AQA AS-Level · Thinka-original Practice Paper

2022 AQA AS-Level Mathematics 7356 Practice Paper with Answers

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

160 marks180 mins2022
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 Question · 53 marks
Question 1 · Multiple choice
1 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.\(\log_3 3\)
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Worked solution

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

Marking scheme

B1 for \(\log_3 10\).
Question 2 · Multiple choice
1 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
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Worked solution

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

Marking scheme

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

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

Marking scheme

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

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 \(x=60^\circ,300^\circ\).

Marking scheme

M1 use of identity; M1 form quadratic in \(\cos x\); A1 \(\cos x=\tfrac12\); A1 60°; A1 300°.
Question 5 · Structured
6 marks
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.
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Worked solution

(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.

Marking scheme

B1 centre; M1A1 radius via distance; B1 equation; M1 substitute (6,6); A1 inside.
Question 6 · Structured
7 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.
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Worked solution

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

Marking scheme

M1A1 intercepts; M1 set up definite integral; A1 integrate; M1 limits; A1 36.
Question 7 · Structured
8 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.
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Worked solution

(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).

Marking scheme

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

\(n^3-n=n(n^2-1)=(n-1)n(n+1)\), 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 \(2\times3=6\).

Marking scheme

M1 factorise; A1 three consecutive integers; A1 divisible by 2; A1 divisible by 3; A1 conclude divisible by 6.
Question 9 · Structured
9 marks
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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Worked solution

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 \(x=\tfrac12\), \(y=\tfrac{13}{4}\).

Marking scheme

M1A1 tangent; M1A1 normal; M1 equate to curve; M1 solve quadratic; A1 other point.
Question 10 · Structured
8 marks
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.
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Worked solution

(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.

Marking scheme

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 Question · 27 marks
Question 1 · Multiple choice
1 marks
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
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Worked solution

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

Marking scheme

B1 straight line through the origin.
Question 2 · Structured
4 marks
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.
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Worked solution

(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 \(=2\times2.143=4.29\,\text{s}\).

Marking scheme

M1A1 max height; M1 time up; A1 total time.
Question 3 · Structured
4 marks
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.
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Worked solution

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

Marking scheme

B1 resultant; M1 magnitude; M1 divide by mass; A1 4.12.
Question 4 · Structured
6 marks
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.
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Worked solution

(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: \(T-3g=3a\Rightarrow T=3(9.8)+3(2.45)=36.75\,\text{N}\).

Marking scheme

M1 equation of motion for system; A1 a=2.45; M1 equation for one mass; A1 T=36.75; A1 consistent check.
Question 5 · Structured
6 marks
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.
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Worked solution

(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) \(s=\int_0^4 v\,dt=[t^3-6t^2+9t]_0^4=64-96+36=4\,\text{m}\).

Marking scheme

M1 set v=0; A1 both times; M1A1 acceleration; M1 integrate; A1 4 m.
Question 6 · Structured
6 marks
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.
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Worked solution

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

Marking scheme

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 Question · 53 marks
Question 1 · Multiple choice
1 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.\(3x^2+c\)
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Worked solution

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

Marking scheme

B1 for \(2x^3+c\).
Question 2 · Multiple choice
1 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.\(320^\circ\)
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Worked solution

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

Marking scheme

B1 for 40°.
Question 3 · Structured
5 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.
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Worked solution

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

Marking scheme

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

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

Marking scheme

M1 discriminant =0; A1 \((k+2)^2=36\); A1 k=4; A1 k=-8.
Question 5 · Structured
8 marks
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 \(30^\circ\text{C}\).
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Worked solution

(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.

Marking scheme

B1 A=60; M1 form equation; A1 k=0.139; M1A1 (c); M1 set =30; A1 ≈13 min.
Question 6 · Structured
7 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.
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Worked solution

(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) \(\tfrac{\sin BAC}{8}=\tfrac{\sin60^\circ}{7.55}\Rightarrow \sin BAC=0.9176\Rightarrow BAC=66.6^\circ\).

Marking scheme

M1A1 cosine rule; A1 AC; M1A1 area; M1 sine rule; A1 66.6°.
Question 7 · Structured
10 marks
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.
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Worked solution

(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 \(=\int_{-1}^{4}\big((x+1)-(x^2-2x-3)\big)dx=\int_{-1}^{4}(-x^2+3x+4)dx=\tfrac{125}{6}\).

Marking scheme

M1 equate; A1 quadratic; A1 both points; B1 inequalities; M1 integrate difference; M1 limits; A1 125/6.
Question 8 · Structured
5 marks
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.
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Worked solution

(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).

Marking scheme

M1A1 derivative; M1 points; A1 both; M1 second derivative; A1 natures.
Question 9 · Structured
8 marks
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.
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Worked solution

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) \(r=\tfrac13\Rightarrow a=18\); \(S_5=18\cdot\tfrac{1-(1/3)^5}{1-1/3}=27\cdot\tfrac{242}{243}\approx26.9\).

Marking scheme

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

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

Marking scheme

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 Question · 27 marks
Question 1 · Multiple choice
1 marks
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
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Worked solution

Selecting whole intact groups at random is cluster sampling.

Marking scheme

B1 cluster.
Question 2 · Multiple choice
1 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
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Worked solution

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

Marking scheme

B1 positively skewed.
Question 3 · Structured
5 marks
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.
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Worked solution

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

Marking scheme

M1A1 mean; M1 squared deviations; M1 variance; A1 3.16.
Question 4 · Structured
7 marks
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.
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Worked solution

(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) \(\binom{4}{2}(0.3)^2(0.25)^2=6(0.09)(0.0625)=0.0759\).

Marking scheme

M1A1 (a); M1A1 (b); M1 arrangement count; A1 (c).
Question 5 · Structured
6 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 \(P(X\ge3)\).
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Worked solution

(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) \(P(X\ge3)=c(3+4)=0.7\).

Marking scheme

M1 sum to 1; A1 c=0.1; M1A1 E(X); A1 0.7.
Question 6 · Structured
7 marks
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.
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Worked solution

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.

Marking scheme

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

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