2023 Cambridge IGCSE Mathematics (0580) 模擬試題連答案詳解

Factorise the numerator:
\[2x^2 - 5x - 12 = (2x + 3)(x - 4)\]
Factorise the denominator:
\[3x^2 - 14x + 8 = (3x - 2)(x - 4)\]
Simplify by cancelling the common factor \((x - 4)\):
\[\frac{(2x + 3)(x - 4)}{(3x - 2)(x - 4)} = \frac{2x + 3}{3x - 2}\]

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

Express both sides as powers of 2:
\[16 = 2^4 \quad \text{and} \quad 8 = 2^3\]
\[(2^4)^{3x - 1} = (2^3)^{2x + 5}\]
\[2^{4(3x - 1)} = 2^{3(2x + 5)}\]
Equate the exponents:
\[4(3x - 1) = 3(2x + 5)\]
\[12x - 4 = 6x + 15\]
\[12x - 6x = 15 + 4\]
\[6x = 19\]
\[x = \frac{19}{6} = 3\frac{1}{6}\]

M1 for expressing both sides in terms of base 2 (or a common base), e.g. \(2^{4(3x-1)}\) and \(2^{3(2x+5)}\)
M1 for equating indices and linear expansion: \(12x - 4 = 6x + 15\) oe
A1 for \(x = \frac{19}{6}\) or \(3\frac{1}{6}\) or \(3.17\) or \(3.166\ldots\)

Upper bound of length = \(84 + 0.5 = 84.5\text{ m}\)
Upper bound of width = \(55 + 0.5 = 55.5\text{ m}\)

Upper bound of perimeter:
\[2 \times (84.5 + 55.5) = 2 \times 140 = 280\text{ m}\]

B1 for length UB = \(84.5\) or width UB = \(55.5\) seen
B1 for \(280\) cao

The gradient of the given line is \(-\frac{1}{4}\).
The gradient of the perpendicular line is \(m = -\frac{1}{-\frac{1}{4}} = 4\).

Using the point-slope form with \((3, -2)\):
\[y - (-2) = 4(x - 3)\]
\[y + 2 = 4x - 12\]
\[y = 4x - 14\]

M1 for gradient \(m = 4\) soi
A1 for \(y = 4x - 14\) cao

Use the cosine rule:
\[AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\]
\[AC^2 = 7.4^2 + 9.6^2 - 2(7.4)(9.6)\cos(62^\circ)\]
\[AC^2 = 54.76 + 92.16 - 142.08\cos(62^\circ)\]
\[AC^2 = 146.92 - 66.7025\ldots = 80.2174\ldots\]
\[AC = \sqrt{80.2174\ldots} \approx 8.9564\text{ cm}\]
Giving the answer correct to 3 significant figures: \(8.96\text{ cm}\).

M1 for correct substitution into cosine rule: \(7.4^2 + 9.6^2 - 2(7.4)(9.6)\cos(62)\)
A1 for \(80.2\) or \(80.21\) to \(80.22\)
A1 for \(8.96\) or \(8.956\ldots\)

1. Factorise the numerator quadratic:
$$2x^2 - x - 15 = (2x + 5)(x - 3)$$

2. Factorise the denominator as a difference of two squares:
$$4x^2 - 25 = (2x - 5)(2x + 5)$$

3. Cancel the common factor of \((2x + 5)\):
$$\frac{(2x + 5)(x - 3)}{(2x - 5)(2x + 5)} = \frac{x - 3}{2x - 5}$$

M1 for \((2x + 5)(x - 3)\)
M1 for \((2x - 5)(2x + 5)\)
A1 for \(\dfrac{x - 3}{2x - 5}\) or \(\dfrac{-x + 3}{5 - 2x}\) final answer

1. Find the upper bounds for the dimensions:
$$\text{Upper bound of length} = 84 + 0.5 = 84.5\text{ m}$$
$$\text{Upper bound of width} = 55 + 0.5 = 55.5\text{ m}$$

2. Calculate the upper bound of the perimeter:
$$\text{Perimeter} = 2 \times (84.5 + 55.5) = 2 \times 140 = 280\text{ m}$$

B1 for length upper bound \(84.5\) or width upper bound \(55.5\) seen
B1 for \(280\) cao

1. Express both sides as powers of base 2:
$$32 = 2^5 \implies 32^p = 2^{5p}$$
$$8 = 2^3 \implies 8^{p+2} = (2^3)^{p+2} = 2^{3(p+2)} = 2^{3p+6}$$

2. Equate the exponents:
$$5p = 3p + 6$$
$$2p = 6$$
$$p = 3$$

M1 for writing both sides with a common base, e.g. \(2^{5p} = 2^{3(p+2)}\) or \(5p = 3(p+2)\) oe
A1 for \(p = 3\) cao

1. Use the cosine rule:
$$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)$$
$$AC^2 = 7.4^2 + 9.6^2 - 2(7.4)(9.6)\cos(64^\circ)$$

2. Evaluate each term:
$$7.4^2 = 54.76$$
$$9.6^2 = 92.16$$
$$2 \times 7.4 \times 9.6 \times \cos(64^\circ) \approx 62.2838$$

3. Calculate \(AC^2\) and \(AC\):
$$AC^2 = 54.76 + 92.16 - 62.2838 = 84.6362$$
$$AC = \sqrt{84.6362} \approx 9.1998\text{ cm}$$

Correct to 3 significant figures, \(AC = 9.20\text{ cm}\).

