2023 Cambridge IGCSE Mathematics (0580) 模拟试题及答案详解

(a) The total volume is the sum of the volume of the cylinder and the volume of the cone:
\(V = V_{\text{cylinder}} + V_{\text{cone}} = \pi r^2 (3r) + \frac{1}{3}\pi r^2 h = \pi r^2 \left( 3r + \frac{1}{3}h \right)\)

(b)(i) The volume of the cone is \(\frac{1}{3}\pi r^2 h\).
The volume of the cylinder is \(\pi r^2 (3r) = 3\pi r^3\).
Given:
\(\frac{1}{3}\pi r^2 h = \frac{1}{4} (3\pi r^3)\)
\(\frac{1}{3} h = \frac{3}{4} r\)
\(h = \frac{9}{4}r = 2.25r\)

(b)(ii) Substituting \(h = \frac{9}{4}r\) into the formula for \(V\):
\(V = \pi r^2 \left( 3r + \frac{1}{3}\left(\frac{9}{4}r\right) \right)\)
\(V = \pi r^2 \left( 3r + \frac{3}{4}r \right)\)
\(V = \pi r^2 \left( \frac{15}{4}r \right) = \frac{15}{4}\pi r^3\)

(c) Volume of the sphere is \(\frac{4}{3}\pi R^3\).
Equating the volumes:
\(\frac{4}{3}\pi R^3 = \frac{15}{4}\pi r^3\)
\(\frac{4}{3} R^3 = \frac{15}{4} r^3\)
\(R^3 = \frac{45}{16} r^3\)
\(R = \sqrt[3]{\frac{45}{16}} r\)
\(R \approx 1.41 r\) (since \(\sqrt[3]{2.8125} \approx 1.4116\))

(d) If \(r = 4.2\) cm, then:
\(h = 2.25 \times 4.2 = 9.45\) cm.
The slant height \(l\) of the cone is:
\(l = \sqrt{r^2 + h^2} = \sqrt{4.2^2 + 9.45^2} = \sqrt{17.64 + 89.3025} = \sqrt{106.9425} \approx 10.3413\) cm.
The total surface area \(A\) is the sum of the circular base, the curved surface of the cylinder, and the curved surface of the cone:
\(A = \pi r^2 + 2\pi r (3r) + \pi r l\)
\(A = 7\pi r^2 + \pi r l\)
\(A = 7\pi (4.2)^2 + \pi (4.2)(10.3413)\)
\(A = 123.48\pi + 43.4335\pi = 166.9135\pi \approx 524.37 \approx 524\) cm\(^2\).

(a)
M1: For writing a correct expression for the volume of the cylinder (\(3\pi r^3\)) or cone (\(\frac{1}{3}\pi r^2 h\)).
A1: For reaching the shown result with clear steps.

(b)(i)
M1: For equating \(\frac{1}{3}\pi r^2 h = \frac{1}{4} \times \text{their cylinder volume}\).
M1: For isolating \(h\) correctly.
A1: For \(h = 2.25r\) or \(h = \frac{9}{4}r\).

(b)(ii)
M1: For substituting their expression for \(h\) from (b)(i) into the formula.
A1: For fully simplifying to reach \(\frac{15}{4}\pi r^3\) (must show at least one intermediate step).

(c)
M1: For equating \(\frac{4}{3}\pi R^3 = \frac{15}{4}\pi r^3\).
M1: For isolating \(R^3\) or \(R\).
A1: For \(R = 1.41r\) (accept \(k = 1.41\)).

(d)
M1: For calculating the slant height \(l \approx 10.3\) cm using Pythagoras.
M1: For a correct expression for the total surface area: \(\pi r^2 + 6\pi r^2 + \pi r l\) (or substitute values).
A1: For \(524\) (accept \(524.0\) to \(524.4\)).

