ICSE Class 10 Mathematics 2023 Solved Board Paper

Step 1 Multiplying the matrices on LHS: $\begin{pmatrix} 2x + 0 \\ 0 + 4y \end{pmatrix} = \begin{pmatrix} 2 \\ -8 \end{pmatrix}$.
Step 2 Equating corresponding elements:
$2x = 2 \implies x = 1$
$4y = -8 \implies y = -2$
Therefore, $x = 1, y = -2$. Correct Option: (a) [1 Mark]

Step 1 By Factor Theorem, if $(x - 2)$ is a factor of $P(x) = x^3 - kx - 12$, then $P(2) = 0$.
Step 2 Substitute $x = 2$:
$2^3 - k(2) - 12 = 0 \implies 8 - 2k - 12 = 0 \implies -2k - 4 = 0 \implies 2k = -4 \implies k = -2$.
Correct Option: (c) [1 Mark]

Step 1 By the Alternate Segment Theorem, the angle between the tangent $ST$ and chord $SP$ equals the angle subtended by $SP$ in the alternate segment:
$\angle PQS = \angle PST = 30^\circ$.
Step 2 In $\triangle PQS$, the sum of interior angles is $180^\circ$:
$\angle PSQ = 180^\circ - (\angle SPQ + \angle PQS) = 180^\circ - (60^\circ + 30^\circ) = 180^\circ - 90^\circ = 90^\circ$.
Correct Option: (d) [1 Mark]

Total number of letters in English alphabet $n(S) = 26$.
Vowels are $\{A, E, I, O, U\}$, so favorable outcomes $n(E) = 5$.
Probability $P(E) = \frac{n(E)}{n(S)} = \frac{5}{26}$. Correct Option: (b) [1 Mark]

Since 3 is a root of the equation, it must satisfy $x^2 - px + 3 = 0$:
$3^2 - p(3) + 3 = 0 \implies 9 - 3p + 3 = 0 \implies 12 - 3p = 0 \implies 3p = 12 \implies p = 4$.
Correct Option: (a) [1 Mark]

Step 1 Since corresponding angles $\angle BAP = \angle DCP = 70^\circ$, line $CD$ is parallel to line $AB$ ($CD \parallel AB$).
Step 2 In $\triangle PAB$, by the Basic Proportionality Theorem (Thales Theorem):
$\frac{PD}{DB} = \frac{PC}{CA} = \frac{6}{4} = \frac{3}{2}$.
Therefore, $PD : DB = 3 : 2$. Correct Option: (c) [1 Mark]

$\text{GST charged} = 10\% \text{ of } ₹3,080 = \frac{10}{100} \times 3080 = ₹308$.
Correct Option: (b) [1 Mark]

Using identity $(a + b)(a - b) = a^2 - b^2$:
$(1 + \sin A)(1 - \sin A) = 1 - \sin^2 A$.
Using fundamental Pythagorean identity $\sin^2 A + \cos^2 A = 1 \implies 1 - \sin^2 A = \cos^2 A$.
Correct Option: (d) [1 Mark]

Centroid formula: $G(x, y) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$.
$x = \frac{-4 + 6 + 4}{3} = \frac{6}{3} = 2$.
$y = \frac{-2 + 2 + 6}{3} = \frac{6}{3} = 2$.
Coordinates of $G = (2, 2)$. Correct Option: (a) [1 Mark]

Given $T_n = 2n + 5$. For $n = 10$:
$T_{10} = 2(10) + 5 = 20 + 5 = 25$.
Correct Option: (c) [1 Mark]

Mean proportional between $a$ and $b$ is given by $x = \sqrt{ab}$.
$x = \sqrt{4 \times 9} = \sqrt{36} = 6$.
Correct Option: (b) [1 Mark]

Median and Quartiles can be obtained from an Ogive (cumulative frequency curve). Mode is determined from a Histogram. The Mean requires arithmetic computation and cannot be read directly from standard curves. Correct Option: (d) [1 Mark]

