2021 HKDSE Mathematics Practice Paper with Answers

\(\frac{(u^3 v^{-2})^4}{u^{-5} v^3} = \frac{u^{12} v^{-8}}{u^{-5} v^3} = u^{12 - (-5)} v^{-8 - 3} = u^{17} v^{-11} = \frac{u^{17}}{v^{11}}\)

1M for \((u^3 v^{-2})^4 = u^{12} v^{-8}\) or applying index law to numerator; 1M for combining indices \(u^{12 - (-5)}\) or \(v^{-8 - 3}\); 1A for \(\frac{u^{17}}{v^{11}}\)

\(\frac{3p + 2q}{5 - p} = 4 \implies 3p + 2q = 4(5 - p) \implies 3p + 2q = 20 - 4p \implies 3p + 4p = 20 - 2q \implies 7p = 20 - 2q \implies p = \frac{20 - 2q}{7}\)

1M for clearing denominator: \(3p + 2q = 4(5 - p)\); 1M for grouping terms involving \(p\) on one side: \(7p = 20 - 2q\); 1A for \(p = \frac{20 - 2q}{7}\) or equivalent

(a) \(3x^2 - 5xy - 2y^2 = (3x + y)(x - 2y)\). (b) \(3x^2 - 5xy - 2y^2 - 9x - 3y = (3x + y)(x - 2y) - 3(3x + y) = (3x + y)(x - 2y - 3)\).

(a) 1A for \((3x + y)(x - 2y)\); (b) 1M for using the result of (a): \((3x + y)(x - 2y) - 3(3x + y)\); 1A for \((3x + y)(x - 2y - 3)\)

(a) For the first inequality: \(5x - 1 + 6 \ge 6(x - 1) \implies 5x + 5 \ge 6x - 6 \implies x \le 11\). For the second inequality: \(4 - 3x < 16 \implies -3x < 12 \implies x > -4\). Combining both: \(-4 < x \le 11\). (b) The negative integers satisfying the range are \(-3, -2, -1\). Therefore, there are 3 negative integers.

(a) 1M for solving either linear inequality correctly; 1M for obtaining both \(x \le 11\) and \(x > -4\); 1A for \(-4 < x \le 11\); (b) 1A for 3

(a) Let \(\$C\) be the cost of the handbag. Marked price \(= 1.4C\). Selling price \(= 1.4C \times (1 - 25\%) = 1.05C\). Profit percentage \(= \frac{1.05C - C}{C} \times 100\% = 5\%\). (b) Profit \(= 1.05C - C = 0.05C = 45 \implies C = 900\). Marked price \(= 1.4 \times 900 = \$1260\).

(a) 1M for expressing selling price in terms of cost: \(1.4C \times 0.75\); 1A for \(5\%\); (b) 1M for setting up equation to find cost: \(0.05C = 45\) or marked price directly; 1A for \(\$1260\)

(a) \(\angle AOB = 165^\circ - 75^\circ = 90^\circ\). (b) Since \(\triangle OAB\) is a right-angled triangle at \(O\), \(AB = \sqrt{OA^2 + OB^2} = \sqrt{12^2 + 5^2} = 13\). (c) Area of \(\triangle OAB = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times 12 \times 5 = 30\).

(a) 1A for \(90^\circ\); (b) 1M for using Pythagoras' theorem \(\sqrt{12^2 + 5^2}\) or cosine rule; 1A for 13; (c) 1A for 30

(a) In \(\triangle ABC\) and \(\triangle AEB\): \(\angle BAC = \angle EAB\) (common angle). \(\angle ACB = \angle ADB\) (angles in the same segment). Since \(AB = AD\), \(\angle ADB = \angle ABD = \angle ABE\) (base angles, isosceles triangle). Therefore, \(\angle ACB = \angle ABE\). Hence, \(\triangle ABC \sim \triangle AEB\) (AA). (b) Since \(\triangle ABC \sim \triangle AEB\), \(\frac{AC}{AB} = \frac{AB}{AE} \implies \frac{AC}{12} = \frac{12}{8} \implies AC = 18\text{ cm}\). Therefore, \(EC = AC - AE = 18 - 8 = 10\text{ cm}\).

(a) 2 marks for full proof with correct geometric reasons (1 mark if reasons omitted or incomplete); (b) 1M for setting up ratio \(\frac{AC}{AB} = \frac{AB}{AE}\); 1A for \(10\text{ cm}\)

(a) Total number of workers \(N = 20\). The 10th and 11th values are 60 and 62. Median \(= \frac{60 + 62}{2} = 61\). Range \(= 78 - 42 = 36\). Lower quartile \(Q_1 = \frac{50 + 53}{2} = 51.5\). Upper quartile \(Q_3 = \frac{67 + 69}{2} = 68\). Inter-quartile range \(= Q_3 - Q_1 = 68 - 51.5 = 16.5\). (b) The number of workers with wage at least \(\$60\) is \(7 + 4 = 11\). Thus, the required probability \(= \frac{11}{20}\) (or \(0.55\)).

(a) 1A for median \(= 61\); 1A for range \(= 36\); 1M for finding \(Q_1\) and \(Q_3\); 1A for inter-quartile range \(= 16.5\); (b) 1A for \(\frac{11}{20}\) (or 0.55)

(a)
\(\dfrac{5(2-x)}{3} + 4 > 2(x+1)\)
\(5(2-x) + 12 > 6(x+1)\)
\(10 - 5x + 12 > 6x + 6\)
\(22 - 5x > 6x + 6\)
\(-11x > -16\)
\(x < \dfrac{16}{11}\)

\(2x + 7 \ge 0\)
\(2x \ge -7\)
\(x \ge -\dfrac{7}{2}\)

Thus, the required range is \(-\dfrac{7}{2} \le x < \dfrac{16}{11}\).

(b) The integers satisfying the range are \(-3, -2, -1, 0, 1\).
Thus, the number of integers is \(5\).

(a)
\(x < \dfrac{16}{11}\) (1M for putting \(x\) on one side)
\(x \ge -\dfrac{7}{2}\) (1M)
\(-\dfrac{7}{2} \le x < \dfrac{16}{11}\) (1A)

(b)
\(5\) (1A)

(a) Let $C = a + br^2$, where $a$ and $b$ are non-zero constants.
When $r = 3$, $a + 9b = 92$.
When $r = 5$, $a + 25b = 188$.
Subtracting the two equations:
$16b = 96 \implies b = 6$.
Substituting $b = 6$ into the first equation:
$a + 9(6) = 92 \implies a = 38$.
Thus, $C = 38 + 6r^2$.
When $r = 6$,
$C = 38 + 6(6^2) = 38 + 216 = 254$.
The cost is $$254\$.

(b) When \$C = 332\$,
\$38 + 6r^2 = 332\$
\$6r^2 = 294\$
\$r^2 = 49\$
\$r = 7\$ (since \$r > 0\$)
Thus, the radius of the ornament is \$7\text{ cm}\$.

