PAT practice · 20 questions · Free
PAT Exercise Paper 5: 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 cyclist's velocity-time graph is made of three straight segments. From $t = 0$ to $t = 4 \ \text{s}$ the velocity rises steadily from rest to $12 \ \text{m s}^{-1}$; it then holds at $12 \ \text{m s}^{-1}$ until $t = 10 \ \text{s}$; from $t = 10 \ \text{s}$ it falls steadily back to rest at $t = 13 \ \text{s}$. The cyclist travels along a straight road and never reverses, so the distance travelled equals the displacement. What distance is covered in the $13 \ \text{s}$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $114 \ \text{m}$ (option B)
Key idea. On a velocity-time graph the gradient is the acceleration and the area beneath the line is the distance. Three straight segments means two triangles and a rectangle, and no equation of motion is needed at all.
Fastest approach
1. Cut the graph into shapes.
Speeding up: a triangle of base $4 \ \text{s}$ and height $12 \ \text{m s}^{-1}$.
Cruising: a rectangle $6 \ \text{s}$ wide and $12 \ \text{m s}^{-1}$ tall.
Slowing down: a triangle of base $3 \ \text{s}$ and height $12 \ \text{m s}^{-1}$.
2. Add the areas.
$\tfrac{1}{2}(4)(12) + (6)(12) + \tfrac{1}{2}(3)(12) = 24 + 72 + 18 = 114 \ \text{m}$
A useful check: the cyclist is at $12 \ \text{m s}^{-1}$ for less than half the time, so the answer must be comfortably under $12 \times 13 = 156 \ \text{m}$ and comfortably over $12 \times 6 = 72 \ \text{m}$.
Common mistake. Treating a sloping segment as though the speed were constant across it. A ramp from rest to $12 \ \text{m s}^{-1}$ covers half of what $12 \ \text{m s}^{-1}$ would cover in the same time, and every factor of two that goes missing here is worth $12$ or $18 \ \text{m}$.
Why the other options are wrong
- A: Adding the first triangle and the rectangle and stopping, so the last three seconds of travel are left out entirely. Incomplete Calculation
- C: Treating the final segment as a rectangle, counting $36 \ \text{m}$ where the triangle's area is only $18 \ \text{m}$. Formula Misapplication
- D: Treating the first segment as a rectangle, counting $48 \ \text{m}$ where the triangle's area is only $24 \ \text{m}$. Formula Misapplication
- E: Multiplying the top speed by the whole $13 \ \text{s}$. That is a speed times a time and so is dimensionally a perfectly good distance, but it assumes the cyclist held $12 \ \text{m s}^{-1}$ from the first instant to the last. Conceptual Misunderstanding
[02] Electric circuits
Question 2
A $24 \ \text{V}$ battery of negligible internal resistance is connected in a single loop with a $3 \ \Omega$ resistor, a lamp and an ammeter, which reads $2 \ \text{A}$. Take the connecting wires and the ammeter to have no resistance, and take the lamp's resistance to be constant at this current. What is the resistance of the lamp?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $9 \ \Omega$ (option C)
Key idea. In a single loop the current is the same everywhere, so the resistor takes $IR = 2 \times 3 = 6 \ \text{V}$ and the lamp takes whatever is left of the $24 \ \text{V}$. The p.d. across a series component is proportional to its resistance, never shared out equally.
Fastest approach
1. Use the shared current.
The loop carries $2 \ \text{A}$ through every component.
2. Find the p.d. across the known resistor.
$V = IR = 2 \times 3 = 6 \ \text{V}$
3. Give the rest to the lamp.
$24 - 6 = 18 \ \text{V}$, so $R_{\text{lamp}} = \dfrac{18}{2} = 9 \ \Omega$.
Check it the other way round: the whole loop is $\dfrac{24}{2} = 12 \ \Omega$, and $12 - 3 = 9 \ \Omega$.
Common mistake. Splitting the $24 \ \text{V}$ equally between the two components. Series components share the current, not the voltage; the voltage divides in the ratio of the resistances, which is exactly what the question is asking you to find.
Why the other options are wrong
- A: Reporting the resistance the question supplied rather than the one it asked for. Misread Question
- B: Assuming the $24 \ \text{V}$ divides equally, giving the lamp $12 \ \text{V}$ and then $12 \div 2$. The result is dimensionally a resistance and nothing about the units objects; what is wrong is the idea that a series circuit shares voltage equally rather than in proportion to resistance. Conceptual Misunderstanding
- D: Computing $\dfrac{24 - 3}{2}$, which subtracts an ohm from a volt before dividing. Dimensional Error
- E: Dividing the full supply voltage by the current. That is the resistance of the whole loop, with the $3 \ \Omega$ never removed. Incomplete Calculation
[03] Waves
Question 3
A travelling wave runs along a long stretched spring. One point on the spring makes $25$ complete oscillations in $10 \ \text{s}$, and the distance from one crest to the next along the spring is $0.60 \ \text{m}$. Take the wave to travel at a constant speed and to lose no amplitude as it goes. What is the speed of the wave?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $1.5 \ \text{m s}^{-1}$ (option D)
Key idea. A crest advances exactly one wavelength in one period, so the speed is $\lambda f$. The frequency is a count divided by a time, $\frac{25}{10} = 2.5 \ \text{Hz}$, and the crest-to-crest distance is the wavelength by definition.
