IB · Mathematics: Applications & Interpretation SL & HL · N2H

IB Mathematics: Applications & Interpretation Paper 2 · HL (long questions) · 2029 syllabus practice papers

2029 syllabus · HL · 100 marks · 2h · calculator. Each generated paper follows the SL & HL Paper 2 · HL (long questions) · 2029 syllabus format, with the same number and style of questions as the real exam, and comes with an IB-style mark scheme.

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Sample Paper 2 · HL (long questions) · 2029 syllabus questions

Question 1Structured · 13 marks

The table shows the average temperature x °C and the number of ice creams sold y at a kiosk on 8 days.

x	15	17	20	24	28	26	37	36
y	48	61	54	39	79	72	93	76

Table 1.1

(a)

Find Pearson’s product-moment correlation coefficient, r, and describe the correlation.

[3]

(b)

Find the equation of the regression line of y on x.

[2]

(c)

Use the regression line to estimate the number of ice creams sold when the temperature is 25 °C, and comment on the reliability of your estimate.

[3]

(d)

Interpret the gradient of the regression line in this context.

[1]

(e)

Find Spearman’s rank correlation coefficient r_s, and explain why it may be a more suitable measure if the relationship is not linear.

[4]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	r = 0.771 GDC (M1) r = 0.771 A1 strong positive linear correlation A1	3
(b)	y = 1.70x + 22.1 GDC (M1) y = 1.70x + 22.1 A1	2
(c)	64.6 1.70 × 25 + 22.1 (M1) 64.6 A1 reliable: interpolation and strong correlation R1	3
(d)	increase per degree about 1.70 more ice creams are sold for each 1 °C increase R1	1
(e)	0.810 ranking the data (M1) r_s = 0.810 A2 Spearman measures a monotonic (not necessarily linear) relationship R1	4

Question 2Structured · 14 marks

A tent is a triangular prism. Its front is an isosceles triangle with base 2 m and sloping sides 2 m, and the tent is 5 m long. The floor is not made of fabric.

(a)

Find the height of the tent.

[2]

(b)

Find the angle between a sloping side and the floor.

[2]

(c)

Find the volume of the tent.

[2]

(d)

Find the area of fabric needed.

[3]

(e)

A pole runs from a bottom front corner of the tent to the top back corner of the ridge. Find its length and the angle it makes with the floor.

[5]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	1.73 m Pythagoras (M1) 1.73 m A1	2
(b)	60.0° cos θ = 12 (M1) 60.0° A1	2
(c)	8.66 m³ area of triangle × length (M1) 8.66 m³ A1	2
(d)	23.5 m² two rectangles and two triangles (M1) correct expression (A1) 23.5 m² A1	3
(e)	5.39 m; 18.8° horizontal distance √(w/2)² + l² (M1) 5.10 (A1) 5.39 m A1 tan θ = height ÷ horizontal distance (M1) 18.8° A1	5

Question 3Structured · 14 marks

An aircraft’s position is (x, y, z) km, where x and y are displacements east and north of an airport and z is the height. At 13:00 it is at (30, 20, 6) and moves with constant velocity −150−100−20 km h⁻¹. Let t be the time in hours after 13:00.

(a)

Write down a vector equation for the position r of the aircraft.

[2]

(b)

Verify that the aircraft would pass directly over the airport, and find its height at that time.

[4]

(c)

Find the time when the aircraft is 4 km above the ground, and its distance from the airport then.

[4]

(d)

The aircraft now adjusts its descent so that it lands at (0, 0, 0), keeping the same horizontal velocity −150−100. Find the new vertical component of velocity.

[4]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	r = 30206 + t−150−100−20 position + t × velocity (M1) r = 30206 + t−150−100−20 A1	2
(b)	t = 0.200; 2.00 km x = 0 ⇒ t = 0.200 A1 y = 0 at the same t R1 substituting into z (M1) 2.00 km A1	4
(c)	t = 0.100 h; 18.5 km 6 − 20t = 4 (M1) t = 0.100 A1 position and distance (M1) 18.5 km A1	4
(d)	−40.0 km h⁻¹ time to reach the airport horizontally (M1) 0.100 h (A1) −4 ÷ time (M1) −40.0 km h⁻¹ A1	4

Question 4Structured · 12 marks

A sweet company makes star-shaped and moon-shaped sweets. 70% are stars and the rest are moons. 15% of stars and 40% of moons are yellow.

