PAT practice · 20 questions · Free
PAT Exercise Paper 2: Physics and Mathematics
Every question carries a worked solution behind a disclosure, so you commit to an answer before you see the key.
[01] Estimation
Question 1
A car is driven $630 \ \text{km}$ along a motorway. Its wheels have diameter $0.60 \ \text{m}$ and roll without slipping, so one complete turn advances the car by one circumference. Taking $\pi$ as $3$, estimate the number of complete turns one wheel makes.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $3.5 \times 10^{5}$ (option D)
Key idea. A count is a division: how many fit. One turn carries the car forward by $\pi d = 1.8 \ \text{m}$, so the number of turns is the journey divided by that. The only real hazard is the kilometre.
Fastest approach
1. One turn.
$\pi d = 3 \times 0.60 = 1.8 \ \text{m}$
2. How many fit.
$\frac{630 \ \text{km}}{1.8 \ \text{m}} = \frac{6.3 \times 10^{5}}{1.8} = 3.5 \times 10^{5}$
Do the powers of ten first and the digits afterwards: $63 \div 1.8 = 35$, with the exponent carried along separately. At one significant figure this is a few hundred thousand turns, which is the honest precision here.
Common mistake. Working with the wheel's radius rather than its diameter, or leaving the journey in kilometres. Both faults are silent: the number that comes out looks perfectly reasonable, which is why the conversion is worth doing before anything else.
Why the other options are wrong
- A: Dividing $630$ by $1.8$ with no conversion at all, so the journey is treated as $630 \ \text{m}$. Unit Error
- B: Converting $630 \ \text{km}$ as $6.3 \times 10^{4} \ \text{m}$, one power of ten short. Order of Magnitude Error
- C: Treating the stated $0.60 \ \text{m}$ as a radius, so each turn is credited with $3.6 \ \text{m}$ instead of $1.8 \ \text{m}$. Nothing in the units objects; the geometry is simply wrong. Conceptual Misunderstanding
- E: Dividing by the diameter itself, as though the wheel advanced $0.60 \ \text{m}$ per turn rather than one whole circumference. Formula Misapplication
[02] Dimensional analysis
Question 2
A transverse pulse travels along a stretched string. Assume its speed can depend only on the tension $F$ and on the mass per unit length $\mu$, so that $v = kF^{\alpha}\mu^{\beta}$, where $k$ is a dimensionless constant; take $k = 1$, since dimensions fix the exponents but never the constant. The string is $4.0 \ \text{m}$ long, has mass $0.20 \ \text{kg}$ and is held at a tension of $45 \ \text{N}$. Assume the string is uniform, and neglect its stiffness and any sag under its own weight. What is the speed of the pulse?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $30 \ \text{m s}^{-1}$ (option E)
Key idea. Write the product of powers and solve for the exponents. With $[F] = \text{M L T}^{-2}$ and $[\mu] = \text{M L}^{-1}$, matching mass, length and time gives $\alpha = \tfrac{1}{2}$ and $\beta = -\tfrac{1}{2}$, that is $v = \sqrt{F/\mu}$.
Fastest approach
1. Match the dimensions.
$\text{L T}^{-1} = \left(\text{M L T}^{-2}\right)^{\alpha}\left(\text{M L}^{-1}\right)^{\beta}$
Mass: $\alpha + \beta = 0$. Time: $-2\alpha = -1$. Length: $\alpha - \beta = 1$.
So $\alpha = \tfrac{1}{2}$, $\beta = -\tfrac{1}{2}$, and $v = \sqrt{F/\mu}$.
2. Substitute.
$\mu = \dfrac{0.20 \ \text{kg}}{4.0 \ \text{m}} = 0.050 \ \text{kg m}^{-1}$
$v = \sqrt{\dfrac{45}{0.050}} = \sqrt{900} = 30 \ \text{m s}^{-1}$
Three of the five values on offer here come from routes whose units are not those of a speed at all, so checking the dimensions before the arithmetic clears most of the page.
Common mistake. Using the string's total mass where the derivation asks for its mass per unit length. $F/m$ has the dimensions of an acceleration, so its square root cannot be a speed, and the dimensions catch that before any arithmetic is done.
Why the other options are wrong
- A: Multiplying where the dimensions demand division: $\sqrt{F\mu}$ carries units of $\text{kg s}^{-1}$, a mass per unit time rather than a speed. Dimensional Error
- B: Multiplying and dropping the square root as well, leaving the product $F\mu$. Formula Misapplication
- C: Computing the mass per unit length as $5 \ \text{kg m}^{-1}$ rather than $0.050 \ \text{kg m}^{-1}$, a decimal point two places out. Order of Magnitude Error
- D: Using the string's total mass of $0.20 \ \text{kg}$ in place of its mass per unit length, so the expression is $\sqrt{F/m}$, the square root of an acceleration. Dimensional Error
[03] Kinematics
Question 3
A cyclist rides $24 \ \text{km}$ out along a straight road at a steady $20 \ \text{km h}^{-1}$, turns round, and rides the same $24 \ \text{km}$ back at a steady $30 \ \text{km h}^{-1}$. Neglect the time spent turning round, and take each leg as covered at exactly its stated speed. What is the average speed for the whole ride?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $24 \ \text{km h}^{-1}$ (option A)
Key idea. An average speed is a rate, so the question to ask is what is it per: total distance per total time. The slow leg lasts longer than the fast one, so it occupies more of the ride and must count for more.
