PAT practice · 20 questions · Free
PAT Exercise Paper 10: Physics and Mathematics
Every question carries a worked solution behind a disclosure, so you commit to an answer before you see the key.
[01] Momentum and energy
Question 1
Two trolleys stand at rest on a straight horizontal track, held together against a compressed spring that is attached to neither of them. One trolley has mass $0.20 \ \text{kg}$ and the other $0.60 \ \text{kg}$. They are released, the spring pushes them apart and drops away, and the $4.8 \ \text{J}$ it had stored becomes kinetic energy of the two trolleys. Friction on the track is negligible, the track is horizontal, and the spring's own mass may be ignored. What is the speed of the lighter trolley just after they separate?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $6.0 \ \text{m s}^{-1}$ (option B)
Key idea. Nothing was moving before the release, so the total momentum stays zero and the two trolleys carry equal and opposite momenta. That fixes the ratio of the speeds at $3:1$ before any energy is considered, and the stored energy then fixes their size. Because the kinetic energy of a body is $p^{2}/2m$, the lighter trolley takes the larger share.
Fastest approach
1. Momentum
Total momentum is zero before and after, so the two trolleys leave with momenta of equal size $p$, and
$0.20\,v_{1} = 0.60\,v_{2} \implies v_{1} = 3v_{2}$
2. Energy
$\tfrac{1}{2}(0.20)v_{1}^{2} + \tfrac{1}{2}(0.60)v_{2}^{2} = 4.8$
Substituting $v_{2} = v_{1}/3$:
$0.10\,v_{1}^{2} + 0.30\left(\tfrac{v_{1}}{3}\right)^{2} = 0.10\,v_{1}^{2} + 0.0333\,v_{1}^{2}$
which is $\tfrac{2}{15}v_{1}^{2} = 4.8$, so $v_{1}^{2} = 36$ and $v_{1} = 6.0 \ \text{m s}^{-1}$.
3. Check
The heavy trolley then moves at $2.0 \ \text{m s}^{-1}$, and the energies are $3.6 \ \text{J}$ and $1.2 \ \text{J}$, which add back to $4.8 \ \text{J}$ and are in the ratio $3:1$, the inverse of the mass ratio.
Common mistake. Splitting the released energy equally between the two trolleys. What is shared equally is the momentum, and since kinetic energy is $p^{2}/2m$, equal momenta hand the lighter body three quarters of the energy here.
Why the other options are wrong
- A: Sharing the $4.8 \ \text{J}$ equally, giving $\sqrt{2 \times 2.4/0.20} \approx 4.9$. Zero total momentum forces equal momenta, not equal energies. Conceptual Misunderstanding
- C: Giving the whole $4.8 \ \text{J}$ to the light trolley, $\sqrt{2 \times 4.8/0.20} \approx 6.9$, which leaves the heavy trolley moving with no energy at all. Incomplete Calculation
- D: Finding the heavy trolley's speed from the whole $4.8 \ \text{J}$, $\sqrt{2 \times 4.8/0.60} = 4.0$, and then applying the correct $3:1$ speed ratio to it. Formula Misapplication
- E: Dividing the energy by the mass and stopping there. $4.8/0.20$ has units of $\text{m}^{2}\text{s}^{-2}$, so it is the square of a speed and the square root is not optional. Dimensional Error
[02] Matter and pressure
Question 2
A bench of mass $24 \ \text{kg}$ stands on four legs, each ending in a flat square foot of side $4.0 \ \text{cm}$. It stands level on a rigid floor, so its weight is shared equally between the four feet, and each foot presses evenly over the whole of its square face. Take $g = 10 \ \text{m s}^{-2}$ and take the bench to be carrying nothing. What pressure does each foot exert on the floor?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $37\,500 \ \text{Pa}$ (option C)
Key idea. Pressure is force per unit area, and the two have to be talking about the same thing: one foot's share of the weight over one foot's area, or the whole weight over the area of all four feet. Mixing a per-foot force with a four-foot area, or the reverse, moves the answer by a factor of four in either direction.
Fastest approach
1. The force on one foot
Weight of the bench: $24 \times 10 = 240 \ \text{N}$, so each foot carries $60 \ \text{N}$.
2. The area of one foot
$4.0 \ \text{cm} = 4.0 \times 10^{-2} \ \text{m}$, so the square foot has area
$(4.0 \times 10^{-2})^{2} = 1.6 \times 10^{-3} \ \text{m}^{2}$
3. The pressure
$p = \frac{60}{1.6 \times 10^{-3}} = 37\,500 \ \text{Pa}$
The whole-bench route gives the same number: $240 / (4 \times 1.6 \times 10^{-3}) = 37\,500 \ \text{Pa}$, which is a free check that the force and the area were matched.
Common mistake. Converting square centimetres to square metres by dividing by $100$. A centimetre is $10^{-2} \ \text{m}$, so a square centimetre is $10^{-4} \ \text{m}^{2}$, and $16 \ \text{cm}^{2}$ is $1.6 \times 10^{-3} \ \text{m}^{2}$.
