How many QCAA Specialist Mathematics questions cover Rates of change and differential equations?

AusGrader has 111 QCAA Specialist Mathematics questions on Rates of change and differential equations, all with instant AI grading and detailed marking feedback.

QCAA Specialist Mathematics — Rates of change and differential equations Questions & Answers

Q6

2025

QCAA

Paper 2

1 mark

Q6

1 mark

Determine the gradient of the tangent to $y^2 = 4x$ when $y = 1$ .

A

4

B

2

C

0.5

D

0.25

Reveal Answer

A

4

Incorrect. This is the derivative of the right side ( $4x$ ) with respect to $x$ , but it fails to apply implicit differentiation to the $y^2$ term.

B

2

Correct Answer

Correct. Using implicit differentiation, $2y \frac{dy}{dx} = 4$ , which simplifies to $\frac{dy}{dx} = \frac{2}{y}$ . Substituting $y = 1$ gives a gradient of 2.

C

0.5

Incorrect. This calculates $\frac{dx}{dy} = \frac{y}{2} = 0.5$ , which is the reciprocal of the gradient. The gradient of the tangent must be $\frac{dy}{dx}$ .

D

0.25

Incorrect. This is the $x$ -coordinate of the point on the curve when $y = 1$ (since $1^2 = 4x \implies x = 0.25$ ), not the gradient of the tangent line.

Q18

2021

QCAA

Paper 1

6 marks

Q18

6 marks

This differential equation can be used to determine the current $I$ (amperes) at time $t$ (seconds) with voltage $V$ (volts) in an electric circuit containing a resistance $R$ (ohms):

$k \frac{dI}{dt} + RI = V$

where $k$ , $R$ and $V$ are positive constants and $t \geq 0$ .

Assuming that there is no current in the electric circuit initially, show that the size of the current can never be greater than $\frac{V}{R}$ .

Reveal Answer

$k \frac{dI}{dt} + RI = V \Rightarrow k \frac{dI}{dt} = V - RI$
$\int \frac{k}{V - RI} dI = \int 1 dt$
$-\frac{k}{R} \ln|V - RI| = t + c$
Given $I = 0$ when $t = 0$
$c = -\frac{k}{R} \ln(V)$ (as $V > 0$ )
$-\frac{k}{R} \ln|V - RI| = t - \frac{k}{R} \ln(V)$
$\ln|V - RI| = -\frac{R}{k}t + \ln(V)$
$V - RI = e^{-\frac{R}{k}t + \ln(V)}$
$V - RI = V e^{-\frac{R}{k}t}$
For all $t, e^{-\frac{R}{k}t} > 0 \Rightarrow V - RI > 0 \Rightarrow V > RI \Rightarrow I < \frac{V}{R}$
So, the size of the current can never be greater than $\frac{V}{R}$ .

Marking Criteria

Descriptor	Marks
correctly uses the separation of variables method to set up indefinite integrals	1
develops a general solution of the differential equation	1
uses the given condition to determine expression for the constant of integration	1
rearranges relationship to express $\ln\|V - RI\|$ as the subject of the equation	1
expresses relationship as an exponential function	1
considers value of $I$ over time to determine the required limit	1

Q17

2025

QCAA

Paper 1

7 marks

Q17

7 marks

The radius of a cylinder decreases at a constant rate of $0.5 \text{ m s}^{-1}$ , while maintaining a constant height of four metres.

Given that the cylinder has an initial volume of $100\pi \text{ m}^3$ , determine the rate of change of the volume ( $\text{m}^3 \text{ s}^{-1}$ ) of the cylinder after four seconds.

Reveal Answer

Given $V = \pi r^2 h \Rightarrow V = 4\pi r^2$

Given $\frac{dr}{dt} = -0.5 \text{ m s}^{-1}$

$V = 4\pi r^2 \Rightarrow \frac{dV}{dt} = 8\pi r \frac{dr}{dt}$

At $t = 0$ , $V = 100\pi$ (given)

100\pi = 4\pi r^2 \Rightarrow r = 5 \text{ m} (\text{as } r > 0)

At $t = 4$ ,

r = 5 - 4 \times 0.5 = 3 \text{ m}

When $r = 3$ :