M1 for implicit cosine rule statement: \(AC^2 = 7.4^2 + 9.6^2 - 2(7.4)(9.6)\cos(64)\) oe
A1 for \(84.6\) or \(84.63\) to \(84.64\)
A1 for \(9.20\) or \(9.199\) to \(9.200\)

1. The gradient of the given line is \(3\).
2. The gradient of the perpendicular line is the negative reciprocal:
$$m = -\frac{1}{3}$$

3. Use the point-slope formula with \((x_1, y_1) = (6, 2)\):
$$y - 2 = -\frac{1}{3}(x - 6)$$
$$y - 2 = -\frac{1}{3}x + 2$$
$$y = -\frac{1}{3}x + 4$$

M1 for gradient \(m = -\frac{1}{3}\) soi
M1 for substituting \((6, 2)\) into \(y = (\text{their } m)x + c\) oe
A1 for \(y = -\frac{1}{3}x + 4\) or \(y = -0.333x + 4\) oe

Factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Factorise the denominator using difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Simplify by cancelling the common factor \((2x + 1)\): \(\dfrac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \dfrac{x - 3}{2x - 1}\).

M1 for \((2x + 1)(x - 3)\) seen. M1 for \((2x - 1)(2x + 1)\) seen. A1 for \(\dfrac{x - 3}{2x - 1}\) cao.

Upper bound of mass = \(650 + 5 = 655\text{ g}\). Lower bound of volume = \(80 - 1 = 79\text{ cm}^3\). Upper bound of density = \(\dfrac{\text{Upper bound of mass}}{\text{Lower bound of volume}} = \dfrac{655}{79} \approx 8.291139...\text{ g/cm}^3\).

M1 for \(655\) seen or \(79\) seen, or \(\dfrac{\text{UB of mass}}{\text{LB of volume}}\) with one correct bound. A1 for \(8.29\) or \(8.291\text{ to }8.29114\) or \(\dfrac{655}{79}\) or \(8\dfrac{23}{79}\).

1. Find the total surface area of the solid cone:
\[\text{Total Surface Area}_{\text{cone}} = \pi r^2 + \pi r l = \pi r^2 + \pi r (3r) = 4\pi r^2\]

2. Find the total surface area of the solid hemisphere (curved surface plus circular base):
\[\text{Total Surface Area}_{\text{hemisphere}} = 2\pi R^2 + \pi R^2 = 3\pi R^2\]

3. Equate the two surface areas and solve for \(R\):
\[3\pi R^2 = 4\pi r^2\]
\[3R^2 = 4r^2\]
\[R^2 = \frac{4}{3}r^2\]
\[R = \sqrt{\frac{4}{3}}r = \frac{2}{\sqrt{3}}r = \frac{2\sqrt{3}}{3}r\]

M1 for total surface area of cone: \(\pi r^2 + \pi r(3r) [= 4\pi r^2]\)
M1 for total surface area of hemisphere: \(2\pi R^2 + \pi R^2 [= 3\pi R^2]\)
A1 for \(R = \frac{2\sqrt{3}}{3}r\) oe (e.g. \(\frac{2}{\sqrt{3}}r\) or \(\sqrt{\frac{4}{3}}r\))

1. Use the area formula for a triangle to find \(\sin(\angle PQR)\):
\[\text{Area} = \frac{1}{2} a b \sin(\angle PQR)\]
\[35.2 = \frac{1}{2} \times 8 \times 11 \times \sin(\angle PQR) = 44 \sin(\angle PQR)\]
\[\sin(\angle PQR) = \frac{35.2}{44} = 0.8\]

2. Find the obtuse angle \(\angle PQR\):
\[\angle PQR = 180^\circ - \arcsin(0.8) = 180^\circ - 53.1301^\circ = 126.8699^\circ\]

3. Use the cosine rule to calculate \(PR\):
\[PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR)\]
\[PR^2 = 8^2 + 11^2 - 2(8)(11)\cos(126.8699^\circ)\]
Since \(\cos(126.8699^\circ) = -0.6\):
\[PR^2 = 64 + 121 - 176(-0.6) = 185 + 105.6 = 290.6\]
\[PR = \sqrt{290.6} \approx 17.04699\text{ cm}\]
Correct to 3 significant figures, \(PR = 17.0\text{ cm}\).

M1 for \(\frac{1}{2} \times 8 \times 11 \times \sin(\angle PQR) = 35.2\) oe
B1 for \(\angle PQR = 126.87^\circ\) or \(126.9^\circ\) or \(126.8699...^\circ\)
M1 for correct use of cosine rule: \(8^2 + 11^2 - 2(8)(11)\cos(\text{their obtuse angle})\)
A1 for \(17.0\) or \(17.04\) to \(17.05\) nfww

1. Find the midpoint of \(AB\):
\[\text{Midpoint} = \left(\frac{-2 + 4}{2}, \frac{9 + (-3)}{2}\right) = \left(\frac{2}{2}, \frac{6}{2}\right) = (1, 3)\]

2. Find the gradient of line \(L_1\):
\[m_1 = \frac{-3 - 9}{4 - (-2)} = \frac{-12}{6} = -2\]

3. Determine the gradient of \(L_2\) (perpendicular to \(L_1\)):
\[m_2 = -\frac{1}{-2} = \frac{1}{2}\]

4. Use the point-slope form with \((1, 3)\) and \(m = \frac{1}{2}\):
\[y - 3 = \frac{1}{2}(x - 1)\]
\[y = \frac{1}{2}x - \frac{1}{2} + 3\]
\[y = \frac{1}{2}x + \frac{5}{2} \quad \text{or} \quad y = 0.5x + 2.5\]

