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Thinka Jun 2022 AQA AS Level-Style Mock — Mathematics 7356

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An original Thinka practice paper modelled on the structure and difficulty of the Jun 2022 AQA AS Level Mathematics 7356 paper. Not affiliated with or reproduced from AQA.

Paper 1 — Section A: Pure Mathematics

Answer all questions. Pure content; calculator and AQA formulae booklet permitted.
10 PastPaper.question · 53 PastPaper.marks
PastPaper.question 1 · Multiple choice
1 PastPaper.marks
Express \(\log_3 5 + \log_3 2\) as a single logarithm. Circle your answer.
  1. A.\(\log_3 7\)
  2. B.\(\log_3 10\)
  3. C.\(\log_3 2.5\)
  4. D.\(\log_3 3\)
PastPaper.showAnswers

PastPaper.workedSolution

\(\log_3 5 + \log_3 2 = \log_3(5\times 2) = \log_3 10\).

PastPaper.markingScheme

B1 for \(\log_3 10\).
PastPaper.question 2 · Multiple choice
1 PastPaper.marks
Which single transformation maps \(y=\cos x\) onto \(y=\cos 2x\)? Circle your answer.
  1. A.Stretch scale factor 2 in the x-direction
  2. B.Stretch scale factor \(\tfrac12\) in the x-direction
  3. C.Stretch scale factor 2 in the y-direction
  4. D.Translation by \(2\) in the x-direction
PastPaper.showAnswers

PastPaper.workedSolution

Replacing \(x\) by \(2x\) is a horizontal stretch with scale factor \(\tfrac12\).

PastPaper.markingScheme

B1 for the correct stretch.
PastPaper.question 3 · Short
3 PastPaper.marks
Find the coefficient of \(x^2\) in the binomial expansion of \((2+3x)^5\).
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PastPaper.workedSolution

The \(x^2\) term is \(\binom{5}{2}(2)^3(3x)^2 = 10\times 8\times 9\,x^2 = 720x^2\).

PastPaper.markingScheme

M1 correct binomial term structure; M1 correct values \(10,8,9\); A1 720.
PastPaper.question 4 · Short
5 PastPaper.marks
Solve \(2\sin^2 x = 3\cos x\) for \(0^\circ \le x \le 360^\circ\).
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PastPaper.workedSolution

Using \(\sin^2 x = 1-\cos^2 x\): \(2-2\cos^2 x = 3\cos x\Rightarrow 2\cos^2 x+3\cos x-2=0\Rightarrow (2\cos x-1)(\cos x+2)=0\). So \(\cos x=\tfrac12\) (reject \(\cos x=-2\)), giving \(x=60^\circ,300^\circ\).

PastPaper.markingScheme

M1 use of identity; M1 form quadratic in \(\cos x\); A1 \(\cos x=\tfrac12\); A1 60°; A1 300°.
PastPaper.question 5 · Structured
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A circle has the points \(A(-1,2)\) and \(B(5,10)\) as the ends of a diameter. (a) Write down the centre. (b) Show that the radius is 5. (c) Find the equation of the circle. (d) Determine whether the point \((6,6)\) lies inside, on, or outside the circle.
PastPaper.showAnswers

PastPaper.workedSolution

(a) Centre = midpoint = \((2,6)\). (b) \(AB=\sqrt{6^2+8^2}=10\), so radius \(=5\). (c) \((x-2)^2+(y-6)^2=25\). (d) \((6-2)^2+(6-6)^2=16<25\), so the point is inside.

PastPaper.markingScheme

B1 centre; M1A1 radius via distance; B1 equation; M1 substitute (6,6); A1 inside.
PastPaper.question 6 · Structured
7 PastPaper.marks
The curve \(y=9-x^2\) meets the x-axis at two points. (a) Find the x-coordinates of these points. (b) Find the exact area enclosed between the curve and the x-axis.
PastPaper.showAnswers

PastPaper.workedSolution

(a) \(9-x^2=0\Rightarrow x=\pm3\). (b) Area \(=\int_{-3}^{3}(9-x^2)\,dx = \left[9x-\tfrac{x^3}{3}\right]_{-3}^{3} = (27-9)-(-27+9)=18-(-18)=36\).

