2024 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

(a) To complete the cumulative frequency table, we sum the frequencies progressively:
\(t \le 10\): 8
\(t \le 20\): 8 + 16 = 24
\(t \le 30\): 24 + 34 = 58
\(t \le 40\): 58 + 38 = 96
\(t \le 50\): 96 + 18 = 114
\(t \le 60\): 114 + 6 = 120

(b)(i) The median position is \(120 / 2 = 60\). The 60th student lies in the interval \(30 < t \le 40\).
Median = \(30 + \frac{60 - 58}{38} \times 10 = 30 + \frac{2}{38} \times 10 \approx 30.5\) minutes.

(b)(ii) The lower quartile (LQ) is at position 30, which lies in the interval \(20 < t \le 30\).
LQ = \(20 + \frac{30 - 24}{34} \times 10 = 20 + 1.765 = 21.76\) minutes.
The upper quartile (UQ) is at position 90, which lies in the interval \(30 < t \le 40\).
UQ = \(30 + \frac{90 - 58}{38} \times 10 = 30 + 8.421 = 38.42\) minutes.
Interquartile range (IQR) = \(38.421 - 21.765 = 16.656 \approx 16.7\) minutes.

(c) Using linear interpolation, we find the number of students taking 35 minutes or less. Since 35 is halfway in the \(30 < t \le 40\) interval, the number of students in the range \(30 < t \le 35\) is \(\frac{35-30}{10} \times 38 = 19\).
Cumulative frequency up to 35 minutes = \(58 + 19 = 77\).
The number of students taking more than 35 minutes = \(120 - 77 = 43\).
Probability = \(\frac{43}{120} \approx 0.358\).

Part (a): [2 marks]
- M1 for at least 3 correct cumulative values
- A1 for all 6 correct (8, 24, 58, 96, 114, 120)

Part (b)(i): [2 marks]
- M1 for choosing correct interval and showing interpolation method: \(30 + \frac{60 - 58}{38} \times 10\)
- A1 for 30.5

Part (b)(ii): [3 marks]
- M1 for calculating LQ = 21.8 or UQ = 38.4
- M1 for \(UQ - LQ\)
- A1 for 16.7 (accept 16.6 to 16.8)

Part (c): [3 marks]
- M1 for finding number of students taking \(\le 35\) minutes is 77, or taking \(>35\) is 43
- M1 for dividing their count of students by 120
- A1 for 43/120 or 0.358

(a) The reflection in the line \(y = -x\) maps each point \((x, y)\) to \((-y, -x)\).
Applying this to the vertices of triangle \(A\):
- \((1, 1) \to (-1, -1)\)
- \((3, 1) \to (-1, -3)\)
- \((3, 4) \to (-4, -3)\)
Thus, the vertices of triangle \(B\) are \((-1, -1)\), \((-1, -3)\), and \((-4, -3)\).

(b) A shear with the \(x\)-axis invariant and shear factor 2 maps each point \((x, y)\) to \((x + 2y, y)\).
Applying this to the vertices of triangle \(A\):
- \((1, 1) \to (1 + 2(1), 1) = (3, 1)\)
- \((3, 1) \to (3 + 2(1), 1) = (5, 1)\)
- \((3, 4) \to (3 + 2(4), 4) = (11, 4)\)
Thus, the vertices of triangle \(C\) are \((3, 1)\), \((5, 1)\), and \((11, 4)\).

(c) Since triangle \(B\) is obtained from triangle \(A\) by reflection in the line \(y = -x\), and a reflection is its own inverse, the single transformation mapping \(B\) back to \(A\) is also a reflection in the line \(y = -x\).

(d) The standard matrix representing a shear with the \(x\)-axis invariant and shear factor \(k\) is \(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\). For \(k = 2\), this is \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\).

Part (a): [3 marks]
- B3 for all three vertices correct (B1 for each correct vertex)

Part (b): [3 marks]
- B3 for all three vertices correct (B1 for each correct vertex)

Part (c): [2 marks]
- M1 for 'reflection'
- A1 for 'in the line \(y = -x\)'

Part (d): [2 marks]
- B2 for the correct matrix (B1 if one element is incorrect or transposed)

(a) The total volume is the sum of the volumes of the cylinder, the hemisphere, and the cone:
- Volume of cylinder = \(\pi r^2 h = \pi r^2 (3r) = 3\pi r^3\)
- Volume of hemisphere = \(\frac{2}{3}\pi r^3\)
- Volume of cone = \(\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3\)
Total Volume = \(3\pi r^3 + \frac{2}{3}\pi r^3 + \frac{2}{3}\pi r^3 = \left(3 + \frac{4}{3}\right)\pi r^3 = \frac{13}{3}\pi r^3\) cm\(^3\).

