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Worked solutions

IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2 TZ3

Session: May 2026
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 5

Q1 Q2 Q3 Q4 Q5

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–5 · 80 marks

Question 1

14 marks

Square-based pyramid: slant height, area, volume, medals

Geometry and trigonometry

(a)

4 marks

(i) $P$ is the midpoint of a $4.4$ cm edge, so $A P$ is half of it.

(ii) Triangle $N A P$ is right-angled at $P$ , with $A P$ adjacent to the $6 6^{\circ}$ angle and $N P$ opposite it, so use tangent.

(iii) $N P$ is the height of triangle $N D A$ and $A D$ is its base.

A P = \frac{4.4}{2} = 2.2 cm N P = 2.2 tan 6 6^{\circ} = 4.94 cm (4.94128 \dots) Area_{N D A} = \frac{1}{2} (4.4) (4.94128 \dots) = 10.9 cm^{2}

Mentor tip

Check your calculator is in degrees before you press tan. A tiny answer like $0.058$ means it is in radians.

(b)

2 marks

The surface is four identical triangular faces plus the square base. Add them up, using the unrounded area from (a).

S A = 4 (10.8708 \dots) + 4. 4^{2} S A = 62.8 cm^{2}

Mentor tip

The base is easy to forget. A solid pyramid has five faces, not four.

(c)

4 marks

(i) Triangle $N M P$ is right-angled at $M$ , with $M P = 2.2$ (half the base width) and hypotenuse $N P$ . Use Pythagoras.

(ii) Put the base area and the height $M N$ into $V = \frac{1}{3} \times base area \times height$ .

M N = 4.94128 \dots^{2} - 2. 2^{2} = 4.42 cm (4.42450 \dots) V = \frac{1}{3} (4.4) (4.4) (4.42450 \dots) = 28.6 cm^{3}

Mentor tip

The slant height $N P$ is not the vertical height. Use $M N$ in the volume formula.

(d)

4 marks

Each medal is a cylinder with $r = 0.9$ and height $0.2$ , so find its volume.
Divide the pyramid's volume by one medal's volume.
Round down: a part-medal cannot be made.

V_{medal} = π (0.9)^{2} (0.2) = 0.508938 \dots cm^{3} \frac{28.5528 \dots}{0.508938 \dots} = 56.10 \dots 56 medals

Mentor tip

“Maximum number” means round down, even if the decimal is above one half. Keep full calculator values until the end.

Question 2

15 marks

Parabolic gate: trapezoidal rule versus integration

Functions
Calculus
Number and algebra

(a)

2 marks

Graph $h (x)$ on the GDC for $0 < x < 6$ and find the minimum, or use $x = - \frac{b}{2 a}$ .
The height is the $h$ -value at the vertex.

x = - \frac{- 2/3}{2 ( 1/9 )} = 3 h (3) = 1 - 2 + 2.5 = 1.5 m

Mentor tip

The question wants a height, so give $1.5$ m, not the point $(3, 1.5)$ .

(b)

1 mark

The parabola opens upwards, so it rises from the vertex to the right edge of the gate.

a = 3, b = 6

(c)

5 marks

(i) $p$ is the height at the vertex, and by symmetry about $x = 3$ the height at $x = 5$ equals the height at $x = 1$ .

(ii) Use the trapezoidal rule with strip width $1$ : the two end heights once, all the inner heights twice.

p = 1.5, q = \frac{35}{18} = 1.94 (1.94444 \dots) A \approx \frac{1}{2} (1) [2.5 + 2.5 + 2 (\frac{35}{18} + \frac{29}{18} + 1.5 + \frac{29}{18} + \frac{35}{18})] A \approx 11.1 m^{2} (\frac{100}{9})

Mentor tip

Count the heights: seven values for six strips. A common slip is doubling the end values too.

(d)

1 mark

The curve is concave up (a U-shaped parabola), so the straight top of each trapezium lies above the curve and every strip adds a little extra area.

concave up ⟹ chords above the curve ⟹ overestimate

Mentor tip

Do not use the exact area from (e) here. The reason must come from the shape of the curve.

(e)

4 marks

(i) The area under the top edge is the integral of $h (x)$ from one side of the gate to the other.

