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Worked solutions

IB Maths AI SL · Paper 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2TZ3

Session: May 2026
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

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 1–12 · 80 marks

Question 1

9 marks

Discrete data, a mean and comparing box plots

Statistics and probability

(a)

1 mark

A number of guesses can only be a whole number, so the data are counted, not measured.

discrete

Mentor tip

Counts are discrete; measurements such as time or length are continuous.

(b)

2 marks

Read the frequency of each score off the bar chart: $2, 5, 9, 14, 7, 4$ for $1$ to $6$ guesses.
Multiply each score by its frequency, add, and divide by the $41$ games.

\overset{x}{ˉ} = \frac{1 ( 2 ) + 2 ( 5 ) + 3 ( 9 ) + 4 ( 14 ) + 5 ( 7 ) + 6 ( 4 )}{41} \overset{x}{ˉ} = \frac{154}{41} = 3.76 (3 s.f.)

Mentor tip

Check that your frequencies add to $41$ before dividing. Entering the scores and frequencies as two GDC lists gives the mean directly.

(c)

3 marks

Kyle: build the running totals $2, 7, 16, 30, 37, 41$ . The median is the $21$ st value, which falls in the $4$ group.
The lower half is the first $20$ values; its middle ( $10$ th and $11$ th) are both $3$ . The upper half is values $22$ to $41$ ; its middle ( $31$ st and $32$ nd) are both $5$ .
Hansol: read the three box lines straight off the given plot.

Kyle: Q_{1} = 3, Q_{2} = 4, Q_{3} = 5 Hansol: Q_{1} = 3.5, Q_{2} = 4.5, Q_{3} = 5

Another way. Type Kyle's scores and frequencies into the GDC's one-variable statistics and read off the quartiles.

Mentor tip

With $41$ values the median is the $21$ st; for the quartiles, leave the median out and split the rest into two halves of $20$ .

(d)

1 mark

Draw Kyle's box on his row from $3$ to $5$ with a line at $4$ .
Whiskers run from the box out to his smallest score, $1$ , and his largest, $6$ . No line should pass through the box itself.

min = 1, Q_{1} = 3, Q_{2} = 4, Q_{3} = 5, max = 6

Mentor tip

Line your plot up with the same scale as Hansol's so the two can be compared by eye.

(e)

2 marks

(i) Fewer guesses is better. Kyle's median ( $4$ ) is lower than Hansol's ( $4.5$ ), so Kyle typically needs fewer guesses.

(ii) Hansol's results are more consistent: his range ( $4$ against $5$ ) and interquartile range ( $1.5$ against $2$ ) are smaller, and his worst game took $5$ guesses while Kyle's took $6$ .

(i) median: 4 < 4.5 (ii) range: 5 - 1 = 4 < 6 - 1 = 5

Mentor tip

The reason must come from the box plots, so quote the median, spread or extremes. The mean is not shown on a box plot, so it does not count here.

Question 2

7 marks

Volume of a cuboid and percentage error

Geometry and trigonometry
Number and algebra

(a)

2 marks

Treat the container as a cuboid and multiply the three given dimensions.

V = 299 \times 218 \times 318 V = 20727876 mm^{3}

Mentor tip

The paper wants the nearest cubic millimetre here, so write every digit rather than rounding to $3$ s.f.

(b)

4 marks

Change your volume into litres: $1 mm^{3} = 0.001$ ml and $1000$ ml $= 1$ litre.
Use percentage error $= \frac{v _{A} - v _{E}}{v _{E}} \times 100%$ , with the stated $20$ litres as the exact value.

20727876 \times 0.001 = 20 727.876 ml = 20.727876 L ε = \frac{20.727876 - 20}{20} \times 100% ε = 3.64% (3 s.f.)

Mentor tip

Both volumes must be in the same unit before you compare them, and the exact (true) value always goes on the bottom.

(c)

1 mark

Think about how the real container differs from a perfect box of those dimensions.

e.g. rounded edges or cut-outs mean it holds less than a cuboid

Mentor tip

Other sensible reasons: the dimensions are outside measurements, so the wall thickness is counted as space. Give a physical reason about the shape, not a comment on rounding.

Question 3

6 marks

Two-way table probabilities, without replacement

Statistics and probability

(a)

1 mark

Take the single cell for left-handed and geometry, over the total of $200$ .

P = \frac{16}{200} = \frac{2}{25} = 0.08

(b)

2 marks

Given left-handed, the sample space shrinks to the left-handed row only: $16 + 6 = 22$ students.

P (G ∣ L) = \frac{16}{22} = \frac{8}{11} = 0.727 (3 s.f.)

Mentor tip

In a conditional probability the denominator is the total of the group you are told about, not $200$ .

