PAT practice · 20 questions · Free
PAT Exercise Paper 4: 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 stone is released from rest and falls freely. Taking $g = 10 \ \text{m s}^{-2}$, ignoring air resistance and assuming the stone has not yet landed, how far does it fall during the third second of its motion, that is, between $t = 2 \ \text{s}$ and $t = 3 \ \text{s}$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $25 \ \text{m}$ (option A)
Key idea. Distance covered during the $n$th second is not a suvat answer on its own. It is the difference of two of them: $\tfrac{1}{2}gn^{2} - \tfrac{1}{2}g(n-1)^{2}$. Write that general expression once and every such question falls out of it.
Fastest approach
1. Displacement from release, as a function of time
Released from rest, so $s = \tfrac{1}{2}gt^{2}$.
$s(2) = \tfrac{1}{2}(10)(4) = 20 \ \text{m}$ $s(3) = \tfrac{1}{2}(10)(9) = 45 \ \text{m}$
2. Subtract
$45 - 20 = 25 \ \text{m}$
3. Or in general
$\tfrac{1}{2}g\left(n^{2} - (n-1)^{2}\right) = \tfrac{1}{2}g(2n-1)$, which for $n = 3$ gives $5 \times 5 = 25 \ \text{m}$. The distances covered in successive seconds are $5, 15, 25, 35, \ldots$, in the ratio of the odd numbers, which is worth recognising on sight.
Common mistake. Multiplying the speed at the end of the interval by one second. The stone is speeding up throughout that second, so no single speed multiplied by one second gives the distance; only the difference of two displacements does.
Why the other options are wrong
- B: Taking the speed at $t = 3 \ \text{s}$, which is $30 \ \text{m s}^{-1}$, and multiplying it by one second. That treats an accelerating body as though it had moved at its final speed for the whole interval, and a speed times a time is a length, so the units raise no objection. Conceptual Misunderstanding
- C: Reading 'the third second' as everything after the first second, giving $\tfrac{1}{2}g(3^{2} - 1^{2}) = 40 \ \text{m}$. Misread Question
- D: Reporting the total distance fallen in three seconds, $\tfrac{1}{2}g(3)^{2}$, rather than the distance covered in the last of them. Misread Question
- E: Using $s = gt^{2}$ over the whole three seconds, so the factor of one half is missing as well as the subtraction. Formula Misapplication
[02] Electric circuits
Question 2
A $12 \ \text{V}$ battery of negligible internal resistance is connected in series with a fixed $2.0 \ \text{k}\Omega$ resistor and a thermistor, using wires of negligible resistance. At the temperature of the room the thermistor has resistance $4.0 \ \text{k}\Omega$. A diode is connected in parallel with the fixed resistor, oriented so that it is reverse biased, and a reverse biased diode may be taken as an ideal open circuit. What is the potential difference across the fixed resistor?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $4 \ \text{V}$ (option B)
Key idea. Two components in series carry the same current, so they share the supply in the ratio of their resistances. The diode is reverse biased, carries nothing and can be deleted from the circuit before any arithmetic starts.
Fastest approach
1. Deal with the diode first
Reverse biased, so an ideal open circuit. It is a wire that is not there, and the circuit is a plain series pair.
2. Current round the loop
$I = \frac{12}{2000 + 4000} = 2 \times 10^{-3} \ \text{A}$
3. Potential difference across the fixed resistor
$V = IR = 2 \times 10^{-3} \times 2000 = 4 \ \text{V}$
As a check, the thermistor takes the other $8 \ \text{V}$, and $4 + 8 = 12$.
Common mistake. Assuming that a diode drawn in a circuit must be carrying current. Its orientation decides that, and one connected against the flow changes nothing at all. The other frequent error is splitting the supply evenly between two components whose resistances differ.
Why the other options are wrong
- A: Treating the reverse biased diode as a short circuit, which would hold both ends of the fixed resistor at the same potential. A diode conducts one way only, and this one is connected against it. Conceptual Misunderstanding
- C: Splitting the supply equally between the two components. That is right only when their resistances are equal, and here the thermistor's is twice the resistor's. Nothing about the units objects to a $6 \ \text{V}$ answer, so only the physics catches it. Conceptual Misunderstanding
- D: The potential difference across the thermistor rather than across the fixed resistor. Misread Question
- E: Finding the current from the fixed resistor alone, $12/2000$, then multiplying by $2000$ again, which returns the supply voltage whatever the resistance was. Formula Misapplication
[03] Waves
Question 3
A loudspeaker sounds a steady note of frequency $170 \ \text{Hz}$ into still air, in which sound travels at $340 \ \text{m s}^{-1}$. The wave is longitudinal: the air oscillates back and forth along the direction of travel, and no air is carried along with the wave, so energy moves outwards while the matter only moves to and fro about fixed positions. Measured along the direction of travel, what is the distance from a compression to the nearest rarefaction?
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 C)
Key idea. Compressions repeat once per wavelength, and a rarefaction sits exactly halfway between two neighbouring compressions. So the spacing asked for is $\tfrac{1}{2}\lambda$, with $\lambda = v/f$.