Volume of cylinder $V = \pi r^2 h$.
Given $h = 3\text{ cm}$ and $V = 48\pi$:
$\pi r^2 (3) = 48\pi \implies 3r^2 = 48 \implies r^2 = 16 \implies r = 4\text{ cm}$.
Correct Option: (c) [1 Mark]

Total principal deposited $P \times n = 800 \times 6 = ₹4,800$.
Maturity Value $(MV) = ₹4,884$.
Interest earned = $MV - (P \times n) = 4884 - 4800 = ₹84$.
Correct Option: (a) [1 Mark]

$2x + 4 \le 14 \implies 2x \le 10 \implies x \le 5$.
Since $x \in W$ (set of whole numbers, which starts from 0):
Solution set = $\{0, 1, 2, 3, 4, 5\}$.
Correct Option: (b) [1 Mark]

Step 1 Let $f(x) = 3x^3 + x^2 - ax - 81$. Since $(x - a)$ is a factor, by the Factor Theorem, $f(a) = 0$.1 Mark
Step 2 Substitute $x = a$:
$3(a)^3 + (a)^2 - a(a) - 81 = 0$
$3a^3 + a^2 - a^2 - 81 = 0$1 Mark
Step 3 Simplify algebraic terms:
$3a^3 - 81 = 0 \implies 3a^3 = 81$1 Mark
$a^3 = 27 \implies a = \sqrt[3]{27} = 3$.
Final Answer: $a = 3$.1 Mark

Monthly deposit $P = ₹1,000$. Time period $n = 2\text{ years} = 24\text{ months}$.
Maturity value $MV = ₹26,000$.

Part (a) Total deposited amount $= P \times n = 1000 \times 24 = ₹24,000$.
Total Interest earned $I = MV - (P \times n) = 26,000 - 24,000 = \mathbf{₹2,000}$.2 Marks

Part (b) Recurring deposit interest formula:
$I = \frac{P \times n(n + 1)}{2 \times 12} \times \frac{r}{100}$
$2000 = \frac{1000 \times 24 \times 25}{24} \times \frac{r}{100}$1 Mark
$2000 = 10 \times 25 \times r \implies 2000 = 250r \implies r = \frac{2000}{250} = \mathbf{8\%}$.
Rate of interest = 8% per annum.1 Mark

Since segment $AB$ passes through the centre $O$, $AB$ is a diameter.

(a) $\angle BDA$ Angle in a semicircle is a right angle: $\angle BDA = 90^\circ$.1 Mark

(b) $\angle BAD$ In right-angled $\triangle ABD$, the acute angles are complementary:
$\angle BAD = 90^\circ - \angle ABD = 90^\circ - 26^\circ = \mathbf{64^\circ}$.1 Mark

(c) $\angle CAD$ By the Alternate Segment Theorem, the angle between tangent $AC$ and chord $AD$ is equal to the angle subtended by chord $AD$ in the alternate segment:
$\angle CAD = \angle ABD = \mathbf{26^\circ}$.
(Alternatively: radius $OA \perp CE \implies \angle CAO = 90^\circ$. Thus $\angle CAD = 90^\circ - 64^\circ = 26^\circ$).1 Mark

(d) $\angle ODB$ In $\triangle OBD$, $OB = OD$ (radii of the same circle).
Therefore, $\triangle OBD$ is isosceles with $\angle ODB = \angle OBD = \mathbf{26^\circ}$.1 Mark

Step 1 Comparing with $ax^2 + bx + c = 0$: $a = 1, b = 4, c = -8$.
Discriminant $D = b^2 - 4ac = 4^2 - 4(1)(-8) = 16 + 32 = 48$.1 Mark
Step 2 Applying quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$x = \frac{-4 \pm \sqrt{48}}{2(1)} = \frac{-4 \pm 4\sqrt{3}}{2} = -2 \pm 2\sqrt{3}$.1 Mark
Step 3 Taking $\sqrt{3} \approx 1.732$ (or $\sqrt{48} \approx 6.928$):
$x_1 = -2 + 2(1.732) = -2 + 3.464 = 1.464 \approx \mathbf{1.5}$1 Mark
$x_2 = -2 - 2(1.732) = -2 - 3.464 = -5.464 \approx \mathbf{-5.5}$
Answers: $x = 1.5$ or $x = -5.5$.1 Mark

