6x)/(x² − 4). (2 marks)
- Write down the vertical and horizontal asymptotes of y = 4/(x + 1) − 5. (2 marks)
- Write 2/(x + 3) + 1/(x − 1) as a single fraction. (3 marks)
- y is inversely proportional to x, and y = 8 when x = 3. Find y when x = 12. (2 marks)
Section 2. More complex questions
- Solve 3/(x − 1) = 2/(x + 2). (3 marks)
- Solve 2/x + 3/(x + 1) = 2. (4 marks)
- Write y = (2x + 1)/(x − 3) in the form k/(x − p) + q, and state its two asymptotes. (3 marks)
- A community hall in Cebu costs 300 pesos to hire, plus 12 pesos for each guest. (a) Write the average cost per guest for n guests as a rule in n. (b) What is the least whole number of guests for which the average cost per guest is no more than 20 pesos? (3 marks)
Section 3. Unfamiliar questions
- One pipe fills a tank in x hours working alone. A second pipe fills it in 3 hours longer. Together they fill it in 2 hours. Find how long each pipe takes alone, and explain why one solution of your equation is not used. (5 marks)
- Gunther claims: “Plot y = (x² − 1)/(x − 1) and you will see the straight line y = x + 1, nothing else.” Is he right? Explain. (3 marks)
How your work is judged
| Level | Criterion A in task A2: what it looks like |
|---|---|
| 1–2 | You answer some Section 1 questions correctly, for example an asymptote or a simplification. |
| 3–4 | You answer most of Section 1 correctly and make progress in Section 2, solving at least one equation with fractions. |
| 5–6 | You answer Section 1 and most of Section 2 correctly with each step shown, checking answers in the original equation, and make a start in Section 3. |
| 7–8 | You answer Sections 1 and 2 correctly and solve most of Section 3, rejecting values that make a bottom zero and explaining why your answers make sense. |
Markscheme
| Task A2 question | Answer and marking |
|---|---|
| 1 | 3x/(x − 2). 3x(x + 2)/((x − 2)(x + 2)): 1 mark for factorizing both, 1 for the answer. |
| 2 | Vertical x = −1, horizontal y = −5. 1 mark each. x = 1 for the vertical is the sign slip. |
| 3 | (3x + 1)/((x + 3)(x − 1)). 2(x − 1) + (x + 3) = 3x + 1. 1 mark for the common denominator, 1 for the numerator, 1 for the answer. |
| 4 | y = 2. k = xy = 24, then y = 24/12. 1 mark for k, 1 for y. 32 comes from treating it as direct proportion (8 × 4). |
| 5 | x = −8. 3(x + 2) = 2(x − 1), so 3x + 6 = 2x − 2. 1 mark for clearing, 1 for x = −8, 1 for the check: 3/(−9) = −1/3 and 2/(−6) = −1/3. |
| 6 | x = 2 or x = −1/2. 2(x + 1) + 3x = 2x(x + 1) gives 2x² − 3x − 2 = 0, then (2x + 1)(x − 2) = 0. 1 mark for clearing, 1 for the quadratic, 1 for both roots, 1 for checking both: x = 2 gives 1 + 1 = 2; x = −1/2 gives −4 + 6 = 2. |
| 7 | y = 7/(x − 3) + 2; vertical x = 3, horizontal y = 2. 2(x − 3) + 7 = 2x + 1. 1 mark for the form, 1 mark for each asymptote. |
| 8 | (a) C = (300 + 12n)/n = 300/n + 12. (b) 38 guests. 300/n + 12 ≤ 20 gives n ≥ 37.5, so n = 38. 1 mark for (a), 1 for 37.5, 1 for rounding up: 37 guests cost 20.11 each, which is too much. |
| 9 | The first pipe takes 3 hours and the second 6 hours. 1/x + 1/(x + 3) = 1/2 gives 2(x + 3) + 2x = x(x + 3), then x² − x − 6 = 0, (x − 3)(x + 2) = 0. x = −2 is rejected: a time cannot be negative. 2 marks for the equation, 1 for the quadratic, 1 for rejecting −2 with the reason, 1 for both times and a check: 1/3 + 1/6 = 1/2. |
| 10 | Not quite. (x − 1)(x + 1)/(x − 1) = x + 1 for every x except 1, but at x = 1 the original has no value. 1 mark for simplifying, 1 for the excluded value, 1 for the conclusion: the same line with no point at x = 1. |
| Marks to a level | Total 30. 1 to 6 marks: 1–2. 7 to 13: 3–4. 14 to 22: 5–6. 23 to 30, with at least 5 of Section 3's 8 marks: 7–8. Check the profile against the clarification before settling the level. |
Exemplar answers
| Task A2 response | What it looks like |
|---|---|
