AQA A-Level · Thinka-original Practice Paper

2022 AQA A-Level Mathematics 7357 Practice Paper with Answers

Thinka Jun 2022 AQA A Level-Style Mock — Mathematics 7357

300 marks360 mins2022
AnalysisSample paper
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2022 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 Question · 100 marks
Question 1 · Multiple choice
1 marks
Differentiate \(y=\ln(3x)\). Circle your answer.
  1. A.\(\tfrac{3}{x}\)
  2. B.\(\tfrac{1}{x}\)
  3. C.\(\tfrac{1}{3x}\)
  4. D.\(\ln 3\)
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Worked solution

\(\tfrac{d}{dx}\ln(3x)=\tfrac{3}{3x}=\tfrac1x\).

Marking scheme

B1 for \(\tfrac1x\).
Question 2 · Multiple choice
1 marks
Find \(\displaystyle\int e^{2x}\,dx\). Circle your answer.
  1. A.\(2e^{2x}+c\)
  2. B.\(\tfrac12 e^{2x}+c\)
  3. C.\(e^{2x}+c\)
  4. D.\(2e^{x}+c\)
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Worked solution

\(\int e^{2x}dx=\tfrac12 e^{2x}+c\).

Marking scheme

B1 for \(\tfrac12 e^{2x}+c\).
Question 3 · Short
5 marks
Find the first three terms, in ascending powers of \(x\), of the binomial expansion of \((2-3x)^7\).
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Worked solution

\((2-3x)^7=2^7+\binom71 2^6(-3x)+\binom72 2^5(-3x)^2+\dots=128-7(64)(3)x+21(32)(9)x^2=128-1344x+6048x^2\).

Marking scheme

M1 binomial structure; A1 128; A1 -1344x; M1 third-term values; A1 6048x².
Question 4 · Structured
12 marks
A circle \(C\) has equation \(x^2+y^2-6x+4y-12=0\). (a) Find the centre and radius. (b) Show that the point \((7,1)\) lies on \(C\). (c) Find the equation of the tangent to \(C\) at \((7,1)\). (d) Find the length of the tangent to \(C\) drawn from the external point \((10,6)\).
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Worked solution

(a) \((x-3)^2+(y+2)^2=25\): centre \((3,-2)\), radius \(5\). (b) \(49+1-42+4-12=0\) ✓. (c) Radius to \((7,1)\) has gradient \(\tfrac{1-(-2)}{7-3}=\tfrac34\), so tangent gradient \(-\tfrac43\): \(y-1=-\tfrac43(x-7)\Rightarrow 4x+3y=31\). (d) Tangent length \(=\sqrt{(10-3)^2+(6+2)^2-5^2}=\sqrt{49+64-25}=\sqrt{88}=9.38\).

Marking scheme

M1A1 centre & radius; M1A1 (b); M1 radius gradient; M1 tangent gradient; A1 tangent equation; M1A1 tangent length.
Question 5 · Structured
10 marks
(a) An arithmetic series has first term 5 and common difference 3. Find (i) the 20th term and (ii) the sum of the first 20 terms. (b) A geometric series has first term 5 and common ratio 1.1. Find (i) the 20th term and (ii) the sum of the first 20 terms. (c) State, with a reason, whether the geometric series has a sum to infinity.
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Worked solution

(a)(i) \(5+19(3)=62\); (ii) \(S_{20}=\tfrac{20}{2}(10+57)=670\). (b)(i) \(5(1.1)^{19}=30.6\); (ii) \(S_{20}=5\cdot\tfrac{1.1^{20}-1}{0.1}=286\). (c) No: \(|r|=1.1>1\), so the series diverges.

