IB DP Mathematics: Applications and Interpretation – Number, approximation, sequences and financial 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 practice questions cover the number, approximation, sequences and financial mathematics unit of IB Diploma Programme Mathematics: Applications and Interpretation. They are aligned to the IB Mathematics: applications and interpretation guide, first assessment 2021, syllabus sections 1.1 to 1.8, which are common content for SL and HL. They follow the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so they apply to the May and November 2026, 2027 and 2028 sessions.

Every AI paper requires a GDC, so every question here is labelled “(calculator allowed)”. Show your working anyway: method marks depend on it. Revise first with the study guide and revision notes. The IB DP Maths AI course hub and printable checklist show the rest of the course.

Give answers exactly or to 3 significant figures, and money to the nearest cent, unless a question says otherwise.

Questions

1. (calculator allowed) Calculate (6.3 × 10⁸) ÷ (4.5 × 10⁻³), giving your answer in the form a × 10ᵏ, where 1 ≤ a < 10 and k is an integer. [2]

2. (calculator allowed) A rectangular tile measures 24 cm by 15 cm, each to the nearest centimetre.

(a) Write down the lower and upper bounds of the length. [1] (b) Find the upper bound of the area. [1] (c) A fitter uses 360 cm² as the area. Find the percentage error if the true dimensions are at their upper bounds. [2]

3. (calculator allowed) A stand has 35 rows of seats. The first row has 28 seats and each row has 3 more seats than the row in front.

(a) Find the number of seats in the last row. [2] (b) Find the total number of seats. [2] (c) Find the first row with more than 100 seats. [2]

4. (calculator allowed) Evaluate the sum of (7r − 3) from r = 1 to r = 25. [3]

5. (calculator allowed) A town had 42 000 people at the start of 2020. The population grows by 2.5% per year.

(a) Find the population at the start of 2030. [2] (b) Find the year in which the population, measured at the start of the year, first exceeds 60 000. [3]

6. (calculator allowed)

(a) Simplify (2x³)⁴ ÷ (4x⁻²). [2] (b) Solve e^(0.2t) = 5. [2]

7. (calculator allowed) Amira invests €8500 at 4.2% per year, compounded quarterly.

(a) Find the value after 6 years. [3] (b) Inflation is 2.6% per year. Find the real value of the investment after 6 years, in today’s money. [2] (c) Find the least number of complete years and quarters until the investment is first worth at least €12 000. [2]

8. (calculator allowed) A machine bought for £64 000 is worth £30 100 after 5 years. It depreciates by the same percentage each year.

(a) Find the annual rate of depreciation. [2] (b) Use your answer to (a), to 3 significant figures, to find the value after 8 years. [2]

9. (calculator allowed) Leo borrows £22 000 at 5.4% per year, compounded monthly. He repays it with equal payments at the end of each month for 6 years.

(a) Find the monthly payment. [3] (b) Find the total interest he pays. [2] (c) Find the amount he still owes after 3 years. [3]

10. (calculator allowed) Nadia pays €250 into a savings account at the end of every month. The account pays 3.9% per year, compounded monthly.

(a) Find the value of the account after 15 years. [3] (b) Find how much of this value is interest. [2]

11. (calculator allowed)

(a) At a café, 2 large, 1 medium and 3 small drinks cost £14.30; 1 large, 2 medium and 1 small cost £10.00; 3 large, 2 medium and 2 small cost £18.20. Find the price of each size. [4]

(b) An open box is made from a 30 cm by 20 cm sheet by cutting a square of side x cm from each corner. Its volume is V = x(30 − 2x)(20 − 2x). Find all possible values of x when V = 1000 cm³. [4]

12. (calculator allowed) A charity records its number of regular donors.

Year	1	2	3	4	5
Donors	410	447	481	520	553

(a) Explain why an arithmetic model is reasonable, and estimate the common difference using the first and last years. [2] (b) Use the model to predict the number of donors in year 10. [2] (c) The actual number in year 10 was 704. Find the percentage error of your prediction. [2] (d) A geometric model has u₁ = 410 and u₅ = 553. Find the common ratio and this model’s prediction for year 10. [3] (e) State which model predicted year 10 better. [1]

Answers

1. 6.3 ÷ 4.5 = 1.4 and 10^(8 − (−3)) = 10¹¹ [1]; 1.4 × 10¹¹ [1] [2] Examiner insight: A GDC display such as 1.4E11 is calculator notation and does not earn the final mark.

2. (a) 23.5 ≤ L < 24.5 [1] (b) 24.5 × 15.5 = 379.75 cm² [1] (c) ε = |360 − 379.75|/379.75 × 100% [1] = 5.20% [1] Examiner insight: The exact value goes in the denominator; dividing by 360 gives 5.49% and loses the accuracy mark.

3. (a) u₃₅ = 28 + 34 × 3 [1] = 130 [1] (b) S₃₅ = (35/2)(28 + 130) [1] = 2765 [1] (c) 28 + 3(n − 1) > 100 gives n > 25 [1]; u₂₅ = 100 and u₂₆ = 103, so row 26 [1] Examiner insight: In (c), “more than 100” excludes row 25, which has exactly 100; check the boundary term before answering.

