PAT practice · 20 questions · Free
PAT Mock: 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 ball is thrown vertically upwards at $20 \ \text{m s}^{-1}$. Taking $g = 10 \ \text{m s}^{-2}$ and ignoring air resistance, what is the greatest height it reaches?
Reveal the answer and worked solution. Commit to an option first.
Answer: C
Key idea. At the top the vertical velocity is zero, so $v^{2} = u^{2} - 2gh$ gives the height directly. Finding the time first works but adds a step for nothing.
Fastest approach
At the highest point $v = 0$:
$0 = u^{2} - 2gh \implies h = \frac{u^{2}}{2g} = \frac{400}{20} = 20 \ \text{m}$
Equivalently by energy: $\tfrac{1}{2}mu^{2} = mgh$, and the mass cancels, which is why it is not given.
Common mistake. Using $h = ut$ with the total flight time rather than the time to the top, which doubles the answer, or using the time to the top with $h = ut$ and forgetting the deceleration.
Why the other options are wrong
- A: Halving 20 rather than dividing $u^{2}$ by $2g$. Formula Misapplication
- B: Arithmetic slip in $400 \div 20$. Arithmetic Slip
- D: Using $h = ut - \tfrac{1}{2}gt^{2}$ with the wrong time. Formula Misapplication
- E: Using the whole flight time, giving twice the true height. Conceptual Misunderstanding
[02] Momentum and energy
Question 2
A $2 \ \text{kg}$ trolley moving at $3 \ \text{m s}^{-1}$ collides with a stationary $4 \ \text{kg}$ trolley and they move off together. How much kinetic energy is lost in the collision?
Reveal the answer and worked solution. Commit to an option first.
Answer: D
Key idea. Momentum is conserved in every collision; kinetic energy is conserved only in elastic ones. Find the common velocity from momentum, then compare the kinetic energies before and after.
Fastest approach
Momentum before: $2 \times 3 = 6 \ \text{kg m s}^{-1}$.
They move off together with total mass $6 \ \text{kg}$, so $v = \frac{6}{6} = 1 \ \text{m s}^{-1}$.
$E_{\text{before}} = \tfrac{1}{2}(2)(3^{2}) = 9 \ \text{J}$ $E_{\text{after}} = \tfrac{1}{2}(6)(1^{2}) = 3 \ \text{J}$
Lost: $9 - 3 = 6 \ \text{J}$
Common mistake. Conserving kinetic energy to find the final velocity. In a collision where the objects stick together, kinetic energy is always lost, and assuming otherwise makes the question unanswerable.
Why the other options are wrong
- A: Using the final velocity as $2 \ \text{m s}^{-1}$ from dividing the initial speed by the mass ratio. Formula Misapplication
- B: Reporting the kinetic energy after the collision rather than the loss. Misread Question
- C: Arithmetic slip in $9 - 3$. Arithmetic Slip
- E: Reporting the kinetic energy before the collision. Misread Question
[03] Electric circuits
Question 3
Three identical $6 \ \Omega$ resistors are connected so that one is in series with a parallel pair of the other two. What is the total resistance?
Reveal the answer and worked solution. Commit to an option first.
Answer: A
Key idea. Reduce the parallel pair first, then add the series resistor. Two equal resistors in parallel always give half of one of them, which is worth knowing without computing.
Fastest approach
Parallel pair: two equal resistors give $\frac{6}{2} = 3 \ \Omega$.
In series with the third: $6 + 3 = 9 \ \Omega$
Common mistake. Adding all three in series for $18 \ \Omega$, or treating the whole network as parallel for $2 \ \Omega$. Reduce the innermost combination first and work outwards.
Why the other options are wrong
- B: Adding two in series and treating the third as absent. Incomplete Calculation
- C: All three in series. Conceptual Misunderstanding
- D: All three treated as parallel: $\frac{6}{3} = 2$. Conceptual Misunderstanding
- E: Reporting the parallel pair plus nothing, or one resistor alone. Incomplete Calculation
[04] Kinematics
Question 4
A projectile is launched at a fixed angle over level ground. If its launch speed is doubled and air resistance is ignored, by what factor does its horizontal range change?
Reveal the answer and worked solution. Commit to an option first.
Answer: B
Key idea. Range is $\frac{v^{2}\sin 2\theta}{g}$. The angle is fixed, so range goes as $v^{2}$ and the numerical value never has to be computed.
Fastest approach
$R = \frac{v^{2}\sin 2\theta}{g}$
With $\theta$ fixed, $R \propto v^{2}$.
Doubling $v$ multiplies $R$ by $2^{2} = 4$.
