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IB Maths AI SL · Paper 2

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2TZ3

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 TZ2 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

Tiling a rectangle with turf, costs and a perimeter

Number and algebra

(a)

4 marks

(i) Count how many whole 1.4 m lengths fit along the 8 m side, rounding down.

Count how many 0.6 m rows fit up the 4.8 m side, then multiply.

(ii) Five rolls take up 7 m of the 8 m, so a strip 1 m wide is left running the full 4.8 m height.

\frac{8}{1.4} = 5.71 \dots \Rightarrow 5 rolls per row \frac{4.8}{0.6} = 8 rows 5 \times 8 = 40 rolls uncovered area = (8 - 7) \times 4.8 = 4.8 m^{2}

Mentor tip

Rolls cannot be cut, so round down. Dividing the garden's area by a roll's area gives 45, which ignores the shapes and is wrong here.

(b)

2 marks

The leftover strip is 1 m by 4.8 m. Split the 1 m width into $0.6 + 0.4$ and fill each column with flats turned to fit.
Check with areas: each flat covers $0.24 m^{2}$ .

0.6 m column: \frac{4.8}{0.4} = 12 flats 0.4 m column: \frac{4.8}{0.6} = 8 flats 12 + 8 = 20 = \frac{4.8}{0.24}

Mentor tip

The area check only works because the flats tile the strip exactly. Always make sure a real arrangement exists.

(c)

2 marks

Multiply each count by its price and add.

40 (5) + 20 (2) = 200 + 40 $240

(d)

3 marks

Work in the same 1 m by 4.8 m strip. Stand two rolls on end in the 0.6 m column: they fill $2 \times 1.4 = 2.8$ m of its height.
The remaining 2 m of that column takes flats 0.4 m tall, and the 0.4 m column still takes 8 flats.
Cost this layout and compare with the phase 2 cost from part (c).

0.6 m column: 2 (1.4) + 5 (0.4) = 4.8 0.4 m column: 8 (0.6) = 4.8 2 rolls and 13 flats 2 (5) + 13 (2) = $36 < 20 (2) = $40 total 200 + 36 = $236 < $240, so Ayden is right

Mentor tip

Finish with a clear comparison and a sentence saying the claim holds. A labelled sketch of the strip is a quick way to show the layout works.

(e)

3 marks

Find the perimeter in metres.
Convert the spacing to 0.4 m and divide. Around a closed loop the number of gaps equals the number of lights, so there is no extra one to add.

P = 2 (8) + 2 (4.8) = 25.6 m \frac{25.6}{0.4} = 64 lights

Mentor tip

Mixing centimetres and metres is the classic slip. Convert first, then divide.

Question 2

17 marks

Expected value, fairness and a chi-squared fit test

Statistics and probability

(a)

4 marks

(i) Multiply the probability of the 100-dollar prize by 6.

(ii) All the probabilities add to 1, so subtract the known ones from 1.

a = 6 (0.002) = 0.012 b = 1 - (0.085 + 0.012 + 0.004 + 0.002 + 0.001) b = 0.896

(b)

4 marks

Find the expected prize by multiplying each prize by its probability and adding.
A card costs 1 dollar, so the game is fair only if the expected gain is 0 (expected prize equal to the cost).

E (prize) = 0 (0.896) + 2 (0.085) + 5 (0.012) + 10 (0.004) + 100 (0.002) + 500 (0.001) E (prize) = 0.97 E (gain) = 0.97 - 1 = - 0.03 \neq = 0 not a fair game

Mentor tip

Say what you are comparing: fairness means expected gain equals zero once the 1-dollar cost is taken off.

(c)

1 mark

Divide the number of 2-dollar wins by the number of cards.

\frac{48}{750} = 0.064

(d)

2 marks

(i) For: her experimental 0.064 is noticeably lower than the stated 0.085.

(ii) Against: relative frequencies from a sample naturally vary around the true probability, and 750 cards may not be enough (or may not be representative) to judge a claim about all cards.