Why the other options are wrong
- A: Converting $16 \ \text{cm}^{2}$ to $0.16 \ \text{m}^{2}$, dividing by $100$ where the square of the prefix needs $10^{4}$. Unit Error
- B: Dividing one foot's share of the weight, $60 \ \text{N}$, by the area of all four feet. The four is then counted twice. Conceptual Misunderstanding
- D: Putting the whole $240 \ \text{N}$ on a single foot's area, which is the pressure under a bench balanced on one leg. Conceptual Misunderstanding
- E: Treating $16 \ \text{cm}^{2}$ as $16 \times 10^{-6} \ \text{m}^{2}$, which is $16 \ \text{mm}^{2}$. Unit Error
[03] Energy and power
Question 3
A car of mass $1200 \ \text{kg}$ travelling at $20 \ \text{m s}^{-1}$ along a level road brakes to rest. All of its kinetic energy is transferred to the brake discs, whose total mass is $8.0 \ \text{kg}$ and whose specific heat capacity is $500 \ \text{J kg}^{-1}\text{K}^{-1}$. The stop is quick enough that no appreciable heat escapes to the air or to the rest of the car while it happens, and the road is level so no gravitational potential energy is released. By how much does the temperature of the discs rise?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $60 \ \text{K}$ (option D)
Key idea. Two masses appear and they do different jobs. The car's mass sets how much energy there is to dispose of, through $\tfrac{1}{2}mv^{2}$; the discs' mass sets what one kelvin of warming costs, through $Mc$. Setting $\tfrac{1}{2}mv^{2} = Mc\,\Delta T$ keeps them apart.
Fastest approach
1. The energy to dispose of
$\tfrac{1}{2}mv^{2} = \tfrac{1}{2}(1200)(20^{2}) = 240\,000 \ \text{J}$
2. The cost of one kelvin
$Mc = 8.0 \times 500 = 4000 \ \text{J K}^{-1}$
3. The rise
$\Delta T = \frac{240\,000}{4000} = 60 \ \text{K}$
A rise of $60 \ \text{K}$ from a single stop is the plausibility check worth doing: it is large enough to feel through a wheel and small enough not to destroy anything, and it is why repeated heavy braking, which stacks these rises faster than the discs can shed them, ends in brake fade.
Common mistake. Using the same mass on both sides of the energy balance. The car is what is moving and the discs are what is heated, and here they differ by a factor of $150$.
Why the other options are wrong
- A: Using the car's mass in $Mc\,\Delta T$ as well as in the kinetic energy, as though the whole car were being warmed. Formula Misapplication
- B: Forgetting to square the speed. Half the mass times the speed is a momentum, not an energy, so it cannot produce a temperature. Dimensional Error
- C: Assuming half the kinetic energy reaches the discs. The stem says all of it does. Misread Question
- E: Dropping the factor of $\tfrac{1}{2}$ in the kinetic energy. Formula Misapplication
[04] Electric fields and charge
Question 4
An electron starts from rest and is accelerated in a vacuum through a potential difference of $5.0 \ \text{kV}$. Take the magnitude of the electron's charge as $1.6 \times 10^{-19} \ \text{C}$, neglect its weight, and assume it strikes nothing on the way. How much kinetic energy does it gain?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $8.0 \times 10^{-16} \ \text{J}$ (option E)
Key idea. Moving a charge $q$ through a potential difference $V$ does work $qV$ on it, and with the electron starting from rest all of that work appears as kinetic energy. The physics is one line; the prefix is where the marks go, because $5.0 \ \text{kV}$ is $5.0 \times 10^{3} \ \text{V}$.
Fastest approach
$E = qV = (1.6 \times 10^{-19}) \times (5.0 \times 10^{3})$
Digits first, then the exponents: $1.6 \times 5.0 = 8.0$ and $-19 + 3 = -16$.
$E = 8.0 \times 10^{-16} \ \text{J}$
The same statement in the other common unit is $5000 \ \text{eV}$, since an electronvolt is exactly the energy an electron gains through one volt. Reading the answer that way is a check that the arithmetic went the right way.
Common mistake. Reading $5.0 \ \text{kV}$ as $5.0 \ \text{V}$, which loses three powers of ten, or quoting the electronic charge itself as though it were the energy.
Why the other options are wrong
- A: Dividing the charge by the potential difference. A charge divided by a voltage is a capacitance in farads, so this route is dead before its arithmetic matters. Dimensional Error
- B: Quoting the electronic charge as the energy. That number is the energy gained through one volt, not through five thousand. Incomplete Calculation
- C: Reading $5.0 \ \text{kV}$ as $5.0 \ \text{V}$ and losing three powers of ten. Unit Error
- D: Using $1.0 \ \text{kV}$, as though the prefix were the whole potential difference and the $5.0$ belonged elsewhere. Misread Question
[05] Atomic structure
Question 5
In the Bohr model of the hydrogen atom the electron may occupy only levels of energy $E_{n} = -\dfrac{13.6}{n^{2}} \ \text{eV}$, measured relative to a free electron at rest, where $n = 1, 2, 3, \ldots$. An atom has its electron in the $n = 2$ level. Take $1 \ \text{eV} = 1.6 \times 10^{-19} \ \text{J}$, treat the atom as isolated, and assume the electron is removed so gently that it ends up at rest far away. What energy, in joules, must be supplied to remove it?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $5.44 \times 10^{-19} \ \text{J}$ (option A)
Key idea. A level energy is negative because the electron is bound: zero is the energy of an electron at rest and free. Removing the electron means supplying exactly enough to bring the energy up to zero, so the ionisation energy from level $n$ is $\left|E_{n}\right| = 13.6/n^{2} \ \text{eV}$, which for $n = 2$ is $3.4 \ \text{eV}$.
Fastest approach
1. The level
$E_{2} = -\frac{13.6}{2^{2}} = -3.4 \ \text{eV}$
2. The energy needed
Bringing the electron from $-3.4 \ \text{eV}$ up to $0$ takes $3.4 \ \text{eV}$.
3. Into joules
$3.4 \times 1.6 \times 10^{-19} = 5.44 \times 10^{-19} \ \text{J}$
Worth noticing how small that is. A single atom holding anything like a joule is not a thing this universe contains, so an answer of that size in joules would be wrong on inspection.
