An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 AQA A Level Mathematics 7357 paper. Not affiliated with or reproduced from AQA.
Paper 1: Pure Mathematics
Answer all questions. Entirely pure content; calculator and AQA formulae booklet permitted.
11 PastPaper.question · 100 PastPaper.marks
PastPaper.question 1 · Multiple choice
1 PastPaper.marks
Differentiate \(y=\ln(5x)\). Circle your answer.
A.\(\tfrac5x\)
B.\(\tfrac1x\)
C.\(\tfrac{1}{5x}\)
D.\(\ln5\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
\(\tfrac{d}{dx}\ln(5x)=\tfrac{5}{5x}=\tfrac1x\).
PastPaper.markingScheme
B1 for \(\tfrac1x\).
PastPaper.question 2 · Multiple choice
1 PastPaper.marks
Find \(\displaystyle\int e^{3x}\,dx\). Circle your answer.
A.\(3e^{3x}+c\)
B.\(\tfrac13 e^{3x}+c\)
C.\(e^{3x}+c\)
D.\(3e^{x}+c\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
\(\int e^{3x}dx=\tfrac13 e^{3x}+c\).
PastPaper.markingScheme
B1 for \(\tfrac13 e^{3x}+c\).
PastPaper.question 3 · Short
5 PastPaper.marks
Find the first three terms, in ascending powers of \(x\), of the binomial expansion of \((3+2x)^6\).
A circle \(C\) has equation \(x^2+y^2-10x+2y+1=0\). (a) Find the centre and radius. (b) Show that the point \((5,4)\) lies on \(C\). (c) Find the equation of the tangent to \(C\) at \((5,4)\). (d) Find the length of the tangent to \(C\) from the external point \((12,3)\).
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) \((x-5)^2+(y+1)^2=25\): centre \((5,-1)\), radius \(5\). (b) \(25+16-50+8+1=0\) ✓. (c) The radius from \((5,-1)\) to \((5,4)\) is vertical, so the tangent is horizontal: \(y=4\). (d) Tangent length \(=\sqrt{(12-5)^2+(3+1)^2-25}=\sqrt{49+16-25}=\sqrt{40}=6.32\).
(a) An arithmetic series has first term 4 and common difference 6. Find (i) the 15th term and (ii) the sum of the first 15 terms. (b) A geometric series has first term 4 and common ratio 1.05. Find (i) the 10th term and (ii) the sum of the first 10 terms. (c) State, with a reason, whether the geometric series converges.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a)(i) \(4+14(6)=88\); (ii) \(S_{15}=\tfrac{15}{2}(8+84)=690\). (b)(i) \(4(1.05)^9=6.20\); (ii) \(S_{10}=4\cdot\tfrac{1.05^{10}-1}{0.05}=50.3\). (c) No: \(|r|=1.05>1\), so it diverges.
PastPaper.markingScheme
M1A1 (a); M1A1 (b)(i); M1A1 (b)(ii); B1 (c) with reason.
PastPaper.question 6 · Structured
12 PastPaper.marks
\(f(x)=x^3-7x+2\). (a) Show that a root of \(f(x)=0\) lies between 2 and 3. (b) Apply the Newton–Raphson method with \(x_0=2.5\) to find \(x_1\) and \(x_2\). (c) The iteration \(x_{n+1}=\sqrt[3]{7x_n-2}\) is used with \(x_0=2.5\). Find the root to 3 decimal places.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) \(f(2)=-4<0\), \(f(3)=8>0\): sign change implies a root in \((2,3)\). (b) \(f'(x)=3x^2-7\); \(x_1=2.5-\tfrac{0.125}{11.75}=2.4894\); \(x_2=2.4894-\tfrac{f(2.4894)}{f'(2.4894)}=2.4893\). (c) \(x_1=\sqrt[3]{15.5}=2.494,\,x_2=2.490,\,x_3=2.489,\dots\to 2.489\).