PastPaper.markingScheme

M1A1 intercepts; M1 set up definite integral; A1 integrate; M1 limits; A1 36.
PastPaper.question 7 · Structured
8 PastPaper.marks
A curve has equation \(y=2x^3-9x^2+12x\). (a) Find \(\dfrac{dy}{dx}\). (b) Find the coordinates of the stationary points. (c) Determine the nature of each stationary point.
PastPaper.showAnswers

PastPaper.workedSolution

(a) \(\tfrac{dy}{dx}=6x^2-18x+12=6(x-1)(x-2)\). (b) Stationary where \(x=1\) (\(y=5\)) and \(x=2\) (\(y=4\)). (c) \(\tfrac{d^2y}{dx^2}=12x-18\); at \(x=1\) it is \(-6<0\) (maximum), at \(x=2\) it is \(6>0\) (minimum).

PastPaper.markingScheme

M1A1 derivative; M1 solve =0; A1 both points; M1 second derivative; A1 both natures.
PastPaper.question 8 · Proof
5 PastPaper.marks
Prove that, for every integer \(n\), \(n^3-n\) is divisible by 6.
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PastPaper.workedSolution

\(n^3-n=n(n^2-1)=(n-1)n(n+1)\), the product of three consecutive integers. Among any three consecutive integers at least one is divisible by 2 and exactly one is divisible by 3. Hence the product is divisible by \(2\times3=6\).

PastPaper.markingScheme

M1 factorise; A1 three consecutive integers; A1 divisible by 2; A1 divisible by 3; A1 conclude divisible by 6.
PastPaper.question 9 · Structured
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A curve has equation \(y=x^2-4x+5\). (a) Find the equation of the tangent to the curve at \(x=3\). (b) Find the equation of the normal at \(x=3\). (c) The normal meets the curve again; find the coordinates of that point.
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PastPaper.workedSolution

At \(x=3\), \(y=2\). \(\tfrac{dy}{dx}=2x-4=2\). (a) Tangent: \(y-2=2(x-3)\Rightarrow y=2x-4\). (b) Normal gradient \(-\tfrac12\): \(y=-\tfrac12 x+\tfrac72\). (c) Set \(x^2-4x+5=-\tfrac12 x+\tfrac72\Rightarrow 2x^2-7x+3=0\Rightarrow(2x-1)(x-3)=0\), other root \(x=\tfrac12\), \(y=\tfrac{13}{4}\).

PastPaper.markingScheme

M1A1 tangent; M1A1 normal; M1 equate to curve; M1 solve quadratic; A1 other point.
PastPaper.question 10 · Structured
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A population is modelled by \(P=500e^{kt}\), where \(t\) is in years. After 3 years the population is 800. (a) Find \(k\) to 3 significant figures. (b) Estimate the population after 10 years. (c) Find, to the nearest year, when the population reaches 2000.
PastPaper.showAnswers

PastPaper.workedSolution

(a) \(800=500e^{3k}\Rightarrow k=\tfrac{1}{3}\ln 1.6=0.157\). (b) \(P=500e^{0.157\times10}=500e^{1.566}\approx2395\). (c) \(2000=500e^{kt}\Rightarrow e^{kt}=4\Rightarrow t=\tfrac{\ln4}{0.1566}\approx8.85\), so about 9 years.

PastPaper.markingScheme

M1 form equation; A1 k=0.157; M1A1 (b); M1 set =2000; A1 ≈9 years.

Paper 1 — Section B: Mechanics

Answer all questions. Take g = 9.8 m s⁻² unless otherwise stated.
6 PastPaper.question · 27 PastPaper.marks
PastPaper.question 1 · Multiple choice
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A particle starts from rest and moves in a straight line with constant acceleration. Which graph best shows its velocity \(v\) against time \(t\)? Circle your answer.
  1. A.A straight line through the origin with positive gradient
  2. B.A horizontal line
  3. C.A curve of increasing gradient
  4. D.A straight line with positive intercept
PastPaper.showAnswers

PastPaper.workedSolution

Constant acceleration from rest gives \(v=at\): a straight line through the origin with positive gradient.

