PAT practice · 20 questions · Free
PAT Exercise Paper 9: Physics and Mathematics
Every question carries a worked solution behind a disclosure, so you commit to an answer before you see the key.
[01] Energy, work and power
Question 1
A crate of mass $40 \ \text{kg}$ is dragged $10 \ \text{m}$ across a level floor at a steady speed. It is pulled by a rope held at $60^{\circ}$ above the horizontal, and the tension in the rope is $60 \ \text{N}$. The crate stays on the floor throughout, the floor is rough, and all distances are measured in the ground frame. Take $g = 10 \ \text{m s}^{-2}$, $\cos 60^{\circ} = 0.50$ and $\sin 60^{\circ} = 0.87$. How much work does the tension in the rope do on the crate?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $300 \ \text{J}$ (option A)
Key idea. Work is the force multiplied by the distance moved in the direction of that force, so only the horizontal component of the tension counts: $W = Fd\cos\theta$. Two special cases settle which trigonometric function belongs there. A rope pulled straight along the floor, $\theta = 0$, must give the full $Fd$, and $\cos 0 = 1$ does. A rope pulled straight up, $\theta = 90^{\circ}$, must do no work at all on a crate that only moves sideways, and $\cos 90^{\circ} = 0$ does. Sine fails both tests.
Fastest approach
1. Resolve the tension along the motion.
The crate moves horizontally, so the useful component is $T\cos\theta = 60 \times 0.50 = 30 \ \text{N}$.
2. Multiply by the distance moved.
$W = 30 \times 10 = 300 \ \text{J}$
3. Notice what was never needed.
The mass and $g$ appear nowhere in that calculation. The weight of the crate points straight down, the crate moves straight sideways, and a force perpendicular to the displacement does no work whatever its size. The normal contact force is in the same position. Both are in the stem so that the route which multiplies them together has a number to land on.
Common mistake. Reaching for $\sin\theta$ because the angle is measured from the horizontal. The rule is not about which line the angle is measured from: it is that the work done uses the component of the force along the displacement, and here that component is adjacent to the angle.
Why the other options are wrong
- B: Using $\sin 60^{\circ}$ instead of $\cos 60^{\circ}$, so the vertical component of the tension is credited with the work: $60 \times 0.87 \times 10$. That component is perpendicular to the motion and does none. Formula Misapplication
- C: Multiplying the full tension by the full distance and ignoring the angle altogether: $60 \times 10$. Incomplete Calculation
- D: Dividing by $\cos 60^{\circ}$ rather than multiplying by it, giving $600 / 0.50$. The answer then exceeds what the rope could do even if it were pulled straight along the floor, which is impossible. Formula Misapplication
- E: Taking the weight as the force that does the work: $mgd = 40 \times 10 \times 10$. Newtons times metres is joules, so the units raise no objection, but the weight acts vertically while the crate moves horizontally, and a force at right angles to the displacement does no work at all. Conceptual Misunderstanding
[02] Momentum and collisions
Question 2
On a horizontal air track a $2.0 \ \text{kg}$ glider moving at $6.0 \ \text{m s}^{-1}$ strikes a stationary $1.0 \ \text{kg}$ glider head on. After the collision the $2.0 \ \text{kg}$ glider is still moving in its original direction at $3.0 \ \text{m s}^{-1}$. Take the track as frictionless, ignore air resistance, and measure every velocity in the track's frame. What fraction of the total kinetic energy is lost in the collision?
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 B)
Key idea. Momentum is conserved in every collision and kinetic energy is not, so momentum is what recovers the missing velocity. Once the second glider's speed is known, the two kinetic energy totals can be compared and the question answered. The audit also settles what kind of collision this is: the energy after is less than the energy before, so it is inelastic, but the gliders did not stick together, because they are moving at different speeds afterwards.
Fastest approach
1. Use momentum to find the missing velocity.
Before: $2.0 \times 6.0 = 12 \ \text{kg m s}^{-1}$.
After: $2.0 \times 3.0 + 1.0 \times v = 12$, so $v = 6.0 \ \text{m s}^{-1}$ for the lighter glider.
That is faster than the $3.0 \ \text{m s}^{-1}$ of the glider behind it, so the two separate rather than collide again. The model is consistent.
2. Compare the kinetic energies.
$E_{\text{before}} = \tfrac{1}{2}(2.0)(6.0)^{2} = 36 \ \text{J}$
$E_{\text{after}} = \tfrac{1}{2}(2.0)(3.0)^{2} + \tfrac{1}{2}(1.0)(6.0)^{2} = 9 + 18 = 27 \ \text{J}$
3. Take the fraction.
Lost: $36 - 27 = 9 \ \text{J}$, and $\dfrac{9}{36} = \dfrac{1}{4}$.
Common mistake. Reading the incident glider's speed halving as the total kinetic energy halving. The lighter glider leaves with $18 \ \text{J}$ of kinetic energy that has to be counted, and once it is, three quarters of the original energy is still kinetic.
Why the other options are wrong
- A: Dividing the $9 \ \text{J}$ lost by $\tfrac{1}{2}(2.0 + 1.0)(6.0)^{2} = 54 \ \text{J}$, using the combined mass in the energy the collision started with. Only the moving glider had kinetic energy before the collision. Formula Misapplication
- C: Assuming the gliders move off together, which gives a common velocity of $4.0 \ \text{m s}^{-1}$ and a loss of $12 \ \text{J}$ out of $36 \ \text{J}$. A fraction is a fraction whichever route reaches it, so nothing dimensional objects; the stem simply says the heavier glider is still moving at $3.0 \ \text{m s}^{-1}$, so they did not stick. Conceptual Misunderstanding
- D: Taking the halving of the incident glider's speed as a halving of the total energy, so the energy carried away by the lighter glider is never counted. Conceptual Misunderstanding
- E: Reporting the fraction of the kinetic energy that remains, $27/36$, rather than the fraction that is lost. Misread Question
[03] Charge and electron flow
Question 3
A copper wire of cross-sectional area $2.5 \ \text{mm}^{2}$ carries a steady current of $3.2 \ \text{A}$. Take the number of free electrons per unit volume in copper as $8.0 \times 10^{28} \ \text{m}^{-3}$ and the magnitude of the charge on an electron as $1.6 \times 10^{-19} \ \text{C}$. Assume every free electron drifts along the wire at the same steady speed and that the wire has a uniform cross-section. What is that drift 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: $1.0 \times 10^{-4} \ \text{m s}^{-1}$ (option C)
Key idea. In one second the electrons that cross a given section are those within a cylinder of length $v$ and cross-sectional area $A$, so the charge crossing per second is $nAve$ and the current is $I = nAve$. Rearranged, $v = \dfrac{I}{nAe}$. Convert the square millimetres to square metres before any of this, because an area scales as the square of a length: $1 \ \text{mm}^{2} = 10^{-6} \ \text{m}^{2}$, not $10^{-3}$ and not $10^{-9}$.
