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

(a) The interquartile range is the difference between the upper quartile and the lower quartile.
\(\text{IQR} = 41 - 24 = 17\text{ cm}\).

(b)(i) Since 41 cm is the upper quartile (75th percentile), \(25\%\) of the plants are taller than 41 cm.
Number of plants \(= 0.25 \times 200 = 50\).

(b)(ii) 15 cm is the 10th percentile and 52 cm is the 90th percentile.
The percentage of plants between these heights is \(90\% - 10\% = 80\%\).
Number of plants \(= 0.80 \times 200 = 160\).

(c) Since 24 cm is the lower quartile (25th percentile), \(25\%\) of the plants (which is 50 plants) have a height of 24 cm or less.
The probability that both of the chosen plants have a height of 24 cm or less (without replacement) is:
\(P = \frac{50}{200} \times \frac{49}{199} = \frac{1}{4} \times \frac{49}{199} = \frac{49}{796} \approx 0.0616\).

(a) [2 marks]:
M1 for \(41 - 24\)
A1 for 17

(b)(i) [2 marks]:
M1 for \(0.25 \times 200\)
A1 for 50

(b)(ii) [3 marks]:
M1 for identifying the percentiles: \(90\%\) and \(10\%\)
M1 for \(0.80 \times 200\)
A1 for 160

(c) [4 marks]:
M1 for identifying 50 plants
M1 for \(P = \frac{50}{200} \times \frac{49}{199}\)
A1 for \(\frac{49}{796}\) or equivalent fraction (accept decimal \(0.0616\) or \(0.061557...\))

(a) Total value \(= 8000 \times (1 + 0.024)^5 = 8000 \times (1.024)^5 \approx 9007.199\).
To the nearest dollar, this is $9007.

(b) Value after 3 years \(= 8000 \times (1 + \frac{r}{100})^3 = 8741.81\)
\((1 + \frac{r}{100})^3 = \frac{8741.81}{8000} \approx 1.09272625\)
\(1 + \frac{r}{100} = \sqrt[3]{1.09272625} \approx 1.03\)
\(\frac{r}{100} = 0.03 \implies r = 3\).

(c) Let \(x\) be the original price.
\(0.85x = 714\)
\(x = \frac{714}{0.85} = 840\).
The original price of the laptop was $840.

(d) Total number of boys \(= \frac{2}{5} \times 450 = 180\).
Total number of girls \(= 450 - 180 = 270\).
Boys playing tennis \(= 0.30 \times 180 = 54\).
Girls playing tennis \(= 0.40 \times 270 = 108\).
Total tennis players \(= 54 + 108 = 162\).

(a) [3 marks]:
M1 for \(8000 \times 1.024^5\)
A1 for \(9007.199...\)
A1 for 9007 (nearest dollar)

(b) [3 marks]:
M1 for setting up \(8000 \times (1 + \frac{r}{100})^3 = 8741.81\)
M1 for finding \(\sqrt[3]{\frac{8741.81}{8000}}\)
A1 for 3 (or 3.00)

(c) [2 marks]:
M1 for \(714 / 0.85\)
A1 for 840

(d) [3 marks]:
M1 for finding 180 boys or 270 girls
M1 for \(0.3 \times 180 + 0.4 \times 270\)
A1 for 162

(a) Midpoint \(= \left(\frac{-3 + 5}{2}, \frac{8 + (-2)}{2}\right) = \left(\frac{2}{2}, \frac{6}{2}\right) = (1, 3)\).

(b) Length \(AB = \sqrt{(5 - (-3))^2 + (-2 - 8)^2} = \sqrt{8^2 + (-10)^2} = \sqrt{64 + 100} = \sqrt{164}\).
Since \(164 = 4 \times 41\), we have:
\(AB = \sqrt{4 \times 41} = 2\sqrt{41}\).
Thus, \(k = 2\).

(c)(i) The centre of circle \(C\) is the midpoint of the diameter \(AB\), which is \((1, 3)\).

(c)(ii) The radius \(r\) is half the length of \(AB\):
\(r = \frac{2\sqrt{41}}{2} = \sqrt{41}\).
\(\text{Area} = \pi r^2 = \pi (\sqrt{41})^2 = 41\pi\).

(d) Let \(D\) have coordinates \((x, y)\).
\(\frac{-3 + x}{2} = 1 \implies -3 + x = 2 \implies x = 5\).
\(\frac{8 + y}{2} = 11 \implies 8 + y = 22 \implies y = 14\).
Therefore, the coordinates of \(D\) are \((5, 14)\).