PastPaper.markingScheme

B1 straight line through the origin.
PastPaper.question 2 · Structured
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A ball is projected vertically upwards from ground level at \(21\,\text{m s}^{-1}\) (take \(g=9.8\)). (a) Find the maximum height reached. (b) Find the total time before it returns to the ground.
PastPaper.showAnswers

PastPaper.workedSolution

(a) \(v^2=u^2-2gs\): \(0=21^2-2(9.8)s\Rightarrow s=\tfrac{441}{19.6}=22.5\,\text{m}\). (b) Time up: \(0=21-9.8t\Rightarrow t=2.143\,\text{s}\); total \(=2\times2.143=4.29\,\text{s}\).

PastPaper.markingScheme

M1A1 max height; M1 time up; A1 total time.
PastPaper.question 3 · Structured
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A particle of mass \(2\,\text{kg}\) is acted on by forces \(\mathbf{F_1}=(3\mathbf{i}+4\mathbf{j})\,\text{N}\) and \(\mathbf{F_2}=(5\mathbf{i}-2\mathbf{j})\,\text{N}\). (a) Find the resultant force. (b) Find the magnitude of the acceleration.
PastPaper.showAnswers

PastPaper.workedSolution

(a) Resultant \(=(8\mathbf{i}+2\mathbf{j})\,\text{N}\). (b) \(|\mathbf{F}|=\sqrt{8^2+2^2}=\sqrt{68}\); \(a=\tfrac{\sqrt{68}}{2}=4.12\,\text{m s}^{-2}\).

PastPaper.markingScheme

B1 resultant; M1 magnitude; M1 divide by mass; A1 4.12.
PastPaper.question 4 · Structured
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Two particles of masses \(3\,\text{kg}\) and \(5\,\text{kg}\) are connected by a light inextensible string passing over a smooth fixed pulley, and released from rest. (a) Find the acceleration of the system. (b) Find the tension in the string.
PastPaper.showAnswers

PastPaper.workedSolution

(a) For the system, \((5-3)g=(5+3)a\Rightarrow a=\tfrac{2\times9.8}{8}=2.45\,\text{m s}^{-2}\). (b) For the 3 kg mass: \(T-3g=3a\Rightarrow T=3(9.8)+3(2.45)=36.75\,\text{N}\).

PastPaper.markingScheme

M1 equation of motion for system; A1 a=2.45; M1 equation for one mass; A1 T=36.75; A1 consistent check.
PastPaper.question 5 · Structured
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A particle moves in a straight line with velocity \(v=3t^2-12t+9\;\text{m s}^{-1}\). (a) Find the times when the particle is instantaneously at rest. (b) Find its acceleration when \(t=3\). (c) Find its displacement during the first 4 seconds.
PastPaper.showAnswers

PastPaper.workedSolution

(a) \(3t^2-12t+9=0\Rightarrow t^2-4t+3=0\Rightarrow(t-1)(t-3)=0\), \(t=1,3\). (b) \(a=\tfrac{dv}{dt}=6t-12\); at \(t=3\), \(a=6\). (c) \(s=\int_0^4 v\,dt=[t^3-6t^2+9t]_0^4=64-96+36=4\,\text{m}\).

PastPaper.markingScheme

M1 set v=0; A1 both times; M1A1 acceleration; M1 integrate; A1 4 m.
PastPaper.question 6 · Structured
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A car of mass \(1200\,\text{kg}\) tows a trailer of mass \(400\,\text{kg}\) by a light rigid tow-bar along a straight horizontal road. The driving force is \(4000\,\text{N}\); the resistances are \(450\,\text{N}\) on the car and \(150\,\text{N}\) on the trailer. (a) Find the acceleration. (b) Find the tension in the tow-bar.
PastPaper.showAnswers

PastPaper.workedSolution

(a) For the whole system: \(4000-(450+150)=1600a\Rightarrow a=\tfrac{3400}{1600}=2.125\,\text{m s}^{-2}\). (b) For the trailer: \(T-150=400a\Rightarrow T=150+400(2.125)=1000\,\text{N}\).