Fastest approach
1. Convert the area once, at the start.
$2.5 \ \text{mm}^{2} = 2.5 \times (10^{-3})^{2} \ \text{m}^{2} = 2.5 \times 10^{-6} \ \text{m}^{2}$
2. Assemble the denominator, digits and powers of ten separately.
$nAe = (8.0 \times 2.5 \times 1.6) \times 10^{28 - 6 - 19} = 32 \times 10^{3} = 3.2 \times 10^{4}$
3. Divide.
$v = \dfrac{3.2}{3.2 \times 10^{4}} = 1.0 \times 10^{-4} \ \text{m s}^{-1}$
That is a tenth of a millimetre per second, roughly a metre every three hours. A lamp still lights the instant the switch closes, because the electric field that starts every electron moving travels along the wire at close to the speed of light while the electrons themselves barely crawl.
Common mistake. Converting the area as though it were a length, $2.5 \ \text{mm}^{2} \to 2.5 \times 10^{-3} \ \text{m}^{2}$, or as though it were a volume, $2.5 \times 10^{-9} \ \text{m}^{2}$. A prefix inside a squared unit is squared with it, and getting that wrong shifts the answer by three orders of magnitude in either direction.
Why the other options are wrong
- A: Leaving the electron charge out: $I / (nA)$. That is an ampere divided by a reciprocal length, which is an ampere metre, and no amount of arithmetic turns it into a speed. Dimensional Error
- B: Leaving the area in square millimetres and treating the number $2.5$ as though it were already in square metres. Unit Error
- D: Converting the area with the factor that belongs to a volume, $10^{-9}$ rather than $10^{-6}$. The result is a perfectly respectable speed in metres per second, which is exactly why no dimensional check catches it, and it is a thousand times too fast. Conceptual Misunderstanding
- E: Dividing the wrong way up, $nAe / I$, which carries seconds per metre rather than metres per second. It is also a speed no electron in a wire could have: a metre of wire would be crossed ten thousand times a second. Conceptual Misunderstanding
[04] Forces and Newton's laws
Question 4
A puck of mass $2.5 \ \text{kg}$ slides on a horizontal sheet of ice. Two horizontal forces act on it at the same time: $12 \ \text{N}$ directed north and $5.0 \ \text{N}$ directed east. Take the ice as frictionless, ignore air resistance, and take the vertical forces on the puck as balanced. What is the magnitude of the puck'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: $5.2 \ \text{m s}^{-2}$ (option D)
Key idea. Forces add as vectors, so the resultant of two perpendicular forces is the hypotenuse of the triangle they make: $\sqrt{12^{2} + 5^{2}} = 13 \ \text{N}$. Before computing anything, note the bracket: the resultant of two forces can never be smaller than their difference or larger than their sum, so it lies between $7 \ \text{N}$ and $17 \ \text{N}$, and the acceleration lies between $2.8$ and $6.8 \ \text{m s}^{-2}$. That alone removes the routes which used only one of the two forces.
Fastest approach
1. Bracket the answer.
Two forces of $12 \ \text{N}$ and $5.0 \ \text{N}$ have a resultant somewhere between $12 - 5 = 7 \ \text{N}$ and $12 + 5 = 17 \ \text{N}$, whatever the angle between them.
2. Add them as vectors.
They are at right angles, so the $5$, $12$, $13$ triple gives the resultant immediately:
$F = \sqrt{12^{2} + 5^{2}} = \sqrt{169} = 13 \ \text{N}$
That sits comfortably inside the bracket.
3. Apply Newton's second law.
$a = \dfrac{F}{m} = \dfrac{13}{2.5} = 5.2 \ \text{m s}^{-2}$
The direction is along the resultant, which is east of north, but the question asks only for the magnitude.
Common mistake. Adding the two force magnitudes as though the forces pointed the same way, or subtracting them as though they opposed. Neither is true of forces at right angles, and both answers sit at the very edge of the bracket rather than inside it.
Why the other options are wrong
- A: Using the $5.0 \ \text{N}$ force alone and ignoring the larger one. Incomplete Calculation
- B: Subtracting the magnitudes, $(12 - 5)/2.5$, as though the two forces opposed each other. Newtons divided by kilograms is still an acceleration, so the units cannot object; the geometry can, because a resultant only equals the difference when the forces are antiparallel. Conceptual Misunderstanding
- C: Using the $12 \ \text{N}$ force alone and ignoring the smaller one. Incomplete Calculation
- E: Adding the magnitudes, $(12 + 5)/2.5$, as though the two forces pointed the same way. That is the largest resultant two such forces could ever have, and it needs them parallel, not perpendicular. Conceptual Misunderstanding
[05] Estimation
Question 5
Estimate the volume of water the showers in a school use in one day. Assume that $200$ pupils each take exactly one shower, that each shower runs for $5.0$ minutes, and that a shower head delivers $12$ litres of water per minute while it is running. Take $1000$ litres as $1 \ \text{m}^{3}$, and ignore water used for anything other than showering.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $12 \ \text{m}^{3}$ (option E)
Key idea. An estimate like this is a product of factors that can each be guessed on their own: how many showers, how long each one lasts, and how fast the water comes out. Write it as $\text{showers} \times \text{minutes per shower} \times \text{litres per minute}$ and watch the minutes cancel, leaving litres. If they do not cancel, a factor has been dropped or misread.