Fastest approach
1. Wavelength
$\lambda = \frac{v}{f} = \frac{340}{170} = 2.0 \ \text{m}$
2. Compression to the nearest rarefaction
Air is bunched at a compression and spread out at a rarefaction, and these alternate, so consecutive ones are half a wavelength apart.
$\tfrac{1}{2}\lambda = 1.0 \ \text{m}$
Nothing in this depends on how far the sound has travelled: the air itself only shuffles back and forth by a fraction of a millimetre, and it is the pattern that moves at $340 \ \text{m s}^{-1}$.
Common mistake. Reporting the wavelength itself, which is the distance from one compression to the next compression. The rarefaction lies between them.
Why the other options are wrong
- A: Inverting the wave equation to get $\lambda = f/v = 0.50 \ \text{m}$ and then halving it. A wavelength is a speed divided by a frequency; the other way round is not a length at all, so the units settle this one before the arithmetic does. Dimensional Error
- B: Taking a quarter of a wavelength, which is the node to antinode spacing on a standing wave. This wave is travelling, and a compression and the neighbouring rarefaction are half a wavelength apart. The answer is a perfectly respectable length, so only the picture of the wave rules it out. Conceptual Misunderstanding
- D: The wavelength itself, which is compression to the next compression rather than compression to the nearest rarefaction. Misread Question
- E: Using $\lambda = 2v/f$, borrowing the factor of two from the standing wave relation $\lambda = 2L/n$, which belongs to a string fixed at both ends and not to a travelling wave. Formula Misapplication
[04] Energy and power
Question 4
An electric pump raises $400 \ \text{kg}$ of water through a vertical height of $12 \ \text{m}$ in $60 \ \text{s}$, drawing $1.0 \ \text{kW}$ of electrical power while it does so. Take $g = 10 \ \text{m s}^{-2}$, assume the water starts and finishes at rest, and neglect friction in the pipework. What is the efficiency of the pump?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $80\%$ (option D)
Key idea. Efficiency is useful power out divided by total power in. The useful output here is the rate at which gravitational potential energy is gained, $mgh/t$, and it must be a power before it can be compared with a power.
Fastest approach
1. Useful energy transferred
$mgh = 400 \times 10 \times 12 = 48\,000 \ \text{J}$
2. Turn it into a power
$P_{\text{out}} = \frac{48\,000}{60} = 800 \ \text{W}$
3. Compare with the power drawn
$\frac{800}{1000} = 0.80 = 80\%$
The remaining $200 \ \text{W}$ leaves as heat and sound in the motor and the water.
Common mistake. Dividing input by output. The tell is immediate: the number comes out above $100\%$, and no machine returns more than it is given.
Why the other options are wrong
- A: Using the mass in place of the weight, so the useful power comes out as $400 \times 12 / 60 = 80 \ \text{W}$. A mass times a height is not an energy until it has been multiplied by $g$. Dimensional Error
- B: The percentage wasted rather than the percentage usefully transferred. Misread Question
- C: Dividing the energy transferred, $48 \ \text{kJ}$, by the power drawn, $1.0 \ \text{kW}$, and calling the result a percentage. An energy divided by a power is a time in seconds, not a fraction. Dimensional Error
- E: Dividing the input power by the useful output instead of the other way round. Both are powers, so the ratio is dimensionally faultless, and only the fact that it exceeds $100\%$ shows that the physics has been inverted. Conceptual Misunderstanding
[05] Gravity and orbits
Question 5
Two small satellites orbit the same planet in circular orbits. Satellite P has orbital radius $r$ and orbital period $2.0$ days. Satellite Q has orbital radius $4r$. Assume the planet's gravity is the only force acting on either satellite, and that each satellite's mass is far smaller than the planet's. What is the orbital period of satellite Q?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $16$ days (option E)
Key idea. Equating the gravitational pull to what circular motion demands, $\frac{GMm}{r^{2}} = \frac{4\pi^{2}mr}{T^{2}}$, gives $T^{2} = \frac{4\pi^{2}r^{3}}{GM}$. So $T \propto r^{3/2}$, and comparing two orbits about the same planet removes $G$, $M$ and the constant together.
Fastest approach
1. Set gravity equal to the centripetal requirement
$\frac{GMm}{r^{2}} = \frac{mv^{2}}{r}$, and with $v = \frac{2\pi r}{T}$ this rearranges to $T^{2} = \frac{4\pi^{2}r^{3}}{GM}$.
2. Take the ratio of the two orbits
$\frac{T_{Q}}{T_{P}} = \left(\frac{4r}{r}\right)^{3/2} = 4^{3/2} = 8$
3. Substitute once
$T_{Q} = 8 \times 2.0 = 16$ days
The satellite further out is slower and has further to go, and both effects lengthen the period, so the answer had to be more than four times two.
Common mistake. Reasoning that four times the radius means four times the period. The orbital speed falls as $r$ grows, so the period grows faster than the radius does, not in step with it.
