PAT practice · 20 questions · Free
PAT Exercise Paper 7: Physics and Mathematics
Every question carries a worked solution behind a disclosure, so you commit to an answer before you see the key.
[01] Kinematics
Question 1
A ball is thrown horizontally at $20 \ \text{m s}^{-1}$ from the edge of a vertical cliff, from a point $45 \ \text{m}$ above level ground. Take $g = 10 \ \text{m s}^{-2}$, ignore air resistance, treat the ball as a point mass, and assume it lands on level ground at the foot of the cliff. How far from the base of the cliff does it land?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $60 \ \text{m}$ (option D)
Key idea. Gravity acts vertically, so it changes the vertical velocity and leaves the horizontal one alone. The horizontal equation $x = ut$ has two unknowns in it and cannot be solved by itself; the vertical equation $h = \tfrac{1}{2}gt^{2}$ has only one, so it is the one to write first.
Fastest approach
1. Count the unknowns before choosing an equation.
Horizontally the ball moves at a constant $20 \ \text{m s}^{-1}$, so $x = 20t$: two unknowns, one equation, and nothing can be solved yet.
Vertically the ball starts with no downward velocity, so $h = \tfrac{1}{2}gt^{2}$: one unknown. Start here.
2. Time from the vertical drop.
$45 = \tfrac{1}{2}(10)t^{2} \implies t^{2} = 9 \implies t = 3 \ \text{s}$
3. Horizontal distance from that time.
$x = 20 \times 3 = 60 \ \text{m}$
Worth noticing what never entered: the mass of the ball, and the horizontal speed while finding the time. The two motions share only the clock.
Common mistake. Letting the horizontal speed influence the fall time. It cannot: the ball hits the ground exactly when a ball simply dropped from the same point would, however fast it was thrown sideways. That is the whole content of treating the two motions independently.
Why the other options are wrong
- A: Treating the horizontal motion as if it also started from rest and accelerated, so using a mean horizontal speed of $10 \ \text{m s}^{-1}$ over the $3 \ \text{s}$. Nothing pushes the ball horizontally, so its horizontal speed never changes and there is no mean to take. Conceptual Misunderstanding
- B: Using $t = u/g = 2.0 \ \text{s}$, the time a ball thrown vertically upwards at $20 \ \text{m s}^{-1}$ takes to reach its highest point, in place of the time to fall $45 \ \text{m}$. Formula Misapplication
- C: Reporting the height of the cliff, that is, the vertical distance fallen rather than the horizontal distance travelled. Misread Question
- E: Taking the fall time as $2h/g = 9 \ \text{s}$ with the square root forgotten. The quantity $2h/g$ has the dimensions of time squared, so this route is wrong before its arithmetic is even looked at. Dimensional Error
[02] Electric fields and charge
Question 2
Two small charged spheres are held $30 \ \text{cm}$ apart in air, one carrying $+2.0 \ \mu\text{C}$ and the other $+3.0 \ \mu\text{C}$. Treat each sphere as a point charge, take the Coulomb constant as $9.0 \times 10^{9} \ \text{N m}^{2} \ \text{C}^{-2}$, and neglect any effect of the air and of the supports. What is the magnitude of the electrostatic force on the $2.0 \ \mu\text{C}$ sphere?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $0.60 \ \text{N}$ (option E)
Key idea. $F = \dfrac{kq_{1}q_{2}}{r^{2}}$. Convert the microcoulombs and the centimetres once, at the start, and the rest is one line. The two spheres attract or repel each other with forces of equal magnitude, so which sphere the question asks about makes no difference to the size of the answer.
Fastest approach
1. Convert every prefix first.
$q_{1} = 2.0 \times 10^{-6} \ \text{C}$, $q_{2} = 3.0 \times 10^{-6} \ \text{C}$, $r = 0.30 \ \text{m}$
2. Numerator, then denominator.
$kq_{1}q_{2} = 9.0 \times 10^{9} \times 6.0 \times 10^{-12} = 5.4 \times 10^{-2}$
$r^{2} = 0.30^{2} = 9.0 \times 10^{-2}$
3. Divide.
$F = \dfrac{5.4 \times 10^{-2}}{9.0 \times 10^{-2}} = 0.60 \ \text{N}$
The forces on the two spheres are a Newton's third law pair, so both have magnitude $0.60 \ \text{N}$ whichever sphere is named.
Common mistake. Leaving the separation in centimetres. Dividing by $30^{2} = 900$ instead of by $0.30^{2} = 0.09$ shrinks the answer by a factor of ten thousand, and the mistake is invisible in the final number because a force of any size looks respectable.
Why the other options are wrong
- A: Leaving the separation in centimetres, so dividing by $30^{2} = 900$ rather than by $0.30^{2} = 0.09$. Unit Error
- B: Computing $kq_{1}q_{2}$ and stopping, so the separation never enters. That product has the units $\text{N m}^{2}$, not newtons, which is the tell. Incomplete Calculation
- C: Dividing by $r$ rather than by $r^{2}$, giving $5.4 \times 10^{-2} / 0.30$. The result carries units of $\text{N m}$, so the route fails on dimensions whatever the arithmetic does. Dimensional Error
- D: Finding $0.60 \ \text{N}$ and then scaling it by $2/3$, on the view that the smaller charge feels the smaller share of the force. The two forces are a third law pair and are equal in magnitude however unequal the charges are. Conceptual Misunderstanding
[03] Electric circuits
Question 3
A lamp is marked $240 \ \text{V}, 60 \ \text{W}$, meaning that it dissipates $60 \ \text{W}$ when $240 \ \text{V}$ is placed across it. The same lamp is instead connected across a steady $120 \ \text{V}$ supply. Assume the resistance of the filament stays at the value it has at the marked rating, so the drop in filament temperature at the lower voltage is deliberately neglected, and take the connecting leads to have no resistance. What power does the lamp then dissipate?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $15 \ \text{W}$ (option A)
Key idea. With the resistance fixed, $P = \dfrac{V^{2}}{R}$, so $P \propto V^{2}$. Halving the voltage halves the current as well, and the power is the product of the two, so it falls by a factor of four rather than by a factor of two.
