PAT practice · 20 questions · Free
PAT Exercise Paper 3: Physics and Mathematics
Every question carries a worked solution behind a disclosure, so you commit to an answer before you see the key.
[01] Kinematics
Question 1
A drone flies at a constant speed of $10 \ \text{m s}^{-1}$ and at a constant height throughout. It flies due north for $30 \ \text{s}$, then due east for $40 \ \text{s}$, then due south for $30 \ \text{s}$. Treat the turns as taking no time and ignore any wind. What total distance does the drone 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: $1000 \ \text{m}$ (option E)
Key idea. Distance is the length of the path and adds arithmetically leg by leg. Displacement is the vector from start to finish and the northward and southward legs cancel in it. The question asks for the path length, so nothing cancels.
Fastest approach
Each leg is a speed times a time.
1. Leg lengths
North: $10 \times 30 = 300 \ \text{m}$. East: $10 \times 40 = 400 \ \text{m}$. South: $10 \times 30 = 300 \ \text{m}$.
2. Add them as lengths
$300 + 400 + 300 = 1000 \ \text{m}$
Equivalently, the speed never changes, so the distance is just $10 \times (30 + 40 + 30)$ and the directions do not enter at all.
Worth doing alongside it: the displacement is $400 \ \text{m}$ due east, because the two north and south legs cancel exactly. Two different questions, two different answers, from the same journey.
Common mistake. Answering with the magnitude of the displacement. A sketch settles it in seconds: the drone ends up $400 \ \text{m}$ east of where it started, but it has flown a path four hundred metres longer than that on the way.
Why the other options are wrong
- A: The magnitude of the displacement from start to finish, which is a vector answer given to a scalar question. Conceptual Misunderstanding
- B: The straight-line distance from the start to the corner reached after the first two legs, $\sqrt{300^{2} + 400^{2}}$, with the third leg never used. Incomplete Calculation
- C: Adding the first two legs and forgetting the southward one. Incomplete Calculation
- D: Adding the resultant of the first two legs, $500 \ \text{m}$, to the length of the third, which mixes a displacement and a distance in one sum. Conceptual Misunderstanding
[02] Electric circuits
Question 2
A resistor transfers $90 \ \text{J}$ of energy to its surroundings in $30 \ \text{s}$ while the potential difference across it is $6.0 \ \text{V}$. Assume the current and the potential difference are both steady over that time and that all of the energy is transferred in the resistor itself. What is the current in the resistor?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $0.50 \ \text{A}$ (option A)
Key idea. Potential difference is energy per unit charge, so the charge that flowed is $Q = \frac{W}{V}$. Current is charge per unit time, so $I = \frac{Q}{t}$. Neither definition needs a resistance, and the whole question is those two divisions in the right order.
Fastest approach
1. Charge, from the definition of potential difference
$V = \dfrac{W}{Q} \implies Q = \dfrac{W}{V} = \dfrac{90}{6} = 15 \ \text{C}$
2. Current, from the definition of current
$I = \dfrac{Q}{t} = \dfrac{15}{30} = 0.50 \ \text{A}$
The same answer arrives one line quicker through power: $P = \frac{90}{30} = 3.0 \ \text{W}$ and $I = \frac{P}{V} = \frac{3.0}{6.0} = 0.50 \ \text{A}$. Both routes are the two definitions rearranged, and neither needs the resistance, which is $12 \ \Omega$ and never asked for.
Common mistake. Dividing the energy by the time and stopping. That is the power in watts, and a watt is not an ampere. Naming the denominator of each quantity as you write it, energy per charge and charge per time, keeps the two apart.
Why the other options are wrong
- B: Dividing the energy by the time, which gives the power of $3.0 \ \text{W}$ and is then quoted as though watts were amperes. Dimensional Error
- C: Dividing the time by the potential difference. The two numbers given are the ones left over after the energy is used, but the combination has no physical meaning. Formula Misapplication
- D: Stopping at the charge, $15 \ \text{C}$, and reporting coulombs as amperes. Incomplete Calculation
- E: Using $I = Qt$ rather than $I = \frac{Q}{t}$, so the charge is multiplied by the thirty seconds instead of shared out over them. Formula Misapplication
[03] Matter and thermal energy
Question 3
$0.20 \ \text{kg}$ of ice at $0 \ ^{\circ}\text{C}$ is dropped into $0.80 \ \text{kg}$ of water at $25 \ ^{\circ}\text{C}$ in a perfectly insulated container of negligible heat capacity. Take the specific heat capacity of liquid water as $4200 \ \text{J kg}^{-1}\text{K}^{-1}$ and the specific latent heat of fusion of ice as $3.36 \times 10^{5} \ \text{J kg}^{-1}$, and assume no water evaporates. What is the final temperature of the contents?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $4.0 \ ^{\circ}\text{C}$ (option B)
Key idea. Melting ice absorbs energy at constant temperature, and that latent heat is far larger per kilogram than the heat capacity suggests: $3.36 \times 10^{5}$ divided by $4200$ is $80$, so melting a kilogram of ice costs as much energy as cooling a kilogram of water by $80 \ \text{K}$. Compare the energy available against the energy needed to melt everything before assuming which way it goes.