Common mistake. Using $13.6 \ \text{eV}$, which is the energy needed to ionise an atom from its ground state and not from the $n = 2$ level, or dividing by $n$ rather than by $n^{2}$.
Why the other options are wrong
- B: Dividing $13.6$ by $n$ rather than by $n^{2}$, giving $6.8 \ \text{eV}$. Formula Misapplication
- C: Using the gap between $n = 1$ and $n = 2$, which is $10.2 \ \text{eV}$. That is the energy to lift a ground-state atom into this level, not the energy to ionise an atom already in it. Conceptual Misunderstanding
- D: Ionising from the ground state, $13.6 \ \text{eV}$, instead of from $n = 2$. Misread Question
- E: Leaving the answer in electronvolts and writing joules beside it. A single atom holding $3.4 \ \text{J}$ would carry about as much energy as a dropped brick, some $10^{19}$ times the true figure. Unit Error
[06] Forces and equilibrium
Question 6
A uniform plank of mass $20 \ \text{kg}$ and length $4.0 \ \text{m}$ rests horizontally on two supports, one at each end. A child of mass $30 \ \text{kg}$ stands on it $1.0 \ \text{m}$ from the left-hand end. Take $g = 10 \ \text{m s}^{-2}$, take the plank as uniform so that its weight acts at its midpoint, and take every force as vertical. What is the upward force from the left-hand support?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $325 \ \text{N}$ (option B)
Key idea. There are two unknown support forces and two equations available: the forces must balance and the moments must balance. Taking moments about one support removes that support's force from the equation, so the other one falls out on its own.
Fastest approach
1. The weights
Plank: $20 \times 10 = 200 \ \text{N}$, acting at the midpoint, $2.0 \ \text{m}$ from either end. Child: $30 \times 10 = 300 \ \text{N}$, acting $1.0 \ \text{m}$ from the left end and so $3.0 \ \text{m}$ from the right.
2. Moments about the right-hand support
The right-hand force has no moment about its own line of action, so it drops out:
$R_{\text{left}} \times 4.0 = 200 \times 2.0 + 300 \times 3.0 = 1300 \ \text{N m}$
$R_{\text{left}} = 325 \ \text{N}$
3. Check
The forces must also balance: the right-hand support then carries $500 - 325 = 175 \ \text{N}$. The child stands nearer the left, so the left support carrying more is the right way round.
Common mistake. Sharing the total weight of $500 \ \text{N}$ equally between the two supports. That is correct only when the load sits at the midpoint, and this child does not.
Why the other options are wrong
- A: Halving the total weight of $500 \ \text{N}$, which is the answer for a load placed centrally and ignores where the child actually stands. Conceptual Misunderstanding
- C: Splitting the plank's weight equally between the supports and giving the whole of the child's weight to the nearer one. Conceptual Misunderstanding
- D: Taking the plank's weight to act at the left-hand end rather than at its midpoint, giving $(800 + 900)/4$. Formula Misapplication
- E: The moment about the right-hand support, $1300 \ \text{N m}$, quoted without dividing by the $4.0 \ \text{m}$ span. A moment is not a force. Dimensional Error
[07] The Solar System and orbits
Question 7
The asteroid belt lies between the orbits of the fourth and fifth planets from the Sun, Mars and Jupiter, whose orbital radii are about $1.5$ and $5.2$ astronomical units. An asteroid in the belt moves on a circular orbit of radius $4.0$ astronomical units about the Sun. The Earth's orbit is a circle of radius $1.0$ astronomical unit and takes $1.0$ year. Treat both orbits as circles about the Sun and neglect the gravitational pull of the planets on the asteroid. What is the asteroid's orbital period?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $8.0 \ \text{years}$ (option C)
Key idea. Kepler's third law says $T^{2} \propto r^{3}$ for bodies orbiting the same central mass. Dividing the asteroid's statement by the Earth's kills the constant, so $T = (r/r_{E})^{3/2}$ in years when $r$ is measured in astronomical units. Neither the Sun's mass nor $G$ is needed, because the Earth supplies the calibration.
Fastest approach
$\left(\frac{T}{1 \ \text{year}}\right)^{2} = \left(\frac{4.0}{1.0}\right)^{3} = 64$
$T = \sqrt{64} = 8.0 \ \text{years}$
Or in one step, $T = 4^{3/2} = \left(\sqrt{4}\right)^{3} = 2^{3} = 8$.
Sanity check against something known: Jupiter sits at $5.2$ astronomical units and takes about $12$ years, so a body a little closer in taking $8$ years is the right size of answer.
Common mistake. Assuming the asteroid travels at the Earth's orbital speed, so that four times the radius means four times the period. It has further to go and it also moves more slowly, and the two effects compound into the power of $\tfrac{3}{2}$.