Both the flight time and the horizontal speed double, which is the same result seen a second way.
Common mistake. Answering 2, on the reasoning that the projectile travels twice as fast. It does, but it is also in the air twice as long, and the two effects multiply.
Why the other options are wrong
- A: Scaling range linearly with speed and ignoring the longer flight time. Incomplete Calculation
- C: Cubing rather than squaring. Formula Misapplication
- D: Applying the square twice. Formula Misapplication
- E: Taking the square root instead of squaring. Formula Misapplication
[05] Matter and pressure
Question 5
What is the pressure due to the water alone at a depth of $20 \ \text{m}$ in a lake? Take the density of water as $1000 \ \text{kg m}^{-3}$ and $g = 10 \ \text{m s}^{-2}$.
Reveal the answer and worked solution. Commit to an option first.
Answer: C
Key idea. $p = \rho g h$. The whole difficulty is arithmetic in standard form, so group the powers of ten before multiplying.
Fastest approach
$p = \rho g h = 1000 \times 10 \times 20$
$= 10^{3} \times 10^{1} \times (2 \times 10^{1}) = 2 \times 10^{5} \ \text{Pa}$
The question says 'due to the water alone', so atmospheric pressure is not added.
Common mistake. Adding atmospheric pressure of about $1 \times 10^{5} \ \text{Pa}$ to get $3 \times 10^{5}$. The question asks for the water's contribution only, which is the distinction between gauge and absolute pressure.
Why the other options are wrong
- A: Dropping two powers of ten. Arithmetic Slip
- B: Dropping one power of ten. Arithmetic Slip
- D: Adding atmospheric pressure, which the question excludes. Misread Question
- E: Gaining one power of ten. Arithmetic Slip
[06] Waves
Question 6
A string of length $1 \ \text{m}$ is fixed at both ends. What is the wavelength of its third harmonic?
Reveal the answer and worked solution. Commit to an option first.
Answer: B
Key idea. A string fixed at both ends fits a whole number of half-wavelengths: $L = \frac{n\lambda}{2}$, so $\lambda = \frac{2L}{n}$.
Fastest approach
$\lambda_{n} = \frac{2L}{n}$
Third harmonic: $n = 3$, $L = 1$.
$\lambda = \frac{2}{3} \ \text{m}$
Sanity check: the fundamental is $2 \ \text{m}$, and the third harmonic must be a third of that.
Common mistake. Using $\lambda = \frac{L}{n}$, which forgets that each loop is half a wavelength, not a whole one.
Why the other options are wrong
- A: Using $\lambda = \frac{L}{n}$, missing the factor of 2. Formula Misapplication
- C: Setting the wavelength equal to the string length. Conceptual Misunderstanding
- D: Inverting the fraction. Formula Misapplication
- E: The fundamental wavelength, which is the first harmonic. Misread Question
[07] Energy and power
Question 7
How much energy does a $60 \ \text{W}$ lamp use in one hour?
Reveal the answer and worked solution. Commit to an option first.
Answer: D
Key idea. $E = Pt$ needs $t$ in seconds for an answer in joules. An hour is $3600 \ \text{s}$, and that conversion is the entire question.
Fastest approach
$E = Pt = 60 \times 3600 = 216\,000 \ \text{J} = 216 \ \text{kJ}$
Common mistake. Leaving the time in hours and reporting $60 \ \text{J}$, or converting but then mislabelling the magnitude. A watt is a joule per second, so seconds are the only unit that works.
Why the other options are wrong
- A: Multiplying by one hour without converting to seconds. Unit Conversion Error
- B: Reporting the number of seconds rather than the energy. Misread Question
- C: Multiplying by 1000 instead of by 3600. Unit Conversion Error
- E: Using $1000 \ \text{W}$ instead of $60 \ \text{W}$, which is a kilowatt-hour. Formula Misapplication
[08] Circular motion
Question 8
A $0.5 \ \text{kg}$ mass moves in a horizontal circle of radius $2 \ \text{m}$ at a constant speed of $4 \ \text{m s}^{-1}$. What is the centripetal force on it?
Reveal the answer and worked solution. Commit to an option first.
Answer: E
Key idea. $F = \frac{mv^{2}}{r}$. Only the speed is squared, and the radius divides rather than multiplies.
Fastest approach
$F = \frac{mv^{2}}{r} = \frac{0.5 \times 16}{2} = \frac{8}{2} = 4 \ \text{N}$
Common mistake. Squaring the radius as well, or multiplying by $r$ instead of dividing. Checking the units settles it: $\text{kg} \times \text{m}^{2}\text{s}^{-2} \div \text{m}$ gives newtons.