(c)

3 marks

There are $22$ left-handed and $178$ right-handed students.
Use the complement: find the chance that neither is left-handed, remembering the second pick is from $199$ students.
Subtract from $1$ .

P (no left-handers) = \frac{178}{200} \times \frac{177}{199} P (at least one) = 1 - \frac{178}{200} \times \frac{177}{199} = \frac{4147}{19900} = 0.208 (3 s.f.)

Another way. Add the three cases directly: L then R, R then L, and L then L, each using $199$ for the second pick.

Mentor tip

The same student cannot be chosen twice, so this is without replacement. A binomial or $(\frac{178}{200})^{2}$ happens to round to the same value here but is the wrong model.

Question 4

5 marks

Independent events and the binomial distribution

Statistics and probability

(a)

2 marks

The attempts are independent, so multiply the probabilities of success.

P = 0.73 \times 0.73 = 0.5329 = 0.533 (3 s.f.)

Mentor tip

Give three significant figures: $0.53$ is too rough.

(b)

3 marks

Let $Y$ be the number of successes: a fixed number of independent trials with the same chance each time, so $Y \sim B (5, 0.73)$ .
At least $3$ means $3$ , $4$ or $5$ , which is $1 - P (Y \leq 2)$ .

Y \sim B (5, 0.73) P (Y \geq 3) = 1 - P (Y \leq 2) = 0.874 (3 s.f.) (GDC)

Mentor tip

Binomial cdf gives $P (Y \leq k)$ , so for “at least $3$ ” subtract $P (Y \leq 2)$ , not $P (Y \leq 3)$ .

Question 5

6 marks

Inverse variation and a cost inequality

Functions

(a)

2 marks

Inverse variation means $c = \frac{k}{n}$ for some constant $k$ .
Use the given pair of values to find $k$ .

187.50 = \frac{k}{4} ⟹ k = 750 c = \frac{750}{n}

(b)

4 marks

(i) Solve $\frac{750}{n} < 100$ . This gives $n > 7.5$ , and $n$ must be a whole number of people.

(ii) Substitute $n = 8$ into your equation from (a).

\frac{750}{n} < 100 ⟹ n > 7.5 n = 8 c = \frac{750}{8} = 93.75 USD

Mentor tip

Round up, not to the nearest whole number: $7$ people would pay about $107$ each, which is over the limit. Check that your final cost really is under $100$ .

Question 6

4 marks

Inverse of a logarithmic function

Functions

(a)

2 marks

$g^{- 1} (3)$ is the input that makes $g$ equal $3$ , so solve $g (x) = 3$ .
That needs $lo g_{10} (x + 1) = 0$ , so $x + 1 = 1$ .

lo g_{10} (x + 1) + 3 = 3 lo g_{10} (x + 1) = 0 ⟹ x + 1 = 1 g^{- 1} (3) = 0

Mentor tip

Write the answer as $g^{- 1} (3) = 0$ so it is clear which value you mean.

(b)

2 marks

$g$ has a vertical asymptote at $x = - 1$ (from the domain $x > - 1$ ).
Reflecting in $y = x$ swaps the roles of $x$ and $y$ , so that vertical asymptote becomes a horizontal one.

g^{- 1} (x) = 1 0^{x - 3} - 1 1 0^{x - 3} \to 0 as x \to - \infty a = - 1

Another way. Sketch $g$ and its reflection in $y = x$ on the GDC and read off where the inverse levels out.

Mentor tip

Features of $f$ and $f^{- 1}$ swap across $y = x$ : a vertical asymptote $x = k$ becomes a horizontal asymptote $y = k$ .

Question 7

10 marks

Fitting a quadratic model to a serve

Number and algebra
Functions

(a)

2 marks

Substitute each point's $x$ and $h$ into $h (x) = a x^{2} + b x + c$ .

a + b + c = 2.8 4 a + 2 b + c = 2.86 9 a + 3 b + c = 2.89

Mentor tip

Each equation uses both coordinates of a point: the $x$ -value goes into the quadratic and the height goes on the other side.

(b)

2 marks

Solve the three simultaneous equations with the GDC's equation solver.

a = - 0.015, b = 0.105, c = 2.71 (GDC) h (x) = - 0.015 x^{2} + 0.105 x + 2.71

Mentor tip

Quadratic regression on the three points gives the same curve, so it makes a quick check of your solver answer.

(c)

1 mark

$c = h (0)$ , and $x = 0$ is where Emily strikes the ball.

h (0) = c = 2.71 m

Mentor tip

Be precise: it is the height of the ball at the instant it is hit, not Emily's height or just “the $y$ -intercept”.

