PAT practice · 20 questions · Free
PAT Exercise Paper 6: Physics and Mathematics
Every question carries a worked solution behind a disclosure, so you commit to an answer before you see the key.
[01] Motion graphs
Question 1
A trolley is pulled along a straight horizontal track by a single horizontal force that stays parallel to the motion throughout. A graph of that force against the distance moved rises in a straight line from $0 \ \text{N}$ at $0 \ \text{m}$ to $12 \ \text{N}$ at $4 \ \text{m}$, and then stays constant at $12 \ \text{N}$ until the trolley has moved $9 \ \text{m}$. How much work does this force do over the $9 \ \text{m}$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $84 \ \text{J}$ (option C)
Key idea. The work done by a force that changes with position is the area under its force-distance graph. Sketch the graph the words describe: a triangle from $0$ to $4 \ \text{m}$ standing next to a rectangle from $4 \ \text{m}$ to $9 \ \text{m}$, then add the two areas.
Fastest approach
The area under a force-distance graph is the work done, so split the described shape in two.
Triangle, from $0$ to $4 \ \text{m}$: $\tfrac{1}{2} \times 4 \times 12 = 24 \ \text{J}$
Rectangle, from $4 \ \text{m}$ to $9 \ \text{m}$: $12 \times 5 = 60 \ \text{J}$
Total: $24 + 60 = 84 \ \text{J}$
Worth checking: the mean force over the whole $9 \ \text{m}$ is $84 / 9 \approx 9.3 \ \text{N}$, which sits sensibly between $0$ and $12 \ \text{N}$.
Common mistake. Multiplying the largest force by the whole distance, which is the area of a rectangle $12 \ \text{N}$ tall and $9 \ \text{m}$ wide. The force is only $12 \ \text{N}$ over the last $5 \ \text{m}$, and over the first $4 \ \text{m}$ it averages half of that.
Why the other options are wrong
- A: The triangle alone, stopping at $4 \ \text{m}$ and ignoring the rest of the journey. Incomplete Calculation
- B: The rectangle alone, ignoring the work done while the force was building up. Incomplete Calculation
- D: Taking the force as a constant $12 \ \text{N}$ over the full $9 \ \text{m}$. Force times distance has the dimensions of work, so the number looks respectable, but it is the area of the wrong shape. Conceptual Misunderstanding
- E: Taking $12 \ \text{N}$ across all $9 \ \text{m}$ and then adding the triangle on top, which counts the first $4 \ \text{m}$ twice. Formula Misapplication
[02] Electric circuits
Question 2
Three resistors of $6 \ \Omega$, $3 \ \Omega$ and $2 \ \Omega$ are connected in parallel across a $24 \ \text{V}$ battery. The battery has negligible internal resistance and the connecting wires have negligible resistance. What is the total current drawn from the battery?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $24 \ \text{A}$ (option D)
Key idea. Every branch is connected straight across the battery, so each carries the full $24 \ \text{V}$ and the branch currents simply add. The total is certainly more than the largest single branch current, which discards most of the alternatives before any arithmetic is done.
Fastest approach
Each branch has the whole $24 \ \text{V}$ across it, so take them one at a time.
$I = \frac{24}{6} + \frac{24}{3} + \frac{24}{2} = 4 + 8 + 12 = 24 \ \text{A}$
The same answer through the combined resistance: $\frac{1}{R} = \frac{1}{6} + \frac{1}{3} + \frac{1}{2} = 1$, so $R = 1 \ \Omega$ and $I = \frac{24}{1} = 24 \ \text{A}$.
Note that $1 \ \Omega$ is below the smallest of the three resistors. That is always true of a parallel combination, and it is the quickest check that the reciprocal formula was used the right way up.
Common mistake. Adding the three resistances as though they were in series. In parallel it is the reciprocals that add, and the combined resistance always comes out below the smallest resistor in the group.
Why the other options are wrong
- A: Adding the resistances in series, $6 + 3 + 2 = 11 \ \Omega$, and dividing: $24 / 11 \approx 2.2 \ \text{A}$. Formula Misapplication
- B: The current in the $6 \ \Omega$ branch alone, with the other two branches never counted. Incomplete Calculation
- C: The current in the $2 \ \Omega$ branch alone, on the view that the smallest resistor takes all of the current. It takes the largest share, not all of it. Conceptual Misunderstanding
- E: Adding the resistances and then taking the reciprocal of the sum as the combined resistance, giving $\tfrac{1}{11} \ \Omega$ and a current of $264 \ \text{A}$. Amperes are the right unit, and a domestic circuit carrying $264 \ \text{A}$ is not the right world. Conceptual Misunderstanding
[03] Waves
Question 3
A stone dropped into a still pond sends out ripples that travel outwards at $0.30 \ \text{m s}^{-1}$. A cork floating on the surface bobs up and down, completing $24$ full oscillations in one minute. Treat the ripples as a single travelling wave of one wavelength, and ignore the way real ripples spread out and die away. What is the distance between successive crests?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $0.75 \ \text{m}$ (option E)
Key idea. $v = f\lambda$, so $\lambda = v / f$. The only trap is the frequency: $24$ oscillations in a minute is $0.40 \ \text{Hz}$, not $24 \ \text{Hz}$, so convert before substituting.