Why the other options are wrong
- A: Reporting $mv^{2}$ without dividing by the radius. Incomplete Calculation
- B: Multiplying by the radius instead of dividing. Formula Misapplication
- C: Dividing by $r^{2}$ rather than by $r$. Formula Misapplication
- D: Forgetting to square the speed: $\frac{0.5 \times 4}{2} \times 2$. Formula Misapplication
[09] Radioactivity
Question 9
A radioactive sample has a half-life of $5$ days. What fraction of the original activity remains after $15$ days?
Reveal the answer and worked solution. Commit to an option first.
Answer: C
Key idea. Count how many half-lives fit into the elapsed time, then halve that many times. Decay is exponential, so the fraction is $\left(\tfrac{1}{2}\right)^{n}$, never $1$ minus something linear.
Fastest approach
$\frac{15}{5} = 3$ half-lives.
$\left(\tfrac{1}{2}\right)^{3} = \tfrac{1}{8}$
Common mistake. Treating decay as linear and answering $\tfrac{1}{3}$, as though three half-lives removed three equal thirds. Each half-life removes half of what is left, not half of the original.
Why the other options are wrong
- A: Counting four half-lives. Miscount
- B: Dividing by the number of days. Conceptual Misunderstanding
- D: Counting two half-lives instead of three. Miscount
- E: Treating the decay as linear in the number of half-lives. Conceptual Misunderstanding
[10] Optics
Question 10
Light travels inside a medium of refractive index $1.5$ towards a boundary with air. What is the sine of the critical angle?
Reveal the answer and worked solution. Commit to an option first.
Answer: A
Key idea. At the critical angle the refracted ray grazes the boundary at $90^{\circ}$, so $n \sin\theta_{c} = 1$ and $\sin\theta_{c} = \frac{1}{n}$.
Fastest approach
$n_{1}\sin\theta_{c} = n_{2}\sin 90^{\circ} = 1$
$\sin\theta_{c} = \frac{1}{1.5} = \frac{2}{3}$
Common mistake. Writing $\sin\theta_{c} = n = 1.5$, which is impossible: no sine exceeds $1$. That check alone eliminates the wrong direction instantly.
Why the other options are wrong
- B: Using $\frac{n}{2}$. Formula Misapplication
- C: Reporting $n$ itself, which cannot be a sine. Conceptual Misunderstanding
- D: Using $\frac{1}{n^{2}}$ rather than $\frac{1}{n}$, roughly. Formula Misapplication
- E: Approximating $\frac{1}{1.5}$ as $\frac{1}{2}$. Estimation Error
[11] Series
Question 11
What is the value of $1 + 2 + 3 + \cdots + 100$?
Reveal the answer and worked solution. Commit to an option first.
Answer: E
Key idea. $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$. Pairing the ends is the same argument: fifty pairs each summing to $101$.
Fastest approach
$\frac{100 \times 101}{2} = 50 \times 101 = 5050$
Pairing: $(1+100), (2+99), \ldots$ gives $50$ pairs of $101$.
Common mistake. Using $\frac{n^{2}}{2}$ or $\frac{n(n-1)}{2}$. The second is the sum up to $99$, which is $4950$ and is offered here.
Why the other options are wrong
- A: $\frac{100 \times 102}{2}$, an off-by-one the other way. Boundary Error
- B: $100 \times 101$, without halving. Incomplete Calculation
- C: $\frac{100 \times 99}{2}$, the sum to 99. Boundary Error
- D: $\frac{100^{2}}{2}$. Formula Misapplication
[12] Differentiation
Question 12
If $y = x^{2}\mathrm{e}^{x}$, what is $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ at $x = 1$?
Reveal the answer and worked solution. Commit to an option first.
Answer: B
Key idea. $\frac{d}{dx}\left(x^{2}\mathrm{e}^{x}\right) = (2x + x^{2})\mathrm{e}^{x}$. The exponential factors out, which makes the substitution trivial.
Fastest approach
$\frac{dy}{dx} = 2x\mathrm{e}^{x} + x^{2}\mathrm{e}^{x} = (2x + x^{2})\mathrm{e}^{x}$
At $x = 1$: $(2 + 1)\mathrm{e} = 3\mathrm{e}$
Common mistake. Differentiating each factor and multiplying: $2x \times \mathrm{e}^{x}$ alone gives $2\mathrm{e}$. The product rule adds two terms.