B1 for midpoint \((1, 3)\) soi
M1 for gradient of \(AB = \frac{-3 - 9}{4 - (-2)} [= -2]\)
M1 for perpendicular gradient \(= -\frac{1}{\text{their } m_1} [= \frac{1}{2}]\)
A1 for \(y = \frac{1}{2}x + \frac{5}{2}\) or \(y = 0.5x + 2.5\) oe

1. Determine the upper and lower bounds of distance and time:
- Distance: \(450\text{ m}\) to nearest \(10\text{ m} \implies \text{Upper Bound } d_{\text{UB}} = 450 + 5 = 455\text{ m}\)
- Time: \(18.4\text{ s}\) to \(1\text{ d.p.} \implies \text{Lower Bound } t_{\text{LB}} = 18.4 - 0.05 = 18.35\text{ s}\)

2. Find the upper bound for average speed in \(\text{m/s}\):
\[\text{Speed}_{\text{UB}} = \frac{d_{\text{UB}}}{t_{\text{LB}}} = \frac{455}{18.35}\text{ m/s}\]

3. Convert the speed from \(\text{m/s}\) to \(\text{km/h}\) by multiplying by \(3.6\):
\[\text{Speed}_{\text{UB}} = \frac{455}{18.35} \times 3.6 = \frac{1638}{18.35} \approx 89.2643...\text{ km/h}\]

4. Give the answer to 3 significant figures: \(89.3\text{ km/h}\).

B1 for \(455\) or \(18.35\) seen
M1 for \(\frac{\text{their } 455}{\text{their } 18.35} \times 3.6\) or \(\frac{\text{their } 455 / 1000}{\text{their } 18.35 / 3600}\)
A1 for \(89.3\) or \(89.26\) to \(89.27\) nfww

1. Express the vector \(\mathbf{p} - \mathbf{q}\) in component form:
\[\mathbf{p} - \mathbf{q} = \begin{pmatrix} 2k - 3 \\ 7 - (k + 1) \end{pmatrix} = \begin{pmatrix} 2k - 3 \\ 6 - k \end{pmatrix}\]

2. Use the magnitude formula:
\[|\mathbf{p} - \mathbf{q}|^2 = (2k - 3)^2 + (6 - k)^2 = 50\]

3. Expand and simplify the quadratic equation:
\[(4k^2 - 12k + 9) + (36 - 12k + k^2) = 50\]
\[5k^2 - 24k + 45 = 50\]
\[5k^2 - 24k - 5 = 0\]

4. Factorise the quadratic equation:
\[(5k + 1)(k - 5) = 0\]
\[k = -\frac{1}{5} \quad \text{or} \quad k = 5\]

Since \(k > 0\), \(k = 5\).

B1 for \(\mathbf{p} - \mathbf{q} = \begin{pmatrix} 2k - 3 \\ 6 - k \end{pmatrix}\) seen or implied
M1 for \((2k - 3)^2 + (6 - k)^2 = (\sqrt{50})^2\)
M1 for simplifying to \(5k^2 - 24k - 5 = 0\) and attempting to solve by factorisation or quadratic formula
A1 for \(k = 5\) (and rejecting \(k = -0.2\))

To find the lower bound for density, we divide the lower bound of the mass by the upper bound of the volume:

1. Find the bounds for each measurement:
- Mass: \(240\text{ g}\) to the nearest \(10\text{ g} \implies \text{Lower bound} = 240 - 5 = 235\text{ g}\)
- Length: \(8.4\text{ cm}\) to 1 d.p. \(\implies \text{Upper bound} = 8.4 + 0.05 = 8.45\text{ cm}\)
- Width: \(5.6\text{ cm}\) to 1 d.p. \(\implies \text{Upper bound} = 5.6 + 0.05 = 5.65\text{ cm}\)
- Height: \(0.5\text{ cm}\) to 1 d.p. \(\implies \text{Upper bound} = 0.5 + 0.05 = 0.55\text{ cm}\)

2. Calculate the upper bound of the volume:
\[\text{Volume}_{\text{upper}} = 8.45 \times 5.65 \times 0.55 = 26.258375\text{ cm}^3\]

3. Calculate the lower bound for density:
\[\text{Density}_{\text{lower}} = \dfrac{235}{26.258375} \approx 8.9495256\dots\text{ g/cm}^3\]

Correct to 3 significant figures, the density is \(8.95\text{ g/cm}^3\).

B1 for at least two of the correct bounds seen: \(235\), \(8.45\), \(5.65\), \(0.55\)
M1 for \(\dfrac{\text{LB mass}}{\text{UB length} \times \text{UB width} \times \text{UB height}}\) with \(235 \le \text{LB mass} < 240\) and \(8.4 < \text{UB length} \le 8.45\), \(5.6 < \text{UB width} \le 5.65\), \(0.5 < \text{UB height} \le 0.55\)
A1 for \(8.95\) or \(8.949\dots\) to \(8.950\dots\)

1. Find the derivative \(\frac{\mathrm{d}y}{\mathrm{d}x}\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 18x + 12\]

2. Set the derivative equal to 0 for stationary points:
\[6x^2 - 18x + 12 = 0\]
\[x^2 - 3x + 2 = 0\]
\[(x - 1)(x - 2) = 0\]
\[x = 1 \quad \text{or} \quad x = 2\]

3. Find the corresponding \(y\)-coordinates:
For \(x = 1\):
\[y = 2(1)^3 - 9(1)^2 + 12(1) - 5 = 2 - 9 + 12 - 5 = 0\]
For \(x = 2\):
\[y = 2(2)^3 - 9(2)^2 + 12(2) - 5 = 16 - 36 + 24 - 5 = -1\]

The coordinates of the stationary points are \((1, 0)\) and \((2, -1)\).

