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(a) Height greater than 80 cm includes \(80 < h \le 90\) and \(90 < h \le 100\). Total number of such saplings is \(18 + 8 = 26\). Probability = \(\frac{26}{120} = \frac{13}{60} \approx 0.217\).

(b) Find the midpoints: 45, 55, 65, 75, 85, 95. Calculate sum of products: \(\Sigma f x = 8(45) + 16(55) + 32(65) + 38(75) + 18(85) + 8(95) = 360 + 880 + 2080 + 2850 + 1530 + 760 = 8460\). Estimate of mean = \(\frac{8460}{120} = 70.5\) cm.

(c) Total saplings with height greater than 60 cm is \(120 - (8 + 16) = 96\). Saplings with height greater than 80 cm is 26. Probability of choosing two without replacement: \(\frac{26}{96} \times \frac{25}{95} = \frac{13}{48} \times \frac{5}{19} = \frac{65}{912} \approx 0.0713\).

(d)(i) Upper Quartile = Lower Quartile + IQR = \(62 + 16 = 78\) cm.
(d)(ii) Since the lower quartile is 62 cm, 25% of the saplings lie at or below this value. Number of saplings = \(0.25 \times 120 = 30\).

(a) M1 for 26/120, A1 for 13/60 or 0.217
(b) M1 for identifying correct midpoints, M1 for \(\Sigma f x = 8460\), M1 for dividing by 120, A1 for 70.5
(c) M1 for 96, M1 for product of probabilities without replacement, A1 for 65/912 or 0.0713
(d)(i) A1 for 78
(d)(ii) M1 for \(0.25 \times 120\), A1 for 30

(a) Numerator: \(3(x^2 - 4) = 3(x-2)(x+2)\). Denominator: \((x+3)(x-2)\). Dividing both: \(\frac{3(x-2)(x+2)}{(x+3)(x-2)} = \frac{3(x+2)}{x+3}\).

(b) Rearrange the second equation to get \(x = y + 1\). Substitute this into the first: \((y+1)^2 + 2y^2 = 17 \implies y^2 + 2y + 1 + 2y^2 = 17 \implies 3y^2 + 2y - 16 = 0\). Factoring the quadratic: \((3y + 8)(y - 2) = 0\). Hence, \(y = 2\) or \(y = -\frac{8}{3}\). Corresponding \(x\) values are: \(x = 2 + 1 = 3\) and \(x = -\frac{8}{3} + 1 = -\frac{5}{3}\).

(c) Square both sides: \(T^2 = 25\left(\frac{w+h}{w-h}\right) \implies T^2(w-h) = 25(w+h) \implies T^2 w - T^2 h = 25w + 25h \implies T^2 w - 25w = T^2 h + 25h \implies w(T^2 - 25) = h(T^2 + 25) \implies w = \frac{h(T^2 + 25)}{T^2 - 25}\).

(a) M1 for factoring out 3, M1 for difference of two squares \((x-2)(x+2)\), M1 for factoring denominator to \((x+3)(x-2)\), A1 for final fraction
(b) M1 for substituting \(x = y+1\), M1 for writing quadratic as \(3y^2+2y-16=0\), M1 for factorizing/solving, A1 for \(x=3, y=2\), A1 for \(x=-5/3, y=-8/3\)
(c) M1 for squaring both sides, M1 for clearing denominator, M1 for grouping \(w\) terms, A1 for correct formula

(a) Total height is \(h + r = 24\), so \(h = 24 - r\). Volume of cylinder = \(\pi r^2 h = \pi r^2(24 - r) = 24\pi r^2 - \pi r^3\). Volume of hemisphere = \(\frac{2}{3}\pi r^3\). Total Volume = \(24\pi r^2 - \pi r^3 + \frac{2}{3}\pi r^3 = 24\pi r^2 - \frac{1}{3}\pi r^3\).

(b) Equating volume to \(1500\pi\): \(24\pi r^2 - \frac{1}{3}\pi r^3 = 1500\pi \implies 24r^2 - \frac{1}{3}r^3 = 1500\). Multiplying by \(-3\): \(r^3 - 72r^2 = -4500 \implies r^3 - 72r^2 + 4500 = 0\).

(c)(i) Total Surface Area = circular base area + cylinder curved surface + hemisphere curved surface = \(\pi R^2 + 2\pi R H + 2\pi R^2 = 3\pi R^2 + 2\pi R H\). For \(R=6\): \(3\pi(36) + 12\pi H = 220\pi \implies 108\pi + 12\pi H = 220\pi \implies 12H = 112 \implies H = \frac{112}{12} = 9.33\) cm.

