A café records the outside temperature 𝑇 (°C) and the number of iced coffees sold 𝑁 on 7 mornings.

(a) GDC linear regression: r≈0.996 — strong positive. (b) a≈2.93,b≈−30.5, so N=2.93T−30.5. (c) N≈2.93(27)−30.5≈48.6⇒49 iced coffees. (d) T=45°C is outside the data range (18–34 °C) — no, extrapolation is unreliable.

A logistics company has three warehouses at coordinates 𝐴 ( 2 , 6 ) , 𝐵 ( 8 , 10 ) and 𝐶 ( 6 , 2 ) (units in km).

(a) AB=62+42=52≈7.21 km. (b) Midpoint of AB=(5,8); gradient of AB=10−68−2=23; perpendicular gradient =−32. Line: y−8=−32(x−5)⇒y=−1.5x+15.5. So m=−1.5,c=15.5. (c) Repeat for perpendicular bisector of BC: midpoint (7,6), gradient of BC=2−106−8=4, perpendicular gradient =−14; line y=−0.25x+7.75. Solve simultaneously with y=−1.5x+15.5: −1.5x+15.5=−0.25x+7.75⇒−1.25x=−7.75⇒x=6.2, and y=−1.5(6.2)+15.5=6.2. Verify: distance from (6.2,6.2) to each of A,B,C equals 17.68≈4.20 km. ✓

SL AI · Practice Paper 1 · Interactive

IB Maths Applications & Interpretation — Interactive Practice Paper 1

Paper 1 for SL AI is short-response with a GDC. Real-world data, financial calculations, geometry and statistics. Two modes: instant per-part feedback, or a full 90-minute timed exam.

40 marks · 5 questions · GDC required · 3 s.f. answers

Choose your mode

Duration:

Progress: 0 / 20 correct

Question 1

7 marks

Financial mathematics — compound interest

Priya invests €4000 in an account paying 3.2% interest per year, compounded monthly.

(a) Find the value of her investment after 5 years. Give your answer to the nearest cent (2 d.p.). [3]

Value

=

€

(b) How many complete years must Priya wait for her investment to first exceed €6000? [3]

years

=

(c) Comment on whether "3.2% per year, compounded monthly" is a better deal than "3.25% per year, compounded annually". Type a for monthly, b for annually. [1]

better =

(a)

Math input error

(GDC TVM:

N = 60, I = 3.2, P / Y = C / Y = 12

).
(b) Solve

4000 (1.00267)^{12 n} > 6000 \Rightarrow 12 n > \log_{1.00267} (1.5) \approx 152.2

, so

n > 12.68

; the first whole year is

13

.
(c) Effective annual rate of monthly plan

= (1.00267)^{12} - 1 \approx 3.247 %

, which is less than

3.25 %

. So (b) annually wins by a whisker — a great discrimination question because the nominal 3.2% "sounds better with more compounding" but the maths says otherwise.

Question 2

8 marks

Bivariate data — linear regression

A café records the outside temperature $T$ (°C) and the number of iced coffees sold $N$ on 7 mornings.

$T$ (°C)	18	21	24	26	29	32	34
$N$	22	31	35	44	52	63	71

(a) Find the Pearson product-moment correlation coefficient $r$ . Give your answer to 3 s.f. [2]

r =

(b) Write the regression line of $N$ on $T$ in the form $N = a T + b$ . Give $a$ and $b$ to 3 s.f. [3]

a =

b =

(c) Use your regression line to estimate $N$ when $T = 27$ °C. Round to the nearest whole number. [2]

N \approx

(d) Explain in one word whether extrapolating this model to $T = 45$ °C is reliable. Type yes or no. [1]

reliable?

(a) GDC linear regression:

r \approx 0.996

— strong positive.
(b)

a \approx 2.93, b \approx - 30.5

, so

N = 2.93 T - 30.5

.
(c)

N \approx 2.93 (27) - 30.5 \approx 48.6 \Rightarrow 49

iced coffees.
(d)

T = 45

°C is outside the data range (18–34 °C) — no, extrapolation is unreliable.