Fastest approach
1. Size the two energies against each other
Available from cooling the warm water all the way to $0 \ ^{\circ}\text{C}$:
$0.80 \times 4200 \times 25 = 84\,000 \ \text{J}$
Needed to melt all the ice:
$0.20 \times 3.36 \times 10^{5} = 67\,200 \ \text{J}$
There is more available than needed, so every gram of ice melts and the final temperature is above $0 \ ^{\circ}\text{C}$. That comparison is the decision the question is really asking for.
2. Spend what is left on warming the whole kilogram
$84\,000 - 67\,200 = 16\,800 \ \text{J}$ remains, and the container now holds $1.00 \ \text{kg}$ of liquid water.
$\Delta T = \dfrac{16\,800}{1.00 \times 4200} = 4.0 \ \text{K}$
Starting from $0 \ ^{\circ}\text{C}$, the final temperature is $4.0 \ ^{\circ}\text{C}$.
Common mistake. Treating this as a plain mixing problem and averaging the temperatures by mass, which ignores the latent heat entirely. Melting the ice costs more energy here than warming any of the water does, so leaving it out moves the answer by sixteen degrees.
Why the other options are wrong
- A: Assuming the mixture must settle at the melting point simply because ice was present. That is only true when there is too little energy to melt all of it, which the energy comparison shows is not the case here. Conceptual Misunderstanding
- C: Melting the ice correctly but then warming only the original $0.80 \ \text{kg}$ of water, forgetting that the meltwater must be raised to the final temperature too. Incomplete Calculation
- D: Taking the plain average of $25 \ ^{\circ}\text{C}$ and $0 \ ^{\circ}\text{C}$, which uses neither the masses nor the latent heat. Formula Misapplication
- E: A mass-weighted mixing calculation, $0.80 \times 25 = 1.00 \times T$, with the latent heat of fusion left out altogether. The arithmetic is sound and the physics is not. Conceptual Misunderstanding
[04] Energy and power
Question 4
A $2.0 \ \text{kg}$ block is released from rest at the top of a rough ramp and slides to the bottom, descending $5.0 \ \text{m}$ vertically. It arrives at the bottom moving at $6.0 \ \text{m s}^{-1}$. Take $g = 10 \ \text{m s}^{-2}$, treat the block as a point mass and ignore air resistance. How much energy has been transferred to thermal energy and sound?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $64 \ \text{J}$ (option C)
Key idea. Energy is conserved even when it is not useful. The gravitational store empties by $mgh$; the kinetic store fills by $\tfrac{1}{2}mv^{2}$; whatever the two do not account for has gone to thermal energy in the ramp and the block, and to sound.
Fastest approach
1. Energy released from the gravitational store
$mgh = 2.0 \times 10 \times 5.0 = 100 \ \text{J}$
2. Energy now in the kinetic store
$\tfrac{1}{2}mv^{2} = \tfrac{1}{2} \times 2.0 \times 6.0^{2} = 36 \ \text{J}$
3. The missing entry
$100 - 36 = 64 \ \text{J}$
The block started from rest, so there is no initial kinetic energy to carry into the account. Notice the mass appears in both terms and cancels from neither, because the two terms are subtracted rather than divided.
Common mistake. Reporting the kinetic energy at the bottom, which is what the block has rather than what it has lost. Reading the last sentence of the stem again before writing a number is the cheapest fix there is.
Why the other options are wrong
- A: Using $mv^{2}$ instead of $\tfrac{1}{2}mv^{2}$, so the kinetic store is credited with $72 \ \text{J}$ and only $28 \ \text{J}$ appears to be missing. Formula Misapplication
- B: Reporting the kinetic energy at the bottom rather than the energy dissipated. Misread Question
- D: Assuming all of the gravitational energy released is dissipated, which would require the block to arrive at rest. It arrives at $6.0 \ \text{m s}^{-1}$, so it cannot have. Conceptual Misunderstanding
- E: Adding the two energies rather than subtracting, which credits the dissipation with energy the block still visibly has. Sign Error
[05] Radioactivity
Question 5
A radioactive source has a half-life of $20$ minutes and its activity is measured as $960 \ \text{Bq}$. Assume the source contains only one radioactive isotope and that the background count is negligible. How long after that measurement will its activity have fallen to $30 \ \text{Bq}$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $100 \ \text{minutes}$ (option D)
Key idea. Take the ratio of the two activities and ask how many halvings it represents. $\frac{960}{30} = 32 = 2^{5}$, so five half-lives have passed. The ratio is a pure number; the time comes only when it multiplies the half-life.
Fastest approach
1. Ratio of the two activities
$\dfrac{960}{30} = 32$
2. How many halvings is that?
$32 = 2^{5}$, so five half-lives.
Counting it out is just as fast and less error-prone: $960 \to 480 \to 240 \to 120 \to 60 \to 30$, which is five arrows.
3. Convert to a time
$5 \times 20 = 100$ minutes
The number of half-lives is a pure number and the half-life carries the units, which is the only way the answer can come out in minutes at all.
Common mistake. Multiplying the half-life by the activity ratio of $32$ rather than by the five halvings that ratio represents. The ratio counts how many times the activity has been divided in half, not how many half-lives have passed.