Why the other options are wrong
- A: Reading the law as $T^{2} \propto r$, which gives $\sqrt{4}$. Formula Misapplication
- B: Taking the orbital speed to be the same as the Earth's, so that the period simply follows the circumference. A more distant orbit is also a slower one. Conceptual Misunderstanding
- D: Using $T \propto r^{2}$, squaring the radius ratio instead of raising it to the power $\tfrac{3}{2}$. Formula Misapplication
- E: Taking the cube from $T^{2} \propto r^{3}$ and never taking the square root, which answers with $T^{2}$ rather than $T$. Incomplete Calculation
[08] Dimensional analysis
Question 8
A small drop of liquid held together by surface tension oscillates in shape at a frequency $f$. Assume $f$ depends only on the drop's radius $r$, the liquid's density $\rho$ and its surface tension $\sigma$, which is a force per unit length, through $f = k\rho^{a}\sigma^{b}r^{c}$ with $k$ a dimensionless constant; take $k = 1$, since dimensions fix the exponents but never the constant. For a water drop take $\sigma = 0.080 \ \text{N m}^{-1}$, $\rho = 1000 \ \text{kg m}^{-3}$ and $r = 2.0 \ \text{mm}$, and neglect gravity and the effect of the surrounding air. What is the oscillation frequency?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $100 \ \text{Hz}$ (option D)
Key idea. Surface tension is a force per unit length, so $[\sigma] = \text{M T}^{-2}$, while $[\rho] = \text{M L}^{-3}$ and $[f] = \text{T}^{-1}$. Matching mass gives $a = -b$, matching time gives $b = \tfrac{1}{2}$, and matching length then gives $c = -\tfrac{3}{2}$. Only one combination survives: $f = \sqrt{\sigma / \rho r^{3}}$.
Fastest approach
1. Fix the exponents
$\text{T}: \quad -2b = -1 \implies b = \tfrac{1}{2}$ $\text{M}: \quad a + b = 0 \implies a = -\tfrac{1}{2}$ $\text{L}: \quad -3a + c = 0 \implies c = -\tfrac{3}{2}$
so $f = \sqrt{\dfrac{\sigma}{\rho r^{3}}}$.
2. Put the numbers in
$\rho r^{3} = 1000 \times (2.0 \times 10^{-3})^{3} = 8.0 \times 10^{-6} \ \text{kg}$
$\frac{\sigma}{\rho r^{3}} = \frac{0.080}{8.0 \times 10^{-6}} = 1.0 \times 10^{4} \ \text{s}^{-2}$
$f = 1.0 \times 10^{2} \ \text{Hz}$
3. Check the direction of variation
A bigger drop oscillates more slowly, since $f$ falls as $r^{-3/2}$, and a denser liquid oscillates more slowly too. Both are what a drop wobbling in slow motion looks like, so the exponents are pointing the right way.
Common mistake. Stopping at $\sigma / \rho r^{3}$. That group has units of $\text{s}^{-2}$, so it is the square of a frequency, and the square root is part of the answer rather than a tidying-up step.
Why the other options are wrong
- A: $\sqrt{\rho r^{3} / \sigma}$, the group inverted. That is the oscillation period in seconds, which is a perfectly real quantity and not the one asked for. Conceptual Misunderstanding
- B: Taking $c = -\tfrac{1}{2}$, which is what the length equation gives if $\rho$ is credited with only one power of length instead of three. Formula Misapplication
- C: Taking $c = -1$, dropping one of the three powers of length that a density carries. Formula Misapplication
- E: $\sigma / \rho r^{3}$ without the square root. The number is right but it has units of $\text{s}^{-2}$, so it cannot be a frequency. Dimensional Error
[09] Energy and power
Question 9
A cyclist and bicycle of total mass $80 \ \text{kg}$ travel at a steady $10 \ \text{m s}^{-1}$ along a level road in still air. The air drag on them is $\tfrac{1}{2}\rho A v^{2}$, with air density $\rho = 1.2 \ \text{kg m}^{-3}$ and effective frontal area $A = 0.50 \ \text{m}^{2}$, and the rolling resistance is a constant $8.0 \ \text{N}$ at any speed. Take the chain and bearings as perfectly efficient, and note that the road is level, so no height is gained. What power must the cyclist supply?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $380 \ \text{W}$ (option E)
Key idea. At a steady speed the driving force equals the total resistance, and the power supplied is that force times the speed. The judgement being tested is whether the smaller effect may be dropped: air drag is $30 \ \text{N}$ here and rolling resistance $8.0 \ \text{N}$, so neglecting rolling resistance would throw away a fifth of the answer. Notice too that the mass is never used, because nothing accelerates and no height is gained.
Fastest approach
1. Size the two resistances
Air drag: $\tfrac{1}{2}\rho A v^{2} = \tfrac{1}{2}(1.2)(0.50)(10^{2}) = 30 \ \text{N}$ Rolling: $8.0 \ \text{N}$
They are within a factor of four of each other, so both stay.
2. Total force and power
$F = 30 + 8.0 = 38 \ \text{N}$
$P = Fv = 38 \times 10 = 380 \ \text{W}$
3. Check the magnitude
A few hundred watts is what a fit cyclist sustains, which is the sort of number this ought to be. Note also that the drag term grows as $v^{2}$, so at twice this speed the rolling resistance really would be negligible; whether an effect is small is a question about the numbers, not about the effect.
Common mistake. Deciding that air drag dominates and dropping the rolling resistance without checking. At $10 \ \text{m s}^{-1}$ the two are within a factor of four, and the smaller one still carries a fifth of the total.
Why the other options are wrong
- A: The total resisting force in newtons, offered as a power. A force becomes a power only once it is multiplied by a speed. Dimensional Error
- B: Counting the rolling resistance and forgetting the air drag altogether. Incomplete Calculation
- C: Using half the speed, as though the cyclist had accelerated up from rest and an average were wanted. The speed is steady throughout. Conceptual Misunderstanding
- D: Counting the air drag only. It is the larger of the two, but the rolling resistance is a fifth of the total resistance and cannot be dropped at this speed. Incomplete Calculation
[10] Estimation
Question 10
Estimate the fraction of the Sun's total radiated power that falls on the Earth. Take the Earth as a sphere of radius $6.0 \times 10^{6} \ \text{m}$ moving on a circular orbit of radius $1.5 \times 10^{11} \ \text{m}$, take the Sun to radiate equally in all directions, and neglect any absorption of the light on the way.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $4.0 \times 10^{-10}$ (option A)
Key idea. By the time it reaches the Earth the Sun's output is spread evenly over a sphere of area $4\pi d^{2}$. The Earth removes from that sphere only what falls on the disc it blocks, of area $\pi R^{2}$, not what would fall on its whole surface. The fraction is $\pi R^{2} / 4\pi d^{2} = R^{2}/4d^{2}$, and both factors of $\pi$ cancel before any arithmetic starts.