| Level 5–6 profile | Section 1 all correct (9 marks). Section 2: question 5 right (3); question 6 solves the quadratic but checks only x = 2 (3); question 7 right form, one asymptote (2); question 8 gets the rule and 37.5 but writes 37 (2): 10 marks. Section 3: question 9 sets up the equation and solves it but keeps both answers (3). 22 marks, and the profile matches 5–6: Sections 1 and 2 mostly secure, a start on Section 3. |
| Level 7–8 answer to question 9 | “Let the first pipe take x hours, so the second takes x + 3. In one hour they fill 1/x and 1/(x + 3) of the tank, and together 1/2. So 1/x + 1/(x + 3) = 1/2. Multiply by 2x(x + 3): 2(x + 3) + 2x = x(x + 3), so x² − x − 6 = 0, (x − 3)(x + 2) = 0. x = −2 would be a negative time, so x = 3. The pipes take 3 hours and 6 hours. Check: 1/3 + 1/6 = 1/2.” |
| Level 7–8 answer to question 10 | “The top is (x − 1)(x + 1), so for any x other than 1 the fraction is x + 1. At x = 1 the bottom is 0, so the graph has no point there. It is the line y = x + 1 with one point missing, so Gunther is nearly right but not exactly.” |
Differentiation
A way in for anyone stuck
- Read Section 3 aloud on request: reading is not what is being assessed.
- Offer a blank two-column frame for question 9: “what one pipe does in one hour” and “what they do together in one hour”.
- For question 7, give the line “2x + 1 = 2(x − 3) + ?” as a prompt for the division.
Stretch
- Question 9 with the second pipe taking 5 hours longer and the two together taking 6 hours. Find both times. (10 hours and 15 hours; the equation is x² − 7x − 30 = 0.)
- Write one equation with fractions that has exactly one solution to reject and one to keep, and show both checks.
Students learning in English as an additional language
| Word in task A2 | Plain meaning |
|---|---|
| asymptote | a line that a graph gets closer and closer to but does not reach |
| inversely proportional | when one quantity doubles, the other halves, so their product stays the same |
| average cost | the total cost divided by the number made or served |
| reject | decide that a solution cannot be used, and say why |
| working alone | doing a job without help from anyone or anything else |
Sentence starters for task A2:
- The graph has a vertical asymptote at … because …
- I reject x = … because …
- Gunther is right / not right because …
Task B1: Adding a list of fractions
Why set it. A real pattern hunt built on lesson 3's adding of fractions. The sheet never names the pattern: students organize their sums, find it, describe it as a rule, test it, extend it to a second list and justify it.
When: Mid-unit, after lesson 4. 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
Support you give (a prompt, a table outline) lowers the evidence for the top two bands: note it on the student's work.
When: Mid-unit, after lesson 4. Time: 50 minutes. Calculator: no calculator. Criterion: B, investigating patterns.
Here is a list of fractions, each with 1 on the top: 1/(1 × 2), 1/(2 × 3), 1/(3 × 4), 1/(4 × 5), and so on, where the bottom of each is two whole numbers in a row multiplied together. Call the sum of the first fraction S1, the sum of the first two S2, the sum of the first three S3, and so on. This task is about finding what SN is, without adding every fraction.
Part 1. Collect results
- Work out S1 to S5, each as a single fraction in its simplest form. Record each sum in a table you design yourself.
Part 2. Describe what you find
- Look at your table. Describe what you notice, in words, and write a rule that gives SN for any N.
- Use your rule to predict S6. Then work it out by adding the next fraction to S5, to check your prediction.
Part 3. Use and test it
- Without adding, predict S10 and S99. Show how your rule gives each answer.
- Now change the list to 1/(1 × 3), 1/(2 × 4), 1/(3 × 5), 1/(4 × 6), and so on, and call the sums T1, T2, …. Collect T1 to T4. Say what is the same and what is different, and what you think happens to the sum if the list goes on for ever.
Part 4. Explain
- Explain, using algebra on the fractions and not more examples, why your rule for SN works for every value of N.
- Do you need to know all of S8 to find S9? Explain your answer.