Fastest approach
1. Work with the ratio, not with the resistance.
$P \propto V^{2}$ at fixed $R$, so
$\dfrac{P_{\text{new}}}{P_{\text{rated}}} = \left(\dfrac{120}{240}\right)^{2} = \dfrac{1}{4}$
$P_{\text{new}} = \dfrac{60}{4} = 15 \ \text{W}$
2. The same answer the long way, as a check.
$R = \dfrac{V^{2}}{P} = \dfrac{240^{2}}{60} = 960 \ \Omega$, and then $P = \dfrac{120^{2}}{960} = 15 \ \text{W}$.
The marked rating is a statement about the lamp at $240 \ \text{V}$, not a property it carries with it onto any supply.
Common mistake. Halving the power because the voltage was halved. Two things halve at once, the current as well as the voltage, and power is their product, so the fall is by a factor of four.
Why the other options are wrong
- B: Scaling by the voltage ratio without squaring it, $60 \times \tfrac{1}{2}$. Equivalently, assuming the current stays at its rated $0.25 \ \text{A}$ when the voltage is halved. A watt is the right unit and the physics is still wrong: the current halves too. Conceptual Misunderstanding
- C: Quoting the marked power unchanged, treating $60 \ \text{W}$ as a fixed property of the lamp rather than as what it dissipates at its marked voltage. Conceptual Misunderstanding
- D: Inverting the voltage ratio and not squaring it, $60 \times \tfrac{240}{120}$. Formula Misapplication
- E: Inverting the voltage ratio and squaring it, $60 \times \left(\tfrac{240}{120}\right)^{2}$. Lowering the supply voltage cannot raise the power, which settles the direction before any arithmetic. Formula Misapplication
[04] Optics
Question 4
A transparent material has refractive index $2.0$, defined as the speed of light in a vacuum divided by the speed of light in the material. Take the speed of light in a vacuum as $3.0 \times 10^{8} \ \text{m s}^{-1}$, take the material to be uniform, and take the light to be a single colour so that dispersion can be ignored. What is the speed of light inside the material?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $1.5 \times 10^{8} \ \text{m s}^{-1}$ (option B)
Key idea. The definition is $n = \dfrac{c}{v}$, so $v = \dfrac{c}{n}$. Because $n$ is greater than one for any transparent material, the answer must come out below $c$, and any route producing a speed above $c$ can be discarded without further thought.
Fastest approach
1. Rearrange the definition given in the question.
$n = \dfrac{c}{v} \implies v = \dfrac{c}{n}$
2. Substitute.
$v = \dfrac{3.0 \times 10^{8}}{2.0} = 1.5 \times 10^{8} \ \text{m s}^{-1}$
3. Check it against the bound.
Light is slower in a material than in a vacuum, and $1.5 \times 10^{8}$ is comfortably below $3.0 \times 10^{8}$. Any candidate answer above $c$ is not a near miss, it is impossible.
Common mistake. Multiplying by the refractive index instead of dividing. The definition is written the other way up, and the result gives light travelling faster inside glass than in empty space, which the bound on $c$ rules out at a glance.
Why the other options are wrong
- A: Dividing by the refractive index twice, once for entering the material and once for leaving it. The index describes the speed inside the material, not a discount applied at each surface. Formula Misapplication
- C: Taking the speed as unchanged, on the view that the refractive index describes only how much a ray bends. It is defined by the speeds, and the bending is a consequence of the speeds differing. Conceptual Misunderstanding
- D: Multiplying by the index rather than dividing, $3.0 \times 10^{8} \times 2.0$. This puts light inside the material faster than light in a vacuum. Formula Misapplication
- E: Multiplying by the square of the index, $3.0 \times 10^{8} \times 2.0^{2}$. Formula Misapplication
[05] Circular motion
Question 5
A cyclist and her bicycle have a combined mass of $80 \ \text{kg}$. She rides round a bend of radius $20 \ \text{m}$ at a steady $5.0 \ \text{m s}^{-1}$. The road is horizontal and unbanked, so the only horizontal force on her is the friction between the tyres and the road surface. Take $g = 10 \ \text{m s}^{-2}$, treat cyclist and bicycle together as a single point mass, and neglect air resistance. What is the magnitude of that frictional force?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $100 \ \text{N}$ (option C)
Key idea. Centripetal force is a requirement, not a new kind of force: something real must supply it, and on a flat road that something is friction. So the frictional force equals $\dfrac{mv^{2}}{r}$, and checking that combination gives newtons is the fastest way to be sure the speed was squared and the radius divided.
Fastest approach
1. Name the real force first.
She turns, so she accelerates towards the centre of the bend. The road is flat, so the weight is vertical and is balanced by the normal contact force. The only horizontal force available is friction, and friction therefore supplies the whole centripetal requirement.
2. Check the dimensions before the value.
$\dfrac{\text{kg} \times (\text{m s}^{-1})^{2}}{\text{m}} = \text{kg m s}^{-2} = \text{N}$
Squaring the speed and dividing by the radius is what makes that work, so the arrangement is confirmed before a single digit is written.