Why the other options are wrong
- A: Assuming the more distant satellite is the faster one, so four times the radius means a quarter of the period. The orbital speed does fall with radius, but the circumference grows faster still, so the period must rise. A time is a perfectly good answer dimensionally, so only the physics rules this out. Conceptual Misunderstanding
- B: Using $T \propto r^{-1/2}$, which is how the orbital speed scales, in place of how the period scales. Formula Misapplication
- C: Using $T \propto \sqrt{r}$, which is what $T^{2} \propto r$ would give. The cube in Kepler's third law has been lost somewhere in the rearrangement. Formula Misapplication
- D: Taking the period to be proportional to the radius, so that four times the radius gives four times the period. That would need the orbital speed to be the same in both orbits. Conceptual Misunderstanding
[06] Radioactivity
Question 6
A thorium-232 nucleus, of proton number $90$, decays through a chain of alpha and beta-minus emissions and ends as lead-208, of proton number $82$. Assume the chain contains no other kind of emission, and that nucleon number and charge are conserved at every step. How many beta-minus particles are emitted 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: $4$ (option A)
Key idea. Only the alphas change the nucleon number, so nucleon number fixes how many there are before anything else is decided. Charge is then balanced separately: each alpha removes two protons and each beta-minus puts one back.
Fastest approach
1. Count the alphas from the nucleon number
A beta-minus emission leaves the nucleon number unchanged, so the whole drop belongs to the alphas.
$\frac{232 - 208}{4} = 6$ alpha particles
2. Balance the proton number
Six alphas remove $2 \times 6 = 12$ protons: $90 - 12 = 78$.
The nucleus actually ends with $82$, so the betas must supply $82 - 78 = 4$ protons, one each.
$4$ beta-minus particles
3. Check the whole equation
$^{232}_{\ 90}\text{Th} \to \ ^{208}_{\ 82}\text{Pb} + 6\,^{4}_{2}\alpha + 4\,^{\ \ 0}_{-1}\beta$, and both the top row and the bottom row balance.
Common mistake. Balancing the proton number on its own, as though the drop from $90$ to $82$ were the work of the betas. Beta-minus emission raises the proton number, so it cannot account for a fall in it.
Why the other options are wrong
- B: The number of alpha particles, $\frac{232-208}{4}$, rather than the number of beta particles. Misread Question
- C: Reading the drop in proton number, $90 - 82$, as the number of beta emissions. Beta-minus decay raises the proton number rather than lowering it, and the fall is the alphas' doing, so this counts the wrong particles in the wrong direction. It is a plain count either way, so nothing but the physics catches it. Conceptual Misunderstanding
- D: The number of protons carried away by the six alpha particles, $2 \times 6$, with the return step never taken. Incomplete Calculation
- E: The change in nucleon number, $232 - 208$, which counts nucleons rather than emissions. Misread Question
[07] Simple machines
Question 7
A light rigid plank rests across a pivot. A load of mass $60 \ \text{kg}$ sits on the plank $0.25 \ \text{m}$ from the pivot, and a person presses vertically downwards on the other side, $1.25 \ \text{m}$ from the pivot. Take $g = 10 \ \text{m s}^{-2}$, treat the plank itself as having negligible mass, and take both forces as vertical. What is the smallest downward force that holds the load in balance?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $120 \ \text{N}$ (option B)
Key idea. In equilibrium the moments about the pivot balance: effort times effort arm equals load times load arm. The longer the effort arm, the smaller the effort, which is the whole point of a lever.
Fastest approach
1. Weight of the load
$W = mg = 60 \times 10 = 600 \ \text{N}$
2. Moment of the load about the pivot
$600 \times 0.25 = 150 \ \text{N m}$
3. Divide by the effort arm
$F = \frac{150}{1.25} = 120 \ \text{N}$
Equivalently, the arms are in the ratio $1 : 5$, so the effort is a fifth of the load. A lever multiplies force by exactly the ratio of the arms and never by more.
Common mistake. Stopping at the moment, $150 \ \text{N m}$, and quoting it as the force. Dividing by the effort arm is the step that turns a moment back into a force, and the units say plainly whether it has been taken.
Why the other options are wrong
- A: Taking moments with the load's mass rather than its weight, giving $60 \times 0.25 / 1.25$. A moment is a force times a distance, and a mass times a distance is not one. Dimensional Error
- C: Reporting the load's moment about the pivot, $600 \times 0.25 = 150 \ \text{N m}$, as though it were the effort. The number is a moment, and it is one division away from being an answer. Dimensional Error
- D: The weight of the load itself, which is the force needed to hold it with no lever at all. Incomplete Calculation
- E: Multiplying by the ratio of the arms instead of dividing, $600 \times \frac{1.25}{0.25}$. That is a force, and a large one, so the units allow it; what it says is that pushing further from the pivot is harder, which is the opposite of what a lever does. Conceptual Misunderstanding
[08] Problem solving and estimation
Question 8
The speed $v$ of a transverse wave on a stretched string is believed to depend only on the tension $T$ in the string and on its mass per unit length $\mu$, through a relation of the form $v = kT^{a}\mu^{b}$ in which $k$ is a dimensionless constant. Experiment gives $k = 1$. A uniform string, whose stiffness may be neglected, is held at a tension of $40 \ \text{N}$ and has a mass per unit length of $0.40 \ \text{kg m}^{-1}$. What is the wave 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: $10 \ \text{m s}^{-1}$ (option C)
Key idea. $[T] = \text{M L T}^{-2}$ and $[\mu] = \text{M L}^{-1}$, so $T^{a}\mu^{b}$ carries $\text{M}^{a+b}\text{L}^{a-b}\text{T}^{-2a}$. Matching that to a speed, $\text{L T}^{-1}$, forces $a = \tfrac{1}{2}$ and $b = -\tfrac{1}{2}$, so $v = \sqrt{T/\mu}$.