(b) Setting the volume equation to 250:
\(\frac{13}{3}\pi r^3 = 250 \implies r^3 = \frac{750}{13\pi} \approx 18.3639\)
\(r = \sqrt[3]{18.3639} \approx 2.638 \approx 2.64\) cm.

(c) The total surface area of the toy consists of:
- Curved surface of hemisphere = \(2\pi r^2\)
- Curved surface of cylinder = \(2\pi r h = 2\pi r (3r) = 6\pi r^2\)
- Curved surface of cone = \(\pi r l\), where \(l\) is the slant height.
\(l = \sqrt{r^2 + h^2} = \sqrt{r^2 + (2r)^2} = \sqrt{5r^2} = r\sqrt{5}\).
Curved surface of cone = \(\pi r (r\sqrt{5}) = \sqrt{5}\pi r^2\).
Total Surface Area = \(2\pi r^2 + 6\pi r^2 + \sqrt{5}\pi r^2 = (8 + \sqrt{5})\pi r^2\).
Using \(r = 2.6382\):
Total Surface Area = \((8 + \sqrt{5})\pi (2.6382)^2 \approx 10.236 \times \pi \times 6.960 \approx 223.8 \approx 224\) cm\(^2\).

Part (a): [3 marks]
- M1 for writing cylinder volume \(3\pi r^3\)
- M1 for writing cone volume \(\frac{2}{3}\pi r^3\)
- A1 for summing to \(\frac{13}{3}\pi r^3\) with clear working

Part (b): [3 marks]
- M1 for setting up equation \(\frac{13}{3}\pi r^3 = 250\)
- M1 for rearranging to make \(r^3\) or \(r\) the subject
- A1 for 2.64

Part (c): [4 marks]
- M1 for identifying the three surface components: \(2\pi r^2 + 6\pi r^2 + \pi r l\)
- M1 for calculating slant height \(l = r\sqrt{5}\)
- M1 for substituting their value of \(r\) from part (b)
- A1 for 224 (accept 223 to 225)

(a) Read directly from the given points:
(i) At mass 100g, the cumulative frequency is 30.
(ii) At mass 140g, the cumulative frequency is 105.

(b)(i) The median is at cumulative frequency \(\frac{120}{2} = 60\).
The value 60 lies in the interval between the points \((100, 30)\) and \((120, 75)\).
Using linear interpolation:
\(\frac{x - 100}{120 - 100} =
\frac{60 - 30}{75 - 30}\)
\(\frac{x - 100}{20} = \frac{30}{45}\)
\(x - 100 = 13.33\)
\(x \approx 113.3\text{ g}\) (or 113g to 3 s.f.).

(ii) The lower quartile (\(Q_1\)) is at cumulative frequency \(120 \times 0.25 = 30\).
From the point \((100, 30)\), \(Q_1 = 100\text{ g}\).
The upper quartile (\(Q_3\)) is at cumulative frequency \(120 \times 0.75 = 90\).
The value 90 lies between \((120, 75)\) and \((140, 105)\).
Using linear interpolation:
\(\frac{x - 120}{140 - 120} = \frac{90 - 75}{105 - 75}\)
\(\frac{x - 120}{20} = \frac{15}{30}\)
\(x - 120 = 10\)
\(Q_3 = 130\text{ g}\).
The interquartile range is \(IQR = Q_3 - Q_1 = 130 - 100 = 30\text{ g}\).

(iii) A mass of 135g lies between the points \((120, 75)\) and \((140, 105)\).
Using linear interpolation to find the cumulative frequency \(CF\) at 135g:
\(\frac{CF - 75}{105 - 75} = \frac{135 - 120}{140 - 120}\)
\(\frac{CF - 75}{30} = \frac{15}{20}\)
\(CF - 75 = 22.5\)
\(CF = 97.5\).
The number of apples with a mass greater than 135g is \(120 - 97.5 = 22.5\).
The percentage is \(\frac{22.5}{120} \times 100\% = 18.75\%\).