4. Arithmetic with u₁ = 4, d = 7, n = 25, u₂₅ = 172 [1]; S₂₅ = (25/2)(4 + 172) [1] = 2200 [1] Examiner insight: A bare GDC sum of 2200 risks losing method marks; state the first term, common difference and number of terms.

5. (a) 42 000 × 1.025¹⁰ [1] = 53 763.5… ≈ 53 800 [1] (b) 42 000 × 1.025ⁿ > 60 000 [1]; n = 14 gives 59 345 and n = 15 gives 60 829 [1]; 2035 [1] Examiner insight: Showing the table values either side of 60 000 earns the method mark even if you slip converting n to a year.

6. (a) 16x¹² ÷ 4x⁻² [1] = 4x¹⁴ [1] (b) 0.2t = ln 5 [1]; t = 8.047… = 8.05 [1] Examiner insight: Writing 0.2t = ln 5 is the method mark; an answer of 8.05 with no working may get only the accuracy mark.

7. (a) FV = 8500(1 + 4.2/400)^(4 × 6) [1]; N = 24 quarters [1]; €10 921.71 [1] (b) 10 921.707… ÷ 1.026⁶ [1] = €9362.81 [1] (c) 8500(1 + 4.2/400)^q ≥ 12 000 gives q = 33.01 [1]; after 33 quarters the value is €11 998.24, so 34 quarters (8 years 6 months) [1] Examiner insight: Interest is added only at the end of each quarter, so 33.01 must be rounded up; giving 33 loses the final mark.

8. (a) 64 000(1 − r/100)⁵ = 30 100 [1]; r = 14.004… ≈ 14.0% [1] (b) 64 000 × 0.86⁸ [1] = 19 149.9… ≈ £19 100 [1] Examiner insight: Using the multiplier 1.14 instead of 0.86 treats depreciation as growth, which is a method error, not a slip.

9. (a) N = 72, I% = 5.4, P/Y = C/Y = 12 [1]; PV = 22 000, FV = 0 [1]; PMT = −358.40, so £358.40 [1] (b) 72 × 358.40 = 25 804.80 [1]; interest = £3804.80 [1] (c) N = 36, PMT = −358.40, PV = 22 000 [1]; solve for FV [1]; he owes £11 887.26 [1] Examiner insight: Writing your TVM entries earns the method marks; a wrong answer with no entries shown cannot earn them.

10. (a) N = 180, I% = 3.9, P/Y = C/Y = 12 [1]; PV = 0, PMT = −250 [1]; FV = €61 022.24 [1] (b) Total paid in = 180 × 250 = 45 000 [1]; interest = €16 022.24 [1] Examiner insight: N counts payments, not years; N = 15 is a method error.

11. (a) 2L + M + 3S = 14.30, L + 2M + S = 10.00 [1]; 3L + 2M + 2S = 18.20 [1]; GDC solver: L = 3.20 [1]; large £3.20, medium £2.50, small £1.80 [1] (b) x(30 − 2x)(20 − 2x) = 1000, with 0 < x < 10 because 20 − 2x > 0 [1]; GDC polynomial solver gives x = 2.93, 5 or 17.1 [1]; 17.1 is outside the domain, so x = 2.93 [1] or x = 5 [1] Examiner insight: Quoting x = 17.1 as a valid answer shows the domain was ignored, and the final mark is withheld.

12. (a) Differences 37, 34, 39, 33 are roughly constant [1]; d ≈ (553 − 410)/4 = 35.75 [1] (b) u₁₀ ≈ 553 + 5 × 35.75 [1] = 731.75 ≈ 732 donors [1] (c) ε = |732 − 704|/704 × 100% [1] = 3.98% [1] (d) 410r⁴ = 553 [1]; r = 1.0777 [1]; u₁₀ = 410 × 1.0777⁹ ≈ 804 donors [1] (e) The arithmetic model, because 732 is closer to 704 than 804 is [1] Examiner insight: In (d), r must use the power 4 (four steps from u₁ to u₅); using r⁵ is a method error.

Where marks are usually lost

• Writing GDC notation (1.4E11) instead of 1.4 × 10¹¹.
• Using the approximate value as the denominator in percentage error.
• Using rⁿ instead of rⁿ⁻¹, or counting years from the wrong start point.
• Entering years instead of periods for N when interest is compounded quarterly or monthly.
• Rounding a number of periods down when the target is reached part-way through a period.
• Giving PV and PMT the same sign in the TVM solver.
• Leaving out TVM entries or the system of equations, so a slip loses every mark.
• Keeping a root outside the domain of a real context.
• Giving a count of people or seats as a decimal.

Next steps

• Revision notes for this unit
• Study guide for full explanations
• IB DP Maths AI course hub
• Printable syllabus checklist
• More AI practice: statistics and probability and geometry and trigonometry
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: applications and interpretation guide, first assessment 2021.