PastPaper.markingScheme

M1 whole-system equation; A1 a=2.125; M1 trailer equation; A1 T=1000; B1 valid method shown.

Paper 2 — Section A: Pure Mathematics

Answer all questions. Pure content; calculator and AQA formulae booklet permitted.
10 PastPaper.question · 53 PastPaper.marks
PastPaper.question 1 · Multiple choice
1 PastPaper.marks
Find \(\displaystyle\int 6x^2\,dx\). Circle your answer.
  1. A.\(12x+c\)
  2. B.\(2x^3+c\)
  3. C.\(6x^3+c\)
  4. D.\(3x^2+c\)
PastPaper.showAnswers

PastPaper.workedSolution

\(\int 6x^2\,dx = 2x^3 + c\).

PastPaper.markingScheme

B1 for \(2x^3+c\).
PastPaper.question 2 · Multiple choice
1 PastPaper.marks
Given that \(\sin\theta=\cos 50^\circ\), a possible value of \(\theta\) is: circle your answer.
  1. A.\(40^\circ\)
  2. B.\(50^\circ\)
  3. C.\(130^\circ\)
  4. D.\(320^\circ\)
PastPaper.showAnswers

PastPaper.workedSolution

\(\cos 50^\circ=\sin 40^\circ\), so \(\theta=40^\circ\) (or \(140^\circ\)).

PastPaper.markingScheme

B1 for 40°.
PastPaper.question 3 · Structured
5 PastPaper.marks
\(f(x)=2x^3+x^2-13x+6\). (a) Show that \((x-2)\) is a factor of \(f(x)\). (b) Factorise \(f(x)\) completely.
PastPaper.showAnswers

PastPaper.workedSolution

(a) \(f(2)=16+4-26+6=0\), so \((x-2)\) is a factor. (b) Dividing, \(f(x)=(x-2)(2x^2+5x-3)=(x-2)(2x-1)(x+3)\).

PastPaper.markingScheme

M1 evaluate f(2); A1 =0 conclude factor; M1 divide; A1 quadratic factor; A1 full factorisation.
PastPaper.question 4 · Structured
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The equation \(x^2+(k+2)x+9=0\) has equal roots. Find the possible values of \(k\).
PastPaper.showAnswers

PastPaper.workedSolution

Equal roots \(\Rightarrow\) discriminant \(=0\): \((k+2)^2-36=0\Rightarrow k+2=\pm6\Rightarrow k=4\) or \(k=-8\).

PastPaper.markingScheme

M1 discriminant =0; A1 \((k+2)^2=36\); A1 k=4; A1 k=-8.
PastPaper.question 5 · Structured
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A bottle of water cools according to \(T=20+Ae^{-kt}\), where \(T\,^\circ\text{C}\) is the temperature \(t\) minutes after it leaves a fridge. Initially \(T=80\), and after 5 minutes \(T=50\). (a) Find \(A\). (b) Find \(k\) to 3 significant figures. (c) Find the temperature after 12 minutes. (d) Find, to the nearest minute, the time taken to reach \(30^\circ\text{C}\).
PastPaper.showAnswers

PastPaper.workedSolution

(a) At \(t=0\): \(80=20+A\Rightarrow A=60\). (b) \(50=20+60e^{-5k}\Rightarrow e^{-5k}=\tfrac12\Rightarrow k=\tfrac{\ln2}{5}=0.139\). (c) \(T=20+60e^{-0.1386\times12}=20+60(0.1895)=31.4^\circ\text{C}\). (d) \(30=20+60e^{-kt}\Rightarrow e^{-kt}=\tfrac16\Rightarrow t=\tfrac{\ln6}{0.1386}=12.9\), about 13 min.