Fastest approach
1. Get the frequency from the count.
$f = \dfrac{25}{10} = 2.5 \ \text{Hz}$, so the period is $T = 0.40 \ \text{s}$.
2. Multiply by the wavelength.
$v = \lambda f = 0.60 \times 2.5 = 1.5 \ \text{m s}^{-1}$
Or read it straight off the definition: a crest moves $0.60 \ \text{m}$ in $0.40 \ \text{s}$, which is $1.5 \ \text{m s}^{-1}$. If you cannot remember whether to multiply or divide, the units settle it: metres times per-second gives metres per second.
Common mistake. Dividing the wavelength by the frequency. That produces metre seconds, which is not a speed of anything, and the check costs less time than the arithmetic did.
Why the other options are wrong
- A: Computing $\lambda / f$ rather than $\lambda f$, which gives metre seconds rather than metres per second. Dimensional Error
- B: Quoting the wavelength as the speed. A crest really does advance $0.60 \ \text{m}$, but it takes a whole period of $0.40 \ \text{s}$ to do it. Incomplete Calculation
- C: Halving the given distance, on the idea that a wavelength runs from a crest to the next trough. That distance is half a wavelength, and the question already gives crest to crest. Metres times per-second still comes out as a speed, so only the definition catches this. Conceptual Misunderstanding
- E: Reading the $25$ oscillations as the frequency and never dividing by the $10 \ \text{s}$. Misread Question
[04] Circular motion
Question 4
A playground roundabout turns steadily, completing one revolution every $4.0 \ \text{s}$. A child sits $1.5 \ \text{m}$ from the axis. Treat the child as a point moving on a horizontal circle at constant speed, and take the rotation rate as unchanging. What is the child's 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: $\frac{3\pi}{4} \ \text{m s}^{-1}$ (option E)
Key idea. In one period the child travels once round the circle, a distance $2\pi r$, so $v = \frac{2\pi r}{T}$. The same statement written with the angular speed is $v = \omega r$ with $\omega = \frac{2\pi}{T}$.
Fastest approach
1. Find the distance once round.
$2\pi r = 2\pi (1.5) = 3\pi \ \text{m}$
2. Divide by the time it takes.
$v = \dfrac{3\pi}{4} \approx 2.4 \ \text{m s}^{-1}$
Then ask whether that number could be true of the world you live in. A brisk walk is about $2 \ \text{m s}^{-1}$, and a roundabout turning once every four seconds does feel roughly like walking pace at arm's length from the middle. An answer of $0.4$ or of $24$ would not.
Common mistake. Losing the factor of $2\pi$, which is what converts one revolution into a distance. A revolution is not a length until the radius has been multiplied in.
Why the other options are wrong
- A: Dividing the radius by the period, which drops the factor $2\pi$ altogether. Formula Misapplication
- B: Using the diameter, $2r$, as the distance once round instead of the circumference. Conceptual Misunderstanding
- C: Taking the distance once round as $\pi r$ rather than $2\pi r$. That is still a length divided by a time, so it is dimensionally a perfectly good speed; the error is geometric, since $\pi r$ is only half the way round. Conceptual Misunderstanding
- D: Quoting the radius in metres as though it were the speed. Dimensional Error
[05] Pulleys and connected bodies
Question 5
Blocks of mass $3 \ \text{kg}$ and $5 \ \text{kg}$ hang from the two ends of a light inextensible string that passes over a smooth pulley of negligible mass. The system is released from rest and the string stays taut. Take $g = 10 \ \text{m s}^{-2}$ and ignore air resistance. What is the magnitude of the acceleration of each block?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $2.5 \ \text{m s}^{-2}$ (option A)
Key idea. Write Newton's second law for each block and there are two equations and two unknowns, the acceleration and the tension. Treating the pair as a single body of mass $8 \ \text{kg}$ pulled by the $20 \ \text{N}$ difference in weights removes the tension before it is ever introduced, because it is internal to that body.
Fastest approach
1. Count the unknowns.
Two blocks, two equations, two unknowns: the shared acceleration and the tension. Only one of them is asked for.
2. Take the two blocks as one system.
The string is inextensible, so both blocks have the same speed at every instant and therefore the same magnitude of acceleration. The tension pulls the system forward on one side and back on the other by equal amounts, so it does not appear.
$a = \dfrac{\text{resultant force}}{\text{total mass}} = \dfrac{(5 - 3)(10)}{5 + 3} = \dfrac{20}{8} = 2.5 \ \text{m s}^{-2}$
The result must be smaller than $g$, because the lighter block has to be hauled upwards by the same string that lets the heavier one fall.
Common mistake. Dividing the resultant force by only one of the two masses. Whatever the string does, it makes both blocks change speed together, so the mass being accelerated is the total.