Fastest approach
1. Write the estimate as a product of factors.
$V = 200 \ \text{showers} \times 5.0 \ \dfrac{\text{min}}{\text{shower}} \times 12 \ \dfrac{\text{litre}}{\text{min}}$
The showers cancel, the minutes cancel, and litres are left. That is the check worth doing before any multiplication.
2. Multiply.
$200 \times 5.0 \times 12 = 12\,000$ litres
3. Convert to cubic metres.
$\dfrac{12\,000}{1000} = 12 \ \text{m}^{3}$
Worth a moment: at $1000 \ \text{kg m}^{-3}$ that is twelve tonnes of water a day, and it all has to be heated. It is why a school's hot water bill is a serious number and why the shower timer is the single factor most worth attacking.
Common mistake. Working out one shower and stopping, or multiplying two of the three factors and forgetting the third. Writing the units alongside each factor catches both: a product that comes out in litres per minute, or in pupil-minutes, has lost something.
Why the other options are wrong
- A: Working out a single shower, $5.0 \times 12 = 60$ litres, and stopping there, or assuming the stated flow is shared by the whole school at once. Either way this is one shower's worth of water, and the school has two hundred of them. Conceptual Misunderstanding
- B: Reading the flow as $12$ litres per hour rather than per minute, so each shower delivers one litre. Misread Question
- C: Multiplying the number of pupils by the number of minutes, $200 \times 5.0$, and reading the product as litres. Pupil-minutes are not a volume, and the flow rate is what would have turned them into one. Dimensional Error
- D: Multiplying the number of pupils by the flow rate, $200 \times 12$, so how long each shower runs never enters the estimate at all. Incomplete Calculation
[06] Matter and density
Question 6
Two liquids are stirred together: $0.60 \ \text{kg}$ of the first, whose density is $1200 \ \text{kg m}^{-3}$, and $0.60 \ \text{kg}$ of the second, whose density is $400 \ \text{kg m}^{-3}$. Assume the two mix completely, that the volume of the mixture is the sum of the two separate volumes, and that none of either liquid evaporates. What is the density of the mixture?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $600 \ \text{kg m}^{-3}$ (option A)
Key idea. Density is mass per unit volume, so the denominator has to be a volume: the total volume, which means finding each liquid's volume separately from $V = m/\rho$ and adding them. Averaging the two densities silently assumes the two liquids contribute equal volumes, and here they do not. Equal masses of a dense liquid and a light one give a small volume of the first and a large volume of the second, so the mixture sits nearer the lighter density than halfway.
Fastest approach
1. Find each volume from its own density.
$V_{1} = \dfrac{0.60}{1200} = 5.0 \times 10^{-4} \ \text{m}^{3}$
$V_{2} = \dfrac{0.60}{400} = 1.5 \times 10^{-3} \ \text{m}^{3}$
The lighter liquid occupies three times the volume of the denser one, which is the whole reason the answer is not the midpoint.
2. Add the masses and add the volumes.
$m = 1.2 \ \text{kg}$, $V = 2.0 \times 10^{-3} \ \text{m}^{3}$
3. Divide.
$\rho = \dfrac{1.2}{2.0 \times 10^{-3}} = 600 \ \text{kg m}^{-3}$
Sanity check: the answer must lie between $400$ and $1200 \ \text{kg m}^{-3}$, and it does, sitting a quarter of the way up because three quarters of the mixture by volume is the lighter liquid.
Common mistake. Averaging the two densities. That is only correct when the two liquids contribute equal volumes; for equal masses the answer is pulled towards whichever liquid takes up more room, which is always the lighter one.
Why the other options are wrong
- B: Averaging the two densities, $(1200 + 400)/2$. Kilograms per cubic metre in, kilograms per cubic metre out, so nothing dimensional objects, but a mean of densities weights the two liquids by volume and here the volumes are $5.0 \times 10^{-4}$ and $1.5 \times 10^{-3} \ \text{m}^{3}$. Conceptual Misunderstanding
- C: Quoting the density of the denser liquid, on the view that a mixture cannot be lighter than its heaviest ingredient. It can, and here three quarters of the volume is the lighter liquid. Conceptual Misunderstanding
- D: Adding the two densities, $1200 + 400$. Densities are per unit volume and do not add unless the volumes are the same, which would then need dividing by two anyway. Formula Misapplication
- E: Dividing the total mass of $1.2 \ \text{kg}$ by the volume of the denser liquid alone, $5.0 \times 10^{-4} \ \text{m}^{3}$, so the volume the second liquid occupies is never added. Incomplete Calculation
[07] Dimensional reasoning
Question 7
The speed $v$ of a wave on the surface of deep water depends only on its wavelength $\lambda$ and on the gravitational field strength $g$, so that $v = k\lambda^{a}g^{b}$ for some dimensionless constant $k$. Assume surface tension and the depth of the water play no part and that the wave is small enough for its amplitude not to matter. Waves of wavelength $4.0 \ \text{m}$ are measured travelling at $2.5 \ \text{m s}^{-1}$. In the same water, how fast do waves of wavelength $16 \ \text{m}$ travel?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $5.0 \ \text{m s}^{-1}$ (option B)
Key idea. Write the proportionality first and leave $k$ alone. Matching dimensions in $v = k\lambda^{a}g^{b}$ gives $\text{L}\,\text{T}^{-1} = \text{L}^{a}(\text{L}\,\text{T}^{-2})^{b}$, so $a + b = 1$ from the lengths and $-2b = -1$ from the times. Hence $b = \tfrac{1}{2}$, $a = \tfrac{1}{2}$ and $v = k\sqrt{g\lambda}$. Speed goes as the square root of wavelength, and taking the ratio of the two cases makes $k$ and $g$ walk out together, so neither ever has to be known.