\frac{dV}{dt} = 8\pi \times 3 \times (-0.5) = -12\pi \text{ m}^3 \text{ s}^{-1}

Marking Criteria

Descriptor	Marks
correctly determines the rule for the volume of cylinder in terms of $r$	1
correctly expresses the rate of change of the radius as a mathematical expression	1
determines a general expression for $\frac{dV}{dt}$	1
determines initial value of $r$	1
determines value of $r$ after 4 seconds	1
determines value of $\frac{dV}{dt}$ when $t = 4$	1
shows logical organisation, having fully attempted the question	1

Q17

2020

QCAA

Paper 2

7 marks

Q17

7 marks

An object is released from rest at a height of 100 m above the ground.
The motion of the vertical descent of the object is modelled by

$v\frac{dv}{dx} = 9.8 - 0.004v^2 \quad (v \ge 0)$

where $v$ is the velocity (m s $^{-1}$ ) and $x$ is the displacement from the ground (m).
Determine the velocity of the object when it strikes the ground.

Reveal Answer

$v\frac{dv}{dx} = 9.8 - 0.004v^2, v > 0$
$\int \frac{v}{9.8 - 0.004v^2} dv = \int dx$
$\frac{-1}{0.008} \int \frac{-0.008v}{9.8 - 0.004v^2} dv = \int dx$
$-125 \ln|9.8 - 0.004v^2| = x + c$

Given $v = 0$ when $x = -100$
$-125 \ln|9.8| = -100 + c$
$c \approx -185.298$

$-125 \ln|9.8 - 0.004v^2| = x - 185.298$
Determining $v$ when $x = 0$
$-125 \ln|9.8 - 0.004v^2| = -185.298$

Using graph facility of GDC
$v \approx -36.7 \text{ ms}^{-1}$ or $v \approx 36.7 \text{ ms}^{-1}$
As $v > 0$ , the negative solution is rejected
$\therefore v \approx 36.7 \text{ ms}^{-1}$

Marking Criteria

Descriptor	Marks
correctly uses separation of variables	1
correctly develops the general solution of the differential equation	1
correctly uses the given position of the origin	1
uses the given condition to determine value for c	1
substitutes the displacement at impact to form an equation in terms of v	1
determines one reasonable solution of v	1
shows logical organisation communicating key steps	1

Q16

2023

QCAA

Paper 2

6 marks

Q16

6 marks

A curve modelled by the relation $xy^2 - y + \cos^{-1}(2x) = 1$ , where $-0.35 \leq x \leq 0.27$ and $0 \leq y \leq 1$ , intersects the $y$ -axis at point $A$ .

Determine the equation of the tangent to the curve at point $A$ .

Reveal Answer

Given $xy^2 - y + \cos^{-1}(2x) = 1$

Determining y-coordinate of A
$0 - y + \cos^{-1}(0) = 1 \Rightarrow y = 0.57$

$\frac{d}{dx}\cos^{-1}(2x) = \frac{-1}{\sqrt{0.25-x^2}}$

$\frac{d}{dx}(xy^2) = y^2 + 2xy\frac{dy}{dx}$

Determining $\frac{dy}{dx}$
$\frac{d}{dx}\left(xy^2 - y + \cos^{-1}(2x)\right) = \frac{d}{dx}(1)$
$y^2 + 2xy\frac{dy}{dx} - \frac{dy}{dx} + \frac{-1}{\sqrt{0.25-x^2}} = 0$
$\frac{dy}{dx}(2xy - 1) = \frac{1}{\sqrt{0.25-x^2}} - y^2$
$\frac{dy}{dx} = \frac{\frac{1}{\sqrt{0.25-x^2}} - y^2}{2xy - 1}$

Determining $\frac{dy}{dx}$ at A.
$\frac{dy}{dx} = \frac{\frac{1}{\sqrt{0.25}} - (0.571)^2}{-1} = -1.67$

Determining equation of tangent at A
$y = mx + c$
$y = -1.67x + 0.57$

Marking Criteria

Descriptor	Marks
correctly determines y-intercept	1
correctly determines $\frac{d}{dx}\cos^{-1}(2x)$	1
correctly determines $\frac{d}{dx}xy^2$	1
determines an expression for $\frac{dy}{dx}$ using a common factor	1
determines a value for $\frac{dy}{dx}$ at A	1
determines equation of the tangent at A	1

Q18

2023

QCAA

Paper 1

6 marks

Q18

6 marks

A particular solution to the differential equation $\frac{dy}{dx} = \frac{x}{(x^2+1)\tan(y)}$ , where $x \ge 0$ and $-\frac{\pi}{2} < y \le 0$ , passes through the origin.