(d)

5 marks

Net: find the height of the ball at $x = 9$ and compare it with $2.24$ m.
End line: find where the ball lands by solving $h (x) = 0$ (take the positive root), and compare with $18$ m.
The serve counts only if both conditions hold, so state a conclusion that covers both.

h (9) = - 0.015 (81) + 0.105 (9) + 2.71 = 2.44 > 2.24 h (x) = 0 ⟹ x = 17.4 (GDC) < 18 clears the net and lands before the end line, so the serve is in

Another way. Instead of solving $h (x) = 0$ , evaluate $h (18) = - 0.26 < 0$ : the ball is already below the ground by the end line, so it has landed in the court.

Mentor tip

Two checks, one conclusion. Show both numbers with their comparison, then finish with a clear sentence.

Question 8

5 marks

A two-sample t-test on climbing times

Statistics and probability

(a)

1 mark

The test is two-tailed, so the alternative just says the two means differ.

H_{1} : μ_{S} \neq = μ_{G}

Mentor tip

In words, say “population mean”, since the hypotheses are about the populations, not these twelve students.

(b)

2 marks

Enter the two sets of times as lists and run a two-sample $t$ -test, two-tailed, with the pooled (equal variance) option on.

p = 0.638821 \dots p = 0.639 (3 s.f.) (GDC)

Mentor tip

Make sure “pooled” is set to yes; the question tells you to assume equal variances.

(c)

2 marks

Compare the $p$ -value with the $5%$ significance level.
Since it is bigger, do not reject $H_{0}$ , then say what that means for the mean times.

0.639 > 0.05 do not reject H_{0} insufficient evidence of a difference in the mean climbing times

Mentor tip

Give the comparison as your reason and mention the mean climbing times in the conclusion. Do not claim the means are proved equal.

Question 9

5 marks

Differentiating a power function and reading its graph

Calculus

(a)

2 marks

Rewrite $\frac{1}{x ^{2}}$ as $x^{- 2}$ , then differentiate each term. The constant disappears.

f (x) = x^{- 2} + 2 x - 5 f^{'} (x) = - 2 x^{- 3} + 2 = - \frac{2}{x ^{3}} + 2

Mentor tip

Powers of $x$ on the bottom of a fraction must go up as negative powers before you use the power rule.

(b)

3 marks

Asymptote: as $x \to 0$ the graph shoots up, so the asymptote is vertical ( $x = 0$ ), and there is no horizontal one at $f (x) = 0$ . False.
$f (- 2) = \frac{1}{4} - 4 - 5 = - 8.75$ , not $1$ , so $f^{- 1} (1) = - 2$ is false.
$f^{'} (0)$ does not exist, since $f$ is undefined at $0$ . False.
$f (3) = \frac{1}{9} + 1 > 0$ , and the curve is rising at $x = 3$ with $f^{'} (3) = 2 - \frac{2}{27} > 0$ . Both true.

horizontal asymptote at f (x) = 0 : false f^{- 1} (1) = - 2 : false f^{'} (0) = 0 : false f (3) > 0 : true f^{'} (3) > 0 : true

Mentor tip

The minimum near $x = 1$ is where $f^{'} (x) = 0$ , not $x = 0$ . Check each statement with a quick calculation rather than by eye.

Question 10

8 marks

A geometric sequence of flow rates

Number and algebra

(a)

2 marks

Divide each rate by the one before it. If both ratios are the same, the sequence is geometric.

\frac{3.5}{2.5} = 1.4, \frac{4.9}{3.5} = 1.4 constant ratio r = 1.4, so geometric

Mentor tip

Show both ratios. Checking one pair of terms is not enough to establish a common ratio.

(b)

2 marks

Use $u_{n} = u_{1} r^{n - 1}$ and find the first $n$ that takes the rate past $25$ .

2.5 (1.4)^{n - 1} > 25 n > 7.84 (GDC) n = 8

Another way. List the terms on the GDC: day $7$ gives about $18.8$ and day $8$ about $26.4$ , so day $8$ is the first over $25$ .

Mentor tip

$n$ is a day number, so take the next whole number above $7.84$ .

(c)

4 marks

(i) Use the geometric series formula with $u_{1} = 2.5$ , $r = 1.4$ and $n = 8$ .

(ii) Each day's rate lasts $24 \times 60 = 1440$ minutes, so multiply the sum of the rates by $1440$ .

S_{8} = \frac{2.5 ( 1. 4 ^{8} - 1 )}{1.4 - 1} = 85.9868 \dots S_{8} = 86.0 m^{3} min^{- 1} (3 s.f.) V = 85.9868 \dots \times 60 \times 24 = 123821 \dots V = 124000 m^{3} (3 s.f.)

Mentor tip

Carry the unrounded sum into (ii). Using $86.0$ gives $123840$ , which still rounds to $124000$ , but it is a good habit.