(a)(i) B1 for 30
(a)(ii) B1 for 105
(b)(i) M1 for attempting to find the mass at cumulative frequency 60. M1 for setting up a correct interpolation expression, e.g. \(100 + 20 \times \frac{30}{45}\). A1 for 113 (or 113.3).
(b)(ii) B1 for \(Q_1 = 100\). M1 for finding \(Q_3 = 130\) by interpolation or reading. A1 for \(IQR = 30\).
(b)(iii) M1 for finding the cumulative frequency at 135g as 97.5. M1 for calculating \(120 - 97.5\) to find the number of apples (22.5). A1 for 18.75% (or 18.8%).

(a) The total volume of the toy is the sum of the volumes of the cone and the cylinder:
\(V = V_{\text{cone}} + V_{\text{cylinder}}\)
\(500 = \frac{1}{3}\pi r^2 h + \pi r^2 (3r)\)
\(500 = \frac{1}{3}\pi r^2 h + 3\pi r^3\)
Multiply the entire equation by 3 to clear the fraction:
\(1500 = \pi r^2 h + 9\pi r^3\)
Isolate the term containing \(h\):
\(\pi r^2 h = 1500 - 9\pi r^3\)
Divide by \(\pi r^2\):
\(h = \frac{1500 - 9\pi r^3}{\pi r^2} = \frac{1500}{\pi r^2} - 9r\).

(b)(i) When \(h = r\), substitute this into the volume formula:
\(500 = \frac{1}{3}\pi r^3 + 3\pi r^3\)
\(500 = \frac{10}{3}\pi r^3\)
\(\pi r^3 = 150\)
\(r^3 = \frac{150}{\pi}\)
\(r = \sqrt[3]{\frac{150}{\pi}} \approx 3.6278\text{ cm}\) (or 3.63 cm to 3 s.f.).

(ii) The total surface area consists of:
1. Curved surface of cone: \(\pi r l\), where \(l = \sqrt{r^2 + h^2}\). Since \(h = r\), \(l = \sqrt{r^2 + r^2} = r\sqrt{2}\).
Area \(= \pi r (r\sqrt{2}) = \pi r^2 \sqrt{2}\).
2. Curved surface of cylinder: \(2\pi r (3r) = 6\pi r^2\).
3. Circular base of cylinder: \(\pi r^2\).

Total Surface Area \(A = \pi r^2 \sqrt{2} + 6\pi r^2 + \pi r^2 = (7 + \sqrt{2})\pi r^2\).
Using \(r = 3.6278\):
\(A = (7 + \sqrt{2}) \pi (3.6278)^2 \approx 8.4142 \times \pi \times 13.161 \approx 347.9\text{ cm}^2\) (or 348 \(\text{cm}^2\) to 3 s.f.).

(a) M1 for writing total volume equation: \(\frac{1}{3}\pi r^2 h + 3\pi r^3 = 500\). M1 for isolating the term with \(h\): \(\pi r^2 h = 1500 - 9\pi r^3\). A1 for correct steps leading to the given expression \(h = \frac{1500}{\pi r^2} - 9r\).
(b)(i) M1 for setting up \(\frac{10}{3}\pi r^3 = 500\). M1 for isolating \(r^3 = \frac{150}{\pi}\). A1 for 3.63 (accept 3.628).
(b)(ii) M1 for calculating slant height \(l = r\sqrt{2}\). M1 for cylinder curved surface area \(6\pi r^2\). M1 for identifying correct total area expression \((7 + \sqrt{2})\pi r^2\) (or summing the parts individually). M1 for substituting their value of \(r\) into the correct formula. A1 for 348 (accept 347.8 to 348.0).

(a) First, find angle \(\angle BAC\):
\(\angle BAC = 130^\circ - 55^\circ = 75^\circ\).
Use the Cosine Rule in \(\triangle ABC\) to find the length of \(BC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC)\)
\(BC^2 = 18^2 + 25^2 - 2(18)(25)\cos(75^\circ)\)
\(BC^2 = 324 + 625 - 900(0.258819)\)
\(BC^2 = 949 - 232.937 = 716.063\)
\(BC = \sqrt{716.063} \approx 26.759\text{ km}\) (or 26.8 km to 3 s.f.).