(a) Factorise the numerator:
\(2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\)
Factorise the denominator:
\(4x^2 - 1 = (2x - 1)(2x + 1)\)
Simplify the fraction:
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\)

(b) Notice that \(x^2 - 4 = (x - 2)(x + 2)\).
The common denominator is \((x - 2)(x + 2)\).
\(\frac{3}{x+2} - \frac{2x-1}{(x-2)(x+2)} = \frac{3(x-2) - (2x-1)}{(x-2)(x+2)}\)
\(= \frac{3x - 6 - 2x + 1}{x^2 - 4}\)
\(= \frac{x - 5}{x^2 - 4}\) (or \(\frac{x - 5}{(x - 2)(x + 2)}\))

(c) Multiply both sides by \((2 - w)\):
\(t(2 - w) = 3w + 5\)
Expand the brackets:
\(2t - tw = 3w + 5\)
Collect all terms containing \(w\) on one side:
\(2t - 5 = 3w + tw\)
Factorise \(w\):
\(2t - 5 = w(3 + t)\)
Divide by \((3 + t)\):
\(w = \frac{2t - 5}{t + 3}\)

(a)
M1: For factorising the numerator to \((2x + 1)(x - 3)\).
M1: For factorising the denominator to \((2x - 1)(2x + 1)\).
M1: For cancelling the common factor \((2x + 1)\).
A1: For the final simplified answer \(\frac{x - 3}{2x - 1}\).

(b)
M1: For writing \(x^2 - 4\) as \((x - 2)(x + 2)\) (or identifying it as the common denominator).
M1: For expressing the numerator as \(3(x - 2) - (2x - 1)\) over a common denominator.
M1: For expanding and simplifying the numerator to \(x - 5\).
A1: For the final simplified fraction \(\frac{x - 5}{x^2 - 4}\) or \(\frac{x - 5}{(x - 2)(x + 2)}\).

(c)
M1: For clearing the fraction: \(t(2 - w) = 3w + 5\).
M1: For expanding: \(2t - tw = 3w + 5\).
M1: For rearranging terms to isolate \(w\) terms: \(2t - 5 = 3w + tw\) (or equivalent).
M1: For factorising: \(w(3 + t) = 2t - 5\).
A1: For the final answer \(w = \frac{2t - 5}{t + 3}\) (or equivalent).

(a) Frequencies are calculated by subtracting consecutive cumulative frequencies:
- \(0 < t \le 10\): \(15\)
- \(10 < t \le 20\): \(42 - 15 = 27\)
- \(20 < t \le 30\): \(78 - 42 = 36\)
- \(30 < t \le 40\): \(104 - 78 = 26\)
- \(40 < t \le 50\): \(115 - 104 = 11\)
- \(50 < t \le 60\): \(120 - 115 = 5\)

(b) The midpoints of the groups are: \(5, 15, 25, 35, 45, 55\).
Estimate of \(\sum fx\):
\(\sum fx = (15 \times 5) + (27 \times 15) + (36 \times 25) + (26 \times 35) + (11 \times 45) + (5 \times 55)\)
\(\sum fx = 75 + 405 + 900 + 910 + 495 + 275 = 3060\)
Estimate of the mean:
\(\text{Mean} = \frac{\sum fx}{\sum f} = \frac{3060}{120} = 25.5\) minutes.

(c) First, determine the number of students who took more than 30 minutes:
\(120 - 78 = 42\) students.
Next, determine how many of these students took more than 40 minutes:
\(120 - 104 = 16\) students.
We choose two students from the 42 students.
The probability that both students took more than 40 minutes is:
\(P(\text{both } > 40) = \frac{16}{42} \times \frac{15}{41} = \frac{8}{21} \times \frac{15}{41} = \frac{120}{861} = \frac{40}{287} \approx 0.139\)

(d) The exact individual times are not known because the data is presented in grouped intervals.

(a)
B3: For all 6 correct frequencies (15, 27, 36, 26, 11, 5).
(If B3 not awarded, award B2 for 4 or 5 correct, or B1 for 2 or 3 correct).

(b)
M1: For finding the midpoints of the intervals (at least 4 correct).
M1: For calculating \(\sum fx\) using midpoints (at least 4 correct products shown).
M1: For dividing their \(\sum fx\) by 120.
A1: For \(25.5\) (or equivalent fraction).

(c)
B1: For identifying 42 students took more than 30 minutes.
B1: For identifying 16 students took more than 40 minutes.
M1: For the product of probabilities: \(\frac{16}{42} \times \frac{15}{41}\) (or their \(\frac{a}{b} \times \frac{a-1}{b-1}\)).
A1: For \(\frac{40}{287}\) or \(0.139\) (accept \(0.1393\) to \(0.1394\)).

(d)
B2: For explaining that individual raw data values are not known / the times are grouped.
(Award B1 for a partially complete explanation, such as 'because it is a cumulative frequency table' without mentioning lack of individual values).