Question 3

8 marks

Geometry — 3-D right triangle & volume

A grain silo consists of a right cylinder of radius $r = 3$ m and height $h = 8$ m, topped by a hemisphere of the same radius.

(a) Find the total volume of the silo. Give your answer to 3 s.f. [3]

Volume

\approx

m³

(b) Find the total outer surface area of the silo, including the flat base but excluding the top of the cylinder (which is covered by the hemisphere). Give your answer to 3 s.f. [3]

Area

\approx

m²

(c) The silo is currently 30% full. How many m³ of grain does it contain? [2]

Grain

\approx

m³

(a)

V = π r^{2} h + \frac{2}{3} π r^{3} = π (9) (8) + \frac{2}{3} π (27) = 72 π + 18 π = 90 π \approx 283 m^{3}

.
(b) Curved cylinder

+

base

+

hemisphere

= 2 π r h + π r^{2} + 2 π r^{2} = 48 π + 9 π + 18 π = 75 π \approx 235 m^{2}

.
(c)

0.30 \times 283 \approx 84.8 m^{3}

Question 4

9 marks

Statistics — normal distribution

The mass $M$ (grams) of eggs from a farm is normally distributed with mean $62$ g and standard deviation $4.5$ g.

(a) Find $P (M > 66.5)$ . Give your answer to 3 s.f. [2]

P (M > 66.5) \approx

(b) An egg is classed "large" if $M > 68$ g. Out of a random batch of 500 eggs, find the expected number of large eggs. [3]

expected

\approx

(c) Find the mass $k$ (to 3 s.f.) such that only 5% of eggs weigh less than $k$ . [3]

k \approx

(d) Two eggs are chosen at random. Find the probability that both are large. Give your answer to 3 s.f. [1]

P(both)

\approx

(a)

z = \frac{66.5 - 62}{4.5} = 1

;

P (Z > 1) \approx 0.159

(GDC: normalcdf).
(b)

P (M > 68) = P (Z > 1.333) \approx 0.0912

;

500 \times 0.0912 \approx 45.7

eggs.
(c) InvNorm(0.05, 62, 4.5)

\approx 54.6

g.
(d)

{0.0912}^{2} \approx 0.00838

Question 5

8 marks

Voronoi diagram & toolkit

A logistics company has three warehouses at coordinates $A (2, 6)$ , $B (8, 10)$ and $C (6, 2)$ (units in km).

(a) Find the distance $A B$ . Give your answer to 3 s.f. [2]

A B \approx

(b) Find the equation of the perpendicular bisector of $A B$ , giving your answer in the form $y = m x + c$ with $m$ and $c$ to 3 s.f. [3]

m =

c =

(c) A new warehouse $D$ is to be built at the point equidistant from all three existing warehouses (the "toxic-waste" / Voronoi vertex). Find its coordinates. Give each to 3 s.f. [3]

x =

y =

(a)

A B = \sqrt{6^{2} + 4^{2}} = \sqrt{52} \approx 7.21

km.
(b) Midpoint of

A B = (5, 8)

; gradient of

A B = \frac{10 - 6}{8 - 2} = \frac{2}{3}

; perpendicular gradient

= - \frac{3}{2}

. Line:

y - 8 = - \frac{3}{2} (x - 5) \Rightarrow y = - 1.5 x + 15.5

. So

m = - 1.5, c = 15.5

.
(c) Repeat for perpendicular bisector of

B C

: midpoint

(7, 6)

, gradient of

B C = \frac{2 - 10}{6 - 8} = 4

, perpendicular gradient

= - \frac{1}{4}

; line

y = - 0.25 x + 7.75

. Solve simultaneously with

y = - 1.5 x + 15.5

- 1.5 x + 15.5 = - 0.25 x + 7.75 \Rightarrow - 1.25 x = - 7.75 \Rightarrow x = 6.2

, and

y = - 1.5 (6.2) + 15.5 = 6.2

. Verify: distance from

(6.2, 6.2)

to each of

A, B, C

equals

\sqrt{17.68} \approx 4.20

km. ✓

📊 Paper submitted — here's your score

0 / 20 parts correct 0%

← Back to all papers · HL AI Paper 3 →