(a) [2 marks]:
M1 for \(\frac{-3+5}{2}\) or \(\frac{8-2}{2}\)
A1 for \((1, 3)\)

(b) [3 marks]:
M1 for \((5 - (-3))^2 + (-2 - 8)^2\)
A1 for \(\sqrt{164}\)
A1 for \(2\sqrt{41}\)

(c)(i) [1 mark]:
B1 for \((1, 3)\) (or follow through from part (a))

(c)(ii) [3 marks]:
M1 for \(r = \frac{\text{their (b)}}{2}\)
M1 for \(\text{Area} = \pi \times r^2\)
A1 for \(41\pi\)

(d) [2 marks]:
M1 for setting up \(\frac{-3+x}{2} = 1\) or \(\frac{8+y}{2} = 11\)
A1 for \((5, 14)\)

(a) When \(x = 0\), \(f(0) = 5\). The \(y\)-intercept is \((0, 5)\).

(b) Using the GDC's local maximum finder:
\(x \approx -0.120\), \(y \approx 5.06\).
Coordinates are \((-0.120, 5.06)\).

(c) Using the GDC's local minimum finder:
\(x \approx 2.79\), \(y \approx -7.21\).
Coordinates are \((2.79, -7.21)\).

(d) Setting \(f(x) = 0\) and finding the roots using the GDC:
\(x \approx -1.09\), \(x \approx 1.16\), \(x \approx 3.93\).

(e) Solving \(x^3 - 4x^2 - x + 5 = 2x - 1\) is equivalent to solving \(x^3 - 4x^2 - 3x + 6 = 0\).
Using the GDC polynomial root finder, the largest solution is \(x \approx 4.38\) (accept \(4.37\)).

(a) [1 mark]:
B1 for \((0, 5)\)

(b) [3 marks]:
M1 for using derivative or GDC maximum tool
A1 for \(x \approx -0.120\)
A1 for \(y \approx 5.06\)

(c) [3 marks]:
M1 for using derivative or GDC minimum tool
A1 for \(x \approx 2.79\)
A1 for \(y \approx -7.21\)

(d) [3 marks]:
B1 for each correct root: \(-1.09\), \(1.16\), and \(3.93\) (each rounded to 3 s.f.)

(e) [1 mark]:
B1 for \(4.38\) (or \(4.37\))

(a) The mapping rule for reflection in the line \(y = -x\) is \((x, y) \to (-y, -x)\).
Applying this to the vertices of \(T\):
\(A(1, 1) \to (-1, -1)\)
\(B(3, 1) \to (-1, -3)\)
\(C(1, 4) \to (-4, -1)\)
So the vertices of \(U\) are \((-1, -1), (-1, -3), (-4, -1)\).

(b) Rotating \(90^\circ\) clockwise about the centre \((0, 2)\):
Subtract the centre: \((x, y) \to (x, y - 2)\).
Rotate \(90^\circ\) clockwise: \((X, Y) \to (Y, -X) \implies (y - 2, -x)\).
Add the centre back: \((y - 2, 2 - x)\).
Applying this to \(T\):
\(A(1, 1) \to (1 - 2, 2 - 1) = (-1, 1)\)
\(B(3, 1) \to (1 - 2, 2 - 3) = (-1, -1)\)
\(C(1, 4) \to (4 - 2, 2 - 1) = (2, 1)\)
So the vertices of \(V\) are \((-1, 1), (-1, -1), (2, 1)\).

(c) Enlarging by factor \(-2\) with centre \((2, 2)\):
New coordinates are \((2 - 2(x - 2), 2 - 2(y - 2)) = (6 - 2x, 6 - 2y)\).
Applying this to \(T\):
\(A(1, 1) \to (6 - 2, 6 - 2) = (4, 4)\)
\(B(3, 1) \to (6 - 6, 6 - 2) = (0, 4)\)
\(C(1, 4) \to (6 - 2, 6 - 8) = (4, -2)\)
So the vertices of \(W\) are \((4, 4), (0, 4), (4, -2)\).

(d) In the transformation from \((1, 1), (3, 1), (1, 4)\) to \((2, 1), (6, 1), (2, 4)\), the \(y\)-coordinates remain unchanged, while the \(x\)-coordinates are multiplied by 2.
This is a stretch with:
- stretch factor 2,
- invariant line \(y\)-axis (or the line \(x = 0\)).