M1 for derivative with at least two terms correct: \(6x^2 - 18x + 12\)
M1 for setting their derivative to 0 and solving for \(x\) (giving \(x = 1\) and \(x = 2\))
A1 for both points correct: \((1, 0)\) and \((2, -1)\) cao

1. Find the \(y\)-coordinate at \(x = 3\):
\[y = (3)^3 - 4(3)^2 + 7 = 27 - 36 + 7 = -2\]
So the point of tangency is \((3, -2)\).

2. Differentiate the equation to find the gradient function:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 8x\]

3. Evaluate the gradient at \(x = 3\):
\[m = 3(3)^2 - 8(3) = 27 - 24 = 3\]

4. Use the point-slope form \(y - y_1 = m(x - x_1)\):
\[y - (-2) = 3(x - 3)\]
\[y + 2 = 3x - 9\]
\[y = 3x - 11\]

M1 for correct differentiation: \(3x^2 - 8x\) (or at least one term correct)
M1 for substituting \(x = 3\) into their derivative to find \(m = 3\) AND finding \(y = -2\)
A1 for \(y = 3x - 11\) cao

1. Rewrite the equation in index form:
\[y = 16x^{-1} + x^2\]

2. Differentiate with respect to \(x\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -16x^{-2} + 2x = -\frac{16}{x^2} + 2x\]

3. Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) to find the turning point:
\[-\frac{16}{x^2} + 2x = 0\]
\[2x = \frac{16}{x^2}\]
\[2x^3 = 16\]
\[x^3 = 8 \implies x = 2\]

4. Calculate the corresponding \(y\)-coordinate:
\[y = \frac{16}{2} + 2^2 = 8 + 4 = 12\]

The coordinates of the turning point are \((2, 12)\).

M1 for correct differentiation of at least one term: \(-16x^{-2}\) or \(2x\)
M1 for equating their derivative to 0 and solving for \(x\) to obtain \(x = 2\)
A1 for \((2, 12)\) cao

(a)(i) The total height is \(h + r = 18\), so \(h = 18 - r\).

(a)(ii) Volume of cylinder \(= \pi r^2 h = \pi r^2 (18 - r) = 18\pi r^2 - \pi r^3\).
Volume of hemisphere \(= \frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3\).
Total Volume \(V = (18\pi r^2 - \pi r^3) + \frac{2}{3}\pi r^3 = 18\pi r^2 - \frac{1}{3}\pi r^3\).

(b)(i) For \(r = 6\):
\(V = 18\pi(6)^2 - \frac{1}{3}\pi(6)^3 = 18\pi(36) - \frac{1}{3}\pi(216) = 648\pi - 72\pi = 576\pi\text{ cm}^3\).

(b)(ii) \(h = 18 - 6 = 12\text{ cm}\).
Surface area consists of:
- Flat circular base: \(\pi r^2 = \pi (6^2) = 36\pi\)
- Curved surface of cylinder: \(2\pi r h = 2\pi(6)(12) = 144\pi\)
- Curved surface of hemisphere: \(2\pi r^2 = 2\pi(6^2) = 72\pi\)
Total surface area \(= 36\pi + 144\pi + 72\pi = 252\pi \approx 791.68\text{ cm}^2\) (or \(792\text{ cm}^2\) to 3 s.f.).

(c) Volume of one small sphere \(= \frac{4}{3}\pi (1.8)^3 = \frac{4}{3}\pi (5.832) = 7.776\pi\text{ cm}^3\).
Number of spheres \(= \frac{576\pi}{7.776\pi} = \frac{576}{7.776} \approx 74.07\).
Maximum complete spheres \(= 74\).

(a)(i) B1 for \(h = 18 - r\) oe
(a)(ii) M1 for cylinder volume \(= \pi r^2(18 - r)\) soi
M1 for hemisphere volume \(= \frac{2}{3}\pi r^3\) soi
A1 for fully correct algebraic simplification leading to \(18\pi r^2 - \frac{1}{3}\pi r^3\) with no errors seen
(b)(i) M1 for substituting \(r = 6\) into formula
A1 for \(576\pi\) (or \(1809.5...\) to 1810)
(b)(ii) M1 for curved surface of cylinder \(= 2\pi(6)(12)\) or \(144\pi\)
M1 for curved hemisphere \(= 2\pi(6)^2\) or \(72\pi\)
M1 for base area \(= \pi(6)^2\) or \(36\pi\)
A1 for \(792\) or \(791.68\dots\) or \(252\pi\)
(c) M1 for \(\frac{4}{3}\pi(1.8)^3\) soi (\(24.429\dots\) or \(7.776\pi\))
M1 for \(\frac{\text{their } 576\pi}{\text{their volume of small sphere}}\)
M1 for obtaining \(74.07\dots\)
A1.5 for \(74\) (must truncate to an integer)

(a) \(\text{Angle } BAC = 140^\circ - 65^\circ = 75^\circ\).

(b) Using the cosine rule in \(\triangle ABC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(75^\circ)\)
\(BC^2 = 42^2 + 68^2 - 2(42)(68)\cos(75^\circ) = 1764 + 4624 - 5712(0.258819) = 6388 - 1478.374 = 4909.626\)
\(BC = \sqrt{4909.626} \approx 70.0687\text{ km}\) (or \(70.1\text{ km}\) to 3 s.f.).

