2025 CCEA GCSE Further Mathematics 2330 Practice Paper with Answers

Factorise the numerator and denominator: \( x^2-9=(x-3)(x+3) \); \( 2x^2-5x-3=(2x+1)(x-3) \) (check: \( (2x+1)(x-3)=2x^2-6x+x-3=2x^2-5x-3 \) ✓). So \( \dfrac{x^2-9}{2x^2-5x-3}=\dfrac{(x-3)(x+3)}{(2x+1)(x-3)} \). Cancelling the common factor \( (x-3) \) (valid for \( x\ne3 \)) gives \( \dfrac{x+3}{2x+1} \).

[1] numerator correctly factorised, \( (x-3)(x+3) \); [1] denominator correctly factorised, \( (2x+1)(x-3) \); [1] common factor \( (x-3) \) correctly identified; [1] correctly cancelled; [1] final simplified answer \( \dfrac{x+3}{2x+1} \).

\( x^2+6x+11 = (x+3)^2-9+11 = (x+3)^2+2 \). So \( p=3,\ q=2 \). Since \( (x+3)^2\ge0 \) for all x, the minimum value of the expression is \( q=2 \), occurring when \( (x+3)^2=0 \), i.e. \( x=-3 \).

[1] correctly completes the square, \( (x+3)^2-9 \); [1] correct final form \( (x+3)^2+2 \) with p, q correctly stated; [1] correct minimum value 2; [1] correct value \( x=-3 \) at which it occurs (ECF from the completed-square form).

First solve \( x^2+3x-10=0 \): factorising, \( (x+5)(x-2)=0 \), so \( x=-5 \) or \( x=2 \). Since the coefficient of \( x^2 \) is positive, the graph of \( y=x^2+3x-10 \) is an upward-opening parabola, so \( y>0 \) outside the roots (to the left of the smaller root and to the right of the larger root). Hence the solution is \( x<-5 \) or \( x>2 \).

[1] correct factorisation \( (x+5)(x-2) \); [1] correct roots \( x=-5,\ x=2 \); [1] correctly identifies that the region outside the roots satisfies the inequality (upward parabola, \( >0 \)); [1] both correct inequalities stated; [1] correctly written using 'or' (not 'and') between the two regions, since no single value satisfies both. Reject \( -50 \).

Rearranging, \( \tan\theta=\dfrac{1}{\sqrt3} \). The principal solution is \( \theta=\tan^{-1}\left(\dfrac{1}{\sqrt3}\right)=30^{\circ} \). Since tangent has period \( 180^{\circ} \) (and is positive in the first and third quadrants), a second solution in the given range is \( \theta=30^{\circ}+180^{\circ}=210^{\circ} \). Both lie within \( 0^{\circ}\le\theta\le360^{\circ} \), so the solutions are \( \theta=30^{\circ} \) and \( \theta=210^{\circ} \).

[1] correct rearrangement \( \tan\theta=1/\sqrt3 \); [1] correct principal solution \( \theta=30^{\circ} \); [1] correct use of the \( 180^{\circ} \) periodicity of tangent; [1] both correct solutions given, \( 30^{\circ} \) and \( 210^{\circ} \), and no extra incorrect solutions included.

Setting the expressions for y equal: \( x+1=x^2-4x+5 \), so \( x^2-5x+4=0 \). Factorising: \( (x-1)(x-4)=0 \), so \( x=1 \) or \( x=4 \). Using \( y=x+1 \): when \( x=1 \), \( y=2 \); when \( x=4 \), \( y=5 \). Check: at \( x=1 \), \( x^2-4x+5=1-4+5=2 \) ✓; at \( x=4 \), \( x^2-4x+5=16-16+5=5 \) ✓.

[1] correctly equates the two expressions for y; [1] correctly rearranged to \( x^2-5x+4=0 \); [1] correctly factorised/solved, \( x=1 \) or \( x=4 \); [1] both corresponding y-values correctly found; [1] both solution pairs stated correctly, with a check shown against the quadratic equation.

Converting from logarithmic to index form: \( 2x+1=3^2=9 \). So \( 2x=8 \), giving \( x=4 \). Check: \( \log_3(2(4)+1)=\log_3(9)=2 \) ✓.

[1] correct conversion to index form, \( 2x+1=9 \); [1] correct rearrangement \( 2x=8 \); [1] \( x=4 \); [1] correct check/verification shown in the original equation.

(a) \( \dfrac{dy}{dx}=3x^2-6x-9 \).
(b) At \( x=0 \): gradient \( =3(0)^2-6(0)-9=-9 \). \( y=0^3-3(0)^2-9(0)+5=5 \). So the point is \( (0,5) \) with gradient \( -9 \).
(c) Using \( y-y_1=m(x-x_1) \) with \( m=-9 \) at \( (0,5) \): \( y-5=-9(x-0) \), so \( y=-9x+5 \).
(d) The normal is perpendicular to the tangent, so its gradient is \( -\dfrac{1}{-9}=\dfrac{1}{9} \). Using \( y-5=\dfrac{1}{9}(x-0) \): \( 9(y-5)=x \), so \( 9y-45=x \), giving \( x-9y+45=0 \).

(a) [1] correct differentiation of each term; [1] fully correct \( dy/dx=3x^2-6x-9 \). (b) [1] correct gradient at x=0 (-9); [1] correct y-coordinate at x=0 (5). (c) [1] correct use of \( y-y_1=m(x-x_1) \) with their gradient and point; [1] correct tangent equation \( y=-9x+5 \). (d) [1] correct perpendicular gradient \( 1/9 \) (using \( m_1m_2=-1 \)); [1] correct normal equation \( x-9y+45=0 \) (or an equivalent correct form), ECF from (b).

