AQA A-Level · Thinka-original Practice Paper

2023 AQA A-Level Mathematics 7357 Practice Paper with Answers

Thinka Jun 2023 AQA A Level-Style Mock — Mathematics 7357

300 marks360 mins2023
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 Question · 100 marks
Question 1 · Multiple choice
1 marks
Differentiate \(y=\ln(5x)\). Circle your answer.
  1. A.\(\tfrac5x\)
  2. B.\(\tfrac1x\)
  3. C.\(\tfrac{1}{5x}\)
  4. D.\(\ln5\)
Show answer & marking scheme

Worked solution

\(\tfrac{d}{dx}\ln(5x)=\tfrac{5}{5x}=\tfrac1x\).

Marking scheme

B1 for \(\tfrac1x\).
Question 2 · Multiple choice
1 marks
Find \(\displaystyle\int e^{3x}\,dx\). Circle your answer.
  1. A.\(3e^{3x}+c\)
  2. B.\(\tfrac13 e^{3x}+c\)
  3. C.\(e^{3x}+c\)
  4. D.\(3e^{x}+c\)
Show answer & marking scheme

Worked solution

\(\int e^{3x}dx=\tfrac13 e^{3x}+c\).

Marking scheme

B1 for \(\tfrac13 e^{3x}+c\).
Question 3 · Short
5 marks
Find the first three terms, in ascending powers of \(x\), of the binomial expansion of \((3+2x)^6\).
Show answer & marking scheme

Worked solution

\((3+2x)^6=3^6+\binom61 3^5(2x)+\binom62 3^4(2x)^2+\dots=729+6(243)(2)x+15(81)(4)x^2=729+2916x+4860x^2\).

Marking scheme

M1 binomial structure; A1 729; A1 2916x; M1 third-term values; A1 4860x².
Question 4 · Structured
12 marks
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 \((12,3)\).
Show answer & marking scheme

Worked solution

(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 \(=\sqrt{(12-5)^2+(3+1)^2-25}=\sqrt{49+16-25}=\sqrt{40}=6.32\).

Marking scheme

M1A1 centre & radius; M1A1 (b); M1 radius direction; A1 tangent y=4; M1A1 tangent length.
Question 5 · Structured
10 marks
(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.
Show answer & marking scheme

Worked solution

(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.

Marking scheme

M1A1 (a); M1A1 (b)(i); M1A1 (b)(ii); B1 (c) with reason.
Question 6 · Structured
12 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.
Show answer & marking scheme

Worked solution

(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) \(x_1=\sqrt[3]{15.5}=2.494,\,x_2=2.490,\,x_3=2.489,\dots\to 2.489\).

Marking scheme

M1 evaluate ends; A1 sign change; M1 NR formula; A1 x₁; A1 x₂; M1 iterate; A1 2.489.
Question 7 · Structured
10 marks
\(p(x)=2x^3+ax^2+bx-6\). Given that \((x-2)\) and \((x+1)\) are factors: (a) find \(a\) and \(b\); (b) factorise \(p(x)\) completely; (c) solve \(p(x)=0\).
Show answer & marking scheme

Worked solution

(a) \(p(2)=16+4a+2b-6=0\Rightarrow2a+b=-5\); \(p(-1)=-2+a-b-6=0\Rightarrow a-b=8\). Adding, \(3a=3\Rightarrow a=1,\,b=-7\). (b) \(p(x)=2x^3+x^2-7x-6=(x-2)(x+1)(2x+3)\). (c) \(x=2,-1,-\tfrac32\).

Marking scheme

M1 p(2)=0; M1 p(-1)=0; A1 a,b; M1 divide; A1 full factorisation; A1 three roots.
Question 8 · Structured
8 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.
Show answer & marking scheme

Worked solution

(a) \(\tfrac{6}{1-2/3}=18\). (b) \(2x=\tfrac{\pi}{4},\tfrac{5\pi}{4}\Rightarrow x=\tfrac{\pi}{8},\tfrac{5\pi}{8}\). (c) \(\tan x=\tfrac34\Rightarrow x=0.6435\) or \(0.6435+\pi=3.785\).

