IB MYP Mathematics: Algorithms, networks and linear programming in MYP (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 exemplars available through your school or the IB store.

These questions practise the algorithm, network and linear programming content of the Thinking with models branch of the IB Mathematics guide (for use from September 2020/January 2021), MYP Mathematics, first assessment 2022. Algorithms are MYP 4-5 standard; linear programming and networks are MYP 4-5 extended, as are questions 4 to 12 where they use those ideas. The marks in brackets are for your own self-marking.

Before you start, read the study guide or the revision notes. Hub: MYP Mathematics. Checklist: printable checklist.

Questions

1. An algorithm starts with a = 84 and b = 36. Repeat: if a is bigger than b, replace a by a − b; otherwise replace b by b − a; stop when a = b and output a. Trace the algorithm in a table and say what the output represents. [4]

2. A shop reduces a price of 80 by 15% each week. After each reduction the algorithm asks whether the price is below 40. If it is, it outputs the number of reductions; if not, it reduces again. Trace the algorithm, giving prices to 2 decimal places, and state the output. [4]

3. A sorted list has 500 names. (a) State the greatest number of checks a linear search needs. [1] (b) Find the greatest number of checks a halving search needs, showing your reasoning. [2] (c) State one condition the list must meet for a halving search to work. [1]

4. A club sells x tickets in advance and y tickets on the door. (a) Write inequalities for: at least 6 advance tickets; no more than 30 tickets in total; door tickets at most three times the advance tickets. [3] (b) Decide whether 7 advance and 20 door tickets satisfies all three. [1]

5. Maximise P = 5x + 4y subject to x + y ≤ 9, x + 3y ≤ 15, x ≥ 0 and y ≥ 0. [6]

6. A farmer mixes x kg of feed A and y kg of feed B. The mix must satisfy 2x + y ≥ 12 and x + 3y ≥ 15, and must weigh no more than 10 kg, so x + y ≤ 10. Feed A costs £5 per kg and feed B costs £4 per kg. Find the cheapest mix and its cost. [5]

7. Maximise P = 7x + 5y subject to 2x + 3y ≤ 19 and 4x + y ≤ 21, where x and y are whole numbers that are not negative. [5]

8. A network has vertices A to F and edges AB, AC, BC, CD, CE, DE and EF. (a) Give the degree of each vertex. [2] (b) Use the degrees to check that the network has 7 edges. [1] (c) List every path from A to F. [2]

9. In a weighted network the edge weights (in minutes) are PQ 5, PR 3, QR 2, QS 6, RS 7, RT 4, ST 1, SU 3, TU 6, TV 5 and UV 2. (a) Find the length of the path P-Q-S-U-V. [1] (b) Use a labelling method to find the shortest path from P to V and its length. [5]

10. A directed network has arcs S to A (6), S to B (3), B to A (2), A to C (4), B to C (7), C to T (1), A to T (9) and B to T (12). (a) Explain why you cannot travel from A to B. [1] (b) Find the shortest path from S to T. [3] (c) List all paths from S to T. [2]

11. A school makes x small gift boxes and y large gift boxes. A small box takes 20 minutes to assemble and a large box takes 30 minutes. There are 600 minutes available. At most 25 boxes can be stored, and at most 18 large boxes are wanted. Profit is £7 per small box and £10 per large box. (a) Write all constraints. [3] (b) Find the corners of the feasible region. [3] (c) Find the plan with the greatest profit and that profit. [2] (d) The large-box profit falls to £9. Does the best plan change? [2]

12. Use the network from question 9. Roadworks raise the weight of RT from 4 to 9. (a) Find the new shortest path from P to V and its length. [3] (b) State how much longer the journey is. [1] (c) Give two reasons why the model’s answer may differ from a real journey. [2] (d) Suggest one change to the model and the extra information it would need. [2]

Answers

1.

a	b
84	36
48	36
12	36
12	24
12	12

Correct first three passes [1]. Correct last two passes [1]. Output 12 [1]. It is the highest common factor of 84 and 36 [1]. [4] Examiner insight: Write a new row for every pass; skipping one makes later values wrong.

2. Prices after each reduction: 68.00, 57.80, 49.13 [1], 41.76, 35.50 [1]. After the fifth reduction 35.50 is below 40 [1]. Output 5 [1]. [4] Examiner insight: Multiply the current price by 0.85 each time rather than subtracting 15% of 80 repeatedly; the second method gives different values.

3. (a) 500 checks [1]. (b) A halving search with k checks can cover up to 2ᵏ − 1 names [1]. 2⁸ − 1 = 255 is too few and 2⁹ − 1 = 511 is enough, so 9 checks [1]. (c) The list must be sorted [1]. [4] Examiner insight: Show the reasoning behind 9; the number alone hides how you reached it.

4. (a) x ≥ 6 [1]; x + y ≤ 30 [1]; y ≤ 3x [1]. (b) x = 7: 7 ≥ 6 true; 27 ≤ 30 true; 20 ≤ 21 true. So yes, it is allowed [1]. [4] Examiner insight: When you check a point, substitute into every inequality and write the result of each, rather than just saying “yes”.

5. The corners come from the axes and the boundary lines. Where x + y = 9 meets x + 3y = 15: subtracting gives 2y = 6, so y = 3 and x = 6 [1]. Corners: (0, 0), (9, 0), (6, 3), (0, 5) [2]. Values of P: 0, 45, 42, 20 [1]. Greatest is P = 45 at (9, 0) [1]. Check: at (9, 0), 9 ≤ 9 and 9 ≤ 15, so the point is allowed [1]. [6] Examiner insight: The maximum can sit on an axis, so check the axis corners too.

