IB · Mathematics: Applications & Interpretation SL & HL · P2H

IB Mathematics: Applications & Interpretation Paper 2 · HL (long questions) · exams to Nov 2028 practice papers

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Sample Paper 2 · HL (long questions) · exams to Nov 2028 questions

Question 1Structured · 14 marksA* challenge

A biologist models the number of fish in a lake t years after they are introduced by N(t) = 10001 + 19e^−0.5t.

(a)

Find the initial number of fish.

[1]

(b)

Find the number of fish after 2 years.

[2]

(c)

Find the time at which the population is growing fastest, and the population at that time.

[4]

(d)

An alternative model is P(t) = 50e^mt, which agrees with N at t = 0 and t = 2. Find m.

[3]

(e)

Compare the predictions of the two models after 8 years, and explain which model is more realistic.

[4]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	50 50 A1	1
(b)	125 N(2) (M1) 125 A1	2
(c)	t = 5.89, N = 500 inflexion point / maximum of N′(t) (M1) N = half the carrying capacity (A1) t = 5.89 A1 N = 500 A1	4
(d)	m = 0.459 50e^2m = 125 (M1) taking logs (M1) m = 0.459 A1	3
(e)	N = 742, P = 1963 N(8) = 742 A1 P(8) = 1963 A1 the exponential model grows without limit R1 the logistic model is more realistic as the lake has limited resources R1	4

Question 2Structured · 12 marks

A sweet company makes star-shaped and moon-shaped sweets. 75% are stars and the rest are moons. 10% of stars and 35% 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 80 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	11	20	16	22	9	2

Table 2.1

(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)	0.163; 0.462 0.75 × 0.1 + 0.25 × 0.35 (M1) 0.163 A1 0.07500.163 (M1) 0.462 A1	4
(b)	colours follow the claimed distribution the colours are distributed as the company claims A1	1
(c)	12.0, 20.0, 16.0, 16.0, 8.00, 8.00 percentage × 80 (M1) 12.0, 20.0, 16.0, 16.0, 8.00, 8.00 A1	2
(d)	5 6 − 1 = 5 A1	1
(e)	0.224 χ² = 6.96 (A1) 0.224 A1	2
(f)	do not reject H₀ 0.224 > 0.05 R1 do not reject H₀: the data are consistent with the claim A1	2

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 (50, 10, 6) and moves with constant velocity −150−30−15 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 −150−30. Find the new vertical component of velocity.

[4]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	r = 50106 + t−150−30−15 position + t × velocity (M1) r = 50106 + t−150−30−15 A1	2
(b)	t = 0.333; 1.00 km x = 0 ⇒ t = 0.333 A1 y = 0 at the same t R1 substituting into z (M1) 1.00 km A1	4
(c)	t = 0.133 h; 30.9 km 6 − 15t = 4 (M1) t = 0.133 A1 position and distance (M1) 30.9 km A1	4
(d)	−20.0 km h⁻¹ time to reach the airport horizontally (M1) 0.200 h (A1) −4 ÷ time (M1) −20.0 km h⁻¹ A1	4

Question 4Structured · 14 marks

An engineer records the time y seconds for a component to fail under a load of x tonnes.

x	2	4	6	8	10	12
y	6.11	16.3	29.3	43.1	61.5	76.6

Table 4.1

(a)

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

[2]

(b)

Find the sum of the squares of the residuals, SS_res, for this linear model.

[2]

(c)

The engineer believes a model of the form y = ax^b is more appropriate. Using a log-log transformation, find a and b.

[5]

(d)

Find SS_res for the model in (c), and hence state which model fits the data better.

[3]

(e)

Use the power model to estimate y when x = 7.

[2]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	y = 7.17x − 11.4 GDC (M1) y = 7.17x − 11.4 A1	2
(b)	29.8 Σ(y − ŷ)² (M1) 29.8 A1	2
(c)	a = 2.29, b = 1.42 ln y = ln a + b ln x (M1) regression of ln y on ln x (M1) ln y = 1.42ln x + 0.827 A1 b = 1.42 A1 a = 2.29 A1	5
(d)	3.94 predicted values from y = ax^b (M1) 3.94 A1 the power model has the smaller SS_res, so fits better R1	3
(e)	36.2 2.29 × 7^1.42 (M1) 36.2 A1	2

Question 5Structured · 14 marks

Each year 10% of the customers of company X move to company Y, and 15% of the customers of company Y move to company X. Initially X has 1500 customers and Y has 1800.

(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 0.150.1 0.85 columns sum to 1 (M1) 0.9 0.150.1 0.85 A1	2
(b)	λ = 1, λ = 0.75 det(T − λI) = 0 (M1) λ = 1, λ = 0.75 A1 eigenvector for 1: 0.150.1 (or a multiple) A1 eigenvector for 0.75: 1−1 A1	4
(c)	P, D P = 0.15 10.1 −1 A1 D = 1 00 0.75 A1	2
(d)	1980 − 480(0.75)ⁿ use of Tⁿ = PDⁿP⁻¹ M1 correct P⁻¹ A1 x_n = 1980 − 480(0.75)ⁿ A2 form A + B(λ)ⁿ A1	5
(e)	1980 1980 A1	1

Question 6Structured · 13 marks

The volume of drink in bottles has mean 250 ml and standard deviation 15 ml. The distribution of the volumes is not known.

(a)

A sample of 40 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 249 ml.

[2]

(c)

A different sample of 40 bottles has mean 247 ml. Find a 95% confidence interval for the population mean, assuming σ = 15.

[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, 5.63) central limit theorem, n large R1 mean 250 A1 variance 15²40 = 5.63 A1	3
(b)	0.337 normal CDF with standard error (M1) 0.337 A1	2
(c)	242.4 < μ < 251.6 z-interval (M1) 247 ± 1.96 × 15√40 (A1) 242.4 < μ < 251.6 A1	3
(d)	supported 250 is in the interval R1 the claim is supported A1	2
(e)	217 2 × 1.96 × 15√n ≤ 4 (M1) solving for n (M1) 217 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