Fastest approach
1. Time for each leg.
Out: $\dfrac{24}{20} = 1.2 \ \text{h}$. Back: $\dfrac{24}{30} = 0.8 \ \text{h}$.
2. Divide the whole by the whole.
$\bar{v} = \dfrac{48 \ \text{km}}{2.0 \ \text{h}} = 24 \ \text{km h}^{-1}$
Worth knowing as a check: for equal distances the answer is the harmonic mean $\frac{2v_{1}v_{2}}{v_{1}+v_{2}}$, which is always below the ordinary mean unless the two speeds are equal. So an answer at or above $25$ is wrong before it is examined.
Common mistake. Averaging the two speeds. That would be right if equal times were spent at each, and here it is equal distances that are covered, which is a different weighting entirely.
Why the other options are wrong
- B: Averaging the two speeds, $\tfrac{1}{2}(20 + 30)$. Equal distances mean unequal times, and the slower leg lasts longer, which drags the true average below the midpoint. Conceptual Misunderstanding
- C: Dividing the whole $48 \ \text{km}$ by the outward leg's $1.2 \ \text{h}$ alone. Incomplete Calculation
- D: Adding the two speeds, as though the legs combined rather than followed one another. Formula Misapplication
- E: Dividing the whole $48 \ \text{km}$ by the return leg's $0.8 \ \text{h}$ alone. Incomplete Calculation
[04] Resistive forces
Question 4
A sphere of mass $0.60 \ \text{kg}$ falls vertically from rest through still air. The air resistance on it is $F = kv^{2}$, directed opposite to the motion, with $k = 0.24 \ \text{kg m}^{-1}$. Take $g = 10 \ \text{m s}^{-2}$, neglect the upthrust of the air, and assume the sphere has been falling for $5.0 \ \text{s}$, which is long enough for it to be moving at its terminal speed. What is that terminal 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: $5 \ \text{m s}^{-1}$ (option B)
Key idea. Terminal speed is a balance of two effects: weight down, drag up, equal and opposite once the acceleration has died away. So $mg = kv^{2}$ and $v = \sqrt{mg/k}$.
Fastest approach
1. Balance the two forces.
$mg = kv^{2}$
2. Solve.
$v^{2} = \dfrac{mg}{k} = \dfrac{0.60 \times 10}{0.24} = \dfrac{6.0}{0.24} = 25$
$v = 5 \ \text{m s}^{-1}$
The square root is not optional. With $k$ measured in $\text{kg m}^{-1}$, the quotient $mg/k$ has the dimensions of a speed squared, so stopping at $25$ leaves an answer that cannot be a speed.
Common mistake. Taking the drag as proportional to $v$ rather than to $v^{2}$, which turns the balance into $mg = kv$ and stops one step early. The stem gives the drag law explicitly, so this is a matter of reading it rather than of recalling one.
Why the other options are wrong
- A: Multiplying by $k$ instead of dividing by it: $\sqrt{mgk}$ carries units of $\text{kg s}^{-1}$. Dimensional Error
- C: Solving $mg = kv$, that is taking the drag to grow in proportion to the speed. The number is $mg/k$, which with $k$ in $\text{kg m}^{-1}$ has the dimensions of a speed squared. Dimensional Error
- D: The speed after $5.0 \ \text{s}$ of free fall, $gt = 50 \ \text{m s}^{-1}$, which is what the sphere would reach if the air were not there at all. Conceptual Misunderstanding
- E: Squaring $mg/k$ instead of taking its square root. Formula Misapplication
[05] Estimation
Question 5
Estimate the area of solar panel a house would need to match its own electricity use. Assume the house draws electrical energy at an average rate of $600 \ \text{W}$; that, averaged over a whole year with nights and cloud included, sunlight delivers $150 \ \text{W}$ to each square metre of roof; that the panels turn $20\%$ of the light falling on them into electricity; and that the electricity can be stored between day and night without loss.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $20 \ \text{m}^{2}$ (option C)
Key idea. Break the estimate into factors you can each defend. One square metre of panel delivers $150 \times 0.20 = 30 \ \text{W}$ of electricity, and the house wants $600 \ \text{W}$, so it needs $600/30$ square metres.
Fastest approach
1. Electricity from one square metre.
$150 \times 0.20 = 30 \ \text{W m}^{-2}$
2. Divide the demand by it.
$A = \dfrac{600 \ \text{W}}{30 \ \text{W m}^{-2}} = 20 \ \text{m}^{2}$
The units do the checking: watts divided by watts per square metre leaves square metres. Any route that never divides by the supply per square metre has an answer still measured in watts.
