ICSE Class 10 Mathematics 2020 Solved Paper

Step 1 Comparing with $ax^2 + bx + c = 0$: $a = 1, b = -7, c = 3$.
Discriminant $D = b^2 - 4ac = (-7)^2 - 4(1)(3) = 49 - 12 = 37$. [1 Mark]
Step 2 Applying quadratic formula: $x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-7) \pm \sqrt{37}}{2(1)} = \frac{7 \pm 6.083}{2}$. [1 Mark]
Step 3 Evaluating roots:
$x_1 = \frac{7 + 6.083}{2} = \frac{13.083}{2} = 6.5415 \approx \mathbf{6.54}$
$x_2 = \frac{7 - 6.083}{2} = \frac{0.917}{2} = 0.4585 \approx \mathbf{0.46}$
Answer: $x = 6.54$ or $0.46$. [1 Mark]

Step 1 Compute $A^2$:
$$A^2 = \begin{pmatrix} x & 3 \\ y & 3 \end{pmatrix} \begin{pmatrix} x & 3 \\ y & 3 \end{pmatrix} = \begin{pmatrix} x^2 + 3y & 3x + 9 \\ xy + 3y & 3y + 9 \end{pmatrix}$$ [1 Mark]
Step 2 Given $A^2 = 3I = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$. Equating corresponding elements:
$3x + 9 = 0 \implies 3x = -9 \implies \mathbf{x = -3}$.
$3y + 9 = 3 \implies 3y = -6 \implies \mathbf{y = -2}$. [1 Mark]
Step 3 Verification:
$x^2 + 3y = (-3)^2 + 3(-2) = 9 - 6 = 3$ (True).
$xy + 3y = (-3)(-2) + 3(-2) = 6 - 6 = 0$ (True).
Answer: $x = -3, y = -2$. [1 Mark]

Step 1 Draw base segment $BC = 4\text{ cm}$. At $B$, construct $\angle ABC = 90^\circ$ using compass. Cut off $AB = 3\text{ cm}$. Join $AC$ to form $\triangle ABC$. [1.5 Marks]
Step 2 Draw the perpendicular bisectors of $AB$ and $BC$. Their point of intersection is $O$, which lies exactly on the midpoint of hypotenuse $AC$. [1.5 Marks]
Step 3 With $O$ as centre and radius $OA = OB = OC$, draw the circumcircle.
Theoretical calculation: In right $\triangle ABC$, $AC = \sqrt{3^2 + 4^2} = 5\text{ cm}$.
Radius $R = \frac{AC}{2} = \frac{5}{2} = \mathbf{2.5\text{ cm}}$.
Measured radius: $2.5\text{ cm}$. [1 Mark]

Step 1 Let $P(x) = 6x^3 + 17x^2 + 4x - 12$.
Test integer factors of $-12$: for $x = -2$:
$P(-2) = 6(-2)^3 + 17(-2)^2 + 4(-2) - 12 = 6(-8) + 17(4) - 8 - 12 = -48 + 68 - 20 = 0$.
Therefore, by Factor Theorem, $(x + 2)$ is a factor of $P(x)$. [1 Mark]
Step 2 Divide $P(x)$ by $(x + 2)$:
$6x^3 + 17x^2 + 4x - 12 = (x + 2)(6x^2 + 5x - 6)$. [1 Mark]
Step 3 Factorising the quotient quadratic $6x^2 + 5x - 6$:
Product $= 6 \times (-6) = -36$, Sum $= 5 \implies +9$ and $-4$.
$6x^2 + 9x - 4x - 6 = 3x(2x + 3) - 2(2x + 3) = (2x + 3)(3x - 2)$.
Complete Factorisation: $(x + 2)(2x + 3)(3x - 2)$. [1 Mark]

Part 1 $\frac{3x}{5} + 2 < x + 4 \implies 2 - 4 < x - \frac{3x}{5} \implies -2 < \frac{2x}{5} \implies -10 < 2x \implies \mathbf{x > -5}$. [1 Mark]
Part 2 $x + 4 \le \frac{x}{2} + 5 \implies x - \frac{x}{2} \le 5 - 4 \implies \frac{x}{2} \le 1 \implies \mathbf{x \le 2}$. [1 Mark]
Step 3 Combining both parts: $-5 < x \le 2, \quad x \in \mathbb{R}$.
Solution set $= \{x : -5 < x \le 2, x \in \mathbb{R}\}$. Correct plot with hollow circle at $-5$ and solid dot at $2$. [1 Mark]