Step 1 State standard Pythagorean identities:
$\sin^2 \theta + \cos^2 \theta = 1 \implies \sin^2 \theta - 1 = -\cos^2 \theta$.1 Mark
$1 + \tan^2 \theta = \sec^2 \theta$.1 Mark
Step 2 Substitute into LHS:
$\text{LHS} = (-\cos^2 \theta)(\sec^2 \theta) + 1$.1 Mark
Step 3 Since $\sec \theta = \frac{1}{\cos \theta}$, $\cos^2 \theta \cdot \sec^2 \theta = 1$:
$\text{LHS} = -(1) + 1 = 0 = \text{RHS}$.
Hence Proved!1 Mark

(a) Correctly plotting points $A(0, 4)$, $B(1, 1)$, $C(4, 0)$ on graph scale.1 Mark

(b) Reflection in x-axis: $(x, y) \to (x, -y)$
Image of $A(0, 4)$ is $\mathbf{E(0, -4)}$.
Image of $B(1, 1)$ is $\mathbf{D(1, -1)}$.1 Mark

(c) Reflection through Origin: $(x, y) \to (-x, -y)$
Image of $B(1, 1)$ through origin is $\mathbf{F(-1, -1)}$.1 Mark

(d) Reflection in y-axis: $(x, y) \to (-x, y)$
Image of $B(1, 1)$ is $\mathbf{H(-1, 1)}$.
Image of $C(4, 0)$ is $\mathbf{G(-4, 0)}$.1 Mark

(e) Geometrical Name Joining $A \to B \to C \to D \to E \to F \to G \to H \to A$ forms an 8-sided closed figure with symmetrical diagonal axes.
Geometrical name: Octagon (or 8-pointed / 8-sided star polygon / Star Octagon).1 Mark

Step 1 Add matrices $B$ and $C$:
$B + C = \begin{pmatrix} 1+4 & 2+1 \\ 2+1 & 4+5 \end{pmatrix} = \begin{pmatrix} 5 & 3 \\ 3 & 9 \end{pmatrix}$.1 Mark

Step 2 Matrix multiplication $A(B + C)$:
$A(B + C) = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} 5 & 3 \\ 3 & 9 \end{pmatrix} = \begin{pmatrix} 1(5) + 3(3) & 1(3) + 3(9) \\ 2(5) + 4(3) & 2(3) + 4(9) \end{pmatrix} = \begin{pmatrix} 14 & 30 \\ 22 & 42 \end{pmatrix}$.1 Mark

Step 3 Subtract $14I$ where $14I = \begin{pmatrix} 14 & 0 \\ 0 & 14 \end{pmatrix}$:
$A(B + C) - 14I = \begin{pmatrix} 14 - 14 & 30 - 0 \\ 22 - 0 & 42 - 14 \end{pmatrix} = \mathbf{\begin{pmatrix} 0 & 30 \\ 22 & 28 \end{pmatrix}}$.1 Mark

Part (a) Midpoint Formula: $D(x, y) = \left(\frac{x_B + x_C}{2}, \frac{y_B + y_C}{2}\right)$.
$D = \left(\frac{0 + (-6)}{2}, \frac{4 + 4}{2}\right) = \left(\frac{-6}{2}, \frac{8}{2}\right) = \mathbf{(-3, 4)}$.1.5 Marks

Part (b) Slope of median $AD$ passing through $A(1, -1)$ and $D(-3, 4)$:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-1)}{-3 - 1} = \frac{5}{-4} = -\frac{5}{4}$.
Equation using point-slope form $y - y_1 = m(x - x_1)$ with $A(1, -1)$:
$y - (-1) = -\frac{5}{4}(x - 1) \implies 4(y + 1) = -5(x - 1) \implies 4y + 4 = -5x + 5$.
Equation of median $AD$: $5x + 4y - 1 = 0$ (or $5x + 4y = 1$).1.5 Marks