(ii) Evaluate it on the GDC, or integrate by hand.

A = \int_{0}^{6} (\frac{1}{9} x^{2} - \frac{2}{3} x + 2.5) d x = [\frac{x ^{3}}{27} - \frac{x ^{2}}{3} + 2.5 x]_{0}^{6} = 8 - 12 + 15 A = 11 m^{2}

Mentor tip

Write the limits and the $d x$ in full. An integral without them is incomplete.

(f)

2 marks

Percentage error is the size of the difference divided by the exact value, times $100$ .

ε = \frac{\frac{100}{9} - 11}{11} \times 100% ε = 1.01%

Mentor tip

Divide by the exact value $A$ , not the estimate. Using $11.1$ instead of $\frac{100}{9}$ gives $0.909%$ , so keep the full value.

Question 3

17 marks

Archery: tree diagram, binomial and normal models

Statistics and probability

(a)

3 marks

First stage: hit $0.34$ , miss $0.66$ .
After a hit: hit $0.72$ , miss $0.28$ . After a miss: hit $0.68$ , miss $0.32$ .
Label every branch with its outcome as well as its probability.

H_{1} : 0.34 \to H_{2} : 0.72, M_{2} : 0.28 M_{1} : 0.66 \to H_{2} : 0.68, M_{2} : 0.32

Mentor tip

Each pair of branches from one point must add to $1$ . That is a quick check on your complements.

(b)

3 marks

At most once means zero or one hit, which is everything except two hits.
Subtract the probability of hitting with both arrows from $1$ .

P (at most one) = 1 - (0.34) (0.72) = 1 - 0.2448 = 0.755 (0.7552)

Another way. Add the three paths directly: $0.34 (0.28) + 0.66 (0.68) + 0.66 (0.32)$ .

Mentor tip

At most once includes zero hits. Finding only exactly one hit gives $0.544$ .

(c)

2 marks

Ten independent shots with a fixed success probability: use $X \sim B (10, 0.94)$ and the binomial pdf on the GDC.

X \sim B (10, 0.94) P (X = 9) = 0.344 (0.343796 \dots) (GDC)

(d)

2 marks

At least $7$ is the complement of $6$ or fewer. Use the binomial cdf.

P (X \geq 7) = 1 - P (X \leq 6) = 0.998 (0.997970 \dots) (GDC)

Mentor tip

The classic slip is $1 - P (X \leq 7)$ , which drops the case of exactly $7$ .

(e)

2 marks

Let $D \sim N (26, 7^{2})$ be the distance from the centre. Hitting the inner circle means $D$ lies between $0$ and $10$ .

P (0 < D < 10) = 0.0110 (0.0110335 \dots) (GDC)

Mentor tip

The GDC asks for the standard deviation, $7$ , not the variance, $49$ .

(f)

3 marks

Missing the target means the arrow lands more than $50$ cm from the centre.
Use the normal cdf from $50$ to a very large number, then write the result in standard form.

P (D > 50) = 0.000303 \dots (GDC) = 3.03 \times 1 0^{- 4}

Another way. Treating distances as starting at $0$ , $1 - P (0 < D < 50) = 4.05 \times 1 0^{- 4}$ is also a fair reading of the model.

Mentor tip

Standard form needs $1 \leq a < 10$ , so $0.303 \times 1 0^{- 3}$ is not finished.

(g)

2 marks

Use inverse normal with left-tail area $0.475$ , since distance $p$ from the centre is a lower tail.

P (D < p) = 0.475 p = 25.6 cm (25.5611 \dots) (GDC)

Mentor tip

Sense check: $0.475$ is just under a half, so $p$ should be just under the mean of $26$ .

Question 4

14 marks

Clock hands: cosine rule and a trigonometric model

Geometry and trigonometry
Functions

(a)

4 marks

Each hour mark is $36 0^{\circ} \div 12 = 3 0^{\circ}$ apart. At 8:00 the hands are four marks apart, so the angle between them is $12 0^{\circ}$ .
The hands and the line joining their tips form a triangle with sides $9$ and $15$ around that angle. Use the cosine rule.

d^{2} = 9^{2} + 1 5^{2} - 2 (9) (15) cos 12 0^{\circ} d^{2} = 306 + 135 = 441 d = 21 cm

Mentor tip

Measure the angle the short way round: from 8 to 12 is four gaps, not eight.