Question 11

8 marks

Bearings, an isosceles triangle and average speed

Geometry and trigonometry

(a)

1 mark

Bearings are measured clockwise from north. Magnetic north is $6°$ anticlockwise (west) of true north.

360° - 6° = 354°

Mentor tip

Bearings are always three figures and measured clockwise, so $6°$ west of north is $354°$ , not $6°$ .

(b)

2 marks

$A B = A C = 5$ km, so triangle $A B C$ is isosceles with the $6°$ angle at $A$ .
The two base angles are equal and share the rest of $180°$ .

\hat{A C B} = \frac{180° - 6°}{2} = 87°

(c)

3 marks

Use the cosine rule with the two $5$ km sides and the $6°$ angle between them.

C B^{2} = 5^{2} + 5^{2} - 2 (5) (5) cos 6° C B = 0.5233595 \dots C B = 0.523 km (3 s.f.)

Another way. Use the sine rule with the $87°$ angle from (b): $\frac{C B}{s i n 6°} = \frac{5}{s i n 87°}$ .

Mentor tip

Make sure your GDC is in degrees; in radians the answer comes out very different.

(d)

2 marks

Total distance is $5$ km plus your $C B$ . The time is $2$ hours $15$ minutes, which is $2.25$ hours.

speed = \frac{5 + 0.5233595 \dots}{2.25} = 2.45 km h^{- 1} (3 s.f.)

Mentor tip

$15$ minutes is $0.25$ hours, not $0.15$ . Convert minutes to hours before dividing.

Question 12

7 marks

The normal to a cubic and where it meets again

Calculus
Geometry and trigonometry
Functions

(a)

2 marks

Differentiate the cubic (or use the GDC's derivative at a point) and evaluate at $x = 6$ .
The normal is perpendicular to the tangent, so its gradient is the negative reciprocal.

f^{'} (x) = \frac{3}{2} x^{2} - \frac{27}{2} x + \frac{51}{2} f^{'} (6) = 54 - 81 + 25.5 = - 1.5 m_{normal} = - \frac{1}{- 1.5} = \frac{2}{3}

Mentor tip

Flip and change the sign. Using $- 1.5$ itself gives the tangent, not the normal.

(b)

1 mark

Use the point $(6, 5)$ with gradient $\frac{2}{3}$ .

y - 5 = \frac{2}{3} (x - 6) y = \frac{2}{3} x + 1

(c)

2 marks

Draw a straight line through $A (6, 5)$ that crosses the $y$ -axis at $(0, 1)$ .
It should rise gently, $2$ up for every $3$ across (for example through $(3, 3)$ and $(9, 7)$ ), and cut across the curve at right angles to it at $A$ .

line through (0, 1), (3, 3), (6, 5), (9, 7)

Mentor tip

Plot two or three exact points from your equation and join them with a ruler, rather than guessing a direction.

(d)

2 marks

Set the cubic equal to the normal and solve with the GDC (or use the intersect tool on both graphs).
One root is $x = 6$ , which is $A$ ; the other two give the points you need. Find each $y$ from the line.

\frac{1}{2} x^{3} - \frac{27}{4} x^{2} + \frac{51}{2} x - 13 = \frac{2}{3} x + 1 x = 0.684737 \dots, 6, 6.81526 \dots (GDC) (0.685, 1.46) and (6.82, 5.54)

Mentor tip

The normal crosses the curve twice more, once very close to $A$ . Zoom in near $x = 6$ so you do not miss the second point.

Our analysis

Where the marks are

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

Statistics and probability25 marks · 31%
Q1, Q3, Q4, Q8
Functions18 marks · 23%
Q5, Q6, Q7, Q12
Number and algebra16 marks · 20%
Q2, Q7, Q10
Geometry and trigonometry14 marks · 18%
Q2, Q11, Q12
Calculus7 marks · 9%
Q9, Q12

Mentor insights

Traps to watch for

01
Choosing two without replacement · Q3
Once one student is picked, only 199 remain. Squaring 178/200 treats it as with replacement, which is the wrong model even when the rounding hides it.
02
Rounding people up, not off · Q5
n > 7.5 means 8 people. Rounding to the nearest whole number or taking 7 breaks the cost condition.
03
One condition is not enough · Q7
The serve must clear the net and land before the end line. Check both with numbers, then give one clear conclusion.
04
Minutes are not decimals of an hour · Q11
2 hours 15 minutes is 2.25 hours. Writing 2.15 gives the wrong speed.
05
Tangent gradient versus normal gradient · Q12
f'(6) = -1.5 is the tangent. The normal needs the negative reciprocal, 2/3, before you write its equation.