Marking scheme

M1A1 (a)(i,ii); M1A1 (b)(i,ii); M1A1 (b)(ii); B1 (c) with reason.
Question 6 · Structured
12 marks
\(f(x)=x^3-5x+1\). (a) Show that a root of \(f(x)=0\) lies between 2 and 3. (b) Apply the Newton–Raphson method with \(x_0=2\) to find \(x_1\) and \(x_2\). (c) The iteration \(x_{n+1}=\sqrt[3]{5x_n-1}\) is used with \(x_0=2\). Find the root to 3 decimal places.
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Worked solution

(a) \(f(2)=-1<0\), \(f(3)=13>0\): a sign change implies a root in \((2,3)\). (b) \(f'(x)=3x^2-5\); \(x_1=2-\tfrac{-1}{7}=2.143\); \(x_2=2.143-\tfrac{f(2.143)}{f'(2.143)}=2.129\). (c) \(x_1=\sqrt[3]{9}=2.080,\;x_2=2.111,\;x_3=2.123,\;x_4=2.128,\dots\to 2.128\).

Marking scheme

M1 evaluate ends; A1 sign-change conclusion; M1 NR formula; A1 x₁; A1 x₂; M1 iterate; A1 2.128.
Question 7 · Structured
10 marks
\(p(x)=2x^3+ax^2-7x+b\). Given that \((x-1)\) and \((x+2)\) are factors of \(p(x)\): (a) find \(a\) and \(b\); (b) factorise \(p(x)\) completely; (c) solve \(p(x)=0\).
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Worked solution

(a) \(p(1)=2+a-7+b=0\Rightarrow a+b=5\); \(p(-2)=-16+4a+14+b=0\Rightarrow4a+b=2\). Subtracting, \(3a=-3\Rightarrow a=-1,\,b=6\). (b) \(p(x)=2x^3-x^2-7x+6=(x-1)(x+2)(2x-3)\). (c) \(x=1,-2,\tfrac32\).

Marking scheme

M1 p(1)=0; M1 p(-2)=0; A1 a,b; M1 divide; A1 full factorisation; A1 three roots.
Question 8 · Structured
8 marks
(a) A geometric series has first term 8 and common ratio \(\tfrac34\). Find its sum to infinity. (b) Solve \(\tan 2x=\sqrt3\) for \(0\le x\le\pi\). (c) Solve \(3\sin x=2\cos x\) for \(0\le x\le2\pi\), giving answers to 2 decimal places.
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Worked solution

(a) \(\tfrac{8}{1-3/4}=32\). (b) \(2x=\tfrac{\pi}{3},\tfrac{4\pi}{3}\Rightarrow x=\tfrac{\pi}{6},\tfrac{2\pi}{3}\). (c) \(\tan x=\tfrac23\Rightarrow x=0.588\) or \(0.588+\pi=3.73\).

Marking scheme

B1 (a); M1 2x values; A1 both x; M1 tan x; A1 0.59; A1 3.73.
Question 9 · Structured
10 marks
(a) Prove the identity \(\dfrac{\sin\theta}{1+\cos\theta}+\dfrac{1+\cos\theta}{\sin\theta}\equiv\dfrac{2}{\sin\theta}\). (b) Hence solve \(\dfrac{\sin\theta}{1+\cos\theta}+\dfrac{1+\cos\theta}{\sin\theta}=4\) for \(0<\theta<2\pi\). (c) Show that \(\dfrac{1-\cos2x}{\sin2x}\equiv\tan x\).
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Worked solution

(a) Common denominator: \(\tfrac{\sin^2\theta+(1+\cos\theta)^2}{\sin\theta(1+\cos\theta)}=\tfrac{\sin^2\theta+1+2\cos\theta+\cos^2\theta}{\sin\theta(1+\cos\theta)}=\tfrac{2+2\cos\theta}{\sin\theta(1+\cos\theta)}=\tfrac{2}{\sin\theta}\). (b) \(\tfrac{2}{\sin\theta}=4\Rightarrow\sin\theta=\tfrac12\Rightarrow\theta=\tfrac{\pi}{6},\tfrac{5\pi}{6}\). (c) \(\tfrac{1-\cos2x}{\sin2x}=\tfrac{2\sin^2x}{2\sin x\cos x}=\tan x\).