(c) Using the sine rule:
\(\frac{\sin(\angle ABC)}{68} = \frac{\sin(75^\circ)}{70.0687}\)
\(\sin(\angle ABC) = \frac{68 \times \sin(75^\circ)}{70.0687} = \frac{68 \times 0.965926}{70.0687} \approx 0.9374\)
\(\angle ABC = \sin^{-1}(0.9374) \approx 69.62^\circ\) (or \(69.6^\circ\)).

(d) Back-bearing from \(B\) to \(A = 65^\circ + 180^\circ = 245^\circ\).
Bearing of \(C\) from \(B = 245^\circ - 69.62^\circ = 175.38^\circ \approx 175.4^\circ\) (or \(175^\circ\)).

(e) Height \(H = AB \times \tan(14^\circ) = 42\text{ km} \times \tan(14^\circ) = 42 \times 0.249328 = 10.4718\text{ km}\).
In metres: \(10.4718 \times 1000 = 10472\text{ m} \approx 10500\text{ m}\) (to 3 s.f.) or \(10470\text{ m}\).

(a) B1 for \(140 - 65 = 75^\circ\)
(b) M1 for \(42^2 + 68^2 - 2(42)(68)\cos(75^\circ)\)
M1 for correct order of operations (\(6388 - 1478.37... = 4909.6...\))
M1 for \(\sqrt{\text{their } 4909.6}\)
A1 for \(70.1\) or \(70.068\dots\) to \(70.07\)
(c) M1 for \(\frac{\sin(\angle ABC)}{68} = \frac{\sin(75^\circ)}{\text{their } BC}\) or correct cosine rule for angle
M1.5 for \(\sin(\angle ABC) = 0.937\dots\)
A1 for \(69.6^\circ\) or \(69.62^\circ\dots\)
(d) M1 for back bearing of \(A\) from \(B = 245^\circ\) soi
M1 for \(245^\circ - \text{their } 69.62^\circ\)
A1 for \(175.4^\circ\) or \(175.38^\circ\dots\)
(e) M1 for \(\tan(14^\circ) = \frac{H}{42}\) oe
M1 for multiplying by \(1000\) to convert \(\text{km}\) to \(\text{m}\)
A1 for \(10500\) or \(10470\) to \(10472\)

(a)(i) For \(N = 120\), median is at cumulative frequency \(60\).
Between \((90, 38)\) and \((110, 76)\):
Using linear interpolation: \(t = 90 + \frac{60 - 38}{76 - 38} \times (110 - 90) = 90 + \frac{22}{38} \times 20 = 90 + 11.58 = 101.58 \approx 102\text{ mins}\).

(a)(ii) Lower quartile \(Q_1\) at CF \(= 30\):
Between \((60, 12)\) and \((90, 38)\): \(Q_1 = 60 + \frac{30 - 12}{38 - 12} \times 30 = 60 + \frac{18}{26} \times 30 = 60 + 20.77 = 80.77\text{ mins}\).
Upper quartile \(Q_3\) at CF \(= 90\):
Between \((110, 76)\) and \((130, 102)\): \(Q_3 = 110 + \frac{90 - 76}{102 - 76} \times 20 = 110 + \frac{14}{26} \times 20 = 110 + 10.77 = 120.77\text{ mins}\).
\(\text{IQR} = Q_3 - Q_1 = 120.77 - 80.77 = 40.0\text{ mins}\) (or \(36\text{ to } 42\text{ mins}\)).

(a)(iii) At \(t = 140\), CF is between \((130, 102)\) and \((150, 114)\):
\(\text{CF} = 102 + \frac{140 - 130}{150 - 130} \times (114 - 102) = 102 + \frac{10}{20} \times 12 = 102 + 6 = 108\).
Number greater than \(140 = 120 - 108 = 12\).

(b)(i) Frequencies:
\(60 < t \leqslant 90: 38 - 12 = 26\)
\(90 < t \leqslant 110: 76 - 38 = 38\)
\(110 < t \leqslant 130: 102 - 76 = 26\)
\(130 < t \leqslant 180: 120 - 102 = 18\).

(b)(ii) Midpoints \(x\): \(30, 75, 100, 120, 155\).
\(\sum f x = 12(30) + 26(75) + 38(100) + 26(120) + 18(155) = 360 + 1950 + 3800 + 3120 + 2790 = 12020\).
\(\text{Mean} = \frac{12020}{120} \approx 100.17 \approx 100\text{ mins}\) (or \(100.2\)).

(c) \(P(\text{both } \leqslant 60) = \frac{12}{120} \times \frac{11}{119} = \frac{1}{10} \times \frac{11}{119} = \frac{11}{1190} \approx 0.00924\).

(a)(i) M1 for finding value at cumulative frequency \(60\)
A1 for answer in range \([101, 103]\)
(a)(ii) M1 for \(Q_1\) at CF \(30\) and \(Q_3\) at CF \(90\)
M1 for subtraction \(Q_3 - Q_1\)
A1 for answer in range \([38, 42]\)
(a)(iii) M1.5 for identifying cumulative frequency at \(t = 140\) as \(108\)
A1 for \(120 - 108 = 12\)
(b)(i) B2 for all three missing values correct: \(26, 38, 18\) (B1 for 1 or 2 correct)
(b)(ii) M1 for at least 3 correct midpoints seen (\(30, 75, 100, 120, 155\))
M1 for \(\sum f x\) with their midpoints and frequencies
M1 for \(\frac{\sum f x}{120}\)
A1 for \(100\) or \(100.2\) nfww
(c) B1 for \(\frac{11}{1190}\) or \(0.00924\) oe

(a)(i) \(\text{Time} = \frac{180}{x}\) hours.

(a)(ii) \(\text{Time} = \frac{180}{x - 15}\) hours.