Fastest approach
1. Let the dimensions fix the exponents.
$[v] = \text{L}\,\text{T}^{-1}$, $[\lambda] = \text{L}$, $[g] = \text{L}\,\text{T}^{-2}$.
Lengths: $a + b = 1$. Times: $-2b = -1$, so $b = \tfrac{1}{2}$ and then $a = \tfrac{1}{2}$.
$v = k\sqrt{g\lambda}$
2. Take the ratio of the two cases.
$\dfrac{v_{2}}{v_{1}} = \sqrt{\dfrac{\lambda_{2}}{\lambda_{1}}} = \sqrt{\dfrac{16}{4}} = 2$
3. Scale the measured speed.
$v_{2} = 2 \times 2.5 = 5.0 \ \text{m s}^{-1}$
Neither $k$ nor $g$ was ever needed. Dimensional analysis gives the form, never the numerical factor, and the ratio of two cases is what removes the need for it.
Common mistake. Scaling the speed in direct proportion to the wavelength. That would need $v = k\lambda$ with $k$ dimensionless, which makes the right hand side a length rather than a speed, and ruling exactly that out is what the dimensional argument is for.
Why the other options are wrong
- A: Solving the two dimensional equations with the signs reversed, giving $v \propto \lambda^{-1/2}$ so that the longer wave arrives more slowly. Long swell outrunning short chop is the everyday observation that says otherwise. Sign Error
- C: Holding the frequency fixed and using $v = f\lambda$, which makes speed proportional to wavelength. The relation $v = f\lambda$ is always true and perfectly consistent dimensionally, so nothing catches this on units; what is wrong is the physics, because on deep water the frequency is not free to stay put. It is fixed by the wavelength through $v = k\sqrt{g\lambda}$. Conceptual Misunderstanding
- D: Using $a = \tfrac{3}{2}$, which comes from balancing the length equation without halving the exponent the time equation already fixed. Formula Misapplication
- E: Multiplying the measured speed by the new wavelength in metres instead of by the ratio of the two wavelengths. Misread Question
[08] Energy, work and power
Question 8
A person of mass $60 \ \text{kg}$ walks at a steady pace up a straight flight of stairs, taking $8.0 \ \text{s}$ to do it. The flight rises $4.0 \ \text{m}$ vertically over a horizontal distance of $3.0 \ \text{m}$, so its sloping length is $5.0 \ \text{m}$. Take $g = 10 \ \text{m s}^{-2}$. Count only the work done in raising the person's weight, ignoring the work done in swinging the limbs, any energy lost as heat and any change in kinetic energy. What is the person's average useful power output?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $300 \ \text{W}$ (option C)
Key idea. Power is work per unit time, and the work done against gravity is the weight multiplied by the vertical rise: $P = \dfrac{mgh}{t}$. Two of the wrong routes here can be thrown out on dimensions alone, before any arithmetic, because dropping $g$ or dropping $h$ leaves something that is not a watt. The remaining trap survives that test, because using the sloping length instead of the vertical rise still gives a perfectly good power: only the physics rules it out, since gravity acts vertically and only vertical displacement does work against it.
Fastest approach
1. Check the shape of each candidate before its value.
A watt is a joule per second, so the expression must contain a mass, an acceleration, a length and a time in the combination $\dfrac{mgh}{t}$. Anything missing $g$ or missing $h$ is not a power and can be discarded on sight.
2. Use the vertical rise, not the sloping length.
The force being worked against is the weight, which points straight down, so the distance moved in the direction of that force is $4.0 \ \text{m}$, not $5.0 \ \text{m}$. The $3.0 \ \text{m}$ of horizontal travel is done against nothing.
3. Evaluate.
$P = \dfrac{60 \times 10 \times 4.0}{8.0} = \dfrac{2400}{8.0} = 300 \ \text{W}$
A sustained $300 \ \text{W}$ is brisk but ordinary for a person on stairs, which is the check worth making on the number itself.
Common mistake. Using the $5.0 \ \text{m}$ measured along the flight. That is the distance travelled, not the distance moved in the direction of the force being worked against, and the whole of the horizontal part of it costs no work against gravity at all.
Why the other options are wrong
- A: Leaving $g$ out: $60 \times 4.0 / 8.0$. That is a mass times a speed, which is a momentum, not a power. Dimensional Error
- B: Leaving the height out: $60 \times 10 / 8.0$. That is a force divided by a time, which is a rate of change of force, not a power. Dimensional Error
- D: Using the $5.0 \ \text{m}$ sloping length of the flight instead of its $4.0 \ \text{m}$ vertical rise. The result is in watts and survives every dimensional check, but gravity pulls straight down, so only the vertical part of the journey does work against it. Conceptual Misunderstanding
- E: Reporting $mgh$, the work done, and never dividing by the time taken. That is $2400 \ \text{J}$, not $2400 \ \text{W}$. Incomplete Calculation
[09] Optics
Question 9
A step-index light guide has a core of refractive index $2.00$ surrounded by a cladding of refractive index $1.60$. A ray travelling inside the core meets the flat boundary with the cladding. Assume both materials are transparent and do not absorb light, and that the guide is straight. What is the sine of the critical angle for that boundary?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $0.80$ (option D)
Key idea. At the critical angle the refracted ray grazes along the boundary at $90^{\circ}$, so Snell's law $n_{1}\sin\theta_{c} = n_{2}\sin 90^{\circ}$ gives $\sin\theta_{c} = \dfrac{n_{2}}{n_{1}}$, with $n_{1}$ the index of the medium the light is in. The bound does half the work: a sine can never exceed $1$, so any route producing a number above $1$ is wrong before it is checked, and the ratio must therefore be the smaller index over the larger.
Fastest approach
1. Identify which medium the light is travelling in.
The ray is inside the core, so $n_{1} = 2.00$, and it is trying to cross into the cladding, so $n_{2} = 1.60$. Total internal reflection can happen at all only because the ray is going from the denser medium to the less dense one.