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 1

Mathematics: Applications & Interpretation, Standard Level

Timezone

TZ1TZ2TZ3

Session: May 2026
Marks: 80
Duration: 1 hour 30 min
Calculator: Allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

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 1–12 · 80 marks

Question 1

9 marks

Discrete data, a mean and comparing box plots

Statistics and probability

(a)

1 mark

A number of guesses can only be a whole number, so the data are counted, not measured.

discrete

Mentor tip

Counts are discrete; measurements such as time or length are continuous.

(b)

2 marks

Read the frequency of each score off the bar chart: $2, 5, 9, 14, 7, 4$ for $1$ to $6$ guesses.
Multiply each score by its frequency, add, and divide by the $41$ games.

\overset{x}{ˉ} = \frac{1 ( 2 ) + 2 ( 5 ) + 3 ( 9 ) + 4 ( 14 ) + 5 ( 7 ) + 6 ( 4 )}{41} \overset{x}{ˉ} = \frac{154}{41} = 3.76 (3 s.f.)

Mentor tip

Check that your frequencies add to $41$ before dividing. Entering the scores and frequencies as two GDC lists gives the mean directly.

(c)

3 marks

Kyle: build the running totals $2, 7, 16, 30, 37, 41$ . The median is the $21$ st value, which falls in the $4$ group.
The lower half is the first $20$ values; its middle ( $10$ th and $11$ th) are both $3$ . The upper half is values $22$ to $41$ ; its middle ( $31$ st and $32$ nd) are both $5$ .
Hansol: read the three box lines straight off the given plot.

Kyle: Q_{1} = 3, Q_{2} = 4, Q_{3} = 5 Hansol: Q_{1} = 3.5, Q_{2} = 4.5, Q_{3} = 5

Another way. Type Kyle's scores and frequencies into the GDC's one-variable statistics and read off the quartiles.

Mentor tip

With $41$ values the median is the $21$ st; for the quartiles, leave the median out and split the rest into two halves of $20$ .

(d)

1 mark

Draw Kyle's box on his row from $3$ to $5$ with a line at $4$ .
Whiskers run from the box out to his smallest score, $1$ , and his largest, $6$ . No line should pass through the box itself.

min = 1, Q_{1} = 3, Q_{2} = 4, Q_{3} = 5, max = 6

Mentor tip

Line your plot up with the same scale as Hansol's so the two can be compared by eye.

(e)

2 marks

(i) Fewer guesses is better. Kyle's median ( $4$ ) is lower than Hansol's ( $4.5$ ), so Kyle typically needs fewer guesses.

(ii) Hansol's results are more consistent: his range ( $4$ against $5$ ) and interquartile range ( $1.5$ against $2$ ) are smaller, and his worst game took $5$ guesses while Kyle's took $6$ .

(i) median: 4 < 4.5 (ii) range: 5 - 1 = 4 < 6 - 1 = 5

Mentor tip

The reason must come from the box plots, so quote the median, spread or extremes. The mean is not shown on a box plot, so it does not count here.

Question 2

7 marks

Volume of a cuboid and percentage error

Geometry and trigonometry
Number and algebra

(a)

2 marks

Treat the container as a cuboid and multiply the three given dimensions.

V = 299 \times 218 \times 318 V = 20727876 mm^{3}

Mentor tip

The paper wants the nearest cubic millimetre here, so write every digit rather than rounding to $3$ s.f.

(b)

4 marks

Change your volume into litres: $1 mm^{3} = 0.001$ ml and $1000$ ml $= 1$ litre.
Use percentage error $= \frac{v _{A} - v _{E}}{v _{E}} \times 100%$ , with the stated $20$ litres as the exact value.

20727876 \times 0.001 = 20 727.876 ml = 20.727876 L ε = \frac{20.727876 - 20}{20} \times 100% ε = 3.64% (3 s.f.)

Mentor tip

Both volumes must be in the same unit before you compare them, and the exact (true) value always goes on the bottom.

(c)

1 mark

Think about how the real container differs from a perfect box of those dimensions.

e.g. rounded edges or cut-outs mean it holds less than a cuboid

Mentor tip

Other sensible reasons: the dimensions are outside measurements, so the wall thickness is counted as space. Give a physical reason about the shape, not a comment on rounding.

Question 3

6 marks

Two-way table probabilities, without replacement

Statistics and probability

(a)

1 mark

Take the single cell for left-handed and geometry, over the total of $200$ .

P = \frac{16}{200} = \frac{2}{25} = 0.08

(b)

2 marks

Given left-handed, the sample space shrinks to the left-handed row only: $16 + 6 = 22$ students.

P (G ∣ L) = \frac{16}{22} = \frac{8}{11} = 0.727 (3 s.f.)