Fastest approach
1. The two areas
Sphere the power has spread over: $4\pi d^{2}$. Disc the Earth blocks: $\pi R^{2}$, its shadow, since the light arrives as an effectively parallel beam.
2. The ratio
$\frac{\pi R^{2}}{4\pi d^{2}} = \frac{R^{2}}{4d^{2}} = \frac{(6.0 \times 10^{6})^{2}}{4(1.5 \times 10^{11})^{2}}$
Digits first, then exponents: $36 / (4 \times 2.25) = 4$, and $10^{13} / 10^{22} = 10^{-9}$.
$= 4.0 \times 10^{-10}$
3. Check the size
A fraction has to lie between $0$ and $1$, which rules out anything of order one immediately. And the Earth is a $10^{-5}$ fraction of the orbit radius across, so a $10^{-10}$ fraction of the sphere's area is the expected order of magnitude for something two-dimensional.
Common mistake. Using the Earth's whole surface area, $4\pi R^{2}$, in the numerator. A sphere placed in a parallel beam intercepts what falls on its shadow, a disc of radius $R$, and the night side receives nothing.
Why the other options are wrong
- B: Spreading the Sun's output over a hemisphere of area $2\pi d^{2}$ rather than over the whole sphere. Nothing about the Sun singles out the Earth's half of the sky. Conceptual Misunderstanding
- C: Using the Earth's whole surface area $4\pi R^{2}$ instead of the disc it actually blocks, which overcounts by four. Conceptual Misunderstanding
- D: Comparing $R$ with $d$ rather than $R^{2}$ with $d^{2}$, which answers a question about lengths where the question is about areas. Formula Misapplication
- E: Forgetting to square the distance, leaving $R^{2}/4d$. That has units of metres, and a fraction of a whole cannot be a length, nor can it exceed one. Dimensional Error
[11] Indices
Question 11
The expression $\dfrac{4^{5} \times 8^{3}}{2 \times 16^{2}}$ is equal to $2^{k}$ for exactly one value of $k$. What is $k$?
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 B)
Key idea. Every number here is a power of $2$, so rewrite each one and the whole expression collapses to a single exponent. Multiplication adds exponents, division subtracts them, and a power of a power multiplies them. Evaluating $4^{5}$ as $1024$ first throws away exactly the structure the question is about.
Fastest approach
### 1. Rewrite each factor as a power of $2$:
$4^{5} = \left(2^{2}\right)^{5} = 2^{10}$, $\quad 8^{3} = \left(2^{3}\right)^{3} = 2^{9}$, $\quad 16^{2} = \left(2^{4}\right)^{2} = 2^{8}$
### 2. Numerator: $2^{10} \times 2^{9} = 2^{19}$. Denominator: $2^{1} \times 2^{8} = 2^{9}$.
### 3. $\dfrac{2^{19}}{2^{9}} = 2^{10}$, so $k = 10$.
Keeping the intermediates as powers rather than as numbers is what makes this three lines. The same route through decimals is $524288 \div 512$, which is correct and slower and hides the answer until the last division.
Common mistake. Adding the exponents inside a power of a power, so $\left(2^{3}\right)^{3}$ becomes $2^{6}$ instead of $2^{9}$. The rule for a power raised to a power multiplies, and it is worth checking on a case small enough to count out: $\left(2^{3}\right)^{3} = 8 \times 8 \times 8 = 512 = 2^{9}$, whereas $2^{6}$ is only $64$.
Why the other options are wrong
- A: Evaluating $\left(2^{3}\right)^{3}$ as $2^{3+3} = 2^{6}$, adding where the power of a power rule multiplies, which gives $10 + 6 - 9$. Formula Misapplication
- C: Subtracting only $16^{2} = 2^{8}$ and never dividing by the lone factor of $2$, giving $19 - 8$. Incomplete Calculation
- D: Reading $4^{5}$ as $2^{4 \times 5} = 2^{20}$, using the base $4$ as a multiplier instead of writing $4 = 2^{2}$. Formula Misapplication
- E: Adding the denominator's exponent instead of subtracting it, giving $19 + 9$. Sign Error
[12] Coordinate geometry
Question 12
For which value of $a$ is the line $ax + 3y = 5$ perpendicular to the line $2x - y = 1$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\frac{3}{2}$ (option C)
Key idea. Rearrange each equation into the form $y = mx + c$ before comparing anything, then use the fact that two perpendicular lines have gradients whose product is $-1$. Neither gradient is visible until the rearranging is done, and reading them off the coefficients by eye is where this goes wrong.
Fastest approach
### 1. $ax + 3y = 5$ rearranges to $y = -\frac{a}{3}x + \frac{5}{3}$, so its gradient is $-\frac{a}{3}$.
### 2. $2x - y = 1$ rearranges to $y = 2x - 1$, so its gradient is $2$.
### 3. Perpendicular means the product of the gradients is $-1$:
$-\frac{a}{3} \times 2 = -1 \implies \frac{2a}{3} = 1 \implies a = \frac{3}{2}$
### 4. Sketch the two lines to check. The second climbs steeply to the right, so the first must fall gently to the left of vertical, meaning its gradient is negative and small in size. With $a = \frac{3}{2}$ the first gradient is $-\frac{1}{2}$, which is exactly that. A negative value of $a$ would make the first line climb as well, and two climbing lines can never be perpendicular.