2. Apply the critical angle condition.
$n_{1}\sin\theta_{c} = n_{2}\sin 90^{\circ} = n_{2}$
$\sin\theta_{c} = \dfrac{n_{2}}{n_{1}} = \dfrac{1.60}{2.00} = 0.80$
3. Check the bound.
$0.80$ is a legal sine, so $\theta_{c} \approx 53^{\circ}$: rays meeting the wall at more than about $53^{\circ}$ from the normal stay in the core, and this is what keeps light inside a fibre over many kilometres. Had the ratio come out above $1$, no angle would satisfy it and the arithmetic would have been the wrong way up.
Common mistake. Comparing the core with air instead of with the cladding. The cladding is the medium on the far side of the boundary, and replacing it with air changes both the critical angle and the range of rays the guide can carry.
Why the other options are wrong
- A: Subtracting the two indices, $2.00 - 1.60$, instead of dividing one by the other. A difference of two pure numbers is still a pure number, so the answer looks like a legal sine, but Snell's law relates the indices by a ratio and never by a difference. Formula Misapplication
- B: Using $1/n_{\text{core}}$, which is the condition for a core surrounded by air. The cladding is there and its index is what belongs in the numerator. Conceptual Misunderstanding
- C: Using $1/n_{\text{cladding}}$, which is the critical angle for a ray inside the cladding meeting air, a boundary this question never mentions. Conceptual Misunderstanding
- E: Taking the ratio the wrong way up, $2.00/1.60$. No sine can exceed $1$, so this can be discarded without any thought about which medium is which. Conceptual Misunderstanding
[10] Atomic structure
Question 10
The nucleus of a neutral atom carries a charge of $+3.2 \times 10^{-18} \ \text{C}$ and contains $24$ neutrons. Take the elementary charge as $1.6 \times 10^{-19} \ \text{C}$, and assume the atom's nucleus is not altered by anything that follows. The atom then loses two of its electrons and becomes an ion. What is the mass number of that ion?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $44$ (option E)
Key idea. The atomic number counts protons and the mass number counts nucleons, protons and neutrons together. Electrons are counted by neither. The reason is a comparison of magnitudes: an electron has about $\tfrac{1}{1800}$ of the mass of a nucleon, so stripping two of them changes the atom's mass by roughly one part in forty thousand and changes the count of nucleons not at all. The nuclear charge is the route to the proton number, since each proton carries $+e$ and the neutrons carry none.
Fastest approach
1. Get the proton number from the nuclear charge.
Only protons contribute to the charge of a nucleus, each carrying $+e$:
$Z = \dfrac{3.2 \times 10^{-18}}{1.6 \times 10^{-19}} = 20$
2. Add the neutrons.
The mass number counts nucleons:
$A = Z + N = 20 + 24 = 44$
3. Ask what losing two electrons changed.
Nothing that either number counts. The ion now has $18$ electrons rather than $20$, so it carries a charge of $+2e$, but its nucleus is untouched: still $20$ protons and $24$ neutrons, still a mass number of $44$. Its actual mass falls by about two electron masses, which is around one part in forty thousand and is why the mass number ignores them.
Common mistake. Subtracting the lost electrons from the mass number. Ionisation removes electrons from outside the nucleus and leaves the nucleus alone, so neither the atomic number nor the mass number changes.
Why the other options are wrong
- A: Counting the electrons the ion has left, $20 - 2$. Electrons are counted by neither the atomic number nor the mass number. Conceptual Misunderstanding
- B: Reporting the proton number. That is the atomic number, which identifies the element; the mass number also counts the neutrons. Misread Question
- C: Reporting the neutron count alone, without adding the protons. Incomplete Calculation
- D: Subtracting the two lost electrons from $44$. It is a perfectly good count of particles, which is why nothing about its form gives it away, but the mass number never included the electrons in the first place: two of them weigh about one part in forty thousand of this atom. Conceptual Misunderstanding
[11] Integration
Question 11
Evaluate $\displaystyle\int_{0}^{3} \frac{x}{\sqrt{x^{2}+16}} \,\mathrm{d}x$.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $1$ (option A)
Key idea. The awkward part of the integrand is the cluster $x^{2}+16$, so give it one name: $u = x^{2}+16$. Then $\mathrm{d}u = 2x\,\mathrm{d}x$, and the stray $x$ on the top is exactly what pays for that substitution, up to a factor of $\frac{1}{2}$. Changing the limits at the same time as the variable is what stops the factor and the endpoints drifting apart.
Fastest approach
### 1. Name the cluster: let $u = x^{2}+16$, so $\mathrm{d}u = 2x\,\mathrm{d}x$ and $x\,\mathrm{d}x = \tfrac{1}{2}\,\mathrm{d}u$.
### 2. Change the limits with the variable: $x = 0$ gives $u = 16$, and $x = 3$ gives $u = 25$.
### 3. The integral becomes
$\tfrac{1}{2}\displaystyle\int_{16}^{25} u^{-1/2} \,\mathrm{d}u = \tfrac{1}{2}\left[2\sqrt{u}\right]_{16}^{25} = \left[\sqrt{u}\right]_{16}^{25}$
### 4. $\sqrt{25} - \sqrt{16} = 5 - 4 = 1$
The two factors of $2$ cancel, which is worth noticing in advance: the antiderivative of $\frac{x}{\sqrt{x^{2}+16}}$ is simply $\sqrt{x^{2}+16}$, and differentiating that back is a five second check.
Common mistake. Changing the variable but leaving the limits attached to $x$, so a $u$ antiderivative is evaluated between $0$ and $3$. Either convert the limits with the variable or convert back to $x$ before substituting, but never mix the two.