Mentor tip

In a conditional probability the denominator is the total of the group you are told about, not $200$ .

(c)

3 marks

There are $22$ left-handed and $178$ right-handed students.
Use the complement: find the chance that neither is left-handed, remembering the second pick is from $199$ students.
Subtract from $1$ .

P (no left-handers) = \frac{178}{200} \times \frac{177}{199} P (at least one) = 1 - \frac{178}{200} \times \frac{177}{199} = \frac{4147}{19900} = 0.208 (3 s.f.)

Another way. Add the three cases directly: L then R, R then L, and L then L, each using $199$ for the second pick.

Mentor tip

The same student cannot be chosen twice, so this is without replacement. A binomial or $(\frac{178}{200})^{2}$ happens to round to the same value here but is the wrong model.

Question 4

5 marks

Independent events and the binomial distribution

Statistics and probability

(a)

2 marks

The attempts are independent, so multiply the probabilities of success.

P = 0.73 \times 0.73 = 0.5329 = 0.533 (3 s.f.)

Mentor tip

Give three significant figures: $0.53$ is too rough.

(b)

3 marks

Let $Y$ be the number of successes: a fixed number of independent trials with the same chance each time, so $Y \sim B (5, 0.73)$ .
At least $3$ means $3$ , $4$ or $5$ , which is $1 - P (Y \leq 2)$ .

Y \sim B (5, 0.73) P (Y \geq 3) = 1 - P (Y \leq 2) = 0.874 (3 s.f.) (GDC)

Mentor tip

Binomial cdf gives $P (Y \leq k)$ , so for “at least $3$ ” subtract $P (Y \leq 2)$ , not $P (Y \leq 3)$ .

Question 5

6 marks

Inverse variation and a cost inequality

Functions

(a)

2 marks

Inverse variation means $c = \frac{k}{n}$ for some constant $k$ .
Use the given pair of values to find $k$ .

187.50 = \frac{k}{4} ⟹ k = 750 c = \frac{750}{n}

(b)

4 marks

(i) Solve $\frac{750}{n} < 100$ . This gives $n > 7.5$ , and $n$ must be a whole number of people.

(ii) Substitute $n = 8$ into your equation from (a).

\frac{750}{n} < 100 ⟹ n > 7.5 n = 8 c = \frac{750}{8} = 93.75 USD

Mentor tip

Round up, not to the nearest whole number: $7$ people would pay about $107$ each, which is over the limit. Check that your final cost really is under $100$ .

Question 6

4 marks

Inverse of a logarithmic function

Functions

(a)

2 marks

$g^{- 1} (3)$ is the input that makes $g$ equal $3$ , so solve $g (x) = 3$ .
That needs $lo g_{10} (x + 1) = 0$ , so $x + 1 = 1$ .

lo g_{10} (x + 1) + 3 = 3 lo g_{10} (x + 1) = 0 ⟹ x + 1 = 1 g^{- 1} (3) = 0

Mentor tip

Write the answer as $g^{- 1} (3) = 0$ so it is clear which value you mean.

(b)

2 marks

$g$ has a vertical asymptote at $x = - 1$ (from the domain $x > - 1$ ).
Reflecting in $y = x$ swaps the roles of $x$ and $y$ , so that vertical asymptote becomes a horizontal one.

g^{- 1} (x) = 1 0^{x - 3} - 1 1 0^{x - 3} \to 0 as x \to - \infty a = - 1

Another way. Sketch $g$ and its reflection in $y = x$ on the GDC and read off where the inverse levels out.

Mentor tip

Features of $f$ and $f^{- 1}$ swap across $y = x$ : a vertical asymptote $x = k$ becomes a horizontal asymptote $y = k$ .

Question 7

10 marks

Fitting a quadratic model to a serve

Number and algebra
Functions

(a)

2 marks

Substitute each point's $x$ and $h$ into $h (x) = a x^{2} + b x + c$ .

a + b + c = 2.8 4 a + 2 b + c = 2.86 9 a + 3 b + c = 2.89

Mentor tip

Each equation uses both coordinates of a point: the $x$ -value goes into the quadratic and the height goes on the other side.

(b)

2 marks

Solve the three simultaneous equations with the GDC's equation solver.

a = - 0.015, b = 0.105, c = 2.71 (GDC) h (x) = - 0.015 x^{2} + 0.105 x + 2.71

Mentor tip

Quadratic regression on the three points gives the same curve, so it makes a quick check of your solver answer.

(c)

1 mark

$c = h (0)$ , and $x = 0$ is where Emily strikes the ball.

h (0) = c = 2.71 m

Mentor tip

Be precise: it is the height of the ball at the instant it is hit, not Emily's height or just “the $y$ -intercept”.