Determine this solution in the form $x = f(y)$ . Leave your answer in simplified form.

Reveal Answer

$\frac{dy}{dx} = \frac{x}{(x^2+1)\tan(y)}$

$\int \tan(y) dy = \int \frac{x}{x^2+1} dx$

$-\int \frac{-\sin(y)}{\cos(y)} dy = \frac{1}{2} \int \frac{2x}{x^2+1} dx$

$-\ln|\cos(y)| = \frac{1}{2}\ln|x^2+1| + c$

Given $y=0$ when $x=0$ ,
$-\ln|\cos(0)| = \frac{1}{2}\ln|1| + c \implies c=0$

$\therefore \frac{1}{2}\ln|x^2+1| = -\ln|\cos(y)|$

$\ln\sqrt{x^2+1} = \ln\left|\frac{1}{\cos(y)}\right|$

$\sqrt{x^2+1} = |\sec(y)|$
$x^2+1 = \sec^2(y)$
$x^2 = \tan^2(y)$
$x = \pm \tan(y)$

As $x \ge 0, -\frac{\pi}{2} < y \le 0$
$x = -\tan(y)$

Marking Criteria

Descriptor	Marks
Correctly separates the variables	1
Applies suitable integration methods	1
Determines a value for the constant of integration	1
Determines an expression for a solution that does not contain logarithms	1
Expresses $x$ in terms of $y$	1
Evaluates the reasonableness of the results and expresses the solution in the form of $x = f(y)$ in simplified form	1

Q5

2020

QCAA

Paper 2

1 mark

Q5

1 mark

The gradient of the tangent at point A on the curve $y^2 = 4x$ is 1.36
The $x$ -coordinate of A is

A

0.12

B

0.46

C

0.54

D

1.47

Reveal Answer

A

0.12

This value is incorrect and does not satisfy the relationship between the gradient and the coordinates on the curve.

B

0.46

This incorrect value likely results from evaluating $\frac{m^2}{4}$ (approx $0.46$ ) instead of the correct relationship derived from the derivative.

C

0.54

Correct Answer

Differentiating $y^2=4x$ gives $\frac{dy}{dx} = \frac{2}{y}$ . Setting $\frac{2}{y} = 1.36$ yields $y \approx 1.47$ , and substituting this back into $x = \frac{y^2}{4}$ gives $x \approx 0.54$ .

D

1.47

This is the $y$ -coordinate of point A ( $y = \frac{2}{1.36} \approx 1.47$ ), but the question asks for the $x$ -coordinate.

Q7

2024

VCAA

Paper 1

4 marks

Q7

4 marks

Solve the differential equation $x + 2y\sqrt{x^2 + 1}\frac{dy}{dx} = 0$ , expressing $y$ as a function of $x$ , given that $y(0) = -2$ .

Reveal Answer

$2y\frac{dy}{dx} = \frac{-x}{\sqrt{x^2 + 1}}, 2y dy = -\frac{x}{\sqrt{x^2 + 1}} dx$

\begin{align*} \int 2y dy &= \int \frac{-x}{\sqrt{x^2 + 1}} dx\\ y^2 &= -\sqrt{x^2 + 1} + c\\ y(0) &= -2 \Rightarrow c = 5\\ y^2 &= 5 - \sqrt{x^2 + 1} \end{align*}

Thus $y = \pm\sqrt{5 - \sqrt{x^2 + 1}}$ . Since $y(0) = -2$ then $y = -\sqrt{5 - \sqrt{x^2 + 1}}$ .

Marking Criteria

Descriptor	Marks
Provides the correct expression for $y$	4
Performs correct integration and finds the constant of integration $c$	3
Performs correct integration of both sides	2
Separates variables correctly	1
None of the above	0

Q10

2022

VCAA

Paper 2

1 mark

Q10

1 mark

Consider the curve given by $5x^2y - 3xy + y^2 = 10$ .