(c)(ii) Total volume of this solid = \(\pi R^2 H + \frac{2}{3}\pi R^3 = \pi(36)(28/3) + \frac{2}{3}\pi(216) = 336\pi + 144\pi = 480\pi\) \(\text{cm}^3\). Equating to sphere volume: \(\frac{4}{3}\pi r_s^3 = 480\pi \implies r_s^3 = 360 \implies r_s = \sqrt[3]{360} \approx 7.11\) cm.

(a) M1 for \(h = 24 - r\), M1 for sum of correct volume formulas, A1 for correct simplification to target formula
(b) M1 for setting volume equal to \(1500\pi\) and dividing by \(\pi\), A1 for algebraic step showing the target equation
(c)(i) M1 for correct TSA formula \(3\pi R^2 + 2\pi R H\), M1 for substituting \(R=6\), M1 for equating to \(220\pi\) and solving, A1 for 9.33
(c)(ii) M1 for finding solid volume \(480\pi\), M1 for equating to \(\frac{4}{3}\pi r_s^3\), M1 for isolating \(r_s^3 = 360\), A1 for 7.11

(a) Multiply all terms by 12: \(4(2x - 5) \le 3(3x + 1) - 12 \implies 8x - 20 \le 9x + 3 - 12 \implies 8x - 20 \le 9x - 9 \implies -11 \le x \implies x \ge -11\).

(b) Subtract 1: \(-6 < 2y \le 8\). Divide by 2: \(-3 < y \le 4\). The integers in this interval are \(-2, -1, 0, 1, 2, 3, 4\).

(c)(i) The conditions give: 1. \(x + y \ge 15\), 2. \(y \le 2x\), 3. \(200x + 300y \le 4500 \implies 2x + 3y \le 45\).
(c)(ii) From \(y \le 2x \implies x \ge y/2\). Substitute into the budget constraint: \(2(y/2) + 3y \le 45 \implies 4y \le 45 \implies y \le 11.25\). Testing the maximum integer \(y = 11\): then \(x \ge 11/2 = 5.5 \implies x = 6\). Check if \((6, 11)\) satisfies all: \(6+11=17 \ge 15\), \(11 \le 12\), \(2(6)+3(11)=45 \le 45\). All satisfied, so the maximum number of Model B tablets is 11.

(a) M1 for multiplying by 12, M1 for expanding brackets correctly, M1 for isolating terms, A1 for \(x \ge -11\)
(b) M1 for isolating term in \(y\), A1 for correct bounds \(-3 < y \le 4\), A1 for correct list of integers
(c)(i) B3 for all three correct inequalities (B1 for each)
(c)(ii) M1 for substituting \(y \le 2x\) into the budget inequality, M1 for analyzing integer solutions near the boundary, A1 for 11

(a) Time = \(\frac{36}{x}\) hours.

(b) Speed is \(x - 3\) km/h. Time = \(\frac{36}{x - 3}\) hours.

(c)(i) Convert 24 minutes to hours: \(\frac{24}{60} = \frac{2}{5}\) hours. The equation is: \(\frac{36}{x-3} - \frac{36}{x} = \frac{2}{5}\). Multiply by \(5x(x-3)\): \(180x - 180(x-3) = 2x(x-3) \implies 180x - 180x + 540 = 2x^2 - 6x \implies 2x^2 - 6x - 540 = 0\). Dividing by 2: \(x^2 - 3x - 270 = 0\).

(c)(ii) Factor the quadratic: \((x - 18)(x + 15) = 0\). Hence, \(x = 18\) or \(x = -15\).

(c)(iii) Speed must be positive, so \(x = 18\). The return speed is \(18 - 3 = 15\) km/h. Return time = \(\frac{36}{15} = 2.4\) hours. Since \(0.4 \times 60 = 24\) minutes, the time taken is 2 hours and 24 minutes.