\(p(x)=2x^3+ax^2+bx-6\). Given that \((x-2)\) and \((x+1)\) are factors: (a) find \(a\) and \(b\); (b) factorise \(p(x)\) completely; (c) solve \(p(x)=0\).
M1 p(2)=0; M1 p(-1)=0; A1 a,b; M1 divide; A1 full factorisation; A1 three roots.
PastPaper.question 8 · Structured
8 PastPaper.marks
(a) A geometric series has first term 6 and common ratio \(\tfrac23\). Find its sum to infinity. (b) Solve \(\tan 2x=1\) for \(0\le x\le\pi\). (c) Solve \(4\sin x=3\cos x\) for \(0\le x\le2\pi\), to 2 decimal places.
(a) Find \(\displaystyle\int\left(6x^2-\tfrac{4}{x^2}+e^{3x}\right)dx\). (b) Use the substitution \(u=x^2+4\) to find \(\displaystyle\int\tfrac{2x}{\sqrt{x^2+4}}\,dx\). (c) Find the area enclosed between the curve \(y=x^2+3\) and the line \(y=4x\).
A curve has equation \(y=2x^3-9x^2+12x-3\). (a) Find \(\dfrac{dy}{dx}\). (b) Find the stationary points and determine their nature. (c) Find the coordinates of the point of inflection. (d) Find the equation of the tangent at \(x=0\). (e) State the set of values of \(x\) for which the curve is increasing.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) \(\tfrac{dy}{dx}=6x^2-18x+12=6(x-1)(x-2)\). (b) \(x=1\Rightarrow(1,2)\); \(x=2\Rightarrow(2,1)\). \(\tfrac{d^2y}{dx^2}=12x-18\): at \(x=1\) negative (max), at \(x=2\) positive (min). (c) \(12x-18=0\Rightarrow x=1.5,\,y=1.5\). (d) At \(x=0\), \(y=-3\), gradient \(12\): \(y=12x-3\). (e) Increasing where \(\tfrac{dy}{dx}>0\): \(x<1\) or \(x>2\).
Answer all questions. Section A is pure; Section B is mechanics (take g = 9.8 m s⁻²).
14 PastPaper.question · 100 PastPaper.marks
PastPaper.question 1 · Multiple choice
1 PastPaper.marks
Find \(\displaystyle\int\tfrac1x\,dx\). Circle your answer.
A.\(\ln|x|+c\)
B.\(-\tfrac{1}{x^2}+c\)
C.\(\tfrac{1}{2}x^2+c\)
D.\(x\ln x+c\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
\(\int\tfrac1x dx=\ln|x|+c\).
PastPaper.markingScheme
B1 for \(\ln|x|+c\).
PastPaper.question 2 · Short
3 PastPaper.marks
Express \(\dfrac{4x+5}{(x+1)(x-2)}\) in partial fractions.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Let \(\tfrac{4x+5}{(x+1)(x-2)}=\tfrac{A}{x+1}+\tfrac{B}{x-2}\). Then \(4x+5=A(x-2)+B(x+1)\). \(x=-1:1=-3A\Rightarrow A=-\tfrac13\); \(x=2:13=3B\Rightarrow B=\tfrac{13}{3}\).
PastPaper.markingScheme
M1 set up identity; M1 substitute; A1 A=-1/3, B=13/3.
PastPaper.question 3 · Proof
6 PastPaper.marks
Prove by contradiction that \(\sqrt5\) is irrational.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Assume \(\sqrt5=\tfrac pq\) in lowest terms. Then \(p^2=5q^2\), so \(5\mid p^2\Rightarrow5\mid p\). Write \(p=5k\): \(25k^2=5q^2\Rightarrow q^2=5k^2\), so \(5\mid q\). Then 5 divides both \(p\) and \(q\), contradicting lowest terms. Hence \(\sqrt5\) is irrational.