(a) Volume of cylinder = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3. Volume of cone = \frac{1}{3}\pi r^2 H = \frac{1}{3}\pi r^2 (3r) = \pi r^3. Total volume = 2\pi r^3 + \pi r^3 = 3\pi r^3. (b)(i) 3\pi r^3 = 120 \Rightarrow r^3 = \frac{40}{\pi} \Rightarrow r \approx 2.335... \approx 2.34\text{ cm}. (b)(ii) The slant height of the cone is l = \sqrt{r^2 + (3r)^2} = r\sqrt{10} \approx 7.38397\text{ cm}. Total surface area = \text{Base area} + \text{Cylinder curved area} + \text{Cone curved area} = \pi r^2 + 2\pi r(2r) + \pi r l = 5\pi r^2 + \pi r (r\sqrt{10}) = (5 + \sqrt{10})\pi r^2. Since r \approx 2.33507\text{ cm}, Area \approx (5 + \sqrt{10})\pi (2.33507)^2 \approx 139.825... \approx 140\text{ cm}^2. (c) Volume of 1 sphere = 120 / 8 = 15\text{ cm}^3. \frac{4}{3}\pi R^3 = 15 \Rightarrow R^3 = \frac{45}{4\pi} \Rightarrow R \approx 1.5309... \approx 1.53\text{ cm}.

(a) M1 for Volume of cylinder = 2\pi r^3 or Volume of cone = \pi r^3. A1 for completing the proof. (b)(i) M1 for 3\pi r^3 = 120, M1 for r^3 = 120/(3\pi), A1 for 2.34. (b)(ii) B1 for l = \sqrt{r^2 + (3r)^2} or 7.38, M1 for cone curved area = \pi r l, M1 for 5\pi r^2 or base + cylinder curved area = 17.13 + 68.52 = 85.65, M1 for summing three parts, A1 for 140 or 139.8. (c) M1 for volume of one sphere = 15, M1 for 4/3 * \pi * R^3 = 15, A1 for 1.53.

(a) Factorise the numerator: 3x^2 - 12 = 3(x^2 - 4) = 3(x-2)(x+2). Factorise the denominator: 2x^2 + 7x + 6 = 2x^2 + 4x + 3x + 6 = 2x(x+2) + 3(x+2) = (2x+3)(x+2). Cancel the common factor (x+2) to get \frac{3(x-2)}{2x+3}. (b) \frac{4(y+2) - 3(y-3)}{(y-3)(y+2)} = \frac{4y+8-3y+9}{(y-3)(y+2)} = \frac{y+17}{(y-3)(y+2)}. (c) Multiply by x(2x+1): 5x + (2x+1) = 3x(2x+1) \Rightarrow 7x + 1 = 6x^2 + 3x \Rightarrow 6x^2 - 4x - 1 = 0. Use the quadratic formula: x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(6)(-1)}}{2(6)} = \frac{4 \pm \sqrt{16+24}}{12} = \frac{4 \pm \sqrt{40}}{12}. x = \frac{4 + 6.32456}{12} \approx 0.860 and x = \frac{4 - 6.32456}{12} \approx -0.194.

(a) M1 for 3(x^2 - 4) or 3(x-2)(x+2), M1 for splitting middle term in denominator (e.g., 4x and 3x), A1 for (2x+3)(x+2), A1 for final answer. (b) M1 for correct common denominator, M1 for expanding 4(y+2) - 3(y-3) with correct signs, A1 for final answer. (c) M1 for 5x + 2x + 1, M1 for = 3x(2x+1), A1 for 6x^2 - 4x - 1 = 0, M1 for quadratic formula with their a, b, c, A1 for 0.860, A1 for -0.194.

(a) Cumulative frequencies are calculated by running totals: 8, 8+14=22, 22+26=48, 48+20=68, 68+12=80. So [A] = 22, [B] = 48, [C] = 68. (b) Midpoints of intervals: 5, 15, 25, 35, 45. Estimated mean = \frac{(8 \times 5) + (14 \times 15) + (26 \times 25) + (20 \times 35) + (12 \times 45)}{80} = \frac{40 + 210 + 650 + 700 + 540}{80} = \frac{2140}{80} = 26.75\text{ cm}. (c) The class interval with highest frequency (26) is 20 < h \le 30. (d) Total number of plants taller than 30 cm is 20 + 12 = 32. Number of plants taller than 40 cm is 12. Probability of picking one taller than 40 cm first is 12/32. Probability of picking second taller than 40 cm without replacement is 11/31. Total probability = \frac{12}{32} \times \frac{11}{31} = \frac{132}{992} = \frac{33}{248} \approx 0.133. (e) Tallest 25% represents 25% of 80 = 20 plants. We need the height corresponding to the cumulative frequency of 80 - 20 = 60. This falls in the interval 30 < h \le 40, where cumulative frequency goes from 48 to 68. Using linear interpolation: Height = 30 + \frac{60 - 48}{68 - 48} \times (40 - 30) = 30 + \frac{12}{20} \times 10 = 30 + 6 = 36\text{ cm}.