3. Substitute.
$F = \dfrac{mv^{2}}{r} = \dfrac{80 \times 25}{20} = 100 \ \text{N}$
For comparison, her weight is $800 \ \text{N}$, so the tyres are being asked for about an eighth of it sideways. That is an easy demand on dry tarmac, which is the right feel for a gentle bend taken at $5 \ \text{m s}^{-1}$.
Common mistake. Reaching for the weight. The weight acts downwards and is cancelled by the road pushing up; it plays no part in turning the bicycle. What matters is the horizontal force, and on a flat road friction is the only candidate.
Why the other options are wrong
- A: Using $\dfrac{mv}{r}$ with the speed not squared, $\dfrac{80 \times 5}{20}$. That combination has the units $\text{kg s}^{-1}$, a mass flow rate, so it cannot be a force at all. Dimensional Error
- B: Carrying a factor of one half over from $\tfrac{1}{2}mv^{2}$, giving $\dfrac{mv^{2}}{2r}$. Kinetic energy has the one half; the centripetal force does not. Formula Misapplication
- D: Quoting the weight, $mg = 80 \times 10$. It is a force in newtons and it is the wrong force: it is vertical, it is balanced by the road, and it does nothing to turn the bicycle. Conceptual Misunderstanding
- E: Computing $mv^{2}$ and stopping, so the radius of the bend never enters. A bend of any radius would then need the same force, which cannot be right. Incomplete Calculation
[06] Problem solving and estimation
Question 6
The power $P$ a car must deliver against air resistance is modelled as depending only on the density $\rho$ of the air, the car's frontal cross-sectional area $A$ and its speed $v$, in the form $P = k\rho^{a}A^{b}v^{c}$, where $k$ is a dimensionless constant. Dimensions fix $a$, $b$ and $c$ but say nothing about $k$, and for this model $k = \tfrac{1}{2}$. A car of frontal area $2.0 \ \text{m}^{2}$ travels at a steady $30 \ \text{m s}^{-1}$ through air of density $1.2 \ \text{kg m}^{-3}$ along a level road. Take air resistance as the only resistance being considered, so rolling resistance and losses in the transmission are excluded. What power must be delivered against the air?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $32400 \ \text{W}$ (option D)
Key idea. Write the dimensions of both sides and match them one base quantity at a time. Power is $\text{M L}^{2}\text{T}^{-3}$, density is $\text{M L}^{-3}$, area is $\text{L}^{2}$ and speed is $\text{L T}^{-1}$, which forces $a = 1$, $b = 1$ and $c = 3$. The factor of $\tfrac{1}{2}$ cannot be recovered this way and has to be given, which is exactly what the question does.
Fastest approach
1. Match the dimensions, one base quantity at a time.
$\left[\rho^{a}A^{b}v^{c}\right] = \text{M}^{a}\,\text{L}^{-3a+2b+c}\,\text{T}^{-c}$, and $[P] = \text{M L}^{2}\text{T}^{-3}$.
Mass: $a = 1$. Time: $-c = -3$, so $c = 3$. Length: $-3 + 2b + 3 = 2$, so $b = 1$.
2. Assemble.
$P = \tfrac{1}{2}\rho A v^{3}$
3. Substitute.
$P = \tfrac{1}{2} \times 1.2 \times 2.0 \times 30^{3} = 0.6 \times 2.0 \times 27000 = 32400 \ \text{W}$
About $32 \ \text{kW}$, which is a plausible fraction of a family car's engine output at motorway speed. Note the cube: the same car at $60 \ \text{m s}^{-1}$ would need eight times this.
Common mistake. Believing the dimensional argument delivers the whole formula. It delivers the exponents and nothing else, because a dimensionless factor is invisible to it. Any numerical constant has to arrive from an experiment or from a derivation, and here it is handed over in the question.
Why the other options are wrong
- A: Using $v^{2}$ rather than $v^{3}$, which gives $\tfrac{1}{2}\rho A v^{2}$. That is the drag force in newtons, not the power, and the exponent matching rules it out at the time-dimension step. Dimensional Error
- B: Dropping the area, giving $\tfrac{1}{2}\rho v^{3}$. The result is a power per unit area in $\text{W m}^{-2}$, so a car of any size would need the same power. Incomplete Calculation
- C: Dropping the density, giving $\tfrac{1}{2}Av^{3}$. The mass dimension is then missing entirely, and the answer would be the same in air and in water. Incomplete Calculation
- E: Taking $\rho A v^{3}$ with the factor of $\tfrac{1}{2}$ discarded, on the view that dimensional analysis has already produced the complete formula. The dimensions are impeccable either way, which is precisely why they cannot settle it. Conceptual Misunderstanding
[07] Problem solving and estimation
Question 7
A sphere moving slowly through a liquid feels a drag force $F = 6\pi\eta rv$ opposing its motion, where $\eta$ is the viscosity of the liquid, $r$ is the radius of the sphere and $v$ is its speed. A small metal sphere of radius $2.0 \ \text{mm}$ and weight $3.6 \times 10^{-3} \ \text{N}$ is released in oil of viscosity $\eta = 0.50 \ \text{Pa s}$. Take $\pi$ as $3$, neglect the upthrust of the oil, and assume the sphere has been falling long enough to be moving at a steady speed. What is that steady speed?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $0.20 \ \text{m s}^{-1}$ (option E)
Key idea. Steady speed means zero acceleration, so the drag exactly balances the weight. Rearrange $W = 6\pi\eta rv$ symbolically to $v = \dfrac{W}{6\pi\eta r}$ before any number is substituted, because that final line can be checked dimensionally and a decimal cannot.