A closed box has a square base of side \(x\,\text{cm}\) and volume \(1000\,\text{cm}^3\). (a) Show that the total surface area is \(S=2x^2+\dfrac{4000}{x}\). (b) Find the value of \(x\) that minimises \(S\). (c) Find the minimum surface area.
(a) A population grows so that \(\dfrac{dP}{dt}=0.4P\). Show that \(P=Ae^{0.4t}\). (b) Given \(P=150\) when \(t=0\), find \(P\) when \(t=5\). (c) Solve the differential equation \(\dfrac{dy}{dx}=4x^3y\), giving \(y\) in terms of \(x\).
(a) Express \(5\sin x-12\cos x\) in the form \(R\sin(x-\alpha)\), \(R>0\), \(0<\alpha<90^\circ\). (b) State the maximum value and the value of \(x\) in \(0\le x\le360^\circ\) at which it occurs. (c) Solve \(5\sin x-12\cos x=6.5\) for \(0\le x\le360^\circ\).
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) \(R=\sqrt{5^2+12^2}=13\), \(\tan\alpha=\tfrac{12}{5}\Rightarrow\alpha=67.38^\circ\): \(13\sin(x-67.38^\circ)\). (b) Max \(13\) when \(x-67.38^\circ=90^\circ\Rightarrow x=157.4^\circ\). (c) \(\sin(x-67.38^\circ)=0.5\Rightarrow x-67.38^\circ=30^\circ,150^\circ\Rightarrow x=97.4^\circ,217.4^\circ\).
Find the weight of a particle of mass \(8\,\text{kg}\) (take \(g=9.8\)). Circle your answer.
A.\(8\,\text{N}\)
B.\(0.82\,\text{N}\)
C.\(78.4\,\text{N}\)
D.\(9.8\,\text{N}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Weight \(=mg=8\times9.8=78.4\,\text{N}\).
PastPaper.markingScheme
B1 for 78.4 N.
PastPaper.question 9 · Structured
6 PastPaper.marks
A particle moves in a straight line with acceleration \(a=(12t-6)\,\text{m s}^{-2}\) and velocity \(4\,\text{m s}^{-1}\) at \(t=0\). (a) Find \(v\) in terms of \(t\). (b) Find \(v\) when \(t=2\). (c) Find the displacement during the first 2 seconds.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) \(v=\int(12t-6)dt=6t^2-6t+C\); \(v(0)=4\Rightarrow C=4\), so \(v=6t^2-6t+4\). (b) \(v(2)=24-12+4=16\). (c) \(s=\int_0^2(6t^2-6t+4)dt=[2t^3-3t^2+4t]_0^2=16-12+8=12\,\text{m}\).
A particle of mass \(3\,\text{kg}\) lies on a smooth plane inclined at \(30^\circ\). It is connected by a light inextensible string over a smooth pulley at the top of the plane to a mass of \(5\,\text{kg}\) hanging freely. The system is released from rest. (a) Find the acceleration. (b) Find the tension. (c) State one assumption about the string.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
For the hanging mass: \(5g-T=5a\). On the plane: \(T-3g\sin30^\circ=3a\). Adding: \(5g-1.5g=8a\Rightarrow a=\tfrac{34.3}{8}=4.29\,\text{m s}^{-2}\). Then \(T=5g-5a=49-21.44=27.6\,\text{N}\). (c) The string is light and inextensible.
A uniform beam \(AB\) of length \(10\,\text{m}\) and weight \(500\,\text{N}\) rests horizontally on supports at \(A\) and \(B\). A load of \(300\,\text{N}\) is placed \(3\,\text{m}\) from \(A\). (a) Find the reaction at each support. (b) Find how far from \(A\) the load must be placed for the reactions to be equal.
M1 moments about A; A1 R_B; M1 resolve; A1 R_A; M1 set equal; A1 d=5.