(a) [3 marks]:
B1 for each correct vertex: \((-1, -1)\), \((-1, -3)\), \((-4, -1)\)

(b) [3 marks]:
B1 for each correct vertex: \((-1, 1)\), \((-1, -1)\), \((2, 1)\)

(c) [3 marks]:
B1 for each correct vertex: \((4, 4)\), \((0, 4)\), \((4, -2)\)

(d) [2 marks]:
B1 for 'Stretch'
B1 for 'factor 2, invariant line \(y\)-axis' (or \(x = 0\))

(a) First find \(f(4)\):
\(f(4) = \frac{2(4) + 1}{4 - 3} = \frac{9}{1} = 9\).
Then substitute into \(g(x)\):
\(g(9) = 3(9) - 5 = 22\).

(b) Let \(y = \frac{2x + 1}{x - 3}\).
Multiply by \(x - 3\):
\(y(x - 3) = 2x + 1 \implies yx - 3y = 2x + 1\).
Rearrange to isolate \(x\):
\(yx - 2x = 3y + 1 \implies x(y - 2) = 3y + 1\).
\(x = \frac{3y + 1}{y - 2}\).
Thus, \(f^{-1}(x) = \frac{3x + 1}{x - 2}\).

(c) Substitute \(f(x)\) into itself:
\(f(f(x)) = f\left(\frac{2x + 1}{x - 3}\right) = \frac{2\left(\frac{2x+1}{x-3}\right) + 1}{\frac{2x+1}{x-3} - 3}\).
Multiply numerator and denominator by \(x - 3\):
\(f(f(x)) = \frac{2(2x+1) + (x-3)}{(2x+1) - 3(x-3)} = \frac{4x + 2 + x - 3}{2x + 1 - 3x + 9} = \frac{5x - 1}{10 - x}\).

(d) Equate \(g(x)\) and \(f(x)\):
\(3x - 5 = \frac{2x + 1}{x - 3}\).
Multiply by \(x - 3\):
\((3x - 5)(x - 3) = 2x + 1 \implies 3x^2 - 14x + 15 = 2x + 1\).
Rearrange to standard form:
\(3x^2 - 16x + 14 = 0\).
Using the quadratic formula:
\(x = \frac{16 \pm \sqrt{(-16)^2 - 4(3)(14)}}{6} = \frac{16 \pm \sqrt{256 - 168}}{6} = \frac{16 \pm \sqrt{88}}{6}\).
\(x_1 \approx 4.23\) and \(x_2 \approx 1.10\).

(a) [2 marks]:
M1 for \(f(4) = 9\)
A1 for 22

(b) [3 marks]:
M1 for expanding \(y(x-3) = 2x+1\) or \(x(y-3) = 2y+1\)
M1 for factorising \(x\) (or \(y\)) to obtain \(x(y-2) = 3y+1\) or equivalent
A1 for \(\frac{3x + 1}{x - 2}\)

(c) [3 marks]:
M1 for writing \(\frac{2(\frac{2x+1}{x-3}) + 1}{\frac{2x+1}{x-3} - 3}\)
M1 for clearing the denominators of the fractions
A1 for \(\frac{5x - 1}{10 - x}\)

(d) [3 marks]:
M1 for simplifying to \(3x^2 - 16x + 14 = 0\)
M1 for substituting correctly into the quadratic formula or using the GDC equation solver
A1 for both \(1.10\) and \(4.23\) (correct to 3 s.f.)

(a)(i) This is an arithmetic sequence with first term \(a = 3\) and common difference \(d = 4\).
\(n\)-th term \(= 3 + 4(n-1) = 4n - 1\).

(a)(ii) Analyzing Sequence B:
First differences: \(7, 11, 15, 19\)
Second differences: \(4, 4, 4\)
Since the second difference is constant, it is a quadratic sequence in the form \(an^2 + bn + c\) where \(a = \frac{4}{2} = 2\).
Subtract \(2n^2\) from the terms of Sequence B:
\(n=1: 2 - 2(1)^2 = 0\)
\(n=2: 9 - 2(2)^2 = 1\)
\(n=3: 20 - 2(3)^2 = 2\)
\(n=4: 35 - 2(4)^2 = 3\)
This remaining sequence is \(0, 1, 2, 3, \dots\), which has the linear formula \(n - 1\).
Thus, the \(n\)-th term of Sequence B is \(2n^2 + n - 1\).

(a)(iii) This is a geometric sequence with first term \(a = 24\) and common ratio \(r = 0.5\).
\(n\)-th term \(= 24 \times 0.5^{n-1}\) (or \(3 \times 2^{4-n}\)).