Sanity check against the world: $20 \ \text{m}^{2}$ is roughly half the roof of a small house, which is why domestic arrays look the size they do.
Common mistake. Multiplying by the efficiency instead of dividing by it. A panel that wastes four fifths of the light it receives needs more roof, not less, and checking the direction of that effect catches the error without any arithmetic.
Why the other options are wrong
- A: Multiplying by the efficiency rather than dividing by it, $600 \times 0.20 / 150$, which has a worse panel needing less roof than a perfect one. Conceptual Misunderstanding
- B: Ignoring the efficiency, so every watt landing on the panel is credited as electricity delivered. Incomplete Calculation
- D: Slipping a power of ten in the supply and using $15 \ \text{W m}^{-2}$. Order of Magnitude Error
- E: Dividing the demand by the efficiency and never by the power per square metre. $600/0.20$ is still a power in watts, not an area. Dimensional Error
[06] Dimensional analysis
Question 6
Two satellites are in circular orbits around the same planet, one of radius $7.0 \times 10^{6} \ \text{m}$ and the other of radius $2.8 \times 10^{7} \ \text{m}$. Assume the orbital period can depend only on the orbital radius, the planet's mass and the gravitational constant $G$, whose dimensions follow from $F = Gm_{1}m_{2}/r^{2}$. The satellites' own masses are negligible and there is no atmospheric drag. By what factor is the outer satellite's period longer than the inner satellite'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: $8$ (option D)
Key idea. Read the dimensions off the equation that defines $G$: rearranging $F = Gm_{1}m_{2}/r^{2}$ gives $[G] = \text{M}^{-1}\text{L}^{3}\text{T}^{-2}$. Writing $T = kr^{a}M^{b}G^{c}$ and matching the three base dimensions gives $a = \tfrac{3}{2}$, which is Kepler's third law recovered without a line of mechanics.
Fastest approach
1. Dimensions of $G$.
$G = \dfrac{Fr^{2}}{m_{1}m_{2}} \implies [G] = \dfrac{\left(\text{M L T}^{-2}\right)\text{L}^{2}}{\text{M}^{2}} = \text{M}^{-1}\text{L}^{3}\text{T}^{-2}$
2. Match the exponents in $T = kr^{a}M^{b}G^{c}$.
Time: $1 = -2c$, so $c = -\tfrac{1}{2}$. Mass: $0 = b - c$, so $b = -\tfrac{1}{2}$. Length: $0 = a + 3c$, so $a = \tfrac{3}{2}$.
Hence $T \propto r^{3/2}$, that is $T^{2} \propto r^{3}$.
3. Take the ratio.
The radii are in the ratio $\frac{2.8 \times 10^{7}}{7.0 \times 10^{6}} = 4$, so the periods are in the ratio $4^{3/2} = 8$.
Dimensions give the scaling and never the constant out front, so this route settles the factor of $8$ while saying nothing about the $2\pi$ that a full derivation supplies.
Common mistake. Assuming both satellites travel at the same speed, so that the period simply follows the circumference and quadruples. Orbital speed falls as $1/\sqrt{r}$, so the outer satellite has both further to go and less speed to do it with.
Why the other options are wrong
- A: Carrying over $T \propto \sqrt{l}$ from a pendulum, where the only length in the problem is the pendulum's own and no central mass appears anywhere. Conceptual Misunderstanding
- B: Assuming the two satellites move at the same speed, so that the period is proportional to the circumference. Nothing in the units objects to that; the physics does, because orbital speed falls as $1/\sqrt{r}$. Conceptual Misunderstanding
- C: Multiplying by the exponent instead of raising to it: $4 \times \tfrac{3}{2}$ rather than $4^{3/2}$. Arithmetic Slip
- E: Quoting the ratio of the squares of the periods, $4^{3}$, which is the ratio of the cubes of the radii. Incomplete Calculation
[07] Density and buoyancy
Question 7
A block of metal hangs from a spring balance. In air the balance reads $50 \ \text{N}$; with the block fully immersed in water, touching neither the sides nor the bottom of the tank, it reads $40 \ \text{N}$. Take the density of water as $1000 \ \text{kg m}^{-3}$, take $g = 10 \ \text{m s}^{-2}$, and neglect the upthrust of the air on the block. What is the density of the metal?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $5000 \ \text{kg m}^{-3}$ (option E)
Key idea. The missing $10 \ \text{N}$ is the weight of the water pushed aside, which fixes the block's volume. Compare the two situations and the constants cancel in the ratio: the density is the weight in air divided by the weight lost, times the density of water.
Fastest approach
1. Upthrust.
$50 - 40 = 10 \ \text{N}$ of water displaced.