(a) The matrix \( \begin{pmatrix}0&-1\\1&0\end{pmatrix} \) is the standard matrix for a rotation of \( 90^{\circ} \) anticlockwise about the origin.
(b) \( T\binom{3}{-2}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\binom{3}{-2}=\binom{(0)(3)+(-1)(-2)}{(1)(3)+(0)(-2)}=\binom{2}{3} \). So the image is \( (2,3) \).
(c) \( T^2=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\begin{pmatrix}(0)(0)+(-1)(1) & (0)(-1)+(-1)(0)\\ (1)(0)+(0)(1) & (1)(-1)+(0)(0)\end{pmatrix}=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}=-I \). Applying a \( 90^{\circ} \) rotation twice is equivalent to a single rotation of \( 180^{\circ} \) about the origin, which is consistent with \( T^2=-I \) (since a \( 180^{\circ} \) rotation sends every point \( (x,y) \) to \( (-x,-y) \)).
(d) \( \det(T)=(0)(0)-(-1)(1)=0+1=1 \). Since \( \det(T)=1 \), the transformation preserves area — the area of any shape is unchanged (scale factor 1) after the transformation, which is consistent with T representing a rotation (rotations do not stretch or shrink shapes).

(a) [1] correctly identifies a rotation about the origin; [1] correctly states 90° anticlockwise. (b) [1] correct matrix multiplication method shown; [1] correct image \( (2,3) \). (c) [1] correct matrix multiplication \( T^2=\begin{pmatrix}-1&0\\0&-1\end{pmatrix} \); [1] correctly describes this as a 180° rotation about the origin. (d) [1] correct value \( \det(T)=1 \); [1] correctly interprets this as areas being preserved/unchanged (scale factor 1).

(a) The curve crosses the x-axis where \( y=0 \): \( 4x-x^2=0 \), i.e. \( x(4-x)=0 \), so \( x=0 \) or \( x=4 \). The other point is at \( x=4 \).
(b) The required area is \( \displaystyle\int_0^4(4x-x^2)\,dx \). The indefinite integral is \( 2x^2-\dfrac{x^3}{3} \ (+C) \). Evaluating: at \( x=4 \): \( 2(16)-\dfrac{64}{3}=32-\dfrac{64}{3}=\dfrac{96-64}{3}=\dfrac{32}{3} \). At \( x=0 \): \( 0 \). So the area \( = \dfrac{32}{3}-0=\dfrac{32}{3}\approx10.7 \) (3 s.f.) square units.

(a) [1] correct factorisation \( x(4-x)=0 \); [1] \( x=4 \). (b) [1] correct integral expression set up, \( \int_0^4(4x-x^2)\,dx \); [1] correct integration of \( 4x \) to \( 2x^2 \); [1] correct integration of \( -x^2 \) to \( -x^3/3 \); [1] correct evaluation at the upper limit x=4, giving \( 32/3 \); [1] correct evaluation at the lower limit x=0, giving 0; [1] correct final area \( 32/3 \) (or 10.7, ECF from (a)).

(a) Taking logarithms of both sides: \( (2x-1)\ln3=\ln20 \), so \( 2x-1=\dfrac{\ln20}{\ln3}=2.727 \) (4 s.f.). Then \( 2x=3.727 \), so \( x=1.86 \) (3 s.f.).
(b) Using the multiplication law of logarithms: \( \log_5[x(x-4)]=1 \). Converting to index form: \( x(x-4)=5^1=5 \), so \( x^2-4x-5=0 \). Factorising: \( (x-5)(x+1)=0 \), so \( x=5 \) or \( x=-1 \). Since \( \log_5(x) \) requires \( x>0 \), and \( \log_5(x-4) \) requires \( x>4 \), the overall restriction is \( x>4 \). \( x=-1 \) does not satisfy \( x>4 \) (in fact both logarithms would be undefined, since \( x \) and \( x-4 \) would both be negative), so it is rejected. \( x=5 \) satisfies \( x>4 \), so it is the only valid solution. Check: \( \log_5(5)+\log_5(1)=1+0=1 \) ✓.

(a) [1] correct method, taking logs of both sides; [1] correct rearrangement \( 2x-1=\ln20/\ln3 \); [1] correct value \( 2x-1=2.73 \) (3 s.f.); [1] \( x=1.86 \) (3 s.f., accept 1.85-1.86). (b) [1] correct combination of logs using the multiplication law; [1] correct index form and quadratic \( x^2-4x-5=0 \); [1] correctly factorised/solved, \( x=5 \) or \( x=-1 \); [1] correctly rejects \( x=-1 \) with a valid reason (domain restriction \( x>4 \)) and confirms \( x=5 \) as the valid solution.