(b) Use the Sine Rule to find \(\angle ABC\):
\(\frac{\sin(\angle ABC)}{AC} = \frac{\sin(\angle BAC)}{BC}\)
\(\frac{\sin(\angle ABC)}{25} = \frac{\sin(75^\circ)}{26.759}\)
\(\sin(\angle ABC) = \frac{25 \sin(75^\circ)}{26.759} \approx \frac{25 \times 0.965926}{26.759} \approx 0.90243\)
\(\angle ABC = \arcsin(0.90243) \approx 64.48^\circ\).

Now, calculate the bearing:
The bearing of \(B\) from \(A\) is \(055^\circ\). Thus, the line \(AB\) has a back-bearing of \(180^\circ + 55^\circ = 235^\circ\) (measured clockwise from North at \(B\)).
Since \(C\) lies to the east of the line \(AB\) from \(B\)'s perspective, the bearing of \(C\) from \(B\) is:
\(235^\circ - \angle ABC = 235^\circ - 64.48^\circ = 170.52^\circ\).
To the nearest degree, the bearing is \(171^\circ\).

(c) The shortest distance from port \(A\) to the path \(BC\) is the perpendicular distance \(d\) from \(A\) to the line \(BC\).
In \(\triangle ABC\):
\(d = AB \sin(\angle ABC)\)
\(d = 18 \sin(64.48^\circ)\)
\(d \approx 18 \times 0.90243 \approx 16.24\text{ km}\) (or 16.2 km to 3 s.f.).

(a) B1 for finding \(\angle BAC = 75^\circ\). M1 for correct substitution into the Cosine Rule: \(18^2 + 25^2 - 2(18)(25)\cos(75^\circ)\). A1 for evaluating \(BC^2 \approx 716.06\). A1 for \(BC \approx 26.8\text{ km}\) (or 26.76).
(b) M1 for applying the Sine Rule: \(\frac{\sin(\angle ABC)}{25} = \frac{\sin(75^\circ)}{26.759}\). A1 for \(\angle ABC \approx 64.5^\circ\) (or 64.48°). M1 for utilizing the back-bearing concept, e.g. \((180+55) - 64.5\). A1 for bearing of 171° (accept 170.5° to 171°).
(c) M1 for recognizing that the shortest distance is represented by the perpendicular height. M1 for calculating \(18 \sin(64.48^\circ)\) or equivalent correct method. A1 for 16.2 (accept 16.2 to 16.3).

(a) Cumulative frequencies are calculated by adding the frequencies sequentially:
- \(\le 10\): \(12\)
- \(\le 20\): \(12 + 26 = 38\)
- \(\le 30\): \(38 + 54 = 92\)
- \(\le 40\): \(92 + 40 = 132\)
- \(\le 50\): \(132 + 20 = 152\)
- \(\le 60\): \(152 + 8 = 160\)

(b) Estimate of the mean is calculated using the midpoint of each interval:
- \(0 < t \le 10\): midpoint = 5, product = \(5 \times 12 = 60\)
- \(10 < t \le 20\): midpoint = 15, product = \(15 \times 26 = 390\)
- \(20 < t \le 30\): midpoint = 25, product = \(25 \times 54 = 1350\)
- \(30 < t \le 40\): midpoint = 35, product = \(35 \times 40 = 1400\)
- \(40 < t \le 50\): midpoint = 45, product = \(45 \times 20 = 900\)
- \(50 < t \le 60\): midpoint = 55, product = \(55 \times 8 = 440\)

Sum of products = \(60 + 390 + 1350 + 1400 + 900 + 440 = 4540\).
Estimated Mean = \(\frac{4540}{160} = 28.375\) minutes.

(c) For the interval \(30 < t \le 40\), the frequency is 40.
Using linear interpolation, the number of students taking between 30 and 38 minutes is:
\(\frac{38 - 30}{40 - 30} \times 40 = \frac{8}{10} \times 40 = 32\) students.
Thus, the number of students in this interval taking more than 38 minutes is \(40 - 32 = 8\) students.
Additionally, all students in intervals \(40 < t \le 50\) and \(50 < t \le 60\) took more than 38 minutes, which is \(20 + 8 = 28\) students.
Total students taking more than 38 minutes = \(8 + 28 = 36\).
Percentage = \(\frac{36}{160} \times 100 = 22.5\%\).

(d) There are 12 students who took 10 minutes or less.
Probability of first student taking \(\le 10\) mins = \(\frac{12}{160}\).
Probability of second student taking \(\le 10\) mins (without replacement) = \(\frac{11}{159}\).
Combined probability = \(\frac{12}{160} \times \frac{11}{159} = \frac{132}{25440} = \frac{11}{2120} \approx 0.00519\).