Step 1 Plot class intervals along the horizontal axis (taking a kink between 0 and 3000) and number of people along the vertical axis ($1\text{ cm} = 2\text{ people}$). Construct rectangles for each class interval. [2 Marks]
Step 2 The highest bar corresponds to the modal class 5000 – 6000 (frequency = 18). Join the top-left vertex to top-left of following bar and top-right vertex to top-right of preceding bar. [1 Mark]
Step 3 Draw a vertical line from the intersection point to the horizontal axis. Reading on the x-axis gives:
Mode ≈ ₹5,400 (acceptable range: ₹5,350 to ₹5,450). [1 Mark]

(i) Arc $AC$ subtends $\angle AOC = 72^\circ$ at the centre and $\angle ABC$ at the circumference.
$\angle ABC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 72^\circ = \mathbf{36^\circ}$. (Angle at circumference is half angle at centre). [1 Mark]
(ii) Given chord $AC = \text{chord } BD$. Equal chords subtend equal angles at the centre:
$\angle BOD = \angle AOC = 72^\circ$.
Therefore, $\angle BAD = \frac{1}{2} \angle BOD = \frac{1}{2} \times 72^\circ = \mathbf{36^\circ}$. [1 Mark]
(iii) Since $AB$ is a diameter, $\angle ADB = 90^\circ$ (Angle in a semicircle is a right angle).
In right $\triangle ABD$: $\angle ABD = 180^\circ - (90^\circ + \angle BAD) = 90^\circ - 36^\circ = \mathbf{54^\circ}$. [1 Mark]

Step 1 Express $\cot A$ and $\tan A$ in terms of $\sin A$ and $\cos A$:
$$\text{LHS} = \frac{\sin A}{1 + \frac{\cos A}{\sin A}} - \frac{\cos A}{1 + \frac{\sin A}{\cos A}} = \frac{\sin A}{\frac{\sin A + \cos A}{\sin A}} - \frac{\cos A}{\frac{\cos A + \sin A}{\cos A}}$$ [1 Mark]
Step 2 Simplify the compound fractions:
$$= \frac{\sin^2 A}{\sin A + \cos A} - \frac{\cos^2 A}{\sin A + \cos A} = \frac{\sin^2 A - \cos^2 A}{\sin A + \cos A}$$ [1 Mark]
Step 3 Factor the difference of squares in the numerator:
$$= \frac{(\sin A - \cos A)(\sin A + \cos A)}{\sin A + \cos A} = \sin A - \cos A = \text{RHS}$$
Hence Proved. [1 Mark]

Step 1 Let the y-axis divide the segment joining $P(5, 3)$ and $Q(-5, 3)$ in the ratio $k : 1$ at point $R(0, y)$ (since any point on the y-axis has $x = 0$). [1 Mark]
Step 2 By section formula for the x-coordinate:
$$x = \frac{k(-5) + 1(5)}{k + 1} \implies 0 = \frac{-5k + 5}{k + 1} \implies -5k + 5 = 0 \implies 5k = 5 \implies k = 1$$
Thus, the required ratio is $1 : 1$ (the segment is bisected by the y-axis). [1.5 Marks]
Step 3 By section formula for the y-coordinate:
$$y = \frac{1(3) + 1(3)}{1 + 1} = \frac{6}{2} = 3$$
Answer: Ratio = $1 : 1$, Point of intersection = $(0, 3)$. [1.5 Marks]

Step 1 Radius of big spherical ball $R = 6\text{ cm}$.
Volume of big sphere $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (6)^3\text{ cm}^3$. [1 Mark]
Step 2 Let $r$ be the radius of each small spherical marble.
Total volume of 64 marbles $= 64 \times \left(\frac{4}{3} \pi r^3\right)$.
By conservation of volume during recasting:
$$\frac{4}{3} \pi (6)^3 = 64 \times \frac{4}{3} \pi r^3 \implies 216 = 64 r^3$$ [1 Mark]
Step 3 Solving for $r$:
$$r^3 = \frac{216}{64} = \frac{27}{8} \implies r = \sqrt[3]{\frac{27}{8}} = \frac{3}{2} = \mathbf{1.5\text{ cm}}$$
Answer: The radius of each marble is $1.5\text{ cm}$. [1 Mark]