2. Volume.
$U = \rho_{w}Vg \implies V = \dfrac{10}{1000 \times 10} = 1.0 \times 10^{-3} \ \text{m}^{3}$
3. Mass, then density.
$m = \dfrac{50}{10} = 5.0 \ \text{kg}$, so $\rho = \dfrac{5.0}{1.0 \times 10^{-3}} = 5000 \ \text{kg m}^{-3}$
In one line, and with $g$ cancelling: $\rho = \rho_{w}\dfrac{W_{\text{air}}}{W_{\text{air}} - W_{\text{water}}} = 1000 \times \dfrac{50}{10}$.
Common mistake. Dividing the weight in air by the weight in water. The denominator of a relative density is the weight lost, not the weight that remains, and the two differ by exactly the upthrust.
Why the other options are wrong
- A: Taking the ratio upside down, upthrust over weight in air, which gives the reciprocal of the relative density. Formula Misapplication
- B: Dividing the immersed reading by the reading in air, $40/50$, a ratio that is not the relative density of anything. Conceptual Misunderstanding
- C: Dividing the weight in air by the weight in water, $50/40$, as though the water supported the block in proportion to what the balance still reads. The upthrust is the difference between the two readings, not their ratio. Conceptual Misunderstanding
- D: Using the immersed reading rather than the weight in air above the upthrust. The block's mass is fixed by what it weighs in air. Misread Question
[08] Energy and power
Question 8
A spring gun uses a spring of stiffness $800 \ \text{N m}^{-1}$, compressed by $2.0 \ \text{cm}$ from its natural length. Assume the spring obeys Hooke's law throughout this range and that its own mass is negligible. How much elastic potential energy is stored in 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: $0.16 \ \text{J}$ (option A)
Key idea. The force grows from zero to $kx$, so the average force is $\tfrac{1}{2}kx$ and the work stored is $\tfrac{1}{2}kx^{2}$, the triangle under the force-extension line. Convert before substituting: $2.0 \ \text{cm}$ is $0.020 \ \text{m}$.
Fastest approach
$x = 2.0 \ \text{cm} = 0.020 \ \text{m}$
$E = \tfrac{1}{2}kx^{2} = \tfrac{1}{2}(800)(0.020)^{2} = 400 \times 4.0 \times 10^{-4} = 0.16 \ \text{J}$
Two of the values offered are forces rather than energies, and one line of dimensional checking removes them: $kx$ is in newtons, $\tfrac{1}{2}kx^{2}$ is in joules.
Common mistake. Substituting the compression in centimetres. The stiffness is quoted per metre, and because the length is squared a stray factor of $100$ becomes a factor of $10^{4}$ in the energy.
Why the other options are wrong
- B: Omitting the factor of $\tfrac{1}{2}$, which is the whole difference between the peak force and the average force during the compression. Formula Misapplication
- C: Using the compression rather than its square, giving $\tfrac{1}{2}kx$. That is half the final force, measured in newtons, not an energy. Dimensional Error
- D: Reporting $kx$, the force needed to hold the spring compressed, in place of the energy stored in it. Dimensional Error
- E: Substituting $x = 2.0$ with the stiffness still per metre, so the compression is a hundred times too large and the energy ten thousand times. Unit Error
[09] Scaling arguments
Question 9
Two solid spheres of the same metal, one of radius $2.0 \ \text{cm}$ and one of radius $4.0 \ \text{cm}$, are heated to the same temperature and left to cool in the same still room. Take the rate at which each loses heat to be proportional to its surface area, with the same constant for both, and the heat it must lose in order to cool by one degree to be proportional to its volume, again with the same constant for both. The larger sphere's initial rate of temperature fall is what multiple of the smaller sphere'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: $\frac{1}{2}$ (option B)
Key idea. The rate of temperature fall goes as surface area over volume, and for a sphere that is $\frac{4\pi r^{2}}{\frac{4}{3}\pi r^{3}} = \frac{3}{r}$. It falls as $1/r$, so doubling the radius halves it. This is the square-cube law: area grows as the square, volume as the cube, and their ratio as the inverse first power.
Fastest approach
Rate of temperature fall $\propto \dfrac{\text{area}}{\text{volume}} = \dfrac{4\pi r^{2}}{\frac{4}{3}\pi r^{3}} = \dfrac{3}{r}$
Doubling the radius therefore halves it: the factor is $\tfrac{1}{2}$.
The radius itself never has to be substituted, which is the point of a scaling argument. Both stated radii are there only to say that one is twice the other.
Sanity check in a limit: a very small sphere is nearly all surface and cools almost at once, while a very large one is nearly all interior and cools slowly. The answer must fall with radius, and it does.
Common mistake. Reporting the factor of $4$ by which the surface area, and so the rate of heat loss in watts, grows. The larger sphere does lose heat faster; it simply has eight times as much heat to lose, so its temperature falls more slowly.