(c) The cost of an ornament of radius \$3\text{ cm}\$ is $$92$.
Two times this cost is $2 \times $92 = $184$.
However, the cost of an ornament of radius $6\text{ cm}$ is $$254
eq $184$.
Thus, the claim is not agreed.

(a) 1M for setting up $C = a + br^2$
1M for finding $a = 38$ and $b = 6$
1A for $254$ or $$254$

(b) 1M for substituting $C = 332$ into $38 + 6r^2 = 332$
1A for $r = 7\text{ cm}$

(c) 1M for comparing $2 \times 92 = 184$ with $254$
1A (f.t.) for correct conclusion with valid reason

(a) Since $x - 2$ is a factor of $p(x)$, by the Factor Theorem,
$p(2) = 0$
$2(2)^3 + a(2)^2 - 11(2) + b = 0$
$16 + 4a - 22 + b = 0$
$4a + b = 6$ ... (1)

By the Remainder Theorem,
$p(-1) = 18$
$2(-1)^3 + a(-1)^2 - 11(-1) + b = 18$
$-2 + a + 11 + b = 18$
$a + b = 9$ ... (2)

Subtracting (2) from (1):
$3a = -3 \implies a = -1$.
Substituting $a = -1$ into (2):
$-1 + b = 9 \implies b = 10$.

(b) Substituting $a = -1$ and $b = 10$,
$p(x) = 2x^3 - x^2 - 11x + 10$.
Since $x - 2$ is a factor of $p(x)$,
$p(x) = (x - 2)(2x^2 + 3x - 5)$
$= (x - 2)(2x + 5)(x - 1)$.

For $p(x) = 0$,
$(x - 2)(2x + 5)(x - 1) = 0$
$x = 2$, $x = -\frac{5}{2}$, or $x = 1$.
Since $2$, $-\frac{5}{2}$, and $1$ are all rational numbers, the claim is correct.

(a) 1M for applying $p(2) = 0$
1M for applying $p(-1) = 18$
1A for $a = -1$
1A for $b = 10$

(b) 1M for factorizing $p(x)$ into $(x - 2)(2x^2 + 3x - 5)$ or $(x - 2)(2x + 5)(x - 1)$
1M for solving to find roots $2, -\frac{5}{2}, 1$
1A (f.t.) for concluding the claim is correct with justification

(a) The data in ascending order are:
$52, 55, 58, 60, 62, 62, 65, 67, 68, 71, 73, 74, 76, 78, 78, 79, 80, 82, 85, 86$.

Median $= \frac{71 + 73}{2} = 72$.
Lower quartile $Q_1 = \frac{62 + 62}{2} = 62$.
Upper quartile $Q_3 = \frac{78 + 79}{2} = 78.5$.
Inter-quartile range $= 78.5 - 62 = 16.5$.
Standard deviation $\approx 9.8614 \approx 9.86$.

(b) (i) Let the new hourly wage be $Y = 1.1X + 5$.
New median $= 1.1 \times 72 + 5 = 79.2 + 5 = 84.2$.
New inter-quartile range $= 1.1 \times 16.5 = 18.15$.

(ii) New standard deviation $= 1.1 \times 9.861415 = 10.84755... \approx 10.8$ (or $10.85$).

(a) 1A for median $= 72$
1A for inter-quartile range $= 16.5$
1A for standard deviation $\approx 9.86$ (accept $9.86$ to $9.861$)

(b)(i) 1A (f.t.) for new median $= 84.2$
1A (f.t.) for new IQR $= 18.15$

(b)(ii) 1M for new $\text{SD} = 1.1 \times \text{old SD}$
1A (f.t.) for new $\text{SD} \approx 10.8$ (accept $10.85$)

(a) Rewriting the equation in standard form:
$(x - 4)^2 + (y + 2)^2 = 5 + 4^2 + (-2)^2 = 25$
Coordinates of the centre $G = (4, -2)$.
Radius $r = \sqrt{25} = 5$.

(b) Since $L$ is tangent to $C$, the perpendicular distance from the centre $(4, -2)$ to the line $3x - 4y + k = 0$ is equal to the radius $5$.
$\frac{|3(4) - 4(-2) + k|}{\sqrt{3^2 + (-4)^2}} = 5$
$\frac{|12 + 8 + k|}{5} = 5$
$|20 + k| = 25$
$20 + k = 25 \implies k = 5$
$20 + k = -25 \implies k = -45$
Thus, the values of $k$ are $5$ and $-45$.

(c) The larger value of $k$ is $5$, so $L_1: 3x - 4y + 5 = 0$.
The line passing through the centre $(4, -2)$ perpendicular to $L_1$ has slope $-\frac{4}{3}$.
Its equation is:
$y - (-2) = -\frac{4}{3}(x - 4)$
$3(y + 2) = -4x + 16$
$4x + 3y - 10 = 0$

Solving the system:
$\begin{cases} 3x - 4y + 5 = 0 \\ 4x + 3y - 10 = 0 \end{cases}$
From the first equation, $y = \frac{3x + 5}{4}$.
Substitute into the second equation:
$4x + 3\left(\frac{3x + 5}{4}\right) - 10 = 0$
$16x + 9x + 15 - 40 = 0$
$25x - 25 = 0 \implies x = 1$.
$y = \frac{3(1) + 5}{4} = 2$.
Thus, the coordinates of the point of contact are $(1, 2)$.

(a) 1A for centre $(4, -2)$
1A for radius $5$

(b) 1M for using perpendicular distance formula or discriminant $\Delta = 0$
1M for setting up $|20 + k| = 25$
1A for $k = 5$ or $k = -45$

(c) 1M for finding the equation of the normal line $4x + 3y - 10 = 0$ or solving by substitution into $C$
1A for $(1, 2)$

(a) Volume of cylinder $P = \pi (6^2)(10) = 360\pi\text{ cm}^3$.
Volume of cone $Q = \frac{1}{3}\pi (8^2)(15) = 320\pi\text{ cm}^3$.
Ratio of volume of $P$ to volume of $Q = 360\pi : 320\pi = 9 : 8$.

(b) (i) Since cone $R$ is similar to cone $Q$, the ratio of linear dimensions is:
$k = \sqrt{\frac{9}{16}} = \frac{3}{4}$.
$\frac{\text{Volume of } R}{\text{Volume of } Q} = \left(\frac{3}{4}\right)^3 = \frac{27}{64}$.
Volume of cone $R = \frac{27}{64} \times 320\pi = 135\pi\text{ cm}^3$.

(ii) Total volume of sphere $S = \text{Volume of } P + \text{Volume of } R$
$= 360\pi + 135\pi = 495\pi\text{ cm}^3$.

Let $r_S$ be the radius of sphere $S$.
$\frac{4}{3}\pi r_S^3 = 495\pi$
$r_S^3 = \frac{495 \times 3}{4} = 371.25$
$r_S = \sqrt[3]{371.25} \approx 7.1868\text{ cm} > 7\text{ cm}$.
(Alternatively: Volume of sphere with radius $7\text{ cm} = \frac{4}{3}\pi(7^3) = \frac{1372\pi}{3} \approx 457.33\pi < 495\pi$.)
Thus, the claim is correct.