Marking scheme

M1 combine fractions; M1 use \(\sin^2+\cos^2=1\); A1 result; M1A1 (b); M1A1 (c).
Question 10 · Structured
15 marks
(a) Find \(\displaystyle\int\left(4x^3-\tfrac{2}{x^2}+e^{2x}\right)dx\). (b) Use the substitution \(u=x^2+1\) to find \(\displaystyle\int 2x\sqrt{x^2+1}\,dx\). (c) Find the area enclosed between the curves \(y=x^2\) and \(y=2x\).
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Worked solution

(a) \(x^4+\tfrac{2}{x}+\tfrac12 e^{2x}+c\). (b) \(u=x^2+1,\,du=2x\,dx\Rightarrow\int\sqrt u\,du=\tfrac23 u^{3/2}=\tfrac23(x^2+1)^{3/2}+c\). (c) Intersect at \(x=0,2\); area \(=\int_0^2(2x-x^2)dx=[x^2-\tfrac{x^3}{3}]_0^2=4-\tfrac83=\tfrac43\).

Marking scheme

M1A1A1 (a); M1 substitution; A1 (b); M1 limits 0,2; M1 integrate difference; A1 4/3.
Question 11 · Structured
16 marks
A curve has equation \(y=x^3-3x^2-9x+5\). (a) Find \(\dfrac{dy}{dx}\). (b) Find the coordinates of the stationary points and determine their nature. (c) Find the coordinates of the point of inflection. (d) Find the equation of the tangent at \(x=-1\). (e) State the set of values of \(x\) for which the curve is increasing.
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Worked solution

(a) \(\tfrac{dy}{dx}=3x^2-6x-9=3(x-3)(x+1)\). (b) \(x=-1\Rightarrow(−1,10)\); \(x=3\Rightarrow(3,-22)\). \(\tfrac{d^2y}{dx^2}=6x-6\): at \(x=-1\) negative (max), at \(x=3\) positive (min). (c) \(6x-6=0\Rightarrow x=1,\,y=-6\). (d) \(x=-1\) is a stationary point, so the tangent is \(y=10\). (e) Increasing where \(\tfrac{dy}{dx}>0\): \(x<-1\) or \(x>3\).

Marking scheme

M1A1 derivative; M1 stationary; A1 both points; M1A1 natures; M1A1 inflection; B1 tangent; M1A1 increasing set.

Paper 2: Pure & Mechanics

Answer all questions. Section A is pure; Section B is mechanics (take g = 9.8 m s⁻²).
14 Question · 100 marks
Question 1 · Multiple choice
1 marks
Find \(\displaystyle\int\tfrac1x\,dx\). Circle your answer.
  1. A.\(\ln|x|+c\)
  2. B.\(-\tfrac{1}{x^2}+c\)
  3. C.\(\tfrac{1}{2}x^2+c\)
  4. D.\(x\ln x+c\)
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Worked solution

\(\int\tfrac1x dx=\ln|x|+c\).

Marking scheme

B1 for \(\ln|x|+c\).
Question 2 · Short
3 marks
Express \(\dfrac{5x-7}{(x-1)(x-3)}\) in partial fractions.
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Worked solution

Let \(\tfrac{5x-7}{(x-1)(x-3)}=\tfrac{A}{x-1}+\tfrac{B}{x-3}\). Then \(5x-7=A(x-3)+B(x-1)\). \(x=1:-2=-2A\Rightarrow A=1\); \(x=3:8=2B\Rightarrow B=4\).

Marking scheme

M1 set up identity; M1 substitute values; A1 A=1, B=4.
Question 3 · Proof
6 marks
Prove by contradiction that \(\sqrt3\) is irrational.
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Worked solution

Assume \(\sqrt3=\tfrac pq\) in lowest terms with \(q\ne0\). Then \(p^2=3q^2\), so \(3\mid p^2\Rightarrow3\mid p\). Write \(p=3k\): \(9k^2=3q^2\Rightarrow q^2=3k^2\), so \(3\mid q\). Then \(3\) divides both \(p\) and \(q\), contradicting lowest terms. Hence \(\sqrt3\) is irrational.

