IB Mathematics: Applications and Interpretation SL Paper 2 Mock Paper (10 questions, 80 marks, 90 minutes)

2026-11-01

IB Mathematics AI SL

IB Mathematics AI SL Paper 2 mock paper: 10 questions, 80 marks, 90 minutes, calculator allowed, with full worked solutions and marking notes.

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Section A (Questions 1 to 5, 40 marks)

Answer ALL questions. A calculator and the formula booklet may be used. Show your working, give numerical answers to the accuracy required, and state units in every final answer. All data are examples and this is not a past paper of any examination board.

#	Question	Answer and marking notes
1	A class of 40 students scored a mean of 62 marks with a standard deviation of 10 marks; the marks are approximately normally distributed. (a) Find the z-score of 75 marks. (b) Estimate the percentage of students scoring above 75 marks and the number of such students. (c) Find the range containing the middle 95% of the marks.	(a) z = (75 – 62) / 10 = 1.30. (b) P(Z above 1.30) = 1 – 0.9032 = 0.0968 = 9.7%; students = 40 x 0.0968 = 3.9, about 4. (c) 62 +/- 1.96 x 10 = 42.4 to 81.6 marks. Marking notes: (a) 2 marks, (b) 3 marks, (c) 3 marks; total 8 marks.
2	A person deposits 50 000 dollars at 4.2% per annum compounded quarterly for 3 years. (a) Find the final amount and the interest earned. (b) A loan of 200 000 dollars at 6% per annum compounded monthly is repaid by equal monthly instalments over 5 years. Find the monthly payment. (c) Find the total interest paid on the loan in (b).	(a) Quarterly rate = 4.2% / 4 = 1.05%, periods = 12; amount = 50 000 x (1 + 0.0105)^12 = 56,676.86 dollars, interest = 6,676.86 dollars. (b) Monthly rate = 0.5%, periods = 60; payment = 200 000 x 0.005 / (1 – 1.005^-60) = 3,866.56 dollars. (c) Total paid = 3,866.56 x 60 = 231,993.62 dollars; total interest = 31,993.62 dollars. Marking notes: (a) 3 marks, (b) 3 marks, (c) 2 marks; total 8 marks.
3	A workshop makes x units of product A and y units of product B each day, subject to x + y ≤ 60; 2x + 3y ≤ 150; x ≤ 40; x ≥ 0, y ≥ 0. Profit is 25 dollars per unit of A and 40 dollars per unit of B. (a) List the vertices of the feasible region. (b) Find the maximum profit and the production quantities. (c) If the profit on B rises to 60 dollars, does the maximum profit change? Justify with calculations.	(a) The vertices are (0, 0), (0, 50), (30, 30), (40, 20) and (40, 0). (b) Evaluate P = 25x + 40y: (0,0) gives 0; (0,50) gives 2 000; (30,30) gives 1 950; (40,20) gives 1 800; (40,0) gives 1 000. Maximum profit is 2 000 dollars at x = 0, y = 50. (c) With 60 dollars per unit of B: (0,50) gives 3 000; (30,30) gives 750 + 1 800 = 2 550; (40,20) gives 1 000 + 1 200 = 2 200. The maximum is still at (0, 50) but rises to 3 000 dollars, an increase of 1 000 dollars. Marking notes: (a) 3 marks, (b) 3 marks, (c) 2 marks; total 8 marks.
4	A town has a population of 1 500 people growing at 8% per year. (a) Write a model for the population N after t years and find the population after 10 years. (b) Find the number of years needed for the population to reach 3 000 (1 decimal place). (c) A sample of 500 mg has a half-life of 12 days. Find the mass remaining after 30 days.	(a) N = 1 500 x 1.08^t; at t = 10, N = 3,238.4 people. (b) 1.08^t = 2, so t = ln 2 / ln 1.08 = 9.006, about 9.0 years. (c) 30 days is 2.5 half-lives, so the mass is 500 x 0.5^2.5 = 88.4 mg. Marking notes: (a) 3 marks, (b) 2 marks, (c) 3 marks; total 8 marks.
5	In triangle ABC, a = 7 cm, b = 9 cm and the included angle C = 52 degrees. (a) Find the length of c (3 decimal places). (b) Find the area of the triangle (2 decimal places). (c) Find angles A and B (1 decimal place).	(a) c^2 = 49 + 81 – 2 x 7 x 9 x cos 52 = 52.427, so c = 7.241 cm. (b) Area = 0.5 x 7 x 9 x sin 52 = 24.82 cm squared. (c) By the sine rule sin A = 7 sin 52 / 7.241 = 0.7618, so A = 49.6 degrees and B = 180 – 52 – 49.6 = 78.4 degrees. Marking notes: (a) 3 marks, (b) 2 marks, (c) 3 marks; total 8 marks.