(a) B1 for 22 and 48, B1 for 68. (b) M1 for correct midpoints 5, 15, 25, 35, 45, M1 for sum of f*x calculation (2140), M1 for dividing by 80, A1 for 26.75. (c) B1 for 20 < h \le 30. (d) M1 for identifying 32 and 12, M1 for 12/32 * 11/31, A1 for 33/248 or 0.133. (e) M1 for finding target cumulative frequency of 60, M1 for setting up interpolation fraction 12/20, A1 for 36.

(a) (i) The median is the 40th value. Since \(t = 14\) has a cumulative frequency of 24 and \(t = 15\) has 56:
\text{Median} = 14 + \frac{40 - 24}{56 - 24} \times (15 - 14) = 14 + \frac{16}{32} = 14.5\text{ seconds}.

(ii) The lower quartile (\(Q_1\)) is the 20th value:
\(Q_1 = 13 + \frac{20 - 8}{24 - 8} \times 1 = 13.75\text{ seconds}.

The upper quartile (\)Q_3\)) is the 60th value:
\(Q_3 = 15 + \frac{60 - 56}{72 - 56} \times 1 = 15.25\text{ seconds}.

\text{Interquartile Range (IQR)} = Q_3 - Q_1 = 15.25 - 13.75 = 1.5\text{ seconds}.

(iii) At \)t = 14.25\), the cumulative frequency is:
\text{CF} = 24 + 0.25 \times (56 - 24) = 32.
The number of students who took more than 14.25 seconds is \(80 - 32 = 48\).

(b) (i) The frequencies are found by subtracting successive cumulative frequencies:
\(p = 24 - 8 = 16\)
\(q = 56 - 24 = 32\)
\(r = 72 - 56 = 16\)

(ii) Midpoints of each class interval:
\(12.5, 13.5, 14.5, 15.5, 16.5\)

\sum f x = 8(12.5) + 16(13.5) + 32(14.5) + 16(15.5) + 8(16.5) = 100 + 216 + 464 + 248 + 132 = 1160

\text{Estimate of mean} = \frac{1160}{80} = 14.5\text{ seconds}.

(a)(i) [2 marks]
M1 for finding 40th value or for reading 14.5
A1 for 14.5

(a)(ii) [3 marks]
M1 for finding UQ or LQ (e.g. 15.25 or 13.75)
M1 for both correct UQ and LQ
A1 for 1.5

(a)(iii) [2 marks]
M1 for finding cumulative frequency of 32 at t = 14.25
A1 for 48

(b)(i) [2 marks]
B2 for all three correct: p = 16, q = 32, r = 16 (B1 for any two correct)

(b)(ii) [4 marks]
M1 for identifying interval midpoints (12.5, 13.5, etc.)
M1 for calculating \sum f x (at least 3 correct products shown)
M1 for dividing their sum by 80
A1 for 14.5

(a) The radius \(r = 6\text{ cm}\).
Since the total height of the toy is 14 cm and the radius of the hemisphere is 6 cm, the height of the cone is:
\(h = 14 - 6 = 8\text{ cm}\).

Volume of hemisphere:
\(V_{\text{hemi}} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (6)^3 = 144\pi \approx 452.39\text{ cm}^3\).

Volume of cone:
\(V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (6)^2 (8) = 96\pi \approx 301.59\text{ cm}^3\).

Total Volume:
\(V_{\text{total}} = 144\pi + 96\pi = 240\pi \approx 753.98 \approx 754\text{ cm}^3\).

(b) Total surface area comprises the curved surface area of the hemisphere and the curved surface area of the cone.

Curved surface area of hemisphere:
\(A_{\text{hemi}} = 2 \pi r^2 = 2 \pi (6)^2 = 72\pi \approx 226.19\text{ cm}^2\).

Slant height of the cone, \(l\):
\(l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = 10\text{ cm}\).