Fastest approach
1. Say what steady speed means.
No acceleration means no resultant force, so the upward drag equals the downward weight:
$W = 6\pi\eta rv$
2. Rearrange first, substitute second.
$v = \dfrac{W}{6\pi\eta r}$
Check the units of that line: $\dfrac{\text{N}}{\text{Pa s} \cdot \text{m}} = \dfrac{\text{N}}{\text{N m}^{-2}\,\text{s}\,\text{m}} = \text{m s}^{-1}$. The rearrangement is sound.
3. Substitute once, with the radius in metres.
$6\pi\eta r = 6 \times 3 \times 0.50 \times 2.0 \times 10^{-3} = 1.8 \times 10^{-2}$
$v = \dfrac{3.6 \times 10^{-3}}{1.8 \times 10^{-2}} = 0.20 \ \text{m s}^{-1}$
Nothing here needed any prior knowledge of viscous drag. The law was handed over in the question, and the physics used was that a body moving steadily has no resultant force on it.
Common mistake. Substituting the radius in millimetres. Every other quantity is in base units, so the one that is not drops the answer by a factor of a thousand, and there is nothing in the final number to say so.
Why the other options are wrong
- A: Multiplying by the radius instead of dividing by it, giving $\dfrac{Wr}{6\pi\eta}$. That expression has the units $\text{m}^{3}\text{s}^{-1}$, a volume flow rate, so it cannot be a speed. Dimensional Error
- B: Leaving the radius in millimetres, dividing by $2.0$ rather than by $2.0 \times 10^{-3}$. Unit Error
- C: Leaving the radius out of the denominator altogether, giving $\dfrac{W}{6\pi\eta}$. Incomplete Calculation
- D: Using the diameter of $4.0 \ \text{mm}$ where the law asks for the radius. The answer is a speed in the right units and is exactly half the truth, so only reading the law carefully catches it. Conceptual Misunderstanding
[08] Problem solving and estimation
Question 8
A car tyre wears down by $6.0 \ \text{mm}$ of tread over a tyre life of $60\,000 \ \text{km}$. Take the tyre's rolling circumference as $2.0 \ \text{m}$, assume it rolls without slipping so that one revolution advances the car by one circumference, and assume the wear is spread evenly over the whole life. Estimate the thickness of rubber lost from the tread in a single revolution.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $2.0 \times 10^{-10} \ \text{m}$ (option A)
Key idea. The question asks for a thickness per revolution, so the denominator is a number of revolutions and nothing else. Convert the life to metres, divide by the circumference to get that number, and only then divide the tread depth by it.
Fastest approach
1. Name the denominator.
Thickness per revolution. So the number of revolutions is what the tread depth must be divided by, not the number of kilometres and not the number of metres.
2. Count the revolutions, with both lengths in metres.
$60\,000 \ \text{km} = 6.0 \times 10^{7} \ \text{m}$
$N = \dfrac{6.0 \times 10^{7}}{2.0} = 3.0 \times 10^{7}$ revolutions
3. Divide.
$\dfrac{6.0 \times 10^{-3}}{3.0 \times 10^{7}} = 2.0 \times 10^{-10} \ \text{m}$
That is about the diameter of an atom, and it should be: rubber comes off a tyre a molecular layer at a time, not in visible flakes. An estimate landing on a scale you can name is an estimate worth some confidence.
Common mistake. Dividing by the wrong count. Dividing the tread depth by $60\,000$ answers a different question, the wear per kilometre, and the number that comes out is a thousand times too big for anything happening in one turn of a wheel.
Why the other options are wrong
- B: Slipping one power of ten in the revolution count, dividing by $3.0 \times 10^{6}$ instead of $3.0 \times 10^{7}$. Order of Magnitude Error
- C: Dividing the tread depth by $60\,000$, the number of kilometres, rather than by the number of revolutions. The answer is a length and it is the wear per kilometre, which is a different quantity from the one asked for. Conceptual Misunderstanding
- D: Reading $60\,000 \ \text{km}$ as $60\,000 \ \text{m}$, so counting $3.0 \times 10^{4}$ revolutions instead of $3.0 \times 10^{7}$. Unit Error
- E: Multiplying the tread depth by the number of revolutions instead of dividing. The result is $180 \ \text{km}$ of rubber from one tyre, which no amount of arithmetic can make plausible. Formula Misapplication
[09] Waves
Question 9
Two loudspeakers are driven in phase by the same signal generator, so they emit sound of the same single frequency exactly in step with one another. A listener stands at a point where the path from one loudspeaker is $0.25 \ \text{m}$ longer than the path from the other. Take the speed of sound in the air as $340 \ \text{m s}^{-1}$, assume the two waves arrive with equal amplitude, and ignore reflections from the walls and the floor. What is the lowest frequency at which the listener hears a minimum of sound?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $680 \ \text{Hz}$ (option B)
Key idea. Two sources in step cancel where one wave arrives half a cycle behind the other, that is where the path difference is an odd number of half wavelengths: $\Delta = \left(m + \tfrac{1}{2}\right)\lambda$. Write that general condition once, then read off the case that makes the wavelength largest, because the largest wavelength is the lowest frequency.
Fastest approach
1. Write the general condition.
The sources are in step, so a minimum needs
$\Delta = \left(m + \tfrac{1}{2}\right)\lambda, \qquad m = 0, 1, 2, \ldots$
2. Read off the special case wanted.
Lowest frequency means longest wavelength, which means the smallest allowed multiple, so $m = 0$ and
$\lambda = 2\Delta = 0.50 \ \text{m}$
3. Convert to a frequency.
$f = \dfrac{v}{\lambda} = \dfrac{340}{0.50} = 680 \ \text{Hz}$
Every other minimum comes from $m = 1, 2, \ldots$, and each of those gives a shorter wavelength and so a higher frequency. Writing the general condition first is what makes that obvious rather than something to be checked case by case.