Fastest approach
1. Write the dimensions of each side
$\text{L T}^{-1} = \left(\text{M L T}^{-2}\right)^{a}\left(\text{M L}^{-1}\right)^{b} = \text{M}^{a+b}\,\text{L}^{a-b}\,\text{T}^{-2a}$
2. Match each base dimension
Mass: $a + b = 0$ Length: $a - b = 1$ Time: $-2a = -1$
The time equation alone gives $a = \tfrac{1}{2}$, and then $b = -\tfrac{1}{2}$. The length equation is satisfied too, which is the sign that the assumed form was possible at all.
3. Substitute once, at the end
$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{40}{0.40}} = \sqrt{100} = 10 \ \text{m s}^{-1}$
Three unknowns and three equations is why this works, and why adding a fourth variable would leave the exponents undetermined.
Common mistake. Guessing $\sqrt{T\mu}$ because it looks symmetric. Dimensions settle it in one line: that combination carries a mass per unit time, which is not a speed, so it can be abandoned before any arithmetic is done.
Why the other options are wrong
- A: Solving the exponents with the roles of $T$ and $\mu$ exchanged, giving $\sqrt{\mu/T}$. That is the reciprocal of a speed, a time per unit length, so the units reject it outright. Dimensional Error
- B: Taking $\sqrt{T\mu}$, so both exponents come out positive. Its dimensions are those of a mass per unit time, which is why writing the dimensions down first is quicker than the arithmetic. Dimensional Error
- D: The right combination with a stray factor of two attached, borrowed from a formula such as $\lambda = 2L/n$. Dimensions fix the exponents and never the number in front, which is exactly why the question has to tell you that $k = 1$; a factor of two survives every dimensional check there is. Conceptual Misunderstanding
- E: Stopping at $T/\mu$ without taking the square root. That quantity is $100 \ \text{m}^{2}\text{s}^{-2}$, a speed squared, which the units announce. Formula Misapplication
[09] Friction and terminal velocity
Question 9
A spherical hailstone of mass $10 \ \text{g}$ falls vertically through still air. The area it presents to the oncoming air, its cross-sectional area, is $2.0 \ \text{cm}^{2}$, and the air has density $1.25 \ \text{kg m}^{-3}$. At these speeds the drag force is modelled as $F = \rho A v^{2}$, where $\rho$ is the density of the air, $A$ is that cross-sectional area and $v$ is the speed. Take $g = 10 \ \text{m s}^{-2}$, neglect upthrust, and assume the hailstone neither melts nor spins. What steady speed does it settle at?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $20 \ \text{m s}^{-1}$ (option D)
Key idea. The hailstone stops accelerating when the drag has grown to match the weight: $\rho A v^{2} = mg$. Everything after that is one rearrangement, one unit conversion and one square root.
Fastest approach
1. Convert before substituting
$m = 10 \ \text{g} = 0.010 \ \text{kg}$ and $A = 2.0 \ \text{cm}^{2} = 2.0 \times 10^{-4} \ \text{m}^{2}$.
2. Balance the two forces
$\rho A v^{2} = mg \implies v = \sqrt{\frac{mg}{\rho A}}$
3. Substitute once
$v = \sqrt{\frac{0.010 \times 10}{1.25 \times 2.0 \times 10^{-4}}} = \sqrt{\frac{0.10}{2.5 \times 10^{-4}}} = \sqrt{400} = 20 \ \text{m s}^{-1}$
About $45$ miles per hour, which is the right order for falling hail and is worth noticing as a check.
Common mistake. Leaving the area in square centimetres. The conversion is a factor of $10^{4}$ and it sits under a square root, so the answer comes out a hundred times too small, which is well outside anything a hailstone does.
Why the other options are wrong
- A: Leaving the area as $2.0$ in the formula, so square centimetres have been treated as square metres. The factor of $10^{4}$ enters under the square root, which turns it into a factor of a hundred in the speed. Unit Error
- B: Writing the weight as $m/g$ rather than $mg$. A mass divided by an acceleration is not a force, and it is being set equal to one. Dimensional Error
- C: Using the sphere's whole surface area, four times the cross-sectional area, in place of the area the oncoming air actually meets. Every dimension is untouched and the number is entirely plausible for a hailstone, so nothing but the physics catches this one. Conceptual Misunderstanding
- E: Stopping at $mg/\rho A$, which is $v^{2}$ rather than $v$. A hailstone at $400 \ \text{m s}^{-1}$ would be faster than a rifle bullet, so the magnitude gives it away as well as the units. Incomplete Calculation
[10] Problem solving and estimation
Question 10
Estimate the number of times a human heart beats in a lifetime. Take the rate as $70$ beats per minute throughout, take a life as $80$ years, and take a year as $5 \times 10^{5}$ minutes.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $2.8 \times 10^{9}$ (option E)
Key idea. Anything that accumulates is a rate multiplied by a duration. Here that is $70 \ \text{min}^{-1} \times 80 \times 5 \times 10^{5} \ \text{min}$, and the minutes cancelling is the check that the right two factors were multiplied.