0.064 \neq = 0.085

Mentor tip

For the counter-argument, explain why the sample might mislead. Just writing "small sample" on its own is thin.

(e)

6 marks

(i) $H_{0}$ : the prize outcomes follow the distribution stated by the lottery company. $H_{1}$ : they do not follow that distribution.

(ii) Expected frequencies are $750 \times$ each stated probability, with the top three prizes combined ( $0.007$ ). Four categories give $4 - 1 = 3$ degrees of freedom.

Run the goodness of fit test on the GDC.

(iii) Compare the $p$ -value with 0.10 and state the conclusion about the company's probabilities.

E = 672, 63.75, 9, 5.25 ν = 3 p = 0.117 (GDC) 0.117 > 0.10 \Rightarrow do not reject H_{0} not enough evidence that the prizes differ from the company’s stated distribution

Mentor tip

Write hypotheses about the whole distribution, not just the 2-dollar prize, and phrase the conclusion in context rather than only "do not reject $H_{0}$ ".

Question 3

18 marks

Area under a parabola, volume, mass and arc length

Calculus
Geometry and trigonometry

(a)

3 marks

(i) Differentiate $f (x) = - \frac{1}{4} x^{2} + x + 2$ term by term.

(ii) Substitute $x = 0$ .

(iii) Take the inverse tangent of that gradient.

f^{'} (x) = - \frac{1}{2} x + 1 f^{'} (0) = 1 θ = tan^{- 1} (1) = 4 5^{\circ} (\frac{π}{4})

(b)

4 marks

(i) The area is the integral of $f$ from 0 to 4. Keep the brackets and the $d x$ .

(ii) Evaluate it on the GDC, then multiply by the thickness in metres, $3 mm = 0.003 m$ .

A = \int_{0}^{4} (- \frac{1}{4} x^{2} + x + 2) d x A = \frac{32}{3} = 10.666 \dots m^{2} V = \frac{32}{3} \times 0.003 = 0.032 m^{3}

Mentor tip

On a show-that, use the unrounded area. $10.7 \times 0.003$ does not give exactly 0.032.

(c)

2 marks

Convert the volume to cubic centimetres ( $1 m^{3} = 1 0^{6} cm^{3}$ ), then multiply by the density.

0.032 \times 1 0^{6} = 32000 cm^{3} 32000 \times 0.00119 = 38.08 \approx 38.1 kg

Mentor tip

Cubic units scale by $10 0^{3}$ , not 100. Converting by 100 gives about 0.004 kg, and a wall weighing a few grams should make you suspicious.

(d)

5 marks

(i) Integrate the new curve $g$ from 0 to 4 on the GDC.

(ii) Work out the percentage decrease from the original area.

Compare with 15% and give a verdict.

\int_{0}^{4} (- \frac{1}{8} x^{2} + \frac{1}{2} x + 2) d x = 9.33 m^{2} (GDC) \frac{10.666 \dots - 9.333 \dots}{10.666 \dots} \times 100 = 12.5% 12.5% < 15% \Rightarrow the claim is not correct

Another way. Find 85% of the original area, $0.85 \times 10.666 \dots = 9.07$ , and note that the new area 9.33 is larger, so the cut is less than 15%.

Mentor tip

Divide by the original area, not the new one, when finding a percentage change.

(e)

4 marks

(i) Differentiate $g$ , substitute into the arc length integral and evaluate on the GDC.

(ii) Add the curved top to the base (4 m) and the two vertical sides (2 m each).

g^{'} (x) = - \frac{1}{4} x + \frac{1}{2} l = \int_{0}^{4} 1 + (- \frac{1}{4} x + \frac{1}{2})^{2} d x = 4.16 m (GDC) perimeter = 4.16 + 4 + 2 + 2 = 12.2 m

Mentor tip

Sense check: the arc should be only a little longer than the 4 m base, since the curve is quite flat.