Why the other options are wrong
- B: Substituting $u = x^{2}+16$ and then evaluating $\left[\sqrt{u}\right]$ between $u = 0$ and $u = 3$, the old limits left in place. Conceptual Misunderstanding
- C: Missing the factor of $\frac{1}{2}$ that $\mathrm{d}u = 2x\,\mathrm{d}x$ carries, which doubles the whole answer. Formula Misapplication
- D: Evaluating $\sqrt{x^{2}+16}$ at the upper limit only, so the value $4$ at the lower limit is never subtracted. Incomplete Calculation
- E: Both slips at once: the factor of $\frac{1}{2}$ dropped and only the upper limit used, giving $2 \times 5$. Formula Misapplication
[12] Ratio and proportion
Question 12
In a school orchestra the ratio of string players to wind players is $3 : 5$. Four more string players join, nobody leaves, and the ratio of string players to wind players becomes $2 : 3$. How many wind players are in the orchestra?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $60$ (option B)
Key idea. A ratio of $3 : 5$ does not say there are $3$ and $5$ of them, only that the two counts are $3k$ and $5k$ for some common multiplier $k$. Writing both sections in terms of that single unknown turns the second ratio into one linear equation, and the wind section is untouched by the four newcomers, so it stays $5k$ throughout.
Fastest approach
### 1. Let the string players be $3k$ and the wind players be $5k$.
### 2. The four newcomers are string players, so afterwards the sections are $3k+4$ and $5k$, and
$\frac{3k+4}{5k} = \frac{2}{3}$
### 3. Cross multiply: $3(3k+4) = 2(5k)$, so $9k + 12 = 10k$ and $k = 12$.
### 4. Wind players: $5k = 60$.
Check: the sections were $36$ and $60$, and $40 : 60$ does reduce to $2 : 3$.
Common mistake. Reading $3 : 5$ as the counts themselves and trying to test $3$ and $5$ against the second condition. A ratio fixes the proportion and nothing else, so it needs a multiplier before it can be used in an equation.
Why the other options are wrong
- A: Solving correctly for $k$ and then reporting $3k+4 = 40$, the number of string players after the four join. Misread Question
- C: Adding the four newcomers to the wind section as well, though they are string players and the wind section never changes. Conceptual Misunderstanding
- D: Reporting the size of the whole orchestra before the newcomers arrive, $3k + 5k = 96$. Misread Question
- E: Reporting the size of the whole orchestra after the newcomers arrive, $40 + 60 = 100$. Misread Question
[13] Quadratics
Question 13
For how many positive integer values of $c$ does the equation $x^{2}+8x+c = 0$ have two distinct real roots?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $15$ (option C)
Key idea. Two distinct real roots means $b^{2}-4ac > 0$, strictly. Here that is $64 - 4c > 0$, so $c < 16$, and the question then stops being about quadratics: it asks how many positive integers are strictly below $16$.
Fastest approach
### 1. Discriminant: $b^{2}-4ac = 8^{2} - 4(1)(c) = 64 - 4c$.
### 2. Two distinct real roots need it strictly positive:
$64 - 4c > 0 \implies 4c < 64 \implies c < 16$
### 3. The positive integers strictly below $16$ are $1, 2, \ldots, 15$, so there are $15$ of them.
Test the boundary rather than trusting it: $c = 16$ gives $x^{2}+8x+16 = (x+4)^{2}$, one repeated root, so $16$ is correctly excluded.
Common mistake. Using $b^{2}-4ac \geq 0$, which is the condition for real roots rather than for distinct ones, and so counting $c = 16$ as well. The word distinct is what makes the inequality strict.
Why the other options are wrong
- A: Counting only the values that also make the roots themselves integers, $c = 7, 12, 15$, a condition the question never imposes. Misread Question
- B: Solving $64 - 4c > 0$ by dividing $64$ by $8$ instead of by $4$, giving $c < 8$ and seven values. Arithmetic Slip
- D: Allowing $b^{2}-4ac = 0$, so $c = 16$ is counted even though the two roots coincide there. Conceptual Misunderstanding
- E: Using $b^{2}-2ac$ in place of $b^{2}-4ac$, giving $c < 32$ and thirty-one values. Formula Misapplication
[14] Probability
Question 14
A machine makes components. Each component is faulty with probability $0.1$, independently of every other component. A sample of four components is taken. What is the probability that exactly one of the four is faulty?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $0.2916$ (option D)
Key idea. Exactly one faulty component means one specific pattern of outcomes, faulty and then three sound, but the faulty one could be any of the four. Nothing distinguishes the four positions, so each has the same probability $0.1 \times 0.9^{3}$ and the answer is four times it.
Fastest approach
### 1. One particular component faulty, the other three sound:
$0.1 \times 0.9^{3} = 0.1 \times 0.729 = 0.0729$
### 2. The faulty one can be any of the four, and those four cases cannot happen together, so add them:
$4 \times 0.0729 = 0.2916$
### 3. As a binomial term this is $\binom{4}{1}(0.1)^{1}(0.9)^{3}$, which is the same arithmetic with the counting done by the coefficient.
Sanity check: the four probabilities for zero, one, two, three and four faults must add to $1$, and $0.6561 + 0.2916 + 0.0486 + 0.0036 + 0.0001 = 1$.
Common mistake. Computing the probability of one particular arrangement and stopping there. The number of arrangements is half of every binomial calculation, and here it is the factor of four.
Why the other options are wrong
- A: Computing $0.1^{4}$, which is the probability that all four are faulty, by raising $p$ to the sample size instead of using one $p$ and three $(1-p)$. Formula Misapplication
- B: $0.1 \times 0.9^{3}$, the probability that one named component is the faulty one, with the four possible positions never counted. Incomplete Calculation
- C: Quoting the probability that a single component is faulty, as though the size of the sample made no difference. Conceptual Misunderstanding
- E: $1 - 0.9^{4}$, the probability that at least one is faulty, which also counts samples containing two, three or four faulty components. Misread Question
[15] Differentiation
Question 15
The curve $y = x^{3}-6x^{2}+9x+2$ has one local maximum and one local minimum. What is the $y$-coordinate of the local maximum?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $6$ (option E)
Key idea. Set $\frac{\mathrm{d}y}{\mathrm{d}x}$ to zero to find where the turning points are, then use the shape to decide which is which. A cubic with a positive $x^{3}$ coefficient rises, turns over, comes back down and rises again, so the left stationary point is always the local maximum. The question asks for the height there, not for the position.