Section B (Questions 6 to 10, 40 marks)

#	Question	Answer and marking notes
6	A bag contains 5 red and 3 blue balls. Two balls are drawn at random without replacement. (a) Find the probability that both are red. (b) Find the probability that the balls are of different colours. (c) In a game a player wins 3 dollars if both are red and loses 1 dollar otherwise. Find the expected value and the fair winning amount.	(a) P(both red) = (5/8) x (4/7) = 20/56 = 0.3571. (b) P(different) = (5/8)(3/7) + (3/8)(5/7) = 30/56 = 0.5357. (c) E = 3 x 0.3571 – 1 x 0.6429 = 0.429 dollars, favourable to the player; a fair game needs w x 0.3571 = 1 x 0.6429, so w = 7.2 dollars. Marking notes: (a) 2 marks, (b) 3 marks, (c) 3 marks; total 8 marks.
7	Let f(x) = x cubed – 6x squared + 9x + 2. (a) Find f prime of x and the stationary points, stating the nature of each. (b) Find the equation of the tangent at x = 0. (c) Find the area enclosed between the curve and the x-axis from x = 0 to x = 3.	(a) f prime = 3x^2 – 12x + 9 = 3(x – 1)(x – 3), so stationary points at x = 1 and x = 3. f double prime = 6x – 12; at x = 1 it is -6 (negative) so (1, 6) is a local maximum; at x = 3 it is 6 (positive) so (3, 2) is a local minimum. (b) f(0) = 2 and f prime(0) = 9, so the tangent is y = 9x + 2. (c) An antiderivative is x^4/4 – 2x^3 + 4.5x^2 + 2x; at x = 3 it equals 12.75 and at x = 0 it equals 0, so the area is 12.75 square units (the curve stays above the x-axis). Marking notes: (a) 4 marks, (b) 2 marks, (c) 2 marks; total 8 marks.
8	Consider the quadratic function y = -2x squared + 12x – 10. (a) Find the vertex and the axis of symmetry. (b) Find the x-intercepts and the value of the discriminant. (c) Write the function in vertex form and describe the transformations from y = -2x squared.	(a) x = -12 / (2 x -2) = 3, y = 8, so the vertex is (3, 8) and the axis is x = 3. (b) Discriminant = 144 – 80 = 64; the intercepts are x = 1 and x = 5, that is (1, 0) and (5, 0). (c) y = -2(x – 3)^2 + 8: take y = -2x squared, translate 3 units right and 8 units up; the curve still opens downwards. Marking notes: (a) 3 marks, (b) 3 marks, (c) 2 marks; total 8 marks.
9	A data set: 12, 15, 15, 18, 20, 22, 25, 28, 30, 42. (a) Find the median, lower quartile, upper quartile and interquartile range. (b) Use 1.5 x IQR to test whether 42 is an outlier. (c) Find the mean and the sample standard deviation (2 decimal places) and comment on the difference between the mean and the median.	(a) Median = (20 + 22) / 2 = 21; lower quartile = 15; upper quartile = 28; IQR = 13. (b) Upper fence = 28 + 1.5 x 13 = 47.5; 42 is below the fence, so it is not an outlier. (c) Mean = 22.7, sample standard deviation = 8.96; the mean is above the median because the large value 42 pulls the mean up while the median is resistant to extremes. Marking notes: (a) 3 marks, (b) 2 marks, (c) 3 marks; total 8 marks.
10	A right circular cone has radius 5 cm and height 12 cm. (a) Find the slant height and the volume (2 decimal places). (b) Find the total surface area (2 decimal places). (c) A similar smaller cone has a length ratio of 1 : 3. Find its volume and total surface area.	(a) Slant height = square root of (25 + 144) = 13 cm; volume = pi x 25 x 12 / 3 = 314.16 cm cubed. (b) Curved area = pi x 5 x 13 = 204.20; base = pi x 25 = 78.54; total surface area = 282.74 cm squared. (c) Length ratio 1 : 3 gives volume ratio 1 : 27, so 11.64 cm cubed; area ratio 1 : 9, so 31.42 cm squared. Marking notes: (a) 3 marks, (b) 3 marks, (c) 2 marks; total 8 marks.

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