(a)

A sweet is chosen at random.
(i) Find the probability that it is yellow.
(ii) Given that it is yellow, find the probability that it is a star.

[4]

The company claims the colours in each bag follow the distribution below. A bag of 100 sweets is tested with a χ² goodness of fit test at the 5% level.

Colour	Brown	Red	Green	Orange	Yellow	Purple
Claimed %	15	25	20	20	10	10
Observed	21	14	24	23	9	9

Table 4.1

(b)

Write down the null hypothesis.

[1]

(c)

Find the expected frequencies.

[2]

(d)

Write down the number of degrees of freedom.

[1]

(e)

Find the p-value.

[2]

(f)

State the conclusion of the test, giving a reason.

[2]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	0.225; 0.467 0.7 × 0.15 + 0.3 × 0.4 (M1) 0.225 A1 0.1050.225 (M1) 0.467 A1	4
(b)	colours follow the claimed distribution the colours are distributed as the company claims A1	1
(c)	15.0, 25.0, 20.0, 20.0, 10.0, 10.0 percentage × 100 (M1) 15.0, 25.0, 20.0, 20.0, 10.0, 10.0 A1	2
(d)	5 6 − 1 = 5 A1	1
(e)	0.122 χ² = 8.69 (A1) 0.122 A1	2
(f)	do not reject H₀ 0.122 > 0.05 R1 do not reject H₀: the data are consistent with the claim A1	2

Question 5Structured · 14 marks

Each year 10% of the customers of company X move to company Y, and 10% of the customers of company Y move to company X. Initially X has 1500 customers and Y has 2500.

(a)

Write down a transition matrix T for the movement of customers.

[2]

(b)

Find the eigenvalues and corresponding eigenvectors of T.

[4]

(c)

Hence write down matrices P and D such that T = PDP⁻¹.

[2]

(d)

Find an expression for the number of customers of X after n years.

[5]

(e)

Write down the number of customers X can expect in the long term.

[1]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	0.9 0.10.1 0.9 columns sum to 1 (M1) 0.9 0.10.1 0.9 A1	2
(b)	λ = 1, λ = 0.8 det(T − λI) = 0 (M1) λ = 1, λ = 0.8 A1 eigenvector for 1: 0.10.1 (or a multiple) A1 eigenvector for 0.8: 1−1 A1	4
(c)	P, D P = 0.1 10.1 −1 A1 D = 1 00 0.8 A1	2
(d)	2000 − 500(0.8)ⁿ use of Tⁿ = PDⁿP⁻¹ M1 correct P⁻¹ A1 x_n = 2000 − 500(0.8)ⁿ A2 form A + B(λ)ⁿ A1	5
(e)	2000 2000 A1	1

Question 6Structured · 13 marks

The volume of drink in bottles has mean 250 ml and standard deviation 20 ml. The distribution of the volumes is not known.

(a)

A sample of 25 bottles is taken. State the approximate distribution of the sample mean X̄, giving a reason.

[3]

(b)

Find the probability that the sample mean is less than 248 ml.

[2]

(c)

A different sample of 25 bottles has mean 247 ml. Find a 95% confidence interval for the population mean, assuming σ = 20.

[3]

(d)

Hence comment on the claim that the mean volume is 250 ml.

[2]

(e)

Find the least sample size needed for the width of a 95% confidence interval to be at most 4 ml.

[3]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	X̄ ~ N(250, 16.0) central limit theorem, n large R1 mean 250 A1 variance 20²25 = 16.0 A1	3
(b)	0.309 normal CDF with standard error (M1) 0.309 A1	2
(c)	239.2 < μ < 254.8 z-interval (M1) 247 ± 1.96 × 20√25 (A1) 239.2 < μ < 254.8 A1	3
(d)	supported 250 is in the interval R1 the claim is supported A1	2
(e)	385 2 × 1.96 × 20√n ≤ 4 (M1) solving for n (M1) 385 A1	3

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in