Fastest approach
### 1. $\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2}-12x+9 = 3(x-1)(x-3)$, so the stationary points are at $x = 1$ and $x = 3$.
### 2. The coefficient of $x^{3}$ is positive, so the curve comes up from below, turns over at the smaller root and turns back up at the larger one. The maximum is at $x = 1$.
### 3. $y(1) = 1 - 6 + 9 + 2 = 6$
Confirming with the second derivative rather than the sketch: $\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 6x-12$, which is $-6$ at $x = 1$, so that point is indeed a maximum.
Common mistake. Stopping at the $x$ values. Solving $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ says where the turning points are, and the height still has to be found by substituting back into $y$, not into the derivative.
Why the other options are wrong
- A: Substituting the stationary value into $\frac{\mathrm{d}y}{\mathrm{d}x}$ rather than into $y$, which returns zero by construction. Formula Misapplication
- B: Reporting $x = 1$, the position of the maximum, instead of the height of the curve there. Misread Question
- C: Classifying the two stationary points the wrong way round and giving $y(3) = 2$, which is the local minimum. Conceptual Misunderstanding
- D: Substituting $x = 2$, the point midway between the two stationary points where $\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 0$. That is the point of inflection, not a turning point. Conceptual Misunderstanding
[16] Percentages
Question 16
A retailer raises the price of a coat by $20\%$. In a later sale the retailer takes $25\%$ off the increased price. What is the overall percentage change from the original price, where a negative value means a fall?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $-10\%$ (option A)
Key idea. A percentage change is a multiplying factor: a rise of $20\%$ is $\times 1.2$ and a reduction of $25\%$ is $\times 0.75$. Successive changes multiply, so the overall factor is $1.2 \times 0.75 = 0.9$, and the original price never has to be named. Percentages may only be added when they act on the same base, which successive changes never do.
Fastest approach
### 1. Write each change as a factor: $+20\%$ is $\times \frac{6}{5}$, and $-25\%$ is $\times \frac{3}{4}$.
### 2. Multiply them:
$\frac{6}{5} \times \frac{3}{4} = \frac{18}{20} = 0.9$
### 3. A factor of $0.9$ is a fall of $10\%$.
Check with a convenient price. Start at $100$: the rise gives $120$, and a quarter off $120$ is $90$, which is $10$ below where it started.
Common mistake. Adding the two percentages to get $20 - 25 = -5$. The reduction acts on the raised price, which is larger than the original, so a quarter off it removes more than five per cent of the starting price.
Why the other options are wrong
- B: Adding the percentage changes, $+20 - 25 = -5$, as though both acted on the original price. Formula Misapplication
- C: Reaching the factor $0.9$ correctly and then reading it as a rise of $10\%$ rather than a fall. Sign Error
- D: Reporting the first change and never applying the sale reduction at all. Incomplete Calculation
- E: Applying the sale as a second increase, $\frac{6}{5} \times \frac{5}{4} = \frac{3}{2}$. Sign Error
[17] Probability
Question 17
A player scores a free throw with probability $\frac{1}{3}$, independently on each attempt. She takes five free throws. What is the probability that she scores at least four of them?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\frac{11}{243}$ (option B)
Key idea. Write the general term $\binom{5}{r}\left(\frac{1}{3}\right)^{r}\left(\frac{2}{3}\right)^{5-r}$ once, then read off the cases the question asks for. At least four means four or five, so two terms are added, and both have the same denominator $3^{5} = 243$, which keeps the arithmetic in whole numbers.
Fastest approach
### 1. General term for $r$ successes out of five:
$P(r) = \binom{5}{r}\left(\frac{1}{3}\right)^{r}\left(\frac{2}{3}\right)^{5-r}$
### 2. Exactly four:
$P(4) = 5 \times \frac{1}{81} \times \frac{2}{3} = \frac{10}{243}$
### 3. Exactly five:
$P(5) = \frac{1}{243}$
### 4. At least four is four or five, and the two cannot both happen, so
$P(\text{at least } 4) = \frac{10}{243} + \frac{1}{243} = \frac{11}{243}$
Sanity check: scoring is the less likely outcome on each throw, so four or more successes out of five ought to be rare, and $\frac{11}{243}$ is about $4.5\%$.
Common mistake. Reading at least four as exactly four and stopping at $\frac{10}{243}$. The phrase at least always covers more than one case, and each of those cases has to be added in.
Why the other options are wrong
- A: $\binom{5}{4}\left(\frac{1}{3}\right)^{4}\frac{2}{3}$ alone, so the case of all five scoring is never added. Incomplete Calculation
- C: $5\left(\frac{1}{3}\right)^{4}$, counting the five orders but forgetting that the remaining throw has to miss, which carries the factor $\frac{2}{3}$. Formula Misapplication
- D: Including exactly three successes as well, which computes at least three rather than at least four. Misread Question
- E: Swapping the scoring and missing probabilities, so this is the probability of at least four misses. Conceptual Misunderstanding
[18] Differentiation
Question 18
The curve $y = x + \dfrac{4}{x}$ is drawn for positive $x$ only, and on that part it has a single turning point. How high is the curve there, that is, what is the $y$-coordinate of the turning point?
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 C)
Key idea. Write $\frac{4}{x}$ as $4x^{-1}$ before differentiating, so the derivative is $1 - 4x^{-2}$. Setting it to zero gives $x^{2} = 4$, and the restriction $x > 0$ is what leaves a single stationary point rather than two. The limits settle its nature without any second derivative: as $x \to 0^{+}$ the $\frac{4}{x}$ term blows up, and as $x \to \infty$ the $x$ term does, so the one stationary point in between must be a minimum.
Fastest approach
### 1. $y = x + 4x^{-1}$, so $\frac{\mathrm{d}y}{\mathrm{d}x} = 1 - 4x^{-2}$.
### 2. Set it to zero: $1 = \frac{4}{x^{2}}$, so $x^{2} = 4$ and $x = 2$, the negative root being excluded by $x > 0$.
### 3. $y(2) = 2 + \frac{4}{2} = 4$
The two halves are equal at the minimum, which is worth remembering: $x + \frac{a}{x}$ always bottoms out where $x = \sqrt{a}$, at the value $2\sqrt{a}$, and here $2\sqrt{4} = 4$.