Question 4

16 marks

Matched savings and annuity growth with the TVM solver

Number and algebra

(a)

2 marks

Find the monthly salary, then take 4% of it. Depositing that much gets the largest possible match.

\frac{60 000}{12} = 5000 0.04 \times 5000 = $200

(b)

3 marks

Each month Kelley puts in 200 dollars and the company matches it, so the payment is 400 dollars.
Use the TVM solver with 40 years of monthly payments and solve for FV.

N = 480, I % = 9.5, P V = 0, P M T = - 400, P / Y = C / Y = 12 F V = 2174 516.77 \dots (GDC) $2 174517

Mentor tip

Forgetting the employer's matching 200 dollars halves the payment. Also round to the nearest dollar, as the question asks.

(c)

2 marks

(i) Less: for example, the account could earn less than 9.5% a year, or she could lose her job or leave before 40 years.

(ii) More (using a different assumption): for example, her salary could rise, so 4% of it (and the match) grows over time.

(i) rate < 9.5% \Rightarrow less (ii) salary rises \Rightarrow larger deposits \Rightarrow more

Mentor tip

Each reason must name one of the listed assumptions and say how breaking it changes the total. Use a different assumption for each part.

(d)

6 marks

Find the balance after 10 years (120 payments of 400 dollars).
Take off the 50 000 dollar withdrawal. That becomes the present value for the remaining 30 years.
Solve for the new final value with 360 more payments.
Subtract it from the answer to part (b).

N = 120, P M T = - 400, P V = 0 \Rightarrow F V = 79 632.27 \dots 79 632.27 \dots - 50000 = 29 632.27 \dots N = 360, P V = - 29 632.27 \dots, P M T = - 400 \Rightarrow F V = 1319 773.68 \dots (GDC) 2174 516.77 \dots - 1319 773.68 \dots = 854 743.09 \dots $854 743 less

Mentor tip

Carry full calculator values between steps. Rounding the balances to 3 s.f. throws the final dollar figure out by thousands.

(e)

3 marks

Keep 360 months and the reduced starting balance, set FV to 2 000 000 and solve for PMT.
The increase is the new payment minus the current 400 dollars.

N = 360, P V = - 29 632.27 \dots, F V = 2000000 \Rightarrow P M T = - 734.59 (GDC) 734.59 - 400 = 334.59 increase \approx $335

Mentor tip

Subtract 400 dollars, the total monthly payment the solver was using, not Kelley's 200-dollar share.

Question 5

15 marks

A cosine model for tide depth and waiting times

Functions

(a)

1 mark

Minimum to the next maximum is half a cycle, so double 6.25.

period = 2 \times 6.25 = 12.5 hours

(b)

5 marks

Find $b$ from the period (degrees or radians).
The principal axis $d$ is halfway between the maximum and minimum depths; the amplitude is half their difference.
The model starts at a minimum, so the cosine is flipped: $a$ is negative.

b = \frac{360}{12.5} = 28.8 or b = \frac{2 π}{12.5} = 0.503 d = \frac{57.5 + 50.5}{2} = 54 ∣ a ∣ = \frac{57.5 - 50.5}{2} = 3.5 w (t) = - 3.5 cos (28.8 t) + 54

Mentor tip

Check $t = 0$ : $- 3.5 cos 0 + 54 = 50.5$ , the minimum. A positive $a$ would start at the maximum.

(c)

2 marks

Graph $w (t)$ and $y = 53$ and find the first intersection, where the depth rises through 53.

- 3.5 cos (28.8 t) + 54 = 53 t = 2.55 hours (GDC)

Mentor tip

Set the GDC to degrees if you used 28.8, radians if you used 0.503. A mismatched mode is the usual reason for strange answers.

(d)

2 marks

Find where the depth next falls back to 53, which is before $t = 12$ .
By $t = 12$ the depth is below 53 again, so there is only one window. Subtract the two times.

t = 9.95 (GDC) 9.951 \dots - 2.548 \dots = 7.40 hours

Another way. By symmetry the second crossing is $12.5 - 2.548 \dots = 9.951 \dots$ , since the window is centred on the maximum at 6.25.