How your work is judged
| Level | Criterion B in task B1: what it looks like |
|---|---|
| 1–2 | With help, you add some of the fractions, write the sums as fractions and say something about them. |
| 3–4 | With help, you find S1 to S5 correctly as simplest fractions and suggest a rule in words for the sums. |
| 5–6 | You organize your sums, describe a rule that fits them, check it on S6, and use it for S10 and S99. You collect T1 to T4 and say what is the same and what differs. |
| 7–8 | You organize your sums, describe a rule that fits all your findings, check and use it, investigate the second list, and justify your rule for every N with algebra rather than examples. |
Markscheme
| Task B1 question | Answer and marking |
|---|---|
| The pattern (do not share it before the task) | Each fraction is 1/n − 1/(n + 1), so the sum of the first N fractions is 1 − 1/(N + 1) = N/(N + 1). The second list gives 3/4 minus two small fractions. |
| 1 | The fractions are 1/2, 1/6, 1/12, 1/20, 1/30. S1 = 1/2, S2 = 2/3, S3 = 3/4, S4 = 4/5, S5 = 5/6. For example 1/2 + 1/6 = 3/6 + 1/6 = 4/6 = 2/3. Unsimplified sums (4/6, 9/12) are fine if the pattern is found. |
| 2 | Any correct description: the top is N and the bottom is one more, so SN = N/(N + 1). A rule for the first few only (“it goes 1/2, 2/3, 3/4”) is evidence for 3–4. |
| 3 | S6 = 6/7. 5/6 + 1/42 = 35/42 + 1/42 = 36/42 = 6/7. The next fraction is 1/(6 × 7) = 1/42. |
| 4 | S10 = 10/11 and S99 = 99/100. |
| 5 | T1 = 1/3, T2 = 11/24, T3 = 21/40, T4 = 17/30. The fractions are 1/3, 1/8, 1/15, 1/24. For example 11/24 + 1/15 = 55/120 + 8/120 = 63/120 = 21/40. The rule N/(N + 1) does not carry over; the sums rise more slowly and seem to level off near 3/4 (0.75). Credit at least four sums and a clear statement of what held and what did not. |
| 6 | At 7–8 expect: 1/(n(n + 1)) = 1/n − 1/(n + 1), checked by combining the right side; then adding the first N terms, every middle term cancels and SN = 1 − 1/(N + 1) = N/(N + 1). Examples alone, however many, stay at 5–6. |
| 7 | No. Adding 1/90 to S8 would give S9, but the rule gives S9 = 9/10 directly from N = 9, with no earlier sum needed. Credit an argument that the rule needs only N. |
Exemplar answers
| Task B1 response | What it looks like |
|---|---|
| Level 3–4 | “S1 = 1/2, S2 = 2/3, S3 = 3/4, S4 = 4/5, S5 = 5/6. The tops go up by 1 and the bottoms go up by 1. S6 will be 6/7.” Both the sums and a rule in words are right, but there is no check by adding, S10 is not given, and the second list is not tried. |
| Level 7–8 | “SN = N/(N + 1). S10 = 10/11 and S99 = 99/100. For the second list I got 1/3, 11/24, 21/40, 17/30: also rising, but slower, and 17/30 is below 3/4, so I think it levels off near 3/4 rather than 1. Why the rule works: 1/n − 1/(n + 1) = ((n + 1) − n)/(n(n + 1)) = 1/(n(n + 1)). So SN = (1 − 1/2) + (1/2 − 1/3) + … + (1/N − 1/(N + 1)). Each fraction after the first cancels with the one before it, leaving 1 − 1/(N + 1) = N/(N + 1).” |
Differentiation
A way in for anyone stuck
- Give a two-column table outline for Part 1 and let the student write the headings.
- Suggest getting each sum by adding one new fraction to the one before, using a common denominator as in lesson 3, rather than starting again.
- Prompt with a question, never the rule: “What is the same about the tops? What is different about the bottoms?”
Stretch
- Investigate 1/(1 × 4), 1/(2 × 5), 1/(3 × 6), …. What number do the sums seem to level off near? (11/18, about 0.61.)
- For which whole numbers N is SN greater than 0.99? (N = 100 and above: S99 is exactly 0.99.)