(a) 1M for calculating volumes of $P$ and $Q$
1A for $9 : 8$ (or $\frac{9}{8}$)

(b)(i) 1M for using $\left(\sqrt{\frac{9}{16}}\right)^3 = \frac{27}{64}$
1A for $135\pi\text{ cm}^3$

(b)(ii) 1M for finding total volume $= 495\pi$
1M for setting $\frac{4}{3}\pi r_S^3 = 495\pi$ or calculating volume for $r = 7$
1A (f.t.) for correct conclusion with reasoning ($r_S \approx 7.19 > 7$)

(a) Treating the 3 girls as 1 single unit, there are 6 units (5 boys and 1 girl-group) to arrange.
Number of ways = \(6! \times 3! = 720 \times 6 = 4320\).

(b) (i) Total number of possible arrangements = \(8! = 40320\).
First arrange the 5 boys in a line: \(5! = 120\) ways.
There are 6 available positions (including the ends) to insert the 3 girls such that no two girls are adjacent.
Number of ways to place the girls = \(P^6_3 = 6 \times 5 \times 4 = 120\).
Number of favorable arrangements = \(120 \times 120 = 14400\).
Probability = \(\frac{14400}{40320} = \frac{5}{14}\).

(ii) Consider the condition that both ends of the line are occupied by boys.
Number of ways to choose and arrange 2 boys at the ends = \(P^5_2 = 20\).
Number of ways to arrange the remaining 6 students between the ends = \(6! = 720\).
Total arrangements with boys at both ends = \(20 \times 720 = 14400\).

Among these, find the number of arrangements where no two girls are adjacent:
The 5 boys are arranged in \(5! = 120\) ways. Since both ends must be boys, the 3 girls must be placed into the 4 spaces strictly between the boys: \(P^4_3 = 24\) ways.
Number of arrangements where both ends are boys and no two girls are adjacent = \(120 \times 24 = 2880\).

Therefore, the number of favorable arrangements where both ends are boys and at least two girls are adjacent = \(14400 - 2880 = 11520\).
Probability = \(\frac{11520}{40320} = \frac{2}{7}\).

(a) \(6! \times 3!\) (1M)
\(4320\) (1A)

(b)(i) Number of arrangements with no two girls adjacent = \(5! \times P^6_3 = 14400\) (1M)
\(\frac{5! \times P^6_3}{8!}\) (1M)
\(\frac{5}{14}\) (or approx. 0.357) (1A)

(b)(ii) Total arrangements with boys at ends = \(P^5_2 \times 6! = 14400\) (1M for finding arrangements with boys at ends or subtracting non-adjacent)
\(\frac{14400 - 5! \times P^4_3}{8!} = \frac{11520}{40320} = \frac{2}{7}\) (or approx. 0.286) (1A)

(a) Let \(a\) be the first term and \(d\) be the common difference.
\(T(3) = a + 2d = 19\)
\(T(7) = a + 6d = 43\)
Subtracting the first equation from the second:
\(4d = 24 \implies d = 6\)
\(a = 19 - 2(6) = 7\).

(b) \(\sum_{n=1}^k T(n) = \frac{k}{2}\big(2(7) + (k-1)(6)\big) = \frac{k}{2}(6k + 8) = 3k^2 + 4k\).
Set \(3k^2 + 4k > 2024 \iff 3k^2 + 4k - 2024 > 0\).
Solving the quadratic equation \(3k^2 + 4k - 2024 = 0\):
\(k = \frac{-4 \pm \sqrt{4^2 - 4(3)(-2024)}}{2(3)} = \frac{-4 \pm \sqrt{24304}}{6} \approx 25.32\) or \(-26.65\).
Since \(k\) must be a positive integer, \(k > 25.32\).
Thus, the least value of \(k\) is 26.

(c) \(G(1) \cdot G(2) \dotsm G(10) = 2^{T(1)} \cdot 2^{T(2)} \dotsm 2^{T(10)} = 2^{\sum_{n=1}^{10} T(n)}\).
Using the formula from (b):
\(\sum_{n=1}^{10} T(n) = 3(10)^2 + 4(10) = 340\).
So \(G(1) \cdot G(2) \dotsm G(10) = 2^{340}\).
\(\log_4(2^{340}) = \log_4\big((4^{1/2})^{340}\big) = \log_4(4^{170}) = 170\).

(a) For setting up equations \(a+2d=19\) and \(a+6d=43\) (1M)
\(a = 7\) and \(d = 6\) (1A)

(b) \(\sum_{n=1}^k T(n) = 3k^2 + 4k\) (1M)
For solving \(3k^2 + 4k - 2024 > 0\) (1M)
\(k = 26\) (1A)

(c) \(\sum_{n=1}^{10} T(n) = 340\) or \(G(1)\dotsm G(10) = 2^{340}\) (1M)
\(170\) (1A)

(a) Slope of \(L_1 = \frac{4 - 0}{0 - (-2)} = 2\).
Equation of \(L_1\): \(y - 4 = 2(x - 0) \implies 2x - y + 4 = 0\).
Since \(L_2 \perp L_1\), the slope of \(L_2 = -\frac{1}{2}\).
Equation of \(L_2\): \(y - 2 = -\frac{1}{2}(x - 4) \implies x + 2y - 8 = 0\).

(b) (i) Solving \(2x - y + 4 = 0\) and \(x + 2y - 8 = 0\):
\(y = 2x + 4 \implies x + 2(2x + 4) - 8 = 0 \implies 5x = 0 \implies x = 0, y = 4\).
The intersection point of \(L_1\) and \(L_2\) is \((0, 4)\).
The \(x\)-intercept of \(L_1\) is \((-2, 0)\).
The \(x\)-intercept of \(L_2\) is \((8, 0)\).
The region \(R\) is bounded by \((-2, 0)\), \((8, 0)\), and \((0, 4)\).
Testing an interior point, say \((0, 1)\):
\(2(0) - (1) + 4 = 3 \ge 0\),
\(0 + 2(1) - 8 = -6 \le 0\),
\(1 \ge 0\).
Thus, the system of inequalities is:
\(\begin{cases} 2x - y + 4 \ge 0 \\ x + 2y - 8 \le 0 \\ y \ge 0 \end{cases}\)

(ii) Let \(P(x, y) = 5x - 2y + 7\).
Evaluating \(P(x, y)\) at the three vertices of \(R\):
At \((-2, 0)\): \(P(-2, 0) = 5(-2) - 2(0) + 7 = -3\).
At \((8, 0)\): \(P(8, 0) = 5(8) - 2(0) + 7 = 47\).
At \((0, 4)\): \(P(0, 4) = 5(0) - 2(4) + 7 = -1\).
Therefore, the maximum value of \(5x - 2y + 7\) is 47, and the minimum value is \(-3\).