(a) Length of PT By the Tangent-Secant Theorem, $PT^2 = PA \times PB$.
Here $PA = AB + BP = 9 + 16 = 25\text{ cm}$, and $PB = 16\text{ cm}$.
$PT^2 = 25 \times 16 = 400 \implies PT = \sqrt{400} = \mathbf{20\text{ cm}}$.1 Mark

(b) $\angle BAT$ By the Alternate Segment Theorem, the angle between tangent $PT$ and chord $BT$ equals the angle subtended by $BT$ in the alternate segment:
$\angle BAT = \angle PTB = \mathbf{50^\circ}$.1 Mark

(c) $\angle BOT$ The angle subtended by arc $BT$ at the centre is twice the angle subtended by it at any point on the remaining part of the circle:
$\angle BOT = 2 \times \angle BAT = 2 \times 50^\circ = \mathbf{100^\circ}$.1 Mark

(d) $\angle ABT$ In $\triangle OBT$, $OB = OT$ (radii). Thus $\triangle OBT$ is isosceles:
$\angle OBT = \frac{180^\circ - \angle BOT}{2} = \frac{180^\circ - 100^\circ}{2} = 40^\circ$.
Radius $OB$ divides angle $\angle ABT$ into $\angle OBA$ and $\angle OBT$:
$\angle ABT = \angle OBA + \angle OBT = 45^\circ + 40^\circ = \mathbf{85^\circ}$.1 Mark

Item 1: Hair Oil
Taxable Selling Price = $1,200 - 100 = ₹1,100$.
GST on hair oil = $18\% \text{ of } ₹1,100 = 0.18 \times 1100 = ₹198$.

Item 2: Cashew Nuts
Taxable Selling Price = ₹600.
GST on cashew nuts = $12\% \text{ of } ₹600 = 0.12 \times 600 = ₹72$.

Part (a) Total GST $= 198 + 72 = \mathbf{₹270}$.1.5 Marks

Part (b) Total Bill $= \text{Total Taxable Price} + \text{Total GST} = (1100 + 600) + 270 = 1700 + 270 = \mathbf{₹1,970}$.1.5 Marks

Part 1 Split into two simultaneous inequations:
Left part: $-5(x - 9) \ge 17 - 9x$
$-5x + 45 \ge 17 - 9x \implies -5x + 9x \ge 17 - 45 \implies 4x \ge -28 \implies x \ge -7$.1 Mark

Right part: $17 - 9x > x + 2$
$-9x - x > 2 - 17 \implies -10x > -15 \implies x < \frac{15}{10} = 1.5$.1 Mark

Part 2 Combining solutions for $x \in R$:
$$-7 \le x < 1.5$$ Solution Set: $\{x : -7 \le x < 1.5, x \in R\}$.
Number Line: Closed dot at $-7$, thick bold segment up to $1.5$, hollow circle at $1.5$.1 Mark

Part (a) Similarity Proof In $\triangle GED$ and $\triangle GBF$:
1. $\angle EGD = \angle BGF$ (Vertically opposite angles)
2. $\angle GED = \angle GBF$ (Alternate interior angles, since $DE \parallel BF$ and $CB$ is transversal)
3. $\angle GDE = \angle GFB$ (Alternate interior angles, since $DE \parallel BF$ and $DF$ is transversal)
Therefore, by AA similarity criterion: $\triangle GED \sim \triangle GBF$.1 Mark

Part (b) Find DE Since $\triangle GED \sim \triangle GBF$, corresponding sides are proportional:
$\frac{DE}{BF} = \frac{EG}{GB} \implies \frac{DE}{30} = \frac{8}{16} = \frac{1}{2}$
$DE = \frac{1}{2} \times 30 = \mathbf{15\text{ cm}}$.1.5 Marks

Part (c) Find Ratio DB : AB In $\triangle BAC$, $DE \parallel AC$. By Basic Proportionality Theorem / similarity $\triangle BDE \sim \triangle BAC$:
$\frac{DB}{AB} = \frac{DE}{AC}$
Given $DE = 15\text{ cm}$ and $AC = 24\text{ cm}$:
$\frac{DB}{AB} = \frac{15}{24} = \frac{5}{8}$.
Ratio $DB : AB = 5 : 8$.1.5 Marks

