IB MYP Mathematics: Algebra, equations and sequences (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 materials. Use these alongside the sample materials and tasks your school provides.

These questions cover the Algebra rows of the numerical and abstract reasoning branch in the IB Mathematics guide for the MYP (first assessment 2022). The marks are Marlbridge practice tariffs, used to show how much working each question needs. Question 12 uses enrichment and extended content and is labelled. Read the study guide and revision notes first. The MYP Mathematics hub and printable checklist show the rest of the course.

Questions

1. (a) Expand and simplify 5(x − 2) − 2(3x − 4). [2] (b) Find the value of 4a² − 3ab when a = −3 and b = 2. [2]

2. Gym A charges £25 to join plus £18 each month. Gym B charges £40 to join plus £13 each month. Form an equation and find after how many months the total cost is the same, and what that cost is. [4]

3. The sequence 4, 11, 18, 25, … continues in the same way. (a) Find a rule for the nth term. [2] (b) Decide whether 207 and 300 are terms of the sequence. [2]

4. The first five terms of a sequence are 3, 9, 19, 33, 51. (a) Find a general rule, showing how you found it. [3] (b) Prove that the difference between the (n + 1)th term and the nth term is 4n + 2. [3]

5. Solve the simultaneous equations 2x + 5y = 13 and 3x − y = 11. [4]

6. (a) Solve 3 − 4x > 19. [2] (b) Solve −3 ≤ 3x + 4 < 16 and list the integers that satisfy it. [3]

7. (a) Factorize 14m²n − 21mn². [2] (b) Factorize 5x² + 8x − 4. [3]

8. (a) Solve x² − 2x − 35 = 0 by factorization. [3] (b) Use the quadratic formula to solve 2x² − 5x − 1 = 0. Give answers to 3 significant figures. [3] (c) The graph of y = x² + 2x − 15 passes through (−5, 0) and (3, 0). Write down the solutions of x² + 2x − 15 = 0 and the equation of the line of symmetry. [2]

9. A rectangular banner is 6 m longer than it is wide. Its area is 112 m². Form and solve a quadratic equation to find the width, then find the perimeter. [6]

10. (a) Make h the subject of S = 2πr² + 2πrh. [3] (b) Make x the subject of y = (4x + 3)/(x − 2). [4]

11. An algorithm starts with D = 1000 and k = 0. It repeats these steps: replace D with 0.7 × D, add 1 to k, and stop if D is less than 200. At the end it outputs k. (a) Complete a trace table for D and k. [3] (b) State the output. [1] (c) Describe the type of sequence D follows and give its common ratio. [1]

12. Extended and enrichment. An arithmetic sequence has first term 9 and common difference 4. A geometric sequence has first term 96 and common ratio 1/2. (a) Find the 30th term of the arithmetic sequence. [2] (b) Find the sum of its first 30 terms. [2] (c) Find the 7th term of the geometric sequence. [2] (d) Find the sum to infinity of the geometric series and explain why it exists. [2]

Answers

1. (a) 5x − 10 − 6x + 8 [1] = −x − 2 [1] (b) 4(−3)² = 36 and −3(−3)(2) = +18 [1]; total 54 [1] [4] Examiner insight: Keep brackets around negative values when substituting. Writing −3² instead of (−3)² is a common way to lose the first step.

2. Let m be the number of months. 25 + 18m = 40 + 13m [1]. 5m = 15 [1]. m = 3 months [1]. Cost: 25 + 18(3) = £79, and Gym B gives 40 + 39 = £79, so £79 [1] [4] Examiner insight: State the unknown first. Checking the answer in both expressions shows the cost really is the same and is worth including.

3. (a) The difference is 7, so the rule is 7n + c [1]. At n = 1, 7 + c = 4, so 7n − 3 [1] (b) 7n − 3 = 207 gives n = 30, a whole number, so 207 is a term [1]. 7n − 3 = 300 gives n = 303/7 = 43.29…, not a whole number, so 300 is not a term [1] [4] Examiner insight: A reason such as “n must be a whole number” is needed; a yes/no on its own gives little to credit.

4. (a) First differences 6, 10, 14, 18; second differences are all 4 [1]. So the rule contains 2n², which gives 2, 8, 18, 32, 50 [1]. Each term is 1 more: 2n² + 1 [1] (b) The nth term is 2n² + 1 and the (n + 1)th is 2(n + 1)² + 1 [1]. Difference: 2(n² + 2n + 1) + 1 − (2n² + 1) [1] = 4n + 2 [1] [6] Examiner insight: A proof must work for every n. Testing n = 1 and n = 2 only verifies; the algebra is what proves.