(a) Equation of \(L_1\): \(2x - y + 4 = 0\) (1A)
Equation of \(L_2\): \(x + 2y - 8 = 0\) (1A)

(b)(i) Intersection point \((0, 4)\) (1M)
System of inequalities: \(2x - y + 4 \ge 0\), \(x + 2y - 8 \le 0\), and \(y \ge 0\) (1A)

(b)(ii) Testing vertices \((-2,0)\), \((8,0)\), and \((0,4)\) (1M)
Maximum value = \(47\) (1A)
Minimum value = \(-3\) (1A)

(a) Centre of \(C = \left(-\frac{-8}{2}, -\frac{4}{2}\right) = (4, -2)\).
Radius of \(C = \sqrt{4^2 + (-2)^2 - (-5)} = \sqrt{16 + 4 + 5} = \sqrt{25} = 5\).

(b) (i) Since \(L\) is tangent to \(C\), the perpendicular distance from the centre \((4, -2)\) to \(L\) equals the radius 5.
\(\frac{|3(4) - 4(-2) + k|}{\sqrt{3^2 + (-4)^2}} = 5\)
\(\frac{|12 + 8 + k|}{5} = 5\)
\(|k + 20| = 25\)
\(k + 20 = 25 \implies k = 5\)
or \(k + 20 = -25 \implies k = -45\).

(ii) The \(y\)-intercept of \(3x - 4y + k = 0\) is \(-\frac{k}{-4} = \frac{k}{4}\).
The smaller \(y\)-intercept occurs when \(k = -45\), so \(L_1: 3x - 4y - 45 = 0\).
The line perpendicular to \(L_1\) passing through the centre \((4, -2)\) has slope \(-\frac{4}{3}\).
Equation of this normal line:
\(y - (-2) = -\frac{4}{3}(x - 4) \implies 4x + 3y - 10 = 0\).
Solving the system:
\(\begin{cases} 3x - 4y = 45 \\ 4x + 3y = 10 \end{cases}\)
Multiply the first by 3 and the second by 4:
\(9x - 12y = 135\)
\(16x + 12y = 40\)
Adding them: \(25x = 175 \implies x = 7\).
Substitute \(x = 7\) into \(4(7) + 3y = 10 \implies 28 + 3y = 10 \implies 3y = -18 \implies y = -6\).
Thus, the point of contact is \((7, -6)\).

(a) Centre \((4, -2)\) (1A)
Radius \(5\) (1A)

(b)(i) \(\frac{|3(4) - 4(-2) + k|}{\sqrt{3^2 + (-4)^2}} = 5\) (1M)
\(|k + 20| = 25\) (1M)
\(k = 5\) or \(k = -45\) (1A)

(b)(ii) Equation of the line through centre perpendicular to \(L_1\): \(4x + 3y - 10 = 0\) (1M)
Point of contact = \((7, -6)\) (1A)

(a) Since \(VB\) is vertically above \(B\), \(VB \perp AB\) and \(VB \perp BC\).
\(VA = \sqrt{VB^2 + AB^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15\text{ cm}\).
\(VC = \sqrt{VB^2 + BC^2} = \sqrt{9^2 + 16^2} = \sqrt{81 + 256} = \sqrt{337}\text{ cm} \approx 18.3575\text{ cm}\).

(b) (i) In \(\triangle ABC\), \(AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 16^2} = 20\text{ cm}\).
In \(\triangle VAC\), by the cosine rule:
\(\cos \angle VAC = \frac{VA^2 + AC^2 - VC^2}{2(VA)(AC)} = \frac{15^2 + 20^2 - 337}{2(15)(20)} = \frac{225 + 400 - 337}{600} = \frac{288}{600} = 0.48\).
\(\angle VAC = \arccos(0.48) \approx 61.3146^\circ \approx 61.3^\circ\).

(ii) Let \(N\) be the foot of the perpendicular from \(B\) to \(AC\).
Since \(VB \perp\) plane \(ABC\) and \(BN \perp AC\), by the theorem of three perpendiculars, \(VN \perp AC\).
Thus, the angle between plane \(VAC\) and the horizontal ground \(ABC\) is \(\angle VNB\).
Area of \(\triangle ABC = \frac{1}{2} \times 12 \times 16 = 96\text{ cm}^2\).
Also, Area \(= \frac{1}{2} \times AC \times BN = \frac{1}{2} \times 20 \times BN = 10BN\).
\(10BN = 96 \implies BN = 9.6\text{ cm}\).
In the right-angled \(\triangle VBN\):
\(\tan \angle VNB = \frac{VB}{BN} = \frac{9}{9.6} = 0.9375\).
\(\angle VNB = \arctan(0.9375) \approx 43.1524^\circ \approx 43.2^\circ\).

(a) \(VA = 15\text{ cm}\) (1A)
\(VC = \sqrt{337}\text{ cm}\) (or approx. 18.4 cm) (1A)

(b)(i) \(AC = 20\text{ cm}\) (1M)
\(\cos \angle VAC = \frac{15^2 + 20^2 - 337}{2(15)(20)}\) (1M)
\(\angle VAC \approx 61.3^\circ\) (1A, accept 61.31°)

(b)(ii) Finding \(BN = 9.6\text{ cm}\) (1M)
\(\tan \angle VNB = \frac{9}{9.6}\) and \(\angle VNB \approx 43.2^\circ\) (1A, accept 43.15°)

Express all terms with base 3:
\begin{aligned}
\frac{(3^{2n})(9^{n+1})}{27^{n-1}} &= \frac{3^{2n} \cdot (3^2)^{n+1}}{(3^3)^{n-1}} \\
&= \frac{3^{2n} \cdot 3^{2n+2}}{3^{3n-3}} \\
&= 3^{2n + (2n+2) - (3n-3)} \\
&= 3^{4n+2 - 3n + 3} \\
&= 3^{n+5}
\end{aligned}

Correct answer: B (1 mark)

Multiply both sides by \( x + 4y \):
\begin{aligned}
3x - 2y &= k(x + 4y) \\
3x - 2y &= kx + 4ky \\
3x - kx &= 4ky + 2y \\
x(3 - k) &= 2y(2k + 1) \\
x &= \frac{2(2k + 1)y}{3 - k}
\end{aligned}

Correct answer: C (1 mark)

By the Remainder Theorem:
\( f(1) = 2(1)^3 + a(1)^2 - 5(1) + b = a + b - 3 = 4 \implies a + b = 7 \).
We can verify using \( f(-2) = -20 \):
\( f(-2) = 2(-8) + 4a + 10 + b = 4a + b - 6 = -20 \implies 4a + b = -14 \).
Subtracting the first from the second: \( 3a = -21 \implies a = -7 \), then \( b = 14 \).
Thus, \( a + b = 7 \).

Correct answer: A (1 mark)

Solve the first inequality:
\begin{aligned}
2x + 5 &\ge 3(x - 1) \\
2x + 5 &\ge 3x - 3 \\
-x &\ge -8 \\
x &\le 8
\end{aligned}

Solve the second inequality:
\begin{aligned}
4 - 3x &< 13 \\
-3x &< 9 \\
x &> -3
\end{aligned}

Combining both gives \( -3 < x \le 8 \).
The integers satisfying this range are \( -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8 \).
There are \( 8 - (-2) + 1 = 11 \) integers.