Marking scheme

B1 (a); M1 2x values; A1 both x; M1 tan x; A1 0.64; A1 3.79.
Question 9 · Structured
10 marks
(a) Prove the identity \(\dfrac{1+\sin x}{\cos x}+\dfrac{\cos x}{1+\sin x}\equiv\dfrac{2}{\cos x}\). (b) Hence solve \(\dfrac{1+\sin x}{\cos x}+\dfrac{\cos x}{1+\sin x}=4\) for \(0
Show answer & marking scheme

Worked solution

(a) Common denominator: \(\tfrac{(1+\sin x)^2+\cos^2x}{\cos x(1+\sin x)}=\tfrac{1+2\sin x+\sin^2x+\cos^2x}{\cos x(1+\sin x)}=\tfrac{2+2\sin x}{\cos x(1+\sin x)}=\tfrac{2}{\cos x}\). (b) \(\tfrac{2}{\cos x}=4\Rightarrow\cos x=\tfrac12\Rightarrow x=\tfrac{\pi}{3},\tfrac{5\pi}{3}\). (c) \(\tfrac{1+\cos2x}{\sin2x}=\tfrac{2\cos^2x}{2\sin x\cos x}=\cot x\).

Marking scheme

M1 combine; M1 use identity; A1 result; M1A1 (b); M1A1 (c).
Question 10 · Structured
15 marks
(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 \(y=4x\).
Show answer & marking scheme

Worked solution

(a) \(2x^3+\tfrac{4}{x}+\tfrac13 e^{3x}+c\). (b) \(u=x^2+4,\,du=2x\,dx\Rightarrow\int u^{-1/2}du=2\sqrt u=2\sqrt{x^2+4}+c\). (c) Intersect: \(x^2+3=4x\Rightarrow x=1,3\); area \(=\int_1^3(4x-x^2-3)dx=[2x^2-\tfrac{x^3}{3}-3x]_1^3=0-(-\tfrac43)=\tfrac43\).

Marking scheme

M1A1A1 (a); M1 substitution; A1 (b); M1 limits 1,3; M1 integrate difference; A1 4/3.
Question 11 · Structured
16 marks
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.
Show answer & marking scheme

Worked solution

(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 \(x>2\).

Marking scheme

M1A1 derivative; M1 stationary; A1 both; M1A1 natures; M1A1 inflection; M1A1 tangent; B1 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.\(x\ln x+c\)
Show answer & marking scheme

Worked solution

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

Marking scheme

B1 for \(\ln|x|+c\).
Question 2 · Short
3 marks
Express \(\dfrac{4x+5}{(x+1)(x-2)}\) in partial fractions.
Show answer & marking scheme

Worked solution

Let \(\tfrac{4x+5}{(x+1)(x-2)}=\tfrac{A}{x+1}+\tfrac{B}{x-2}\). Then \(4x+5=A(x-2)+B(x+1)\). \(x=-1:1=-3A\Rightarrow A=-\tfrac13\); \(x=2:13=3B\Rightarrow B=\tfrac{13}{3}\).

Marking scheme

M1 set up identity; M1 substitute; A1 A=-1/3, B=13/3.
Question 3 · Proof
6 marks
Prove by contradiction that \(\sqrt5\) is irrational.
Show answer & marking scheme

Worked solution

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.

