IB · Mathematics: Applications & Interpretation SL & HL · N2S

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

2029 syllabus · SL · 75 marks · 1h 30m · calculator. Each generated paper follows the SL & HL Paper 2 · SL (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 · SL (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	12	17	24	25	30	32	30	40
y	26	33	67	35	78	87	80	100

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 26 °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 = 0.920 GDC (M1) r = 0.920 A1 strong positive linear correlation A1	3
(b)	y = 2.93x − 13.6 GDC (M1) y = 2.93x − 13.6 A1	2
(c)	62.5 2.93 × 26 − 13.6 (M1) 62.5 A1 reliable: interpolation and strong correlation R1	3
(d)	increase per degree about 2.93 more ice creams are sold for each 1 °C increase R1	1
(e)	0.970 ranking the data (M1) r_s = 0.970 A2 Spearman measures a monotonic (not necessarily linear) relationship R1	4

Question 2Structured · 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, 20, 5) and moves with constant velocity −250−100−20 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 −250−100. Find the new vertical component of velocity.

[4]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	r = 50205 + t−250−100−20 position + t × velocity (M1) r = 50205 + t−250−100−20 A1	2
(b)	t = 0.200; 1.00 km x = 0 ⇒ t = 0.200 A1 y = 0 at the same t R1 substituting into z (M1) 1.00 km A1	4
(c)	t = 0.0500 h; 40.6 km 5 − 20t = 4 (M1) t = 0.0500 A1 position and distance (M1) 40.6 km A1	4
(d)	−26.7 km h⁻¹ time to reach the airport horizontally (M1) 0.150 h (A1) −4 ÷ time (M1) −26.7 km h⁻¹ A1	4

Question 3Structured · 12 marks

A sweet company makes star-shaped and moon-shaped sweets. 75% are stars and the rest are moons. 10% 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 120 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	14	32	28	22	9	15

Table 3.1

(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)	0.175; 0.429 0.75 × 0.1 + 0.25 × 0.4 (M1) 0.175 A1 0.07500.175 (M1) 0.429 A1	4
(b)	colours follow the claimed distribution the colours are distributed as the company claims A1	1
(c)	18.0, 30.0, 24.0, 24.0, 12.0, 12.0 percentage × 120 (M1) 18.0, 30.0, 24.0, 24.0, 12.0, 12.0 A1	2
(d)	5 6 − 1 = 5 A1	1
(e)	0.645 χ² = 3.36 (A1) 0.645 A1	2
(f)	do not reject H₀ 0.645 > 0.05 R1 do not reject H₀: the data are consistent with the claim A1	2

Question 4Structured · 14 marks

A tent is a triangular prism. Its front is an isosceles triangle with base 2.4 m and sloping sides 2 m, and the tent is 3 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)	1.60 m Pythagoras (M1) 1.60 m A1	2
(b)	53.1° cos θ = 1.22 (M1) 53.1° A1	2
(c)	5.76 m³ area of triangle × length (M1) 5.76 m³ A1	2
(d)	15.8 m² two rectangles and two triangles (M1) correct expression (A1) 15.8 m² A1	3
(e)	3.61 m; 26.3° horizontal distance √(w/2)² + l² (M1) 3.23 (A1) 3.61 m A1 tan θ = height ÷ horizontal distance (M1) 26.3° A1	5

Question 5Structured · 13 marks

The volume of drink in bottles has mean 1000 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 999 ml.

[2]

(c)

A different sample of 40 bottles has mean 997 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 1000 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(1000, 5.63) central limit theorem, n large R1 mean 1000 A1 variance 15²40 = 5.63 A1	3
(b)	0.337 normal CDF with standard error (M1) 0.337 A1	2
(c)	992.4 < μ < 1002 z-interval (M1) 997 ± 1.96 × 15√40 (A1) 992.4 < μ < 1002 A1	3
(d)	supported 1000 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

Question 6Structured · 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	3.48	4.73	6.04	6.65	7.87	8.21

Table 6.1

(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 = 0.481x + 2.80 GDC (M1) y = 0.481x + 2.80 A1	2
(b)	0.403 Σ(y − ŷ)² (M1) 0.403 A1	2
(c)	a = 2.45, b = 0.492 ln y = ln a + b ln x (M1) regression of ln y on ln x (M1) ln y = 0.492ln x + 0.896 A1 b = 0.492 A1 a = 2.45 A1	5
(d)	0.139 predicted values from y = ax^b (M1) 0.139 A1 the power model has the smaller SS_res, so fits better R1	3
(e)	6.38 2.45 × 7^0.492 (M1) 6.38 A1	2

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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