Fastest approach
1. Total time lived, in minutes
$80 \times 5 \times 10^{5} = 4 \times 10^{7} \ \text{min}$
2. Multiply by the rate
$70 \times 4 \times 10^{7} = 2.8 \times 10^{9}$
3. Do the powers of ten separately from the digits
$7 \times 4 = 28$, and $10^{1} \times 10^{7} = 10^{8}$, giving $28 \times 10^{8} = 2.8 \times 10^{9}$. Keeping the exponent in its own column is what stops a factor of ten going missing in the last line.
A year is really about $5.3 \times 10^{5}$ minutes, so the figure to quote is a few times $10^{9}$, not $2.8 \times 10^{9}$ exactly.
Common mistake. Quoting the result to three or four figures. Every input is a round number chosen for convenience, so one significant figure, about $3 \times 10^{9}$, is all the estimate can support.
Why the other options are wrong
- A: The beats in a single year, $70 \times 5 \times 10^{5}$, with the multiplication by the eighty years never done. Incomplete Calculation
- B: The number of minutes in a lifetime, $5 \times 10^{5} \times 80$. That is the time lived rather than the number of beats, and the rate has been left out. Misread Question
- C: One power of ten lost in the last step: $3.5 \times 10^{7}$ multiplied by $80$ is $2.8 \times 10^{9}$, not $2.8 \times 10^{8}$. Order of Magnitude Error
- D: Using sixty years rather than the eighty the question states. The route and the units are right and only the input is wrong, which is why an estimate should always be repeated with the numbers read back off the question. Conceptual Misunderstanding
[11] Graphs
Question 11
A cubic curve crosses the $x$-axis at $x = -3$, $x = 1$ and $x = 4$, and it passes through the point $(0, 24)$. What is the value of $y$ on the curve 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: $-20$ (option A)
Key idea. Roots at $x = -3$, $x = 1$ and $x = 4$ force the form $y = k(x+3)(x-1)(x-4)$, and the point $(0, 24)$ fixes $k$. A sketch settles the sign before any arithmetic: between the roots at $1$ and $4$ the curve is below the axis, so the value at $x = 2$ has to be negative.
Fastest approach
### 1. Write the curve from its roots. A root at $x = a$ means a factor of $(x - a)$:
$y = k(x+3)(x-1)(x-4)$
### 2. Use the given point. At $x = 0$, $k(3)(-1)(-4) = 12k = 24$, so $k = 2$.
### 3. Substitute $x = 2$:
$y = 2(5)(1)(-2) = -20$
The sketch is the check. The leading coefficient is positive, so the curve rises for large $x$, and $x = 2$ lies between the roots at $1$ and $4$, where it dips below the axis.
Common mistake. Taking the curve to be $y = (x+3)(x-1)(x-4)$ and never using the point $(0, 24)$. Three roots fix the shape but not the vertical scale, and one extra point is exactly what fixes it.
Why the other options are wrong
- B: Leaving the scale factor out and evaluating $(x+3)(x-1)(x-4)$ at $x = 2$, which ignores the point the curve is told to pass through. Incomplete Calculation
- C: Scale factor found correctly, then $(2-4)$ evaluated as $+2$, which flips the sign of the whole product. Sign Error
- D: Quoting the value at $x = 0$ that the question supplies rather than the value at $x = 2$. Misread Question
- E: Writing the factors as $(x-3)(x+1)(x+4)$, copying the sign of each root straight into its bracket. That curve needs $k = -2$ to pass through $(0, 24)$ and gives $36$ at $x = 2$. Conceptual Misunderstanding
[12] Trigonometry
Question 12
You are given that $\sin(A - B) = \sin A \cos B - \cos A \sin B$. What is the exact value of $\sin 15^{\circ}$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\frac{\sqrt{6} - \sqrt{2}}{4}$ (option B)
Key idea. $15^{\circ} = 45^{\circ} - 30^{\circ}$, and both of those have exact values worth knowing. The formula then does the rest, and nothing has to leave surd form.
Fastest approach
### 1. Write $15^{\circ}$ as $45^{\circ} - 30^{\circ}$, since both are standard angles.
### 2. Substitute the exact values:
$\sin 15^{\circ} = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot\frac{1}{2}$
### 3. Collect over the common denominator $4$:
$\sin 15^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}$
Sanity check: $\sqrt{6} \approx 2.45$ and $\sqrt{2} \approx 1.41$, so the value is about $0.26$, which is the right size for a sine of a small angle.