Why the other options are wrong
- A: Quoting the half-life itself, which the question supplies rather than asks for. Misread Question
- B: Reporting the activity ratio $\frac{960}{30} = 32$ as a number of minutes. The ratio is a pure number and cannot be a time. Dimensional Error
- C: Counting four halvings rather than five, which reaches $60 \ \text{Bq}$ and stops one step early. Incomplete Calculation
- E: Multiplying the half-life by the activity ratio, $20 \times 32$, instead of by the number of halvings. Formula Misapplication
[06] Gravitation and orbits
Question 6
Two sacks of sand, each of mass $50 \ \text{kg}$, rest on a floor with their centres $0.50 \ \text{m}$ apart. Take the gravitational constant as $G = 6.7 \times 10^{-11} \ \text{N m}^{2}\text{kg}^{-2}$. Treating each sack as a point mass at its centre and ignoring the gravitational pull of everything else, what is the gravitational force each sack exerts on the other?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $6.7 \times 10^{-7} \ \text{N}$ (option E)
Key idea. $F = \dfrac{Gm_{1}m_{2}}{r^{2}}$. The two masses multiply, they never add, and the separation is squared. Because $r$ is less than a metre here, dividing by $r$ instead of by $r^{2}$ makes the force smaller rather than larger, so a rough check on size will not rescue that slip.
Fastest approach
1. Assemble the numerator and the denominator separately
$m_{1}m_{2} = 50 \times 50 = 2.5 \times 10^{3} \ \text{kg}^{2}$, and $r^{2} = 0.50^{2} = 0.25 \ \text{m}^{2}$.
2. Divide, then apply the constant
$\dfrac{2.5 \times 10^{3}}{0.25} = 1.0 \times 10^{4}$
$F = 6.7 \times 10^{-11} \times 1.0 \times 10^{4} = 6.7 \times 10^{-7} \ \text{N}$
3. Ask whether that number could be true
Under a millionth of a newton is about the weight of a grain of sand, and nobody has ever watched two sacks in a shed drift together. A force of that size is exactly what everyday experience demands, which is the reassurance the arithmetic on its own cannot give.
Common mistake. Adding the two masses instead of multiplying them. That leaves kilograms where the formula needs kilograms squared, so the result is not a force at all, whatever units get written after it.
Why the other options are wrong
- A: Using only one of the two masses, as though gravity depended on the attracting body alone. Incomplete Calculation
- B: Adding the masses instead of multiplying them. The numerator then carries kilograms rather than kilograms squared, and the expression cannot produce a force. Dimensional Error
- C: Multiplying by $r^{2}$ instead of dividing by it, so the attraction grows with separation rather than falling. Formula Misapplication
- D: Dividing by the separation rather than by its square, which halves the answer here rather than doubling it because the separation is less than a metre. Formula Misapplication
[07] Friction, springs and machines
Question 7
A light spring obeys Hooke's law with a spring constant of $200 \ \text{N m}^{-1}$ and is never stretched beyond its elastic limit. Its extension is increased from $0.30 \ \text{m}$ to $0.40 \ \text{m}$. How much additional elastic potential energy is stored in the spring?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $7.0 \ \text{J}$ (option A)
Key idea. Write it in symbols first: $\Delta E = \tfrac{1}{2}k x_{2}^{2} - \tfrac{1}{2}k x_{1}^{2} = \tfrac{1}{2}k\left(x_{2}^{2} - x_{1}^{2}\right)$. Because the energy is quadratic in the extension, the extra energy depends on where the stretch starts, not only on how long it is.
Fastest approach
1. Difference of two squares, in symbols
$\Delta E = \tfrac{1}{2}k\left(x_{2}^{2} - x_{1}^{2}\right)$
2. Substitute once, at the end
$x_{2}^{2} - x_{1}^{2} = 0.40^{2} - 0.30^{2} = 0.16 - 0.09 = 0.07 \ \text{m}^{2}$
$\Delta E = \tfrac{1}{2} \times 200 \times 0.07 = 7.0 \ \text{J}$
3. Check it against a case you can see
The same extra $0.10 \ \text{m}$ applied from rest, from $0$ to $0.10 \ \text{m}$, would store only $\tfrac{1}{2} \times 200 \times 0.01 = 1.0 \ \text{J}$. Seven times as much energy for the same extra stretch, because the spring is already pulling hard before it starts. That is the whole content of the quadratic.
Common mistake. Using $\tfrac{1}{2}k(\Delta x)^{2}$ with $\Delta x = 0.10 \ \text{m}$, which answers a different question: the energy stored by stretching a relaxed spring by ten centimetres. Energies subtract, extensions do not.