Fastest approach
Frequency first: $24$ oscillations in $60 \ \text{s}$ gives $f = 0.40 \ \text{Hz}$.
$v = f\lambda \implies \lambda = \frac{v}{f} = \frac{0.30}{0.40} = 0.75 \ \text{m}$
Check through the period instead: $T = 1/f = 2.5 \ \text{s}$, and in $2.5 \ \text{s}$ the wave travels $0.30 \times 2.5 = 0.75 \ \text{m}$, which is one wavelength by definition.
Common mistake. Using $24$ as the frequency. The count is per minute, and a speed in metres per second needs a frequency in per second before the two are combined.
Why the other options are wrong
- A: Dividing the speed by the raw count, $0.30 / 24$, leaving the frequency in oscillations per minute. Unit Error
- B: Multiplying by the frequency instead of dividing. Metres per second times per second gives metres per second squared, which is an acceleration, not a length. Dimensional Error
- C: Reading the wavelength straight off the speed, which is only right if exactly one crest passes each second. Conceptual Misunderstanding
- D: Counting each up-and-down movement as two oscillations, giving $0.80 \ \text{Hz}$ and halving the wavelength. Misread Question
[04] Circular motion
Question 4
A grinding wheel of diameter $0.50 \ \text{m}$ spins steadily about its axis at an angular speed of $6.0 \ \text{rad s}^{-1}$. What is the magnitude of the centripetal acceleration of a point on its rim?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $9.0 \ \text{m s}^{-2}$ (option A)
Key idea. With an angular speed the centripetal acceleration is $a = \omega^{2} r$: the radius multiplies, it does not divide. Check the units of the answer before anything else, because most of the wrong routes here do not even come out in metres per second squared.
Fastest approach
The rim sits at $r = \frac{0.50}{2} = 0.25 \ \text{m}$.
$a = \omega^{2} r = (6.0)^{2} \times 0.25 = 36 \times 0.25 = 9.0 \ \text{m s}^{-2}$
The units settle the form: $\text{s}^{-2} \times \text{m}$ gives $\text{m s}^{-2}$, so the radius has to multiply.
The other form agrees. The rim speed is $v = \omega r = 1.5 \ \text{m s}^{-1}$, and $\frac{v^{2}}{r} = \frac{2.25}{0.25} = 9.0 \ \text{m s}^{-2}$.
Common mistake. Reaching for $v^{2}/r$ and putting the angular speed in place of the speed. The two forms are $a = v^{2}/r$ and $a = \omega^{2} r$, and mixing them divides by the radius where it should multiply, which is a factor of $r^{2}$ adrift.
Why the other options are wrong
- B: Using the diameter where the formula asks for the radius, so the answer is doubled. Misread Question
- C: Dividing the angular speed by the radius, $6.0 / 0.25$, which is neither of the two standard forms. Formula Misapplication
- D: Stopping at $\omega^{2} = 36$ and never multiplying by the radius. That leaves per second squared rather than metres per second squared, so it cannot be an acceleration whatever its size. Dimensional Error
- E: Using $v^{2}/r$ with the angular speed substituted for the speed: $36 / 0.25$. Conceptual Misunderstanding
[05] Optics
Question 5
A person stands $2.0 \ \text{m}$ in front of a flat vertical mirror fixed to a wall, then walks $0.50 \ \text{m}$ straight towards it along the normal to the mirror. By how much does the distance between the person and their image change?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $1.0 \ \text{m}$ (option B)
Key idea. A plane mirror places the image as far behind the glass as the object stands in front of it, so the person and their image are always separated by twice the person's distance from the mirror. That mirror symmetry does the whole job, and no ray tracing is needed.
Fastest approach
The image is always as far behind the mirror as the person is in front, so the person-to-image distance is twice the person-to-mirror distance.
Before: $2 \times 2.0 = 4.0 \ \text{m}$
After: $2 \times 1.5 = 3.0 \ \text{m}$
The gap closes by $1.0 \ \text{m}$, which is twice the $0.50 \ \text{m}$ walked: the person and the image approach the glass at the same rate, from opposite sides.