(b)

4 marks

(i) The tip swings $15$ cm above and below the centre, so the amplitude is $15$ .

(ii) One full turn takes $60$ minutes, so $b = 360 \div 60$ .

(iii) The centre of the clock is $15$ cm above the bottom, which is the principal axis.

a = 15 b = \frac{360}{60} = 6 d = 15

Mentor tip

Check with $t = 0$ : the minute hand points to 12, and $15 cos 0^{\circ} + 15 = 30$ is the top of the clock.

(c)

2 marks

Substitute $t = 20$ into the model, with the calculator in degrees.

h (20) = 15 cos (12 0^{\circ}) + 15 h (20) = 7.5 cm

Another way. At 8:20 the minute hand points at 4, which is $3 0^{\circ}$ below the horizontal, so the height is $15 - 15 sin 3 0^{\circ}$ .

(d)

1 mark

The hour hand does not stay on the 8: in 20 minutes it moves on a little, while the minute hand swings round to 4. The angle between the hands is now larger than $12 0^{\circ}$ , and with the same two hand lengths a larger angle gives a longer third side.

angle between hands > 12 0^{\circ} ⟹ distance > 21 cm

Mentor tip

Mention the hour hand moving. Saying “the angle is bigger” without a reason is thin.

(e)

3 marks

The hour hand turns $0. 5^{\circ}$ per minute, so after 20 minutes it is $1 0^{\circ}$ past the 8, which puts it $2 0^{\circ}$ below the horizontal.
Its tip is $9 sin 2 0^{\circ}$ below the centre, so its height is $15 - 9 sin 2 0^{\circ}$ .
Subtract the minute hand height from (c).

h_{hour} = 15 - 9 sin 2 0^{\circ} = 11.9218 \dots difference = 11.9218 \dots - 7.5 = 4.42 cm

Mentor tip

Draw a quick sketch of the clock at 8:20 before working out angles. It is easy to measure from the wrong line.

Question 5

20 marks

Trays: quadratic area and minimising a perimeter

Number and algebra
Functions
Calculus

(a)

4 marks

The two perimeters use the whole strip: $4 x + 2 (y + 3 y) = 200$ , which simplifies to $y = 25 - 0.5 x$ .
The combined area is $x^{2} + 3 y^{2}$ .
Substitute for $y$ , expand and collect terms.

4 x + 8 y = 200 ⟹ y = 25 - 0.5 x A = x^{2} + 3 (25 - 0.5 x)^{2} A = x^{2} + 3 (625 - 25 x + 0.25 x^{2}) A = x^{2} + 1875 - 75 x + 0.75 x^{2} = 1.75 x^{2} - 75 x + 1875

Mentor tip

On a show-that, write every line of the expansion. Jumping straight to the given result proves nothing.

(b)

4 marks

Set the area equal to $1075$ and solve the quadratic (GDC or formula).
Keep the root that makes both $x$ and $y$ whole numbers.
Find $y$ and state both trays' dimensions.

1.75 x^{2} - 75 x + 800 = 0 x = 20 or x = 22.857 \dots x = 20 ⟹ y = 25 - 10 = 15 20 \times 20 cm and 15 \times 45 cm

Mentor tip

The other root is not a whole number, so reject it. The lengths are integers.

(c)

1 mark

The ceramic rectangle has area $l \times w$ .

l w = 150

(d)

4 marks

The $2$ cm borders sit on both sides of the $l$ edges, adding $4$ to the $w$ direction. The $3$ cm borders add $6$ to the $l$ direction.
So the whole tray is $(w + 4)$ by $(l + 6)$ .
Replace $l$ with $\frac{150}{w}$ from (c) and simplify.

P = 2 (w + 4) + 2 (l + 6) P = 2 (w + 4) + 2 (\frac{150}{w} + 6) P = 2 w + \frac{300}{w} + 20

Mentor tip

Read the diagram carefully: the border along the $l$ sides makes the tray wider, not longer.

(e)

5 marks

(i) Write $\frac{300}{w}$ as $300 w^{- 1}$ and differentiate term by term. The constant disappears.