(e)

5 marks

Arrival time: distance divided by speed.
Find the depth pattern around that time. A minimum falls at $t = 25$ , so the depth is below 53 when the ship arrives.
Find the next time after arrival when $w (t) = 53$ and subtract the arrival time.

\frac{550}{21} = 26.19 \dots hours w (t) = 53, t > 26.19 \Rightarrow t = 27.548 \dots (GDC) 27.548 \dots - 26.190 \dots = 1.36 hours (\approx 81.5 minutes)

Mentor tip

Extend the GDC window past $t = 26$ . The earlier high-water window (about $t = 15.05$ to $22.45$ ) has already closed when the ship turns up.

Our analysis

Where the marks are

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

Number and algebra30 marks · 38%
Q1, Q4
Statistics and probability17 marks · 21%
Q2
Calculus16 marks · 20%
Q3
Functions15 marks · 19%
Q5
Geometry and trigonometry2 marks · 3%
Q3

Mentor insights

Traps to watch for

01
Using areas instead of layouts · Q1
Rolls and flats cannot be cut, so count whole pieces along each side. Dividing total area by one roll's area overcounts.
02
Unit conversion for volume and density · Q3
A cubic metre is a million cubic centimetres. Converting by 100 gives a mass that is far too small.
03
Forgetting the matched deposit · Q4
The account receives 400 dollars a month: Kelley's 200 plus the employer's 200. The increase in (e) is measured from that 400 dollars.
04
Sign of a when the model starts at a minimum · Q5
A cosine starts at its maximum, so a model starting at a minimum needs a negative amplitude. Check $w (0)$ before moving on.
05
Hypotheses about one prize only · Q2
The goodness of fit test is about the whole distribution. Hypotheses and conclusions that mention only the 2-dollar prize miss the point.

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

TZ1TZ2TZ3

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 TZ2 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

Tiling a rectangle with turf, costs and a perimeter

Number and algebra

(a)

4 marks

(i) Count how many whole 1.4 m lengths fit along the 8 m side, rounding down.

Count how many 0.6 m rows fit up the 4.8 m side, then multiply.

(ii) Five rolls take up 7 m of the 8 m, so a strip 1 m wide is left running the full 4.8 m height.

\frac{8}{1.4} = 5.71 \dots \Rightarrow 5 rolls per row \frac{4.8}{0.6} = 8 rows 5 \times 8 = 40 rolls uncovered area = (8 - 7) \times 4.8 = 4.8 m^{2}

Mentor tip

Rolls cannot be cut, so round down. Dividing the garden's area by a roll's area gives 45, which ignores the shapes and is wrong here.

(b)

2 marks

The leftover strip is 1 m by 4.8 m. Split the 1 m width into $0.6 + 0.4$ and fill each column with flats turned to fit.
Check with areas: each flat covers $0.24 m^{2}$ .

0.6 m column: \frac{4.8}{0.4} = 12 flats 0.4 m column: \frac{4.8}{0.6} = 8 flats 12 + 8 = 20 = \frac{4.8}{0.24}

Mentor tip

The area check only works because the flats tile the strip exactly. Always make sure a real arrangement exists.

(c)

2 marks

Multiply each count by its price and add.

40 (5) + 20 (2) = 200 + 40 $240

(d)

3 marks

Work in the same 1 m by 4.8 m strip. Stand two rolls on end in the 0.6 m column: they fill $2 \times 1.4 = 2.8$ m of its height.
The remaining 2 m of that column takes flats 0.4 m tall, and the 0.4 m column still takes 8 flats.
Cost this layout and compare with the phase 2 cost from part (c).