Common mistake. Assuming the image stays put while the person walks. Both move, towards each other, so the separation closes at twice the walking speed.
Why the other options are wrong
- A: Moving the person but leaving the image where it was, so the gap closes by only the $0.50 \ \text{m}$ walked. The image advances by the same amount at the same time. Conceptual Misunderstanding
- C: Reporting the person's new distance from the mirror rather than the change in the person-to-image distance. Misread Question
- D: Reporting the new person-to-image distance instead of how much it changed. Incomplete Calculation
- E: Reporting the original person-to-image distance instead of the change. Misread Question
[06] Forces and Newton's laws
Question 6
A block of mass $2.0 \ \text{kg}$ is pulled along a rough horizontal table by a light horizontal string. The tension in the string is $9.0 \ \text{N}$ and the table exerts a frictional force of $3.0 \ \text{N}$ on the block, opposing its motion. The block stays on the table throughout. Take $g = 10 \ \text{m s}^{-2}$. What is the magnitude of the block's acceleration?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $3.0 \ \text{m s}^{-2}$ (option C)
Key idea. Newton's second law takes the resultant force. Vertically the $20 \ \text{N}$ weight and the $20 \ \text{N}$ normal contact force cancel, which is the first law doing its work. Horizontally the two forces oppose one another, so check the sign before the magnitude: the resultant is $9.0 - 3.0 = 6.0 \ \text{N}$.
Fastest approach
1. Vertically.
Nothing accelerates up or down, so the weight $2.0 \times 10 = 20 \ \text{N}$ is balanced by a normal contact force of $20 \ \text{N}$.
2. Horizontally.
The tension pulls forwards and the friction pulls backwards, so the resultant is $9.0 - 3.0 = 6.0 \ \text{N}$ in the direction of motion.
3. Second law.
$a = \frac{F}{m} = \frac{6.0}{2.0} = 3.0 \ \text{m s}^{-2}$
Common mistake. Applying the second law to the tension alone. Friction acts on the block at the same time, and it is the resultant of all the horizontal forces that sets the acceleration.
Why the other options are wrong
- A: Dividing the resultant force of $6.0 \ \text{N}$ by the weight of $20 \ \text{N}$ rather than by the mass. A force divided by a force is a pure number and cannot be an acceleration. Dimensional Error
- B: Using the friction alone, $3.0 / 2.0$, which is the deceleration friction would produce with the string cut. Formula Misapplication
- D: Using the tension alone, $9.0 / 2.0$, so the second law is applied to one force instead of to the resultant. Incomplete Calculation
- E: Quoting the resultant force of $6.0 \ \text{N}$ as though it were the acceleration, which skips the division by the mass entirely. Conceptual Misunderstanding
[07] Simple machines
Question 7
A load of $600 \ \text{N}$ hangs from the lower block of a pulley system in which four sections of rope share the load. The free end of that rope is tied to a light rigid lever at a point $0.20 \ \text{m}$ from its pivot, and the effort is applied to the other end of the lever, $0.60 \ \text{m}$ from the pivot on the opposite side. The pulleys and the lever are light, the rope is inextensible, and friction everywhere is negligible. Mechanical advantage means the load divided by the effort. What is the mechanical advantage of the whole machine?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $12$ (option D)
Key idea. Stages of a machine multiply, because the second stage acts on what the first has already produced. Four rope sections share the load, so the rope tension is a quarter of it, and the lever then multiplies force again by the ratio of its arms. The energy audit is the check: an ideal machine gives out exactly the energy put in, so the effort must move twelve times as far as the load rises.
Fastest approach
1. The pulley stage.
Four sections of rope share the $600 \ \text{N}$ load, so the tension throughout the rope is $\frac{600}{4} = 150 \ \text{N}$.
2. The lever stage.
Take moments about the pivot, with the effort arm $0.60 \ \text{m}$ and the rope's arm $0.20 \ \text{m}$:
$E \times 0.60 = 150 \times 0.20 \implies E = 50 \ \text{N}$
3. The whole machine.
$\text{MA} = \frac{600}{50} = 12$, which is the pulley system's $4$ times the lever's $3$.
Energy check: with no friction the machine gives out exactly what is put in, so an effort twelve times smaller must move twelve times further. Nothing has been created, only traded.
Common mistake. Adding the two mechanical advantages instead of multiplying them. The lever acts on the force the pulley system has already reduced, so the two factors compound rather than accumulate.