(b)(i) \(36\text{ minutes} = \frac{36}{60} = \frac{3}{5}\text{ hours}\).
\(\frac{180}{x - 15} - \frac{180}{x} = \frac{3}{5}\)
Multiply throughout by \(5x(x - 15)\):
\(5 \times 180 x - 5 \times 180 (x - 15) = 3 x (x - 15)\)
\(900x - 900x + 13500 = 3x^2 - 45x\)
\(3x^2 - 45x - 13500 = 0\)
Divide throughout by 3:
\(x^2 - 15x - 4500 = 0\).

(b)(ii) Two numbers that multiply to \(-4500\) and add to \(-15\) are \(-75\) and \(60\).
\((x - 75)(x + 60)\).

(b)(iii) \(x - 75 = 0 \implies x = 75\) or \(x + 60 = 0 \implies x = -60\).
Since speed must be positive, \(x = 75\text{ km/h}\).

(c) Outward time \(= \frac{180}{75} = 2.4\text{ hours}\).
Return time \(= \frac{180}{75 - 15} = \frac{180}{60} = 3\text{ hours}\).
Total time \(= 2.4 + 3 = 5.4\text{ hours}\).
\(0.4\text{ hours} = 0.4 \times 60 = 24\text{ minutes}\).
Total time \(= 5\text{ hours } 24\text{ minutes}\).

(a)(i) B1 for \(\frac{180}{x}\)
(a)(ii) B1 for \(\frac{180}{x - 15}\)
(b)(i) M1 for \(\frac{36}{60}\) or \(\frac{3}{5}\) or \(0.6\) seen
M1 for \(\frac{180}{x - 15} - \frac{180}{x} = \frac{3}{5}\) oe
M1.5 for correctly clearing fractions: \(180x - 180(x - 15) = 0.6x(x - 15)\) or \(900x - 900(x - 15) = 3x(x - 15)\)
A1 for fully correct algebraic working leading to \(x^2 - 15x - 4500 = 0\) with no steps omitted
(b)(ii) M1 for \((x + a)(x + b)\) where \(ab = -4500\) or \(a + b = -15\)
A1 for \((x - 75)(x + 60)\)
(b)(iii) B1 for \(x = 75\) (and \(x = -60\))
B1 for selecting positive value \(x = 75\) with reason or rejecting \(-60\)
(c) M1 for \(\frac{180}{\text{their } 75}\) (\(= 2.4\))
M1 for \(\frac{180}{\text{their } 60}\) (\(= 3\))
M1 for adding times to get \(5.4\text{ hours}\)
A1 for \(5\text{ hours } 24\text{ minutes}\)

(a) \(P(\text{Red from } A) = \frac{5}{8}\).

(b)(i) \(P(\text{Bag } A) = \frac{2}{6} = \frac{1}{3}\).
\(P(\text{Bag } B) = \frac{4}{6} = \frac{2}{3}\).
\(P(\text{Red}) = P(A) \times P(\text{Red}|A) + P(B) \times P(\text{Red}|B) = \left(\frac{1}{3} \times \frac{5}{8}\right) + \left(\frac{2}{3} \times \frac{4}{10}\right) = \frac{5}{24} + \frac{8}{30} = \frac{5}{24} + \frac{4}{15} = \frac{25 + 32}{120} = \frac{57}{120} = \frac{19}{40}\) (or \(0.475\)).

(b)(ii) \(P(A | \text{Red}) = \frac{P(A \cap \text{Red})}{P(\text{Red})} = \frac{\frac{5}{24}}{\frac{19}{40}} = \frac{5}{24} \times \frac{40}{19} = \frac{5 \times 5}{3 \times 19} = \frac{25}{57} \approx 0.439\).

(c)(i) Total marbles in Bag \(B = 10\).
\(P(\text{Both Red}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}\).
\(P(\text{Both Blue}) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{5}{15}\).
\(P(\text{Same Colour}) = \frac{2}{15} + \frac{5}{15} = \frac{7}{15} \approx 0.467\).

(c)(ii) \(P(\text{At least one blue}) = 1 - P(\text{Both Red}) = 1 - \frac{2}{15} = \frac{13}{15} \approx 0.867\).

(a) B1 for \(\frac{5}{8}\) or \(0.625\)
(b)(i) M1 for \(\frac{1}{3} \times \frac{5}{8}\) (\(=\frac{5}{24}\))
M1 for \(\frac{2}{3} \times \frac{4}{10}\) (\(=\frac{4}{15}\))
M0.5 for adding the two probabilities
A1 for \(\frac{19}{40}\) or \(0.475\)
(b)(ii) M1.5 for \(\frac{\text{their } (5/24)}{\text{their } (19/40)}\)
M0.5 for simplifying fraction
A1 for \(\frac{25}{57}\) or \(0.439\) or \(0.4385\dots\)
(c)(i) M1 for \(\frac{4}{10} \times \frac{3}{9}\) (\(=\frac{2}{15}\))
M1 for \(\frac{6}{10} \times \frac{5}{9}\) (\(=\frac{5}{15}\))
M0.5 for adding their two probabilities
A1 for \(\frac{7}{15}\) or \(0.467\) or \(0.4666\dots\)
(c)(ii) M2 for \(1 - \text{their } P(\text{Both Red})\) or \(\left(\frac{4}{10}\times\frac{6}{9}\right) + \left(\frac{6}{10}\times\frac{4}{9}\right) + \left(\frac{6}{10}\times\frac{5}{9}\right)\)
M0.5 for correct computation
A1 for \(\frac{13}{15}\) or \(0.867\) or \(0.8666\dots\)

(a) Using Pythagoras' theorem on the right-angled cross-section of the cone:
\(h_{\text{cone}} = \sqrt{l^2 - r^2} = \sqrt{7.5^2 - 4.5^2} = \sqrt{56.25 - 20.25} = \sqrt{36} = 6.0\text{ cm}\).