Step 1 Highest frequency bar corresponds to class 500–600 (frequency 16).
Draw diagonals connecting upper corners of the modal bar to the adjacent bars.1.5 Marks

Step 2 Drop perpendicular from intersection point to horizontal axis.
Reading from the graph: $\mathbf{\text{Mode} = ₹525}$ (acceptable range ₹520 to ₹530).
Formula Check: $\text{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 500 + \frac{16 - 14}{32 - 14 - 10} \times 100 = 500 + \frac{2}{8} \times 100 = 500 + 25 = ₹525$.1.5 Marks

Let the first term be $a$ and common difference be $d$.
$T_5 = a + 4d = 4$ — (Eq. 1)
$T_9 = a + 8d = -12$ — (Eq. 2)

Subtract Eq. 1 from Eq. 2
$(a + 8d) - (a + 4d) = -12 - 4 \implies 4d = -16 \implies \mathbf{d = -4}$. (Common Difference)1 Mark

Find First Term
Substitute $d = -4$ into Eq. 1:
$a + 4(-4) = 4 \implies a - 16 = 4 \implies \mathbf{a = 20}$.1 Mark

Sum of 16 Terms
$S_n = \frac{n}{2}[2a + (n - 1)d]$
$S_{16} = \frac{16}{2}[2(20) + (16 - 1)(-4)] = 8[40 + 15(-4)] = 8[40 - 60] = 8(-20) = \mathbf{-160}$.1 Mark

(a) Coordinates of A and B From the graph, $A$ lies on the $x$-axis at 4 units and $B$ lies on the $y$-axis at 4 units:
$A(4, 0)$ and $B(0, 4)$.1 Mark

(b) Coordinates of P Point $P$ divides $AB$ in the ratio $m_1 : m_2 = 3 : 1$ from $A(4, 0)$ to $B(0, 4)$:
$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{3(0) + 1(4)}{3 + 1} = \frac{4}{4} = 1$
$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{3(4) + 1(0)}{3 + 1} = \frac{12}{4} = 3$
Coordinates of $P = (1, 3)$.1.5 Marks

(c) Perpendicular Line Equation Slope of line segment $AB$:
$m_{AB} = \frac{4 - 0}{0 - 4} = \frac{4}{-4} = -1$.
Since the required line is perpendicular to $AB$, its slope $m_\perp = -\frac{1}{m_{AB}} = -\frac{1}{-1} = 1$.
Equation of line with slope $m = 1$ passing through $P(1, 3)$:
$y - y_1 = m(x - x_1) \implies y - 3 = 1(x - 1) \implies y - 3 = x - 1$.
Equation of required line: $x - y + 2 = 0$ (or $y = x + 2$).1.5 Marks

Sample space $S = \{1, 2, 3, \dots, 25\} \implies n(S) = 25$.

(a) Multiple of 5 $E_1 = \{5, 10, 15, 20, 25\} \implies n(E_1) = 5$.
$P(E_1) = \frac{5}{25} = \mathbf{\frac{1}{5}}$.1 Mark

(b) Perfect Square $E_2 = \{1, 4, 9, 16, 25\} \implies n(E_2) = 5$.
$P(E_2) = \frac{5}{25} = \mathbf{\frac{1}{5}}$.1 Mark

(c) Prime Number $E_3 = \{2, 3, 5, 7, 11, 13, 17, 19, 23\} \implies n(E_3) = 9$.
$P(E_3) = \mathbf{\frac{9}{25}}$.1 Mark

Step 1 Set up time expressions:
Initial time $t_1 = \frac{100}{x}\text{ hours}$.
Increased speed time $t_2 = \frac{100}{x + 5}\text{ hours}$.1 Mark

Step 2 According to given condition $t_1 - t_2 = 1$:
$\frac{100}{x} - \frac{100}{x + 5} = 1$
$100\left(\frac{x + 5 - x}{x(x + 5)}\right) = 1 \implies \frac{500}{x^2 + 5x} = 1$
$x^2 + 5x - 500 = 0$.1 Mark