Why the other options are wrong
- B: Taking the work as the final force, $200 \times 0.40 = 80 \ \text{N}$, times the extra extension of $0.10 \ \text{m}$. A force times a distance is an energy, so nothing looks wrong, but the force is not constant over that stretch and using its largest value overstates the work. Conceptual Misunderstanding
- C: Reporting the energy already stored at $0.30 \ \text{m}$ rather than the increase. Misread Question
- D: Computing $k\left(x_{2}^{2} - x_{1}^{2}\right)$ with the factor of one half dropped. Formula Misapplication
- E: Reporting the total energy stored at $0.40 \ \text{m}$, without subtracting what the spring already held. Incomplete Calculation
[08] Problem solving and estimation
Question 8
On a stretched string the speed of a transverse wave is $v = \sqrt{\dfrac{T}{\mu}}$, where $T$ is the tension in the string and $\mu$ is its mass per unit length. A uniform string of mass $20 \ \text{g}$ and length $5.0 \ \text{m}$ is to carry transverse waves at $200 \ \text{m s}^{-1}$. Assume the string stays uniform and that its mass does not change as it is stretched. What tension is required?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $160 \ \text{N}$ (option B)
Key idea. Nothing here has to be remembered: the relationship is given. Square both sides and rearrange to $T = \mu v^{2}$, then check that rearrangement against its units before putting numbers in. $\text{kg m}^{-1} \times \text{m}^{2}\text{s}^{-2} = \text{kg m s}^{-2}$, which is a newton, so the rearrangement is sound.
Fastest approach
1. Mass per unit length
$\mu = \dfrac{20 \times 10^{-3}}{5.0} = 4.0 \times 10^{-3} \ \text{kg m}^{-1}$
The grams must become kilograms here, before anything else happens.
2. Rearrange the supplied relationship
$v = \sqrt{\dfrac{T}{\mu}} \implies v^{2} = \dfrac{T}{\mu} \implies T = \mu v^{2}$
Check the units of that last form: $\text{kg m}^{-1}$ times $\text{m}^{2}\text{s}^{-2}$ gives $\text{kg m s}^{-2}$, a newton. The other rearrangement someone might write, $T = \frac{v^{2}}{\mu}$, gives $\text{m}^{3}\text{s}^{-2}\text{kg}^{-1}$, which is not a force, and the check rules it out in seconds.
3. Substitute
$T = 4.0 \times 10^{-3} \times 200^{2} = 4.0 \times 10^{-3} \times 4.0 \times 10^{4} = 160 \ \text{N}$
Common mistake. Using the string's total mass in place of its mass per unit length. The symbol $\mu$ is defined in the stem as a mass divided by a length, and the length of the string is given for exactly that reason.
Why the other options are wrong
- A: Using $\mu v$ rather than $\mu v^{2}$, so the speed is never squared. That product carries $\text{kg s}^{-1}$, which is not a force, and a units check kills it before any arithmetic. Dimensional Error
- C: Using the string's total mass of $0.020 \ \text{kg}$ in place of its mass per unit length, so the five metres of string are never used. Conceptual Misunderstanding
- D: Leaving the mass in grams, so $\mu$ is taken as $4.0 \ \text{g m}^{-1}$ rather than $4.0 \times 10^{-3} \ \text{kg m}^{-1}$. Unit Error
- E: Rearranging to $T = \frac{v^{2}}{\mu}$ rather than $T = \mu v^{2}$, inverting the role of the mass per unit length. Formula Misapplication
[09] Problem solving and estimation
Question 9
A storeroom measures $5.0 \ \text{m}$ by $4.0 \ \text{m}$ by $2.5 \ \text{m}$ and is otherwise empty. Table tennis balls, each a sphere of diameter $40 \ \text{mm}$, are poured in until it is full. Assume the balls occupy $60\%$ of the room's volume once poured, that none is crushed, and that the volume of a sphere is $\tfrac{4}{3}\pi r^{3}$. Which of these is the best estimate of the number of balls in the room?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $9.0 \times 10^{5}$ (option C)
Key idea. A counting estimate is one division: the volume actually available divided by the volume of one ball. The two places it goes wrong are the packing fraction, which the stem supplies, and the radius, which is half the diameter the stem supplies.
Fastest approach
1. Volume available to the balls
Room: $5.0 \times 4.0 \times 2.5 = 50 \ \text{m}^{3}$. At $60\%$: $30 \ \text{m}^{3}$.
2. Volume of one ball
Diameter $40 \ \text{mm}$ means radius $0.020 \ \text{m}$.
$\tfrac{4}{3}\pi (0.020)^{3} = \tfrac{4}{3}\pi \times 8.0 \times 10^{-6} \approx 3.35 \times 10^{-5} \ \text{m}^{3}$
3. Divide the big by the small
$\dfrac{30}{3.35 \times 10^{-5}} \approx 9.0 \times 10^{5}$
About nine hundred thousand balls. Every input here was supplied, so the only judgement left is arithmetic, but note where the precision really lives: the packing fraction is the one number a different person would have chosen differently, and quoting the result to two significant figures rather than to six is what that uncertainty deserves.
Common mistake. Putting the diameter into the sphere formula in place of the radius, which multiplies the volume of a single ball by eight and divides the count by eight. Halving the diameter before anything else is written down avoids it.