PastPaper.markingScheme

B1 A=60; M1 form equation; A1 k=0.139; M1A1 (c); M1 set =30; A1 ≈13 min.
PastPaper.question 6 · Structured
7 PastPaper.marks
In triangle \(ABC\), \(AB=7\,\text{cm}\), \(BC=8\,\text{cm}\) and angle \(ABC=60^\circ\). (a) Find the length \(AC\). (b) Find the area of the triangle. (c) Find angle \(BAC\) to 1 decimal place.
PastPaper.showAnswers

PastPaper.workedSolution

(a) \(AC^2=7^2+8^2-2(7)(8)\cos60^\circ=113-56=57\Rightarrow AC=7.55\,\text{cm}\). (b) Area \(=\tfrac12(7)(8)\sin60^\circ=24.2\,\text{cm}^2\). (c) \(\tfrac{\sin BAC}{8}=\tfrac{\sin60^\circ}{7.55}\Rightarrow \sin BAC=0.9176\Rightarrow BAC=66.6^\circ\).

PastPaper.markingScheme

M1A1 cosine rule; A1 AC; M1A1 area; M1 sine rule; A1 66.6°.
PastPaper.question 7 · Structured
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The line \(y=x+1\) and the curve \(y=x^2-2x-3\) intersect at two points. (a) Find the coordinates of the points of intersection. (b) Write down the inequalities that define the region enclosed between the line and the curve. (c) Find the exact area of the enclosed region.
PastPaper.showAnswers

PastPaper.workedSolution

(a) \(x^2-2x-3=x+1\Rightarrow x^2-3x-4=0\Rightarrow(x-4)(x+1)=0\); points \((-1,0),(4,5)\). (b) \(-1\le x\le4\) and \(x^2-2x-3\le y\le x+1\). (c) Area \(=\int_{-1}^{4}\big((x+1)-(x^2-2x-3)\big)dx=\int_{-1}^{4}(-x^2+3x+4)dx=\tfrac{125}{6}\).

PastPaper.markingScheme

M1 equate; A1 quadratic; A1 both points; B1 inequalities; M1 integrate difference; M1 limits; A1 125/6.
PastPaper.question 8 · Structured
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A curve has equation \(y=2x^3-3x^2-12x\). (a) Find \(\dfrac{dy}{dx}\). (b) Find the coordinates of the stationary points. (c) Determine the nature of each.
PastPaper.showAnswers

PastPaper.workedSolution

(a) \(\tfrac{dy}{dx}=6x^2-6x-12=6(x-2)(x+1)\). (b) \(x=2\Rightarrow y=-20\); \(x=-1\Rightarrow y=7\). (c) \(\tfrac{d^2y}{dx^2}=12x-6\); at \(x=2\) positive (min), at \(x=-1\) negative (max).

PastPaper.markingScheme

M1A1 derivative; M1 points; A1 both; M1 second derivative; A1 natures.
PastPaper.question 9 · Structured
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A geometric series has second term 6 and sum to infinity 27. (a) Show that the common ratio satisfies \(9r^2-9r+2=0\). (b) Find the two possible values of \(r\). (c) For \(r=\tfrac13\), find the first term and the sum of the first 5 terms.
PastPaper.showAnswers

PastPaper.workedSolution

Let first term \(a\). \(ar=6\) and \(\tfrac{a}{1-r}=27\). So \(a=\tfrac6r\) and \(\tfrac{6/r}{1-r}=27\Rightarrow6=27r(1-r)\Rightarrow27r^2-27r+6=0\Rightarrow9r^2-9r+2=0\). (b) \((3r-1)(3r-2)=0\Rightarrow r=\tfrac13,\tfrac23\). (c) \(r=\tfrac13\Rightarrow a=18\); \(S_5=18\cdot\tfrac{1-(1/3)^5}{1-1/3}=27\cdot\tfrac{242}{243}\approx26.9\).

PastPaper.markingScheme

M1 two equations; M1 eliminate a; A1 given quadratic; M1 factorise; A1 both r; M1 a=18; A1 S₅.
PastPaper.question 10 · Structured
4 PastPaper.marks
Express \(\dfrac{5}{2-\sqrt3}\) in the form \(a+b\sqrt3\), where \(a\) and \(b\) are integers.
PastPaper.showAnswers

PastPaper.workedSolution

Multiply numerator and denominator by \(2+\sqrt3\): \(\dfrac{5(2+\sqrt3)}{(2-\sqrt3)(2+\sqrt3)}=\dfrac{5(2+\sqrt3)}{4-3}=5(2+\sqrt3)=10+5\sqrt3\).

