IB MYP Mathematics: Criteria A and B by year (MYP Mathematics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover Criterion A (knowing and understanding) and Criterion B (investigating patterns) from the IB Middle Years Programme Mathematics guide, first assessment 2022, using the Year 1, 3 and 5 descriptor sets. The mark totals are for practice only. Read the study guide first if you need the explanations, and keep the revision notes beside you. Course pages: subject hub and printable checklist.

Questions

1. For Criterion A, state what changes between the 3-4 band and the 5-6 band, and between the 5-6 band and the 7-8 band. [3]

2. (Year 1 level) A box of 15 notebooks costs 48.75. At the same rate, find the cost of 22 notebooks. [3]

3. (Year 5 level) A loading ramp rises 0.36 m over a horizontal distance of 4.32 m.

(a) Find the length of the sloping surface, to 3 significant figures. [3] (b) Write the slope as a ratio 1 : k. [1]

4. (Year 5 level) A bamboo shoot is 30 cm tall and grows 8% taller each week. After how many whole weeks is it first taller than 100 cm? [5]

5. (Year 1 level) Hexagonal tiles with side 1 are joined edge to edge in a straight row. One tile has perimeter 6, two tiles have perimeter 10, and three tiles have perimeter 14.

(a) Describe the pattern as a rule for n tiles. [2] (b) Find the perimeter of 25 tiles, and verify the rule with 4 tiles. [2]

6. (Year 3 level) The first five terms of a pattern are 3, 8, 15, 24, 35.

(a) Find a rule for the nth term. [3] (b) Verify your rule for the 6th term, which is 48, and justify it by rewriting the rule. [2]

7. (Year 3 level) A task asks you to investigate the number of dots in stage n of a growing pattern, where stages 1 to 4 have 4, 7, 10 and 13 dots. Three students write:

• Student A: “It goes up by 3 each time, so stage 5 has 16.”
• Student B: “Dots = 3n + 1, because the table suggests it.”
• Student C: “Dots = 3n + 1. Stage 8 has 25 dots, which I checked by drawing it. The pattern starts with one dot and adds 3 new dots at each stage.”

For each student, say which Criterion B strands are evidenced and what is missing. [4]

8. (Year 5 level) Work out (2k − 1)(2k + 1) + 1 for k = 1, 2, 3 and describe what you notice. Then prove your conjecture for every positive integer k. [6]

9. (Year 5 level) A student’s table contains an arithmetic slip in the fourth row. The quadratic rule she gives fits her table exactly and is as complex as the correct rule. State the highest Criterion B level she can reach, and explain why. [3]

10. (Year 5 level) A tree diagram starts with 3 branches at stage 1. At each later stage every branch from the previous stage splits into 2. Let u(n) be the number of branches at stage n and T(n) be the total number of branches over stages 1 to n.

(a) Complete a table of u(n) and T(n) for n = 1 to 4. [2] (b) Suggest general rules for u(n) and T(n). [2] (c) Test your rule for T(n) with n = 6, which has 189 branches in total. [1] (d) Prove the rule for T(n). [3]

11. (Year 3 or 5 level) A school club hires a hall for 180 and spends 6 on food per ticket sold. Tickets cost 15 each.

(a) Write an expression for the profit P when t tickets are sold. [2] (b) Find the number of tickets needed to break even. [1] (c) Find the smallest number of tickets that gives a profit of at least 400. [3]

Answers

1. From 3-4 to 5-6, the problems change from more complex to challenging, while staying in familiar situations [1]. From 5-6 to 7-8, the problems stay challenging but now include unfamiliar situations as well as familiar ones [1]. The strands (select, apply, solve) are worded the same way in every band [1]. [3] Examiner insight: A good answer names both what changes (problem difficulty, then situation) and what stays the same; listing only the band labels shows less understanding.

2. Unit rate: 48.75 ÷ 15 = 3.25 per notebook [1]. 3.25 × 22 [1] = 71.50 [1]. [3] Examiner insight: Show the unit rate as a separate step. A bare final figure leaves no evidence for the selection and application strands.

3. (a) Select Pythagoras for the right-angled triangle [1]. Length² = 0.36² + 4.32² = 0.1296 + 18.6624 = 18.792 [1]. Length = 4.33 m (3 s.f.) [1]. (b) 4.32 ÷ 0.36 = 12, so the slope is 1 : 12 [1]. [4] Examiner insight: Say why Pythagoras applies, since the problem never names it. Giving the length to more or fewer figures than asked is an avoidable loss.

4. Select exponential growth: height = 30 × 1.08^t [1]. Set up 30 × 1.08^t > 100, so 1.08^t > 3.33… [1]. Take logarithms (extended) to get t > ln(10/3) ÷ ln 1.08 = 15.64, or narrow down by testing whole values [1]. Check: t = 15 gives 95.2 cm (below 100) and t = 16 gives 102.8 cm (above) [1]. First whole number of weeks: 16 weeks [1]. [5] Examiner insight: The answer must be rounded up to 16, not down to 15, because 15 weeks has not yet reached 100 cm. The check line shows you understand why.