Why the other options are wrong
- A: The lever alone, $0.60 / 0.20$, with the pulley system ignored. Incomplete Calculation
- B: The pulley system alone, from the four rope sections, with the lever ignored. Incomplete Calculation
- C: Adding the two stages, $4 + 3$, rather than multiplying them. Both factors are pure numbers so nothing looks wrong, but a machine whose stages merely added would need the second stage to act on the load rather than on the first stage's output. Conceptual Misunderstanding
- E: Measuring the effort arm from the rope's end of the lever rather than from the pivot, giving a lever ratio of $0.80 / 0.20 = 4$. Misread Question
[08] Estimation
Question 8
A school hall measures roughly $10 \ \text{m}$ by $10 \ \text{m}$, with a ceiling $2.5 \ \text{m}$ high. Treat it as an empty rectangular box of still air and take the density of air as $1.2 \ \text{kg m}^{-3}$. Estimate the mass of the air inside it, giving your answer to one significant figure.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $3 \times 10^{2} \ \text{kg}$ (option E)
Key idea. Mass is density times volume, and the volume of a box is the product of its three sides. Every input here is a rough figure, so one significant figure is the honest precision to report: carrying more digits would claim an accuracy the measurements never had.
Fastest approach
$V = 10 \times 10 \times 2.5 = 250 \ \text{m}^{3}$
$m = \rho V = 1.2 \times 250 = 300 \ \text{kg}$
To one significant figure, $3 \times 10^{2} \ \text{kg}$.
Worth pausing on the number: that is roughly the mass of four adults, hanging invisibly in the room. Air feels weightless because it pushes on every side of you at once, not because there is little of it.
Common mistake. Working from the floor area and forgetting the height, or reporting a string of digits the inputs cannot support. The room is measured to the nearest metre and the density is rounded, so one significant figure is as much as this estimate can carry.
Why the other options are wrong
- A: Reading the density as grams per cubic metre, which is a thousand times too small. Unit Error
- B: Adding the three dimensions rather than multiplying them, so the volume becomes $22.5$ instead of $250$. Formula Misapplication
- C: Multiplying the floor area of $100 \ \text{m}^{2}$ by the density. An area times a mass per unit volume is a mass per unit length, so the height has to appear somewhere for the answer to be a mass at all. Dimensional Error
- D: Dividing the volume by the density instead of multiplying, giving $250 / 1.2 \approx 210 \ \text{kg}$. Formula Misapplication
[09] Estimation and dimensional reasoning
Question 9
A drop of oil of volume $4.0 \times 10^{-11} \ \text{m}^{3}$ is placed on a large tray of clean water. It spreads out into a patch of area $2.0 \times 10^{-2} \ \text{m}^{2}$ and stops. Assume the patch is a layer exactly one molecule thick and that the molecules pack together with no gaps. Estimate the size of one oil molecule, giving your answer to one significant figure.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $2 \times 10^{-9} \ \text{m}$ (option A)
Key idea. A layer of volume $V$ spread over an area $A$ has thickness $V / A$, and that is the only way to combine a volume and an area to get a length. When the formula will not come to mind, ask what the units must be and the expression rebuilds itself.
Fastest approach
The oil keeps its volume and simply changes shape, so thickness is volume divided by area.
$d = \frac{4.0 \times 10^{-11}}{2.0 \times 10^{-2}} = 2.0 \times 10^{-9} \ \text{m}$
Do the digits and the powers of ten separately: $\frac{4.0}{2.0} = 2.0$, and $10^{-11 - (-2)} = 10^{-9}$.
A couple of nanometres is the right size for a molecule, and that is the check that matters most. An estimate of this kind earns its keep by landing where molecules actually live, not by its digits.
Common mistake. Dividing the area by the volume. The check takes a second: metres squared over metres cubed is one over metres, so that route is dead before any arithmetic is done.
Why the other options are wrong
- B: The right route with the powers of ten mishandled: $10^{-11}$ divided by $10^{-2}$ is $10^{-9}$, not $10^{-8}$. Order of Magnitude Error
- C: Taking the cube root of the drop's volume, which estimates the size of the drop itself, about a third of a millimetre, rather than the thickness of the layer it spreads into. Conceptual Misunderstanding
- D: Taking the square root of the patch's area, which gives how wide the patch is rather than how thick. Misread Question
- E: Dividing the area by the volume. That returns a reciprocal length, and quoting it in metres puts a molecule further across than the Earth is from the Moon. Dimensional Error
[10] Forces and Newton's laws
Question 10
An astronaut of mass $60 \ \text{kg}$ floats at rest beside a spacecraft of mass $3000 \ \text{kg}$, far from any star or planet, and pushes steadily on its hull for $2.0 \ \text{s}$. She leaves the hull moving at $0.40 \ \text{m s}^{-1}$, measured in the frame in which both were initially at rest. There is no tether and gravitational forces are negligible. What is the average force she exerts on the spacecraft?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $12 \ \text{N}$ (option B)
Key idea. By Newton's third law the force on the spacecraft is equal in size to the force on the astronaut, and that one follows from her own momentum change: $F = m\Delta v / \Delta t$. Several of the numbers on offer are an acceleration or an impulse dressed in newtons, so rule them out by dimensions before any arithmetic.