(a) At \( t=0 \): \( N=150-120e^0=150-120=30 \) (thousand subscribers).
(b) At \( t=6 \): \( N=150-120e^{-0.25\times6}=150-120e^{-1.5}=150-120(0.2231)=150-26.78=123.2 \), so \( N\approx123 \) thousand (nearest thousand).
(c) As \( t\to\infty \), \( e^{-0.25t}\to0 \), so \( N\to150-120(0)=150 \). In context, this means the number of subscribers is predicted to level off (saturate) at a maximum of about 150 000, and will never exceed this value, no matter how long the service has been running.
(d) Setting \( N=100 \): \( 100=150-120e^{-0.25t} \), so \( 120e^{-0.25t}=50 \), giving \( e^{-0.25t}=\dfrac{50}{120}=\dfrac{5}{12} \). Taking natural logs: \( -0.25t=\ln\left(\dfrac{5}{12}\right) \), so \( t=\dfrac{-\ln(5/12)}{0.25}=\dfrac{\ln(12/5)}{0.25}=\dfrac{0.8755}{0.25}=3.50 \) months (3 s.f.).

(a) [1] correct substitution \( t=0 \); [1] \( N=30 \). (b) [1] correct substitution \( t=6 \) and correct evaluation of \( e^{-1.5} \); [1] \( N\approx123 \) (accept 123-124). (c) [1] correctly identifies \( e^{-0.25t}\to0 \) as \( t\to\infty \), so \( N\to150 \); [1] correct contextual interpretation (subscriber numbers level off/saturate at a maximum, do not exceed 150 000). (d) [1] correct equation \( 120e^{-0.25t}=50 \) formed; [1] correctly isolates \( e^{-0.25t}=5/12 \); [1] correctly takes natural logs of both sides; [1] correct rearrangement for t; [1] \( t=3.50 \) months (3 s.f., accept 3.49-3.51).

(a) \( \dfrac{dy}{dx}=3x^2-6x-9=3(x^2-2x-3)=3(x-3)(x+1) \). Stationary points where \( dy/dx=0 \): \( x=3 \) or \( x=-1 \). At \( x=-1 \): \( y=(-1)^3-3(-1)^2-9(-1)+10=-1-3+9+10=15 \). At \( x=3 \): \( y=27-27-27+10=-17 \). So the stationary points are \( (-1,15) \) and \( (3,-17) \).
(b) \( \dfrac{d^2y}{dx^2}=6x-6 \). At \( x=-1 \): \( 6(-1)-6=-12<0 \), so \( (-1,15) \) is a local maximum. At \( x=3 \): \( 6(3)-6=12>0 \), so \( (3,-17) \) is a local minimum.
(c) The curve is increasing where \( dy/dx>0 \), i.e. where \( 3(x-3)(x+1)>0 \). Since this is an upward-opening quadratic in x (as a function of \( dy/dx \) against x) with roots \( x=-1 \) and \( x=3 \), it is positive outside these roots: \( x<-1 \) or \( x>3 \).

(a) [1] correct differentiation \( dy/dx=3x^2-6x-9 \); [1] correctly factorised/solved for x=-1, x=3; [1] correct y-value at x=-1 (15); [1] correct y-value at x=3 (-17). (b) [1] correct second derivative \( 6x-6 \); [1] correct evaluation and 'maximum' correctly stated at x=-1; [1] correct evaluation and 'minimum' correctly stated at x=3. (c) [1] correctly identifies the condition \( dy/dx>0 \) is needed; [1] correctly links this to the region outside the roots (using the shape of the \( dy/dx \) graph or a sign test); [1] correct final answer \( x<-1 \) or \( x>3 \) (ECF from (a)).

From the first equation, \( y=5-x \). Substituting into the second equation: \( x^2+(5-x)^2=17 \). Expanding: \( x^2+25-10x+x^2=17 \), so \( 2x^2-10x+25=17 \), giving \( 2x^2-10x+8=0 \). Dividing by 2: \( x^2-5x+4=0 \). Factorising: \( (x-1)(x-4)=0 \), so \( x=1 \) or \( x=4 \). Using \( y=5-x \): when \( x=1 \), \( y=4 \); when \( x=4 \), \( y=1 \). Check: \( 1^2+4^2=1+16=17 \) ✓; \( 4^2+1^2=16+1=17 \) ✓ (and both pairs sum to 5 ✓).

[1] correct rearrangement of the first equation for y; [1] correct substitution into the second equation; [1] correctly expanded, e.g. \( x^2+25-10x+x^2=17 \); [1] correctly simplified to \( x^2-5x+4=0 \); [1] correctly factorised/solved, \( x=1 \) or \( x=4 \); [1] both y-values correctly found; [1] both solution pairs stated correctly; [1] both pairs verified against the original equations (sum = 5 and sum of squares = 17). Award up to a maximum of 10 for full, clearly-shown working with both correct solution pairs and verification.

(a) The volume of a cylinder is \( V=\pi r^2h \). Since \( V=500\pi \): \( \pi r^2h=500\pi \), so \( r^2h=500 \), giving \( h=\dfrac{500}{r^2} \), as required.
(b) The total surface area of a closed cylinder is the area of the two circular ends plus the curved surface area: \( S=2\pi r^2+2\pi rh \). Substituting \( h=500/r^2 \) from (a): \( S=2\pi r^2+2\pi r\left(\dfrac{500}{r^2}\right)=2\pi r^2+\dfrac{1000\pi r}{r^2}=2\pi r^2+\dfrac{1000\pi}{r} \), as required.
(c) Writing \( S=2\pi r^2+1000\pi r^{-1} \): \( \dfrac{dS}{dr}=4\pi r-1000\pi r^{-2}=4\pi r-\dfrac{1000\pi}{r^2} \). Setting \( dS/dr=0 \): \( 4\pi r=\dfrac{1000\pi}{r^2} \), so \( 4r^3=1000 \), giving \( r^3=250 \), so \( r=\sqrt[3]{250}=6.30\text{ cm} \) (3 s.f.). To confirm this is a minimum, \( \dfrac{d^2S}{dr^2}=4\pi+\dfrac{2000\pi}{r^3} \), which is positive for all \( r>0 \), confirming \( r=6.30\text{ cm} \) gives a minimum surface area.
(d) At \( r=\sqrt[3]{250}=6.2996\text{ cm} \): \( 2\pi r^2=2\pi(6.2996)^2=2\pi(39.68)=79.37\pi \), and \( \dfrac{1000\pi}{r}=\dfrac{1000\pi}{6.2996}=158.7\pi \). Adding: \( S=79.37\pi+158.7\pi=238.1\pi\approx748\text{ cm}^2 \) (3 s.f.). (Exactly, \( S_{min}=150\times2^{2/3}\pi \), which evaluates to the same value, confirming the result.)