Common mistake. Writing $\sin 15^{\circ} = \sin 45^{\circ} - \sin 30^{\circ}$. Sine is not additive over its argument, and that route gives about $0.21$ rather than $0.26$.
Why the other options are wrong
- A: Computing $\sin 45^{\circ} - \sin 30^{\circ}$, as though the sine of a difference were the difference of the sines. Conceptual Misunderstanding
- C: Factoring $\frac{\sqrt{2}}{2}$ out of both terms to leave $\cos 30^{\circ} - \sin 30^{\circ}$, then forgetting to multiply that factor back in. Incomplete Calculation
- D: Using the sum formula instead, which gives $\sin 75^{\circ}$, the sine of the complementary angle. Sign Error
- E: Adding the two sines. The result is bigger than $1$, which no sine can be, so this one can be rejected on sight. Conceptual Misunderstanding
[13] Integration
Question 13
A curve passes through the point $(1, 10)$, and for $x > 0$ its gradient is $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6\sqrt{x} + \dfrac{4}{x^{2}}$. What is the value of $y$ at $x = 4$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $41$ (option C)
Key idea. Integrating gives $y = 4x^{3/2} - \frac{4}{x} + c$: one unknown constant, and exactly one piece of information to fix it. Integrating between the limits $1$ and $4$ and adding the known value at $x = 1$ is the same calculation with the constant never written down.
Fastest approach
### 1. Integrate term by term. Raise each power by one and divide by the new power: $6x^{1/2}$ gives $4x^{3/2}$, and $4x^{-2}$ gives $-4x^{-1}$.
$y = 4x^{3/2} - \frac{4}{x} + c$
### 2. Use the point $(1, 10)$: $4 - 4 + c = 10$, so $c = 10$.
### 3. Substitute $x = 4$, using $4^{3/2} = 8$:
$y = 32 - 1 + 10 = 41$
The definite-integral route reaches the same place without ever naming $c$:
$y(4) = 10 + \int_{1}^{4}\left(6\sqrt{x} + \frac{4}{x^{2}}\right)\mathrm{d}x = 10 + 31 = 41$
Common mistake. Dropping the constant of integration. Without it the curve is forced through a point nobody chose, and the point $(1, 10)$ the question supplies is left unused.
Why the other options are wrong
- A: Integrating correctly but leaving the constant out, which forces the curve through $(1, 0)$ instead of $(1, 10)$. Incomplete Calculation
- B: Integrating $4x^{-2}$ as $+\frac{4}{x}$, losing the sign that dividing by the new power $-1$ produces. The constant then comes out as $2$. Sign Error
- D: Evaluating only the $4x^{3/2}$ term at $x = 4$ and forgetting the $-\frac{4}{x}$ term there, giving $32 + 10$. Arithmetic Slip
- E: Integrating $6x^{1/2}$ as $9x^{3/2}$, dividing by $\frac{2}{3}$ where the rule multiplies by it. Formula Misapplication
[14] Algebraic manipulation
Question 14
Make $u$ the subject of $v = \dfrac{3u + 2}{u - 4}$, and hence find $u$ when $v = 5$.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $11$ (option D)
Key idea. Multiply by the denominator, gather every term containing $u$ on one side and everything else on the other, then factorise: $u = \frac{4v + 2}{v - 3}$. Rearranging first and substituting last answers any other value of $v$ for free.
Fastest approach
### 1. Multiply both sides by $(u - 4)$, keeping the bracket whole:
$v(u - 4) = 3u + 2$
### 2. Expand, then collect the $u$ terms and factorise:
$vu - 4v = 3u + 2 \implies u(v - 3) = 4v + 2$
### 3. Divide, and only now substitute $v = 5$:
$u = \frac{4v + 2}{v - 3} = \frac{22}{2} = 11$
Check by putting $u = 11$ back into the original: $\frac{33 + 2}{11 - 4} = \frac{35}{7} = 5$.
Common mistake. Multiplying only the first term inside the bracket, writing $vu - 4$ instead of $vu - 4v$. Every term inside a bracket is multiplied, not just the one nearest the sign.
Why the other options are wrong
- A: Moving the $-20$ across as though it were positive, giving $2u = 2 - 20$. Sign Error
- B: Reaching $u(v - 3) = 4v + 2$ and then dividing the wrong way round, quoting $\frac{v - 3}{4v + 2}$. Formula Misapplication
- C: Multiplying only the first term in the bracket, so that $5u - 4 = 3u + 2$ and $u = 3$. Formula Misapplication
- E: Reaching $2u = 22$ and then multiplying by $2$ instead of dividing. Arithmetic Slip
[15] Series
Question 15
A geometric progression has first term $5$ and common ratio $2$. What is the sum of its first eight terms?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $1275$ (option E)
Key idea. $S_{n} = \frac{a(r^{n} - 1)}{r - 1}$. With $r = 2$ the denominator is $1$, so the sum is just $5(2^{8} - 1)$ and the whole question reduces to one power of two.