Fastest approach
Work with the astronaut, because all of her numbers are given.
$F = \frac{m\Delta v}{\Delta t} = \frac{60 \times 0.40}{2.0} = 12 \ \text{N}$
By Newton's third law the force she exerts on the hull has the same magnitude, $12 \ \text{N}$, in the opposite direction.
The spacecraft's mass is not needed for the force, and it is worth seeing what it does instead. The same $12 \ \text{N}$ acting for the same $2.0 \ \text{s}$ gives the spacecraft a speed of $\frac{60 \times 0.40}{3000} = 0.008 \ \text{m s}^{-1}$, fifty times less than hers. Equal forces, very unequal velocities.
Common mistake. Reaching for the spacecraft's mass because the question asks about the force on the spacecraft. The third law makes the two forces equal, so the body to work with is the one whose speed and mass are both given.
Why the other options are wrong
- A: Quoting the astronaut's acceleration, $0.40 / 2.0 = 0.20 \ \text{m s}^{-2}$, as the force, so her mass never enters at all. Dimensional Error
- C: Quoting the impulse, $60 \times 0.40 = 24 \ \text{N s}$. That is the momentum change, not the force that produced it, and it needs dividing by the $2.0 \ \text{s}$ the push lasted. Dimensional Error
- D: Multiplying by the contact time instead of dividing by it. Formula Misapplication
- E: Using the spacecraft's mass with the astronaut's change of velocity. The third law pairs the forces, not the velocity changes: the heavier body feels exactly the same force and therefore changes velocity far less. Conceptual Misunderstanding
[11] Series
Question 11
Pipes are stacked in horizontal rows. The bottom row holds $30$ pipes, every row above holds exactly one fewer than the row directly below it, and the top row holds $12$ pipes. Assuming no pipe is missing from any row, how many pipes are in the stack?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $399$ (option C)
Key idea. The row sizes run $30, 29, \ldots, 12$, an arithmetic progression. Pairing the bottom row with the top row gives $42$, and every such pair gives $42$, so the total is the number of rows times the average row of $21$. The step that decides the answer is the number of rows, and from $12$ to $30$ inclusive there are $19$ of them, not $18$.
Fastest approach
### 1. Rows present: $30 - 12 + 1 = 19$.
### 2. Average row: $\frac{30 + 12}{2} = 21$.
### 3. Total: $19 \times 21 = 399$
Test the inclusive count on a case small enough to see whole: rows of $3$, $2$ and $1$ are plainly three rows, and $3 - 1 + 1 = 3$ agrees. The same arithmetic on $30$ and $12$ is then safe.
Common mistake. Counting $30 - 12 = 18$ rows. Subtracting the two labels counts the gaps between rows rather than the rows, and the stack is one row taller than the number of gaps.
Why the other options are wrong
- A: Counting $30 - 12 = 18$ rows and multiplying by the correct average of $21$, one row short. Incomplete Calculation
- B: Computing $1 + 2 + \cdots + 30$ and subtracting $1 + 2 + \cdots + 12$, which strips off the top row of $12$ along with the rows that are not there. Formula Misapplication
- D: Continuing the pattern above the top row all the way to a row of one pipe, which is the sum $1 + 2 + \cdots + 30$. Misread Question
- E: Treating all $19$ rows as holding $30$ pipes each. Conceptual Misunderstanding
[12] Integration
Question 12
A particle moves along a straight line and never reverses. Its speed at time $t$ seconds is $v = 3t^{2} + 2$, measured in $\text{m s}^{-1}$. How far does it travel between $t = 0$ and $t = 3$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $33 \ \text{m}$ (option D)
Key idea. Distance is accumulated speed, so it is $\int_{0}^{3} v \, \mathrm{d}t$. The particle never reverses, which is what makes that integral the distance travelled rather than only the displacement.