Correct answer: D (1 mark)

Let \( C \) be the cost of the item.
Marked price \( M = C \times (1 + 40\%) = 1.4C \).
Selling price \( S = M \times (1 - 25\%) = 1.4C \times 0.75 = 1.05C \).
Given \( S = 420 \):
\( 1.05C = 420 \implies C = \frac{420}{1.05} = 400 \).
Profit \( = S - C = 420 - 400 = \$20 \).

Correct answer: A (1 mark)

Volume of cylinder \( = \pi r^2 h = \pi (6^2)(14) = 504\pi\text{ cm}^3 \).
Let \( H \) be the height of each cone.
Total volume of two cones \( = 2 \times \left(\frac{1}{3}\pi (4^2) H\right) = \frac{32}{3}\pi H \).
Equating the volumes:
\begin{aligned}
\frac{32}{3}\pi H &= 504\pi \\
32H &= 1512 \\
H &= \frac{1512}{32} = 47.25\text{ cm}
\end{aligned}

Correct answer: B (1 mark)

Using trigonometric identities for allied angles:
\( \sin(180^\circ - \theta) = \sin\theta \)
\( \tan(360^\circ - \theta) = -\tan\theta = -\frac{\sin\theta}{\cos\theta} \)
\( \cos(90^\circ + \theta) = -\sin\theta \)

Substitute into the expression:
\begin{aligned}
\frac{\sin\theta}{-\frac{\sin\theta}{\cos\theta}} + (-\sin\theta) &= -\cos\theta - \sin\theta
\end{aligned}

Correct answer: C (1 mark)

Let the original set of \( 8 \) numbers be \( x_1, x_2, \dots, x_8 \).
Mean \( \bar{x} = 15 \), and standard deviation \( \sigma = 4 \).
Sum of squared deviations from the mean for the original data:
\( \sum_{i=1}^8 (x_i - 15)^2 = 8 \times \sigma^2 = 8 \times 4^2 = 128 \).

When a datum \( x_9 = 15 \) is added:
New mean \( \bar{x}_{\text{new}} = \frac{8(15) + 15}{9} = 15 \).
New sum of squared deviations:
\( \sum_{i=1}^9 (x_i - 15)^2 = 128 + (15 - 15)^2 = 128 \).

New standard deviation \( \sigma_{\text{new}} = \sqrt{\frac{128}{9}} = \frac{\sqrt{128}}{3} = \frac{8\sqrt{2}}{3} \approx 3.77 \).

Correct answer: D (1 mark)

Express each base as a power of \(3\):
\[\frac{(3^{2n})((3^2)^{3n})}{(3^3)^n} = \frac{3^{2n} \cdot 3^{6n}}{3^{3n}} = 3^{2n + 6n - 3n} = 3^{5n}\]

1M for simplifying using laws of indices with base 3, 1A for obtaining \(3^{5n}\).

Expand both sides:
\[uv - 2u = 3u + 6v\]
Rearrange terms involving \(v\) to one side:
\[uv - 6v = 3u + 2u\]
\[v(u - 6) = 5u\]
\[v = \frac{5u}{u - 6}\]

1M for expanding and grouping terms in \(v\), 1A for making \(v\) the subject correctly.

Group the terms:
\[4x^2 - 9y^2 - 4x + 6y = (4x^2 - 9y^2) - (4x - 6y)\]
\[= (2x - 3y)(2x + 3y) - 2(2x - 3y)\]
Factor out the common factor \((2x - 3y)\):
\[= (2x - 3y)(2x + 3y - 2)\]

1M for applying difference of squares and factoring out common monomial, 1A for correct final factorized form.

For the first inequality:
\[5 - 2x \le 3 \implies -2x \le -2 \implies x \ge 1\]
For the second inequality:
\[4x + 7 < -5 \implies 4x < -12 \implies x < -3\]
Combining with 'or', the solution is \(x < -3 \text{ or } x \ge 1\).

1M for solving each linear inequality, 1A for the correct combined compound inequality solution.

Let \(\$M\) be the marked price.
\[\text{Selling price} = 400 \times (1 + 20\%) = \$480\]
Since selling price is at a \(20\%\) discount on the marked price:
\[M \times (1 - 20\%) = 480\]
\[0.8M = 480 \implies M = 600\]
Thus, the marked price is \(\$600\).

1M for setting up selling price from cost or marked price, 1A for finding \(\$600\).

Since \(x + 2\) is a factor of \(f(x)\), \(f(-2) = 0\):
\[2(-2)^3 + k(-2)^2 - 13(-2) + 6 = 0\]
\[-16 + 4k + 26 + 6 = 0 \implies 4k + 16 = 0 \implies k = -4\]
So \(f(x) = 2x^3 - 4x^2 - 13x + 6\).
The remainder when \(f(x)\) is divided by \(2x - 1\) is \(f\left(\frac{1}{2}\right)\):
\[f\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) - 4\left(\frac{1}{4}\right) - 13\left(\frac{1}{2}\right) + 6 = \frac{1}{4} - 1 - \frac{13}{2} + 6 = -\frac{5}{4}\]

1M for finding \(k = -4\) using factor theorem, 1A for calculating the remainder \(-\frac{5}{4}\).

The slope of the line \(3x - 4y + 5 = 0\) is \(\frac{3}{4}\).
Since \(L_1\) is perpendicular to this line, the slope of \(L_1\) is \(-\frac{4}{3}\).
The equation of \(L_1\) is:
\[y - (-1) = -\frac{4}{3}(x - 2) \implies y + 1 = -\frac{4}{3}x + \frac{8}{3} \implies y = -\frac{4}{3}x + \frac{5}{3}\]
Thus, the \(y\)-intercept of \(L_1\) is \(\frac{5}{3}\).

1M for finding the perpendicular slope \(-\frac{4}{3}\), 1A for determining the \(y\)-intercept \(\frac{5}{3}\).

Let \(y_i = -2x_i + 5\).
\[\text{New mean} = -2(24) + 5 = -48 + 5 = -43\]
\[\text{New standard deviation} = |-2| \times 6 = 2 \times 6 = 12\]

1M for applying linear transformation rules to mean and standard deviation, 1A for obtaining \(\text{Mean} = -43\) and \(\text{Standard deviation} = 12\).

Express all terms with base 3:
\[ \frac{(9^{2n})(3^{n+1})}{27^{n-1}} = \frac{((3^2)^{2n})(3^{n+1})}{(3^3)^{n-1}} = \frac{3^{4n} \cdot 3^{n+1}}{3^{3n-3}} = \frac{3^{5n+1}}{3^{3n-3}} = 3^{(5n+1)-(3n-3)} = 3^{2n+4} \]

1M for expressing powers in base 3 and simplifying indices, 1A for correct answer A.