Why the other options are wrong
- B: Dividing the $20 \ \text{N}$ resultant by the heavier mass alone, forgetting that the string drags the lighter block along as well, so all $8 \ \text{kg}$ has to be accelerated. Incomplete Calculation
- C: Taking the driving force to be the heavier block's whole weight, $50 \ \text{N}$, over the total mass. A force divided by a mass is dimensionally a perfectly good acceleration, so nothing about the units objects; what is left out is the lighter block's weight pulling back through the string. Conceptual Misunderstanding
- D: Quoting $g$, as though the heavier block were in free fall and the string did nothing at all. Conceptual Misunderstanding
- E: Quoting the $20 \ \text{N}$ resultant force as though it were the acceleration, with the division by the mass never done. Dimensional Error
[06] Orbits and satellites
Question 6
A satellite is in a low circular polar orbit with a period of $96$ minutes, crossing over both poles on every orbit. Take the Earth to be a sphere turning steadily once every $24$ hours about its axis, and take the plane of the orbit to stay fixed relative to the stars. Through what angle does the Earth turn beneath the satellite during one complete orbit?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $24^{\circ}$ (option B)
Key idea. The Earth's rotation is a rate: $360^{\circ}$ per $24$ hours, or $15^{\circ}$ per hour. The angle it turns through is that rate times the length of one orbit, and this is exactly why a polar orbiter sees a fresh strip of ground on every pass while a geostationary satellite, whose period is the full $24$ hours, sees the same one forever.
Fastest approach
1. Put both times in the same unit.
One orbit is $96$ minutes; one rotation is $24 \times 60 = 1440$ minutes.
2. Take the fraction of a rotation.
$\dfrac{96}{1440} = \dfrac{1}{15}$ of a turn.
3. Convert to an angle.
$\dfrac{360^{\circ}}{15} = 24^{\circ}$
Equivalently, $15^{\circ}$ per hour times $1.6$ hours. A geostationary satellite is the limiting case of the same sum: set the period to $24$ hours and the angle becomes $360^{\circ}$, so the satellite comes back over exactly the same longitude, which is what makes the orbit useful.
Common mistake. Comparing the orbit time with $24$ rather than with $1440$, so the answer comes out sixty times too large. Both quantities in a ratio have to be measured in the same unit before the ratio means anything.
Why the other options are wrong
- A: Quoting the Earth's rotation rate of $15$ degrees per hour and stopping there, without multiplying by the $1.6$ hours the orbit lasts. Incomplete Calculation
- C: Using a $12$-hour day, so the Earth is taken through $360^{\circ}$ twice as fast as it really turns. This produces a perfectly respectable angle, which is why only the physics of the day length catches it. Conceptual Misunderstanding
- D: Quoting the $96$ minutes as a number of degrees. A time is not an angle, however close the two numbers look. Dimensional Error
- E: Subtracting from $360^{\circ}$, which answers a different question: how much further the ground has to turn before it comes back under the satellite's starting longitude. Misread Question
[07] Optics
Question 7
A narrow beam of light strikes a flat mirror, making an angle of $20^{\circ}$ with the normal to the mirror. The mirror is then turned through $10^{\circ}$ about an axis lying in its own surface and perpendicular to the plane of the beam, in the sense that increases the angle of incidence. The incoming beam is not moved, the mirror stays flat, and everything remains in one plane. Through what angle does the reflected beam turn?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $20^{\circ}$ (option C)
Key idea. Turning the mirror turns its normal by the same amount. The angle of incidence therefore grows by $10^{\circ}$, and the reflected beam sits that much further from a normal that has itself already moved $10^{\circ}$, so the beam swings by $2 \times 10^{\circ}$. The starting angle of incidence drops out entirely, which is worth noticing: it is given and it is not needed.
Fastest approach
1. Measure everything from the original normal.
Before: the angle of incidence is $20^{\circ}$, so the reflected beam leaves at $20^{\circ}$ on the other side of that normal.
2. Turn the mirror and re-measure.
The normal turns with the mirror, to $10^{\circ}$ from where it was. The angle of incidence is now $20^{\circ} + 10^{\circ} = 30^{\circ}$, and the reflected beam leaves at $30^{\circ}$ from the new normal, which is $10^{\circ} + 30^{\circ} = 40^{\circ}$ from the original one.
3. Subtract.
$40^{\circ} - 20^{\circ} = 20^{\circ}$
Redo it with a starting angle of $35^{\circ}$ and the answer is still $20^{\circ}$. Whenever a given number refuses to appear in the answer, that is the result worth remembering: a mirror turned by $\theta$ swings the reflected beam by $2\theta$.
Common mistake. Turning the beam with the mirror, one degree for one degree, having tracked the surface but not the normal. The law of reflection is stated about the normal, so the normal is the line that has to be followed when anything moves.