Marking scheme

B1 assume rational; M1 \(p^2=5q^2\); A1 5|p; M1 substitute; A1 5|q; A1 contradiction.
Question 4 · Structured
10 marks
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.
Show answer & marking scheme

Worked solution

(a) Height \(h=\tfrac{1000}{x^2}\); \(S=2x^2+4xh=2x^2+\tfrac{4000}{x}\). (b) \(\tfrac{dS}{dx}=4x-\tfrac{4000}{x^2}=0\Rightarrow x^3=1000\Rightarrow x=10\). (c) \(S=200+400=600\,\text{cm}^2\) (\(\tfrac{d^2S}{dx^2}>0\), minimum).

Marking scheme

M1 height; A1 show S; M1 differentiate; M1 solve =0; A1 x=10; A1 S=600.
Question 5 · Structured
10 marks
(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 \(x\).
Show answer & marking scheme

Worked solution

(a) \(\int\tfrac{dP}{P}=\int0.4\,dt\Rightarrow\ln P=0.4t+c\Rightarrow P=Ae^{0.4t}\). (b) \(A=150\Rightarrow P=150e^{2}=1108\). (c) \(\int\tfrac{dy}{y}=\int4x^3dx\Rightarrow\ln y=x^4+c\Rightarrow y=Be^{x^4}\).

Marking scheme

M1 separate; A1 show; M1 A=150; A1 1108; M1 separate; A1 \(y=Be^{x^4}\).
Question 6 · Structured
10 marks
(a) Solve \(3^{2x}-12\cdot3^{x}+27=0\). (b) Solve \(\log_3 x+\log_3(x+6)=3\). (c) Evaluate \(\log_2 32-\log_2 4\).
Show answer & marking scheme

Worked solution

(a) Let \(y=3^x\): \(y^2-12y+27=0\Rightarrow(y-3)(y-9)=0\Rightarrow3^x=3\,(x=1)\) or \(3^x=9\,(x=2)\). (b) \(\log_3 x(x+6)=3\Rightarrow x^2+6x=27\Rightarrow(x+9)(x-3)=0\Rightarrow x=3\) (reject \(-9\)). (c) \(5-2=3\).

Marking scheme

M1 substitution; A1 both x; M1 combine logs; M1 solve; A1 x=3; A1 (c).
Question 7 · Structured
11 marks
(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 \(0\le x\le360^\circ\).
Show answer & marking scheme

Worked solution

(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) \(\sin(x-67.38^\circ)=0.5\Rightarrow x-67.38^\circ=30^\circ,150^\circ\Rightarrow x=97.4^\circ,217.4^\circ\).

Marking scheme

M1 R=13; M1 α; A1 form; M1A1 (b); M1 solve; A1 both x.
Question 8 · Multiple choice
1 marks
Find the weight of a particle of mass \(8\,\text{kg}\) (take \(g=9.8\)). Circle your answer.
  1. A.\(8\,\text{N}\)
  2. B.\(0.82\,\text{N}\)
  3. C.\(78.4\,\text{N}\)
  4. D.\(9.8\,\text{N}\)
Show answer & marking scheme

Worked solution

Weight \(=mg=8\times9.8=78.4\,\text{N}\).

Marking scheme

B1 for 78.4 N.
Question 9 · Structured
6 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.
Show answer & marking scheme

Worked solution

(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) \(s=\int_0^2(6t^2-6t+4)dt=[2t^3-3t^2+4t]_0^2=16-12+8=12\,\text{m}\).

Marking scheme

M1 integrate a; A1 v; B1 (b); M1 integrate v; A1 12 m.
Question 10 · Structured
9 marks
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.
Show answer & marking scheme

Worked solution

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.

Marking scheme

M1 hanging-mass equation; M1 on-plane equation; A1 add; A1 a=4.29; M1 T; A1 27.6; B1 assumption.
Question 11 · Structured
10 marks
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.
Show answer & marking scheme

Worked solution

(a) Moments about \(A\): \(R_B(10)=500(5)+300(3)=3400\Rightarrow R_B=340\,\text{N}\); \(R_A=500+300-340=460\,\text{N}\). (b) Equal reactions \(=400\) each: \(400(10)=500(5)+300d\Rightarrow300d=1500\Rightarrow d=5\,\text{m}\).