Marking scheme

B1 assume rational in lowest terms; M1 \(p^2=3q^2\); A1 3|p; M1 substitute; A1 3|q; A1 contradiction.
Question 4 · Structured
10 marks
An open-topped box has a square base of side \(x\,\text{cm}\) and volume \(500\,\text{cm}^3\). (a) Show that the external surface area is \(S=x^2+\dfrac{2000}{x}\). (b) Find the value of \(x\) that minimises \(S\). (c) Find the minimum surface area.
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Worked solution

(a) Height \(h=\tfrac{500}{x^2}\); \(S=x^2+4xh=x^2+\tfrac{2000}{x}\). (b) \(\tfrac{dS}{dx}=2x-\tfrac{2000}{x^2}=0\Rightarrow x^3=1000\Rightarrow x=10\). (c) \(S=100+200=300\,\text{cm}^2\) (and \(\tfrac{d^2S}{dx^2}>0\) confirms a minimum).

Marking scheme

M1 height; A1 show S; M1 differentiate; M1 solve =0; A1 x=10; A1 S=300.
Question 5 · Structured
10 marks
(a) A population grows so that \(\dfrac{dP}{dt}=0.5P\). Show that \(P=Ae^{0.5t}\). (b) Given \(P=200\) when \(t=0\), find \(P\) when \(t=4\). (c) Solve the differential equation \(\dfrac{dy}{dx}=6x^2y\), giving \(y\) in terms of \(x\).
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Worked solution

(a) \(\int\tfrac{dP}{P}=\int0.5\,dt\Rightarrow\ln P=0.5t+c\Rightarrow P=Ae^{0.5t}\). (b) \(A=200\Rightarrow P=200e^{2}=1478\). (c) \(\int\tfrac{dy}{y}=\int6x^2dx\Rightarrow\ln y=2x^3+c\Rightarrow y=Be^{2x^3}\).

Marking scheme

M1 separate; A1 show; M1 A=200; A1 1478; M1 separate; A1 \(y=Be^{2x^3}\).
Question 6 · Structured
10 marks
(a) Solve \(3^{2x}-10\cdot3^{x}+9=0\). (b) Solve \(\log_2 x+\log_2(x-2)=3\). (c) Evaluate \(\log_3 81-\log_3 9\).
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Worked solution

(a) Let \(y=3^x\): \(y^2-10y+9=0\Rightarrow(y-1)(y-9)=0\Rightarrow3^x=1\,(x=0)\) or \(3^x=9\,(x=2)\). (b) \(\log_2 x(x-2)=3\Rightarrow x^2-2x=8\Rightarrow(x-4)(x+2)=0\Rightarrow x=4\) (reject \(x=-2\)). (c) \(\log_3 81-\log_3 9=4-2=2\).

Marking scheme

M1 substitution; A1 both x; M1 combine logs; M1 solve; A1 x=4; A1 (c).
Question 7 · Structured
10 marks
(a) Express \(3\sin x+4\cos x\) 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 \(3\sin x+4\cos x=2.5\) for \(0\le x\le360^\circ\).
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Worked solution

(a) \(R=\sqrt{3^2+4^2}=5\), \(\tan\alpha=\tfrac43\Rightarrow\alpha=53.13^\circ\): \(5\sin(x+53.13^\circ)\). (b) Max \(5\) when \(x+53.13^\circ=90^\circ\Rightarrow x=36.9^\circ\). (c) \(\sin(x+53.13^\circ)=0.5\Rightarrow x+53.13^\circ=30^\circ,150^\circ\); only \(150^\circ\) is valid \(\Rightarrow x=96.9^\circ\).

Marking scheme

M1 R=5; M1 α; A1 form; M1A1 (b); M1 solve; A1 96.9°.
Question 8 · Multiple choice
1 marks
Find the weight of a particle of mass \(5\,\text{kg}\) (take \(g=9.8\)). Circle your answer.
  1. A.\(5\,\text{N}\)
  2. B.\(0.51\,\text{N}\)
  3. C.\(49\,\text{N}\)
  4. D.\(9.8\,\text{N}\)
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Worked solution

Weight \(=mg=5\times9.8=49\,\text{N}\).