(a) A1 for \(36/x\)
(b) A1 for \(36/(x-3)\)
(c)(i) M1 for \(24/60\) or \(2/5\) hours, M1 for setting up the difference equation, M1 for clearing denominators, A1 for showing the simplification to the given quadratic equation
(c)(ii) M1 for factorizing or using quadratic formula, A1 for 18, A1 for -15
(c)(iii) M1 for identifying correct speed \(x=18\), M1 for dividing 36 by 15, M1 for converting 2.4 hours to hours and minutes, A1 for 2 hours and 24 minutes

(a) Use the cosine rule: \(AC^2 = 110^2 + 150^2 - 2(110)(150)\cos(72^\circ) = 12100 + 22500 - 33000\cos(72^\circ) \approx 34600 - 10197.56 = 24402.44\). So, \(AC = \sqrt{24402.44} \approx 156.21 \approx 156\) m.

(b) Use the sine rule: \(\frac{\sin(ACB)}{110} = \frac{\sin(72^\circ)}{156.21} \implies \sin(ACB) = \frac{110\sin(72^\circ)}{156.21} \approx 0.66967\). Angle \(ACB = \sin^{-1}(0.66967) \approx 42.04^\circ \approx 42.0^\circ\).

(c) Area = \(\frac{1}{2} a b \sin(C) = \frac{1}{2} (110)(150)\sin(72^\circ) = 8250\sin(72^\circ) \approx 7845.7 \approx 7850\) \(\text{m}^2\).

(d) \(ABP\) is a right-angled triangle at \(B\). \(\tan(\text{elevation}) = \frac{BP}{AB} = \frac{18}{110}\). Angle of elevation = \(\tan^{-1}\left(\frac{18}{110}\right) \approx 9.29^\circ\).

(e) The shortest path is the perpendicular height from \(B\) to \(AC\). Let this length be \(d\). Area = \(\frac{1}{2} \times AC \times d \implies 7845.7 = \frac{1}{2} \times 156.21 \times d \implies d = \frac{2 \times 7845.7}{156.21} \approx 100.45 \approx 100\) m (or using trigonometry: \(d = 110 \sin(180 - 72 - 42.04) \approx 100.46\) m).

(a) M1 for correct cosine rule equation, M1 for evaluating right-hand side, A1 for 156
(b) M1 for correct sine rule expression, M1 for isolating \(\sin(ACB)\), A1 for 42.0
(c) M1 for \(\frac{1}{2} a b \sin(C)\) formula, A1 for 7850
(d) M1 for recognizing right-angled triangle, M1 for using tangent ratio, A1 for 9.29
(e) M1 for equating area to \(\frac{1}{2} \times AC \times d\) or equivalent trigonometric method, A1 for 100

(a)(i) The common difference is 3, so the next term is \(17 + 3 = 20\).
(a)(ii) \(n\)-th term = \(a + (n-1)d = 5 + 3(n-1) = 3n + 2\).

(b)(i) This is a geometric sequence with ratio 3, so the next term is \(162 \times 3 = 486\).
(b)(ii) \(n\)-th term = \(a \times r^{n-1} = 2 \times 3^{n-1}\).

(c) The first differences of Sequence C are: 3, 7, 11, 15. The second differences are: 4, 4, 4. Since the second difference is constant, it is quadratic: \(u_n = a n^2 + b n + c\). Here, \(2a = 4 \implies a = 2\). Subtract \(2n^2\) from terms: \(4 - 2(1)^2 = 2\), \(7 - 2(2)^2 = -1\), \(14 - 2(3)^2 = -4\). This linear part decreases by 3 each time, so \(b = -3\). For \(n=1\), \(2(1)^2 - 3(1) + c = 4 \implies c = 5\). Thus, \(u_n = 2n^2 - 3n + 5\).

(d) Set \(n^3 - 6n^2 + 2n = 12 \implies n^3 - 6n^2 + 2n - 12 = 0\). Grouping terms: \(n^2(n-6) + 2(n-6) = 0 \implies (n^2 + 2)(n-6) = 0\). Since \(n^2 + 2 = 0\) has no real solutions, \(n = 6\).

(a)(i) B1 for 20
(a)(ii) B2 for \(3n + 2\) (B1 for \(3n + k\))
(b)(i) B1 for 486
(b)(ii) B2 for \(2 \times 3^{n-1}\) (B1 for any form \(k \times 3^n\))
(c) M1 for realizing second difference is 4, M1 for showing quadratic term coefficient is 2, M1 for linear part analysis, A1 for \(2n^2 - 3n + 5\)
(d) M1 for equating to 12, M1 for grouping/factoring cubic equation, A1 for \(n = 6\)

(a)(i) \(\vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\).
(a)(ii) \(\vec{OQ} = \vec{OA} + \frac{1}{2}\vec{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\).
(a)(iii) Since \(OP:PA = 2:1\), we have \(\vec{OP} = \frac{2}{3}\mathbf{a}\). Therefore, \(\vec{PQ} = \vec{PO} + \vec{OQ} = -\frac{2}{3}\mathbf{a} + \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = -\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b}\).