Fastest approach
### 1. $\int_{0}^{3}\left(3t^{2} + 2\right)\mathrm{d}t = \left[t^{3} + 2t\right]_{0}^{3}$
### 2. $= (27 + 6) - 0 = 33 \ \text{m}$
### 3. Bracket it before trusting it. The speed rises from $2$ to $29 \ \text{m s}^{-1}$, so the distance is certainly more than $2 \times 3 = 6 \ \text{m}$ and certainly less than $29 \times 3 = 87 \ \text{m}$. Any answer outside that range is wrong whatever the algebra said, and the two bounds are cheap enough to write down first.
Common mistake. Multiplying one speed by the time. That is only valid at constant speed, and here it lands on the edge of the bracket rather than inside it.
Why the other options are wrong
- A: Integrating $3t^{2}$ to $\frac{t^{3}}{3}$, dividing by the old power instead of the new one. Formula Misapplication
- B: Differentiating the speed rather than integrating it, which gives the acceleration at $t = 3$. Conceptual Misunderstanding
- C: Integrating only $3t^{2}$ and dropping the constant term. Incomplete Calculation
- E: Taking the final speed of $29 \ \text{m s}^{-1}$ as though it held for all three seconds. Conceptual Misunderstanding
[13] Trigonometry
Question 13
As $\theta$ runs over every angle, the expression $7\sin\theta + 24\cos\theta$ takes a range of values. By how much does its greatest value exceed its least value?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $50$ (option E)
Key idea. Give the messy pair one name: $a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)$ with $R = \sqrt{a^{2} + b^{2}}$. The expression is then a single sine wave of amplitude $R$, running from $-R$ to $R$, so the gap between the extremes is $2R$.
Fastest approach
### 1. $R = \sqrt{7^{2} + 24^{2}} = \sqrt{49 + 576} = \sqrt{625} = 25$
### 2. Greatest value $25$, least value $-25$.
### 3. Difference: $2 \times 25 = 50$
The triple $7$, $24$, $25$ appearing in a question of this shape is the tell that harmonic form is wanted, exactly as $3$, $4$, $5$ is in a right-angled triangle.
Common mistake. Adding or subtracting the coefficients. The two terms peak at different angles, so neither $31$ nor $17$ is ever reached, and the least value is $-25$ rather than $0$.
Why the other options are wrong
- A: Subtracting the coefficients, as though the two terms partly cancelled at the maximum. Formula Misapplication
- B: Giving the greatest value alone, having taken the least value to be $0$. Incomplete Calculation
- C: Adding the coefficients, which assumes both terms peak at the same angle. Conceptual Misunderstanding
- D: Taking the amplitude to be the larger coefficient and doubling that. Conceptual Misunderstanding
[14] Probability
Question 14
A red die and a blue die, both fair and six-sided, are thrown together. What is the probability that the product of the two scores is odd?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\frac{1}{4}$ (option A)
Key idea. A product is odd only when every factor is odd, so the event is simply that both dice show odd numbers. The dice are distinguishable and fair, so all $36$ ordered outcomes are equally likely and nothing distinguishes one die from the other.
Fastest approach
### 1. A single even factor makes the product even, so both scores must be odd.
### 2. Each die is odd with probability $\frac{3}{6} = \frac{1}{2}$, and the two dice are independent.
### 3. $P = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
Counting instead gives the same thing: $3 \times 3 = 9$ favourable ordered pairs out of $6 \times 6 = 36$.
Common mistake. Assuming a product is as likely to be odd as even. Of the four parity combinations, three contain an even factor, so even products outnumber odd ones three to one.
Why the other options are wrong
- B: Treating the dice as indistinguishable, which gives $21$ unordered pairs with six of them odd and odd. Conceptual Misunderstanding
- C: Counting three kinds of outcome, odd with odd, one of each, and even with even, and treating those three as equally likely. Conceptual Misunderstanding
- D: Assuming a product is as likely to be odd as to be even. Conceptual Misunderstanding
- E: Finding the probability that the product is even. Misread Question
[15] Probability
Question 15
A biased spinner has four sectors labelled $1$, $2$, $3$ and $4$, and every spin lands on exactly one of them. The probability of landing on $1$ is $\frac{2}{5}$ and the probability of landing on $2$ is $\frac{1}{4}$. Landing on $3$ and landing on $4$ are equally likely. What is the probability of landing on $3$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\frac{7}{40}$ (option B)
Key idea. The four probabilities must total $1$, which fixes how much is left for sectors $3$ and $4$ together. Equal likelihood then splits that remainder into two.
Fastest approach
### 1. $P(1) + P(2) = \frac{2}{5} + \frac{1}{4} = \frac{8}{20} + \frac{5}{20} = \frac{13}{20}$
### 2. Left over: $1 - \frac{13}{20} = \frac{7}{20}$, shared between two equally likely sectors.
### 3. $P(3) = \frac{7}{40}$
Audit the whole thing against the total that has to hold: $\frac{16}{40} + \frac{10}{40} + \frac{7}{40} + \frac{7}{40} = \frac{40}{40} = 1$.