Part (a): [2 marks]
- B1 for any 3 correct cumulative values.
- B1 for all 6 correct: 12, 38, 92, 132, 152, 160.

Part (b): [4 marks]
- M1 for finding the midpoints of the intervals (at least 4 correct).
- M1 for calculating \(\sum f \times x\) (allow one arithmetic error).
- M1 for division of their sum by 160.
- A1 for 28.375 or 28.4.

Part (c): [3 marks]
- M1 for finding the proportion of the \(30 < t \le 40\) class: \(\frac{40 - 38}{10} \times 40 = 8\).
- M1 for finding the total number of students: \(8 + 20 + 8 = 36\).
- A1 for 22.5%.

Part (d): [2 marks]
- M1 for product \(\frac{12}{160} \times \frac{11}{159}\).
- A1 for \(\frac{11}{2120}\) or 0.00519 (accept 3 significant figures).

(a) To find local extrema, we take the derivative of \( f(x) \):
\( f'(x) = 3x^2 - 6x - 4 \).
Setting \( f'(x) = 0 \):
\( x = \frac{6 \pm \sqrt{(-6)^2 - 4(3)(-4)}}{6} = \frac{6 \pm \sqrt{84}}{6} = 1 \pm \frac{\sqrt{21}}{3} \).
This gives the \(x\)-coordinates:
- \( x_1 \approx -0.5275\) (local maximum)
- \( x_2 \approx 2.5275\) (local minimum)

Substitute these back into \( f(x) \):
- For \( x = -0.528\): \( f(-0.528) \approx 6.13\).
- For \( x = 2.53\): \( f(2.53) \approx -8.13\).

So, local maximum is \( (-0.528, 6.13) \) and local minimum is \( (2.53, -8.13) \).

(b) The function \( g(x) = 2x - 3 \) is a linear function in the form \( y = mx + c \).
Therefore, the gradient is 2.

(c) To find the points of intersection, set \( f(x) = g(x) \):
\( x^3 - 3x^2 - 4x + 5 = 2x - 3 \)
\( x^3 - 3x^2 - 6x + 8 = 0 \).

Using factor theorem or a GDC, we test small integer values:
- If \( x = 1 \): \( 1^3 - 3(1)^2 - 6(1) + 8 = 1 - 3 - 6 + 8 = 0 \). So \( (x - 1) \) is a factor.

Factorising the cubic:
\( (x - 1)(x^2 - 2x - 8) = 0 \)
\( (x - 1)(x - 4)(x + 2) = 0 \).

Thus, the \(x\)-coordinates of the points of intersection are \( x = -2 \), \( x = 1 \), and \( x = 4 \).

(d) We want to solve \( f(x) > g(x) \), which simplifies to:
\( x^3 - 3x^2 - 6x + 8 > 0 \).

Referring to the roots from part (c), we determine the sign of the expression in each interval:
- For \( x < -2 \), the expression is negative.
- For \( -2 < x < 1 \), the expression is positive.
- For \( 1 < x < 4 \), the expression is negative.
- For \( x > 4 \), the expression is positive.

Therefore, the solution to the inequality is \( -2 < x < 1 \) or \( x > 4 \).

Part (a): [4 marks]
- M1 for differentiating \( f(x) \) correctly: \( 3x^2 - 6x - 4 \).
- M1 for setting their derivative to 0 and solving for \( x \) (using GDC or formula).
- A1 for maximum coordinates: \( (-0.528, 6.13) \) (accept answers rounding to these values).
- A1 for minimum coordinates: \( (2.53, -8.13) \) (accept answers rounding to these values).

Part (b): [1 mark]
- B1 for gradient = 2.

Part (c): [3 marks]
- M1 for setting up the equation \( x^3 - 3x^2 - 6x + 8 = 0 \).
- M1 for finding at least one correct root (e.g., \( x = 1 \)).
- A1 for all three correct values: \( x = -2, x = 1, x = 4 \).

Part (d): [3 marks]
- M1 for identifying critical regions using their roots from (c).
- A1 for \( -2 < x < 1 \).
- A1 for \( x > 4 \).

(a) The volume of the hemisphere is given by:
\( V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 \).

The volume of the cone is given by:
\( V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (15 - r) \).

The total volume \( V \) is:
\( V = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 (15 - r) = 252\pi \).