The equation of the tangent to this curve at the point $(1, m)$ , where $m$ is a real constant, will have a negative gradient when

A

$m \in R \setminus [-1, 0]$

B

$m = -\sqrt{11} - 1$ only

C

$m \in R \setminus (-1, 0]$

D

$m = \sqrt{11} - 1$ only

E

$m = -\sqrt{11} - 1$ or $m = \sqrt{11} - 1$

Reveal Answer

A

$m \in R \setminus [-1, 0]$

While this is the range of $m$ values that yield a negative gradient, it fails to account for the fact that the point $(1, m)$ must actually lie on the given curve.

B

$m = -\sqrt{11} - 1$ only

This is only one of the two valid values for $m$ that lie on the curve and produce a negative gradient.

C

$m \in R \setminus (-1, 0]$

This inequality is incorrect for the gradient condition, and it also ignores that the point $(1, m)$ must satisfy the curve's equation.

D

$m = \sqrt{11} - 1$ only

This is only one of the two valid values for $m$ that lie on the curve and produce a negative gradient.

E

$m = -\sqrt{11} - 1$ or $m = \sqrt{11} - 1$

Correct Answer

Substituting $(1, m)$ into the curve's equation gives $m^2 + 2m - 10 = 0$ , yielding $m = -1 \pm \sqrt{11}$ . Both of these values satisfy the condition for a negative gradient, which is $\frac{dy}{dx} = \frac{-7m}{2m+2} < 0$ .

Q16

2025

QCAA

Paper 2

6 marks

Q16

6 marks

A certain population can be approximately modelled by the differential equation

\frac{dP}{dt} = 0.5P(1 - 0.2P)

where $P$ is the population in millions and $t$ is the number of years since 1 January 2025.

Given that the population on 1 January 2025 was estimated at 0.3 million, use a calculus approach to estimate the population on 1 January 2030.

Reveal Answer

Given $\frac{dP}{dt} = 0.5P(1 - 0.2P)$ :

\int \frac{1}{P(1 - 0.2P)} \ dP = \int 0.5 \ dt \quad \dots (1)

\begin{align*} \frac{1}{P(1 - 0.2P)} &= \frac{A}{P} + \frac{B}{(1 - 0.2P)} \quad \dots (2)\\ A(1 - 0.2P) + BP &= 1 \end{align*}

$P = 0 : A = 1$
$P = 5 : B = 0.2$

From (1) and (2):

\begin{align*} \int \frac{1}{P} + \frac{0.2}{(1 - 0.2P)} \ dP = \int 0.5 \ dt\\ \ln|P| - \ln|1 - 0.2P| = 0.5t + c \end{align*}

Given $P = 0.3$ when $t = 0$

\ln(0.3) - \ln(1 - 0.2 \times 0.3) = c \Rightarrow c \approx -1.142

$\ln(P) - \ln(1 - 0.2P) = 0.5t - 1.142$
When $t = 5$
$\ln(P) - \ln(1 - 0.2P) = 0.5 \times 5 - 1.142 = 1.358$

Using GDC: $P \approx 2.19$
The estimated population on 1 January 2030 is 2.2 million.

Marking Criteria

Descriptor	Marks
correctly separates the variables	1
uses partial fractions	1
uses suitable integration methods to determine a solution to the given differential equation	1
determines constant of integration	1
determines an equation in terms of $P$ whose solution represents the required population	1
estimates the required population	1

Q8

2022

QCAA

Paper 2

1 mark

Q8

1 mark

Determine the gradient of the tangent to the curve $y^2 - 3x = 5$ at the point $(1, 2\sqrt{2})$ .

A

0.41

B

0.53

C

1.06

D

8.49

Reveal Answer

A

0.41

This value is incorrect. It does not match the result derived from implicit differentiation.

B

0.53

Correct Answer

Differentiating $y^2 - 3x = 5$ implicitly gives $2y \frac{dy}{dx} - 3 = 0$ . Solving for the gradient yields $\frac{dy}{dx} = \frac{3}{2y}$ . Substituting $y = 2\sqrt{2}$ gives $\frac{3}{4\sqrt{2}} \approx 0.53$ .

C

1.06

This answer results from incorrectly differentiating $y^2$ as $y \frac{dy}{dx}$ instead of $2y \frac{dy}{dx}$ , leading to $\frac{3}{y} \approx 1.06$ .