Common mistake. Reporting $x = 2$. Solving the derivative gives the position of the stationary point, and the question asks how high the curve is there, so the value has to be put back into $y$.
Why the other options are wrong
- A: Taking $x = -2$, the root that the stated restriction $x > 0$ removes; that stationary point exists on the other branch and is a maximum at $y = -4$. Misread Question
- B: Reporting $x = 2$, the position of the stationary point, rather than the height of the curve there. Incomplete Calculation
- D: Solving $x^{2} = 4$ as $x = 4$, with no square root taken, giving $4 + 1$. Formula Misapplication
- E: Inverting the equation $\frac{4}{x^{2}} = 1$ into $x^{2} = \frac{1}{4}$, so $x = \frac{1}{2}$ and $y = \frac{1}{2} + 8$. Formula Misapplication
[19] Coordinate geometry
Question 19
A straight fence runs from the point $P(-3, 2)$ to the point $Q(5, 6)$. A post is to stand on the perpendicular bisector of $PQ$, at the point where that bisector crosses the $y$-axis. What is the $y$-coordinate of the post?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $6$ (option D)
Key idea. A perpendicular bisector needs two things and both come from $P$ and $Q$: it passes through the midpoint of $PQ$, and its gradient is the negative reciprocal of the gradient of $PQ$. Missing either one gives a line that is perpendicular but in the wrong place, or in the right place but at the wrong angle.
Fastest approach
### 1. Midpoint of $PQ$: $\left(\frac{-3+5}{2}, \frac{2+6}{2}\right) = (1, 4)$.
### 2. Gradient of $PQ$: $\frac{6-2}{5-(-3)} = \frac{4}{8} = \frac{1}{2}$, so the perpendicular gradient is $-2$.
### 3. Through $(1,4)$ with gradient $-2$:
$y - 4 = -2(x-1) \implies y = -2x + 6$
### 4. At $x = 0$ this gives $y = 6$.
Check it against what a perpendicular bisector means, which is the set of points equidistant from $P$ and $Q$. From $(0,6)$ to $P(-3,2)$ is $\sqrt{9+16} = 5$, and from $(0,6)$ to $Q(5,6)$ is $\sqrt{25+0} = 5$. Equal, as required.
Common mistake. Taking the negative reciprocal of the gradient and then drawing the line through $P$ or through $Q$ instead of through the midpoint. Perpendicular is only half of what the word bisector asks for.
Why the other options are wrong
- A: Inverting the gradient of $PQ$ without negating it, so the line through the midpoint is taken as $y = 2x+2$. Formula Misapplication
- B: Finding where $PQ$ itself crosses the $y$-axis, $y = \frac{1}{2}x + \frac{7}{2}$, rather than where its perpendicular bisector does. Misread Question
- C: Negating the gradient of $PQ$ without inverting it, so the bisector is taken as $y - 4 = -\frac{1}{2}(x-1)$. Formula Misapplication
- E: Using the correct gradient of $-2$ but running the line through $Q(5,6)$ instead of through the midpoint, giving $y = -2x+16$. Conceptual Misunderstanding
[20] Coordinate geometry
Question 20
The points $A(1, 3)$ and $B(7, 5)$ both lie above the $x$-axis. A point $P$ is chosen on the $x$-axis, and the walker goes from $A$ to $P$ and then from $P$ to $B$ in straight lines. What is the smallest possible value of the total distance $AP + PB$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $10$ (option E)
Key idea. Reflecting $A$ in the $x$-axis to $A'(1,-3)$ leaves every length $AP$ unchanged, because $P$ is on the mirror line. So $AP + PB = A'P + PB$, which is a journey from $A'$ to $B$ by way of a point on the axis, and that is shortest when it is straight. The minimum is therefore the plain distance $A'B$, and the constraint disappears.
Fastest approach
### 1. Reflect $A$ in the $x$-axis: $A' = (1, -3)$. For any $P$ on that axis, $AP = A'P$, since the reflection fixes $P$ and preserves distance.
### 2. So $AP + PB = A'P + PB \geq A'B$, with equality exactly when $P$ lies on the straight segment $A'B$. Because $A'$ and $B$ are on opposite sides of the axis, that segment really does cross it, so the minimum is attained.
### 3. $A'B = \sqrt{(7-1)^{2} + \left(5-(-3)\right)^{2}} = \sqrt{36+64} = 10$
### 4. Confirming with calculus instead of geometry: minimising $\sqrt{(x-1)^{2}+9} + \sqrt{(x-7)^{2}+25}$ gives $x = \frac{13}{4}$, where the two legs measure $\frac{15}{4}$ and $\frac{25}{4}$ and total $10$. The two methods agree, and the geometric one needed no differentiation.
Worth noting for the sanity check: any allowed path has to reach the axis and come back, so it can never be shorter than the direct distance $AB = 2\sqrt{10} \approx 6.3$, and $10$ sits comfortably above that floor.
Common mistake. Quoting the straight distance from $A$ to $B$. That path never touches the $x$-axis, so it is not one of the journeys on offer; the reflection is what turns the allowed journeys into straight lines that can be compared.
Why the other options are wrong
- A: The horizontal separation of $A$ and $B$, which counts the run of the journey and neither the descent to the axis nor the climb back. Incomplete Calculation
- B: $AB = \sqrt{6^{2}+2^{2}}$, the direct distance, ignoring the requirement that the walker touch the $x$-axis on the way. Conceptual Misunderstanding
- C: Dropping straight down from $A$ to the axis and climbing straight up to $B$, giving $3+5$ with no horizontal travel counted at all. Incomplete Calculation
- D: Reflecting $A$ in the $y$-axis instead of the $x$-axis, giving $(-1,3)$ and the distance $\sqrt{8^{2}+2^{2}}$. Formula Misapplication