Why the other options are wrong
- A: Scaling as $1/r^{2}$, from a half-remembered inverse-square law. Area over volume goes as $1/r$, because the cube in the denominator beats the square in the numerator by exactly one power. Conceptual Misunderstanding
- C: Treating the cooling rate as a property of the metal, so that size cannot matter. What differs between these two spheres is the geometry, not the material. Conceptual Misunderstanding
- D: Taking volume over surface area, which doubles, rather than surface area over volume. Formula Misapplication
- E: Reporting the factor by which the rate of heat loss grows, which is the ratio of the surface areas, rather than the rate of temperature fall. Misread Question
[10] Estimation
Question 10
Estimate the pressure an adult standing still exerts on the floor. Take the person's mass as $70 \ \text{kg}$, take $g = 10 \ \text{m s}^{-2}$, take the total area of the two shoe soles in contact with the floor as $350 \ \text{cm}^{2}$, and assume the weight is spread evenly over that area.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $2 \times 10^{4} \ \text{Pa}$ (option C)
Key idea. $p = F/A$, with $F$ the weight of $700 \ \text{N}$ and $A$ in square metres. Handle the powers of ten first: a square metre holds $10^{4}$ square centimetres, so $350 \ \text{cm}^{2}$ is $3.5 \times 10^{-2} \ \text{m}^{2}$.
Fastest approach
1. Force.
$F = mg = 70 \times 10 = 700 \ \text{N}$
2. Area.
$350 \ \text{cm}^{2} = 350 \times 10^{-4} \ \text{m}^{2} = 3.5 \times 10^{-2} \ \text{m}^{2}$
3. Divide.
$p = \dfrac{7.0 \times 10^{2}}{3.5 \times 10^{-2}} = 2.0 \times 10^{4} \ \text{Pa}$
Worth comparing with something known: atmospheric pressure is about $1 \times 10^{5} \ \text{Pa}$, so standing on the floor presses on it about a fifth as hard as the air already does. That is why a person leaves no mark on a wooden floor and a stiletto heel, spreading the same weight over a hundredth of the area, does.
Common mistake. Dividing by $100$ rather than by $10^{4}$ when converting the area. A centimetre is a hundredth of a metre, so a square centimetre is a ten-thousandth of a square metre, and squaring the conversion is the step usually missed.
Why the other options are wrong
- A: Dividing the weight by the area still in square centimetres, $700/350$. Unit Error
- B: Dividing the mass rather than the weight by the area. Kilograms per square metre is not a pressure: the $g$ is what turns a mass into the force the floor actually feels. Dimensional Error
- D: Using the area of one shoe while still loading it with the whole weight of the person. Conceptual Misunderstanding
- E: Slipping one power of ten in the conversion and using $3.5 \times 10^{-3} \ \text{m}^{2}$. Order of Magnitude Error
[11] Indices
Question 11
Evaluate $\left(\dfrac{8}{27}\right)^{-2/3}$, giving your answer as a fraction in its lowest 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: $\frac{9}{4}$ (option D)
Key idea. A negative index inverts the fraction, the denominator of the index takes a root, and the numerator takes a power. Doing them in the order invert, cube root, square keeps every number in single digits, because $8$ and $27$ are both cubes.
Fastest approach
The index $-\tfrac{2}{3}$ does three separate things, so do them one at a time.
Invert, because the index is negative:
$\left(\dfrac{8}{27}\right)^{-2/3} = \left(\dfrac{27}{8}\right)^{2/3}$
Take the cube root, because the denominator of the index is $3$:
$\sqrt[3]{\dfrac{27}{8}} = \dfrac{3}{2}$
Square, because the numerator of the index is $2$:
$\left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}$
Rooting before powering is what keeps the arithmetic small. Powering first means cube rooting $\frac{729}{64}$, which is the same answer reached the slow way.
Common mistake. Carrying out only part of what the index asks for. A negative fractional index is an inversion, a root and a power together, and stopping after one or two of the three is the commonest way to lose this question.
Why the other options are wrong
- A: Dropping the minus sign and evaluating $\left(\frac{8}{27}\right)^{2/3}$ instead. Sign Error
- B: Taking the cube root of $\frac{8}{27}$ and stopping, so both the inversion and the squaring are missed. Incomplete Calculation
- C: Inverting and cube rooting correctly, then never squaring. Incomplete Calculation
- E: Treating the index as $-1$, so the fraction is inverted and the $\frac{2}{3}$ is discarded. Formula Misapplication
[12] Surds
Question 12
Write $\dfrac{6}{3-\sqrt{3}}$ in the form $a + b\sqrt{3}$, where $a$ and $b$ are rational.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $3+\sqrt{3}$ (option E)
Key idea. Multiply top and bottom by the conjugate $3+\sqrt{3}$. The denominator becomes a difference of two squares, $9-3=6$, which is the whole point of the conjugate: it removes the surd from the bottom, and here it cancels the $6$ on top exactly.