Why the other options are wrong
- A: Differentiating only the polynomial factor. Formula Misapplication
- C: Slip in $2x + x^{2}$ at $x = 1$. Arithmetic Slip
- D: Multiplying the exponentials rather than keeping one factor. Formula Misapplication
- E: Differentiating only the exponential. Incomplete Calculation
[13] Integration
Question 13
Evaluate $\displaystyle\int_{0}^{1} x\,\mathrm{e}^{x^{2}} \,\mathrm{d}x$.
Reveal the answer and worked solution. Commit to an option first.
Answer: A
Key idea. The derivative of $x^{2}$ is $2x$, and the integrand has an $x$. So this is a reverse chain rule needing a factor of $\tfrac{1}{2}$, not integration by parts.
Fastest approach
Let $u = x^{2}$, so $\mathrm{d}u = 2x\,\mathrm{d}x$ and $x\,\mathrm{d}x = \tfrac{1}{2}\mathrm{d}u$.
Limits: $x=0 \to u=0$, $x=1 \to u=1$.
$\tfrac{1}{2}\int_{0}^{1}\mathrm{e}^{u}\,\mathrm{d}u = \tfrac{1}{2}\left[\mathrm{e}^{u}\right]_{0}^{1} = \frac{\mathrm{e}-1}{2}$
Common mistake. Forgetting the factor of $\tfrac{1}{2}$, which is what the extra $x$ in the integrand pays for. Reaching for integration by parts also works far more slowly.
Why the other options are wrong
- B: Forgetting to subtract the value at the lower limit. Boundary Error
- C: Correct integral with the factor of $\tfrac{1}{2}$ dropped. Incomplete Calculation
- D: Multiplying by 2 instead of by $\tfrac{1}{2}$. Formula Misapplication
- E: Substituting the limits into $\mathrm{e}^{u}$ with $u$ read as $2x$. Formula Misapplication
[14] Exponentials and logarithms
Question 14
Solve $3^{2x} = 27^{\,x-1}$.
Reveal the answer and worked solution. Commit to an option first.
Answer: D
Key idea. $27 = 3^{3}$. Once both sides share a base, the exponents can be equated directly and no logarithms are needed.
Fastest approach
$27^{x-1} = \left(3^{3}\right)^{x-1} = 3^{3(x-1)}$
$3^{2x} = 3^{3x-3} \implies 2x = 3x - 3 \implies x = 3$
Common mistake. Taking logs immediately. It works, but rewriting $27$ as $3^{3}$ turns the problem into one line of linear algebra.
Why the other options are wrong
- A: Sign error in $2x = 3x - 3$. Sign Error
- B: Solving $2x = x - 1$, having forgotten to multiply the exponent by 3. Formula Misapplication
- C: Arithmetic slip when rearranging. Arithmetic Slip
- E: Solving $2x = 3x - 3$ and then doubling. Arithmetic Slip
[15] Geometry
Question 15
A chord of length $6$ is drawn in a circle of radius $5$. How far is the chord from the centre?
Reveal the answer and worked solution. Commit to an option first.
Answer: D
Key idea. The perpendicular from the centre to a chord bisects it. That gives a right-angled triangle with the radius as hypotenuse and half the chord as one leg.
Fastest approach
Half-chord: $3$. Radius: $5$.
$d = \sqrt{5^{2} - 3^{2}} = \sqrt{25 - 9} = \sqrt{16} = 4$
The 3-4-5 triangle is the tell.
Common mistake. Using the full chord length of $6$ as a leg, giving $\sqrt{25-36}$, which is not a real number and should immediately signal the error.
Why the other options are wrong
- A: Adding with the full chord: $\sqrt{25+36}$. Formula Misapplication
- B: Subtracting half the chord from the radius: $5 - 3$. Conceptual Misunderstanding
- C: Reporting the half-chord rather than the distance. Misread Question
- E: Adding rather than subtracting: $\sqrt{25+9}$. Sign Error
[16] Binomial expansion
Question 16
What is the coefficient of $x^{2}$ in the expansion of $(1 + 2x)^{5}$?
Reveal the answer and worked solution. Commit to an option first.
Answer: E
Key idea. The $x^{2}$ term is $\binom{5}{2}(2x)^{2}$. Both the binomial coefficient and the squared $2$ contribute, and forgetting either is the usual error.
Fastest approach
$\binom{5}{2}(2x)^{2} = 10 \times 4x^{2} = 40x^{2}$
Coefficient: $40$
Common mistake. Using $\binom{5}{2}$ alone and answering $10$, or multiplying by $2$ rather than by $2^{2}$.