(b)(i) \(\vec{PR} = \vec{PO} + \vec{OR} = -\frac{2}{3}\mathbf{a} + k\mathbf{b}\).
(b)(ii) Since \(PR\) is parallel to \(\mathbf{b} - 2\mathbf{a}\), we have \(\vec{PR} = m(\mathbf{b} - 2\mathbf{a}) = -2m\mathbf{a} + m\mathbf{b}\) for some scalar \(m\). Comparing coefficients of \(\mathbf{a}\): \(-2m = -\frac{2}{3} \implies m = \frac{1}{3}\). Comparing coefficients of \(\mathbf{b}\): \(k = m = \frac{1}{3}\).

(c)(i) The position vector of \(A\) is \(\begin{pmatrix} 6 \\ 3 \end{pmatrix}\), so the coordinates are \((6, 3)\).
(c)(ii) \(\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 6 \\ 3 \end{pmatrix} = \begin{pmatrix} -8 \\ 2 \end{pmatrix}\). Magnitude is \(|\vec{AB}| = \sqrt{(-8)^2 + 2^2} = \sqrt{68} \approx 8.25\) units.

(a)(i) B1 for \(\mathbf{b} - \mathbf{a}\)
(a)(ii) M1 for correct vector sum path, A1 for \(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\)
(a)(iii) M1 for using \(\vec{OP} = \frac{2}{3}\mathbf{a}\), A1 for \(-\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b}\)
(b)(i) M1 for correct path, A1 for \(-\frac{2}{3}\mathbf{a} + k\mathbf{b}\)
(b)(ii) M1 for equating to \(m(\mathbf{b} - 2\mathbf{a})\), M1 for finding \(m = 1/3\), A1 for \(k = 1/3\)
(c)(i) B1 for \((6, 3)\)
(c)(ii) M1 for finding \(\vec{AB} = \begin{pmatrix} -8 \\ 2 \end{pmatrix}\) or using Pythagoras, A1 for 8.25

(a) (i)
Using Pythagoras' theorem for the cone:
\(\text{height}^2 + 6^2 = 10^2\)
\(\text{height}^2 = 100 - 36 = 64\)
\(\text{height} = \sqrt{64} = 8\text{ cm}\). (Shown)

(a) (ii)
\(V = \frac{1}{3}\pi r^2 H = \frac{1}{3} \times \pi \times 6^2 \times 8 = 96\pi\text{ cm}^3\).

(b) (i)
The volume of the cylinder is \(V_{\text{cylinder}} = \pi r^2 h = 36\pi h\).
Total volume of the toy is:
\(V_{\text{total}} = 96\pi + 36\pi h = 496\pi\)
Divide by \(\pi\):
\(96 + 36h = 496\)
\(36h = 400\)
\(h = \frac{400}{36} = \frac{100}{9} \approx 11.111...\text{ cm}\)
Thus, \(h = 11.1\text{ cm}\) correct to 3 significant figures. (Shown)

(b) (ii)
The total surface area includes:
- Curved surface area of the cone: \(\pi r l = \pi \times 6 \times 10 = 60\pi\text{ cm}^2\)
- Curved surface area of the cylinder: \(2\pi r h = 2 \times \pi \times 6 \times \frac{100}{9} = \frac{400\pi}{3}\text{ cm}^2\)
- Circular base of the cylinder: \(\pi r^2 = \pi \times 6^2 = 36\pi\text{ cm}^2\)
Total surface area:
\(A = 60\pi + \frac{400\pi}{3} + 36\pi = 96\pi + \frac{400\pi}{3} = \frac{688\pi}{3} \approx 720.47\text{ cm}^2\).
Using the 3 s.f. value \(h = 11.1\):
\(A = 60\pi + 2\pi(6)(11.1) + 36\pi = 229.2\pi \approx 720.05\text{ cm}^2\).
To 3 significant figures, both calculations give \(720\text{ cm}^2\).

(c)
Total volume of the toy: \(496\pi \approx 1558.23\text{ cm}^3\).
\(\text{Mass} = \text{Volume} \times \text{Density} = 1558.23\text{ cm}^3 \times 0.75\text{ g/cm}^3 = 1168.67\text{ g}\).
Converting to kilograms:
\(\text{Mass} = \frac{1168.67}{1000} \approx 1.17\text{ kg}\) (to 3 s.f.).