Curved surface area of cone:
\(A_{\text{cone}} = \pi r l = \pi (6)(10) = 60\pi \approx 188.50\text{ cm}^2\).

Total surface area:
\(A_{\text{total}} = 72\pi + 60\pi = 132\pi \approx 414.69 \approx 415\text{ cm}^2\).

(c) (i) Mass = Volume \(\times\) Density
Using \(V_{\text{total}} = 240\pi\):
\(\text{Mass} = 240\pi \times 0.85 = 204\pi \approx 640.88 \approx 641\text{ g}\) (or \(754 \times 0.85 = 640.9\text{ g}\)).

(ii) One tin covers \(1.5\text{ m}^2 = 1.5 \times 10,000\text{ cm}^2 = 15,000\text{ cm}^2\).
Number of toys = \(\frac{15000}{132\pi} \approx \frac{15000}{414.69} \approx 36.17\).

Since we need the maximum number of whole toys that can be painted:
Maximum number of toys = 36.

(a) [4 marks]
M1 for finding height of cone h = 8 cm
M1 for \(\frac{2}{3} \pi \times 6^3\) or \(\frac{1}{3} \pi \times 6^2 \times 8\)
M1 for adding hemisphere volume and cone volume
A1 for 754 or 240\(\pi\) or 753.98

(b) [4 marks]
M1 for slant height of cone l = 10 cm
M1 for curved surface area of hemisphere \(2\pi \times 6^2\) or cone \(\pi \times 6 \times 10\)
M1 for adding both curved surface areas
A1 for 415 or 132\(\pi\) or 414.69

(c)(i) [2 marks]
M1 for multiplying volume by 0.85
A1 for 641 or 640.9

(c)(ii) [3 marks]
M1 for converting 1.5 m\(^2\) to 15,000 cm\(^2\)
M1 for 15,000 / (their surface area from b)
A1 for 36 (must be integer, do not accept 37)

(a) Split the inequality into two parts:
Part 1:
\(4x - 3 \le 3(2x + 1)\)
\(4x - 3 \le 6x + 3\)
\(-6 \le 2x \implies x \ge -3\)

Part 2:
\(3(2x + 1) < 2x + 15\)
\(6x + 3 < 2x + 15\)
\(4x < 12 \implies x < 3\)

Combining the two solutions gives:
\(-3 \le x < 3\).

(b) Factorise the numerator and the denominator:
Numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\).
Denominator: \(x^2 - 9 = (x - 3)(x + 3)\).

Simplify by canceling common terms:
\(\frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} = \frac{2x + 1}{x + 3}\).

(c) Express with a common denominator:
\(\frac{3(y + 4) - 2(y - 2)}{(y - 2)(y + 4)}\)
\(= \frac{3y + 12 - 2y + 4}{(y - 2)(y + 4)}\)
\(= \frac{y + 16}{(y - 2)(y + 4)}\)

(d) Use the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(z = \frac{8 \pm \sqrt{(-8)^2 - 4(3)(2)}}{2(3)}\)
\(z = \frac{8 \pm \sqrt{64 - 24}}{6}\)
\(z = \frac{8 \pm \sqrt{40}}{6}\)

\(z_1 = \frac{8 + 6.32456}{6} \approx 2.39\)
\(z_2 = \frac{8 - 6.32456}{6} \approx 0.28\)

(a) [4 marks]
M1 for expanding bracket to 6x + 3
M1 for solving first inequality to get x >= -3
M1 for solving second inequality to get x < 3
A1 for -3 <= x < 3

(b) [4 marks]
M1 for factorising the numerator into (2x + a)(x + b)
A1 for correct numerator factorisation: (2x + 1)(x - 3)
B1 for correct denominator factorisation: (x - 3)(x + 3)
A1 for (2x + 1) / (x + 3)

(c) [3 marks]
M1 for correct common denominator: (y - 2)(y + 4)
M1 for 3(y + 4) - 2(y - 2) as the numerator
A1 for (y + 16) / ((y - 2)(y + 4)) or (y + 16) / (y^2 + 2y - 8)

(d) [2 marks]
M1 for correct substitution of values into the quadratic formula (or equivalent completing the square step)
A1 for both 2.39 and 0.28 correct to 2 decimal places

**(a)**
The volume of a hemisphere of radius \(r\) is:
\(V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3\)

The volume of a cone of radius \(r\) and height \(h\) is:
\(V_{\text{cone}} = \frac{1}{3}\pi r^2 h\)

Adding these volumes gives the total volume \(V\):
\(V = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2r + h) = \frac{1}{3}\pi r^2 (h + 2r)\) (as required).