Students learning in English as an additional language
| Word in task B1 | Plain meaning |
|---|---|
| list | a set of things written one after another in order |
| sum | the answer when numbers are added |
| simplest form | a fraction written so the top and bottom have no common factor except 1 |
| predict | say what will happen before you check |
| rule | a sentence or formula that works every time |
| for ever | without stopping |
Sentence starters for task B1:
- My table shows that …
- My rule is: …. I tested it on …
- It works for every N because …
Task C1: Put Tomas's working right and explain it
Why set it. Criterion C on its own: notation, a diagram and a line of reasoning a classmate can follow, built around two real errors from lessons 5 and 6.
When: After lesson 6. Time: 40 minutes. Format: Individual, written, in class or as homework. One page of words, working and a labelled sketch.
The student sheet, with its task-specific clarification
When: After lesson 6. Time: 40 minutes. Calculator: no calculator. Criterion: C, communicating.
Lin was away for lessons 5 and 6. She copied this working from a classmate, Tomás, and it has mistakes in it.
Tomás’s working
(a) Solve 1/(x − 2) + 3 = x/(x − 2).
Multiply by (x − 2): 1 + 3 = x, so x = 4.
(b) The graph of y = 2/(x − 1) + 1.
The vertical asymptote is x = −1 and the horizontal asymptote is y = 1. The curve meets y = 1 when x = 3.
Write a one-page explanation for Lin. She should be able to follow it on her own, without asking anyone.
Part 1. Put it right
- Find every mistake in Tomás’s working. For each one, say what he did, why it is wrong, and write the correct working.
Part 2. Teach the method
- Write a worked solution to (a) that Lin could follow, with each step on its own line and the answer checked in the original equation.
- Draw a sketch of y = 2/(x − 1) + 1 with both asymptotes dashed and labelled and both axis intercepts marked. Under it, write two sentences saying how you found the asymptotes from the rule.
Part 3. Check your explanation
- Make up one equation with a fraction in it that has a value which must be rejected. Solve it and explain the rejection to Lin in two sentences.
- Read your page as if you were Lin. Check that your equals signs join equal things, your brackets are in place, and you used the words asymptote, denominator, check and reject.
How your work is judged
| Level | Criterion C in task C1: what it looks like |
|---|---|
| 1–2 | Some of your steps can be followed, but the order is hard to see and the sketch or the notation is missing or unclear. |
| 3–4 | Your working for the equation is in a sensible order and your sketch shows both asymptotes; some labels or notation are wrong or missing, so Lin has to guess in places. |
| 5–6 | Each line follows from the one before and the check is written out. Your sketch labels the asymptotes and intercepts and matches your words, and your notation is mostly correct. |
| 7–8 | Your page is complete and concise: every step is justified in words or symbols, the sketch, working and sentences agree, notation is correct throughout, and your rejected-value example is explained so that Lin could use it alone. |
Markscheme
| Task C1 question | Answer and marking |
|---|---|
| The mistake in (a) | Tomás multiplied the two fractions by (x − 2) but not the 3. Correct: 1 + 3(x − 2) = x, so 3x − 5 = x, 2x = 5 and x = 5/2. Check: left side 1/(1/2) + 3 = 5, right side (5/2)/(1/2) = 5. His x = 4 fails the check: 1/2 + 3 = 7/2 but 4/2 = 2. |
| The mistakes in (b) | The vertical asymptote is where the bottom is zero, x = 1, not −1: the bracket x − 1 means the graph has moved 1 to the right. y = 1 is right. The curve never meets y = 1: 2/(x − 1) is never 0, and at x = 3 the value is 2/2 + 1 = 2, not 1. |
| Sketch (Part 2) | Two branches of the reciprocal shape centred on (1, 1): one up and to the right of x = 1, one down and to the left. The y-intercept is (0, −1) and the x-intercept is (−1, 0). Tomás’s wrong vertical line x = −1 passes through the x-intercept, which is why it looks plausible. |
| Own equation (Part 3) | Any equation that clears to a value making a bottom zero, with the check shown. Example: x/(x − 4) = 4/(x − 4) gives x = 4, which makes the bottom 0, so there is no solution at all. A better example keeps a second root, such as x²/(x − 3) = 9/(x − 3): x = 3 is rejected and x = −3 stays. |
| How to level | Level the page as a whole against the clarification, not by counting corrections. Notation, the sketch and the order of steps decide between 5–6 and 7–8 more often than the corrections do. Mathematical slips count only where they break the line of reasoning Lin has to follow; accuracy itself is criterion A. |
Exemplar answers
| Task C1 response | What it looks like |
|---|---|
| Level 3–4 extract | “In (a) Tomás forgot the 3. 1 + 3(x − 2) = x so 1 + 3x − 6 = x = 2x = 5 = 2.5. In (b) the asymptote should be x = 1.” The mistakes are found and the answer is right, but the equals signs join unequal things (x = 2x = 5), no check is shown and the sketch has no labels. |
| Level 7–8 extract | “Tomás multiplied both fractions by (x − 2) but left the 3 as it was. Every term must be multiplied: 1 + 3(x − 2) = x. Expand: 1 + 3x − 6 = x, so 3x − 5 = x, then 2x = 5, x = 5/2. Check in the original: 1/(5/2 − 2) + 3 = 1/(1/2) + 3 = 5 and (5/2)/(1/2) = 5. They match, so x = 5/2.” Each step on its own line, equals signs only between equal values, the check written out, and the word “every” says what rule he broke. |
Differentiation
A way in for anyone stuck
- Give the steps of lesson 6 as a reference card: multiply every term, solve, check each answer in the original.