(a) (i)
M1 for \(10^2 - 6^2\) or \(h^2 + 6^2 = 10^2\)
A1 for 8 with clear working shown.

(a) (ii)
M1 for \(\frac{1}{3} \times \pi \times 6^2 \times 8\)
A1 for \(96\pi\)

(b) (i)
M1 for \(96\pi + \pi \times 6^2 \times h = 496\pi\)
M1 for \(36h = 400\) or \(h = \frac{400}{36}\)
A1 for \(11.11...\) and concluding \(11.1\)

(b) (ii)
M1 for \(\pi \times 6 \times 10\) (\(=60\pi\) or \(188.5\))
M1 for \(2 \times \pi \times 6 \times \frac{100}{9}\) or \(2 \times \pi \times 6 \times 11.1\) (\(= \frac{400\pi}{3}\) or \(133.2\pi\) or \(418.5\) to \(418.9\))
M1 for \(\pi \times 6^2\) (\(=36\pi\) or \(113.1\))
A1 for 720 or 720.05 to 720.5

(c)
M1 for \(496\pi \times 0.75 \div 1000\) or \((1558 \text{ to } 1560) \times 0.75 \div 1000\)
A1 for 1.17 or 1.168 to 1.172

(a)
Factorise the numerator:
\(2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2)\)
Factorise the denominator:
\(x^2 + 5x + 6 = (x + 2)(x + 3)\)
Simplify the fraction:
\(\frac{2(x - 2)(x + 2)}{(x + 2)(x + 3)} = \frac{2(x - 2)}{x + 3} = \frac{2x - 4}{x + 3}\).

(b)
Multiply to find a common denominator:
\(\frac{3(x - 2) - 1(2x + 1)}{(2x + 1)(x - 2)} = 2\)
Expand numerator and denominator:
\(\frac{3x - 6 - 2x - 1}{2x^2 - 4x + x - 2} = 2\)
\(\frac{x - 7}{2x^2 - 3x - 2} = 2\)
Multiply both sides by the denominator:
\(x - 7 = 2(2x^2 - 3x - 2)\)
\(x - 7 = 4x^2 - 6x - 4\)
Rearrange to form a quadratic equation equal to 0:
\(4x^2 - 7x + 3 = 0\)
Factorise the quadratic expression:
\((4x - 3)(x - 1) = 0\)
So, \(4x - 3 = 0 \implies x = 0.75\)
or \(x - 1 = 0 \implies x = 1\).

(c)
Multiply by the denominator \(3t + w\):
\(y(3t + w) = 2t - 5\)
Expand the bracket:
\(3yt + yw = 2t - 5\)
Rearrange to bring all terms containing \(t\) to one side:
\(yw + 5 = 2t - 3yt\)
Factorise out \(t\):
\(yw + 5 = t(2 - 3y)\)
Divide by \((2 - 3y)\):
\(t = \frac{yw + 5}{2 - 3y}\)
(Alternatively: \(t = \frac{-yw - 5}{3y - 2}\)).

(a)
M1 for \(2(x - 2)(x + 2)\) or \(2(x^2 - 4)\)
M1 for \((x + 2)(x + 3)\)
A1 for \(\frac{2(x - 2)}{x + 3}\) or \(\frac{2x - 4}{x + 3}\)

(b)
M1 for \(3(x - 2) - (2x + 1)\) as a numerator on LHS
M1 for \(2x^2 - 3x - 2\) as expanded denominator on LHS
M1 for forming quadratic equation: \(4x^2 - 7x + 3 = 0\) (or equivalent)
M1 for solving their quadratic, e.g., factorising to \((4x - 3)(x - 1)\) or using the quadratic formula correctly
A1 for \(x = 0.75\) (or \(\frac{3}{4}\)) and \(x = 1\) (both answers must be correct)

(c)
M1 for \(y(3t + w) = 2t - 5\)
M1 for \(3yt + yw = 2t - 5\)
M1 for isolating terms with \(t\) on one side, e.g., \(2t - 3yt = yw + 5\) or \(3yt - 2t = -yw - 5\)
M1 for factorising \(t\): \(t(2 - 3y) = yw + 5\) or \(t(3y - 2) = -yw - 5\)
A1 for \(t = \frac{yw + 5}{2 - 3y}\) or \(t = \frac{-yw - 5}{3y - 2}\)