(b)(i) Substitute \(n = 5\) into \(n^3 - 2n^2 + 5\):
\(5^3 - 2(5^2) + 5 = 125 - 50 + 5 = 80\).

(b)(ii) Set \(n^3 - 2n^2 + 5 = 149\):
\(n^3 - 2n^2 - 144 = 0\).
Testing positive integer values of \(n\):
If \(n = 6\):
\(6^3 - 2(6^2) - 144 = 216 - 72 - 144 = 0\).
Thus, 149 is indeed the 6th term.

(a)(i) [2 marks]:
M1 for identifying a common difference of 4
A1 for \(4n - 1\)

(a)(ii) [3 marks]:
M1 for identifying that the quadratic coefficient is \(2\) (i.e. \(2n^2\))
M1 for subtracting \(2n^2\) to find the remaining linear sequence \(n - 1\)
A1 for \(2n^2 + n - 1\)

(a)(iii) [2 marks]:
M1 for identifying the common ratio \(r = 0.5\) (or \(\frac{1}{2}\))
A1 for \(24 \times 0.5^{n-1}\) or equivalent

(b)(i) [2 marks]:
M1 for substituting \(n = 5\) into the given formula
A1 for 80

(b)(ii) [2 marks]:
M1 for setting up the equation \(n^3 - 2n^2 - 144 = 0\)
A1 for showing \(n = 6\) is the correct integer solution

(a) Substitute the second equation into the first:
\(2x - 2 = x^2 - 4x + 3\).
Rearrange into quadratic form:
\(x^2 - 6x + 5 = 0\).
Factorising this quadratic equation:
\((x - 1)(x - 5) = 0\).
So, \(x = 1\) or \(x = 5\).
Find the corresponding \(y\)-values:
If \(x = 1\), \(y = 2(1) - 2 = 0\).
If \(x = 5\), \(y = 2(5) - 2 = 8\).
The solutions are \((1, 0)\) and \((5, 8)\).

(b) Multiply the entire equation by the common denominator \((x + 2)(x - 1)\):
\(5(x - 1) + 3(x + 2) = 2(x + 2)(x - 1)\).
Expand both sides:
\(5x - 5 + 3x + 6 = 2(x^2 + x - 2)\)
\(8x + 1 = 2x^2 + 2x - 4\).
Rearrange into standard quadratic form:
\(2x^2 - 6x - 5 = 0\).
Apply the quadratic formula:
\(x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(-5)}}{2(2)}\)
\(x = \frac{6 \pm \sqrt{36 + 40}}{4} = \frac{6 \pm \sqrt{76}}{4}\).
\(x \approx 3.68\) or \(x \approx -0.679\).

(a) [5 marks]:
M1 for setting \(x^2 - 4x + 3 = 2x - 2\)
M1 for simplifying to \(x^2 - 6x + 5 = 0\)
M1 for factoring to \((x - 1)(x - 5) = 0\) or using quadratic formula
A1 for finding \(x = 1\) and \(x = 5\)
A1 for coordinates \((1, 0)\) and \((5, 8)\)

(b) [6 marks]:
M1 for multiplying to clear denominators: \(5(x-1) + 3(x+2) = 2(x+2)(x-1)\)
M1 for expanding LHS to \(8x + 1\)
M1 for expanding RHS to \(2x^2 + 2x - 4\)
M1 for rearranging to \(2x^2 - 6x - 5 = 0\)
A1 for correct application of quadratic formula \(\frac{6 \pm \sqrt{76}}{4}\)
A1 for the values \(-0.679\) and \(3.68\) (both correct to 3 s.f.)

\(\textbf{Part (a)(i)}\):
To find the estimate of the mean, we find the mid-interval values (\(x\)) for each class:
- For \(0 < t \le 10\), midpoint \(x = 5\)
- For \(10 < t \le 20\), midpoint \(x = 15\)
- For \(20 < t \le 30\), midpoint \(x = 25\)
- For \(30 < t \le 40\), midpoint \(x = 35\)
- For \(40 < t \le 50\), midpoint \(x = 45\)
- For \(50 < t \le 60\), midpoint \(x = 55\)

Next, we calculate the sum of the products of midpoints and frequencies (\(\sum f x\)):
\(\sum f x = (5 \times 12) + (15 \times 28) + (25 \times 38) + (35 \times 22) + (45 \times 15) + (55 \times 5)\)
\(\sum f x = 60 + 420 + 950 + 770 + 675 + 275 = 3150\)

The estimate of the mean is:
\(\text{Mean} = \frac{\sum f x}{\sum f} = \frac{3150}{120} = 26.25\) minutes.