- Let students explain aloud to a partner first, then write down what they said.
- Offer the frame: “Tomás wrote …. This is wrong because …. The correct working is ….”
Stretch
- Add a third part to your page: how could Lin tell, from the rule alone, that y = 2/(x − 1) + 1 can never equal 1?
- Write a version of the page for a student who has not met asymptotes. What did you change, and why?
Students learning in English as an additional language
| Word in task C1 | Plain meaning |
|---|---|
| mistake | a step that is not correct |
| explain | say what you did and why |
| sketch | a quick drawing that shows the shape and the important points, not every point |
| dashed | drawn as a line of short pieces with gaps, not as a solid line |
| intercept | the point where a graph crosses an axis |
Sentence starters for task C1:
- Tomás wrote …, but …
- First I …, because …
- I reject this value because it makes …
Task D1: Printing cloth bags for a market stall
Why set it. Average cost as a rational function, its floor, a comparison of two quotes and the limits of the model, in one real decision. An alternative to D2: set one of them.
When: End of unit, after lesson 7. Time: 50 minutes. Format: Individual, in class. A written solution with explanations and a short note to the group.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 7. Time: 50 minutes. Calculator: no calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.
A group in Arusha prints cloth bags by hand to sell at a weekend market. Setting up the screens and ink costs 240 shillings, however many bags are printed. Each bag then costs 8 shillings in cloth and ink. They will sell every bag at 14 shillings.
Part 1. Average cost
- Write a rule for the total cost of printing n bags. Then write the average cost per bag, C, as a rule in n.
- Find the average cost per bag when 10, 30, 60 and 120 bags are printed. Describe how quickly the average falls as more bags are printed.
Part 2. Reading the rule
- Describe the graph of C against n, including its asymptotes, and say what each asymptote means for the group.
- Can the average cost per bag ever be 8? Can it be 7? Explain using your rule.
Part 3. Decisions
- How many bags must the group print so that the average cost per bag is no more than the selling price of 14?
- A second printer quotes C = 150/n + 9. For which numbers of bags is the second printer cheaper per bag? Write an equation and solve it.
Part 4. Does it make sense?
- The group expects to sell about 55 bags, but it could be 100. Which printer should they use? Work out the total cost at both numbers and say how sure you are.
- Name one real thing the rule C = 240/n + 8 leaves out. Then write a short note to the group with your recommendation, so that someone who has not seen your working can understand it.