(ii) Set the derivative equal to zero and solve for $w > 0$ .

\frac{d P}{d w} = 2 - \frac{300}{w ^{2}} 2 - \frac{300}{w ^{2}} = 0 ⟹ w^{2} = 150 w = 150 = 12.2 cm (12.2474 \dots)

Another way. Graph $P$ on the GDC for $w > 0$ and read off the minimum point.

Mentor tip

Differentiating $300 w^{- 1}$ gives $- 300 w^{- 2}$ . Losing that minus sign is the usual error.

(f)

1 mark

Use $l w = 150$ with $w = 150$ .

l = \frac{150}{150} = 150 = 12.2 cm

Mentor tip

The ceramic part turns out to be a square. That is typical when you minimise a perimeter for a fixed area.

(g)

1 mark

Add the borders: $w + 4$ for one side and $l + 6$ for the other.

w + 4 = 16.2 cm (16.2474 \dots) l + 6 = 18.2 cm (18.2474 \dots)

Mentor tip

The question asks for the whole tray, so add the wood, not just the ceramic.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Geometry and trigonometry carries the most marks on this paper.

Geometry and trigonometry22 marks · 28%
Q1, Q4
Statistics and probability17 marks · 21%
Q3
Calculus15 marks · 19%
Q2, Q5
Number and algebra13 marks · 16%
Q2, Q5
Functions13 marks · 16%
Q2, Q4, Q5

Mentor insights

Traps to watch for

01
Slant height used as vertical height · Q1
$N P$ is along the face, not straight up. Use Pythagoras to get $M N$ before finding the volume.
02
Percentage error over the wrong value · Q2
Divide by the exact area $11$ , and keep $\frac{100}{9}$ unrounded, or the answer drifts.
03
At most once is not exactly once · Q3
Zero hits counts too. The quickest route is one minus the chance of two hits.
04
Forgetting the hour hand moves · Q4
By 8:20 the hour hand has turned $1 0^{\circ}$ past the 8, which changes both the angle and its height.
05
Border widths on the wrong sides · Q5
The $2$ cm strips widen the $w$ direction and the $3$ cm strips lengthen the $l$ direction: $(w + 4)$ by $(l + 6)$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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Worked solutions

IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2 TZ3

Session: May 2026
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 5

Q1 Q2 Q3 Q4 Q5

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–5 · 80 marks

Question 1

14 marks

Square-based pyramid: slant height, area, volume, medals

Geometry and trigonometry

(a)

4 marks

(i) $P$ is the midpoint of a $4.4$ cm edge, so $A P$ is half of it.

(ii) Triangle $N A P$ is right-angled at $P$ , with $A P$ adjacent to the $6 6^{\circ}$ angle and $N P$ opposite it, so use tangent.

(iii) $N P$ is the height of triangle $N D A$ and $A D$ is its base.

A P = \frac{4.4}{2} = 2.2 cm N P = 2.2 tan 6 6^{\circ} = 4.94 cm (4.94128 \dots) Area_{N D A} = \frac{1}{2} (4.4) (4.94128 \dots) = 10.9 cm^{2}

Mentor tip

Check your calculator is in degrees before you press tan. A tiny answer like $0.058$ means it is in radians.

(b)

2 marks

The surface is four identical triangular faces plus the square base. Add them up, using the unrounded area from (a).

S A = 4 (10.8708 \dots) + 4. 4^{2} S A = 62.8 cm^{2}

Mentor tip

The base is easy to forget. A solid pyramid has five faces, not four.

(c)

4 marks

(i) Triangle $N M P$ is right-angled at $M$ , with $M P = 2.2$ (half the base width) and hypotenuse $N P$ . Use Pythagoras.

(ii) Put the base area and the height $M N$ into $V = \frac{1}{3} \times base area \times height$ .

M N = 4.94128 \dots^{2} - 2. 2^{2} = 4.42 cm (4.42450 \dots) V = \frac{1}{3} (4.4) (4.4) (4.42450 \dots) = 28.6 cm^{3}

Mentor tip

The slant height $N P$ is not the vertical height. Use $M N$ in the volume formula.