Step 3 Solve quadratic equation:
$(x + 25)(x - 20) = 0 \implies x = 20$ or $x = -25$.
Since speed cannot be negative, $x = 20$.
Original speed = 20 km/hr.1 Mark

Step 1: Volume of Solid
Solid consists of hemisphere ($r = 7\text{ cm}$) and cone ($r = 7\text{ cm}, h = 4\text{ cm}$):
$V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (7)^3 = \frac{686\pi}{3}\text{ cm}^3$.1 Mark
$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (7^2)(4) = \frac{196\pi}{3}\text{ cm}^3$.1 Mark
Total Volume of Solid $V_{\text{solid}} = \frac{686\pi + 196\pi}{3} = \frac{882\pi}{3} = 294\pi\text{ cm}^3$.

Step 2: Water Level Rise
Let the rise in water level in the cylinder be $H$.
Volume of water displaced = Volume of cylinder of radius $R = 14\text{ cm}$ and height $H$:
$\pi R^2 H = V_{\text{solid}}$1 Mark
$\pi (14)^2 H = 294\pi \implies 196 H = 294 \implies H = \frac{294}{196} = \frac{3}{2} = \mathbf{1.5\text{ cm}}$.
Rise in water level = 1.5 cm.1 Mark

Let Assumed Mean $A = 25$ (midpoint of 20–30). Total frequency $\sum f = 3 + 8 + 14 + 9 + 4 + 2 = 40$.1 Mark
Sum of deviations $\sum fd = -60 - 80 + 0 + 90 + 80 + 60 = 90$.1 Mark
Shortcut formula: $\bar{x} = A + \frac{\sum fd}{\sum f} = 25 + \frac{90}{40} = 25 + 2.25 = \mathbf{27.25\text{ marks}}$.1 Mark

Step 1 Let the required number be $x$.
The resulting numbers $(4 + x), (6 + x), (8 + x), (11 + x)$ are in proportion:
$\frac{4 + x}{6 + x} = \frac{8 + x}{11 + x}$.1 Mark

Step 2 Cross-multiplying:
$(4 + x)(11 + x) = (6 + x)(8 + x)$
$44 + 15x + x^2 = 48 + 14x + x^2$.1 Mark

Step 3 Subtracting $x^2$ from both sides:
$15x - 14x = 48 - 44 \implies x = 4$.
The number to be added is 4.1 Mark

Steps of Construction
1. Draw line segment $AB = 6\text{ cm}$.
2. At point $A$, construct an angle $\angle BAX = 120^\circ$ using a pair of compasses.
3. Cut off an arc of length $5\text{ cm}$ along the ray $AX$ to locate vertex $C$. Join $BC$ to complete $\triangle ABC$.2 Marks
4. Draw the perpendicular bisectors of sides $AB$ and $AC$. Produce them to intersect at point $O$ (circumcentre lies outside because $\angle A$ is obtuse).
5. With $O$ as centre and radius $OA = OB = OC$, draw the circumscribing circle.1 Mark

Radius Measurement
Measuring with ruler: $\mathbf{\text{Radius } R \approx 5.5\text{ cm}}$ (acceptable examination tolerance: $5.3\text{ to } 5.7\text{ cm}$).1 Mark

Step 1 Express RHS as $\frac{3}{1}$ and apply Componendo and Dividendo: $\frac{a + b}{a - b} = \frac{c + d}{c - d}$:
$$\frac{(\sqrt{2x+2} + \sqrt{2x-1}) + (\sqrt{2x+2} - \sqrt{2x-1})}{(\sqrt{2x+2} + \sqrt{2x-1}) - (\sqrt{2x+2} - \sqrt{2x-1})} = \frac{3 + 1}{3 - 1}$$1 Mark
Step 2 Simplify terms:
$$\frac{2\sqrt{2x+2}}{2\sqrt{2x-1}} = \frac{4}{2} \implies \frac{\sqrt{2x+2}}{\sqrt{2x-1}} = 2$$1 Mark
Step 3 Squaring both sides:
$$\frac{2x + 2}{2x - 1} = 4 \implies 2x + 2 = 4(2x - 1) \implies 2x + 2 = 8x - 4$$
$$6x = 6 \implies \mathbf{x = 1}$$.1 Mark

First term $a = 15$. Common difference $d = 30 - 15 = 15$.