How your work is judged
| Level | Criterion D in task D1: what it looks like |
|---|---|
| 1–2 | You find some average costs for given numbers of bags, but you do not link them to a decision. |
| 3–4 | You write the average cost rule, find correct values and the number of bags for 14 per bag, and say whether your answers are sensible. |
| 5–6 | You use the rule correctly for every part, solve for the 14 per bag and the second quote, explain the lowest the average can reach, and say how far the model can be trusted. |
| 7–8 | You choose and use suitable methods throughout, justify each answer, show why the average has a floor, compare the printers across the whole range, and explain what the model leaves out and how that changes your advice. |
| Level | Criterion C in task D1: what it looks like |
|---|---|
| 1–2 | Some working is shown, but it is hard to see how you reached the average costs or the number of bags. |
| 3–4 | Your working for the averages and the comparison can be followed, with some correct notation and units. |
| 5–6 | Your working is clear and complete, usually with correct units (shillings, bags) and notation, and your note to the group is mostly well organized. |
| 7–8 | Your working is clear, complete and well ordered, units and notation are always correct, and your note to the group makes sense on its own. |
Markscheme
| Task D1 question | Answer and marking |
|---|---|
| 1 | Total = 240 + 8n; average C = (240 + 8n)/n = 240/n + 8. 1 mark for each. Writing C = 8n + 240 forgets to divide by n. |
| 2 | 32, 16, 12 and 10 shillings. 240/10 + 8 = 32, 240/30 + 8 = 16, 240/60 + 8 = 12, 240/120 + 8 = 10. The first 30 bags take the average from very large to 16; the next 90 only take it from 16 to 10. Credit a description that the fall slows. |
| 3 | Vertical asymptote n = 0; horizontal asymptote C = 8. n = 0 is not a number of bags, and for few bags the 240 is shared over very few. C = 8 is the floor: the material cost per bag that no number of bags can remove. |
| 4 | No for 8 and no for 7. 240/n is positive, so C is always more than 8; 7 is below the floor. Accept “240/n would have to be 0 or −1”. |
| 5 | 40 bags or more. 240/n + 8 ≤ 14 gives 240/n ≤ 6, so n ≥ 40; at 40 bags the average is exactly 14, the break-even. 1 mark for the inequality, 1 for 40. |
| 6 | Fewer than 90 bags. 150/n + 9 < 240/n + 8 gives 1 < 90/n, so n < 90. At n = 90 both are 32/3 (about 10.67). 1 mark for the equation, 1 for solving, 1 for stating that the first printer is cheaper above 90. |
| 7 | At 55 bags: first printer 240 + 8 × 55 = 680, about 12.36 each; second printer 150 + 9 × 55 = 645, about 11.73 each, so the second is cheaper by 35. At 100 bags: first 1040, second 1050, so the first is cheaper by 10. The switch is at 90, so if the group is unsure whether sales are above or below 90 the saving or loss is small either way (35 against 10). Credit any reasoned recommendation that states this. |
| 8 (criterion C) | A good note gives the rule, the 40 bags break-even, the 90 bags switch, a recommendation, and one limit (for example, the set-up cost may rise for more than one day’s printing, or bags may be spoiled), with units, in sentences that make sense without the working. |
Exemplar answers
| Task D1 response | What it looks like |
|---|---|
| Level 3–4 (D) | “C = 240/n + 8. 10 bags: 32. 30 bags: 16. 60 bags: 12. 120 bags: 10. For 14 they need 40 bags. The second printer is cheaper for 89 bags or fewer.” Correct numbers, but no reason that the average cannot reach 8, no comparison at 55 or 100, and no comment on the model. |
| Level 7–8 (D and C) | “The average can never fall to 8, because 240/n is always positive: that is the cost of cloth and ink for each bag. At 55 bags the second printer is cheaper by 35 in total; at 100 bags the first is cheaper by 10. The two quotes meet at 90 bags, so the choice is not risky either way. I recommend the second printer, since 55 is below 90. The model assumes the set-up cost does not change if bags are printed over two days.” The note to the group is four clear sentences with units. |
Differentiation
A way in for anyone stuck
- Give a table outline for question 2 with columns for n, 240/n and C.
- For question 6, give the first step: “subtract 8 from both sides, then subtract 150/n from both sides”.
- For question 8, ask: “If the set-up cost went up, which printer would look better?”
Stretch
- A third printer quotes C = 100/n + 10. Is it ever the cheapest of all three? (Yes: for fewer than 50 bags, since it beats the second quote when n < 50 and the first when n < 70.) Show your working.
- Write the rule when the set-up cost is S and the cost per bag is c. What do S and c do to the graph? (C = S/n + c: S sets how steeply it falls, c is the horizontal asymptote.)
Students learning in English as an additional language
| Word in task D1 | Plain meaning |
|---|---|
| set-up cost | money spent once, before making anything, however many are made |
| break even | to earn exactly as much as the cost |
| quote | the price a supplier says they will charge |
| recommend | say what someone should do, and why |
Sentence starters for task D1:
- The average cost is … because …
- The average cannot be below … because …
- I recommend … because …
Task D2: A courier's mean speed
Why set it. Mean speed over a return journey as a rational function, its ceiling, and a check on a common belief, in one real decision. An alternative to D1: set one of them.