\(\textbf{Part (a)(ii)}\):
The median is the value at the 50th percentile, which corresponds to the \(\frac{120}{2}\)th = 60th student.
We find the cumulative frequencies:
- For \(t \le 10\): 12
- For \(t \le 20\): \(12 + 28 = 40\)
- For \(t \le 30\): \(40 + 38 = 78\)
Since 60 lies between 40 and 78, the median lies in the interval \(20 < t \le 30\).

\(\textbf{Part (b)(i)}\):
The interquartile range (IQR) is:
\(\text{IQR} = \text{Upper Quartile} - \text{Lower Quartile}\)
\(\text{IQR} = 34.5 - 16.5 = 18\) minutes.

\(\textbf{Part (b)(ii)}\):
The time of 16.5 minutes corresponds to the 25th percentile (lower quartile).
The number of students who completed in 16.5 minutes or less is:
\(25\% \times 120 = 0.25 \times 120 = 30\).

\(\textbf{Part (b)(iii)}\):
The time of 44.0 minutes corresponds to the 90th percentile.
The percentage of students taking more than 44.0 minutes is \(100\% - 90\% = 10\%\).
The number of students is:
\(10\% \times 120 = 0.10 \times 120 = 12\).

\(\textbf{Part (a)(i)}\) [4 marks]:
M1 for finding mid-interval values of at least four intervals.
M1 for calculating \(\sum f x\) with their mid-intervals (at least 4 correct products shown).
A1 for \(\sum f x = 3150\).
A1 for 26.25 (or \(26\frac{1}{4}\)).

\(\textbf{Part (a)(ii)}\) [1 mark]:
B1 for \(20 < t \le 30\).

\(\textbf{Part (b)(i)}\) [2 marks]:
M1 for \(34.5 - 16.5\).
A1 for 18.

\(\textbf{Part (b)(ii)}\) [2 marks]:
M1 for \(\frac{25}{100} \times 120\) or identifying 30th value.
A1 for 30.

\(\textbf{Part (b)(iii)}\) [2 marks]:
M1 for \((1 - 0.90) \times 120\) or identifying 108th value.
A1 for 12.

\(\textbf{Part (a)(i)}\):
Using a graphic display calculator to find the local maximum, or by differentiation:
\(f'(x) = 3x^2 - 6x - 9 = 0 \Rightarrow 3(x-3)(x+1) = 0\).
The local maximum occurs at \(x = -1\).
The y-coordinate is \(f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 15 = -1 - 3 + 9 + 15 = 20\).
So, the local maximum is \((-1, 20)\).

\(\textbf{Part (a)(ii)}\):
The local minimum occurs at \(x = 3\).
The y-coordinate is \(f(3) = (3)^3 - 3(3)^2 - 9(3) + 15 = 27 - 27 - 27 + 15 = -12\).
So, the local minimum is \((3, -12)\).

\(\textbf{Part (a)(iii)}\):
Using the root-finding function on the GDC for \(x^3 - 3x^2 - 9x + 15 = 0\):
- First root: \(x \approx -2.62\)
- Second root: \(x \approx 1.34\)
- Third root: \(x \approx 4.28\)

\(\textbf{Part (b)(i)}\):
To find the intersection of \(y = f(x)\) and \(y = g(x)\), we find the roots of:
\(x^3 - 3x^2 - 9x + 15 = 2x + 3 \Rightarrow x^3 - 3x^2 - 11x + 12 = 0\).
Using a GDC to find the intersections:
- \(x \approx -2.71\)
- \(x \approx 0.929\)
- \(x \approx 4.78\)

\(\textbf{Part (b)(ii)}\):
The inequality \(f(x) > g(x)\) is satisfied where the cubic curve lies above the straight line.
By observing the intersection points on the GDC:
- For \(x < -2.71\), \(f(x) < g(x)\).
- For \(-2.71 < x < 0.929\), the curve is above the line, so \(f(x) > g(x)\).
- For \(0.929 < x < 4.78\), the curve is below the line, so \(f(x) < g(x)\).
- For \(x > 4.78\), the curve is above the line, so \(f(x) > g(x)\).

Thus, the solution is \(-2.71 < x < 0.929\) or \(x > 4.78\).

\(\textbf{Part (a)(i)}\) [2 marks]:
M1 for \(x = -1\) or \(y = 20\) seen.
A1 for \((-1, 20)\).