Common mistake. Setting the two gradients equal, which is the condition for parallel lines, not perpendicular ones. The two conditions are easy to swap under time pressure and give completely different answers, so it is worth writing down which one is wanted before touching the algebra.
Why the other options are wrong
- A: Using the condition for parallel lines and setting $-\frac{a}{3} = 2$. Conceptual Misunderstanding
- B: Taking the product of perpendicular gradients as $+1$ rather than $-1$, giving $-\frac{2a}{3} = 1$. Sign Error
- D: Reaching $\frac{2a}{3} = 1$ correctly and then reading off $a = 3$, dividing by the $3$ but never by the $2$. Arithmetic Slip
- E: Reading the gradient of $ax + 3y = 5$ as $-\frac{3}{a}$, inverting the ratio of the two coefficients. Formula Misapplication
[13] Differentiation
Question 13
The normal to the curve $y = x^{2} - 6x + 13$ at the point where $x = 4$ crosses the $y$-axis. What is the $y$-coordinate of the crossing 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: $7$ (option D)
Key idea. The derivative gives the gradient of the tangent. The normal is perpendicular to it, so its gradient is $-\frac{1}{\text{tangent gradient}}$. Two sign decisions sit in this question, one in the reciprocal and one in the step to $x = 0$, and checking the sign of each before evaluating anything is faster than checking the arithmetic afterwards.
Fastest approach
### 1. Point on the curve: at $x = 4$, $y = 16 - 24 + 13 = 5$.
### 2. Gradient of the tangent: $\frac{\mathrm{d}y}{\mathrm{d}x} = 2x - 6$, which is $2$ at $x = 4$.
### 3. Gradient of the normal: $-\frac{1}{2}$.
### 4. Normal: $y - 5 = -\frac{1}{2}(x - 4)$. At $x = 0$, $y = 5 + 2 = 7$.
Check the sign before trusting the number. Going from $x = 4$ to $x = 0$ moves four to the left, and a line of gradient $-\frac{1}{2}$ rises as it goes left, so the crossing must be above $5$. That discards the two values below the point of contact without a single further calculation.
Common mistake. Using the derivative itself as the gradient of the normal. That is the tangent, and its line falls away from the curve in the opposite direction, which is why the wrong route lands well below the point rather than above it.
Why the other options are wrong
- A: Using the tangent instead of the normal, so $y - 5 = 2(x-4)$ and the crossing is at $5 - 8$. Misread Question
- B: Taking the reciprocal of the tangent gradient without changing its sign, giving a normal of gradient $\frac{1}{2}$ and a crossing at $5 - 2$. Sign Error
- C: Getting the gradient right but writing the point as $(5, 4)$, with the coordinates in the wrong order, giving $4 + \frac{5}{2}$. Arithmetic Slip
- E: Negating the tangent gradient without inverting it, giving a normal of gradient $-2$ and a crossing at $5 + 8$. Formula Misapplication
[14] Quadratics
Question 14
The equation $2x^{2} - 8x - 6 = 0$ has roots $\alpha$ and $\beta$. What is the value of $(\alpha - \beta)^{2}$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $28$ (option E)
Key idea. For $ax^{2}+bx+c = 0$ the roots satisfy $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$, so divide through by the leading coefficient before reading anything off. Then $(\alpha-\beta)^{2} = (\alpha+\beta)^{2} - 4\alpha\beta$, which never needs the roots themselves.
Fastest approach
### 1. Divide by $2$: $x^{2} - 4x - 3 = 0$, so $\alpha + \beta = 4$ and $\alpha\beta = -3$.
### 2. $(\alpha - \beta)^{2} = (\alpha+\beta)^{2} - 4\alpha\beta = 16 - 4(-3) = 16 + 12 = 28$
### 3. Sanity check on the sign. The product of the roots is negative, so the roots lie on opposite sides of zero and are therefore further apart than a pair with the same sign would be. A value smaller than $(\alpha+\beta)^{2} = 16$ would be going the wrong way.
Working symbolically to the last line is what keeps this short: the roots are $2 \pm \sqrt{7}$, and finding them first turns a two-line question into a surd expansion.
Common mistake. Reading the sum and the product straight off $-8$ and $-6$ without dividing by the leading coefficient of $2$. The relationships are $-\frac{b}{a}$ and $\frac{c}{a}$, and they only reduce to $-b$ and $c$ when the quadratic is monic.
Why the other options are wrong
- A: Taking the product of the roots as $+3$ rather than $-3$, so the calculation becomes $16 - 12$. Sign Error
- B: Reporting the correction term $-4\alpha\beta = 12$ on its own, without adding the square of the sum. Incomplete Calculation
- C: Stopping at $(\alpha+\beta)^{2} = 16$, as though squaring a difference gave the same result as squaring a sum. Formula Misapplication
- D: Computing $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2} - 2\alpha\beta = 22$, using the identity for the sum of the squares rather than the one for the square of the difference. Conceptual Misunderstanding
[15] Trigonometry
Question 15
A straight ladder of length $12 \ \text{m}$ leans against a vertical wall, with its foot on horizontal ground and the ladder making an angle of $60^{\circ}$ with the ground. Assuming the ladder does not bend and its foot does not slip, how far is the foot of the ladder from the wall?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $6 \ \text{m}$ (option A)
Key idea. The ladder is the hypotenuse, the distance from the wall is the side adjacent to the $60^{\circ}$ angle, so the ratio is the cosine and $\cos 60^{\circ} = \frac{1}{2}$ exactly. Naming which side is which before choosing between sine, cosine and tangent is the whole question.