Total letters in ‘AUTHORIZES’ $= 10$. Total outcomes $n(S) = 10$.
(i) Vowels in ‘AUTHORIZES’ are: A, U, O, I, E $\implies n(E_1) = 5$.
$P(\text{vowel}) = \frac{5}{10} = \mathbf{\frac{1}{2}}$ (or $0.5$). [1 Mark]
(ii) First 9 letters of English alphabet: {A, B, C, D, E, F, G, H, I}.
Letters in word from this set: A, H, I, E $\implies n(E_2) = 4$.
$P = \frac{4}{10} = \mathbf{\frac{2}{5}}$ (or $0.4$). [1 Mark]
(iii) Last 9 letters of English alphabet (18th to 26th): {R, S, T, U, V, W, X, Y, Z}.
Letters in word from this set: U, T, R, Z, S $\implies n(E_3) = 5$.
$P = \frac{5}{10} = \mathbf{\frac{1}{2}}$ (or $0.5$). [1 Mark]

Item 1 Medicines: Taxable value $= ₹950$. GST $= 5\% \text{ of } 950 = \mathbf{₹47.50}$. [1 Mark]
Item 2 Shoes: Taxable value $= ₹3,000$. GST $= 18\% \text{ of } 3000 = \mathbf{₹540.00}$. [1 Mark]
Item 3 Laptop Bag: Marked price $= ₹1,000$. Discount $= 30\% \text{ of } 1000 = ₹300$.
Taxable Selling Price $= 1000 - 300 = ₹700$.
GST $= 18\% \text{ of } 700 = \mathbf{₹126.00}$. [1 Mark]
Totals
(i) Total GST paid $= 47.50 + 540.00 + 126.00 = \mathbf{₹713.50}$. [0.5 Mark]
(ii) Total taxable amount $= 950 + 3000 + 700 = ₹4,650$.
Total bill amount $= ₹4,650 + ₹713.50 = \mathbf{₹5,363.50}$. [0.5 Mark]

Nominal Value (NV) per share $= ₹120$, Dividend rate $= 15\%$.
(i) Total NV of 500 shares $= 500 \times 120 = ₹60,000$.
Total dividend paid $= 15\% \text{ of } 60,000 = \frac{15}{100} \times 60000 = \mathbf{₹9,000}$. [1 Mark]
(ii) Dividend on 1 share $= 15\% \text{ of } 120 = ₹18$.
Annual income of Mr. Sharma (holding 80 shares) $= 80 \times 18 = \mathbf{₹1,440}$. [1 Mark]
(iii) Return percent on investment $= 10\%$.
Let MV be the market value of each share.
$\text{Return on 1 share} = 10\% \text{ of MV} = \text{Dividend on 1 share} = ₹18$.
$\frac{10}{100} \times \text{MV} = 18 \implies \mathbf{\text{MV} = ₹180}$. [1 Mark]

Step 1 Calculate $\sum f_i$ and $\sum f_i x_i$:
$\sum f_i = 3 + 7 + f + 9 + 6 = 25 + f$. [1 Mark]
$\sum f_i x_i = (5 \times 3) + (10 \times 7) + (15 \times f) + (20 \times 9) + (25 \times 6)$
$= 15 + 70 + 15f + 180 + 150 = 415 + 15f$. [1 Mark]
Step 2 Applying mean formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$:
$$16 = \frac{415 + 15f}{25 + f} \implies 16(25 + f) = 415 + 15f$$
$$400 + 16f = 415 + 15f \implies 16f - 15f = 415 - 400 \implies \mathbf{f = 15}$$
Answer: $f = 15$. [1 Mark]

Step 1 Let the first term be $a$ and common ratio be $r$ ($r > 0$).
$T_4 = a r^3 = 10$
$T_6 = a r^5 = 40$
Dividing: $\frac{T_6}{T_4} = \frac{a r^5}{a r^3} = r^2 = \frac{40}{10} = 4 \implies \mathbf{r = 2}$ (since $r > 0$). [1.5 Marks]
Step 2 Find first term $a$:
$a r^3 = 10 \implies a (2)^3 = 10 \implies 8a = 10 \implies \mathbf{a = \frac{5}{4} = 1.25}$. [1 Mark]
Step 3 Find number of terms $n$:
Last term $T_n = a r^{n-1} = 640 \implies \frac{5}{4} \times 2^{n-1} = 640$
$$2^{n-1} = \frac{640 \times 4}{5} = 128 \times 4 = 512 = 2^9 \implies n - 1 = 9 \implies \mathbf{n = 10}$$
Answer: First term $a = \frac{5}{4}$, common ratio $r = 2$, number of terms $n = 10$. [1.5 Marks]