(a) [1] correct formula \( V=\pi r^2h \) used; [1] correctly rearranged to \( h=500/r^2 \). (b) [1] correct formula for total surface area of a closed cylinder, \( 2\pi r^2+2\pi rh \); [1] correct substitution of h from (a); [1] correctly simplified to the given form. (c) [1] correct differentiation, \( dS/dr=4\pi r-1000\pi/r^2 \); [1] correctly sets \( dS/dr=0 \) and solves \( r^3=250 \); [1] \( r=6.30\text{ cm} \) (3 s.f.); [1] correct use of the second derivative (or another valid method, e.g. sign change) to confirm a minimum. (d) [1] correct substitution of their r into S; [1] correct arithmetic; [1] \( S_{min}=748\text{ cm}^2 \) (3 s.f., accept 746-750, ECF from (c)).

(a) \( \overrightarrow{AB}=\mathbf{b}-\mathbf{a}=(8-2)\mathbf{i}+(-3-5)\mathbf{j}=(6\mathbf{i}-8\mathbf{j})\text{ m} \).
(b) \( AB=|\overrightarrow{AB}|=\sqrt{6^2+(-8)^2}=\sqrt{36+64}=\sqrt{100}=10.0\text{ m} \).
(c) The midpoint of AB has position vector \( \mathbf{m}=\dfrac{\mathbf{a}+\mathbf{b}}{2}=\dfrac{(2+8)\mathbf{i}+(5-3)\mathbf{j}}{2}=\dfrac{10\mathbf{i}+2\mathbf{j}}{2}=(5\mathbf{i}+\mathbf{j})\text{ m} \).

(a) [1] correct i-component (6); [1] correct j-component (-8). (b) [1] correct use of Pythagoras' theorem (ECF); [1] \( AB=10.0\text{ m} \). (c) [1] correct method \( (\mathbf{a}+\mathbf{b})/2 \); [1] correct i-component (5); [1] correct j-component (1).

(a) For motion with constant velocity, \( \mathbf{r}=\mathbf{r_0}+\mathbf{v}t=(1\mathbf{i}+2\mathbf{j})+(3\mathbf{i}-4\mathbf{j})t=(1+3t)\mathbf{i}+(2-4t)\mathbf{j} \).
(b) At \( t=5 \): \( \mathbf{r}=(1+15)\mathbf{i}+(2-20)\mathbf{j}=(16\mathbf{i}-18\mathbf{j})\text{ m} \).
(c) Speed is the magnitude of the (constant) velocity vector: \( |\mathbf{v}|=\sqrt{3^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5.0\text{ m s}^{-1} \).

(a) [1] correct i-component of r(t); [1] correct j-component of r(t). (b) [1] correct substitution t=5 (ECF); [1] correct position vector (16i-18j). (c) [1] correct use of Pythagoras' theorem on the velocity components; [1] correct squares (9 and 16) summed to 25; [1] \( 5.0\text{ m s}^{-1} \).

(a) Acceleration is the rate of change of velocity: \( a=\dfrac{dv}{dt}=6-2t \). At \( t=2 \): \( a=6-2(2)=2\text{ m s}^{-2} \).
(b) The particle is at rest when \( v=0 \): \( 6t-t^2=0 \), i.e. \( t(6-t)=0 \), so \( t=0 \) or \( t=6 \). The value other than \( t=0 \) is \( t=6\text{ s} \).
(c) Since \( v=t(6-t)\ge0 \) throughout \( 0\le t\le6 \) (the particle does not change direction in this interval), the distance travelled equals the displacement: \( \text{distance}=\displaystyle\int_0^3(6t-t^2)\,dt=\left[3t^2-\dfrac{t^3}{3}\right]_0^3=\left(3(9)-\dfrac{27}{3}\right)-0=27-9=18.0\text{ m} \).

(a) [1] correct differentiation \( a=6-2t \); [1] correct substitution t=2; [1] \( a=2\text{ m s}^{-2} \). (b) [1] correct factorisation/method \( t(6-t)=0 \); [1] \( t=6\text{ s} \). (c) [1] correct integral \( \int_0^3(6t-t^2)\,dt \) set up and evaluated; [1] \( 18.0\text{ m} \), with recognition that v does not change sign on [0,3] so distance = displacement.