Why the other options are wrong
- A: Halving the mirror's rotation instead of doubling it. Formula Misapplication
- B: Assuming the reflected beam turns exactly with the mirror. Both quantities are angles, so nothing about the units objects; what is missed is that the normal turns too, so the angle of incidence changes as well as the surface. Conceptual Misunderstanding
- D: Reporting the new angle of incidence, $20^{\circ} + 10^{\circ}$, rather than how far the reflected beam has swung. Misread Question
- E: Doubling twice, once for the reflection and again for the swing, which applies the same factor of two to a result that already carries it. Formula Misapplication
[08] Scaling and estimation
Question 8
A uniform cylindrical wire is stretched until its length has doubled. The stretching is uniform along the wire, the material is unchanged so its resistivity is the same, and no material is added or removed, so the volume of the wire is unchanged. By what factor is the resistance of the wire multiplied?
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. $R = \frac{\rho L}{A}$, and a fixed volume means $LA$ is fixed, so doubling the length halves the cross-section. Both changes push the resistance the same way, and the factors multiply: $2 \times 2$.
Fastest approach
1. Write the formula and the constraint.
$R = \dfrac{\rho L}{A}$, with $\rho$ fixed and $V = LA$ fixed.
2. Apply the constraint.
$L \to 2L$ with $LA$ fixed forces $A \to \dfrac{A}{2}$.
3. Take the ratio of the two cases.
$\dfrac{R_{\text{new}}}{R_{\text{old}}} = \dfrac{2L}{A/2} \times \dfrac{A}{L} = 4$
No length, no area and no resistivity is ever needed, because every one of them cancels in the ratio. The same argument in one line: $R \propto \dfrac{L}{A} = \dfrac{L^{2}}{V}$, and with $V$ fixed the resistance goes as the square of the length.
Common mistake. Doubling the length and leaving the cross-section alone. Stretching a fixed amount of metal into twice the length has to make it thinner, and the thinning matters exactly as much as the lengthening does.
Why the other options are wrong
- A: Using $R \propto \dfrac{A}{L}$, the formula turned upside down. Formula Misapplication
- B: Taking resistance to be a property of the material, so that reshaping the wire cannot change it. Resistivity is such a property; resistance is not, because it also depends on the shape. Conceptual Misunderstanding
- C: Doubling the length and holding the cross-section fixed, which the constant-volume condition forbids. Incomplete Calculation
- E: Taking the cross-section to shrink by the square of the length factor, to $\dfrac{A}{4}$, as an area normally would under a scaling. Here it is the volume that is held fixed, so the area only halves. Conceptual Misunderstanding
[09] Energy and power
Question 9
Water leaves a reservoir and falls $100 \ \text{m}$ to the turbines of a hydroelectric station at a steady rate of $500 \ \text{m}^{3}$ of water each second. Take the density of water as $1000 \ \text{kg m}^{-3}$ and $g = 10 \ \text{m s}^{-2}$. Assume every joule of gravitational potential energy released is delivered to the turbines, and neglect any kinetic energy the water still carries when it arrives. What power is delivered?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $5 \times 10^{8} \ \text{W}$ (option E)
Key idea. Power is energy per second, so ask what arrives each second. Every rate and every density in this question has a denominator, and naming it is the whole method: $500 \ \text{m}^{3}$ per second times $1000 \ \text{kg}$ per cubic metre is $5 \times 10^{5} \ \text{kg}$ per second, and each of those kilograms releases $gh$ joules.
Fastest approach
1. Convert the flow to a mass per second.
$\rho Q = 1000 \times 500 = 5 \times 10^{5} \ \text{kg s}^{-1}$
2. Find the energy each kilogram releases.
$gh = 10 \times 100 = 1000 \ \text{J kg}^{-1}$
3. Multiply.
$P = 5 \times 10^{5} \times 1000 = 5 \times 10^{8} \ \text{W} = 500 \ \text{MW}$
Keeping the powers of ten separate from the digits makes this a one-line sum: $10^{3} \times (5 \times 10^{2}) \times 10^{1} \times 10^{2} = 5 \times 10^{8}$. The result is the right size for a large hydroelectric station, which is the check worth making before writing it down.
Common mistake. Multiplying the four numbers without asking what each of them is per. A density is per cubic metre, a flow rate is per second, and a power is per second; the only combination in which every denominator is used exactly once is the one above.
Why the other options are wrong
- A: Treating $500 \ \text{m}^{3}$ of water per second as $500 \ \text{kg}$ per second, so the density is never used. That is a factor of $1000$ lost, which is the density itself. Unit Error
- B: Stopping at $\rho g h$, the energy released per cubic metre, and quoting it as a power. It still has to be multiplied by the number of cubic metres arriving each second. Incomplete Calculation
- C: Leaving $g$ out. With no acceleration anywhere in it the expression cannot be a power, which is faster to see than the arithmetic is to check. Dimensional Error
- D: Inserting a factor of one half, by analogy with $\tfrac{1}{2}mv^{2}$. The result is still a power, so dimensions raise no objection; the physics does, because the energy released by a mass falling through a height is $mgh$ and carries no such factor. Conceptual Misunderstanding
[10] Waves
Question 10
The speed $v$ of a transverse wave on a stretched string depends only on the tension $T$ in the string, measured in newtons, and on the string's mass per unit length $\mu$, measured in kilograms per metre. The relation is a product of powers of $T$ and $\mu$ multiplied by a dimensionless constant, and for this relation that constant is $1$. A string is held at a tension of $40 \ \text{N}$ and has a mass per unit length of $0.10 \ \text{kg m}^{-1}$. What is the speed of a transverse wave on 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: $20 \ \text{m s}^{-1}$ (option A)
Key idea. A tension is a force, $\text{M L T}^{-2}$, and a mass per unit length is $\text{M L}^{-1}$. Dividing one by the other cancels the mass and leaves $\text{L}^{2}\text{T}^{-2}$, a speed squared, so the only product of powers that can be a speed is $\sqrt{T/\mu}$.