Marking scheme

B1 for 49 N.
Question 9 · Structured
6 marks
A particle moves in a straight line with acceleration \(a=(6t-4)\,\text{m s}^{-2}\) and velocity \(2\,\text{m s}^{-1}\) when \(t=0\). (a) Find \(v\) in terms of \(t\). (b) Find \(v\) when \(t=3\). (c) Find the displacement during the first 3 seconds.
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Worked solution

(a) \(v=\int(6t-4)dt=3t^2-4t+C\); \(v(0)=2\Rightarrow C=2\), so \(v=3t^2-4t+2\). (b) \(v(3)=27-12+2=17\). (c) \(s=\int_0^3(3t^2-4t+2)dt=[t^3-2t^2+2t]_0^3=27-18+6=15\,\text{m}\).

Marking scheme

M1 integrate a; A1 v; B1 (b); M1 integrate v; A1 15 m.
Question 10 · Structured
9 marks
A particle of mass \(4\,\text{kg}\) lies on a smooth plane inclined at \(30^\circ\) to the horizontal. It is connected by a light inextensible string over a smooth pulley at the top of the plane to a particle 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 used about the string.
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Worked solution

For the hanging mass: \(6g-T=6a\). For the mass on the plane: \(T-4g\sin30^\circ=4a\). Adding: \(6g-4g(0.5)=10a\Rightarrow a=\tfrac{58.8-19.6}{10}=3.92\,\text{m s}^{-2}\). Then \(T=6g-6a=58.8-23.52=35.3\,\text{N}\). (c) The string is light and inextensible (so tension is constant and the masses share one acceleration).

Marking scheme

M1 equation hanging mass; M1 equation on plane; A1 add; A1 a=3.92; M1 T; A1 35.3; B1 assumption.
Question 11 · Structured
10 marks
A uniform beam \(AB\) of length \(8\,\text{m}\) and weight \(400\,\text{N}\) rests horizontally on supports at \(A\) and \(B\). A person of weight \(600\,\text{N}\) stands \(2\,\text{m}\) from \(A\). (a) Find the magnitude of the reaction at each support. (b) Find how far from \(A\) the person must stand for the two reactions to be equal.
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Worked solution

(a) Moments about \(A\): \(R_B(8)=400(4)+600(2)=2800\Rightarrow R_B=350\,\text{N}\); \(R_A=400+600-350=650\,\text{N}\). (b) For equal reactions each is \(500\,\text{N}\): \(500(8)=400(4)+600d\Rightarrow600d=2400\Rightarrow d=4\,\text{m}\).

Marking scheme

M1 moments about A; A1 R_B; M1 resolve vertically; A1 R_A; M1 set reactions equal; A1 d=4.
Question 12 · Structured
8 marks
A particle moves so that its position vector is \(\mathbf{r}=(t^2-2t)\mathbf{i}+(t^2-4t)\mathbf{j}\), where \(t\) is in seconds. (a) Find its velocity when \(t=1\). (b) Find its speed when \(t=1\). (c) Find the time at which the particle is moving parallel to \(\mathbf{i}\).
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Worked solution

\(\mathbf{v}=(2t-2)\mathbf{i}+(2t-4)\mathbf{j}\). (a) At \(t=1\): \(\mathbf{v}=0\mathbf{i}-2\mathbf{j}\). (b) Speed \(=\sqrt{0^2+(-2)^2}=2\,\text{m s}^{-1}\). (c) Parallel to \(\mathbf{i}\) when the \(\mathbf{j}\)-component is zero: \(2t-4=0\Rightarrow t=2\).