**(b)(i)**
Given \(V = 150\pi\) and \(h = 8\):
\(\frac{1}{3}\pi r^2 (8 + 2r) = 150\pi\)

Divide both sides by \(\pi\):
\(\frac{1}{3}r^2 (8 + 2r) = 150\)

Multiply both sides by 3:
\(r^2 (8 + 2r) = 450 \implies 2r^3 + 8r^2 = 450\)

Divide by 2:
\(r^3 + 4r^2 = 225 \implies r^3 + 4r^2 - 225 = 0\) (as required).

**(b)(ii)**
Substitute \(r = 5\) into the cubic expression:
\(5^3 + 4(5^2) - 225 = 125 + 4(25) - 225 = 125 + 100 - 225 = 0\).
Since the expression equals 0, \(r = 5\) is a solution.

**(b)(iii)**
Since \(r = 5\) is a solution, \((r - 5)\) is a factor.
By polynomial division or matching coefficients:
\(r^3 + 4r^2 - 225 = (r - 5)(r^2 + 9r + 45) = 0\)

For the other two solutions, solve \(r^2 + 9r + 45 = 0\) using the quadratic formula:
\(r = \frac{-9 \pm \sqrt{9^2 - 4(1)(45)}}{2} = \frac{-9 \pm \sqrt{81 - 180}}{2} = \frac{-9 \pm \sqrt{-99}}{2}\)

Since the discriminant is negative (\(-99 < 0\)), there are no real solutions to this quadratic equation. Thus, the other two solutions are not real numbers and cannot represent the physical radius of the toy.

**(c)**
First, calculate the slant height \(l\) of the cone:
\(l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 8^2} = \sqrt{89} \approx 9.434\text{ cm}\)

The curved surface area of the cone (red surface) is:
\(A_{\text{red}} = \pi r l = \pi \times 5 \times \sqrt{89} \approx 148.19\text{ cm}^2\)

The curved surface area of the hemisphere (blue surface) is:
\(A_{\text{blue}} = 2\pi r^2 = 2\pi \times 5^2 = 50\pi \approx 157.08\text{ cm}^2\)

The total painted surface area is:
\(A_{\text{total}} = 157.08 + 148.19 = 305.27\text{ cm}^2\)

The percentage of the area that is blue is:
\(\text{Percentage} = \frac{157.08}{305.27} \times 100 \approx 51.456\% \approx 51.5\%\) (correct to 3 significant figures).

**(a)**
- **M1**: For showing correct formulas for volume of hemisphere and cone, i.e., \(\frac{2}{3}\pi r^3\) and \(\frac{1}{3}\pi r^2 h\).
- **A1**: For fully correct algebraic steps to arrive at \(\frac{1}{3}\pi r^2 (h + 2r)\).

**(b)(i)**
- **M1**: For equating the volume expression to \(150\pi\) and substituting \(h = 8\): \(\frac{1}{3}\pi r^2 (8 + 2r) = 150\pi\).
- **A1**: For simplifying to \(r^3 + 4r^2 - 225 = 0\) with at least one intermediate step shown.

**(b)(ii)**
- **B1**: For substituting \(r = 5\) into the cubic expression and showing that it equals 0.

**(b)(iii)**
- **M1**: For dividing the cubic polynomial by \((r-5)\) to find the quadratic factor \(r^2 + 9r + 45\).
- **M1**: For calculating the discriminant: \(9^2 - 4(1)(45) = -99\) or using the quadratic formula to show \(\sqrt{-99}\).
- **A1**: For explaining that because the discriminant is negative, the other two solutions are not real and thus cannot be a physical length.

**(c)**
- **M1**: For calculating the slant height \(l = \sqrt{5^2 + 8^2} = \sqrt{89} \approx 9.43\) or better.
- **M1**: For finding the curved surface area of the cone: \(\pi \times 5 \times \sqrt{89} \approx 148.2\) or better.
- **M1**: For finding the curved surface area of the hemisphere: \(2 \times \pi \times 5^2 = 50\pi \approx 157.1\) or better.
- **M1**: For a correct percentage calculation attempt: \(\frac{\text{Blue Area}}{\text{Total Area}} \times 100\).
- **A1**: For \(51.5\%\) (accept \(51.4\% - 51.5\%\)).