(d)

5 marks

Net: find the height of the ball at $x = 9$ and compare it with $2.24$ m.
End line: find where the ball lands by solving $h (x) = 0$ (take the positive root), and compare with $18$ m.
The serve counts only if both conditions hold, so state a conclusion that covers both.

h (9) = - 0.015 (81) + 0.105 (9) + 2.71 = 2.44 > 2.24 h (x) = 0 ⟹ x = 17.4 (GDC) < 18 clears the net and lands before the end line, so the serve is in

Another way. Instead of solving $h (x) = 0$ , evaluate $h (18) = - 0.26 < 0$ : the ball is already below the ground by the end line, so it has landed in the court.

Mentor tip

Two checks, one conclusion. Show both numbers with their comparison, then finish with a clear sentence.

Question 8

5 marks

A two-sample t-test on climbing times

Statistics and probability

(a)

1 mark

The test is two-tailed, so the alternative just says the two means differ.

H_{1} : μ_{S} \neq = μ_{G}

Mentor tip

In words, say “population mean”, since the hypotheses are about the populations, not these twelve students.

(b)

2 marks

Enter the two sets of times as lists and run a two-sample $t$ -test, two-tailed, with the pooled (equal variance) option on.

p = 0.638821 \dots p = 0.639 (3 s.f.) (GDC)

Mentor tip

Make sure “pooled” is set to yes; the question tells you to assume equal variances.

(c)

2 marks

Compare the $p$ -value with the $5%$ significance level.
Since it is bigger, do not reject $H_{0}$ , then say what that means for the mean times.

0.639 > 0.05 do not reject H_{0} insufficient evidence of a difference in the mean climbing times

Mentor tip

Give the comparison as your reason and mention the mean climbing times in the conclusion. Do not claim the means are proved equal.

Question 9

5 marks

Differentiating a power function and reading its graph

Calculus

(a)

2 marks

Rewrite $\frac{1}{x ^{2}}$ as $x^{- 2}$ , then differentiate each term. The constant disappears.

f (x) = x^{- 2} + 2 x - 5 f^{'} (x) = - 2 x^{- 3} + 2 = - \frac{2}{x ^{3}} + 2

Mentor tip

Powers of $x$ on the bottom of a fraction must go up as negative powers before you use the power rule.

(b)

3 marks

Asymptote: as $x \to 0$ the graph shoots up, so the asymptote is vertical ( $x = 0$ ), and there is no horizontal one at $f (x) = 0$ . False.
$f (- 2) = \frac{1}{4} - 4 - 5 = - 8.75$ , not $1$ , so $f^{- 1} (1) = - 2$ is false.
$f^{'} (0)$ does not exist, since $f$ is undefined at $0$ . False.
$f (3) = \frac{1}{9} + 1 > 0$ , and the curve is rising at $x = 3$ with $f^{'} (3) = 2 - \frac{2}{27} > 0$ . Both true.

horizontal asymptote at f (x) = 0 : false f^{- 1} (1) = - 2 : false f^{'} (0) = 0 : false f (3) > 0 : true f^{'} (3) > 0 : true

Mentor tip

The minimum near $x = 1$ is where $f^{'} (x) = 0$ , not $x = 0$ . Check each statement with a quick calculation rather than by eye.

Question 10

8 marks

A geometric sequence of flow rates

Number and algebra

(a)

2 marks

Divide each rate by the one before it. If both ratios are the same, the sequence is geometric.

\frac{3.5}{2.5} = 1.4, \frac{4.9}{3.5} = 1.4 constant ratio r = 1.4, so geometric

Mentor tip

Show both ratios. Checking one pair of terms is not enough to establish a common ratio.

(b)

2 marks

Use $u_{n} = u_{1} r^{n - 1}$ and find the first $n$ that takes the rate past $25$ .

2.5 (1.4)^{n - 1} > 25 n > 7.84 (GDC) n = 8

Another way. List the terms on the GDC: day $7$ gives about $18.8$ and day $8$ about $26.4$ , so day $8$ is the first over $25$ .

Mentor tip

$n$ is a day number, so take the next whole number above $7.84$ .

(c)

4 marks

(i) Use the geometric series formula with $u_{1} = 2.5$ , $r = 1.4$ and $n = 8$ .

(ii) Each day's rate lasts $24 \times 60 = 1440$ minutes, so multiply the sum of the rates by $1440$ .

S_{8} = \frac{2.5 ( 1. 4 ^{8} - 1 )}{1.4 - 1} = 85.9868 \dots S_{8} = 86.0 m^{3} min^{- 1} (3 s.f.) V = 85.9868 \dots \times 60 \times 24 = 123821 \dots V = 124000 m^{3} (3 s.f.)