(b) Volume of cylinder = \(\pi r^2 h = \pi \times 4.5^2 \times 12.0 = 243\pi \approx 763.407\text{ cm}^3\).
Volume of cone = \(\frac{1}{3}\pi r^2 h_{\text{cone}} = \frac{1}{3}\pi \times 4.5^2 \times 6.0 = 40.5\pi \approx 127.235\text{ cm}^3\).
Total volume = \(243\pi + 40.5\pi = 283.5\pi \approx 890.64\text{ cm}^3\).
Correct to 1 decimal place: \(890.6\text{ cm}^3\).

(c) The exterior surface area consists of:
1. Base circle of cylinder: \(\pi r^2 = \pi \times 4.5^2 = 20.25\pi\)
2. Curved surface area of cylinder: \(2\pi r h = 2\pi \times 4.5 \times 12.0 = 108\pi\)
3. Curved surface area of cone: \(\pi r l = \pi \times 4.5 \times 7.5 = 33.75\pi\)
Total exterior surface area = \(20.25\pi + 108\pi + 33.75\pi = 162\pi \approx 508.938\text{ cm}^2\).
Correct to 3 significant figures: \(509\text{ cm}^2\) (or \(508.9\text{ cm}^2\)).

(d) Mass = \(\text{Density} \times \text{Volume} = 8.4 \times 890.6415 = 7481.39\text{ g}\).
In kilograms: \(\frac{7481.39}{1000} \approx 7.48\text{ kg}\).

(e) Let \(k\) be the linear scale factor.
Ratio of areas: \(k^2 = \frac{1100}{162\pi} = \frac{1100}{508.938} \approx 2.16136\).
\(k = \sqrt{2.16136} \approx 1.470156\).
Height of cylinder = \(12.0 \times 1.470156 \approx 17.6\text{ cm}\).

(a) M1 for \(\sqrt{7.5^2 - 4.5^2}\) or \(h^2 + 4.5^2 = 7.5^2\)
A1 for complete evaluation showing \(\sqrt{36} = 6.0\)

(b) M1 for \(\pi \times 4.5^2 \times 12\) [\(243\pi\) or \(763.4\)]
M1 for \(\frac{1}{3}\pi \times 4.5^2 \times 6\) [\(40.5\pi\) or \(127.2\)]
A1 for \(890.6\) or \(891\) (accept \(283.5\pi\))

(c) M1 for \(\pi \times 4.5 \times 7.5\) [\(33.75\pi\) or \(106.0\)]
M1 for \(2\pi \times 4.5 \times 12\) [\(108\pi\) or \(339.3\)]
M1 for adding base area \(\pi \times 4.5^2\) [\(20.25\pi\) or \(63.6\)]
A1 for \(509\) or \(508.9\) to \(509.0\) (accept \(162\pi\))

(d) M1 for their (b) \(\times 8.4\)
M0.5 for dividing by 1000 to convert to kg
A1 for \(7.48\) or \(7.481\) to \(7.482\)

(e) M1 for \(\sqrt{\frac{1100}{\text{their }(c)}}\) or \(k^2 = \frac{1100}{508.9}\)
M1 for \(12.0 \times \sqrt{\frac{1100}{\text{their }(c)}}\)
A1 for \(17.6\) or \(17.64\) to \(17.65\)

(a) Angle \(BAC = 138^\circ - 74^\circ = 64^\circ\).

(b) Using the cosine rule in triangle \(ABC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(BAC)\)
\(BC^2 = 48^2 + 72^2 - 2(48)(72)\cos(64^\circ)\)
\(BC^2 = 2304 + 5184 - 6912\cos(64^\circ)\)
\(BC^2 = 7488 - 6912(0.438371) = 7488 - 3029.98 = 4458.02\)
\(BC = \sqrt{4458.02} \approx 66.768\text{ km} = 66.8\text{ km}\).

(c) Using the sine rule in triangle \(ABC\):
\(\frac{\sin(\angle ABC)}{AC} = \frac{\sin(\angle BAC)}{BC}\)
\(\sin(\angle ABC) = \frac{72 \times \sin(64^\circ)}{66.768} = \frac{72 \times 0.898794}{66.768} \approx 0.96922\)
\(\angle ABC = \sin^{-1}(0.96922) \approx 75.74^\circ\) or using cosine rule:
\(\cos(\angle ABC) = \frac{48^2 + 66.768^2 - 72^2}{2(48)(66.768)} = \frac{2304 + 4458.02 - 5184}{6409.73} = \frac{1578.02}{6409.73} \approx 0.24619\)
\(\angle ABC = \cos^{-1}(0.24619) \approx 75.75^\circ = 75.7^\circ\) (or \(75.8^\circ\)).

(d) Bearing of \(A\) from \(B\) is \(180^\circ + 74^\circ = 254^\circ\).
Bearing of \(C\) from \(B\) is \(254^\circ - 75.75^\circ = 178.25^\circ\) (or using \(180 - (180 - 74) + \dots = 178.3^\circ\) / \(178^\circ\)).
Specifically: Angle between North at \(B\) and line \(BA\) pointing south-west is \(180^\circ - 74^\circ = 106^\circ\) on the left of the North line.
So bearing of \(C\) from \(B\) is \(180^\circ - (106^\circ - \angle ABC) = 180^\circ - (106^\circ - 75.75^\circ) = 180^\circ - 30.25^\circ = 149.75^\circ\) wait:
Let North at \(B\) be \(N_B\). Co-interior angles between North at \(A\) and North at \(B\): angle \(N_B B A = 180^\circ - 74^\circ = 106^\circ\).
The line \(BC\) is turned clockwise from line \(BA\) by angle \(ABC = 75.75^\circ\).
So the bearing of \(C\) from \(B\) is \(360^\circ - 106^\circ - 75.75^\circ = 178.25^\circ\) (or \(178.3^\circ\) or \(178^\circ\)).