Marking scheme

M1 differentiate r; A1 v at t=1; M1A1 speed; M1 set j-comp =0; A1 t=2.
Question 13 · Structured
10 marks
A block of mass \(10\,\text{kg}\) rests on a rough horizontal surface with coefficient of friction \(0.3\). It is pulled by a force of \(50\,\text{N}\) acting at \(20^\circ\) above the horizontal (take \(g=9.8\)). (a) Find the normal reaction. (b) Find the friction force. (c) Find the acceleration of the block.
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Worked solution

(a) Resolve vertically: \(N=10g-50\sin20^\circ=98-17.1=80.9\,\text{N}\). (b) \(F=\mu N=0.3(80.9)=24.3\,\text{N}\). (c) Horizontally: \(50\cos20^\circ-F=10a\Rightarrow46.98-24.27=10a\Rightarrow a=2.27\,\text{m s}^{-2}\).

Marking scheme

M1 resolve vertically; A1 N; M1A1 friction; M1 resolve horizontally; A1 a=2.27.
Question 14 · Structured
6 marks
A ball is thrown horizontally with speed \(15\,\text{m s}^{-1}\) from a point \(20\,\text{m}\) above horizontal ground (take \(g=9.8\), ignore air resistance). (a) Find the time taken to reach the ground. (b) Find the horizontal distance travelled before landing.
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Worked solution

(a) Vertically \(20=\tfrac12(9.8)t^2\Rightarrow t^2=4.08\Rightarrow t=2.02\,\text{s}\). (b) Horizontally (constant speed) \(x=15(2.02)=30.3\,\text{m}\).

Marking scheme

M1 vertical equation; A1 t=2.02; M1 horizontal motion; A1 30.3 m; B1 method.

Paper 3: Pure & Statistics

Answer all questions. Section A is pure; Section B is statistics (Large Data Set context).
13 Question · 100 marks
Question 1 · Multiple choice
1 marks
State the period of \(y=\sin(\pi x)\). Circle your answer.
  1. A.\(\pi\)
  2. B.\(\tfrac12\)
  3. C.\(2\)
  4. D.\(2\pi\)
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Worked solution

Period \(=\tfrac{2\pi}{\pi}=2\).

Marking scheme

B1 for 2.
Question 2 · Structured
10 marks
Differentiate each of the following. (a) \(y=(2x+1)^5\). (b) \(y=x\ln x\). (c) \(y=\dfrac{x}{x^2+1}\). (d) Hence state the gradient of the curve in part (a) at \(x=0\).
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Worked solution

(a) Chain rule: \(10(2x+1)^4\). (b) Product rule: \(\ln x+1\). (c) Quotient rule: \(\tfrac{(x^2+1)-x(2x)}{(x^2+1)^2}=\tfrac{1-x^2}{(x^2+1)^2}\). (d) At \(x=0\): \(10(1)^4=10\).

Marking scheme

M1A1 (a); M1A1 (b); M1A1 (c); M1A1 (d).
Question 3 · Structured
9 marks
Find each integral. (a) \(\displaystyle\int(3x+2)^4\,dx\). (b) \(\displaystyle\int\tfrac{6}{2x-1}\,dx\). (c) \(\displaystyle\int_1^2\tfrac{4}{x^2}\,dx\).
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Worked solution

(a) \(\tfrac{(3x+2)^5}{15}+c\). (b) \(3\ln|2x-1|+c\). (c) \([-\tfrac{4}{x}]_1^2=-2-(-4)=2\).

Marking scheme

M1A1 (a); M1A1 (b); M1 antiderivative; A1 (c)=2.
Question 4 · Structured
10 marks
(a) Solve \(2\cos 2x+1=0\) for \(0\le x\le2\pi\). (b) Prove that \(\sec^2 x-1\equiv\tan^2 x\). (c) Solve \(\sin x=\cos x\) for \(0\le x\le2\pi\).
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Worked solution

(a) \(\cos2x=-\tfrac12\Rightarrow2x=\tfrac{2\pi}{3},\tfrac{4\pi}{3},\tfrac{8\pi}{3},\tfrac{10\pi}{3}\Rightarrow x=\tfrac{\pi}{3},\tfrac{2\pi}{3},\tfrac{4\pi}{3},\tfrac{5\pi}{3}\). (b) \(\sec^2x-1=\tfrac{1}{\cos^2x}-1=\tfrac{1-\cos^2x}{\cos^2x}=\tfrac{\sin^2x}{\cos^2x}=\tan^2x\). (c) \(\tan x=1\Rightarrow x=\tfrac{\pi}{4},\tfrac{5\pi}{4}\).