(d)

4 marks

Each medal is a cylinder with $r = 0.9$ and height $0.2$ , so find its volume.
Divide the pyramid's volume by one medal's volume.
Round down: a part-medal cannot be made.

V_{medal} = π (0.9)^{2} (0.2) = 0.508938 \dots cm^{3} \frac{28.5528 \dots}{0.508938 \dots} = 56.10 \dots 56 medals

Mentor tip

“Maximum number” means round down, even if the decimal is above one half. Keep full calculator values until the end.

Question 2

15 marks

Parabolic gate: trapezoidal rule versus integration

Functions
Calculus
Number and algebra

(a)

2 marks

Graph $h (x)$ on the GDC for $0 < x < 6$ and find the minimum, or use $x = - \frac{b}{2 a}$ .
The height is the $h$ -value at the vertex.

x = - \frac{- 2/3}{2 ( 1/9 )} = 3 h (3) = 1 - 2 + 2.5 = 1.5 m

Mentor tip

The question wants a height, so give $1.5$ m, not the point $(3, 1.5)$ .

(b)

1 mark

The parabola opens upwards, so it rises from the vertex to the right edge of the gate.

a = 3, b = 6

(c)

5 marks

(i) $p$ is the height at the vertex, and by symmetry about $x = 3$ the height at $x = 5$ equals the height at $x = 1$ .

(ii) Use the trapezoidal rule with strip width $1$ : the two end heights once, all the inner heights twice.

p = 1.5, q = \frac{35}{18} = 1.94 (1.94444 \dots) A \approx \frac{1}{2} (1) [2.5 + 2.5 + 2 (\frac{35}{18} + \frac{29}{18} + 1.5 + \frac{29}{18} + \frac{35}{18})] A \approx 11.1 m^{2} (\frac{100}{9})

Mentor tip

Count the heights: seven values for six strips. A common slip is doubling the end values too.

(d)

1 mark

The curve is concave up (a U-shaped parabola), so the straight top of each trapezium lies above the curve and every strip adds a little extra area.

concave up ⟹ chords above the curve ⟹ overestimate

Mentor tip

Do not use the exact area from (e) here. The reason must come from the shape of the curve.

(e)

4 marks

(i) The area under the top edge is the integral of $h (x)$ from one side of the gate to the other.

(ii) Evaluate it on the GDC, or integrate by hand.

A = \int_{0}^{6} (\frac{1}{9} x^{2} - \frac{2}{3} x + 2.5) d x = [\frac{x ^{3}}{27} - \frac{x ^{2}}{3} + 2.5 x]_{0}^{6} = 8 - 12 + 15 A = 11 m^{2}

Mentor tip

Write the limits and the $d x$ in full. An integral without them is incomplete.

(f)

2 marks

Percentage error is the size of the difference divided by the exact value, times $100$ .

ε = \frac{\frac{100}{9} - 11}{11} \times 100% ε = 1.01%

Mentor tip

Divide by the exact value $A$ , not the estimate. Using $11.1$ instead of $\frac{100}{9}$ gives $0.909%$ , so keep the full value.

Question 3

17 marks

Archery: tree diagram, binomial and normal models

Statistics and probability

(a)

3 marks

First stage: hit $0.34$ , miss $0.66$ .
After a hit: hit $0.72$ , miss $0.28$ . After a miss: hit $0.68$ , miss $0.32$ .
Label every branch with its outcome as well as its probability.

H_{1} : 0.34 \to H_{2} : 0.72, M_{2} : 0.28 M_{1} : 0.66 \to H_{2} : 0.68, M_{2} : 0.32

Mentor tip

Each pair of branches from one point must add to $1$ . That is a quick check on your complements.

(b)

3 marks

At most once means zero or one hit, which is everything except two hits.
Subtract the probability of hitting with both arrows from $1$ .

P (at most one) = 1 - (0.34) (0.72) = 1 - 0.2448 = 0.755 (0.7552)

Another way. Add the three paths directly: $0.34 (0.28) + 0.66 (0.68) + 0.66 (0.32)$ .

Mentor tip

At most once includes zero hits. Finding only exactly one hit gives $0.544$ .

(c)

2 marks

Ten independent shots with a fixed success probability: use $X \sim B (10, 0.94)$ and the binomial pdf on the GDC.