Cross-multiply both sides:
\[ 3(3x - 2y) = 5(x + 4y) \]
\[ 9x - 6y = 5x + 20y \]
\[ 9x - 5x = 20y + 6y \]
\[ 4x = 26y \implies y = \frac{4x}{26} = \frac{2x}{13} \]

1M for expanding and grouping terms in y, 1A for correct answer A.

Let \(\$C\) be the cost of the handbag.
Marked price \(= C(1 + 40\%) = 1.4C\).
Selling price \(= 1.4C \times (1 - 25\%) = 1.4C \times 0.75 = 1.05C\).
Given selling price is \(\$630\):
\[ 1.05C = 630 \implies C = \frac{630}{1.05} = 600 \]
Profit \(= 630 - 600 = \$30\).

1M for setting up relation between selling price and cost, 1A for correct answer A.

Solve the first inequality:
\[ 3x - 7 \ge 4x - 18 \implies -x \ge -11 \implies x \le 11 \]
Solve the second inequality:
\[ 5 - x < 12 \implies -x < 7 \implies x > -7 \]
Combining the inequalities:
\[ -7 < x \le 11 \]
The integers satisfying the inequalities are \(-6, -5, \dots, 11\).
Number of integers \(= 11 - (-6) + 1 = 18\).

1M for finding the solution range of x, 1A for correct answer B.

Since \(x - 2\) is a factor of \(f(x)\), \(f(2) = 0\):
\[ 2(2)^3 + a(2)^2 - 11(2) + b = 0 \implies 16 + 4a - 22 + b = 0 \implies 4a + b = 6 \quad \text{--- (1)} \]
Since the remainder when divided by \(x + 1\) is \(-18\), \(f(-1) = -18\):
\[ 2(-1)^3 + a(-1)^2 - 11(-1) + b = -18 \implies -2 + a + 11 + b = -18 \implies a + b = -27 \quad \text{--- (2)} \]
Subtracting (2) from (1):
\[ 3a = 33 \implies a = 11 \]

1M for applying factor/remainder theorem to set up simultaneous equations, 1A for correct answer D.

In the right-angled triangle \(ABD\):
\[ \tan 60^\circ = \frac{AB}{BD} \implies \sqrt{3} = \frac{6}{BD} \implies BD = \frac{6}{\sqrt{3}} = 2\sqrt{3}\text{ cm} \]
Since \(BD : DC = 1 : 2\), \(DC = 2 \times 2\sqrt{3} = 4\sqrt{3}\text{ cm}\).
Thus, \(BC = BD + DC = 6\sqrt{3}\text{ cm}\).
In \(\triangle ABC\), by Pythagoras' theorem:
\[ AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + (6\sqrt{3})^2} = \sqrt{36 + 108} = \sqrt{144} = 12\text{ cm} \]

1M for finding BD and BC, 1A for correct answer B.

Volume of the circular cone:
\[ V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (6^2)(8) = 96\pi\text{ cm}^3 \]
Volume of each sphere:
\[ V_{\text{sphere}} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (2^3) = \frac{32}{3}\pi\text{ cm}^3 \]
Number of spheres:
\[ N = \frac{96\pi}{\frac{32}{3}\pi} = 96 \times \frac{3}{32} = 9 \]

1M for calculating both volumes, 1A for correct answer C.

For a linear transformation \(y = ax + b\):
New mean \(= 3 \times 45 - 8 = 135 - 8 = 127\).
New standard deviation \(= |a| \times \sigma_x = 3 \times 6 = 18\).
New variance \(= (\text{new standard deviation})^2 = 18^2 = 324\).

1M for calculating transformed mean and standard deviation/variance, 1A for correct answer B.

Express all terms with base \(2\):
\[ 4^{n+1} = (2^2)^{n+1} = 2^{2n+2} \]
\[ 8^{2n-1} = (2^3)^{2n-1} = 2^{6n-3} \]
\[ 16^{n+2} = (2^4)^{n+2} = 2^{4n+8} \]

Then:
\[ \frac{(4^{n+1})(8^{2n-1})}{16^{n+2}} = \frac{2^{2n+2} \cdot 2^{6n-3}}{2^{4n+8}} = \frac{2^{(2n+2)+(6n-3)}}{2^{4n+8}} = \frac{2^{8n-1}}{2^{4n+8}} = 2^{(8n-1)-(4n+8)} = 2^{4n-9} \]

1M for converting to base 2 and using laws of indices correctly; 1A for correct answer A.

\[ \frac{2p - 3q}{p + 2} = 5q \]
\[ 2p - 3q = 5q(p + 2) \]
\[ 2p - 3q = 5pq + 10q \]
\[ 2p - 5pq = 13q \]
\[ p(2 - 5q) = 13q \]
\[ p = \frac{13q}{2 - 5q} \]

1M for clearing denominator and collecting terms in \(p\); 1A for correct answer A.

For the first inequality:
\[ \frac{3x - 1}{4} + 2 > x \implies 3x - 1 + 8 > 4x \implies 3x + 7 > 4x \implies x < 7 \]

For the second inequality:
\[ 2x + 9 \ge 5x - 15 \implies 24 \ge 3x \implies x \le 8 \]

Combining both inequalities:
\[ x < 7 \text{ and } x \le 8 \implies x < 7 \]

The positive integers satisfying \(x < 7\) are \(1, 2, 3, 4, 5, 6\).
Thus, there are \(6\) positive integers.

1M for solving both linear inequalities correctly; 1A for identifying 6 positive integers (Option B).

Let \(\$C\) be the cost of the watch.
\[ \text{Marked price} = C(1 + 40\%) = 1.4C \]
\[ \text{Selling price} = 1.4C(1 - 25\%) = 1.4C \times 0.75 = 1.05C \]
\[ \text{Profit} = 1.05C - C = 0.05C \]
Since the profit is \(\$150\):
\[ 0.05C = 150 \implies C = \frac{150}{0.05} = 3000 \]
Therefore, the cost of the watch is \(\$3\,000\).

1M for setting up the equation in terms of cost; 1A for finding cost $3000 (Option C).

Since \(AB\) is a diameter, \(\angle ADB = 90^\circ\) (angle in semi-circle).
In \(\triangle ADE\), \(\angle ADE = \angle ADB = 90^\circ\).
Given \(\angle DAE = \angle CAD = 32^\circ\).
Using the angle sum of \(\triangle ADE\):
\[ \angle AED = 180^\circ - \angle DAE - \angle ADE = 180^\circ - 32^\circ - 90^\circ = 58^\circ \]

1M for using angle in semi-circle or properties of cyclic quads; 1A for getting 58 degrees (Option B).

Let \(X\) be the original variable and \(Y = -3X + 5\) be the transformed variable.

New mean:
\[ \mu_Y = -3\mu_X + 5 = -3(14) + 5 = -42 + 5 = -37 \]

New variance:
\[ \text{Var}(Y) = (-3)^2 \cdot \text{Var}(X) = 9 \times 9 = 81 \]

Therefore, the new mean is \(-37\) and the new variance is \(81\).

1M for applying linear transformation rules to mean and variance; 1A for Option A.