Common mistake. Stopping at $\frac{7}{20}$. That is the probability of landing on $3$ or on $4$, and only one of those two was asked for.
Why the other options are wrong
- A: Halving the remaining $\frac{7}{20}$ twice, once for each of the two sectors. Arithmetic Slip
- C: Assuming all four sectors are equally likely, which the two stated probabilities already rule out. Conceptual Misunderstanding
- D: Giving the probability of landing on $3$ or on $4$ without splitting it between them. Incomplete Calculation
- E: Treating $3$ and $4$ as the only outcomes left and sharing a probability of $1$ between them. Conceptual Misunderstanding
[16] Geometry
Question 16
The floor of a gazebo is a regular polygon with nine equal sides. Joining the centre of the floor to each corner cuts it into nine identical isosceles triangles. What is the size of one interior angle of the polygon?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $140^{\circ}$ (option C)
Key idea. The nine angles at the centre fill one full turn, so each is $40^{\circ}$. Each triangle is isosceles, and two of its base angles meet at every corner of the polygon, which is why an interior angle is twice a base angle rather than equal to one.
Fastest approach
### 1. Angle at the centre of one triangle: $\frac{360^{\circ}}{9} = 40^{\circ}$.
### 2. Its two base angles are equal: $\frac{180^{\circ} - 40^{\circ}}{2} = 70^{\circ}$ each.
### 3. Two base angles meet at each corner: $2 \times 70^{\circ} = 140^{\circ}$.
The angle-sum formula agrees, which is the point of sketching the decomposition rather than memorising it: $\frac{(9 - 2) \times 180^{\circ}}{9} = \frac{1260^{\circ}}{9} = 140^{\circ}$.
Common mistake. Stopping at $70^{\circ}$, the base angle of a single triangle. Every corner of the polygon is shared by two neighbouring triangles, so the interior angle there is the sum of two base angles.
Why the other options are wrong
- A: Giving the angle at the centre rather than the angle at a corner. Conceptual Misunderstanding
- B: Stopping at the base angle of one triangle, which is half the interior angle. Incomplete Calculation
- D: Using $(9 - 1) \times 180^{\circ}$ for the angle sum in place of $(9 - 2) \times 180^{\circ}$. Formula Misapplication
- E: Giving the sum of all nine interior angles instead of one of them. Misread Question
[17] Numerical reasoning
Question 17
A cubical tank of internal side $2 \ \text{m}$ is full of water and is emptied one bucketful at a time with a bucket holding $8$ litres. Take $1 \ \text{m}^{3}$ to be $1000$ litres, assume every bucketful comes out completely full and that nothing is spilled. How many bucketfuls are needed?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $1000$ (option D)
Key idea. Three separate steps, each finished before the next begins: the side is cubed to give $8 \ \text{m}^{3}$, the volume is converted once to $8000$ litres, and only then is it divided by the capacity of the bucket. Mixing any two of those steps in one line is what produces every wrong answer here.
Fastest approach
### 1. Volume: $2^{3} = 8 \ \text{m}^{3}$.
### 2. In litres: $8 \times 1000 = 8000$.
### 3. Bucketfuls: $\frac{8000}{8} = 1000$
The two eights are a coincidence of the numbers and not a cancellation: one is a volume in cubic metres, the other a capacity in litres, and dividing them before converting compares quantities that are a thousand-fold apart.
Common mistake. Dividing $8 \ \text{m}^{3}$ by $8$ litres and answering $1$. Until one of the two is converted, that division has no meaning, and the answer it gives is a thousand times too small.
Why the other options are wrong
- A: Dividing $8 \ \text{m}^{3}$ by $8$ litres with neither quantity converted. Unit Error
- B: Taking the volume as $2 \ \text{m}^{3}$, so that only one of the cube's three dimensions is used. Formula Misapplication
- C: Squaring the side rather than cubing it, giving a volume of $4 \ \text{m}^{3}$. Formula Misapplication
- E: Giving the volume in litres and never dividing by the size of the bucket. Incomplete Calculation
[18] Integration
Question 18
Evaluate $\displaystyle\int_{1}^{2}\left(\frac{6}{x^{2}} + 4x\right)\mathrm{d}x$.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $9$ (option E)
Key idea. Write $\frac{6}{x^{2}}$ as $6x^{-2}$. Raising the power by one gives $-1$ and dividing by $-1$ flips the sign, so the antiderivative is $-\frac{6}{x}$. That minus sign then decides both evaluations, and the value at the lower limit is negative.