Marking scheme

M1 cos2x=-1/2; A1 all four x; M1A1 (b); M1A1 (c).
Question 5 · Structured
10 marks
A radioactive substance decays so that its mass is \(m=m_0e^{-kt}\), with a half-life of 20 days. (a) Show that \(k=\dfrac{\ln2}{20}\). (b) Given \(m_0=50\,\text{g}\), find the mass after 30 days. (c) Find, to the nearest day, the time for the mass to fall to \(10\,\text{g}\). (d) Find the rate of decay (in g per day) at \(t=0\).
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Worked solution

(a) \(\tfrac12 m_0=m_0e^{-20k}\Rightarrow e^{-20k}=\tfrac12\Rightarrow k=\tfrac{\ln2}{20}\). (b) \(k=0.03466\); \(m=50e^{-1.0397}=17.7\,\text{g}\). (c) \(10=50e^{-kt}\Rightarrow t=\tfrac{\ln5}{0.03466}=46\,\text{days}\). (d) \(\tfrac{dm}{dt}=-km_0e^{-kt}\); at \(t=0\), \(=-0.03466(50)=-1.73\,\text{g/day}\).

Marking scheme

M1 half-life equation; A1 show k; M1A1 (b); M1A1 (c); M1A1 (d).
Question 6 · Structured
10 marks
(a) Use the arithmetic-series formula to prove that the sum of the first \(n\) odd numbers is \(n^2\). (b) An arithmetic series is \(7,11,15,\dots\). Find the sum of the first 30 terms. (c) A geometric series is \(100,80,64,\dots\). Find its sum to infinity. (d) Find the first term of the series in (b) that exceeds 200.
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Worked solution

(a) Odd numbers form an AP with \(a=1,d=2\): \(S_n=\tfrac n2(2+(n-1)2)=\tfrac n2(2n)=n^2\). (b) \(a=7,d=4\): \(S_{30}=15(14+29(4))=15(130)=1950\). (c) \(r=0.8\): \(\tfrac{100}{1-0.8}=500\). (d) \(7+(n-1)4>200\Rightarrow n>49.25\Rightarrow n=50\) (term \(=201\)).

Marking scheme

M1A1 (a) proof; M1A1 (b); M1A1 (c); M1A1 (d).
Question 7 · Multiple choice
1 marks
For \(X\sim B(20,0.3)\), state the mean \(E(X)\). Circle your answer.
  1. A.\(0.3\)
  2. B.\(6\)
  3. C.\(4.2\)
  4. D.\(14\)
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Worked solution

\(E(X)=np=20\times0.3=6\).

Marking scheme

B1 for 6.
Question 8 · Structured
6 marks
(a) Define a simple random sample. (b) A company has 500 employees, each with a payroll number. Describe how to select a simple random sample of 50 employees. (c) State one advantage of stratified sampling over simple random sampling here.
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Worked solution

(a) A simple random sample is one in which every possible sample of the required size has an equal chance of being selected (equivalently, every member has an equal chance of being chosen). (b) Number the employees 1–500; use random numbers (or a random-number generator) to select 50 distinct payroll numbers, ignoring repeats and any number above 500. (c) Stratified sampling ensures each subgroup (e.g. department or grade) is represented in proportion, reducing sampling bias.