PastPaper.question 12 · Structured
8 PastPaper.marks
A particle has position vector \(\mathbf{r}=(t^2-4t)\mathbf{i}+(t^2-2t)\mathbf{j}\) (\(t\) in seconds). (a) Find its velocity when \(t=1\). (b) Find its speed when \(t=1\). (c) Find the time at which it is moving parallel to \(\mathbf{j}\).
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
\(\mathbf{v}=(2t-4)\mathbf{i}+(2t-2)\mathbf{j}\). (a) At \(t=1\): \(\mathbf{v}=-2\mathbf{i}+0\mathbf{j}\). (b) Speed \(=\sqrt{(-2)^2+0^2}=2\,\text{m s}^{-1}\). (c) Parallel to \(\mathbf{j}\) when the \(\mathbf{i}\)-component is zero: \(2t-4=0\Rightarrow t=2\).
PastPaper.markingScheme
M1 differentiate; A1 v at t=1; M1A1 speed; M1 set i-comp =0; A1 t=2.
PastPaper.question 13 · Structured
6 PastPaper.marks
A ball is thrown horizontally with speed \(20\,\text{m s}^{-1}\) from a point \(25\,\text{m}\) above horizontal ground (take \(g=9.8\)). (a) Find the time to reach the ground. (b) Find the horizontal distance travelled.
A particle \(P\) of mass \(4\,\text{kg}\) lies on a smooth horizontal table. It is connected by a light inextensible string passing over a smooth pulley at the edge of the table to a particle \(Q\) of mass \(6\,\text{kg}\) hanging freely. The system is released from rest. (a) Find the acceleration. (b) Find the tension in the string. (c) State one assumption you have made about the pulley.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
For \(Q\): \(6g-T=6a\). For \(P\): \(T=4a\). Adding: \(6g=10a\Rightarrow a=\tfrac{58.8}{10}=5.88\,\text{m s}^{-2}\). Then \(T=4(5.88)=23.52\,\text{N}\). (c) The pulley is smooth, so the tension is the same on both sides of the string.
PastPaper.markingScheme
M1 equation for Q; M1 equation for P; A1 add; A1 a=5.88; M1 T; A1 23.52; B1 assumption.
Paper 3: Pure & Statistics
Answer all questions. Section A is pure; Section B is statistics (Large Data Set context).
13 PastPaper.question · 100 PastPaper.marks
PastPaper.question 1 · Multiple choice
1 PastPaper.marks
State the period of \(y=\sin(2\pi x)\). Circle your answer.
A.\(2\pi\)
B.\(\tfrac12\)
C.\(1\)
D.\(\pi\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Period \(=\tfrac{2\pi}{2\pi}=1\).
PastPaper.markingScheme
B1 for 1.
PastPaper.question 2 · Structured
10 PastPaper.marks
Differentiate each of the following. (a) \(y=(3x-2)^4\). (b) \(y=x^2\ln x\). (c) \(y=\dfrac{x}{x^2+4}\). (d) Hence state the gradient of the curve in part (a) at \(x=1\).
M1 cos2x value; A1 all four x; M1A1 (b); M1A1 (c).
PastPaper.question 5 · Structured
10 PastPaper.marks
A radioactive sample decays as \(m=m_0e^{-kt}\), with a half-life of 12 years. (a) Find \(k\) to 3 significant figures. (b) Given \(m_0=80\,\text{g}\), find the mass after 20 years. (c) Find, to the nearest year, the time to fall to \(20\,\text{g}\). (d) Find the rate of decay at \(t=0\).
(a) An arithmetic series is \(5,9,13,\dots\). Find the sum of the first 25 terms. (b) A geometric series is \(200,150,112.5,\dots\). Find its sum to infinity. (c) Use the arithmetic-series formula to prove that \(1+3+5+\dots+(2n-1)=n^2\). (d) Find the first term of the series in (a) that exceeds 100.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) \(a=5,d=4\): \(S_{25}=\tfrac{25}{2}(10+24(4))=\tfrac{25}{2}(106)=1325\). (b) \(r=0.75\): \(\tfrac{200}{1-0.75}=800\). (c) AP with \(a=1,d=2\): \(S_n=\tfrac n2(2+(n-1)2)=n^2\). (d) \(5+(n-1)4>100\Rightarrow n>24.75\Rightarrow n=25\) (term \(=101\)).