(a) Treating the car and trailer as one system of total mass \( 900+300=1200\text{ kg} \), with total driving force \( 2400\text{ N} \) and total resistance \( 150+50=200\text{ N} \): using \( F=ma \), \( 2400-200=1200a \), so \( a=\dfrac{2200}{1200}=1.83\text{ m s}^{-2} \) (3 s.f.).
(b) For the trailer alone (mass 300 kg), the only forward force is the tension T in the tow-bar, opposed by the resistance of 50 N: \( T-50=300a \). Using \( a=1.8333 \) (unrounded): \( T=300(1.8333)+50=550+50=600\text{ N} \).
(c) For the car alone (mass 900 kg), the forces are the driving force forward, resistance backward, and the tow-bar tension acting backward on the car (reaction to the tension pulling the trailer forward): \( 2400-150-T=900a \), so \( T=2400-150-900(1.8333)=2250-1650=600\text{ N} \), which agrees exactly with the value found in (b), confirming the result.

(a) [1] correct total mass (1200 kg); [1] correct total resistance (200 N); [1] correct equation of motion for the system; [1] \( a=1.83\text{ m s}^{-2} \). (b) [1] correct equation of motion for the trailer, \( T-50=300a \); [1] correct substitution of a (using an unrounded or sufficiently accurate value, ECF); [1] correct rearrangement; [1] \( T=600\text{ N} \). (c) [1] correct equation of motion for the car, \( 2400-150-T=900a \); [1] correctly solved to give \( T=600\text{ N} \), matching (b), with an explicit statement that this confirms the answer.

(a) Since the resultant of the three forces is zero: \( \mathbf{F_1}+\mathbf{F_2}+\mathbf{F_3}=\mathbf{0} \), so \( \mathbf{F_3}=-(\mathbf{F_1}+\mathbf{F_2})=-\left[(5-3)\mathbf{i}+(2+4)\mathbf{j}\right]=-(2\mathbf{i}+6\mathbf{j})=(-2\mathbf{i}-6\mathbf{j})\text{ N} \).
(b) \( |\mathbf{F_3}|=\sqrt{(-2)^2+(-6)^2}=\sqrt{4+36}=\sqrt{40}=6.32\text{ N} \) (3 s.f.).
(c) Since both components of \( \mathbf{F_3} \) are negative, the vector lies in the third quadrant. The reference (acute) angle to the i-axis is \( \tan^{-1}\left(\dfrac{6}{2}\right)=\tan^{-1}(3)=71.57^{\circ} \). Measuring anticlockwise from the positive i-direction, the angle to a vector in the third quadrant is \( 180^{\circ}+71.57^{\circ}=251.57^{\circ}\approx252^{\circ} \) (nearest degree).

(a) [1] correctly sets \( \mathbf{F_3}=-(\mathbf{F_1}+\mathbf{F_2}) \); [1] correct i-component (-2); [1] correct j-component (-6). (b) [1] correct use of Pythagoras' theorem (ECF); [1] correct value under the square root (40); [1] \( 6.32\text{ N} \). (c) [1] correctly identifies F₃ lies in the third quadrant (both components negative); [1] correct reference angle \( \tan^{-1}(6/2)=71.6^{\circ} \) (ECF); [1] correct method to convert to a full anticlockwise angle (\( 180^{\circ}+ \) reference angle); [1] \( 252^{\circ} \) (accept 251°-252°).

(a) Since the wall is smooth, it exerts only a horizontal normal reaction S on the ladder at B; there is no vertical force from the wall. Resolving vertically for the ladder: the only vertical forces are the ladder's weight (200 N, down) and the ground's normal reaction R (up), so \( R=200\text{ N} \).
(b) Taking moments about A: the weight (200 N) acts at the ladder's midpoint, a horizontal distance \( 3.0\cos60^{\circ} \) from A, producing a moment tending to rotate the ladder clockwise (downwards) about A. The wall's reaction S is horizontal, acting at B, a vertical height \( 6.0\sin60^{\circ} \) above A, producing an anticlockwise moment. For equilibrium: \( S\times(6.0\sin60^{\circ}) = 200\times(3.0\cos60^{\circ}) \). So \( S = \dfrac{200\times3.0\times\cos60^{\circ}}{6.0\times\sin60^{\circ}} = \dfrac{200\times1.5}{5.196} = \dfrac{300}{5.196} = 57.7\text{ N} \) (3 s.f.).
(c) Resolving horizontally: the friction force F at the ground must balance the wall's reaction S (the only other horizontal force), so \( F=S=57.7\text{ N} \). For the ladder to be in equilibrium without slipping, the friction required must not exceed the maximum available friction, \( \mu R \), so the minimum coefficient of friction needed is \( \mu_{min}=\dfrac{F}{R}=\dfrac{57.7}{200}=0.289 \) (3 s.f.).

(a) [1] correctly identifies the wall exerts no vertical force (smooth, horizontal reaction only); [1] \( R=200\text{ N} \). (b) [1] correct moment arm for the weight (\( 3.0\cos60^{\circ} \)); [1] correct moment arm for S (\( 6.0\sin60^{\circ} \)); [1] correct moments equation formed; [1] \( S=57.7\text{ N} \). (c) [1] correctly identifies \( F=S \) from horizontal equilibrium (ECF); [1] correct formula \( \mu_{min}=F/R \); [1] \( \mu_{min}=0.289 \) (ECF, accept 0.288-0.289).

Arranging the marks in order: 9, 10, 11, 12, 13, 14, 15, 16, 18.
(a) Mean \( = \dfrac{9+10+11+12+13+14+15+16+18}{9} = \dfrac{118}{9} = 13.1 \) (3 s.f.).
(b) With \( n=9 \) values in order, the median is the 5th value: \( \text{median}=13 \).
(c) The lower half (excluding the median) is 9, 10, 11, 12; the lower quartile is the median of these, \( Q_1=\dfrac{10+11}{2}=10.5 \). The upper half is 14, 15, 16, 18; the upper quartile is \( Q_3=\dfrac{15+16}{2}=15.5 \). Interquartile range \( = Q_3-Q_1 = 15.5-10.5 = 5.0 \).