5. (a) Each extra tile adds 4 to the perimeter (6, 10, 14) [1]. Rule: perimeter = 4n + 2 [1]. (b) For n = 25: 4 × 25 + 2 = 102 [1]. Verify with 4 tiles: the rule gives 18, and 14 + 4 = 18 from the pattern [1]. [4] Examiner insight: Strand iii credit comes from testing a case that was not used to build the rule. Reusing n = 1 to 3 does not show this.

6. (a) First differences: 5, 7, 9, 11 [1]. Second differences are constant at 2, so the rule has n² as its leading term [1]. Subtracting n² leaves 2, 4, 6, 8, 10, which is 2n, so u(n) = n² + 2n [1]. (b) n = 6: 36 + 12 = 48 [1]. Justify: n² + 2n = (n + 1)² − 1, so each term is one less than a square, and 4, 9, 16, 25, 36 each less 1 gives the given terms [1]. [5] Examiner insight: A rewrite such as (n + 1)² − 1 explains why the pattern behaves as it does, which goes beyond checking one more value.

7. Student A states a prediction consistent with the pattern (strand ii, lower band) but gives no rule in n and no verification [1]. Student B suggests a general rule (strand ii at a higher band) but shows no test or justification, so strand iii is missing [1]. Student C gives the rule, tests it on a new case (stage 8: 3 × 8 + 1 = 25) and justifies it from the structure of the pattern, so all of strands ii and iii are evidenced [1]. None of the three shows how a technique was chosen (strand i), such as a table or differences, which is also needed; and if the task told them which method to use, the level would be capped at 4 [1]. [4] Examiner insight: Match each statement to a strand and say what extra step would raise it. Avoid assigning a level without naming the evidence.

8. k = 1: 1 × 3 + 1 = 4; k = 2: 3 × 5 + 1 = 16; k = 3: 5 × 7 + 1 = 36 [1]. The results are 4, 16, 36, which are 4k², so the conjecture is that the result is always (2k)², a perfect square and a multiple of 4 [1]. Let the two odd numbers be 2k − 1 and 2k + 1 [1]. Their product is (2k − 1)(2k + 1) = 4k² − 1 [1]. Adding 1 gives 4k² [1], which equals (2k)² for every positive integer k, so it is a perfect square and a multiple of 4 [1]. [6] Examiner insight: Three values suggest the conjecture; the algebra with a general k is what proves it. State the conclusion for every k in words.

9. The highest level is 6 [1]. The rule is consistent with her findings, but the findings are incorrect [1]. Provided the rule is as complex as the correct one, the cap is 6 and no higher [1]. [3] Examiner insight: Mention the condition that the rule is of equivalent complexity; without it, the cap statement is incomplete.

10. (a)

n	1	2	3	4
u(n)	3	6	12	24
T(n)	3	9	21	45

Two correct rows [2]. (b) u(n) doubles each time from 3, so u(n) = 3 × 2^(n−1) [1]. T(n) = 3, 9, 21, 45 fits T(n) = 3(2^n − 1) [1]. (c) n = 6: 3 × (64 − 1) = 189, matching the given value [1]. (d) T(n) = 3(1 + 2 + 4 + … + 2^(n−1)). Let S = 1 + 2 + 4 + … + 2^(n−1) [1]. Then 2S = 2 + 4 + … + 2^n, so 2S − S = 2^n − 1 [1]. Hence S = 2^n − 1 and T(n) = 3(2^n − 1) for every n [1]. [8] Examiner insight: A proof must cover all n. The doubling trick above does, while testing n = 6 only supports the rule.

11. (a) Income 15t and cost 180 + 6t [1]. P = 15t − (180 + 6t) = 9t − 180 [1]. (b) 9t − 180 = 0, so t = 180 ÷ 9 = 20 tickets [1]. (c) 9t − 180 ≥ 400, so 9t ≥ 580 [1]. t ≥ 64.4… [1]. Tickets are whole, so 65 tickets [1]. Check: 9 × 65 − 180 = 405 (at least 400) and 9 × 64 − 180 = 396 (less) (no extra mark). [6] Examiner insight: An inequality with whole items must be rounded up. Rounding 64.4 to 64 gives profit 396, below the target.

Where marks are usually lost

• Not saying why a method was chosen, so strand i has no evidence.
• Rounding the wrong way when a question asks for the first whole value that passes a limit.
• Stating “it goes up by 3” without a rule in n.
• Testing a rule only on the cases used to find it.
• Treating several correct checks as a proof when the question asks you to prove.
• Writing a rule from two data points and not checking it against a third.
• Ignoring units or the number of significant figures asked for.
• Describing a band by its label alone and not by what changes in the descriptor.

Next steps

• Revise with the revision notes.
• Re-read the study guide.
• Visit the MYP Mathematics hub and the printable checklist.
• Try all free 10-minute diagnostics at /diagnostics/.
• Book a free trial class at /trial/.

Official syllabus

International Baccalaureate Organization, Middle Years Programme Mathematics guide, from 2020, first assessment 2022. Criterion A and Criterion B for Years 1, 3 and 5 are paraphrased from this guide.