Why the other options are wrong
- A: Using the diameter of $40 \ \text{mm}$ as the radius, which makes each ball eight times too big. Formula Misapplication
- B: Treating each ball as a cube of side $40 \ \text{mm}$ rather than as a sphere. The dimensions are right and the geometry is not: a cube holds about $1.9$ times the volume of the sphere that fits inside it, so the count comes out too low. Conceptual Misunderstanding
- D: Using $\pi r^{3}$ for the volume of a sphere, dropping the factor of $\tfrac{4}{3}$. Formula Misapplication
- E: Using the whole $50 \ \text{m}^{3}$ of the room, as though spheres could fill a space with no gaps between them. Incomplete Calculation
[10] Problem solving and estimation
Question 10
The speed $v$ of sound in a gas is modelled as depending only on the pressure $p$ of the gas and on its density $\rho$, through $v = k\,p^{a}\rho^{b}$, where $k$ is a dimensionless constant and $a$ and $b$ are fixed numbers. Requiring the expression to be dimensionally consistent fixes $a$ and $b$. Taking $k = 1.2$, $p = 1.0 \times 10^{5} \ \text{Pa}$ and $\rho = 1.25 \ \text{kg m}^{-3}$, what speed does the model predict?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $339 \ \text{m s}^{-1}$ (option D)
Key idea. Write the dimensions of each side and match them. $[p] = \text{M L}^{-1}\text{T}^{-2}$ and $[\rho] = \text{M L}^{-3}$, and the left-hand side is $\text{L T}^{-1}$. Matching the powers of mass, length and time gives three equations for two unknowns, and they are consistent, which is itself a check that the model is not missing a variable.
Fastest approach
1. Match the exponents
$\left[p^{a}\rho^{b}\right] = \left(\text{M L}^{-1}\text{T}^{-2}\right)^{a}\left(\text{M L}^{-3}\right)^{b} = \text{M}^{a+b}\,\text{L}^{-a-3b}\,\text{T}^{-2a}$
Set that equal to $\text{L T}^{-1}$:
Time: $-2a = -1$, so $a = \tfrac{1}{2}$. Mass: $a + b = 0$, so $b = -\tfrac{1}{2}$. Length: $-a - 3b = -\tfrac{1}{2} + \tfrac{3}{2} = 1$, which checks out.
2. So the model is a square root of a ratio
$v = k\sqrt{\dfrac{p}{\rho}}$
3. Evaluate
$\dfrac{p}{\rho} = \dfrac{1.0 \times 10^{5}}{1.25} = 8.0 \times 10^{4} \ \text{m}^{2}\text{s}^{-2}$, and $\sqrt{8.0 \times 10^{4}} \approx 283 \ \text{m s}^{-1}$.
$v = 1.2 \times 283 \approx 339 \ \text{m s}^{-1}$
The measured speed of sound in air near room temperature is about $340 \ \text{m s}^{-1}$, which is why a constant of $1.2$ was chosen. Dimensions gave the form; they could never have given that $1.2$.
Common mistake. Putting the density in the numerator, so the model reads $k\sqrt{p\rho}$. That combination has dimensions $\text{M L}^{-2}\text{T}^{-1}$ and is not a speed at all, so it fails the very consistency requirement the question is built on. It also says sound travels faster in a denser gas, which is backwards.
Why the other options are wrong
- A: Dividing by the constant $k$ instead of multiplying by it. Arithmetic Slip
- B: Evaluating $\sqrt{p/\rho}$ and forgetting to apply the constant $k$ at all. Incomplete Calculation
- C: Placing $k$ inside the square root, giving $\sqrt{kp/\rho}$, so the constant is effectively square-rooted. Formula Misapplication
- E: Taking $b = +\tfrac{1}{2}$, giving $k\sqrt{p\rho}$. That expression carries $\text{M L}^{-2}\text{T}^{-1}$ rather than $\text{L T}^{-1}$, so it is not a speed, and it also has sound speeding up in a denser gas. Dimensional Error
[11] Series
Question 11
An arithmetic progression has first term $7$ and common difference $5$. What is the sum of its first $20$ terms?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $1090$ (option E)
Key idea. $S_{n} = \frac{n}{2}\left(2a + (n-1)d\right)$. The bracket holds $2a$, not $a$, because it is the sum of the first and last terms, and the last term uses $n-1$ steps rather than $n$.
Fastest approach
1.
The last term is $a + (n-1)d = 7 + 19 \times 5 = 102$.
2.
Pair the first term with the last: $7 + 102 = 109$. There are $10$ such pairs among $20$ terms.
$S_{20} = 10 \times 109 = 1090$
3.
The same thing in one line: $S_{20} = \frac{20}{2}\left(2 \times 7 + 19 \times 5\right) = 10(14 + 95) = 1090$.
Common mistake. Writing $19$ steps as $20$, or putting $a$ rather than $2a$ inside the bracket. The bracket is $a + \ell$, the first term plus the last, and $\ell = a + (n-1)d$.
Why the other options are wrong
- A: Reporting the twentieth term, $a + 19d$, rather than the sum of the terms. Misread Question
- B: Computing $2a + 19d = 109$ and stopping there, without the factor of $\frac{n}{2} = 10$ outside the bracket. Incomplete Calculation
- C: Using $n = 19$, which sums only as far as the nineteenth term and leaves out the final $102$. Incomplete Calculation
- D: Writing the bracket as $a + (n-1)d = 102$ instead of $2a + (n-1)d = 109$, giving $10 \times 102$. Formula Misapplication
[12] Differentiation
Question 12
A rectangular sheet of card measures $16 \ \text{cm}$ by $10 \ \text{cm}$. A square of side $x$ is cut from each of the four corners and the flaps are folded up to make an open tray. Which value of $x$ gives the tray its greatest volume?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $2 \ \text{cm}$ (option A)
Key idea. Each side loses $x$ at both ends, so the base is $(16 - 2x)$ by $(10 - 2x)$ and $V = x(16-2x)(10-2x)$. The derivative is a quadratic with two roots, and only one of them describes a tray that exists: $x$ must satisfy $0 < x < 5$, or the shorter side is used up.