Mentor tip

Carry the unrounded sum into (ii). Using $86.0$ gives $123840$ , which still rounds to $124000$ , but it is a good habit.

Question 11

8 marks

Bearings, an isosceles triangle and average speed

Geometry and trigonometry

(a)

1 mark

Bearings are measured clockwise from north. Magnetic north is $6°$ anticlockwise (west) of true north.

360° - 6° = 354°

Mentor tip

Bearings are always three figures and measured clockwise, so $6°$ west of north is $354°$ , not $6°$ .

(b)

2 marks

$A B = A C = 5$ km, so triangle $A B C$ is isosceles with the $6°$ angle at $A$ .
The two base angles are equal and share the rest of $180°$ .

\hat{A C B} = \frac{180° - 6°}{2} = 87°

(c)

3 marks

Use the cosine rule with the two $5$ km sides and the $6°$ angle between them.

C B^{2} = 5^{2} + 5^{2} - 2 (5) (5) cos 6° C B = 0.5233595 \dots C B = 0.523 km (3 s.f.)

Another way. Use the sine rule with the $87°$ angle from (b): $\frac{C B}{s i n 6°} = \frac{5}{s i n 87°}$ .

Mentor tip

Make sure your GDC is in degrees; in radians the answer comes out very different.

(d)

2 marks

Total distance is $5$ km plus your $C B$ . The time is $2$ hours $15$ minutes, which is $2.25$ hours.

speed = \frac{5 + 0.5233595 \dots}{2.25} = 2.45 km h^{- 1} (3 s.f.)

Mentor tip

$15$ minutes is $0.25$ hours, not $0.15$ . Convert minutes to hours before dividing.

Question 12

7 marks

The normal to a cubic and where it meets again

Calculus
Geometry and trigonometry
Functions

(a)

2 marks

Differentiate the cubic (or use the GDC's derivative at a point) and evaluate at $x = 6$ .
The normal is perpendicular to the tangent, so its gradient is the negative reciprocal.

f^{'} (x) = \frac{3}{2} x^{2} - \frac{27}{2} x + \frac{51}{2} f^{'} (6) = 54 - 81 + 25.5 = - 1.5 m_{normal} = - \frac{1}{- 1.5} = \frac{2}{3}

Mentor tip

Flip and change the sign. Using $- 1.5$ itself gives the tangent, not the normal.

(b)

1 mark

Use the point $(6, 5)$ with gradient $\frac{2}{3}$ .

y - 5 = \frac{2}{3} (x - 6) y = \frac{2}{3} x + 1

(c)

2 marks

Draw a straight line through $A (6, 5)$ that crosses the $y$ -axis at $(0, 1)$ .
It should rise gently, $2$ up for every $3$ across (for example through $(3, 3)$ and $(9, 7)$ ), and cut across the curve at right angles to it at $A$ .

line through (0, 1), (3, 3), (6, 5), (9, 7)

Mentor tip

Plot two or three exact points from your equation and join them with a ruler, rather than guessing a direction.

(d)

2 marks

Set the cubic equal to the normal and solve with the GDC (or use the intersect tool on both graphs).
One root is $x = 6$ , which is $A$ ; the other two give the points you need. Find each $y$ from the line.

\frac{1}{2} x^{3} - \frac{27}{4} x^{2} + \frac{51}{2} x - 13 = \frac{2}{3} x + 1 x = 0.684737 \dots, 6, 6.81526 \dots (GDC) (0.685, 1.46) and (6.82, 5.54)

Mentor tip

The normal crosses the curve twice more, once very close to $A$ . Zoom in near $x = 6$ so you do not miss the second point.

Our analysis

Where the marks are

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

Statistics and probability25 marks · 31%
Q1, Q3, Q4, Q8
Functions18 marks · 23%
Q5, Q6, Q7, Q12
Number and algebra16 marks · 20%
Q2, Q7, Q10
Geometry and trigonometry14 marks · 18%
Q2, Q11, Q12
Calculus7 marks · 9%
Q9, Q12

Mentor insights

Traps to watch for

01
Choosing two without replacement · Q3
Once one student is picked, only 199 remain. Squaring 178/200 treats it as with replacement, which is the wrong model even when the rounding hides it.
02
Rounding people up, not off · Q5
n > 7.5 means 8 people. Rounding to the nearest whole number or taking 7 breaks the cost condition.
03
One condition is not enough · Q7
The serve must clear the net and land before the end line. Check both with numbers, then give one clear conclusion.
04
Minutes are not decimals of an hour · Q11
2 hours 15 minutes is 2.25 hours. Writing 2.15 gives the wrong speed.
05
Tangent gradient versus normal gradient · Q12
f'(6) = -1.5 is the tangent. The normal needs the negative reciprocal, 2/3, before you write its equation.

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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