Common mistake. Using the condition for a maximum. Sources in step reinforce where the path difference is a whole number of wavelengths and cancel where it is a half-integer number, and taking the path difference as one whole wavelength gives a frequency at which the listener hears the sound at its loudest, not at its quietest.
Why the other options are wrong
- A: Taking the path difference as a quarter of a wavelength, so $\lambda = 4\Delta = 1.0 \ \text{m}$. A quarter of a cycle is a phase difference of $90^{\circ}$, which does not cancel anything. Formula Misapplication
- C: Using the condition for a maximum, path difference equal to one whole wavelength, so $\lambda = 0.25 \ \text{m}$. The units are right and the frequency is real, but it is the frequency of the loudest point, not the quietest. Conceptual Misunderstanding
- D: Taking the next case up, $\Delta = \tfrac{3}{2}\lambda$. That is a genuine minimum, but the question asks for the lowest frequency and this one is three times it. Incomplete Calculation
- E: Inverting the half wavelength condition, taking the wavelength to be half the path difference rather than the path difference to be half a wavelength. Formula Misapplication
[10] Forces and equilibrium
Question 10
An astronaut together with her equipment has a total mass of $80 \ \text{kg}$. At the Moon's surface the gravitational field strength is $1.6 \ \text{N kg}^{-1}$; at the Earth's surface take it as $10 \ \text{N kg}^{-1}$. She is standing still on the Moon's surface. What is her total weight there?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $128 \ \text{N}$ (option C)
Key idea. Mass is how much matter there is and travels unchanged; weight is the force the local gravitational field exerts on that mass, $W = mg$, and changes with where the body is. The field strength quoted in $\text{N kg}^{-1}$ says outright how many newtons each kilogram attracts.
Fastest approach
1. Read the unit of the field strength.
$1.6 \ \text{N kg}^{-1}$ means each kilogram is pulled with $1.6 \ \text{N}$.
2. Multiply.
$W = mg = 80 \times 1.6 = 128 \ \text{N}$
3. Sanity check the size.
On Earth the same astronaut would weigh $80 \times 10 = 800 \ \text{N}$, so the Moon figure should be roughly a sixth of that. $128$ is close to $800/6 \approx 133$, so the answer sits where experience says it should. Her mass is $80 \ \text{kg}$ in both places: it is the weight that changed, not the amount of her.
Common mistake. Quoting the mass as though it were the weight. A kilogram is not a newton, and the two are different physical quantities rather than two units for one quantity. The giveaway is that the number would then be the same on the Moon, on Earth and in deep space, which is exactly what weight is not.
Why the other options are wrong
- A: Dividing the mass by the field strength, $80 / 1.6$, instead of multiplying. That combination has the units $\text{kg}^{2}\text{N}^{-1}$ and is not a force at all. Formula Misapplication
- B: Reporting the mass with newtons written after it, treating mass and weight as one quantity under two names. Conceptual Misunderstanding
- D: Using the Earth's field strength of $10 \ \text{N kg}^{-1}$, treating weight as a fixed property the astronaut carries with her. Weight is the force the local field exerts, so it is the field strength of the place she is standing that belongs in the calculation. Conceptual Misunderstanding
- E: Multiplying by $16$ rather than by $1.6$, a slipped power of ten. Arithmetic Slip
[11] Differentiation
Question 11
The curve $y = x(2x-3)^{2}$ is a cubic. What is the value of $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ at $x = 2$?
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 D)
Key idea. $x(2x-3)^{2}$ is a cubic in disguise. Multiplying out first gives $4x^{3} - 12x^{2} + 9x$, which differentiates term by term with nothing to remember and no rule to combine with another rule under pressure. Find the derivative as a function of $x$ first, and substitute $x = 2$ once, at the end.
Fastest approach
### 1. Expand the square: $(2x-3)^{2} = 4x^{2} - 12x + 9$.
### 2. Multiply through by $x$: $y = 4x^{3} - 12x^{2} + 9x$.
### 3. Differentiate term by term: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 12x^{2} - 24x + 9$.
### 4. Substitute $x = 2$: $48 - 48 + 9 = 9$.
The product rule gives the same value, provided the inner derivative of $2x-3$ survives: $(2x-3)^{2} + x \cdot 2(2x-3) \cdot 2$, which at $x = 2$ is $1 + 8 = 9$. Expanding first avoids needing it.
Common mistake. Differentiating the two factors separately and multiplying the results together. The derivative of a product is not the product of the derivatives, and multiplying the brackets out first removes the temptation entirely.
Why the other options are wrong
- A: Differentiating the factor $x$ to $1$ and leaving $(2x-3)^{2}$ untouched, so only one of the two product terms is written down: $(2\times 2-3)^{2} = 1$. Incomplete Calculation
- B: Evaluating $y$ at $x = 2$ rather than its derivative: $2 \times (4-3)^{2} = 2$. Misread Question
- C: Using the product rule but differentiating $(2x-3)^{2}$ as $2(2x-3)$, with the inner derivative of $2$ omitted, giving $1 + 4$. Formula Misapplication
- E: Expanding correctly to $4x^{3}-12x^{2}+9x$ and then differentiating the middle term to $-12x$ instead of $-24x$, giving $48 - 24 + 9$. Arithmetic Slip
[12] Polynomials and factorisation
Question 12
What is the value of $87^{2} - 13^{2}$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $7400$ (option E)
Key idea. Work the identity $a^{2} - b^{2} = (a-b)(a+b)$ in symbols and put the numbers in only at the end. Two three-digit squarings collapse into a subtraction and an addition, and here the addition lands on $100$, so the multiplication is free.