D

8.49

This value is incorrect. It appears to result from multiplying $3$ by $y$ ( $3 \times 2\sqrt{2} \approx 8.49$ ) rather than dividing $3$ by $2y$ .

Q10

2020

VCAA

Paper 2

1 mark

Q10

1 mark

A tank initially contains 300 grams of salt that is dissolved in 50 L of water. A solution containing 15 grams of salt per litre of water is poured into the tank at a rate of 2 L per minute and the mixture in the tank is kept well stirred. At the same time, 5 L of the mixture flows out of the tank per minute.

A differential equation representing the mass, $m$ grams, of salt in the tank at time $t$ minutes, for a non-zero volume of mixture, is

A

$\frac{dm}{dt}=0$

B

$\frac{dm}{dt}=-\frac{5m}{50-5t}$

C

$\frac{dm}{dt}=30-\frac{m}{10}$

D

$\frac{dm}{dt}=30-\frac{5m}{50-3t}$

E

$\frac{dm}{dt}=30-\frac{5m}{50-5t}$

Reveal Answer

A

$\frac{dm}{dt}=0$

This implies the mass of salt is constant, completely ignoring both the inflow and outflow of salt from the tank.

B

$\frac{dm}{dt}=-\frac{5m}{50-5t}$

This ignores the inflow of salt ( $30$ g/min) and incorrectly calculates the volume as $50-5t$ instead of $50-3t$ .

C

$\frac{dm}{dt}=30-\frac{m}{10}$

This assumes the volume of the tank is constant at $50$ L, leading to an outflow rate of $\frac{5m}{50} = \frac{m}{10}$ , but the volume is actually decreasing by $3$ L/min.

D

$\frac{dm}{dt}=30-\frac{5m}{50-3t}$

Correct Answer

The rate of salt entering is $15 \times 2 = 30$ g/min, and the rate leaving is the outflow rate ( $5$ L/min) times the concentration $\frac{m}{50-3t}$ .

E

$\frac{dm}{dt}=30-\frac{5m}{50-5t}$

This incorrectly calculates the volume of the mixture at time $t$ as $50-5t$ , which only accounts for the outflow rate and ignores the $2$ L/min inflow.

Q4

2023

VCAA

Paper 1

3 marks

Q4

3 marks

Consider the relation $x \arcsin(y^2) = \pi$ .

Use implicit differentiation to find $\frac{dy}{dx}$ at the point $\left(6, \frac{1}{\sqrt{2}}\right)$ .

Give your answer in the form $-\frac{\pi\sqrt{a}}{b}$ , where $a, b \in Z^+$ .

Reveal Answer

Using implicit differentiation,

\arcsin(y^2) + x \times \frac{2y}{\sqrt{1-y^4}} \frac{dy}{dx} = 0

At the point $\left(6, \frac{1}{\sqrt{2}}\right)$ ,

\begin{align*} \arcsin\left(\frac{1}{2}\right) + 6 \times \frac{\sqrt{2}}{\sqrt{1-\frac{1}{4}}} \frac{dy}{dx} &= 0\\ \Rightarrow \frac{\pi}{6} + \frac{12\sqrt{2}}{\sqrt{3}} \frac{dy}{dx} &= 0 \end{align*}

and so

\frac{dy}{dx} = -\frac{\pi}{6} \times \frac{\sqrt{3}}{12\sqrt{2}} = -\frac{\pi}{6} \times \frac{\sqrt{6}}{24} = -\frac{\pi\sqrt{6}}{144}

Marking Criteria

Descriptor	Marks
Correctly differentiates the relation implicitly, demonstrating appropriate use of the product and chain rules (e.g. $\arcsin(y^2) + x \frac{2y}{\sqrt{1-y^4}} \frac{dy}{dx} = 0$ )	1
Correctly substitutes $x = 6$ and $y = \frac{1}{\sqrt{2}}$ into the differentiated equation	1
Correctly evaluates to find the final answer in the required form, $\frac{dy}{dx} = -\frac{\pi\sqrt{6}}{144}$	1

Q13

2024

SCSA

Paper 2

8 marks

Q13

A brumby is a free-roaming wild horse found in large numbers in parts of Australia. The culling of brumbies was banned in the year 2000. At this time the estimated population of brumbies in Kosciuszko National Park was 1600.