(a) [1] correct sum (118); [1] correct mean 13.1 (accept 13.1 or 118/9). (b) [1] correct median, 13. (c) [1] correct lower quartile 10.5; [1] correct upper quartile 15.5; [1] correct IQR 5.0.

The differences in rank, \( d = \text{Judge 1 rank} - \text{Judge 2 rank} \), are: A: \( 1-2=-1 \); B: \( 2-1=1 \); C: \( 3-4=-1 \); D: \( 4-3=1 \); E: \( 5-6=-1 \); F: \( 6-5=1 \). So \( d^2 \) values are all 1, giving \( \Sigma d^2=6 \). With \( n=6 \): \( r_s=1-\dfrac{6\times6}{6(6^2-1)}=1-\dfrac{36}{6\times35}=1-\dfrac{36}{210}=1-0.171=0.829 \) (3 s.f.). Since \( r_s \) is close to 1, this indicates a strong, positive level of agreement between the two judges' rankings of the competitors.

[1] each correct value of d for at least 4 of the 6 competitors (up to [2] for all correct); [1] correct \( \Sigma d^2=6 \); [1] correct substitution into the formula; [1] \( r_s=0.829 \) (accept 0.83); [1] correct comment on the strength and direction of agreement (strong, positive).

Using the midpoints \( x = 10, 30, 50, 70, 90 \) for each class:
(a) \( \Sigma fx = 4(10)+10(30)+16(50)+7(70)+3(90) = 40+300+800+490+270 = 1900 \). \( \Sigma f=40 \). Mean \( \bar{x}=\dfrac{\Sigma fx}{\Sigma f}=\dfrac{1900}{40}=47.5\text{ minutes} \).
(b) \( \Sigma fx^2 = 4(10)^2+10(30)^2+16(50)^2+7(70)^2+3(90)^2 = 400+9000+40000+34300+24300 = 108\,000 \). \( \sigma=\sqrt{\dfrac{108\,000}{40}-47.5^2}=\sqrt{2700-2256.25}=\sqrt{443.75}=21.1\text{ minutes} \) (3 s.f.).

(a) [1] correct \( \Sigma fx=1900 \); [1] correct \( \Sigma f=40 \); [1] correct mean 47.5. (b) [1] correct \( \Sigma fx^2=108\,000 \); [1] correct substitution into the given formula; [1] correct value under the square root (443.75, accept 443-444); [1] \( \sigma=21.1 \) (accept 21.0-21.1, ECF from (a)).

(a) French only \( = 18-8=10 \) (students who study French, minus those who study both). Spanish only \( = 15-8=7 \).
(b) The number studying at least one language is \( 10\ (\text{French only})+7\ (\text{Spanish only})+8\ (\text{both})=25 \). So the number studying neither is \( 30-25=5 \).
(c) \( P(\text{French or Spanish})=\dfrac{25}{30}=\dfrac{5}{6} \).
(d) \( P(\text{French but not Spanish})=\dfrac{10}{30}=\dfrac{1}{3} \).

(a) [1] correct French only (10); [1] correct Spanish only (7). (b) [1] correct method (25 study at least one); [1] correct answer (5 study neither). (c) [1] correct numerator (25, ECF); [1] correct probability \( 5/6 \). (d) [1] correct numerator (10); [1] correct probability \( 1/3 \).

(a) \( P(\text{RR})=\dfrac{4}{10}\times\dfrac{3}{9}=\dfrac{12}{90} \). \( P(\text{BB})=\dfrac{6}{10}\times\dfrac{5}{9}=\dfrac{30}{90} \). Since these are the only two ways of drawing two counters the same colour (and they are mutually exclusive): \( P(\text{same colour})=\dfrac{12}{90}+\dfrac{30}{90}=\dfrac{42}{90}=\dfrac{7}{15} \).
(b) Using the definition of conditional probability, \( P(\text{RR}\mid\text{same colour})=\dfrac{P(\text{RR})}{P(\text{same colour})}=\dfrac{12/90}{42/90}=\dfrac{12}{42}=\dfrac{2}{7} \).

(a) [1] correct \( P(\text{RR})=12/90 \); [1] correct \( P(\text{BB})=30/90 \); [1] correctly adds the two (mutually exclusive) probabilities; [1] correct final answer \( 7/15 \). (b) [1] correct use of the conditional probability formula \( P(A|B)=P(A\cap B)/P(B) \); [1] correctly identifies the numerator as \( P(\text{RR}) \) (ECF); [1] correct division; [1] final answer \( 2/7 \) (ECF from (a)).

(a) Each question is an independent trial with a fixed probability of success (guessing correctly) of \( \dfrac{1}{4}=0.25 \), repeated 10 times, so \( X\sim B(10, 0.25) \).
(b) \( P(X=3)=\binom{10}{3}(0.25)^3(0.75)^7 = 120\times0.015625\times0.1335 = 0.250 \) (3 s.f.).
(c) \( P(X\le1)=P(X=0)+P(X=1) \). \( P(X=0)=(0.75)^{10}=0.0563 \). \( P(X=1)=\binom{10}{1}(0.25)(0.75)^9=10\times0.25\times0.0751=0.1877 \). So \( P(X\le1)=0.0563+0.1877=0.244 \) (3 s.f.).