PastPaper.markingScheme
M1A1 (a); M1A1 (b); M1A1 (c); M1A1 (d).
PastPaper.question 7 · Multiple choice
1 PastPaper.marks
For \(X\sim B(25,0.2)\), state the mean \(E(X)\). Circle your answer.
A.\(0.2\)
B.\(5\)
C.\(4\)
D.\(20\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
\(E(X)=np=25\times0.2=5\).
PastPaper.markingScheme
B1 for 5.
PastPaper.question 8 · Structured
6 PastPaper.marks
(a) Define a simple random sample. (b) A college has 600 students, each with an ID number. Describe how to take a simple random sample of 60. (c) State one advantage of stratified sampling here.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) A simple random sample is one in which every possible sample of the required size is equally likely (every member has an equal chance of selection). (b) Number students 1–600; use random numbers to choose 60 distinct IDs, ignoring repeats and numbers above 600. (c) Stratified sampling represents each subgroup (e.g. year group) in proportion, reducing bias.
Eight values are \(10,12,14,15,18,20,24,28\). (a) Find the median. (b) Find \(Q_1\), \(Q_3\) and the IQR. (c) Find the mean. (d) Find the standard deviation to 2 decimal places. (e) Using the \(1.5\times\text{IQR}\) rule, determine whether 28 is an outlier.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) Median \(=\tfrac{15+18}{2}=16.5\). (b) \(Q_1=12,\,Q_3=22\) (mean of 20 and 24), \(\text{IQR}=10\). (c) Mean \(=\tfrac{141}{8}=17.625\). (d) \(\sum x^2=2749\); variance \(=\tfrac{2749}{8}-17.625^2=32.98\); s.d. \(=5.74\). (e) \(Q_3+1.5(10)=37>28\), so 28 is not an outlier.
(a) For \(X\sim B(12,0.35)\) find (i) \(P(X=4)\) and (ii) \(P(X\le2)\). (b) A coin-like process is claimed to succeed with probability greater than 0.35. In 12 trials there are 9 successes. Test at the 5% level whether the success probability exceeds 0.35.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a)(i) \(\binom{12}{4}0.35^4 0.65^8=0.2367\). (ii) \(P(X\le2)=0.1513\). (b) \(H_0:p=0.35\), \(H_1:p>0.35\), \(X\sim B(12,0.35)\). \(P(X\ge9)=1-P(X\le8)=0.0028<0.05\), so reject \(H_0\): evidence the success probability exceeds 0.35.
Events \(A\) and \(B\) satisfy \(P(A)=0.6\), \(P(B)=0.3\) and \(P(A\cap B)=0.18\). (a) Find \(P(A\cup B)\). (b) Find \(P(B\mid A)\). (c) Determine, with justification, whether \(A\) and \(B\) are independent.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) \(P(A\cup B)=0.6+0.3-0.18=0.72\). (b) \(P(B\mid A)=\tfrac{0.18}{0.6}=0.3\). (c) \(P(A)P(B)=0.6(0.3)=0.18=P(A\cap B)\), so \(A\) and \(B\) are independent.
PastPaper.markingScheme
M1A1 (a); M1A1 (b); M1 compare; A1 conclusion.
PastPaper.question 13 · Structured
6 PastPaper.marks
The discrete random variable \(X\) has \(P(X=0)=0.1,\,P(X=1)=0.3,\,P(X=2)=0.4,\,P(X=3)=0.2\). (a) Find \(E(X)\). (b) Find \(\text{Var}(X)\). (c) Find \(E(3X-2)\).