Fastest approach
### 1. Both terms are squares, so factorise rather than evaluate:
$a^{2} - b^{2} = (a-b)(a+b)$
### 2. With $a = 87$ and $b = 13$:
$87^{2} - 13^{2} = (87-13)(87+13) = 74 \times 100$
### 3. $= 7400$
The direct route, $7569 - 169$, reaches the same number and offers three more places for a slip to hide.
Common mistake. Writing $a^{2} - b^{2}$ as $(a-b)^{2}$. Squaring the difference gives $a^{2} - 2ab + b^{2}$, which is a different expression and agrees with the difference of two squares only when $b = 0$.
Why the other options are wrong
- A: Stopping at the first factor, $87 - 13$, without multiplying by the second. Incomplete Calculation
- B: Stopping at the second factor, $87 + 13$, without multiplying by the first. Incomplete Calculation
- C: Forming $74 \times 100$ correctly and then dropping a power of ten in the multiplication. Order of Magnitude Error
- D: Reading $a^{2} - b^{2}$ as $(a-b)^{2}$ and computing $74^{2}$. Formula Misapplication
[13] Circle geometry
Question 13
A circle has centre $O$ and radius $6$. The points $A$ and $B$ lie on the circle, and the angle $AOB$ is a right angle. The chord $AB$ divides the circle into two regions. What is the area of the smaller one?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $9\pi - 18$ (option A)
Key idea. A chord cuts a circle into two segments. The smaller one is what is left of the sector once the triangle $OAB$ has been taken out of it, so the answer is a quarter of the circle minus a right-angled triangle whose two perpendicular sides are both radii.
Fastest approach
### 1. A right angle at the centre is a quarter of a full turn, so the sector is a quarter of the circle:
$\frac{1}{4}\pi(6)^{2} = 9\pi$
### 2. Triangle $OAB$ has two perpendicular sides of length $6$, both radii:
$\frac{1}{2}(6)(6) = 18$
### 3. The segment is the sector with that triangle removed:
$9\pi - 18 \approx 10.3$
Sanity check on the size: the answer has to be smaller than the sector's $9\pi \approx 28.3$ and smaller than half the circle, and it is.
Common mistake. Subtracting nothing and reporting the sector. The sector between the two radii includes the triangle they bound, and the chord cuts precisely that triangle off.
Why the other options are wrong
- B: Reporting the area of triangle $OAB$ itself, rather than what is left of the sector when that triangle is removed. Misread Question
- C: Taking the triangle's area as $\frac{1}{2}(6)(3)$, halving one radius as though the height were half the base. Formula Misapplication
- D: Reporting the quarter-circle sector, with the triangle never subtracted. Incomplete Calculation
- E: Using the area of the whole circle, $36\pi$, in place of the quarter that a right angle at the centre cuts off, then subtracting the triangle from it. Conceptual Misunderstanding
[14] Probability
Question 14
A machine has two components. On any one run, component $X$ fails with probability $\frac{1}{5}$ and component $Y$ fails with probability $\frac{1}{4}$, and the two failures are independent. What is the probability that exactly one of the two components fails on a run?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\frac{7}{20}$ (option B)
Key idea. Exactly one failure happens in two ways that cannot both occur: $X$ fails while $Y$ survives, or $Y$ fails while $X$ survives. Independence lets each case be multiplied out, and the two cases exclude one another, so they are added. The answer is bracketed before it is computed: it must be less than the probability that at least one fails, and more than either single case on its own.
Fastest approach
### 1. $X$ fails and $Y$ survives:
$\frac{1}{5} \times \frac{3}{4} = \frac{3}{20}$
### 2. $Y$ fails and $X$ survives:
$\frac{4}{5} \times \frac{1}{4} = \frac{4}{20}$
### 3. These two cases cannot both happen, so add them:
$\frac{3}{20} + \frac{4}{20} = \frac{7}{20}$
Cross-check from the other end: at least one fails with probability $1 - \frac{4}{5}\times\frac{3}{4} = \frac{8}{20}$, and removing the $\frac{1}{20}$ where both fail leaves $\frac{7}{20}$.
Common mistake. Adding $\frac{1}{5}$ and $\frac{1}{4}$ directly. Probabilities may be added only when the events cannot both happen, and these can, so the sum counts the run where both components fail twice over.
Why the other options are wrong
- A: Computing $\frac{1}{5}\times\frac{1}{4}$, the probability that both components fail. Misread Question
- C: Computing $1 - \frac{4}{5}\times\frac{3}{4}$, the probability that at least one fails, which also counts the run where both do. Misread Question
- D: Adding $\frac{1}{5} + \frac{1}{4}$ as though the two failures were mutually exclusive, so the both-fail case is counted twice. Conceptual Misunderstanding
- E: Computing $\frac{4}{5}\times\frac{3}{4}$, the probability that neither component fails. Misread Question
[15] Integration
Question 15
The line $y = 2x$ meets the curve $y = 6x - x^{2}$ at two points. What is the area of the region enclosed between the line and the curve?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\frac{32}{3}$ (option C)
Key idea. The area between two graphs is the integral of the upper one minus the lower one, taken between the $x$-values where they cross. Subtract first and integrate once: the gap $\left(6x-x^{2}\right)-2x = 4x - x^{2}$ is simpler than either graph on its own.