X \sim B (10, 0.94) P (X = 9) = 0.344 (0.343796 \dots) (GDC)

(d)

2 marks

At least $7$ is the complement of $6$ or fewer. Use the binomial cdf.

P (X \geq 7) = 1 - P (X \leq 6) = 0.998 (0.997970 \dots) (GDC)

Mentor tip

The classic slip is $1 - P (X \leq 7)$ , which drops the case of exactly $7$ .

(e)

2 marks

Let $D \sim N (26, 7^{2})$ be the distance from the centre. Hitting the inner circle means $D$ lies between $0$ and $10$ .

P (0 < D < 10) = 0.0110 (0.0110335 \dots) (GDC)

Mentor tip

The GDC asks for the standard deviation, $7$ , not the variance, $49$ .

(f)

3 marks

Missing the target means the arrow lands more than $50$ cm from the centre.
Use the normal cdf from $50$ to a very large number, then write the result in standard form.

P (D > 50) = 0.000303 \dots (GDC) = 3.03 \times 1 0^{- 4}

Another way. Treating distances as starting at $0$ , $1 - P (0 < D < 50) = 4.05 \times 1 0^{- 4}$ is also a fair reading of the model.

Mentor tip

Standard form needs $1 \leq a < 10$ , so $0.303 \times 1 0^{- 3}$ is not finished.

(g)

2 marks

Use inverse normal with left-tail area $0.475$ , since distance $p$ from the centre is a lower tail.

P (D < p) = 0.475 p = 25.6 cm (25.5611 \dots) (GDC)

Mentor tip

Sense check: $0.475$ is just under a half, so $p$ should be just under the mean of $26$ .

Question 4

14 marks

Clock hands: cosine rule and a trigonometric model

Geometry and trigonometry
Functions

(a)

4 marks

Each hour mark is $36 0^{\circ} \div 12 = 3 0^{\circ}$ apart. At 8:00 the hands are four marks apart, so the angle between them is $12 0^{\circ}$ .
The hands and the line joining their tips form a triangle with sides $9$ and $15$ around that angle. Use the cosine rule.

d^{2} = 9^{2} + 1 5^{2} - 2 (9) (15) cos 12 0^{\circ} d^{2} = 306 + 135 = 441 d = 21 cm

Mentor tip

Measure the angle the short way round: from 8 to 12 is four gaps, not eight.

(b)

4 marks

(i) The tip swings $15$ cm above and below the centre, so the amplitude is $15$ .

(ii) One full turn takes $60$ minutes, so $b = 360 \div 60$ .

(iii) The centre of the clock is $15$ cm above the bottom, which is the principal axis.

a = 15 b = \frac{360}{60} = 6 d = 15

Mentor tip

Check with $t = 0$ : the minute hand points to 12, and $15 cos 0^{\circ} + 15 = 30$ is the top of the clock.

(c)

2 marks

Substitute $t = 20$ into the model, with the calculator in degrees.

h (20) = 15 cos (12 0^{\circ}) + 15 h (20) = 7.5 cm

Another way. At 8:20 the minute hand points at 4, which is $3 0^{\circ}$ below the horizontal, so the height is $15 - 15 sin 3 0^{\circ}$ .

(d)

1 mark

angle between hands > 12 0^{\circ} ⟹ distance > 21 cm

Mentor tip

Mention the hour hand moving. Saying “the angle is bigger” without a reason is thin.

(e)

3 marks

The hour hand turns $0. 5^{\circ}$ per minute, so after 20 minutes it is $1 0^{\circ}$ past the 8, which puts it $2 0^{\circ}$ below the horizontal.
Its tip is $9 sin 2 0^{\circ}$ below the centre, so its height is $15 - 9 sin 2 0^{\circ}$ .
Subtract the minute hand height from (c).

h_{hour} = 15 - 9 sin 2 0^{\circ} = 11.9218 \dots difference = 11.9218 \dots - 7.5 = 4.42 cm

Mentor tip

Draw a quick sketch of the clock at 8:20 before working out angles. It is easy to measure from the wrong line.