Fastest approach
1.
$V = x(16 - 2x)(10 - 2x) = 4x^{3} - 52x^{2} + 160x$
2.
$\frac{\mathrm{d}V}{\mathrm{d}x} = 12x^{2} - 104x + 160 = 4(3x - 20)(x - 2)$
3.
The roots are $x = 2$ and $x = \frac{20}{3}$. The card is only $10 \ \text{cm}$ wide, so $x$ must be less than $5$, and $\frac{20}{3}$ is discarded.
$x = 2 \ \text{cm}$, giving $V = 2 \times 12 \times 6 = 144 \ \text{cm}^{3}$.
Common mistake. Solving the quadratic and reporting whichever root comes out first. Both roots are stationary points of the cubic, but only one of them lies in $0 < x < 5$, and outside that range the expression stops describing a tray at all.
Why the other options are wrong
- B: Writing the base as $(16 - x)$ by $(10 - x)$, forgetting that a square is removed from both ends of every side. That cubic has a stationary point at $x = 4$. Formula Misapplication
- C: Taking $x$ as large as the shorter side allows. At $x = 5$ the base has zero width and the tray holds nothing. Conceptual Misunderstanding
- D: The other root of the derivative. It is a genuine stationary point of the cubic but lies outside $0 < x < 5$, so no tray of that shape can be folded. Incomplete Calculation
- E: Treating the $160x$ term as a constant when differentiating, so the derivative loses its $+160$ and becomes $12x^{2} - 104x$. Formula Misapplication
[13] Exponentials and logarithms
Question 13
Solve $\log_{2} x + \log_{2}(x - 2) = 3$.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $x = 4$ (option B)
Key idea. $\log_{2}A + \log_{2}B = \log_{2}(AB)$, so the two logarithms become one and the equation converts to exponential form as $x(x-2) = 2^{3}$. A logarithm only accepts a positive argument, so any root with $x \leq 2$ has to go.
Fastest approach
1.
$\log_{2}\left[x(x-2)\right] = 3$
2.
Convert to exponential form: $x(x - 2) = 2^{3} = 8$.
3.
$x^{2} - 2x - 8 = 0 \implies (x - 4)(x + 2) = 0$
4.
$x = -2$ makes both logarithms undefined, so $x = 4$.
Check: $\log_{2}4 + \log_{2}2 = 2 + 1 = 3$.
Common mistake. Adding the arguments instead of multiplying them, which turns the left side into $\log_{2}(2x - 2)$ and gives $x = 5$. Addition of logarithms corresponds to multiplication inside them, never to addition inside them.
Why the other options are wrong
- A: The second root of the quadratic, kept without checking it. It makes $\log_{2}x$ the logarithm of a negative number, which does not exist. Conceptual Misunderstanding
- C: Adding the arguments rather than multiplying: $\log_{2}(2x - 2) = 3$ gives $2x - 2 = 8$. Formula Misapplication
- D: Solving $\log_{2}x = 3$ alone and ignoring the second logarithm entirely. Incomplete Calculation
- E: Solving $x(x - 2) = 8$ by dropping the factor of $x$ and setting $x - 2 = 8$. Formula Misapplication
[14] Binomial expansion
Question 14
The expression $(1 + 3x)^{4}$ is expanded in full. What is the sum of the five coefficients in that expansion?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $256$ (option C)
Key idea. Every power of $x$ becomes $1$ when $x = 1$, so the value of the polynomial at $x = 1$ is exactly the sum of its coefficients. There is no need to write the expansion out.
Fastest approach
1.
Putting $x = 1$ turns each term $c_{r}x^{r}$ into $c_{r}$, so the sum of the coefficients is the value of $(1 + 3x)^{4}$ at $x = 1$.
2.
$(1 + 3)^{4} = 4^{4} = 256$
3.
The long way agrees. The expansion is $1 + 12x + 54x^{2} + 108x^{3} + 81x^{4}$, and $1 + 12 + 54 + 108 + 81 = 256$.
Common mistake. Expanding all five terms and then slipping in the arithmetic, when a single substitution settles it. The other frequent error is forgetting that the $3$ is raised to a power in every term, so it contributes far more than a factor of $3$ overall.
Why the other options are wrong
- A: Expanding $(1 + x)^{4}$ and losing the $3$ altogether, which gives $2^{4}$. Misread Question
- B: Adding $1^{4}$ and $3^{4}$, as though the expansion were just the two terms of the bracket raised to the fourth power. Conceptual Misunderstanding
- D: Evaluating $4^{5}$, taking the exponent from the five terms in the expansion rather than from the bracket. Formula Misapplication
- E: Substituting $x = 3$ rather than $x = 1$, which evaluates the expansion at a point instead of summing its coefficients: $(1 + 9)^{4}$. Misread Question
[15] Trigonometry
Question 15
An angle $\theta$ is obtuse, so it lies between $90^{\circ}$ and $180^{\circ}$, and $\sin\theta = \frac{3}{5}$. What is $\tan\theta$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $-\frac{3}{4}$ (option D)
Key idea. $\sin^{2}\theta + \cos^{2}\theta = 1$ fixes the size of the cosine but not its sign. The quadrant fixes the sign: between $90^{\circ}$ and $180^{\circ}$ the sine is positive and the cosine is negative, so the tangent is negative.