Fastest approach
Multiply numerator and denominator by the conjugate of the denominator:
$\dfrac{6}{3-\sqrt{3}} \times \dfrac{3+\sqrt{3}}{3+\sqrt{3}} = \dfrac{6\left(3+\sqrt{3}\right)}{(3-\sqrt{3})(3+\sqrt{3})}$
The denominator is a difference of two squares:
$(3-\sqrt{3})(3+\sqrt{3}) = 9 - 3 = 6$
so the $6$ cancels and
$\dfrac{6\left(3+\sqrt{3}\right)}{6} = 3+\sqrt{3}$
Numerical check: $3-\sqrt{3} \approx 1.27$ and $6 \div 1.27 \approx 4.73$, which is $3+\sqrt{3}$.
Common mistake. Multiplying the numerator by one thing and the denominator by another. Whatever multiplies the bottom must multiply the top, or the value of the expression has been changed rather than rewritten.
Why the other options are wrong
- A: Cancelling the $3$ in the numerator against the $3$ in the denominator to get $\frac{2}{1-\sqrt{3}}$, then rationalising that. Only a factor of the whole denominator may be cancelled, and $3$ is not a factor of $3-\sqrt{3}$. Conceptual Misunderstanding
- B: Writing the conjugate with the wrong sign in the numerator, so the top becomes $6(3-\sqrt{3})$ while the bottom is correctly $9-3=6$. Sign Error
- C: Discarding the surd and reading the denominator as $3$, which gives $6 \div 3$. Conceptual Misunderstanding
- D: Expanding $(3-\sqrt{3})(3+\sqrt{3})$ as $9+3=12$, so the answer is halved. Sign Error
[13] Completing the square
Question 13
The curve $y = 2x^{2} - 20x + 53$ has a single 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: $3$ (option A)
Key idea. Take the $2$ out of the first two terms before completing the square, so the constant produced inside the bracket is doubled on its way back out. In the form $y = 2(x-5)^{2}+3$ the minimum is whatever sits outside the square, because a square is never negative.
Fastest approach
1. Factor the leading coefficient out of the $x$ terms only.
$y = 2\left(x^{2}-10x\right)+53$
2. Complete the square inside the bracket.
$x^{2}-10x = (x-5)^{2}-25$
3. Multiply back out and collect the constants.
$y = 2\left[(x-5)^{2}-25\right]+53 = 2(x-5)^{2}-50+53 = 2(x-5)^{2}+3$
A square is never negative, so the smallest value of $y$ is $3$, taken where the bracket is zero, at $x=5$.
Same answer by symmetry: the curve is symmetric about $x = \frac{-b}{2a} = 5$, and substituting gives $50-100+53 = 3$.
Common mistake. Forgetting that the $-25$ inside the bracket is multiplied by the $2$ outside it, which leaves the final constant $25$ too large. Substituting the vertex back into the original expression catches that in one line.
Why the other options are wrong
- B: Giving the $x$-coordinate of the minimum point instead of the $y$-coordinate. Misread Question
- C: Computing $53 - \frac{400}{16}$, dividing $b^{2}$ by $4a^{2}$ rather than by $4a$; equivalently, forgetting to double the $-25$ as it leaves the bracket. Formula Misapplication
- D: Reading the constant term as the minimum. That is the value at $x=0$, which is not the vertex. Conceptual Misunderstanding
- E: Adding $\frac{400}{8}$ to $53$ rather than subtracting it. Sign Error
[14] Logarithms
Question 14
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: $4$ (option B)
Key idea. $\log_{2}A + \log_{2}B = \log_{2}(AB)$, so the equation becomes $x(x-2)=2^{3}$. That quadratic has two roots, but a logarithm accepts only a positive argument, so each root has to be tested in the equation as originally written.
Fastest approach
Combine the logarithms, then undo the logarithm:
$\log_{2}x + \log_{2}(x-2) = \log_{2}\left[x(x-2)\right] = 3$
$x(x-2) = 2^{3} = 8$
$x^{2}-2x-8 = 0 \implies (x-4)(x+2) = 0 \implies x = 4 \text{ or } x = -2$
Now test both against the original equation. At $x=-2$ both $\log_{2}x$ and $\log_{2}(x-2)$ are undefined, so that root is not a solution.
Check the survivor: $\log_{2}4 + \log_{2}2 = 2+1 = 3$.
Common mistake. Offering both roots of the quadratic. Combining the two logarithms quietly widens the domain, because $\log_{2}\left[x(x-2)\right]$ is defined whenever the product is positive, while the original equation needs $x>2$.
Why the other options are wrong
- A: The second root of $x^{2}-2x-8=0$, kept even though $\log_{2}(-2)$ has no value. Conceptual Misunderstanding
- C: Adding the arguments rather than multiplying them, and so solving $2x-2=8$. Formula Misapplication
- D: Solving $\log_{2}x = 3$ and ignoring the second logarithm entirely. Incomplete Calculation
- E: Solving $\log_{2}(x-2) = 3$ and ignoring the first logarithm entirely. Incomplete Calculation
[15] Exponential equations
Question 15
The equation $2^{2x} - 12\left(2^{x}\right) + 32 = 0$ has two real solutions. What is their sum?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $5$ (option C)
Key idea. $2^{2x} = \left(2^{x}\right)^{2}$, so putting $u = 2^{x}$ turns the equation into $u^{2}-12u+32=0$. The roots of that quadratic are values of $u$, not of $x$, and each one still has to be turned back into an $x$.