Question 5

20 marks

Trays: quadratic area and minimising a perimeter

Number and algebra
Functions
Calculus

(a)

4 marks

The two perimeters use the whole strip: $4 x + 2 (y + 3 y) = 200$ , which simplifies to $y = 25 - 0.5 x$ .
The combined area is $x^{2} + 3 y^{2}$ .
Substitute for $y$ , expand and collect terms.

4 x + 8 y = 200 ⟹ y = 25 - 0.5 x A = x^{2} + 3 (25 - 0.5 x)^{2} A = x^{2} + 3 (625 - 25 x + 0.25 x^{2}) A = x^{2} + 1875 - 75 x + 0.75 x^{2} = 1.75 x^{2} - 75 x + 1875

Mentor tip

On a show-that, write every line of the expansion. Jumping straight to the given result proves nothing.

(b)

4 marks

Set the area equal to $1075$ and solve the quadratic (GDC or formula).
Keep the root that makes both $x$ and $y$ whole numbers.
Find $y$ and state both trays' dimensions.

1.75 x^{2} - 75 x + 800 = 0 x = 20 or x = 22.857 \dots x = 20 ⟹ y = 25 - 10 = 15 20 \times 20 cm and 15 \times 45 cm

Mentor tip

The other root is not a whole number, so reject it. The lengths are integers.

(c)

1 mark

The ceramic rectangle has area $l \times w$ .

l w = 150

(d)

4 marks

The $2$ cm borders sit on both sides of the $l$ edges, adding $4$ to the $w$ direction. The $3$ cm borders add $6$ to the $l$ direction.
So the whole tray is $(w + 4)$ by $(l + 6)$ .
Replace $l$ with $\frac{150}{w}$ from (c) and simplify.

P = 2 (w + 4) + 2 (l + 6) P = 2 (w + 4) + 2 (\frac{150}{w} + 6) P = 2 w + \frac{300}{w} + 20

Mentor tip

Read the diagram carefully: the border along the $l$ sides makes the tray wider, not longer.

(e)

5 marks

(i) Write $\frac{300}{w}$ as $300 w^{- 1}$ and differentiate term by term. The constant disappears.

(ii) Set the derivative equal to zero and solve for $w > 0$ .

\frac{d P}{d w} = 2 - \frac{300}{w ^{2}} 2 - \frac{300}{w ^{2}} = 0 ⟹ w^{2} = 150 w = 150 = 12.2 cm (12.2474 \dots)

Another way. Graph $P$ on the GDC for $w > 0$ and read off the minimum point.

Mentor tip

Differentiating $300 w^{- 1}$ gives $- 300 w^{- 2}$ . Losing that minus sign is the usual error.

(f)

1 mark

Use $l w = 150$ with $w = 150$ .

l = \frac{150}{150} = 150 = 12.2 cm

Mentor tip

The ceramic part turns out to be a square. That is typical when you minimise a perimeter for a fixed area.

(g)

1 mark

Add the borders: $w + 4$ for one side and $l + 6$ for the other.

w + 4 = 16.2 cm (16.2474 \dots) l + 6 = 18.2 cm (18.2474 \dots)

Mentor tip

The question asks for the whole tray, so add the wood, not just the ceramic.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Geometry and trigonometry carries the most marks on this paper.

Geometry and trigonometry22 marks · 28%
Q1, Q4
Statistics and probability17 marks · 21%
Q3
Calculus15 marks · 19%
Q2, Q5
Number and algebra13 marks · 16%
Q2, Q5
Functions13 marks · 16%
Q2, Q4, Q5

Mentor insights

Traps to watch for

01
Slant height used as vertical height · Q1
$N P$ is along the face, not straight up. Use Pythagoras to get $M N$ before finding the volume.
02
Percentage error over the wrong value · Q2
Divide by the exact area $11$ , and keep $\frac{100}{9}$ unrounded, or the answer drifts.
03
At most once is not exactly once · Q3
Zero hits counts too. The quickest route is one minus the chance of two hits.
04
Forgetting the hour hand moves · Q4
By 8:20 the hour hand has turned $1 0^{\circ}$ past the 8, which changes both the angle and its height.
05
Border widths on the wrong sides · Q5
The $2$ cm strips widen the $w$ direction and the $3$ cm strips lengthen the $l$ direction: $(w + 4)$ by $(l + 6)$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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