6. The corners: 2x + y = 12 with x + y = 10 gives (2, 8) [1]. 2x + y = 12 with x + 3y = 15: y = 12 − 2x, so x + 36 − 6x = 15, giving x = 4.2 and y = 3.6 [1]. x + 3y = 15 with x + y = 10 gives (7.5, 2.5) [1]. Costs: 5(2) + 4(8) = 42; 5(4.2) + 4(3.6) = 35.4; 5(7.5) + 4(2.5) = 47.5 [1]. The cheapest mix is 4.2 kg of A and 3.6 kg of B, costing £35.40 [1]. [5] Examiner insight: For a minimum, you still compare every corner; the first corner you find is not necessarily the cheapest.

7. Solving 2x + 3y = 19 with 4x + y = 21: y = 21 − 4x, so 2x + 63 − 12x = 19, giving x = 4.4 and y = 3.4 [1]. P = 47.8 at that corner, but it is not whole-number [1]. Nearby whole-number points: (4, 3) gives 43 and is allowed [1]. (4, 4) breaks 2x + 3y ≤ 19 (20) and (5, 2) breaks 4x + y ≤ 21 (22) [1]. (3, 4) gives 41 and (5, 1) gives 40, both smaller. The best plan is x = 4, y = 3 with P = 43 [1]. [5] Examiner insight: Name the nearby points you rejected and why; that justifies the choice.

8. (a) A 2, B 2, C 4 [1]; D 2, E 3, F 1 [1]. (b) The total is 14 and 14 ÷ 2 = 7, so there are 7 edges [1]. (c) A-C-E-F, A-B-C-E-F [1], A-C-D-E-F, A-B-C-D-E-F [1]. [5] Examiner insight: List paths in a systematic order, such as always trying the lower-lettered neighbour first, so that you do not miss one.

9. (a) 5 + 6 + 3 + 2 = 16 minutes [1]. (b) Finish P (0): Q 5, R 3 [1]. Finish R (3): S 10, T 7 and Q stays 5 [1]. Finish Q (5): S stays 10. Finish T (7): S 8, U 13, V 12 [1]. Finish S (8): U 11. Finish U (11): V stays 12 [1]. Trace back: V from T, T from R, R from P. The shortest path is P-R-T-V, 12 minutes [1]. [6] Examiner insight: Show the label changes at each step; a bare path and number cannot show that you ruled out other routes.

10. (a) The only arc between A and B points from B to A [1]. (b) Finish S (0): A 6, B 3. Finish B (3): A 5, C 10, T 15 [1]. Finish A (5): C 9, T 14. Finish C (9): T 10 [1]. The shortest path is S-B-A-C-T, length 10 [1]. (c) S-B-A-C-T, S-A-C-T, S-B-C-T [1]; S-B-A-T, S-A-T, S-B-T [1]. [6] Examiner insight: In a directed network, check every step against the arrow before adding its weight.

11. (a) 20x + 30y ≤ 600 (or 2x + 3y ≤ 60) [1]; x + y ≤ 25 and y ≤ 18 [1]; x ≥ 0 and y ≥ 0 [1]. (b) y = 18 meets 2x + 3y = 60 where 2x = 6, so (3, 18) [1]. x + y = 25 meets 2x + 3y = 60: y = 25 − x, so 2x + 75 − 3x = 60, giving x = 15, y = 10 [1]. The other corners are (0, 0), (0, 18) and (25, 0) [1]. (c) P = 7x + 10y: 0, 180, 201, 205, 175 for (0,0), (0,18), (3,18), (15,10), (25,0) [1]. The best plan is 15 small and 10 large boxes, profit £205 [1]. (d) With P = 7x + 9y the values at (3, 18) and (15, 10) are 183 and 195 [1]. So the best plan does not change, and the profit is £195 [1]. [10] Examiner insight: Part (d) asks for a decision; state it in words.

12. (a) Finish P (0): Q 5, R 3. Finish R (3): S 10, T 12 [1]. Finish Q (5): S stays 10. Finish S (10): T 11, U 13 [1]. Finish T (11): V 16. Finish U (13): V 15. The shortest path is P-R-S-U-V, 15 minutes [1]. (b) 15 − 12 = 3 minutes longer [1]. (c) For example: travel times change with traffic, so the weights are not fixed [1]; the model ignores waiting at junctions or road type [1]. (d) For example, use time-of-day weights (peak and off-peak) [1], which would need recorded journey times at different hours [1]. [8] Examiner insight: Say how the answer could change, not only that something is unrealistic.

Where marks are usually lost

• Leaving out a row of a trace table or applying a decision the wrong way.
• Subtracting a percentage from the original instead of applying it to the current value.
• Reading “at least” and “at most” the wrong way round.
• Missing the axis corners of a feasible region.
• Giving the value of the objective without the point where it happens.
• Counting edges instead of adding weights when asked for the shortest path.
• Stopping the labelling as soon as the destination gets a label.
• Writing “not realistic” without saying what changes.

Next steps

• Revision notes
• Study guide
• MYP Mathematics hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

IB Mathematics guide (for use from September 2020/January 2021), MYP Mathematics, first assessment 2022, published by the International Baccalaureate Organization.