Marking scheme

B1 definition; M1 numbering; A1 random selection of 50; B1 advantage; B1 contextual reason; B1 clarity.
Question 9 · Structured
10 marks
Eight recorded times (in minutes) are \(12,15,15,18,20,22,25,30\). (a) Find the median. (b) Find the lower quartile, upper quartile and interquartile range. (c) Find the mean. (d) Find the standard deviation to 2 decimal places. (e) Using the rule that an outlier lies more than \(1.5\times\text{IQR}\) beyond a quartile, determine whether 30 is an outlier.
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Worked solution

(a) Median \(=\tfrac{18+20}{2}=19\). (b) \(Q_1=15,\,Q_3=25,\,\text{IQR}=10\). (c) Mean \(=\tfrac{157}{8}=19.625\). (d) \(\sum x^2=3327\); variance \(=\tfrac{3327}{8}-19.625^2=30.73\); s.d. \(=5.54\). (e) \(Q_3+1.5(10)=40\); since \(30<40\), 30 is not an outlier.

Marking scheme

B1 median; B1 quartiles; B1 IQR; M1A1 mean; M1A1 s.d.; B1 outlier conclusion.
Question 10 · Structured
10 marks
The heights of adult plants are modelled by \(X\sim N(170,8^2)\) (in cm). (a) Find \(P(X>180)\). (b) Find \(P(165
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Worked solution

(a) \(z=\tfrac{180-170}{8}=1.25\Rightarrow P(Z>1.25)=0.1056\). (b) \(z=\pm0.625\Rightarrow\Phi(0.625)-\Phi(-0.625)=0.7340-0.2660=0.4680\). (c) \(P(X>x)=0.10\Rightarrow z=1.2816\Rightarrow x=170+1.2816(8)=180.3\,\text{cm}\).

Marking scheme

M1 standardise; A1 (a); M1A1 (b); M1 z=1.2816; A1 (c).
Question 11 · Structured
10 marks
A spinner is claimed to land on red with probability greater than 0.4. (a) For \(X\sim B(15,0.4)\) find (i) \(P(X=6)\) and (ii) \(P(X\le4)\). (b) In 15 spins the spinner lands on red 10 times. Test, at the 5% significance level, whether the probability of red exceeds 0.4.
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Worked solution

(a)(i) \(\binom{15}{6}0.4^6 0.6^9=0.2066\). (ii) \(P(X\le4)=0.2173\). (b) \(H_0:p=0.4\), \(H_1:p>0.4\), \(X\sim B(15,0.4)\). \(P(X\ge10)=1-P(X\le9)=0.0338<0.05\), so reject \(H_0\): evidence that the probability of red exceeds 0.4.

Marking scheme

M1A1 (a)(i); A1 (a)(ii); B1 hypotheses; M1 P(X≥10); A1 0.0338; A1 conclusion.
Question 12 · Structured
7 marks
Events \(A\) and \(B\) satisfy \(P(A)=0.5\), \(P(B)=0.4\) and \(P(A\cap B)=0.2\). (a) Find \(P(A\cup B)\). (b) Find \(P(A\mid B)\). (c) Determine, with justification, whether \(A\) and \(B\) are independent.
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Worked solution

(a) \(P(A\cup B)=0.5+0.4-0.2=0.7\). (b) \(P(A\mid B)=\tfrac{0.2}{0.4}=0.5\). (c) \(P(A\cap B)=0.2\) and \(P(A)P(B)=0.5(0.4)=0.2\); equal, so \(A\) and \(B\) are independent.

Marking scheme

M1A1 (a); M1A1 (b); M1 compare products; A1 conclusion.
Question 13 · Structured
6 marks
The discrete random variable \(X\) has distribution \(P(X=1)=0.2,\,P(X=2)=0.3,\,P(X=3)=0.4,\,P(X=4)=0.1\). (a) Find \(E(X)\). (b) Find \(\text{Var}(X)\). (c) Find \(E(2X+1)\).
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Worked solution

(a) \(E(X)=1(0.2)+2(0.3)+3(0.4)+4(0.1)=2.4\). (b) \(E(X^2)=1(0.2)+4(0.3)+9(0.4)+16(0.1)=6.6\); \(\text{Var}(X)=6.6-2.4^2=0.84\). (c) \(E(2X+1)=2(2.4)+1=5.8\).

Marking scheme

M1A1 E(X); M1 E(X²); A1 Var; A1 (c).

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