Fastest approach
### 1. $\int\left(6x^{-2} + 4x\right)\mathrm{d}x = -\frac{6}{x} + 2x^{2}$
### 2. At $x = 2$: $-3 + 8 = 5$. At $x = 1$: $-6 + 2 = -4$.
### 3. $5 - (-4) = 9$
Check the sign before admiring the number. The integrand is positive throughout $1 \leqslant x \leqslant 2$, so the answer has to be positive, and subtracting a negative value at the lower limit is what makes it so.
Common mistake. Subtracting $4$ instead of $-4$ at the lower limit. The antiderivative is negative there, so subtracting it adds to the total rather than removing from it.
Why the other options are wrong
- A: Subtracting the value at the lower limit as $+4$ rather than $-4$, so the double negative is lost. Sign Error
- B: Integrating $6x^{-2}$ to $+\frac{6}{x}$, with the sign from the power rule the wrong way round. Formula Misapplication
- C: Evaluating at the upper limit only. Incomplete Calculation
- D: Integrating the $4x$ term and ignoring $\frac{6}{x^{2}}$ altogether. Incomplete Calculation
[19] Series
Question 19
A knockout tournament begins with $64$ players. Every match has exactly one winner, the loser takes no further part, and there are no byes and no drawn matches. How many matches are played altogether?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $63$ (option A)
Key idea. Solve it once in general and the numbers become a detail: every match removes exactly one player, and the tournament stops when all but the champion have been removed, so $n$ players always need $n - 1$ matches. The round-by-round series $32 + 16 + \cdots + 1$ gives the same total the longer way.
Fastest approach
### 1. Each match eliminates one player. Of the $64$ who enter, one is left at the end, so $63$ are eliminated.
### 2. One elimination per match gives $63$ matches.
### 3. By rounds, the same answer: $32 + 16 + 8 + 4 + 2 + 1 = 63$, a geometric series with first term $32$ and ratio $\frac{1}{2}$.
The general result $n - 1$ is the one worth keeping, because it survives byes and entry numbers that are not powers of two, where the round-by-round series does not.
Common mistake. Starting the round-by-round sum at $64$. The first round contains $64$ players and therefore $32$ matches, and adding a phantom round of $64$ matches puts the total one clear of the true value.
Why the other options are wrong
- B: Counting one match for each player, which credits the champion with a defeat that never happens. Conceptual Misunderstanding
- C: Counting each match once for each of the two players in it, so every match is counted twice. Conceptual Misunderstanding
- D: Summing $64 + 32 + 16 + 8 + 4 + 2 + 1$, which invents a first round of $64$ matches. Formula Misapplication
- E: Assuming all six rounds hold $32$ matches, as the first round does. Formula Misapplication
[20] Graphs
Question 20
How many vertical asymptotes does the curve $y = \dfrac{x^{2} - 4}{x^{2} - x - 6}$ have?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $1$ (option B)
Key idea. Factorise both parts before concluding anything: $\frac{(x-2)(x+2)}{(x-3)(x+2)}$. The factor $x + 2$ cancels, and a factor that cancels is one you should ask about rather than discard, because at $x = -2$ the curve has a single missing point rather than an asymptote. Only $x = 3$ sends the curve to infinity.
Fastest approach
### 1. $\frac{x^{2} - 4}{x^{2} - x - 6} = \frac{(x-2)(x+2)}{(x-3)(x+2)} = \frac{x-2}{x-3}$ for $x \neq -2$.
### 2. As $x$ approaches $-2$ the value approaches $\frac{-4}{-5} = \frac{4}{5}$, which is finite. That is a hole in the curve, not an asymptote.
### 3. At $x = 3$ the denominator vanishes while the numerator does not, so the curve runs off to infinity. One vertical asymptote.
The line $y = 1$ is a horizontal asymptote of the same curve, and is a different question.
Common mistake. Solving $x^{2} - x - 6 = 0$ and answering with the number of roots. A zero of the denominator is only an asymptote when the numerator does not vanish there as well.
Why the other options are wrong
- A: Taking a cancelling factor to remove every asymptote, leaving a curve defined everywhere. Conceptual Misunderstanding
- C: Counting both roots of $x^{2} - x - 6$ without noticing that $x + 2$ divides the numerator too. Incomplete Calculation
- D: Counting the two roots of the denominator and adding the horizontal asymptote $y = 1$. Misread Question
- E: Counting the zeros of the numerator as asymptotes alongside the zeros of the denominator. Conceptual Misunderstanding