Fastest approach
1.
$\cos^{2}\theta = 1 - \left(\tfrac{3}{5}\right)^{2} = \tfrac{16}{25}$, so $\cos\theta = \pm\tfrac{4}{5}$.
2.
The angle is obtuse, so the cosine is negative: $\cos\theta = -\tfrac{4}{5}$.
3.
$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{-4/5} = -\frac{3}{4}$
The 3-4-5 triangle supplies the digits; the quadrant supplies the sign.
Common mistake. Taking the positive square root by habit and answering $\frac{3}{4}$. The identity gives two candidate cosines and only the quadrant can choose between them.
Why the other options are wrong
- A: Dividing the cosine by the sine, which is $\cot\theta$ rather than $\tan\theta$. Formula Misapplication
- B: Reporting $\sec\theta = \frac{1}{\cos\theta}$ instead of the tangent. Formula Misapplication
- C: Stopping at the cosine, which is what the identity delivers, rather than forming the ratio of sine to cosine. Incomplete Calculation
- E: Taking $\cos\theta = +\frac{4}{5}$, so the tangent comes out positive. An obtuse angle has a negative cosine. Sign Error
[16] Inequalities
Question 16
How many integers $n$ with $1 \leq n \leq 20$ satisfy $n^{2} - 12n + 20 \geq 0$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $13$ (option E)
Key idea. $n^{2} - 12n + 20 = (n-2)(n-10)$, and the parabola opens upwards, so the expression is negative only strictly between the roots. The solution set is therefore $n \leq 2$ together with $n \geq 10$, which is two separate stretches rather than one interval.
Fastest approach
1.
Factorise: $n^{2} - 12n + 20 = (n - 2)(n - 10)$.
2.
The parabola opens upwards, so it is at or above zero outside the roots: $n \leq 2$ or $n \geq 10$. The sign is $\geq$, so both roots are included.
3.
Count inside the window $1 \leq n \leq 20$:
$n = 1, 2$ gives $2$ values, and $n = 10, 11, \ldots, 20$ gives $11$ values.
$2 + 11 = 13$
Common mistake. Writing the answer as a single interval between the roots. An upward parabola is negative between its roots and positive outside them, so an inequality of this shape splits into two pieces, and the count has to cover both.
Why the other options are wrong
- A: Taking the values strictly between the roots, $n = 3$ to $n = 9$, which is where the expression is negative. Conceptual Misunderstanding
- B: Taking the values between the roots including the roots themselves, $n = 2$ to $n = 10$. Conceptual Misunderstanding
- C: Discarding $n = 2$ and $n = 10$, where the expression is exactly zero. The sign is $\geq$, so those two count. Misread Question
- D: Counting $n = 10$ up to $n = 20$ as ten values rather than eleven. Arithmetic Slip
[17] Units and numerical reasoning
Question 17
A solid metal cube has edges of length $20 \ \text{cm}$. What is its volume in cubic metres?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $0.008 \ \text{m}^{3}$ (option A)
Key idea. A conversion factor between lengths is cubed when it acts on a volume. Since $1 \ \text{m} = 100 \ \text{cm}$, it follows that $1 \ \text{m}^{3} = 100^{3} = 10^{6} \ \text{cm}^{3}$, not $100 \ \text{cm}^{3}$.
Fastest approach
1.
Convert the edge first, then cube: $20 \ \text{cm} = 0.2 \ \text{m}$.
$V = (0.2)^{3} = 0.008 \ \text{m}^{3}$
2.
The other order agrees, provided the factor is cubed: $V = 20^{3} = 8000 \ \text{cm}^{3}$, and
$\frac{8000}{10^{6}} = 8 \times 10^{-3} \ \text{m}^{3}$
3.
A sanity check on the size: the cube is a fifth of a metre on each edge, so it should occupy about a hundredth of a cubic metre, and it does.
Common mistake. Dividing by $100$ or by $1000$ instead of by $10^{6}$. The figure $1000$ is especially tempting because $1000 \ \text{cm}^{3}$ is one litre, and there are a thousand litres in a cubic metre, not one.
Why the other options are wrong
- B: Dividing $8000 \ \text{cm}^{3}$ by $10^{5}$, one power of ten short of the correct $10^{6}$. Order of Magnitude Error
- C: Dividing by $10^{4}$, from squaring the factor of $100$ rather than cubing it, as though the conversion acted on an area. Formula Misapplication
- D: Using $1 \ \text{m}^{3} = 1000 \ \text{cm}^{3}$. That figure is the litre, and a cubic metre holds a thousand of them. Unit Error
- E: Dividing by $100$ just once, converting a single edge to metres and leaving the other two in centimetres. Unit Error
[18] Exponentials and logarithms
Question 18
The equation $2^{2x} - 10 \times 2^{x} + 16 = 0$ has two solutions. What is the larger 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: $3$ (option B)
Key idea. $2^{2x} = \left(2^{x}\right)^{2}$, so writing $u = 2^{x}$ turns the equation into $u^{2} - 10u + 16 = 0$. Each value of $u$ then has to be converted back, because the question asks for $x$ and not for $2^{x}$.