Divide the entire equation by \( \pi \):
\( \frac{2}{3}r^3 + \frac{1}{3}r^2 (15 - r) = 252 \)
\( \frac{2}{3}r^3 + 5r^2 - \frac{1}{3}r^3 = 252 \)
\( \frac{1}{3}r^3 + 5r^2 = 252 \).

Multiply by 3:
\( r^3 + 15r^2 = 756 \)
\( r^3 + 15r^2 - 756 = 0 \).

(b) (i) Substitute \( r = 6 \) into the equation:
\( 6^3 + 15(6)^2 - 756 = 216 + 15(36) - 756 = 216 + 540 - 756 = 756 - 756 = 0 \).
Thus, \( r = 6 \) is indeed a solution.

(ii) Since \( r = 6 \) is a root, \( r - 6 \) is a factor.
Using polynomial division or algebraic matching:
\( r^3 + 15r^2 - 756 = (r - 6)(r^2 + 21r + 126) = 0 \).

To find if there are any other real solutions, analyze the quadratic part \( r^2 + 21r + 126 = 0 \):
Discriminant \( \Delta = b^2 - 4ac = 21^2 - 4(1)(126) = 441 - 504 = -63 \).

Since the discriminant is less than 0, there are no other real roots. Hence, \( r = 6 \) is the only real solution.

(c) The total surface area of the toy consists of the curved surface area of the hemisphere and the curved surface area of the cone.

- Curved surface area of the hemisphere: \( A_{\text{hemisphere}} = 2\pi r^2 = 2\pi(6)^2 = 72\pi \).
- Height of the cone is \( h = 15 - 6 = 9 \) cm.
- Slant height of the cone: \( l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 9^2} = \sqrt{117} \approx 10.817 \) cm.
- Curved surface area of the cone: \( A_{\text{cone}} = \pi r l = \pi (6)(\sqrt{117}) \approx 64.90\pi \).

Total Surface Area = \( 72\pi + 6\sqrt{117}\pi \approx 226.19 + 203.89 = 430.08 \) cm\(^2\).
To 3 significant figures, this is \( 430 \) cm\(^2\).

Part (a): [3 marks]
- M1 for writing down correct expressions for both volumes: \( \frac{2}{3}\pi r^3 \) and \( \frac{1}{3}\pi r^2(15-r) \).
- M1 for setting their sum equal to \( 252\pi \) and eliminating \( \pi \).
- A1 for fully convincing algebraic steps showing \( r^3 + 15r^2 - 756 = 0 \).

Part (b)(i): [1 mark]
- B1 for substitution of \( r=6 \) showing LHS calculation equals 0.

Part (b)(ii): [3 marks]
- M1 for factorising out \( (r - 6) \) to get the quadratic factor \( r^2 + 21r + 126 \).
- M1 for evaluating the discriminant of \( r^2 + 21r + 126 \) (e.g., \( 21^2 - 4 \times 126 = -63 \)).
- A1 for concluding that \( \Delta < 0 \) means no other real solutions exist.

Part (c): [4 marks]
- M1 for calculating the slant height of the cone: \( l = \sqrt{6^2 + 9^2} = \sqrt{117} \).
- M1 for using the curved surface area formulas \( 2\pi r^2 \) and \( \pi r l \).
- M1 for summing the two surface areas: \( 72\pi + 6\pi\sqrt{117} \).
- A1 for 430 (accept 430.1 or 430).

(a) The total frequency is \(N = 160\).
The median is at the \(80\)th cumulative frequency value.
The cumulative frequency value of 80 lies in the interval \(20 \le h \le 30\), between the points \((20, 38)\) and \((30, 92)\).
Using linear interpolation:
\(\text{Median} = 20 + \frac{80 - 38}{92 - 38} \times (30 - 20) = 20 + \frac{42}{54} \times 10 = 20 + 7.78 = 27.78\text{ cm}\).
Correct to 3 significant figures, this is \(27.8\text{ cm}\).

(b) The lower quartile (LQ) is at the \(40\)th cumulative frequency value.
The value 40 lies in the interval \(20 \le h \le 30\), between \((20, 38)\) and \((30, 92)\).
\(LQ = 20 + \frac{40 - 38}{92 - 38} \times 10 = 20 + \frac{2}{54} \times 10 = 20.370\text{ cm}\).