Fastest approach
### 1. Where they meet: $6x - x^{2} = 2x$, so $4x - x^{2} = 0$ and $x = 0$ or $x = 4$.
### 2. Between those values the parabola is above the line, and the vertical gap is $4x - x^{2}$.
### 3. $\displaystyle\int_{0}^{4}\left(4x - x^{2}\right)\mathrm{d}x = \left[2x^{2} - \frac{x^{3}}{3}\right]_{0}^{4} = 32 - \frac{64}{3} = \frac{32}{3}$
Check it against a result worth carrying: the region a chord cuts off a parabola has area $\frac{2}{3}$ of the base times the greatest height. The base here is $4$ and the greatest gap is $4$, reached at $x = 2$, and $\frac{2}{3}(4)(4) = \frac{32}{3}$.
Common mistake. Integrating the curve on its own between the two crossing points and calling that the enclosed area. That measures the region all the way down to the $x$-axis, so it includes the triangle underneath the line as well.
Why the other options are wrong
- A: Integrating the line minus the curve, so the subtraction runs the wrong way round and the answer comes out negative. Sign Error
- B: Treating the region as a triangle of base $4$ and greatest height $4$, giving $\frac{1}{2}(4)(4)$, which understates a region with a curved top. Conceptual Misunderstanding
- D: Reporting $\int_{0}^{4}2x\,\mathrm{d}x$, the area under the line, rather than the area between the two graphs. Misread Question
- E: Integrating $6x - x^{2}$ from $0$ to $4$ and never subtracting the line. Incomplete Calculation
[16] Graphs
Question 16
The curve $y = ab^{x}$, where $a$ and $b$ are positive constants, passes through the points $(0, 3)$ and $(2, 12)$. What is the $y$-coordinate of the point on the curve where $x = 3$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $24$ (option D)
Key idea. At $x = 0$ the power is $1$, so the first point hands over $a = 3$ at once. Dividing the second point by the first removes $a$ altogether and leaves $b^{2} = 4$. Equal steps in $x$ multiply $y$ by the same factor; they do not add the same amount.
Fastest approach
### 1. $x = 0$ gives $y = a$, so $a = 3$.
### 2. Divide the second case by the first, which cancels $a$:
$\frac{12}{3} = b^{2}$, so $b^{2} = 4$ and $b = 2$, taking the positive root.
### 3. $y = 3 \times 2^{3} = 24$
Check: the values at $x = 0, 1, 2, 3$ run $3, 6, 12, 24$, each one double the last, and the given points sit correctly among them.
Common mistake. Reading the factor of $4$ between $x = 0$ and $x = 2$ as the factor for a single step in $x$. It covers two steps, so the one-step factor is its square root.
Why the other options are wrong
- A: Finding $b = 2$ but losing $a$, so the curve is read as $y = 2^{x}$ and $x = 3$ gives $8$. Incomplete Calculation
- B: Joining the two points with a straight line of gradient $\frac{12-3}{2} = 4.5$ and reading off $3 + 3(4.5)$. Conceptual Misunderstanding
- C: Adding the rise of $9$ again, treating equal steps in $x$ as equal additions to $y$ rather than equal multiplications. Conceptual Misunderstanding
- E: Using the two-step factor of $4$ as though it were the one-step factor, giving $12 \times 4$. Formula Misapplication
[17] Probability
Question 17
A biased coin shows heads with probability $\frac{2}{5}$. It is tossed three times, and the tosses are independent. What is the probability of getting at least one head?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\frac{98}{125}$ (option E)
Key idea. At least one head is everything except no heads at all, and no heads is a single case rather than three. So compute $\left(\frac{3}{5}\right)^{3}$ and subtract it from $1$. The limiting behaviour is the check worth making: as tosses are added the answer must climb towards $1$, and $1 - \left(\frac{3}{5}\right)^{n}$ does exactly that.
Fastest approach
### 1. The complement of at least one head is three tails in a row.
### 2. The tosses are independent, so
$P(\text{three tails}) = \left(\frac{3}{5}\right)^{3} = \frac{27}{125}$
### 3. $P(\text{at least one head}) = 1 - \frac{27}{125} = \frac{98}{125}$
The long way round agrees: exactly one, two and three heads give $\frac{54}{125}$, $\frac{36}{125}$ and $\frac{8}{125}$, which add to $\frac{98}{125}$ after three times the work.
Common mistake. Multiplying the single-toss probability by $3$. That gives $\frac{6}{5}$, which is greater than $1$ and so cannot be a probability at all: the three ways of getting a head overlap, so their probabilities may not simply be added.
Why the other options are wrong
- A: Computing $\left(\frac{2}{5}\right)^{3}$, the probability that all three tosses are heads. Misread Question
- B: Computing $\left(\frac{3}{5}\right)^{3}$, the probability of no heads, and stopping there instead of subtracting it from $1$. Incomplete Calculation
- C: Quoting the single-toss probability, as though the extra tosses made no difference. Conceptual Misunderstanding
- D: Computing $3\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)^{2}$, the probability of exactly one head, which leaves out two heads and three heads. Misread Question
[18] Circle geometry
Question 18
The points $A$ and $B$ are the two ends of a diameter of a circle, with $AB = 10$. The point $C$ lies on the circle, with $AC = 6$. What is the area of triangle $ABC$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $24$ (option A)
Key idea. An angle subtended at the circle by a diameter is a right angle, so the angle at $C$ is $90^{\circ}$. That makes $AC$ and $BC$ the two perpendicular sides and $AB$ the hypotenuse. Count what is known against what is needed: two lengths and no angle looks like too little, and the circle theorem is what closes the gap.
Fastest approach
### 1. $AB$ is a diameter, so the angle $ACB$ is a right angle.
### 2. Pythagoras supplies the third side:
$BC = \sqrt{10^{2} - 6^{2}} = \sqrt{64} = 8$
This is the $6, 8, 10$ triangle, a scaled $3, 4, 5$.