PastPaper.markingScheme

M1 multiply by conjugate; A1 denominator =1; A1 \(a=10\); A1 \(b=5\).

Paper 2 — Section B: Statistics

Answer all questions. A calculator with statistical functions may be used.
6 PastPaper.question · 27 PastPaper.marks
PastPaper.question 1 · Multiple choice
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A school has 40 classes of 30 students. A researcher selects 3 whole classes at random and surveys every student in them. This sampling method is: circle your answer.
  1. A.Cluster
  2. B.Stratified
  3. C.Systematic
  4. D.Quota
PastPaper.showAnswers

PastPaper.workedSolution

Selecting whole intact groups at random is cluster sampling.

PastPaper.markingScheme

B1 cluster.
PastPaper.question 2 · Multiple choice
1 PastPaper.marks
On a box plot the median is much closer to the lower quartile than to the upper quartile. The distribution is best described as: circle your answer.
  1. A.Symmetric
  2. B.Positively skewed
  3. C.Negatively skewed
  4. D.Uniform
PastPaper.showAnswers

PastPaper.workedSolution

A longer tail to the right (median near \(Q_1\)) indicates positive skew.

PastPaper.markingScheme

B1 positively skewed.
PastPaper.question 3 · Structured
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The times (in minutes) for six visits are: \(12, 15, 18, 11, 14, 20\). (a) Find the mean. (b) Find the standard deviation to 2 decimal places.
PastPaper.showAnswers

PastPaper.workedSolution

(a) Mean \(=\tfrac{90}{6}=15\). (b) Squared deviations: \(9,0,9,16,1,25\), sum \(=60\); variance \(=\tfrac{60}{6}=10\); s.d. \(=\sqrt{10}=3.16\).

PastPaper.markingScheme

M1A1 mean; M1 squared deviations; M1 variance; A1 3.16.
PastPaper.question 4 · Structured
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On each gym visit a member independently chooses an apple, banana or cake, with \(P(A)=0.3\), \(P(B)=0.45\), \(P(C)=0.25\). For four randomly chosen visits, find the probability that the member chose: (a) at least one apple; (b) the same item on all four visits; (c) apple exactly twice and cake exactly twice.
PastPaper.showAnswers

PastPaper.workedSolution

(a) \(1-(0.7)^4=0.7599\). (b) \(0.3^4+0.45^4+0.25^4=0.0081+0.0410+0.0039=0.0530\). (c) \(\binom{4}{2}(0.3)^2(0.25)^2=6(0.09)(0.0625)=0.0759\).

PastPaper.markingScheme

M1A1 (a); M1A1 (b); M1 arrangement count; A1 (c).
PastPaper.question 5 · Structured
6 PastPaper.marks
The discrete random variable \(X\) has \(P(X=x)=cx\) for \(x=1,2,3,4\) and 0 otherwise. (a) Find \(c\). (b) Find \(E(X)\). (c) Find \(P(X\ge3)\).
PastPaper.showAnswers

PastPaper.workedSolution

(a) \(c(1+2+3+4)=1\Rightarrow10c=1\Rightarrow c=0.1\). (b) \(E(X)=c(1+4+9+16)=0.1\times30=3\). (c) \(P(X\ge3)=c(3+4)=0.7\).

PastPaper.markingScheme

M1 sum to 1; A1 c=0.1; M1A1 E(X); A1 0.7.
PastPaper.question 6 · Structured
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A coin is suspected of being biased towards heads. It is tossed 20 times and 15 heads are obtained. Test, at the 5% significance level, whether there is evidence that the probability of a head exceeds 0.5.
PastPaper.showAnswers

PastPaper.workedSolution

Let \(p=P(\text{head})\). \(H_0:p=0.5\), \(H_1:p>0.5\). Under \(H_0\), \(X\sim B(20,0.5)\). \(P(X\ge15)=1-P(X\le14)=0.0207\). Since \(0.0207<0.05\), reject \(H_0\): there is evidence at the 5% level that the coin is biased towards heads.

PastPaper.markingScheme

B1 hypotheses; B1 distribution; M1 \(P(X\ge15)\); A1 0.0207; M1 compare; A1 conclusion in context.

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