Fastest approach
Let $u = 2^{x}$, so that $2^{2x} = u^{2}$:
$u^{2}-12u+32 = 0 \implies (u-4)(u-8) = 0 \implies u = 4 \text{ or } u = 8$
Convert each back:
$2^{x} = 4 \implies x = 2$, and $2^{x} = 8 \implies x = 3$.
Sum: $2+3 = 5$.
Both roots of the quadratic are positive, so neither is lost. A negative root would have to be discarded, because $2^{x}$ is positive for every real $x$.
Common mistake. Stopping at the values of $2^{x}$. The quadratic is solved in $u$, so $4$ and $8$ are values of $2^{x}$, and their sum of $12$ answers a different question from the one asked.
Why the other options are wrong
- A: Solving $2^{x}=4$ and stopping, so only one of the two solutions is counted. Incomplete Calculation
- B: Solving $2^{x}=8$ and stopping, so only the other solution is counted. Incomplete Calculation
- D: Multiplying the two solutions together rather than adding them. Misread Question
- E: Adding $4$ and $8$, the roots of the quadratic in $2^{x}$, without converting either back into a value of $x$. Conceptual Misunderstanding
[16] Polynomials and the remainder theorem
Question 16
What is the remainder when $x^{3}+2x^{2}-5x+7$ is divided by $x+3$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $13$ (option D)
Key idea. Writing $p(x) = (x+3)q(x) + r$ and putting $x=-3$ kills the first term whatever $q$ is, leaving $p(-3) = r$. So the remainder is the value of the polynomial at the number that makes the divisor zero, and for $x+3$ that number is $-3$.
Fastest approach
$x+3 = 0$ when $x = -3$, so the remainder is $p(-3)$ for $p(x) = x^{3}+2x^{2}-5x+7$.
Term by term, watching every sign:
$(-3)^{3} = -27$
$2(-3)^{2} = 2 \times 9 = 18$
$-5 \times (-3) = 15$
constant: $7$
$-27+18+15+7 = 13$
Long division gives the same $13$ after three rounds of subtraction rather than one line of substitution.
Common mistake. Substituting $x=3$ because the divisor reads $x+3$. The divisor vanishes at $x=-3$, and that is the only value at which $p(x) = (x+3)q(x)+r$ collapses to $p = r$.
Why the other options are wrong
- A: Taking $(-3)^{2}$ as $-9$, so the quadratic term contributes $-18$ instead of $+18$. Arithmetic Slip
- B: Evaluating $-5x$ at $x=-3$ as $-15$, so the linear term is subtracted where it should be added. Sign Error
- C: Leaving the constant term $7$ out of the total. Incomplete Calculation
- E: Substituting $x=3$, reading the root straight off the $+3$ in the divisor. Conceptual Misunderstanding
[17] Simultaneous equations
Question 17
The line $y = 2x$ meets the curve $y = x^{2}-3$ at two points. How far apart are those two points?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $4\sqrt{5}$ (option E)
Key idea. Eliminate $y$ to get one quadratic in $x$. Its two roots are the $x$-coordinates of the intersections, and each gives a $y$ through the line's own equation. The distance then needs both differences, because the points differ in $x$ and in $y$.
Fastest approach
Set the two expressions for $y$ equal to each other:
$x^{2}-3 = 2x \implies x^{2}-2x-3 = 0 \implies (x-3)(x+1) = 0$
$x=3$ gives $y=6$, and $x=-1$ gives $y=-2$, so the points are $(3,6)$ and $(-1,-2)$.
$\Delta x = 3-(-1) = 4$ and $\Delta y = 6-(-2) = 8$, so
$d = \sqrt{4^{2}+8^{2}} = \sqrt{80} = 4\sqrt{5}$
Quick check on the structure: the line has gradient $2$, so any separation along it is $\sqrt{1+2^{2}} = \sqrt{5}$ times the horizontal separation, and $4\sqrt{5}$ is exactly that.
Common mistake. Reporting one of the two separations on its own. The horizontal gap is $4$ and the vertical gap is $8$, and the distance between the points is neither: it is the hypotenuse of the right-angled triangle those two gaps form.
Why the other options are wrong
- A: The horizontal separation of the two roots, with the vertical separation ignored. Incomplete Calculation
- B: Evaluating $y=2x$ at $x=-1$ as $+2$, which puts the second point at $(-1,2)$ and makes the vertical gap $4$. Sign Error
- C: The vertical separation of the two points, with the horizontal separation ignored. Incomplete Calculation
- D: Computing $\sqrt{4+8^{2}}$, squaring the vertical separation but not the horizontal one. Formula Misapplication
[18] Discriminants and tangency
Question 18
The line $y = 4x + c$ is a tangent to the parabola $y = x^{2}$. What is the value of $c$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $-4$ (option A)
Key idea. A tangent meets the parabola exactly once, so eliminating $y$ must leave a quadratic with a repeated root, and that is exactly the statement $b^{2}-4ac = 0$. The condition applies to the quadratic you actually obtain, so everything has to be collected on one side first.