Fastest approach
1. Write the dimensions of each ingredient.
$[T] = \text{M L T}^{-2}$ and $[\mu] = \text{M L}^{-1}$.
2. Look for the combination that has no mass in it.
$\left[\dfrac{T}{\mu}\right] = \dfrac{\text{M L T}^{-2}}{\text{M L}^{-1}} = \text{L}^{2}\text{T}^{-2}$
That is the square of a speed, so $v = \sqrt{T/\mu}$ up to a dimensionless constant, which the question says is $1$.
3. Substitute once, at the end.
$v = \sqrt{\dfrac{40}{0.10}} = \sqrt{400} = 20 \ \text{m s}^{-1}$
Note what this method can and cannot deliver. It fixes the exponents completely, because there is only one way to cancel the mass and reach a length over a time. It can never supply a numerical factor out front, which is why the question had to state that the constant is $1$ rather than leave it to be deduced.
Common mistake. Reporting $T/\mu$ without the square root. Its units are metres squared per second squared, so it is a speed squared and not a speed, and that is visible before any arithmetic is attempted.
Why the other options are wrong
- B: Carrying a factor of $\sqrt{2}$ into the answer, from a half remembered $\tfrac{1}{2}$ in an energy formula. Dimensional analysis fixes the exponents and never the dimensionless factor in front, so a constant cannot be imported from another formula; the question supplies the only one there is, and it is $1$. Conceptual Misunderstanding
- C: Quoting the tension in newtons as though it were the speed. Dimensional Error
- D: Reading the mass per unit length as $0.0010 \ \text{kg m}^{-1}$, a factor of one hundred, which becomes a factor of ten after the square root. Order of Magnitude Error
- E: Computing $T/\mu$ and stopping, so the answer is left with the dimensions of a speed squared. Dimensional Error
[11] Series
Question 11
Every term of an infinite geometric series is positive. The series has sum to infinity $24$, and its first two terms add to $18$. What is the first term?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $12$ (option B)
Key idea. Two unknowns, $a$ and $r$, need two equations, and the question supplies exactly two: $\frac{a}{1-r} = 24$ and $a(1+r) = 18$. Substituting $a = 24(1-r)$ into the second turns them into a difference of two squares, so no quadratic formula is needed. The series converges, which is what lets the sum to infinity be quoted at all, so $|r| < 1$.
Fastest approach
### 1. Sum to infinity: $\frac{a}{1-r} = 24$, so $a = 24(1-r)$.
### 2. First two terms: $a + ar = a(1+r) = 18$.
### 3. Substitute: $24(1-r)(1+r) = 18$, so $1 - r^{2} = \frac{3}{4}$ and $r^{2} = \frac{1}{4}$.
### 4. Every term is positive, so $r = \frac{1}{2}$ and $a = 24 \times \frac{1}{2} = 12$.
Check: $12, 6, 3, \ldots$ adds to $24$, and the first two terms give $18$.
Common mistake. Keeping $r = -\frac{1}{2}$, which satisfies both equations but produces the alternating series $36, -18, 9, \ldots$ and so is ruled out by the stated positivity. A root that solves the algebra still has to survive the conditions in the question.
Why the other options are wrong
- A: Reporting the second term $ar = 6$ rather than the first. Misread Question
- C: Taking the sum of the first two terms to be the first term, which is what $r = 0$ would give. Incomplete Calculation
- D: Reporting the sum to infinity, which the question supplies. Misread Question
- E: Keeping the root $r = -\frac{1}{2}$, which the positive terms rule out, giving $a = 24\left(1 + \frac{1}{2}\right)$. Conceptual Misunderstanding
[12] Integration
Question 12
A curve passes through the point $(2, 5)$, and at every point its gradient is $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2} - 4x + 1$. What is the $y$-coordinate of the point on the curve where $x = 4$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $39$ (option C)
Key idea. Integrating gives a whole family of curves, $y = x^{3} - 2x^{2} + x + c$, one for each value of $c$. The point $(2,5)$ picks out the single member the question is about. Faster still, the change in $y$ between two $x$ values is the definite integral of the gradient between them, so $y(4) = 5 + \int_{2}^{4}\left(3x^{2}-4x+1\right)\mathrm{d}x$ and $c$ never has to be found.