When: End of unit, after lesson 7. Time: 50 minutes. Format: Individual, in class, one lesson. A written solution with explanations and a short reply to the courier.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 7. Time: 50 minutes. Calculator: no calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.
A bicycle courier rides 15 km to a customer at a steady 30 km/h, then rides straight back along the same road at a steady speed of v km/h. The courier wants to know the mean speed for the whole round trip.
Part 1. Set it up
- Write an expression for the total time of the round trip, in hours, in terms of v.
- Show that the mean speed for the round trip is M = 60v/(v + 30) km/h.
- Find M when v = 15, v = 30 and v = 60.
Part 2. Reading the rule
- Write M in the form k/(v + p) + q, and give its asymptotes. Say which of them tells the courier something real.
- Can the courier’s mean speed for the round trip ever reach 60 km/h? Explain using the rule.
- The courier says: “My mean speed is the average of my two speeds.” Test this claim when v = 60.
Part 3. Decisions
- The courier wants a mean speed of 35 km/h. Find the return speed v that is needed.
- On this road a cyclist cannot hold more than about 45 km/h. What is the highest mean speed the courier can reach?
Part 4. Does it make sense?
- The 15 km is only known to the nearest kilometre. Explain whether this changes your mean speeds.
- Write a short reply to the courier giving your answers, how reliable they are, and one real thing the model leaves out.
How your work is judged
| Level | Criterion D in task D2: what it looks like |
|---|---|
| 1–2 | You find some of the right numbers, for example the time for the outward ride, but do not reach a mean speed. |
| 3–4 | You find the mean speed for each return speed, find the return speed for a mean of 35, and say whether your answers are sensible. |
| 5–6 | You use suitable methods for every part, read the asymptotes from the rule, explain why the mean cannot reach 60, test the courier’s claim, and say how reliable your numbers are. |
| 7–8 | You choose and use suitable methods throughout, justify each answer from the rule, explain the ceiling and why the claim fails, show why the distance does not matter, and explain what the model leaves out and how it changes your advice. |
| Level | Criterion C in task D2: what it looks like |
|---|---|
| 1–2 | Some working is shown, but it is hard to see how you found the total time or the mean speeds. |
| 3–4 | Your working for the times and speeds can be followed, with some correct notation and units (h, km/h). |
| 5–6 | Your working is clear and complete, usually with correct units and notation, and your reply to the courier is mostly well organized. |
| 7–8 | Your working is clear, complete and well ordered, units and notation are always correct, and your reply to the courier makes sense on its own. |
Markscheme
| Task D2 question | Answer and marking |
|---|---|
| 1 | Time = 15/30 + 15/v = 1/2 + 15/v hours. 1 mark for each of the two times. |
| 2 | M = 30/(1/2 + 15/v) = 60v/(v + 30). Total distance 30 km. Multiply top and bottom by 2v: 60v/(v + 30). 1 mark for the distance over time, 1 for the algebra. |
| 3 | 20, 30 and 40 km/h. 60 × 15/45 = 20; 60 × 30/60 = 30; 60 × 60/90 = 40. 1 mark for all three; 1 for any two. v = 30 giving 30 is a check on the rule. |
| 4 | M = 60 − 1800/(v + 30). 60(v + 30) − 1800 = 60v. Asymptotes v = −30 and M = 60. Only M = 60 means something real: the vertical asymptote is at a speed that is not possible. 1 mark for the form, 1 for the asymptotes, 1 for choosing M = 60. |
| 5 | No. 1800/(v + 30) is always positive, so M is always less than 60. The outward ride alone takes 1/2 hour, so even with an instant return the mean cannot exceed 30/(1/2) = 60. Credit either reason. |
| 6 | The claim fails. (30 + 60)/2 = 45, but M = 40. The courier spends more time at the slower speed: 1/2 hour out and 1/4 hour back. 1 mark for each number, 1 for the reason. |
| 7 | v = 42 km/h. 60v = 35(v + 30) gives 25v = 1050. 1 mark for the equation, 1 for 42. Check: 60 × 42/72 = 2520/72 = 35. It is below the 45 km/h limit in question 8, so the courier can do it. |