### 3. The two perpendicular sides are $6$ and $8$:
Area $= \frac{1}{2}(6)(8) = 24$
The hypotenuse is $AB$, and a hypotenuse is never one of the two lengths that multiply together in the area of a right-angled triangle.
Common mistake. Using the diameter as one of the two perpendicular sides. It is the hypotenuse of this triangle, so it is the one length of the three that does not appear in $\frac{1}{2} \times \text{base} \times \text{height}$.
Why the other options are wrong
- B: Using $\frac{1}{2}(6)(10)$, multiplying the given side by the hypotenuse instead of by the other perpendicular side. Formula Misapplication
- C: Using $\frac{1}{2}(8)(10)$, multiplying the computed side by the hypotenuse. Formula Misapplication
- D: Computing $6 \times 8$ and forgetting the factor of $\frac{1}{2}$. Incomplete Calculation
- E: Reporting the area of the circle, $\pi(5)^{2}$, rather than of the triangle inside it. Misread Question
[19] Differentiation
Question 19
Let $f(x) = 12\sqrt{x} + \dfrac{8}{\sqrt{x}}$ for $x > 0$. What is $f'(4)$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\frac{5}{2}$ (option B)
Key idea. Nothing here can be differentiated until it is written as a power: $12\sqrt{x} = 12x^{1/2}$ and $\frac{8}{\sqrt{x}} = 8x^{-1/2}$. Then $\frac{\mathrm{d}}{\mathrm{d}x}x^{n} = nx^{n-1}$ handles both. Multiplying a coefficient by a negative exponent flips its sign, so the two terms pull the gradient in opposite directions, and the sign of the second one is what the question is really about.
Fastest approach
### 1. Rewrite both terms as powers: $f(x) = 12x^{1/2} + 8x^{-1/2}$.
### 2. Differentiate, multiplying each coefficient by its exponent and dropping the exponent by one:
$f'(x) = 6x^{-1/2} - 4x^{-3/2}$
### 3. At $x = 4$: $x^{1/2} = 2$, so $x^{-1/2} = \frac{1}{2}$ and $x^{-3/2} = \frac{1}{8}$.
### 4. $f'(4) = 6\left(\frac{1}{2}\right) - 4\left(\frac{1}{8}\right) = 3 - \frac{1}{2} = \frac{5}{2}$
Check the sign before anything else: adding the second term instead of subtracting it moves the answer by a whole unit while leaving it looking perfectly respectable.
Common mistake. Differentiating $8x^{-1/2}$ to $+4x^{-3/2}$. Multiplying the coefficient by the exponent $-\frac{1}{2}$ makes it negative, and the exponent itself drops to $-\frac{3}{2}$ rather than rising.
Why the other options are wrong
- A: Differentiating $12x^{1/2}$ as $\frac{1}{2}x^{-1/2}$, dropping the coefficient $12$, while the second term is handled correctly: $\frac{1}{4} - \frac{1}{2}$. Formula Misapplication
- C: Differentiating the first term only and treating $\frac{8}{\sqrt{x}}$ as a constant. Incomplete Calculation
- D: Taking the derivative of $8x^{-1/2}$ to be $+4x^{-3/2}$, giving $3 + \frac{1}{2}$. Sign Error
- E: Evaluating $f(4) = 24 + 4$ rather than $f'(4)$. Misread Question
[20] Integration
Question 20
What is the area of the region enclosed between the curves $y = x^{2}$ and $y = 8 - x^{2}$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\frac{64}{3}$ (option C)
Key idea. The two parabolas meet where $x^{2} = 8 - x^{2}$, so at $x = \pm 2$. Between those the upper curve is $8 - x^{2}$ and the gap is $8 - 2x^{2}$, which is an even function. Nothing distinguishes the left half of the region from the right half, so integrating from $0$ to $2$ and doubling is the same as integrating from $-2$ to $2$, with smaller numbers.
Fastest approach
### 1. Where they meet: $x^{2} = 8 - x^{2}$, so $x^{2} = 4$ and $x = \pm 2$.
### 2. The vertical gap between them is $\left(8 - x^{2}\right) - x^{2} = 8 - 2x^{2}$.
### 3. The gap is unchanged by replacing $x$ with $-x$, so integrate over half the region and double:
$2\displaystyle\int_{0}^{2}\left(8 - 2x^{2}\right)\mathrm{d}x = 2\left[8x - \frac{2x^{3}}{3}\right]_{0}^{2} = 2\left(16 - \frac{16}{3}\right) = \frac{64}{3}$
Sanity check on the size: the region sits inside a rectangle $4$ wide and $8$ tall, of area $32$, and $\frac{64}{3} \approx 21.3$ fills about two thirds of it, which is what a lens between two parabolas should do.
Common mistake. Integrating from $0$ to $2$ and stopping there. The curves also meet at $x = -2$, and the piece on the left of the $y$-axis is the same size as the piece on the right, so an answer that forgets one of them is exactly half the truth.
Why the other options are wrong
- A: Integrating $x^{2}$ from $-2$ to $2$, which measures the area under the lower curve instead of the gap between the two. Misread Question
- B: Integrating the gap from $0$ to $2$ only, so the half of the region with $x < 0$ is missing. Incomplete Calculation
- D: Integrating $8 - x^{2}$ from $-2$ to $2$, the region under the upper curve alone, with the lower curve never subtracted. Incomplete Calculation
- E: Treating the region as the rectangle $4$ wide and $8$ tall that just contains it. Conceptual Misunderstanding