Fastest approach
### 1. Integrate: $y = x^{3} - 2x^{2} + x + c$.
### 2. Use the point: $8 - 8 + 2 + c = 5$, so $c = 3$.
### 3. Substitute $x = 4$: $64 - 32 + 4 + 3 = 39$.
Or in one line, without ever finding $c$:
$y(4) = y(2) + \int_{2}^{4}\left(3x^{2}-4x+1\right)\mathrm{d}x = 5 + (36 - 2) = 39$
Common mistake. Dropping the constant of integration. Every antiderivative of $3x^{2}-4x+1$ has the same shape, and the constant is the only thing that says which of them passes through $(2,5)$, so leaving it out answers a different question.
Why the other options are wrong
- A: Differentiating $3x^{2}-4x+1$ instead of integrating it, giving $6x - 4$ at $x = 4$. Conceptual Misunderstanding
- B: Evaluating $x^{3}-2x^{2}+x$ at $x = 4$ with the constant of integration omitted. Incomplete Calculation
- D: Taking the constant of integration to be the given $y$-value of $5$ rather than solving for it. Formula Misapplication
- E: Sign slip in $8 - 8 + 2 + c = 5$, giving $c = 7$. Sign Error
[13] Geometry
Question 13
In triangle $ABC$ the angle at $C$ is a right angle, with $AC = 6$ and $BC = 8$. The point $D$ lies on $AB$ so that $CD$ is perpendicular to $AB$. How long is $CD$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $4.8$ (option D)
Key idea. The same triangle has two bases with two matching heights: the legs $6$ and $8$, and the hypotenuse with $CD$. Equating the two expressions for the area gives $CD$ in one line. The similar triangles $ACD$, $CBD$ and $ABC$ give the same result and also say where $D$ sits.
Fastest approach
### 1. Pythagoras: $AB = \sqrt{6^{2}+8^{2}} = \sqrt{100} = 10$.
### 2. Area from the legs: $\tfrac{1}{2}\times 6 \times 8 = 24$.
### 3. Area from the hypotenuse: $\tfrac{1}{2}\times 10 \times CD = 24$, so $CD = \frac{48}{10} = 4.8$.
Check with similar triangles: $AD = \frac{6^{2}}{10} = 3.6$ and $DB = \frac{8^{2}}{10} = 6.4$, which add to $10$ as they must, and $CD = \sqrt{3.6 \times 6.4} = 4.8$.
Common mistake. Reporting one of the two pieces the perpendicular cuts the hypotenuse into, $3.6$ or $6.4$, instead of the perpendicular itself. A rough sketch settles which segment is which: $CD$ is shorter than either leg, so any answer above $6$ is the wrong length.
Why the other options are wrong
- A: Using $\text{area} = \text{base} \times \text{height}$ without the factor of one half, so $CD = \frac{24}{10}$. Incomplete Calculation
- B: Dividing the product of the legs by their sum, $\frac{48}{6+8}$, rather than by the hypotenuse. Formula Misapplication
- C: Reporting $AD = \frac{6^{2}}{10}$, the distance from $A$ to the foot of the perpendicular. Misread Question
- E: Reporting $DB = \frac{8^{2}}{10}$, the projection of the longer leg on the hypotenuse. Conceptual Misunderstanding
[14] Trigonometry
Question 14
Find the sum, in degrees, of every solution of $\sin 2x = \frac{1}{2}$ that lies in the interval $0^{\circ} \leq x \leq 360^{\circ}$.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $540^{\circ}$ (option E)
Key idea. Substituting $u = 2x$ turns the interval $0^{\circ} \leq x \leq 360^{\circ}$ into $0^{\circ} \leq u \leq 720^{\circ}$, which is two full revolutions, so there are four solutions rather than two. Sine takes each value twice per revolution, at $u$ and at $180^{\circ}-u$.
Fastest approach
### 1. Let $u = 2x$. As $x$ runs from $0^{\circ}$ to $360^{\circ}$, $u$ runs from $0^{\circ}$ to $720^{\circ}$.
### 2. $\sin u = \frac{1}{2}$ gives $u = 30^{\circ}, 150^{\circ}$ in the first revolution, and $390^{\circ}, 510^{\circ}$ in the second.
### 3. Halve each: $x = 15^{\circ}, 75^{\circ}, 195^{\circ}, 255^{\circ}$.
### 4. Sum: $15 + 75 + 195 + 255 = 540^{\circ}$.
A sketch of $y = \sin 2x$ over the interval shows two complete waves, so a horizontal line at $y = \frac{1}{2}$ must cut it four times.
Common mistake. Solving over the interval given for $x$ instead of the doubled interval for $2x$, which finds $15^{\circ}$ and $75^{\circ}$ and stops. Whenever the angle inside the function is a multiple of $x$, the search interval is stretched by the same factor.
Why the other options are wrong
- A: Keeping only $15^{\circ}$ and $75^{\circ}$, from letting $2x$ run to $360^{\circ}$ rather than to $720^{\circ}$. Incomplete Calculation
- B: Solving $\sin x = \frac{1}{2}$ instead of $\sin 2x = \frac{1}{2}$, giving $30^{\circ} + 150^{\circ}$. Formula Misapplication
- C: Using only the $30^{\circ} + 360^{\circ}k$ branch, giving $15^{\circ} + 195^{\circ}$ and missing the supplementary angle in each revolution. Incomplete Calculation
- D: Using only the $150^{\circ} + 360^{\circ}k$ branch, giving $75^{\circ} + 255^{\circ}$ and dropping the principal value. Conceptual Misunderstanding
[15] Graphs
Question 15
The curve $y = f(x)$ has exactly one minimum point, at $(2, -3)$. The curve $y = 2f(x+1) - 4$ also has exactly one minimum point. What is the $y$-coordinate of that minimum?