0.6 m column: 2 (1.4) + 5 (0.4) = 4.8 0.4 m column: 8 (0.6) = 4.8 2 rolls and 13 flats 2 (5) + 13 (2) = $36 < 20 (2) = $40 total 200 + 36 = $236 < $240, so Ayden is right

Mentor tip

Finish with a clear comparison and a sentence saying the claim holds. A labelled sketch of the strip is a quick way to show the layout works.

(e)

3 marks

Find the perimeter in metres.
Convert the spacing to 0.4 m and divide. Around a closed loop the number of gaps equals the number of lights, so there is no extra one to add.

P = 2 (8) + 2 (4.8) = 25.6 m \frac{25.6}{0.4} = 64 lights

Mentor tip

Mixing centimetres and metres is the classic slip. Convert first, then divide.

Question 2

17 marks

Expected value, fairness and a chi-squared fit test

Statistics and probability

(a)

4 marks

(i) Multiply the probability of the 100-dollar prize by 6.

(ii) All the probabilities add to 1, so subtract the known ones from 1.

a = 6 (0.002) = 0.012 b = 1 - (0.085 + 0.012 + 0.004 + 0.002 + 0.001) b = 0.896

(b)

4 marks

Find the expected prize by multiplying each prize by its probability and adding.
A card costs 1 dollar, so the game is fair only if the expected gain is 0 (expected prize equal to the cost).

E (prize) = 0 (0.896) + 2 (0.085) + 5 (0.012) + 10 (0.004) + 100 (0.002) + 500 (0.001) E (prize) = 0.97 E (gain) = 0.97 - 1 = - 0.03 \neq = 0 not a fair game

Mentor tip

Say what you are comparing: fairness means expected gain equals zero once the 1-dollar cost is taken off.

(c)

1 mark

Divide the number of 2-dollar wins by the number of cards.

\frac{48}{750} = 0.064

(d)

2 marks

(i) For: her experimental 0.064 is noticeably lower than the stated 0.085.

(ii) Against: relative frequencies from a sample naturally vary around the true probability, and 750 cards may not be enough (or may not be representative) to judge a claim about all cards.

0.064 \neq = 0.085

Mentor tip

For the counter-argument, explain why the sample might mislead. Just writing "small sample" on its own is thin.

(e)

6 marks

(i) $H_{0}$ : the prize outcomes follow the distribution stated by the lottery company. $H_{1}$ : they do not follow that distribution.

(ii) Expected frequencies are $750 \times$ each stated probability, with the top three prizes combined ( $0.007$ ). Four categories give $4 - 1 = 3$ degrees of freedom.

Run the goodness of fit test on the GDC.

(iii) Compare the $p$ -value with 0.10 and state the conclusion about the company's probabilities.

E = 672, 63.75, 9, 5.25 ν = 3 p = 0.117 (GDC) 0.117 > 0.10 \Rightarrow do not reject H_{0} not enough evidence that the prizes differ from the company’s stated distribution

Mentor tip

Write hypotheses about the whole distribution, not just the 2-dollar prize, and phrase the conclusion in context rather than only "do not reject $H_{0}$ ".

Question 3

18 marks

Area under a parabola, volume, mass and arc length

Calculus
Geometry and trigonometry

(a)

3 marks

(i) Differentiate $f (x) = - \frac{1}{4} x^{2} + x + 2$ term by term.

(ii) Substitute $x = 0$ .

(iii) Take the inverse tangent of that gradient.

f^{'} (x) = - \frac{1}{2} x + 1 f^{'} (0) = 1 θ = tan^{- 1} (1) = 4 5^{\circ} (\frac{π}{4})

(b)

4 marks

(i) The area is the integral of $f$ from 0 to 4. Keep the brackets and the $d x$ .

(ii) Evaluate it on the GDC, then multiply by the thickness in metres, $3 mm = 0.003 m$ .

A = \int_{0}^{4} (- \frac{1}{4} x^{2} + x + 2) d x A = \frac{32}{3} = 10.666 \dots m^{2} V = \frac{32}{3} \times 0.003 = 0.032 m^{3}

Mentor tip

On a show-that, use the unrounded area. $10.7 \times 0.003$ does not give exactly 0.032.