Fastest approach
1.
Let $u = 2^{x}$. Then $2^{2x} = u^{2}$ and the equation reads
$u^{2} - 10u + 16 = 0$
2.
$(u - 2)(u - 8) = 0$, so $u = 2$ or $u = 8$.
3.
Convert back: $2^{x} = 2$ gives $x = 1$, and $2^{x} = 8$ gives $x = 3$.
The larger solution is $x = 3$. Both values of $u$ are positive, so neither has to be discarded.
Common mistake. Stopping at the values of $2^{x}$ and quoting one of those. The substitution has to be undone: a value of $8$ for $2^{x}$ means a value of $3$ for $x$.
Why the other options are wrong
- A: The smaller of the two solutions, from $2^{x} = 2$. Misread Question
- C: Reading $2^{2x}$ as $(2x)^{2}$ and $2^{x}$ as $2x$, which gives $4x^{2} - 20x + 16 = 0$ with roots $1$ and $4$. Formula Misapplication
- D: Reporting the larger value of $2^{x}$ rather than the value of $x$ that produces it. Incomplete Calculation
- E: Adding the two values of $2^{x}$, that is $2 + 8$, which is the sum of the roots of the quadratic and not a solution at all. Conceptual Misunderstanding
[19] Trigonometry
Question 19
What is the greatest value taken by $3\sin^{2}\theta + 4\cos^{2}\theta$ as $\theta$ varies over all real values?
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. Replace $\cos^{2}\theta$ by $1 - \sin^{2}\theta$. The expression becomes $4 - \sin^{2}\theta$, and since $\sin^{2}\theta$ runs over $[0, 1]$, the whole expression runs over $[3, 4]$.
Fastest approach
1.
$3\sin^{2}\theta + 4\cos^{2}\theta = 3\sin^{2}\theta + 4\left(1 - \sin^{2}\theta\right) = 4 - \sin^{2}\theta$
2.
$\sin^{2}\theta$ is never negative and never exceeds $1$, so the expression is largest when $\sin^{2}\theta = 0$.
Greatest value: $4$.
3.
Read it as a weighted average. The expression is a mixture of $3$ and $4$ with weights $\sin^{2}\theta$ and $\cos^{2}\theta$ that always add to $1$, so it can never leave the range from $3$ to $4$.
Common mistake. Treating the expression like $a\sin\theta + b\cos\theta$ and reaching for $\sqrt{a^{2}+b^{2}}$. That result is for first powers of sine and cosine, which peak at different angles; here the two terms are squares whose weights are tied together by the identity.
Why the other options are wrong
- A: Subtracting the coefficients, $4 - 3$, as though the two terms opposed each other. Formula Misapplication
- B: Reporting the least value rather than the greatest. It occurs where $\sin^{2}\theta = 1$. Misread Question
- D: Applying the harmonic form $\sqrt{3^{2} + 4^{2}}$, which belongs to $3\sin\theta + 4\cos\theta$ and not to the squares. Formula Misapplication
- E: Adding the coefficients, as though $\sin^{2}\theta$ and $\cos^{2}\theta$ could both equal $1$ at the same angle. Conceptual Misunderstanding
[20] Series
Question 20
Evaluate $\dfrac{1}{1 \times 2} + \dfrac{1}{2 \times 3} + \dfrac{1}{3 \times 4} + \cdots + \dfrac{1}{9 \times 10}$.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\frac{9}{10}$ (option D)
Key idea. $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$. Written that way, every interior fraction appears once with each sign and cancels, so only the first and the last survive.
Fastest approach
1.
Split each term: $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$. Check it on the first term: $1 - \frac{1}{2} = \frac{1}{2}$, which is $\frac{1}{1 \times 2}$.
2.
Write the sum out:
$\left(1 - \tfrac{1}{2}\right) + \left(\tfrac{1}{2} - \tfrac{1}{3}\right) + \cdots + \left(\tfrac{1}{9} - \tfrac{1}{10}\right)$
3.
Everything between the ends cancels in pairs.
$1 - \tfrac{1}{10} = \tfrac{9}{10}$
The last term of the sum is $\frac{1}{9 \times 10}$, so the surviving fraction is $\frac{1}{10}$ and not $\frac{1}{9}$.
Common mistake. Adding nine fractions one at a time and slipping over a common denominator, when the whole sum collapses to two terms. The other frequent error is losing track of which fraction survives at the far end.
Why the other options are wrong
- A: Splitting the term the wrong way round, as $\frac{1}{k+1} - \frac{1}{k}$, which negates every term and so negates the sum. Sign Error
- B: Cancelling everything after the first bracket and leaving $1 - \frac{1}{2}$, rather than cancelling in pairs down the whole sum. Incomplete Calculation
- C: Telescoping to $1 - \frac{1}{n}$ with $n = 9$, the number of terms, when nine terms end at $\frac{1}{9} - \frac{1}{10}$ and leave $1 - \frac{1}{10}$. Formula Misapplication
- E: The sum to infinity, which these terms approach but never reach after finitely many of them. Conceptual Misunderstanding