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. Changes made outside $f$ act on the $y$-coordinate, and in the order written: multiply by $2$ first, then subtract $4$. Changes made inside $f$ act on the $x$-coordinate and act in the opposite sense, so $f(x+1)$ shifts the curve one to the left. Only the outside changes can move a $y$-coordinate.
Fastest approach
### 1. The $x$-coordinate is governed by $x+1$, so the turning point moves from $x = 2$ to $x = 1$. That does not affect the question asked.
### 2. The $y$-coordinate is stretched by a factor of $2$: $-3 \to -6$.
### 3. Then translated $4$ downwards: $-6 - 4 = -10$.
So the minimum is at $(1, -10)$.
Check the direction: the stretch pushes a point below the axis further below, and subtracting $4$ pushes it lower still, so the answer must be well below $-3$.
Common mistake. Applying the vertical stretch to the translation as well, as though the curve were $2\left(f(x+1)-4\right)$, or translating upwards because a minus sign is read as a shift towards zero. Track one point through one operation at a time.
Why the other options are wrong
- B: Translating $4$ downwards but leaving out the stretch: $-3 - 4$. Incomplete Calculation
- C: Applying the stretch but leaving out the translation: $2 \times (-3)$. Incomplete Calculation
- D: Assuming the transformations move the curve sideways only, so the minimum height is unchanged. Conceptual Misunderstanding
- E: Reporting the $x$-coordinate of the new minimum, $2 - 1$, rather than its $y$-coordinate. Misread Question
[16] Binomial expansion
Question 16
A committee of three people is to be chosen from five women and four men. In how many ways can it be chosen if it must contain at least one woman and at least one man?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $70$ (option B)
Key idea. Count everything, then remove what the condition forbids. There are $\binom{9}{3}$ committees in all, and the only forbidden ones are all-women, $\binom{5}{3}$, and all-men, $\binom{4}{3}$. That is one subtraction instead of two separate cases, and it also gives an upper bound to check the answer against.
Fastest approach
### 1. All committees: $\binom{9}{3} = 84$.
### 2. All women: $\binom{5}{3} = 10$. All men: $\binom{4}{3} = 4$.
### 3. $84 - 10 - 4 = 70$.
Adding the two allowed compositions gives the same number: $\binom{5}{1}\binom{4}{2} + \binom{5}{2}\binom{4}{1} = 30 + 40 = 70$.
The answer has to sit below $84$ and above either single composition, and $70$ does.
Common mistake. Choosing one woman, then one man, then anybody from the remaining seven. That gives $5 \times 4 \times 7 = 140$ and counts most committees twice, because the same three people arrive by more than one route.
Why the other options are wrong
- A: Counting only the two women and one man committees, $\binom{5}{2}\binom{4}{1}$, and forgetting the other composition. Incomplete Calculation
- C: $\binom{9}{3}$, the unrestricted total, with the condition never applied. Misread Question
- D: Choosing one woman, one man and then any of the remaining seven: $5 \times 4 \times 7$, which counts committees more than once. Conceptual Misunderstanding
- E: Treating the committee as ordered and using $9 \times 8 \times 7$. Formula Misapplication
[17] Binomial expansion
Question 17
In how many distinguishable orders can the six letters of the word BANANA be written?
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 C)
Key idea. Treat the six letters as distinct and there are $6!$ orders. Nothing distinguishes the three A's from each other, so each genuine word has been counted $3!$ times over, and likewise $2!$ times for the two N's. Divide by both.
Fastest approach
$\dfrac{6!}{3!\,2!} = \dfrac{720}{6 \times 2} = 60$
Counting positions instead gives the same thing: choose $3$ of the $6$ places for the A's, $\binom{6}{3} = 20$, then $2$ of the remaining $3$ for the N's, $\binom{3}{2} = 3$, and the B takes the last place. $20 \times 3 = 60$.
Common mistake. Dividing by $3!$ for the A's and forgetting the two N's, which leaves every word counted twice. Any repeated letter needs its own factorial in the denominator.
Why the other options are wrong
- A: $\binom{6}{2} = 15$, the positions for the two N's alone, with the other four letters never placed. Incomplete Calculation
- B: $\binom{6}{3} = 20$, the positions for the three A's alone, overlooking the three ways of arranging the B and the two N's. Incomplete Calculation
- D: $\frac{6!}{3!}$, dividing for the three A's but not for the two N's. Formula Misapplication
- E: $\frac{6!}{2!}$, dividing for the two N's but still treating the three A's as distinguishable. Conceptual Misunderstanding
[18] Geometry
Question 18
Two similar triangles have perimeters in the ratio $3 : 2$. The smaller triangle has area $36$. What is the area of the larger one?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $81$ (option D)
Key idea. Perimeter is a length, so the ratio $3:2$ is the linear scale factor. Every length in the larger triangle is $\frac{3}{2}$ times its counterpart, and an area is a product of two lengths, so areas scale by $\left(\frac{3}{2}\right)^{2}$.