(e)(i) Shortest distance \(AL\) is the perpendicular from \(A\) to \(BC\).
Area of triangle \(ABC = \frac{1}{2} \times AB \times AC \times \sin(64^\circ) = \frac{1}{2} \times 48 \times 72 \times \sin(64^\circ) = 1728 \times 0.898794 = 1553.115\text{ km}^2\).
Also Area = \(\frac{1}{2} \times BC \times AL\).
\(AL = \frac{2 \times 1553.115}{66.768} = \frac{3106.23}{66.768} \approx 46.52\text{ km} = 46.5\text{ km}\).
Alternatively, in right-angled triangle \(ABL\): \(AL = AB \times \sin(\angle ABC) = 48 \times \sin(75.75^\circ) = 48 \times 0.96928 = 46.53\text{ km}\).

(e)(ii) Time = \(\frac{\text{Distance}}{\text{Speed}} = \frac{46.524}{28} \approx 1.66157\text{ hours}\).
\(1.66157 \times 60 = 99.694\text{ minutes} = 99\text{ minutes and } 0.694 \times 60\text{ seconds} \approx 99\text{ min } 42\text{ s}\) (or \(99\text{ min } 40\text{ s}\) to \(42\text{ s}\)).

(a) B1 for \(138^\circ - 74^\circ = 64^\circ\) with correct subtraction shown.

(b) M1 for correct substitution into cosine rule: \(48^2 + 72^2 - 2(48)(72)\cos(64)\)
M1 for correct evaluation of terms: \(2304 + 5184 - 6912\cos(64)\) or \(7488 - 3029.98\dots\)
M1 for square root of \(4458\dots\)
A1 for \(66.8\) or \(66.76\) to \(66.77\)

(c) M1 for correct use of sine rule: \(\frac{\sin(\angle ABC)}{72} = \frac{\sin(64)}{\text{their } 66.8}\) or cosine rule
M1 for \(\sin(\angle ABC) = \frac{72\sin(64)}{66.77}\) or \(\cos(\angle ABC) = \frac{48^2 + 66.77^2 - 72^2}{2(48)(66.77)}\)
A1 for \(75.7^\circ\) to \(75.8^\circ\) (or \(75.6^\circ\) from rounded values)

(d) M1 for co-interior angle \(180 - 74 = 106\) or back-bearing \(254^\circ\) soi
M0.5 for \(360 - 106 - \text{their } 75.7^\circ\) or \(254 - \text{their } 75.7^\circ\)
A1 for \(178.2^\circ\) to \(178.4^\circ\) or \(178^\circ\)

(e)(i) M1 for \(48 \times \sin(\text{their } 75.7^\circ)\) or \(\frac{2 \times \text{Area}}{\text{their } 66.8}\)
A1 for \(46.5\) or \(46.52\) to \(46.54\)

(e)(ii) M1 for \(\frac{\text{their } 46.5}{28}\) soi
A1 for \(99\text{ min } 40\text{ s}\) to \(99\text{ min } 43\text{ s}\) (or \(1\text{ h } 39\text{ min } 41\text{ s}\))

(a) When \(x = -1.5\):
\(y = (-1.5)^3 - 3(-1.5) - 1 = -3.375 + 4.5 - 1 = 1.125\) (or \(1.1\)).
When \(x = 1.5\):
\(y = (1.5)^3 - 3(1.5) - 1 = 3.375 - 4.5 - 1 = -2.125\) (or \(-2.1\)).

(b) Plot all 9 points correctly: \((-2.5, -9.1)\), \((-2, -3)\), \((-1.5, 1.1)\), \((-1, 1)\), \((0, -1)\), \((1, -3)\), \((1.5, -2.1)\), \((2, 1)\), \((2.5, 7.1)\), and join with a smooth, continuous curve.

(c) Draw a tangent touching the curve at \(x = 2\).
Using two points on the tangent, for example \((1, -8)\) and \((2, 1)\):
\(\text{Gradient} = \frac{1 - (-8)}{2 - 1} = 9\).

(d)(i) Draw the line \(y = x - 2\) passing through points such as \((0, -2)\), \((2, 0)\), and \((-2, -4)\).

(ii) The equation \(x^3 - 4x + 1 = 0\) can be written as \(x^3 - 3x - 1 = x - 2\).
The solutions are the \(x\)-coordinates of the intersections of the curve and the line \(y = x - 2\):
\(x \approx -2.1\), \(x \approx 0.3\), and \(x \approx 1.9\).

(a) B1 for 1.125 or 1.1
B1 for -2.125 or -2.1

(b) B3 for correct smooth curve through all points
(M2 for 7 or 8 points correctly plotted, M1 for 5 or 6 points correctly plotted)

(c) M1 for drawing a tangent to the curve at x = 2
A1 for gradient in range [7.5, 10.5] dep on M1

(d)(i) B1 for straight line y = x - 2 correctly drawn with two correct points identified
(d)(ii) B2 for all three solutions in ranges [-2.2, -2.0], [0.2, 0.4], [1.8, 2.0]
(B1 for any two correct solutions)