(c)

2 marks

Convert the volume to cubic centimetres ( $1 m^{3} = 1 0^{6} cm^{3}$ ), then multiply by the density.

0.032 \times 1 0^{6} = 32000 cm^{3} 32000 \times 0.00119 = 38.08 \approx 38.1 kg

Mentor tip

Cubic units scale by $10 0^{3}$ , not 100. Converting by 100 gives about 0.004 kg, and a wall weighing a few grams should make you suspicious.

(d)

5 marks

(i) Integrate the new curve $g$ from 0 to 4 on the GDC.

(ii) Work out the percentage decrease from the original area.

Compare with 15% and give a verdict.

\int_{0}^{4} (- \frac{1}{8} x^{2} + \frac{1}{2} x + 2) d x = 9.33 m^{2} (GDC) \frac{10.666 \dots - 9.333 \dots}{10.666 \dots} \times 100 = 12.5% 12.5% < 15% \Rightarrow the claim is not correct

Another way. Find 85% of the original area, $0.85 \times 10.666 \dots = 9.07$ , and note that the new area 9.33 is larger, so the cut is less than 15%.

Mentor tip

Divide by the original area, not the new one, when finding a percentage change.

(e)

4 marks

(i) Differentiate $g$ , substitute into the arc length integral and evaluate on the GDC.

(ii) Add the curved top to the base (4 m) and the two vertical sides (2 m each).

g^{'} (x) = - \frac{1}{4} x + \frac{1}{2} l = \int_{0}^{4} 1 + (- \frac{1}{4} x + \frac{1}{2})^{2} d x = 4.16 m (GDC) perimeter = 4.16 + 4 + 2 + 2 = 12.2 m

Mentor tip

Sense check: the arc should be only a little longer than the 4 m base, since the curve is quite flat.

Question 4

16 marks

Matched savings and annuity growth with the TVM solver

Number and algebra

(a)

2 marks

Find the monthly salary, then take 4% of it. Depositing that much gets the largest possible match.

\frac{60 000}{12} = 5000 0.04 \times 5000 = $200

(b)

3 marks

Each month Kelley puts in 200 dollars and the company matches it, so the payment is 400 dollars.
Use the TVM solver with 40 years of monthly payments and solve for FV.

N = 480, I % = 9.5, P V = 0, P M T = - 400, P / Y = C / Y = 12 F V = 2174 516.77 \dots (GDC) $2 174517

Mentor tip

Forgetting the employer's matching 200 dollars halves the payment. Also round to the nearest dollar, as the question asks.

(c)

2 marks

(i) Less: for example, the account could earn less than 9.5% a year, or she could lose her job or leave before 40 years.

(ii) More (using a different assumption): for example, her salary could rise, so 4% of it (and the match) grows over time.

(i) rate < 9.5% \Rightarrow less (ii) salary rises \Rightarrow larger deposits \Rightarrow more

Mentor tip

Each reason must name one of the listed assumptions and say how breaking it changes the total. Use a different assumption for each part.

(d)

6 marks

Find the balance after 10 years (120 payments of 400 dollars).
Take off the 50 000 dollar withdrawal. That becomes the present value for the remaining 30 years.
Solve for the new final value with 360 more payments.
Subtract it from the answer to part (b).

N = 120, P M T = - 400, P V = 0 \Rightarrow F V = 79 632.27 \dots 79 632.27 \dots - 50000 = 29 632.27 \dots N = 360, P V = - 29 632.27 \dots, P M T = - 400 \Rightarrow F V = 1319 773.68 \dots (GDC) 2174 516.77 \dots - 1319 773.68 \dots = 854 743.09 \dots $854 743 less

Mentor tip

Carry full calculator values between steps. Rounding the balances to 3 s.f. throws the final dollar figure out by thousands.