5. Multiply the second equation by 5: 15x − 5y = 55 [1]. Add to the first: 17x = 68 [1]. x = 4 [1]. Substitute: 12 − y = 11, so y = 1 [1] [4] Examiner insight: Once you have x and y, test them in the equation you did not use to find y; a slip in either value shows up straight away.

6. (a) −4x > 16 [1]; dividing by −4 reverses the sign: x < −4 [1] (b) Subtract 4: −7 ≤ 3x < 12 [1]. Divide by 3: −7/3 ≤ x < 4 [1]. Integers: −2, −1, 0, 1, 2, 3 [1] [5] Examiner insight: Listing integers tests the end points. 4 is excluded because of the strict sign, and −7/3 is about −2.33, so −2 is the smallest integer that fits.

7. (a) Common factor 7mn [1]: 7mn(2m − 3n) [1] (b) Two numbers with product 5 × (−4) = −20 and sum 8 are 10 and −2 [1]. 5x² + 10x − 2x − 4 = 5x(x + 2) − 2(x + 2) [1] = (5x − 2)(x + 2) [1] [5] Examiner insight: Take out the highest common factor, not just any factor. Expand your brackets to check.

8. (a) (x − 7)(x + 5) = 0 [1]; x − 7 = 0 or x + 5 = 0 [1]; x = 7 or x = −5 [1] (b) a = 2, b = −5, c = −1 so x = (5 ± √(25 + 8)) / 4 = (5 ± √33) / 4 [1]. x = 2.686… [1] and x = −0.186…, so x = 2.69 or x = −0.186 [1] (c) The solutions are the x-intercepts: x = −5 and x = 3 [1]. The line of symmetry is halfway between: x = −1 [1] [8] Examiner insight: Substitute b = −5 with its bracket so −b becomes +5 and 4ac becomes −4(2)(−1) = +8. Round only at the end.

9. Let w be the width in metres; the length is w + 6. w(w + 6) = 112 [1]. w² + 6w − 112 = 0 [1]. (w + 14)(w − 8) = 0 [1]. w = −14 or 8 [1]. A width cannot be negative, so w = 8 m and the length is 14 m [1]. Perimeter 2(8 + 14) = 44 m [1] [6] Examiner insight: Rejecting the negative root with a reason connected to the context is part of a complete solution. Check that 8 × 14 gives 112.

10. (a) S − 2πr² = 2πrh [1]. Divide by 2πr [1]. h = (S − 2πr²) / (2πr) [1] (b) y(x − 2) = 4x + 3 [1]. yx − 2y = 4x + 3 [1]. yx − 4x = 3 + 2y, so x(y − 4) = 3 + 2y [1]. x = (2y + 3)/(y − 4) [1] [7] Examiner insight: When the subject appears twice, collect those terms on one side and factorize. Test with a value: x = 5 gives y = 23/3, and the formula returns 5.

11. (a) Pass 1: D = 700, k = 1; pass 2: D = 490, k = 2 [1]. Pass 3: D = 343, k = 3; pass 4: D = 240.1, k = 4 [1]. Pass 5: D = 168.07, k = 5, and 168.07 is below 200, so it stops [1] (b) The output is 5 [1] (c) A geometric sequence with common ratio 0.7 [1] [5] Examiner insight: Record each variable after each step and apply the stopping test to the new value. Testing the old value gives an output that is one pass too late.

12. (a) a₃₀ = 9 + 29 × 4 [1] = 125 [1] (b) S₃₀ = (30/2)(9 + 125) [1] = 2010 [1] (c) a₇ = 96 × (1/2)⁶ [1] = 1.5 [1] (d) S∞ = 96 / (1 − 1/2) = 192 [1]. It exists because the common ratio, 1/2, lies between −1 and 1, so the terms shrink towards zero and the series converges [1] [8] Examiner insight: Use the correct power: the nth term uses rⁿ⁻¹, so the 7th term uses the 6th power. Give a reason for convergence that refers to the ratio.

Where marks are usually lost

• Not defining the unknown or not writing the equation before solving it.
• Forgetting to check a context answer, such as a negative width or a non-integer term number.
• Reversing the inequality sign at the wrong time, or not reversing it after dividing by a negative.
• Giving only one root of a quadratic, or not rejecting an impossible one.
• Treating a pattern that fits a few terms as proven.
• Losing a sign when substituting a negative, especially in the quadratic formula.
• Using the nth power instead of the (n − 1)th in a geometric term.
• Stopping a trace table one pass early or late.

Next steps

• Re-read the revision notes for the methods you missed.
• Return to the study guide for the full worked examples.
• Check progress against the MYP Mathematics hub and the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Mathematics guide for the Middle Years Programme (for use from September 2020/January 2021), International Baccalaureate Organization, first assessment 2022, numerical and abstract reasoning, Algebra rows.