Why the other options are wrong
- A: Using $\binom{5}{2} = 10$ and $2^{3}$. Formula Misapplication
- B: Using $\binom{5}{3}$ with $2^{4}$. Formula Misapplication
- C: $\binom{5}{2}$ with the factor of $(2)^{2}$ omitted. Incomplete Calculation
- D: Multiplying by 2 rather than by $2^{2}$. Formula Misapplication
[17] Probability
Question 17
Three fair coins are tossed. What is the probability that exactly two show heads?
Reveal the answer and worked solution. Commit to an option first.
Answer: D
Key idea. Exactly two heads can happen in three ways (HHT, HTH, THH) out of eight equally likely outcomes. Which coin shows tails is the thing being counted.
Fastest approach
Sample space: $2^{3} = 8$ equally likely outcomes.
Exactly two heads: HHT, HTH, THH. Three of them.
$P = \frac{3}{8}$
Common mistake. Answering $\tfrac{1}{8}$, having counted 'two heads' as a single outcome. The two heads can fall on any of three pairs of coins.
Why the other options are wrong
- A: Counting HHT as the only favourable outcome. Sample Space Error
- B: Using a sample space of 4, as though the coins were indistinguishable. Sample Space Error
- C: Reasoning that one of three coins shows tails, so the probability is a third. Conceptual Misunderstanding
- E: Assuming heads and tails split the outcomes evenly regardless of the count. Conceptual Misunderstanding
[18] Trigonometry
Question 18
What is the maximum value of $5\sin\theta - 12\cos\theta$?
Reveal the answer and worked solution. Commit to an option first.
Answer: E
Key idea. $a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)$ with $R = \sqrt{a^{2}+b^{2}}$. The sign of $b$ shifts the phase, not the amplitude.
Fastest approach
$R = \sqrt{5^{2} + 12^{2}} = \sqrt{169} = 13$
Maximum value: $13$.
The minus sign changes where the maximum occurs, not how big it is.
Common mistake. Subtracting the coefficients to get $-7$, or taking the larger coefficient alone. The two terms peak at different values of $\theta$, so neither is the answer.
Why the other options are wrong
- A: Adding the coefficients: $5 + 12$. Formula Misapplication
- B: Taking the sine coefficient alone. Conceptual Misunderstanding
- C: Subtracting: $12 - 5$. Formula Misapplication
- D: Taking the cosine coefficient alone. Conceptual Misunderstanding
[19] Quadratics
Question 19
The roots of $x^{2} - 5x + 6 = 0$ are $\alpha$ and $\beta$. What is $\alpha^{2} + \beta^{2}$?
Reveal the answer and worked solution. Commit to an option first.
Answer: C
Key idea. $\alpha + \beta = 5$ and $\alpha\beta = 6$ can be read straight off the coefficients, and $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2} - 2\alpha\beta$.
Fastest approach
$\alpha + \beta = 5$, $\alpha\beta = 6$
$\alpha^{2} + \beta^{2} = (\alpha+\beta)^{2} - 2\alpha\beta = 25 - 12 = 13$
Check: the roots are $2$ and $3$, and $4 + 9 = 13$.
Common mistake. Answering $25$, having squared the sum without subtracting twice the product. That is $(\alpha+\beta)^{2}$, which counts the cross term.
Why the other options are wrong
- A: Reporting the sum of the roots. Misread Question
- B: Reporting $2\alpha\beta$. Misread Question
- D: $(\alpha+\beta)^{2}$, without subtracting the cross term. Incomplete Calculation
- E: $(\alpha+\beta)^{2} + 2\alpha\beta$, adding rather than subtracting. Sign Error
[20] Graphs
Question 20
What is the equation of the horizontal asymptote of $y = \dfrac{3x^{2}+1}{x^{2}-4}$?
Reveal the answer and worked solution. Commit to an option first.
Answer: A
Key idea. For large $|x|$ only the highest powers matter. Equal degrees top and bottom means the asymptote is the ratio of the leading coefficients.
Fastest approach
Divide numerator and denominator by $x^{2}$:
$y = \frac{3 + \frac{1}{x^{2}}}{1 - \frac{4}{x^{2}}} \to \frac{3}{1} = 3$ as $x \to \pm\infty$
Horizontal asymptote: $y = 3$.
The vertical asymptotes at $x = \pm 2$ come from the denominator and are a separate question.
Common mistake. Answering $y = 0$ by habit. That is right when the denominator has the higher degree, but here the degrees match and the ratio of leading coefficients wins.
Why the other options are wrong
- B: Taking the constant 4 from the denominator. Conceptual Misunderstanding
- C: The rule for a denominator of higher degree, applied where the degrees are equal. Formula Misapplication
- D: Using the denominator's leading coefficient alone. Formula Misapplication
- E: Reading off the vertical asymptote at $x = 2$. Misread Question