Marking scheme

M1 moments about A; A1 R_B; M1 resolve; A1 R_A; M1 set equal; A1 d=5.
Question 12 · Structured
8 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 \(\mathbf{j}\).
Show answer & marking scheme

Worked solution

\(\mathbf{v}=(2t-4)\mathbf{i}+(2t-2)\mathbf{j}\). (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: \(2t-4=0\Rightarrow t=2\).

Marking scheme

M1 differentiate; A1 v at t=1; M1A1 speed; M1 set i-comp =0; A1 t=2.
Question 13 · Structured
6 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.
Show answer & marking scheme

Worked solution

(a) \(25=\tfrac12(9.8)t^2\Rightarrow t^2=5.102\Rightarrow t=2.26\,\text{s}\). (b) \(x=20(2.26)=45.2\,\text{m}\).

Marking scheme

M1 vertical equation; A1 t=2.26; M1 horizontal motion; A1 45.2 m; B1 method.
Question 14 · Structured
9 marks
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.
Show answer & marking scheme

Worked solution

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.

Marking scheme

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 Question · 100 marks
Question 1 · Multiple choice
1 marks
State the period of \(y=\sin(2\pi x)\). Circle your answer.
  1. A.\(2\pi\)
  2. B.\(\tfrac12\)
  3. C.\(1\)
  4. D.\(\pi\)
Show answer & marking scheme

Worked solution

Period \(=\tfrac{2\pi}{2\pi}=1\).

Marking scheme

B1 for 1.
Question 2 · Structured
10 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 \(x=1\).
Show answer & marking scheme

Worked solution

(a) Chain rule: \(12(3x-2)^3\). (b) Product rule: \(2x\ln x+x\). (c) Quotient rule: \(\tfrac{(x^2+4)-x(2x)}{(x^2+4)^2}=\tfrac{4-x^2}{(x^2+4)^2}\). (d) At \(x=1\): \(12(1)^3=12\).

Marking scheme

M1A1 (a); M1A1 (b); M1A1 (c); M1A1 (d).
Question 3 · Structured
9 marks
Find each integral. (a) \(\displaystyle\int(2x+5)^3\,dx\). (b) \(\displaystyle\int\tfrac{9}{3x+1}\,dx\). (c) \(\displaystyle\int_1^2\tfrac{6}{x^2}\,dx\).
Show answer & marking scheme

Worked solution

(a) \(\tfrac{(2x+5)^4}{8}+c\). (b) \(3\ln|3x+1|+c\). (c) \([-\tfrac{6}{x}]_1^2=-3-(-6)=3\).

Marking scheme

M1A1 (a); M1A1 (b); M1 antiderivative; A1 (c)=3.
Question 4 · Structured
10 marks
(a) Solve \(2\cos 2x=\sqrt3\) for \(0\le x\le2\pi\). (b) Prove that \(\tan^2 x+1\equiv\sec^2 x\). (c) Solve \(2\sin x+1=0\) for \(0\le x\le2\pi\).
Show answer & marking scheme

Worked solution

(a) \(\cos2x=\tfrac{\sqrt3}{2}\Rightarrow2x=\tfrac{\pi}{6},\tfrac{11\pi}{6},\tfrac{13\pi}{6},\tfrac{23\pi}{6}\Rightarrow x=\tfrac{\pi}{12},\tfrac{11\pi}{12},\tfrac{13\pi}{12},\tfrac{23\pi}{12}\). (b) \(\tan^2x+1=\tfrac{\sin^2x}{\cos^2x}+1=\tfrac{\sin^2x+\cos^2x}{\cos^2x}=\tfrac{1}{\cos^2x}=\sec^2x\). (c) \(\sin x=-\tfrac12\Rightarrow x=\tfrac{7\pi}{6},\tfrac{11\pi}{6}\).