(a) [1] correctly identifies a binomial distribution; [1] correct parameters n=10, p=0.25. (b) [1] correct use of \( \binom{10}{3} \) (=120); [1] correct substitution into the binomial formula; [1] \( P(X=3)=0.250 \) (accept 0.249-0.250). (c) [1] correct \( P(X=0)=0.0563 \); [1] correct \( P(X=1)=0.188 \); [1] correct sum, \( P(X\le1)=0.244 \) (accept 0.243-0.244).

(a) \( z_1=\dfrac{48.5-50.0}{0.8}=-1.875 \); \( z_2=\dfrac{51.5-50.0}{0.8}=1.875 \). \( P(\text{reject})=P(Z<-1.875)+P(Z>1.875) \). Since the normal distribution is symmetric, \( P(Z<-1.875)=P(Z>1.875) \), so \( P(\text{reject})=2\times P(Z>1.875)=2\times(1-\Phi(1.875))=2\times(1-0.9696)=2\times0.0304=0.0608 \) (3 s.f.).
(b) Expected number rejected \( =500\times0.0608=30.4\approx30 \) bolts.

(a) [1] both z-values correctly calculated (±1.875); [1] correctly uses symmetry to combine both tails; [1] correct value of \( \Phi(1.875) \) or \( 1-\Phi(1.875) \) used; [1] \( P(\text{reject})=0.0608 \) (accept 0.060-0.061). (b) [1] correct method \( 500\times P(\text{reject}) \) (ECF); [1] \( 30.4 \); [1] correctly rounded to a whole number of bolts, 30.

(a) Choosing 4 from 12 with no restriction: \( \binom{12}{4}=495 \).
(b) Choosing 2 boys from 7 and 2 girls from 5, independently: \( \binom{7}{2}\times\binom{5}{2}=21\times10=210 \).
(c) 'At least 3 girls' means exactly 3 girls (and 1 boy), or exactly 4 girls (and 0 boys), which are mutually exclusive cases. Exactly 3 girls, 1 boy: \( \binom{5}{3}\times\binom{7}{1}=10\times7=70 \). Exactly 4 girls, 0 boys: \( \binom{5}{4}\times\binom{7}{0}=5\times1=5 \). Total: \( 70+5=75 \).
(d) The chairperson can be any of the 4 committee members, and the secretary any of the remaining 3 (a different person), so the number of ways is \( 4\times3=12 \) (equivalently \( ^4P_2=12 \)).
(e) By the multiplication principle, the total number of ways is (ways to choose the committee) × (ways to allocate the two roles within it): \( 495\times12=5940 \). (As a check, this can also be found directly: choose the chairperson (12 ways), then the secretary from the remaining students (11 ways), then the remaining 2 (unordered) committee members from the remaining 10 students (\( \binom{10}{2}=45 \) ways): \( 12\times11\times45=5940 \), which agrees.)

(a) [1] correct method \( \binom{12}{4} \); [1] \( 495 \). (b) [1] correct method \( \binom{7}{2}\times\binom{5}{2} \); [1] correct individual values (21, 10); [1] \( 210 \). (c) [1] correctly identifies the two mutually exclusive cases; [1] both cases correctly calculated (70 and 5); [1] correctly summed, \( 75 \). (d) [1] correct method \( 4\times3 \) (or \( ^4P_2 \)); [1] \( 12 \). (e) [1] correct use of the multiplication principle, combining (a) and (d) (ECF); [1] \( 5940 \), with a valid alternative check method shown (e.g. the direct 12×11×45 method) for full credit.

(a) Cutting time: \( 2x+5y\le100 \). Assembly time: \( 3x+4y\le108 \). Non-negativity: \( x\ge0 \), \( y\ge0 \).
(b) \( P=3x+5y \).
(c) Setting \( y=0 \) in each constraint: cutting gives \( x\le50 \), assembly gives \( x\le36 \); the binding constraint is assembly, giving vertex \( (36,0) \) (check cutting: \( 2(36)=72\le100 \) ✓). Setting \( x=0 \): cutting gives \( y\le20 \), assembly gives \( y\le27 \); the binding constraint is cutting, giving vertex \( (0,20) \) (check assembly: \( 4(20)=80\le108 \) ✓). Solving \( 2x+5y=100 \) and \( 3x+4y=108 \) simultaneously: from the first, \( x=50-2.5y \); substituting, \( 3(50-2.5y)+4y=108 \Rightarrow150-7.5y+4y=108\Rightarrow-3.5y=-42\Rightarrow y=12 \), so \( x=50-2.5(12)=50-30=20 \), giving vertex \( (20,12) \). Together with the origin, the vertices are \( (0,0),\ (36,0),\ (20,12),\ (0,20) \).
(d) \( P(0,0)=0 \). \( P(36,0)=3(36)=108 \). \( P(20,12)=3(20)+5(12)=60+60=120 \). \( P(0,20)=5(20)=100 \). The maximum profit is £120, at \( (20,12) \): 20 small boxes and 12 large boxes per day. (Check: at this point, cutting used \( =2(20)+5(12)=40+60=100 \) minutes exactly, and assembly used \( =3(20)+4(12)=60+48=108 \) minutes exactly, confirming both resources are fully used at the optimum.)