| 8 | 36 km/h. M is increasing, so the highest value is at v = 45: 60 × 45/75 = 2700/75 = 36. 1 mark for seeing that M rises with v, 1 for the value. |
| 9 | No. With distance d each way, M = 2d/(d/30 + d/v) = 60v/(v + 30); the d cancels. 1 mark for the cancelling, 1 for the conclusion. Credit a test of d = 14.5 and d = 15.5 giving the same M. |
| 10 (criterion C) | A strong reply gives the mean speeds for the return speeds, the 60 km/h ceiling, the 36 km/h limit, and says that traffic, hills and stops would lower the real means because the rule assumes steady speeds. Units throughout, in sentences that make sense without the working. |
Exemplar answers
| Task D2 response | What it looks like |
|---|---|
| Level 3–4 (D) | “Time out is 15/30 = 0.5 hours. M = 60v/(v + 30). At v = 15, 20 km/h; at v = 60, 40 km/h. For 35 km/h the return needs v = 42.” Correct numbers, but the claim is not tested, the ceiling of 60 is not mentioned and nothing is said about reliability. |
| Level 7–8 (D and C) | “M = 60 − 1800/(v + 30), so M is always below 60: the ceiling. For v = 60 the average of the speeds is 45, but the mean speed is 40 because 1/2 hour was at 30 and only 1/4 hour at 60. The distance cancels, so 15 km to the nearest kilometre does not matter. A cyclist holding 45 km/h gets a mean of 36 km/h. Stops and hills would lower this, because the rule assumes steady speeds.” The reply is four clear sentences with units. |
Differentiation
A way in for anyone stuck
- Give the first line for question 2: “total distance = … km, total time = … hours”.
- For question 4, give the division 60v ÷ (v + 30) as 60(v + 30) − 1800 and ask the student to check it by expanding.
- For question 6, ask: “How long was the courier at 30 km/h, and how long at 60 km/h?”
Stretch
- Write the mean speed when the way out is at a km/h and the way back at b km/h. What does it simplify to? (2ab/(a + b).)
- For which return speed is the mean exactly 45 km/h, the claimed average? (90 km/h.)
Students learning in English as an additional language
| Word in task D2 | Plain meaning |
|---|---|
| mean speed | the total distance divided by the total time |
| round trip | a journey there and back again |
| steady speed | a speed that does not change |
| ceiling | a highest value that a quantity can come close to but never pass |
| claim | something a person says is true |
Sentence starters for task D2:
- The total time is … because …
- The mean speed cannot reach … because …
- The courier’s claim is / is not right because …
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.
To print a sheet, open it on the unit page, then print the page. Every task is written to be done without a calculator; a GDC is not needed for any of them.
The sheets are public, so a student may have seen A1 and A2. For an unseen A1, use this parallel version (same marks, same markscheme notes): Q1 (a) simplify (x² − 16)/(x² + x − 12) giving (x − 4)/(x − 3); (b) excluded values −4 and 3. Q2 (4x + 8)/(x² − 9) × (x − 3)/(x + 2) = 4/(x + 3). Q3 (x² + 2x − 15)/(x + 2) ÷ (x + 5)/(2x + 4) = 2(x − 3). Q4 3/(x + 1) + 2/(x − 2) = (5x − 4)/((x + 1)(x − 2)). Q5 4/(x² − 4) − 1/(x + 2) = (6 − x)/((x − 2)(x + 2)). Q6 Dalia writes (x + 2)/(x + 8) = 2/8 (at x = 1 the sides are 1/3 and 1/4). Q7 (x² + kx + 12)/(x² − 9): k = −7 gives (x − 4)/(x + 3) and k = 7 gives (x + 4)/(x − 3).
For an unseen A2, use this parallel version (same marks): Q1 simplify (4x² + 12x)/(x² − 9) = 4x/(x − 3). Q2 asymptotes of y = 6/(x − 2) + 4: x = 2 and y = 4. Q3 3/(x + 2) + 2/(x − 3) = 5(x − 1)/((x + 2)(x − 3)). Q4 y = 10 when x = 4, so k = 40 and y = 5 when x = 8. Q5 4/(x + 1) = 3/(x − 2) gives x = 11. Q6 3/x + 2/(x − 1) = 2 gives x = 1/2 or x = 3. Q7 (3x + 5)/(x + 1) = 2/(x + 1) + 3, asymptotes x = −1 and y = 3. Q8 hire 450 plus 15 per guest, average at most 25: 45 guests. Q9 pipes taking x and x + 5 hours, together 6 hours: x² − 7x − 30 = 0, so 10 and 15 hours (x = −3 rejected). Q10 (x² − 4)/(x − 2) is the line y = x + 2 with no point at x = 2.
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.