Fastest approach
$S_{8} = \frac{a(r^{8} - 1)}{r - 1} = \frac{5(2^{8} - 1)}{2 - 1} = 5 \times 255 = 1275$
Test the formula on a case you can add up by hand before trusting it on eight terms: for two terms it gives $5(2^{2} - 1) = 15$, and $5 + 10 = 15$.
A second check comes free from the doubling. Each term is bigger than everything before it put together, so the sum must be a little under twice the last term: $2 \times 640 = 1280$, and $1275$ sits just below it.
Common mistake. Quoting the eighth term, $5 \times 2^{7} = 640$, instead of the sum of the first eight. In a doubling series the last term is always about half the total, which is worth knowing as a check in both directions.
Why the other options are wrong
- A: Using the arithmetic sum formula, taking the gap between the first two terms as a common difference of $5$: $\frac{8}{2}(10 + 35) = 180$. Formula Misapplication
- B: Computing $2^{8} - 1$ and never multiplying by the first term. Incomplete Calculation
- C: Summing seven terms, $5(2^{7} - 1)$, from counting the doublings rather than the terms. Arithmetic Slip
- D: Reporting the eighth term, $5 \times 2^{7}$, rather than the sum of the eight. Misread Question
[16] Graphs
Question 16
For how many integer values of $x$ does the curve $y = (x^{2} - 1)(x^{2} - 9)$ lie below the $x$-axis?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $2$ (option A)
Key idea. The curve is even, so the sketch is symmetric about the $y$-axis, and its roots are $x = \pm 1$ and $x = \pm 3$. It is positive for large $|x|$, so the sign alternates inwards: negative between $1$ and $3$, positive between $-1$ and $1$, negative again between $-3$ and $-1$.
Fastest approach
### 1. A product is negative exactly when one factor is negative and the other is positive, which here means $1 < x^{2} < 9$.
### 2. That is $1 < |x| < 3$: two strips, one on each side of the origin.
### 3. The only integers inside them are $x = 2$ and $x = -2$, so the count is $2$.
The symmetry halves the work. Whatever happens to the right of the $y$-axis happens to the left, so count one strip and double.
Common mistake. Assuming the curve stays below the axis all the way from $-3$ to $3$. It comes back above between $-1$ and $1$, where both brackets are negative and their product is positive: at $x = 0$ the value is $9$.
Why the other options are wrong
- B: Counting the four values where the curve meets the axis, $x = \pm 1$ and $x = \pm 3$, rather than the values where it lies below it. Misread Question
- C: Taking the curve to be below the axis everywhere between its outermost roots and counting $-2, -1, 0, 1, 2$. Conceptual Misunderstanding
- D: Counting every integer with $1 \le |x| \le 3$, which sweeps in the four roots, where the curve sits on the axis rather than below it. Misread Question
- E: Counting every integer from $-3$ to $3$ inclusive. Conceptual Misunderstanding
[17] Numerical reasoning
Question 17
A grain of rice has a volume of about $20 \ \text{mm}^{3}$. Assuming the grains pack together with no gaps between them, estimate how many grains would fill a box of volume $1 \ \text{m}^{3}$.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $5 \times 10^{7}$ (option B)
Key idea. Count by dividing the big volume by the small one, but convert first. A metre is $10^{3}$ millimetres, so a cubic metre is $\left(10^{3}\right)^{3} = 10^{9}$ cubic millimetres: the conversion factor is cubed because the quantity is a volume.
Fastest approach
### 1. Convert the box: $1 \ \text{m}^{3} = \left(10^{3} \ \text{mm}\right)^{3} = 10^{9} \ \text{mm}^{3}$.
### 2. Divide the big volume by the small one, taking the powers of ten first:
$\frac{10^{9}}{20} = \frac{10^{9}}{2 \times 10^{1}} = 5 \times 10^{7}$
Worth saying out loud what the no-gaps assumption buys. Real grains poured into a box leave roughly a third of the space empty, so a number obtained this way is an upper bound, and the honest report is a few times $10^{7}$ rather than a figure quoted to three digits.
Common mistake. Converting $1 \ \text{m}^{3}$ into $10^{6} \ \text{mm}^{3}$. That is the conversion for an area, from square metres to square millimetres. A volume has three dimensions, each carrying its own factor of $10^{3}$.
Why the other options are wrong
- A: Using $1 \ \text{m}^{3} = 10^{6} \ \text{mm}^{3}$, cubing $10^{2}$ instead of $10^{3}$. Unit Error
- C: Dividing by $2$ rather than by $20$, which loses one power of ten from the grain volume. Order of Magnitude Error
- D: Quoting the volume of the box in cubic millimetres and never dividing by the volume of a grain at all. Incomplete Calculation
- E: Multiplying by the grain volume instead of dividing by it. Formula Misapplication
[18] Geometry
Question 18
A solid sphere is melted down and recast, with no material lost, as eight identical smaller solid spheres. What is the ratio of the radius of one small sphere to the radius of the original?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\frac{1}{2}$ (option C)
Key idea. Volume scales as the cube of a length, so a sphere holding one eighth of the material has $\sqrt[3]{\tfrac{1}{8}} = \tfrac{1}{2}$ of the radius. The factor $\frac{4}{3}\pi$ is common to both sides and cancels, which is why it never has to be written down at all.