The upper quartile (UQ) is at the \(120\)th cumulative frequency value.
The value 120 lies in the interval \(30 \le h \le 40\), between \((30, 92)\) and \((40, 138)\).
\(UQ = 30 + \frac{120 - 92}{138 - 92} \times (40 - 30) = 30 + \frac{28}{46} \times 10 = 30 + 6.087 = 36.087\text{ cm}\).

The interquartile range (IQR) is:
\(IQR = UQ - LQ = 36.087 - 20.370 = 15.717\text{ cm}\).
Correct to 3 significant figures, this is \(15.7\text{ cm}\).

(c) For a height of 45 cm, we interpolate in the interval \(40 \le h \le 50\), between \((40, 138)\) and \((50, 154)\).
Since 45 is the midpoint of 40 and 50:
\(\text{Cumulative frequency} = 138 + \frac{45 - 40}{50 - 40} \times (154 - 138) = 138 + 0.5 \times 16 = 146\).
Therefore, 146 plants have a height of 45 cm or less.
The number of plants greater than 45 cm is:
\(160 - 146 = 14\).

(a)
M1 for identifying the 80th value
M1 for using linear interpolation on the interval [20, 30] (e.g., \(20 + \frac{42}{54} \times 10\))
A1 for \(27.8\) (accept \(27.78\) or similar)

(b)
M1 for identifying 40th and 120th values
M1 for calculating LQ = \(20.37\) or \(20.4\)
M1 for calculating UQ = \(36.09\) or \(36.1\)
M1 for subtracting their LQ from their UQ
A1 for \(15.7\) (accept \(15.72\))

(c)
M1 for setting up interpolation at 45 (e.g., finding the midpoint of 138 and 154)
A1 for finding the cumulative frequency value of 146
A1 for \(160 - 146 = 14\)

(a) The volume of the hemisphere is:
\(V_{\text{hemisphere}} = \frac{2}{3}\pi r^3\)

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

Since \(h = 3r\):
\(V_{\text{cone}} = \frac{1}{3}\pi r^2 (3r) = \pi r^3\)

The total volume is:
\(V_{\text{total}} = \frac{2}{3}\pi r^3 + \pi r^3 = \frac{5}{3}\pi r^3\)

We are given that the total volume is \(360\pi\text{ cm}^3\):
\(\frac{5}{3}\pi r^3 = 360\pi\)

Divide both sides by \(\pi\):
\(\frac{5}{3}r^3 = 360\)

\(r^3 = 360 \times \frac{3}{5} = 216\)

\(r = \sqrt[3]{216} = 6\text{ cm}\).

(b) The height of the cone is \(h = 3r = 3(6) = 18\text{ cm}\).
The slant height \(l\) of the cone is:
\(l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 18^2} = \sqrt{36 + 324} = \sqrt{360} \approx 18.974\text{ cm}\).

Correct to 3 significant figures, this is \(19.0\text{ cm}\).

(c) The total surface area consists of the curved surface area of the cone and the curved surface area of the hemisphere.

Curved surface area of the cone:
\(A_{\text{cone}} = \pi r l = \pi \times 6 \times \sqrt{360} \approx 357.643\text{ cm}^2\)

Curved surface area of the hemisphere:
\(A_{\text{hemisphere}} = 2\pi r^2 = 2\pi \times 6^2 = 72\pi \approx 226.195\text{ cm}^2\)

Total surface area:
\(A_{\text{total}} = 357.643 + 226.195 = 583.838\text{ cm}^2\).

Correct to 3 significant figures, this is \(584\text{ cm}^2\).

(a)
M1 for hemisphere volume expression: \(\frac{2}{3}\pi r^3\)
M1 for substituting \(h = 3r\) into cone volume to get \(\pi r^3\)
M1 for equating sum to \(360\pi\): \(\frac{5}{3}\pi r^3 = 360\pi\) (or equivalent)
A1 for resolving to \(r^3 = 216\) leading to \(r = 6\)

(b)
M1 for \(\sqrt{6^2 + 18^2}\)
A1 for \(19.0\) (or \(18.97...\))

(c)
M1 for curved surface area of cone: \(\pi \times 6 \times \text{their } l\)
M1 for curved surface area of hemisphere: \(2 \times \pi \times 6^2\) (or \(72\pi\))
M1 for adding their two surface areas together
A1 for showing exact or highly accurate values (e.g. \(357.6\) and \(226.2\))
A1 for \(584\) (accept \(583.8\) - \(584.1\))