Rewrite the expression in terms of powers of 16:
\begin{aligned}
13 \times 16^{12} + 28 \times 16^8 + 5 \times 16^2 + 11
&= 13 \times 16^{12} + (16 + 12) \times 16^8 + 5 \times 16^2 + 11 \\
&= 13 \times 16^{12} + 1 \times 16^9 + 12 \times 16^8 + 5 \times 16^2 + 11 \times 16^0 \\
&= \text{D} \times 16^{12} + 1 \times 16^9 + \text{C} \times 16^8 + 5 \times 16^2 + \text{B} \times 16^0
\end{aligned}
Writing this as a hexadecimal number:
- The coefficient of \(16^{12}\) is \(\text{D}\).
- The coefficients of \(16^{11}\) and \(16^{10}\) are \(0\).
- The coefficient of \(16^9\) is \(1\).
- The coefficient of \(16^8\) is \(\text{C}\).
- The coefficients of \(16^7, 16^6, 16^5, 16^4, 16^3\) are \(0\).
- The coefficient of \(16^2\) is \(5\).
- The coefficient of \(16^1\) is \(0\).
- The coefficient of \(16^0\) is \(\text{B}\).

Thus, the hexadecimal representation is \(\text{D}001\text{C}0000050\text{B}_{16}\).

1M for expressing \(28 \times 16^8\) as \(16^9 + 12 \times 16^8\) and identifying hex digits; 1A for correct answer A.

Rationalize the denominator of the complex fraction:
\[ \frac{k+3i}{1-2i} = \frac{(k+3i)(1+2i)}{(1-2i)(1+2i)} = \frac{k + 2ki + 3i + 6i^2}{1 - 4i^2} = \frac{(k - 6) + (2k + 3)i}{5} \]
The real part is \(\dfrac{k-6}{5}\) and the imaginary part is \(\dfrac{2k+3}{5}\).
Since the real part equals the imaginary part:
\[ \frac{k-6}{5} = \frac{2k+3}{5} \implies k - 6 = 2k + 3 \implies k = -9 \]

1M for rationalizing the complex fraction and equating real and imaginary parts; 1A for correct answer B.

Find the vertices of the feasible region defined by the system of inequalities:
1. Intersection of \(x = 1\) and \(y = 0\): \((1, 0)\)
2. Intersection of \(x = 1\) and \(2x - y = -2\): \(y = 2(1) + 2 = 4 \implies (1, 4)\)
3. Intersection of \(2x - y = -2\) and \(x + y = 8\):
Adding the two equations: \(3x = 6 \implies x = 2\), so \(y = 6 \implies (2, 6)\)
4. Intersection of \(x + y = 8\) and \(y = 0\): \((8, 0)\)

Evaluate \(P(x, y) = 3x - 4y + 10\) at each vertex:
- At \((1, 0)\): \(P(1, 0) = 3(1) - 4(0) + 10 = 13\)
- At \((1, 4)\): \(P(1, 4) = 3(1) - 4(4) + 10 = -3\)
- At \((2, 6)\): \(P(2, 6) = 3(2) - 4(6) + 10 = 6 - 24 + 10 = -8\)
- At \((8, 0)\): \(P(8, 0) = 3(8) - 4(0) + 10 = 34\)

Therefore, the minimum value is \(-8\).

1M for finding the vertices of the feasible region and testing them; 1A for correct answer C.

Let the first term be \(a_1\) and the common ratio be \(r > 0\).
From \(\log_3 a_2 + \log_3 a_8 = 8\):
\[ \log_3 (a_2 a_8) = 8 \implies a_2 a_8 = 3^8 \]
Since \(a_n\) is a geometric sequence, \(a_2 a_8 = (a_1 r)(a_1 r^7) = a_1^2 r^8 = (a_1 r^4)^2 = a_5^2\).
Since all terms are positive, \(a_5 = \sqrt{3^8} = 3^4 = 81\).
Given \(a_4 = 9\), the common ratio is:
\[ r = \frac{a_5}{a_4} = \frac{81}{9} = 9 = 3^2 \]
Then, the 9th term is:
\[ a_9 = a_5 \cdot r^4 = 3^4 \cdot (3^2)^4 = 3^4 \cdot 3^8 = 3^{12} \]
Thus, \(\log_3 a_9 = 12\).

1M for using properties of GP to find \(a_5\) and common ratio \(r\); 1A for correct answer B.

Find the centre and radius of the circle:
Rewrite the equation of the circle \(x^2 + y^2 - 8x + 6y - 75 = 0\):
- Centre \(G = \left(-\dfrac{-8}{2}, -\dfrac{6}{2}\right) = (4, -3)\)
- Radius \(R = \sqrt{4^2 + (-3)^2 - (-75)} = \sqrt{16 + 9 + 75} = \sqrt{100} = 10\)

Since the straight line is tangent to the circle, the perpendicular distance from \(G(4, -3)\) to the line \(3x + 4y + k = 0\) is equal to \(R = 10\):
\[ \frac{|3(4) + 4(-3) + k|}{\sqrt{3^2 + 4^2}} = 10 \implies \frac{|12 - 12 + k|}{5} = 10 \implies \frac{|k|}{5} = 10 \]
\[ |k| = 50 \implies k = 50 \text{ or } k = -50 \]

1M for finding radius/centre and setting up distance formula; 1A for correct answer A.

Let \(M\) be the midpoint of \(AB\) and \(O\) be the centre of the square base \(ABCD\).
Since \(VAB\) is an isosceles triangle with \(VA = VB = 10\text{ cm}\) and \(AB = 12\text{ cm}\):
\[ VM = \sqrt{VA^2 - AM^2} = \sqrt{10^2 - 6^2} = 8\text{ cm} \]
Since \(O\) is the centre of the base and \(M\) is the midpoint of \(AB\), the line \(OM\) is perpendicular to \(AB\), and \(OM = \dfrac{12}{2} = 6\text{ cm}\).
Since \(VO\) is perpendicular to the base plane \(ABCD\), the angle between plane \(VAB\) and plane \(ABCD\) is \(\angle VMO\).
In the right-angled triangle \(\triangle VOM\):
\[ \cos \angle VMO = \frac{OM}{VM} = \frac{6}{8} = 0.75 \]
\[ \angle VMO = \arccos(0.75) \approx 41.41^\circ \approx 41^\circ \]

1M for determining the slant height and applying trigonometry to find the angle between planes; 1A for correct answer B.

The total number of ways to select 4 students from \(6 + 5 = 11\) students is:
\[ \binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330 \]
We consider the complementary events:
- 0 girls (4 boys): \(\binom{6}{4}\binom{5}{0} = 15 \times 1 = 15\)
- 1 girl (3 boys, 1 girl): \(\binom{6}{3}\binom{5}{1} = 20 \times 5 = 100\)

Total number of unfavorable outcomes = \(15 + 100 = 115\).
Thus, the number of ways with at least 2 girls is:
\[ 330 - 115 = 215 \]
The required probability is:
\[ \frac{215}{330} = \frac{43}{66} \]

1M for finding total outcomes and either direct or complementary favorable combinations; 1A for correct answer D.