Scientists have modelled the population, $P(t)$ , of brumbies in Kosciuszko National Park $t$ years since the ban, by

\begin{align*} P(t) = \frac{18000}{10.25e^{-0.15t} + 1} \end{align*}

Q13c

It can be shown that the growth rate of the population of brumbies can be expressed as

$\frac{dP}{dt} = \frac{1}{r}P(k - P)$ .

Q13a

1 mark

Use the model to determine how long it will take the brumbies to increase to a number that is triple the number when the ban came into effect.

Reveal Answer

Solve when $P(t) = 4800\quad$ i.e. $4800 = \frac{18000}{10.25e^{-0.15t} + 1}$

From CAS $t = 8.7711\dots \text{years}$

Marking Criteria

Descriptor	Marks
solves for $t$ correctly	1

Q13b

2 marks

From this model, determine the estimated long run number of brumbies in Kosciuszko National Park.

Reveal Answer

Using $t \to \infty$ then $e^{-0.15t} \to 0,$ hence, $P(t) \to \frac{18000}{10.25(0) + 1} = 18000$ .

Hence in the long term, the limiting population will be 18 000 brumbies.

Marking Criteria

Descriptor	Marks
considers $t \to \infty$ to use $e^{-0.15t} \to 0$	1
determines the long run number of brumbies	1

Q13c

3 marks

Determine the values of the constants $r$ and $k$ .

Reveal Answer

$k = 18000$ as this is the limiting population.

$\therefore \frac{dP}{dt} = \frac{1}{r}P(18000 - P)$

Using $P(t) = \frac{18000}{10.25e^{-0.15t} + 1}$ from CAS $P'(0) = 218.666\dots$ (when $P = 1600$ )

Hence substituting into $\frac{dP}{dt} = \frac{1}{r}P(18000 - P)$

i.e. $218.666\dots = \frac{1}{r}(1600)(18000 - 1600)$

$\therefore r = 120000$

Marking Criteria

Descriptor	Marks
states the value of $k$ correctly	1
states the value of $r$ correctly	1
provides justification for the determination of the value for $r$	1

Q13d

2 marks

Determine the greatest growth rate for the population of brumbies.

Reveal Answer

Greatest rate of growth will occur when $P = 9000$ (half the limiting population)

Using $P = 9000, \quad \frac{dP}{dt} = \frac{1}{120000}(9000)(18000 - 9000) = 675 \text{ brumbies per year}$

Hence greatest growth rate will be 675 brumbies per year.

Marking Criteria

Descriptor	Marks
states that the maximum growth rate occurs when $P = 0.5k = 9000$	1
calculates the maximum growth rate correctly and states the correct units	1

Q1

2025

VCAA

Paper 1

4 marks

Q1

4 marks

Consider the curve with equation $xe^{-2y} + y^2e^x = 8e^4$ .

Find the equation of the tangent to the curve at the point $(4, -2)$ .

Reveal Answer

Method 1
$x\left(-2e^{-2y}\right)\frac{dy}{dx} + e^{-2y} + 2ye^x\frac{dy}{dx} + y^2e^x = 0$

$\frac{dy}{dx} = \frac{-e^{-2y} - y^2e^x}{-2xe^{-2y} + 2ye^x}$

At $(4, -2)$ :

$\frac{dy}{dx} = \frac{-e^4 - 4e^4}{-8e^4 - 4e^4}$

$\frac{dy}{dx} = \frac{5}{2}$

$y = \frac{5}{12}x - \frac{11}{3}$

Marking Criteria

Descriptor	Marks
Differentiates implicitly to obtain $x\left(-2e^{-2y}\right)\frac{dy}{dx} + e^{-2y} + 2ye^x\frac{dy}{dx} + y^2e^x = 0$ or equivalent	1
Rearranges the equation to make $\frac{dy}{dx}$ the subject	1
Substitutes $x=4$ and $y=-2$ to calculate the gradient $\frac{dy}{dx} = \frac{5}{12}$	1
Determines the correct equation of the tangent $y = \frac{5}{12}x - \frac{11}{3}$	1

QCAA Specialist Mathematics Rates of change and differential equations

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