Fastest approach
### 1. No material is lost, so eight small volumes make one original volume:
$8 \times \tfrac{4}{3}\pi r^{3} = \tfrac{4}{3}\pi R^{3}$
### 2. The constant cancels, leaving $8r^{3} = R^{3}$.
### 3. Take the cube root:
$\frac{r}{R} = \frac{1}{\sqrt[3]{8}} = \frac{1}{2}$
Check it from the other end: halving a radius divides the volume by $2^{3} = 8$, and eight of those pieces are exactly what the question makes.
Common mistake. Dividing the radius by $8$. That is the volume ratio applied to a length, and eight spheres of that size would hold only $\frac{1}{64}$ of the original material, so the sanity check fails immediately.
Why the other options are wrong
- A: Scaling the radius by the same factor as the volume. Eight spheres that small hold $\frac{1}{64}$ of the original material. Conceptual Misunderstanding
- B: Treating volume as proportional to the square of the radius, which gives $\frac{1}{\sqrt{8}}$ instead of the cube root. Formula Misapplication
- D: Giving the ratio the other way up, the original radius compared with the small one. Misread Question
- E: Quoting the ratio of the volumes rather than the ratio of the radii. Misread Question
[19] Exponential growth and decay
Question 19
A culture of bacteria doubles in number every $25$ minutes, and the doubling is assumed to continue unchecked with no cells dying. The culture starts with $4 \times 10^{3}$ cells. How many cells does it contain $2$ hours and $5$ minutes later?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $1.28 \times 10^{5}$ (option D)
Key idea. Two hours and five minutes is $125$ minutes, which is exactly five doubling periods, so the population is multiplied by $2^{5} = 32$. Growth like this multiplies by a fixed factor each period; it does not add a fixed amount.
Fastest approach
### 1. Convert the elapsed time to minutes: $2$ hours and $5$ minutes is $125$ minutes.
### 2. Count the doubling periods: $\frac{125}{25} = 5$.
### 3. Multiply, doing the power of ten separately from the digits:
$4 \times 10^{3} \times 2^{5} = (4 \times 32) \times 10^{3} = 128 \times 10^{3} = 1.28 \times 10^{5}$
Splitting the digits from the exponent keeps the arithmetic down to $4 \times 32$, which is the only real calculation in the question.
Common mistake. Multiplying by the number of doubling periods instead of raising two to that power. Five doublings multiply the population by $32$, not by $5$ and not by $10$.
Why the other options are wrong
- A: Treating the growth as linear and adding the starting $4 \times 10^{3}$ cells once every $25$ minutes. Conceptual Misunderstanding
- B: Multiplying by $2$ and then by $5$, rather than by $2^{5}$. Formula Misapplication
- C: Dropping the extra five minutes and using two hours, which contains only four complete doubling periods. Misread Question
- E: Counting six doublings, one for the start of the interval as well as one for the end of each period. Arithmetic Slip
[20] Binomial expansion
Question 20
What is the term independent of $x$ in the expansion of $\left(2x - \dfrac{3}{x^{2}}\right)^{6}$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $2160$ (option E)
Key idea. Write the general term $\binom{6}{k}(2x)^{6-k}\left(-\frac{3}{x^{2}}\right)^{k}$. Its power of $x$ is $(6 - k) - 2k = 6 - 3k$, and setting that to zero picks out $k = 2$, so only one term of the seven ever has to be evaluated.
Fastest approach
### 1. Write the general term:
$\binom{6}{k}(2x)^{6-k}\left(-\frac{3}{x^{2}}\right)^{k}$
### 2. Collect the power of $x$. The first bracket contributes $6 - k$ and the second contributes $-2k$, giving $6 - 3k$, which is zero when $k = 2$.
### 3. Evaluate that single term:
$\binom{6}{2}(2)^{4}(-3)^{2} = 15 \times 16 \times 9 = 2160$
The sign looks after itself: $k$ is even, so the minus inside the second bracket is squared away.
Common mistake. Assuming the constant term must be the middle one. That holds for an expansion where both parts carry the same power of $x$, and it fails as soon as one part carries $x^{-2}$, because each factor taken from that bracket lowers the power by two while each factor from the other raises it by one.
Why the other options are wrong
- A: Taking the middle term of the seven, $k = 3$, instead of solving $6 - 3k = 0$. That term is $\binom{6}{3}(2)^{3}(-3)^{3}$, and it carries $x^{-3}$ rather than no $x$ at all. Conceptual Misunderstanding
- B: Finding $k = 2$ correctly and then evaluating $2^{4} \times (-3)^{2}$ without the binomial coefficient. Incomplete Calculation
- C: Evaluating $\binom{6}{2}(2)^{4}$ and forgetting that the second bracket also contributes $(-3)^{2}$. Incomplete Calculation
- D: Using $(2)^{3}$ rather than $(2)^{4}$, from an off-by-one in the power $6 - k$ left over for the first bracket. Arithmetic Slip