Fastest approach
### 1. The wall is vertical and the ground horizontal, so the ladder, the wall and the ground make a right-angled triangle with the ladder as the hypotenuse.
### 2. The distance from the wall is adjacent to the $60^{\circ}$ angle:
$\text{distance} = 12\cos 60^{\circ} = 12 \times \tfrac{1}{2} = 6 \ \text{m}$
### 3. The bound does the checking for you. A leg of a right-angled triangle can never be longer than the hypotenuse, so no answer above $12 \ \text{m}$ can be right, whatever ratio produced it. That eliminates two of the five before any trigonometry, and knowing $\cos 60^{\circ} = \frac{1}{2}$ exactly settles the rest.
Common mistake. Reaching for the sine because the angle is at the foot of the ladder. The sine gives the side opposite the angle, which here is the height reached up the wall, not the distance along the ground.
Why the other options are wrong
- B: Using $12\tan 30^{\circ}$, applying the tangent to the complementary angle as though the ladder were the vertical side of the triangle. Formula Misapplication
- C: Using $12\sin 60^{\circ}$, which gives the height reached up the wall rather than the distance along the ground. Conceptual Misunderstanding
- D: Using $12\tan 60^{\circ}$, which treats the ladder as the side adjacent to the angle and produces a leg longer than the hypotenuse. Conceptual Misunderstanding
- E: Dividing by $\cos 60^{\circ}$ instead of multiplying by it, so the ladder is placed opposite the angle rather than across from the right angle. Formula Misapplication
[16] Numerical reasoning
Question 16
Estimate the value of $\dfrac{\left(6.1 \times 10^{8}\right) \times \left(3.9 \times 10^{-3}\right)}{2.05 \times 10^{4}}$ by first rounding each of the three numbers to one significant figure. Give the estimate in standard form.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $1.2 \times 10^{2}$ (option B)
Key idea. Do the powers of ten first and the digits afterwards. The exponents combine by addition and subtraction and are where an estimate goes badly wrong; the leading digits only ever move the answer by a factor of a few.
Fastest approach
### 1. Round to one significant figure: $6 \times 10^{8}$, $4 \times 10^{-3}$ and $2 \times 10^{4}$.
### 2. Digits: $\frac{6 \times 4}{2} = 12$.
### 3. Exponents: $8 + (-3) - 4 = 1$.
### 4. $12 \times 10^{1} = 1.2 \times 10^{2}$
The exact value is about $116$, so rounding three numbers to one significant figure has cost about $3\%$. That is the deal an estimate offers: the exponent is exact and the leading digit is approximate.
Common mistake. Adding the denominator's exponent instead of subtracting it. Every wrong answer here is a factor of ten or more away from the right one, which is what makes tracking exponents separately worth the extra line.
Why the other options are wrong
- A: Rounding $3.9 \times 10^{-3}$ to $4 \times 10^{-4}$, losing a whole decade during the rounding step itself. Order of Magnitude Error
- C: Dividing only the powers of ten and never dividing the digits by $2$, giving $24 \times 10^{1}$. Incomplete Calculation
- D: Reading the exponent $-3$ as $+3$, so the exponents give $8 + 3 - 4 = 7$. Misread Question
- E: Adding the denominator's exponent rather than subtracting it, so the exponents give $8 - 3 + 4 = 9$. Sign Error
[17] Quadratics
Question 17
The roots of $x^{2} - 7x + 9 = 0$ are $\alpha$ and $\beta$. What is the value of $(\alpha + 2)(\beta + 2)$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $27$ (option C)
Key idea. Expand first and only then substitute: $(\alpha+2)(\beta+2) = \alpha\beta + 2(\alpha+\beta) + 4$. Every term is one of the two clusters $\alpha+\beta$ and $\alpha\beta$, both of which are visible in the coefficients, so the individual roots never have to be found.
Fastest approach
### 1. From the coefficients: $\alpha + \beta = 7$ and $\alpha\beta = 9$.
### 2. Expand: $(\alpha+2)(\beta+2) = \alpha\beta + 2\alpha + 2\beta + 4 = \alpha\beta + 2(\alpha+\beta) + 4$.
### 3. Substitute: $9 + 2(7) + 4 = 9 + 14 + 4 = 27$
Naming the two clusters is what makes this quick. The roots themselves are $\frac{7 \pm \sqrt{13}}{2}$, so anyone who solves the quadratic first is committed to multiplying out a pair of surds for an answer that turns out to be an integer.
Common mistake. Expanding $(\alpha+2)(\beta+2)$ as $\alpha\beta + 4$ and losing the two cross terms. The same slip in reverse, forgetting the $2 \times 2$, is just as common, and both are caught by expanding the bracket in full before any numbers appear.
Why the other options are wrong
- A: Expanding to $\alpha\beta + 4$, dropping both cross terms, giving $9 + 4$. Incomplete Calculation
- B: Keeping the cross terms but forgetting the constant $2 \times 2$, giving $9 + 14$. Incomplete Calculation
- D: Doubling the product rather than the sum, giving $9 + 2(9) + 4$. Formula Misapplication
- E: Treating 'add 2 to each root' as 'multiply each root by 2', so the product becomes $4\alpha\beta = 36$. Conceptual Misunderstanding
[18] Differentiation
Question 18
The tangent to the curve $y = \dfrac{8}{x}$ at the point where $x = 2$ crosses the $x$-axis. What is the $x$-coordinate of the crossing 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 D)
Key idea. Write $\frac{8}{x}$ as $8x^{-1}$ before differentiating, which gives $-8x^{-2}$. The gradient is negative because the curve falls as $x$ increases, and that sign is worth checking before the arithmetic: a tangent that rises would cross the axis on the wrong side of the point of contact.