(a) [1] correct cutting inequality; [1] correct assembly inequality; [1] both non-negativity constraints. (b) [1] correct profit expression. (c) [1] correct vertex (36,0) found and justified; [1] correct vertex (0,20) found and justified; [1] correct simultaneous solution method; [1] correct vertex (20,12). (d) [1] profit correctly evaluated at all vertices (ECF); [1] correctly identifies (20,12) as the maximum; [1] correct final answer (20 small, 12 large, £120), with a valid check that both constraints are satisfied exactly at the optimum.

(a) \( \text{MA}_1=\dfrac{12+18+26+16}{4}=18.0 \); \( \text{MA}_2=\dfrac{18+26+16+14}{4}=18.5 \); \( \text{MA}_3=\dfrac{26+16+14+20}{4}=19.0 \); \( \text{MA}_4=\dfrac{16+14+20+28}{4}=19.5 \); \( \text{MA}_5=\dfrac{14+20+28+18}{4}=20.0 \).
(b) \( \text{MA}_1 \) is centred between weeks 2 and 3 (at 2.5); \( \text{MA}_2 \) is centred at 3.5; \( \text{MA}_3 \) is centred at 4.5. To find the trend at week 4 exactly, we average the moving averages centred immediately either side of week 4, namely \( \text{MA}_2 \) (centred 3.5) and \( \text{MA}_3 \) (centred 4.5): trend at week \( 4 = \dfrac{18.5+19.0}{2}=18.75 \).
(c) Seasonal variation \( = \text{actual}-\text{trend} = 16-18.75 = -2.75 \). (This suggests week 4 sales are typically about £2.75 (i.e. £275) below the underlying trend.)

(a) [1] each for correct MA1, MA2 (up to 2); [1] correct MA3, MA4 (up to 1 combined, allow ECF); [1] correct MA5 — award up to [4] total for all five values correct (18.0, 18.5, 19.0, 19.5, 20.0), or [2]-[3] for correct method with 1-2 slips. (b) [1] correctly identifies MA2 and MA3 as the pair to average for week 4; [1] correct method (averaging the pair); [1] \( 18.75 \) (ECF from (a)). (c) [1] correct use of the additive model (actual − trend); [1] \( -2.75 \) (ECF from (b)), with correct sign.

(a) The converse of \( p\Rightarrow q \) is \( q\Rightarrow p \): 'If a number n is divisible by 2 and by 3, then n is divisible by 6.'
(b) The truth table for \( p\Leftrightarrow q \):

p q p⇔q
T T T
T F F
F T F
F F T

So \( p\Leftrightarrow q \) is true exactly when p and q share the same truth value — both true, or both false — and false whenever they differ.
(c) The contrapositive of \( p\Rightarrow q \) is \( \lnot q\Rightarrow\lnot p \): 'If n is not divisible by 2 and by 3, then n is not divisible by 6.' Truth table comparing \( p\Rightarrow q \) and \( \lnot q\Rightarrow\lnot p \):

p q p⇒q ¬q⇒¬p
T T T T
T F F F
F T T T
F F T T

In every one of the four rows, the columns for \( p\Rightarrow q \) and \( \lnot q\Rightarrow\lnot p \) match exactly, so the contrapositive is logically equivalent to (always has the same truth value as) the original implication, for all p and q.
(d) Yes, \( p\Leftrightarrow q \) is true for every integer n. This is because \( 6=2\times3 \), and 2 and 3 are coprime (share no common factors other than 1); a standard result in number theory states that an integer n is divisible by the product of two coprime numbers if and only if it is divisible by each of them individually. So n is divisible by 6 if and only if n is divisible by both 2 and 3 — this is not just a logical possibility (as in (b)), but a mathematically true statement for every integer n.

(a) [1] correct converse stated in words. (b) [1] correct p⇔q column, all four rows correct; [1] correct identification that p⇔q true when p,q match (both T or both F); [1] correct identification p⇔q false when they differ; [1] full table clearly presented with all combinations. (c) [1] correct contrapositive stated (¬q⇒¬p, in words or symbols); [1] correct p⇒q column in the truth table; [1] correct ¬q⇒¬p column in the truth table; [1] correct conclusion that the two columns match in every row, confirming logical equivalence. (d) [1] correctly states p⇔q is true for all integers n; [1] correct mathematical justification referencing 6=2×3 with 2, 3 coprime (or an equivalent valid number-theoretic argument).

(a) P and Q have no predecessor: \( ES_P=0,\ EF_P=3 \); \( ES_Q=0,\ EF_Q=5 \). R depends on P: \( ES_R=EF_P=3,\ EF_R=3+4=7 \). S depends on Q: \( ES_S=EF_Q=5,\ EF_S=5+6=11 \). T depends on R and S: \( ES_T=\max(EF_R,EF_S)=\max(7,11)=11,\ EF_T=11+2=13 \).
(b) The project duration is the largest EF value, at T: 13 days. There are two possible paths through the network: P→R→T, with total duration \( 3+4+2=9 \) days; and Q→S→T, with total duration \( 5+6+2=13 \) days, which matches the project duration. So the critical path is Q→S→T, and the minimum project duration is 13 days.

(a) [1] correct ES/EF for P and Q; [1] correct ES/EF for R; [1] correct ES/EF for S; [1] correct ES/EF for T, correctly taking the maximum of EF(R) and EF(S). (b) [1] correctly calculates the duration of path P-R-T (9 days); [1] correctly calculates the duration of path Q-S-T (13 days); [1] correctly identifies Q-S-T as the critical path (the longer path, matching the project's EF at T) and states the minimum project duration as 13 days.