(e)

3 marks

Keep 360 months and the reduced starting balance, set FV to 2 000 000 and solve for PMT.
The increase is the new payment minus the current 400 dollars.

N = 360, P V = - 29 632.27 \dots, F V = 2000000 \Rightarrow P M T = - 734.59 (GDC) 734.59 - 400 = 334.59 increase \approx $335

Mentor tip

Subtract 400 dollars, the total monthly payment the solver was using, not Kelley's 200-dollar share.

Question 5

15 marks

A cosine model for tide depth and waiting times

Functions

(a)

1 mark

Minimum to the next maximum is half a cycle, so double 6.25.

period = 2 \times 6.25 = 12.5 hours

(b)

5 marks

Find $b$ from the period (degrees or radians).
The principal axis $d$ is halfway between the maximum and minimum depths; the amplitude is half their difference.
The model starts at a minimum, so the cosine is flipped: $a$ is negative.

b = \frac{360}{12.5} = 28.8 or b = \frac{2 π}{12.5} = 0.503 d = \frac{57.5 + 50.5}{2} = 54 ∣ a ∣ = \frac{57.5 - 50.5}{2} = 3.5 w (t) = - 3.5 cos (28.8 t) + 54

Mentor tip

Check $t = 0$ : $- 3.5 cos 0 + 54 = 50.5$ , the minimum. A positive $a$ would start at the maximum.

(c)

2 marks

Graph $w (t)$ and $y = 53$ and find the first intersection, where the depth rises through 53.

- 3.5 cos (28.8 t) + 54 = 53 t = 2.55 hours (GDC)

Mentor tip

Set the GDC to degrees if you used 28.8, radians if you used 0.503. A mismatched mode is the usual reason for strange answers.

(d)

2 marks

Find where the depth next falls back to 53, which is before $t = 12$ .
By $t = 12$ the depth is below 53 again, so there is only one window. Subtract the two times.

t = 9.95 (GDC) 9.951 \dots - 2.548 \dots = 7.40 hours

Another way. By symmetry the second crossing is $12.5 - 2.548 \dots = 9.951 \dots$ , since the window is centred on the maximum at 6.25.

(e)

5 marks

Arrival time: distance divided by speed.
Find the depth pattern around that time. A minimum falls at $t = 25$ , so the depth is below 53 when the ship arrives.
Find the next time after arrival when $w (t) = 53$ and subtract the arrival time.

\frac{550}{21} = 26.19 \dots hours w (t) = 53, t > 26.19 \Rightarrow t = 27.548 \dots (GDC) 27.548 \dots - 26.190 \dots = 1.36 hours (\approx 81.5 minutes)

Mentor tip

Extend the GDC window past $t = 26$ . The earlier high-water window (about $t = 15.05$ to $22.45$ ) has already closed when the ship turns up.

Our analysis

Where the marks are

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

Number and algebra30 marks · 38%
Q1, Q4
Statistics and probability17 marks · 21%
Q2
Calculus16 marks · 20%
Q3
Functions15 marks · 19%
Q5
Geometry and trigonometry2 marks · 3%
Q3

Mentor insights

Traps to watch for

01
Using areas instead of layouts · Q1
Rolls and flats cannot be cut, so count whole pieces along each side. Dividing total area by one roll's area overcounts.
02
Unit conversion for volume and density · Q3
A cubic metre is a million cubic centimetres. Converting by 100 gives a mass that is far too small.
03
Forgetting the matched deposit · Q4
The account receives 400 dollars a month: Kelley's 200 plus the employer's 200. The increase in (e) is measured from that 400 dollars.
04
Sign of a when the model starts at a minimum · Q5
A cosine starts at its maximum, so a model starting at a minimum needs a negative amplitude. Check $w (0)$ before moving on.
05
Hypotheses about one prize only · Q2
The goodness of fit test is about the whole distribution. Hypotheses and conclusions that mention only the 2-dollar prize miss the point.

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