Fastest approach
### 1. Point of contact: at $x = 2$, $y = \frac{8}{2} = 4$.
### 2. $y = 8x^{-1}$, so $\frac{\mathrm{d}y}{\mathrm{d}x} = -8x^{-2} = -\frac{8}{x^{2}}$, which is $-2$ at $x = 2$.
### 3. Tangent: $y - 4 = -2(x-2)$, so $y = -2x + 8$ and the crossing is at $x = 4$.
### 4. Check the direction. The curve is decreasing everywhere for positive $x$, so the tangent slopes down and must reach the axis to the right of $x = 2$. Any answer at or to the left of $2$ has the tangent rising, which the curve never does.
Common mistake. Differentiating $\frac{8}{x}$ as though the $x$ were in the numerator, or evaluating $-\frac{8}{x^{2}}$ as $-\frac{8}{x}$ and losing the square. Rewriting as $8x^{-1}$ first removes both possibilities.
Why the other options are wrong
- A: Using the normal instead of the tangent, so the gradient is $\frac{1}{2}$ and the line is $y = \frac{x}{2} + 3$. Misread Question
- B: Taking the gradient as $+2$, missing the minus sign that comes from differentiating a negative power, which gives $y = 2x$. Sign Error
- C: Evaluating $-\frac{8}{x^{2}}$ as $-\frac{8}{2} = -4$, dividing by $x$ rather than by $x^{2}$, which gives $y = -4x + 12$. Formula Misapplication
- E: Reporting where the tangent meets the $y$-axis rather than the $x$-axis. Misread Question
[19] Series
Question 19
An $8$ by $8$ grid of unit cells is drawn, as on a chessboard. Counting squares of every size from $1$ by $1$ up to $8$ by $8$, and counting two squares as different whenever they occupy different cells, how many squares are there in the grid 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: $204$ (option E)
Key idea. Do not count squares, count positions. A $k$ by $k$ square is fixed by the cell at its top left corner, and that corner can sit in any of $9-k$ columns and $9-k$ rows. So there are $(9-k)^{2}$ squares of side $k$, one formula that answers all eight cases at once.
Fastest approach
### 1. General case first. A $k$ by $k$ square has $9-k$ possible horizontal positions and $9-k$ vertical ones, so there are $(9-k)^{2}$ of them.
### 2. Read off the special cases: $64$ of side $1$, $49$ of side $2$, then $36$, $25$, $16$, $9$, $4$ and $1$.
### 3. Total: $8^{2}+7^{2}+\cdots+1^{2} = \frac{8 \times 9 \times 17}{6} = 204$
### 4. Test the general result on a board small enough to check by eye. On a $2$ by $2$ grid the formula gives $4 + 1 = 5$, and there really are four small squares plus the whole board. A formula that survives the smallest case is usually safe on the largest.
Common mistake. Answering $64$, the number of cells. That counts only the squares of side $1$, and the question asks for squares of every size, of which the single largest is the whole grid.
Why the other options are wrong
- A: Summing the first eight integers, $\frac{8 \times 9}{2}$, instead of the first eight squares, so each size contributes its side length rather than its number of positions. Formula Misapplication
- B: Counting the cells of the grid, which is only the squares of side $1$. Misread Question
- C: Counting the $64$ cells and the whole board and stopping there, missing every intermediate size. Conceptual Misunderstanding
- D: Summing $1^{2}+2^{2}+\cdots+7^{2}$, which counts every size except the $64$ squares of side $1$. Incomplete Calculation
[20] Indices
Question 20
Solve $\dfrac{2^{x} \times 8^{\,x+1}}{4^{\,x-1}} = 32$.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $0$ (option A)
Key idea. Every base is a power of $2$, so the equation becomes $2^{\text{something}} = 2^{5}$ and the exponents can be equated directly. No logarithms are needed, and each bracket must be multiplied out in full: $8^{x+1} = 2^{3(x+1)}$, not $2^{3x+1}$.
Fastest approach
### 1. Rewrite each factor: $8^{x+1} = 2^{3(x+1)} = 2^{3x+3}$ and $4^{x-1} = 2^{2(x-1)} = 2^{2x-2}$.
### 2. Combine the exponents: $x + (3x+3) - (2x-2) = 2x + 5$.
### 3. $32 = 2^{5}$, so $2x + 5 = 5$ and $x = 0$.
### 4. Substitute the value back into the original, which is the cheapest possible check:
$\frac{2^{0} \times 8^{1}}{4^{-1}} = \frac{1 \times 8}{\frac{1}{4}} = 32$
That the answer turns out to be zero is not a sign of an error. It simply says the two sides already agree when every exponent is at its base value.
Common mistake. Multiplying only the $x$ by the new exponent, so $8^{x+1}$ becomes $2^{3x+1}$. The exponent of the outer power multiplies the whole inner exponent, brackets included, and substituting the answer back catches the slip immediately.
Why the other options are wrong
- B: Reading $32$ as $2^{6}$ rather than $2^{5}$, so the equation becomes $2x + 5 = 6$. Arithmetic Slip
- C: Adding the denominator's exponent instead of subtracting it, giving $x + 3x + 3 + 2x - 2 = 5$. Sign Error
- D: Expanding $8^{x+1}$ as $2^{3x+1}$, multiplying only the $x$ by $3$ and leaving the $1$ alone, which gives $2x + 3 = 5$. Formula Misapplication
- E: Solving $2^{x} = 32$ and ignoring the other two factors altogether. Incomplete Calculation