\(\textbf{Part (a)(ii)}\) [2 marks]:
M1 for \(x = 3\) or \(y = -12\) seen.
A1 for \((3, -12)\).

\(\textbf{Part (a)(iii)}\) [3 marks]:
B1 for \(-2.62\).
B1 for \(1.34\).
B1 for \(4.28\).
(Accept alternative rounding: e.g. -2.619, 1.337, 4.282).

\(\textbf{Part (b)(i)}\) [2 marks]:
M1 for setting \(x^3 - 3x^2 - 9x + 15 = 2x + 3\).
A1 for \(-2.71, 0.929, 4.78\).

\(\textbf{Part (b)(ii)}\) [2 marks]:
B1 for \(-2.71 < x < 0.929\).
B1 for \(x > 4.78\).

\(\textbf{Part (a)}\):
Let North be represented by a vertical line at each port.
- The bearing of \(B\) from \(A\) is \(055^\circ\).
- Using alternate angles, the angle of the line \(AB\) with the South line at \(B\) is \(55^\circ\).
- The bearing of \(C\) from \(B\) is \(135^\circ\). The angle of \(BC\) with the South line at \(B\) is \(180^\circ - 135^\circ = 45^\circ\) (to the east of South).
- Thus, the total angle \(\angle ABC = 55^\circ + 45^\circ = 100^\circ\).

\(\textbf{Part (b)}\):
Using the Cosine Rule on triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)\)
\(AC^2 = 12.0^2 + 18.5^2 - 2 \cdot 12.0 \cdot 18.5 \cdot \cos(100^\circ)\)
\(AC^2 = 144 + 342.25 - 444 \cdot (-0.173648)\)
\(AC^2 = 486.25 + 77.100\)
\(AC^2 = 563.350\)
\(AC = \sqrt{563.350} \approx 23.735\text{ km}\).
To 3 significant figures, \(AC = 23.7\text{ km}\).

\(\textbf{Part (c)}\):
Using the Sine Rule to find angle \(\angle BAC\):
\(\frac{\sin(\angle BAC)}{BC} = \frac{\sin(\angle ABC)}{AC}\)
\(\sin(\angle BAC) = \frac{18.5 \cdot \sin(100^\circ)}{23.735}\)
\(\sin(\angle BAC) \approx \frac{18.5 \cdot 0.984808}{23.735} \approx 0.767597\)
\(\angle BAC = \sin^{-1}(0.767597) \approx 50.14^\circ\).

Since \(C\) lies to the east of the line \(AB\), the bearing of \(C\) from \(A\) is:
\(\text{Bearing} = 55^\circ + 50.14^\circ = 105.14^\circ\).
To 1 decimal place, the bearing is \(105.1^\circ\).

\(\textbf{Part (d)}\):
The area of triangle \(ABC\) is given by:
\(\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(\angle ABC)\)
\(\text{Area} = \frac{1}{2} \cdot 12.0 \cdot 18.5 \cdot \sin(100^\circ)\)
\(\text{Area} = 111 \cdot \sin(100^\circ) \approx 111 \cdot 0.984808 = 109.31\text{ km}^2\).
To 3 significant figures, the area is \(109\text{ km}^2\).

\(\textbf{Part (a)}\) [2 marks]:
M1 for \(180^\circ - 135^\circ = 45^\circ\) or alternate angle \(55^\circ\) identified.
A1 for a complete logical proof showing \(55^\circ + 45^\circ = 100^\circ\).

\(\textbf{Part (b)}\) [3 marks]:
M1 for substituting correctly into the Cosine Rule: \(12.0^2 + 18.5^2 - 2(12.0)(18.5)\cos(100^\circ)\).
A1 for \(AC^2 \approx 563.35\).
A1 for 23.7 (accept 23.73 to 23.74).

\(\textbf{Part (c)}\) [4 marks]:
M1 for substituting correctly into the Sine Rule: \(\frac{\sin(\angle BAC)}{18.5} = \frac{\sin(100^\circ)}{\text{their } AC}\).
A1 for \(\angle BAC \approx 50.1^\circ\) (accept 50.14).
M1 for adding \(55^\circ\) to their \(\angle BAC\).
A1 for 105.1 (accept 105.14).

\(\textbf{Part (d)}\) [2 marks]:
M1 for \(\frac{1}{2} \times 12.0 \times 18.5 \times \sin(100^\circ)\).
A1 for 109 (accept 109.3).