Fastest approach
Linear scale factor $k = \frac{3}{2}$.
Area scale factor $k^{2} = \frac{9}{4}$.
$36 \times \frac{9}{4} = 81$
Sanity check: the larger triangle is only half as big again in each direction, so its area should be a little over double, and $81$ is.
Common mistake. Multiplying the area by $\frac{3}{2}$, which is the factor for a length and not for an area. The same slip in reverse, cubing the ratio, belongs to volumes.
Why the other options are wrong
- A: Squaring the ratio correctly but applying it the wrong way round: $36 \times \left(\frac{2}{3}\right)^{2}$, which shrinks the larger triangle. Conceptual Misunderstanding
- B: Scaling linearly and the wrong way round: $36 \times \frac{2}{3}$. Formula Misapplication
- C: Scaling the area by the linear ratio: $36 \times \frac{3}{2}$. Conceptual Misunderstanding
- E: Cubing the ratio, $36 \times \left(\frac{3}{2}\right)^{3}$, which is how volumes scale, not areas. Formula Misapplication
[19] Numerical reasoning
Question 19
A sheet of paper is $0.1 \ \text{mm}$ thick. Suppose it could be folded in half twenty times, each fold exactly doubling the thickness and the fold itself taking up no thickness at all. Using the approximation $2^{10} \approx 10^{3}$, estimate the thickness of the folded wad in 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: $100 \ \text{m}$ (option E)
Key idea. Twenty doublings multiply the thickness by $2^{20} = \left(2^{10}\right)^{2} \approx \left(10^{3}\right)^{2} = 10^{6}$. Doing the powers of ten first keeps the whole estimate to one line, and the only remaining care is the millimetre to metre conversion, which is a factor of $10^{3}$.
Fastest approach
### 1. Twenty folds multiply the thickness by $2^{20} = \left(2^{10}\right)^{2} \approx 10^{6}$.
### 2. Thickness $\approx 0.1 \times 10^{6} = 10^{5} \ \text{mm}$.
### 3. Convert once: $10^{5} \ \text{mm} = 10^{5} \div 10^{3} = 10^{2} \ \text{m} = 100 \ \text{m}$.
One significant figure is all this deserves: $2^{20}$ is really $1.05 \times 10^{6}$, so the true value is about $105 \ \text{m}$, and quoting it that precisely would claim an accuracy the approximation never had.
Common mistake. Treating twenty folds as twenty sheets, which turns exponential growth into multiplication by $20$ and gives an answer of a couple of millimetres. Each fold doubles what is already there, so the thickness grows by a factor of a thousand for every ten folds.
Why the other options are wrong
- A: Adding a sheet per fold rather than doubling: $0.1 \times 20 = 2 \ \text{mm}$. Conceptual Misunderstanding
- B: Using $20^{2}$ in place of $2^{20}$: $0.1 \times 400 = 40 \ \text{mm}$. Formula Misapplication
- C: Using $2^{10} \approx 10^{3}$ for the whole wad, which is ten folds rather than twenty. Incomplete Calculation
- D: Converting $10^{5} \ \text{mm}$ to metres by dividing by $10^{4}$ instead of $10^{3}$. Unit Error
[20] Integration
Question 20
A curve has gradient $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4 - 2x$ at every point, and its greatest $y$-value is $7$. At what value of $y$ does the curve cross the $y$-axis?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $3$ (option A)
Key idea. The known value here is not handed over as a point: it has to be found. The gradient is zero at $x = 2$, and since the gradient falls from positive to negative there, that is where the greatest value $7$ occurs. That single point then fixes the constant of integration, exactly as a stated point would.
Fastest approach
### 1. The maximum is where the gradient vanishes: $4 - 2x = 0$, so $x = 2$.
### 2. Integrate: $y = 4x - x^{2} + c$.
### 3. At $x = 2$: $8 - 4 + c = 7$, so $c = 3$.
### 4. The curve meets the $y$-axis at $x = 0$, where $y = c = 3$.
Check it: $y = 4x - x^{2} + 3 = 7 - (x-2)^{2}$, whose greatest value is indeed $7$, at $x = 2$.
Common mistake. Reading the greatest value $7$ as the value at $x = 0$. It is the value at the stationary point, and the whole first step is finding where that point is.
Why the other options are wrong
- B: Reporting the gradient at $x = 0$, which is $4$, rather than the $y$-value there. Misread Question
- C: Integrating $-2x$ to $-2x^{2}$, so the stationary point of the integrated curve moves to $x = 1$ and the constant comes out as $5$. Formula Misapplication
- D: Taking the greatest value to be the intercept, that is, assuming the maximum sits on the $y$-axis. Conceptual Misunderstanding
- E: Sign slip in $8 - 4 + c = 7$, solving it as $c = 7 + 4$. Sign Error