Marking scheme

M1 cos2x value; A1 all four x; M1A1 (b); M1A1 (c).
Question 5 · Structured
10 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 \(t=0\).
Show answer & marking scheme

Worked solution

(a) \(e^{-12k}=\tfrac12\Rightarrow k=\tfrac{\ln2}{12}=0.0578\). (b) \(m=80e^{-0.0578\times20}=80e^{-1.155}=25.2\,\text{g}\). (c) \(20=80e^{-kt}\Rightarrow t=\tfrac{\ln4}{0.0578}=24\,\text{years}\). (d) \(\tfrac{dm}{dt}=-km_0\) at \(t=0=-0.0578(80)=-4.62\,\text{g/year}\).

Marking scheme

M1A1 k; M1A1 (b); M1A1 (c); M1A1 (d).
Question 6 · Structured
10 marks
(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.
Show answer & marking scheme

Worked solution

(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 \(=101\)).

Marking scheme

M1A1 (a); M1A1 (b); M1A1 (c); M1A1 (d).
Question 7 · Multiple choice
1 marks
For \(X\sim B(25,0.2)\), state the mean \(E(X)\). Circle your answer.
  1. A.\(0.2\)
  2. B.\(5\)
  3. C.\(4\)
  4. D.\(20\)
Show answer & marking scheme

Worked solution

\(E(X)=np=25\times0.2=5\).

Marking scheme

B1 for 5.
Question 8 · Structured
6 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.
Show answer & marking scheme

Worked solution

(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.

Marking scheme

B1 definition; M1 numbering; A1 random selection; B1 advantage; B1 reason; B1 clarity.
Question 9 · Structured
10 marks
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.
Show answer & marking scheme

Worked solution

(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.

Marking scheme

B1 median; B1 quartiles; B1 IQR; M1A1 mean; M1A1 s.d.; B1 outlier.
Question 10 · Structured
10 marks
The masses of items are modelled by \(X\sim N(50,6^2)\) (in grams). (a) Find \(P(X>56)\). (b) Find \(P(44
Show answer & marking scheme

Worked solution

(a) \(z=\tfrac{56-50}{6}=1\Rightarrow P(Z>1)=0.1587\). (b) \(z=-1\) and \(z=1.5\Rightarrow\Phi(1.5)-\Phi(-1)=0.9332-0.1587=0.7745\). (c) \(P(X>x)=0.05\Rightarrow z=1.645\Rightarrow x=50+1.645(6)=59.9\,\text{g}\).

Marking scheme

M1 standardise; A1 (a); M1A1 (b); M1 z=1.645; A1 (c).
Question 11 · Structured
10 marks
(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.
Show answer & marking scheme

Worked solution

(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.

Marking scheme

M1A1 (a)(i); A1 (a)(ii); B1 hypotheses; M1 P(X≥9); A1 0.0028; A1 conclusion.
Question 12 · Structured
7 marks
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.
Show answer & marking scheme

Worked solution

(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.

Marking scheme

M1A1 (a); M1A1 (b); M1 compare; A1 conclusion.
Question 13 · Structured
6 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 \(E(3X-2)\).
Show answer & marking scheme

Worked solution

(a) \(E(X)=0(0.1)+1(0.3)+2(0.4)+3(0.2)=1.7\). (b) \(E(X^2)=0+0.3+1.6+1.8=3.7\); \(\text{Var}(X)=3.7-1.7^2=0.81\). (c) \(E(3X-2)=3(1.7)-2=3.1\).

Marking scheme

M1A1 E(X); M1 E(X²); A1 Var; A1 (c).

Wondering how well you actually know this?

Thinka is an AI practice app for DSE students — unlimited questions, instant auto-marking, and detailed step-by-step solutions. 100,000+ students use it to confirm they actually know it, not just think they do.

Want more questions like this? Practice unlimited on Thinka — instant answers included.

Start Practising Free