Using the standard score formula \(z = \dfrac{x - \mu}{\sigma}\):
For Alan:
\[ \frac{78 - \mu}{\sigma} = 1.5 \implies 78 - \mu = 1.5\sigma \quad \text{--- (1)} \]
For Betty:
\[ \frac{54 - \mu}{\sigma} = -0.5 \implies 54 - \mu = -0.5\sigma \quad \text{--- (2)} \]
Subtracting (2) from (1):
\[ 78 - 54 = 1.5\sigma - (-0.5\sigma) \implies 24 = 2.0\sigma \implies \sigma = 12 \]
Substituting \(\sigma = 12\) into (2):
\[ 54 - \mu = -0.5(12) = -6 \implies \mu = 60 \]
For Calvin with \(z = 2.0\):
\[ \text{Score} = \mu + 2.0\sigma = 60 + 2.0(12) = 60 + 24 = 84 \]

1M for setting up simultaneous equations in \(\mu\) and \(\sigma\) and solving; 1A for correct answer C.

\( 7 \times 16^{11} + 11 \times 16^8 + 33 \times 16^2 + 5 = 7 \times 16^{11} + 11 \times 16^8 + (2 \times 16 + 1) \times 16^2 + 5 \)
\( = 7 \times 16^{11} + 11 \times 16^8 + 2 \times 16^3 + 1 \times 16^2 + 5 \times 16^0 \)
In hexadecimal notation:
\( 11 = \text{B}_{16} \)
The coefficients corresponding to powers of 16 from \( 16^{11} \) down to \( 16^0 \) are:
\( 16^{11}: 7 \)
\( 16^{10}: 0 \)
\( 16^{9}: 0 \)
\( 16^{8}: \text{B} \)
\( 16^{7}: 0 \)
\( 16^{6}: 0 \)
\( 16^{5}: 0 \)
\( 16^{4}: 0 \)
\( 16^{3}: 2 \)
\( 16^{2}: 1 \)
\( 16^{1}: 0 \)
\( 16^{0}: 5 \)
Thus, the number is \( 700\text{B}00002105_{16} \).

Correct option B: 1 mark.

Multiply the numerator and the denominator by the complex conjugate \( 2 + i \):
\( z = \dfrac{(a + 3i)(2 + i)}{(2 - i)(2 + i)} = \dfrac{(2a - 3) + (a + 6)i}{4 - i^2} = \dfrac{2a - 3}{5} + \dfrac{a + 6}{5}i \).
Since the imaginary part is \( 1 \):
\( \dfrac{a + 6}{5} = 1 \implies a + 6 = 5 \implies a = -1 \).
Then the real part of \( z \) is:
\( \dfrac{2(-1) - 3}{5} = \dfrac{-5}{5} = -1 \).

Correct option A: 1 mark.

Find the vertices of the polygonal region \( R \):
1. Intersection of \( x = 2 \) and \( y = 1 \): \( (2, 1) \)
2. Intersection of \( y = 1 \) and \( x - 2y = 1 \): \( (3, 1) \)
3. Intersection of \( x - 2y = 1 \) and \( 2x + y = 12 \):
\( x = 2y + 1 \implies 2(2y + 1) + y = 12 \implies 5y = 10 \implies y = 2, x = 5 \), giving \( (5, 2) \)
4. Intersection of \( 2x + y = 12 \) and \( x = 2 \): \( (2, 8) \)

Evaluate \( P(x, y) = 3x - 4y + 5 \) at each vertex:
- \( P(2, 1) = 3(2) - 4(1) + 5 = 7 \)
- \( P(3, 1) = 3(3) - 4(1) + 5 = 10 \)
- \( P(5, 2) = 3(5) - 4(2) + 5 = 12 \)
- \( P(2, 8) = 3(2) - 4(8) + 5 = 6 - 32 + 5 = -21 \)

Hence, the least value is \( -21 \).

Correct option D: 1 mark.

Let \( a_1 \) be the first term and \( r \) be the common ratio.
\( a_3 = a_1 r^2 = 24 \)
\( a_6 = a_1 r^5 = 192 \)
Dividing the two equations gives \( r^3 = \dfrac{192}{24} = 8 \implies r = 2 \).
Then \( a_1 (2^2) = 24 \implies a_1 = 6 \).
The sum of the first \( n \) terms is \( S_n = \dfrac{6(2^n - 1)}{2 - 1} = 6(2^n - 1) \).
We require \( 6(2^n - 1) > 50\,000 \):
\( 2^n - 1 > \dfrac{50\,000}{6} \approx 8333.33 \implies 2^n > 8334.33 \).
Since \( 2^{13} = 8192 \) and \( 2^{14} = 16\,384 \), the smallest integer \( n \) is \( 14 \).

Correct option B: 1 mark.

Let \( M \) be the centre of the rectangle \( ABCD \). The diagonal \( AC = \sqrt{16^2 + 12^2} = 20\text{ cm} \), so \( AM = 10\text{ cm} \).
The height of the pyramid is \( VM = \sqrt{VA^2 - AM^2} = \sqrt{26^2 - 10^2} = \sqrt{576} = 24\text{ cm} \).
Let \( P \) be the midpoint of \( AB \). Since \( \triangle VAB \) is isosceles, \( VP \perp AB \).
Also \( MP \perp AB \) with \( MP = \dfrac{BC}{2} = 6\text{ cm} \).
The angle between the plane \( VAB \) and the plane \( ABCD \) is \( \angle VPM \).
In \( \triangle VPM \), \( \tan\angle VPM = \dfrac{VM}{MP} = \dfrac{24}{6} = 4 \).
\( \angle VPM = \arctan(4) \approx 75.96^\circ \approx 76^\circ \).

Correct option C: 1 mark.

The possible compositions of (boys, girls) for a 5-member committee satisfying the conditions are:
1. 3 boys and 2 girls: \( C_3^6 \times C_2^5 = 20 \times 10 = 200 \)
2. 2 boys and 3 girls: \( C_2^6 \times C_3^5 = 15 \times 10 = 150 \)
3. 1 boy and 4 girls: \( C_1^6 \times C_4^5 = 6 \times 5 = 30 \)
(0 boys and 5 girls is excluded because at least 1 boy is required; 4 boys and 1 girl is excluded because at least 2 girls are required.)

Total number of ways \( = 200 + 150 + 30 = 380 \).

Correct option A: 1 mark.

Let \( X = \{x_1, \dots, x_n\} \) and \( Y = 3 - 2X \).
I. \( \text{Mean}(Y) = 3 - 2\,\text{Mean}(X) = 3 - 2m \). So I is true.
II. \( \text{Var}(Y) = (-2)^2 \text{Var}(X) = 4v \ne 3 - 4v \). So II is false.
III. The standard deviation of \( X \) is \( \sigma_X = \sqrt{v} \). Then \( \sigma_Y = |-2| \sigma_X = 2\sqrt{v} \). So III is true.
Therefore, I and III only are true.

Correct option B: 1 mark.