Fastest approach
Eliminate $y$ and collect on one side:
$x^{2} = 4x + c \implies x^{2}-4x-c = 0$
Exactly one intersection means one repeated root, so the discriminant vanishes:
$b^{2}-4ac = (-4)^{2}-4(1)(-c) = 16+4c = 0 \implies c = -4$
Check it a second way, through the gradient. The parabola has gradient $2x$, which equals $4$ at $x=2$, so the point of contact is $(2,4)$ and $c = y-4x = 4-8 = -4$.
A negative intercept is what a sketch predicts: $y = 4x$ passes through the origin and cuts the parabola twice, so the line has to be lowered before it merely touches.
Common mistake. Leaving the constant on the right-hand side, or losing its sign as it crosses. The quadratic to test is $x^{2}-4x-c = 0$, whose constant term is $-c$, and that enters the discriminant as $-4(1)(-c) = +4c$.
Why the other options are wrong
- B: Using $b$ rather than $b^{2}$ in the discriminant, and so solving $-4+4c = 0$. Formula Misapplication
- C: Giving the $x$-coordinate of the point of contact instead of the intercept the question asks for. Misread Question
- D: A sign slip as the constant crosses the equals sign, which gives $16-4c = 0$. Sign Error
- E: Reaching the point of contact $(2,4)$ by the gradient route and then computing $y+4x = 4+8$ instead of $y-4x$. Sign Error
[19] Modulus inequalities
Question 19
The inequality $\left|3x-4\right| \le 11$ is satisfied exactly when $a \le x \le b$. What is $b$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $5$ (option B)
Key idea. $|A| \le k$ says exactly that $-k \le A \le k$, which is a single chain rather than two separate problems. Then undo what was done to $x$, in reverse order: add the $4$ to all three parts, then divide all three by $3$.
Fastest approach
$\left|3x-4\right| \le 11 \iff -11 \le 3x-4 \le 11$
Add $4$ throughout:
$-7 \le 3x \le 15$
Divide throughout by $3$:
$-\dfrac{7}{3} \le x \le 5$
So the upper end is $b = 5$.
Check the endpoint, which is the cheapest test available: $\left|3(5)-4\right| = |11| = 11$, satisfying the inequality with equality, exactly as an endpoint should.
Common mistake. Doing one of the two operations and stopping, or applying an operation to only part of the chain. Every step has to act on all three parts at once, which is also what keeps the two endpoints the right way round.
Why the other options are wrong
- A: Giving the lower endpoint $a$ where the question asks for the upper endpoint $b$. Misread Question
- C: Dividing the $3x$ by $3$ before adding the $4$, which turns the inequality into $|x-4| \le \frac{11}{3}$ and puts the upper end at $4+\frac{11}{3}$. Formula Misapplication
- D: Reading the bound on $\left|3x-4\right|$ as though it were a bound on $x$ itself. Conceptual Misunderstanding
- E: Reaching $3x \le 15$ and stopping there, so the division by $3$ never happens. Incomplete Calculation
[20] Quadratic inequalities
Question 20
For how many integer values of $x$ is $x^{2}-5x-24$ negative?
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 C)
Key idea. Factorise to find the roots. With a positive coefficient of $x^{2}$ the parabola opens upwards, so the expression is negative exactly between the roots. Counting the integers in that interval is the second half of the question, and the endpoints are excluded because there the expression is zero rather than negative.
Fastest approach
$x^{2}-5x-24 = (x-8)(x+3)$
The parabola opens upwards and crosses the axis at $x=-3$ and $x=8$, so it lies below the axis exactly for
$-3 < x < 8$
The integers strictly inside are $-2, -1, 0, 1, \ldots, 7$, and from $-2$ to $7$ inclusive there are
$7-(-2)+1 = 10$
of them. The endpoints $-3$ and $8$ give zero, which is not negative, so neither is counted.
Common mistake. Using $8-(-3) = 11$ as the count. That is the gap between the endpoints, not the number of integers strictly between them. Writing down the first two and the last two of the list is faster than any formula and does not go wrong.
Why the other options are wrong
- A: Counting only $0$ to $7$ and forgetting that $-2$ and $-1$ lie in the interval too. Incomplete Calculation
- B: Taking $8-(-3) = 11$ and then subtracting $2$ for the two excluded endpoints, where only one subtraction is needed. Arithmetic Slip
- D: Using the difference of the endpoints, $8-(-3)$, as the count. Formula Misapplication
- E: Counting $x=-3$ and $x=8$ as well, where the expression is zero and therefore not negative. Misread Question