Part 1 $n^\text{th}$ term formula: $T_n = a + (n - 1)d = 300$
$15 + (n - 1)15 = 300 \implies 15n = 300 \implies \mathbf{n = 20}$.
Therefore, the $20^\text{th}$ term is 300.1.5 Marks

Part 2 Sum of $n$ terms $S_n = \frac{n}{2}(a + l)$ where last term $l = 300$:
$S_{20} = \frac{20}{2}(15 + 300) = 10 \times 315 = \mathbf{3,150}$.1.5 Marks

Let $CD$ be the tower of height 100 m, with foot $C$.
By alternate interior angles, angle of elevation at ship $A$ is $\angle CAD = 45^\circ$, and at ship $B$ is $\angle CBD = 38^\circ$.

Distance AC In right-angled $\triangle ACD$:
$\tan 45^\circ = \frac{CD}{AC} \implies 1 = \frac{100}{AC} \implies AC = 100\text{ m}$.1.5 Marks

Distance BC In right-angled $\triangle BCD$:
$\tan 38^\circ = \frac{CD}{BC} \implies BC = \frac{100}{\tan 38^\circ}$.
From natural tangents tables: $\tan 38^\circ \approx 0.7813$.
$BC = \frac{100}{0.7813} \approx 127.99\text{ m}$.1.5 Marks

Total Distance AB
$AB = AC + BC = 100 + 127.99 = 227.99\text{ m} \approx \mathbf{228\text{ m}}$ (to the nearest metre).1 Mark

Let $P(x) = 2x^3 - x^2 - 13x - 6$. Factors of the constant term $-6$ are $\pm 1, \pm 2, \pm 3, \pm 6$.

Step 1: Test x = -2
$P(-2) = 2(-2)^3 - (-2)^2 - 13(-2) - 6 = 2(-8) - 4 + 26 - 6 = -16 - 4 + 26 - 6 = 0$.
By Factor Theorem, $(x + 2)$ is a factor of $P(x)$.1.5 Marks

Step 2: Polynomial Division
Dividing $2x^3 - x^2 - 13x - 6$ by $(x + 2)$:
$2x^3 - x^2 - 13x - 6 = (x + 2)(2x^2 - 5x - 3)$.1.5 Marks

Step 3: Factorize Quadratic Quotient
$2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (x - 3)(2x + 1)$.
Complete Factorisation: $(x + 2)(x - 3)(2x + 1)$.1 Mark

Cumulative Frequency Table
Points to plot (Upper Boundary, $cf$): $(28, 0), (30, 2), (32, 6), (34, 16), (36, 29), (38, 44), (40, 53), (42, 58), (44, 60)$.
Total number of students $N = 60$.2 Marks

(a) Median $\frac{N}{2} = \frac{60}{2} = 30^\text{th}$ student.
Locate $cf = 30$ on vertical axis, draw horizontal line to curve and drop perpendicular to weight axis.
Median $\approx 36.1\text{ kg}$ (acceptable range $36.0\text{ to } 36.3\text{ kg}$).1.5 Marks

(b) Upper Quartile $Q_3$ $\frac{3N}{4} = \frac{3 \times 60}{4} = 45^\text{th}$ student.
From $cf = 45$, drop perpendicular from curve to weight axis.
Upper Quartile $Q_3 \approx 38.2\text{ kg}$ (acceptable range $38.0\text{ to } 38.4\text{ kg}$).1.5 Marks

(c) Students Weighing Above 37 kg
From weight $= 37\text{ kg}$ on horizontal axis, draw vertical line to ogive curve.
Corresponding cumulative frequency reading $\approx 36.5$ (approx 36 or 37 students weigh $\le 37\text{ kg}$).
Number of students weighing above $37\text{ kg} = 60 - 36.5 \approx \mathbf{23\text{ to } 24\text{ students}}$.1 Mark