Step 1 Compute $A^2$ and $B^2$:
$$A^2 = \begin{pmatrix} 3 & 0 \\ 5 & 1 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 5 & 1 \end{pmatrix} = \begin{pmatrix} 9 & 0 \\ 20 & 1 \end{pmatrix}$$
$$B^2 = \begin{pmatrix} -4 & 2 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} -4 & 2 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 16 + 2 & -8 + 0 \\ -4 + 0 & 2 + 0 \end{pmatrix} = \begin{pmatrix} 18 & -8 \\ -4 & 2 \end{pmatrix}$$ [1 Mark]
Step 2 Compute $2AB$:
$$AB = \begin{pmatrix} 3 & 0 \\ 5 & 1 \end{pmatrix} \begin{pmatrix} -4 & 2 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -12 & 6 \\ -19 & 10 \end{pmatrix} \implies 2AB = \begin{pmatrix} -24 & 12 \\ -38 & 20 \end{pmatrix}$$ [1 Mark]
Step 3 Combine $A^2 - 2AB + B^2$:
$$= \begin{pmatrix} 9 & 0 \\ 20 & 1 \end{pmatrix} - \begin{pmatrix} -24 & 12 \\ -38 & 20 \end{pmatrix} + \begin{pmatrix} 18 & -8 \\ -4 & 2 \end{pmatrix} = \begin{pmatrix} 9 - (-24) + 18 & 0 - 12 - 8 \\ 20 - (-38) - 4 & 1 - 20 + 2 \end{pmatrix} = \mathbf{\begin{pmatrix} 51 & -20 \\ 54 & -17 \end{pmatrix}}$$ [1 Mark]

(i) In $\triangle PAB$ and $\triangle PCD$:
$\angle APB = \angle CPD$ (Vertically opposite angles)
$\angle PAB = \angle PCD$ (Angles in the same segment subtended by chord $BD$ are equal)
Therefore, by AA similarity criterion, $\triangle PAB \sim \triangle PCD$. [1 Mark]
(ii) Since $\triangle PAB \sim \triangle PCD$, corresponding sides are in proportion:
$$\frac{AB}{CD} = \frac{PA}{PC} \implies \frac{9}{CD} = \frac{7.5}{5} = \frac{3}{2} \implies CD = \frac{9 \times 2}{3} = \mathbf{6\text{ cm}}$$ [1 Mark]
(iii) Ratio of areas of similar triangles equals square of ratio of corresponding sides:
$$\frac{\text{Area}(\triangle PAB)}{\text{Area}(\triangle PCD)} = \left(\frac{PA}{PC}\right)^2 = \left(\frac{7.5}{5}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} = \mathbf{9 : 4}$$ [1 Mark]

Step 1 Let cliff height be $H$ ($AB = H$) and distance between them be $d$ ($BD = EC = d$).
Height of tower $CD = EB = 20\text{ m} \implies AE = H - 20$.
In right $\triangle AEC$: $\tan 45^\circ = \frac{AE}{EC} = \frac{H - 20}{d} \implies 1 = \frac{H - 20}{d} \implies \mathbf{H = d + 20}$. [1.5 Marks]
Step 2 In right $\triangle ABD$: $\tan 60^\circ = \frac{AB}{BD} = \frac{H}{d} \implies \sqrt{3} = \frac{H}{d} \implies \mathbf{H = d\sqrt{3}}$. [1 Mark]
Step 3 Equating expressions for $H$:
$$d\sqrt{3} = d + 20 \implies d(\sqrt{3} - 1) = 20 \implies d = \frac{20}{\sqrt{3} - 1} = \frac{20(\sqrt{3} + 1)}{2} = 10(1.732 + 1) = \mathbf{27.32\text{ m}}$$
Then cliff height $H = d + 20 = 27.32 + 20 = \mathbf{47.32\text{ m}}$.
Answer: (i) Height of cliff $= 47.32\text{ m}$, (ii) Distance $= 27.32\text{ m}$. [1.5 Marks]

Step 1 Slope of Line 1 ($5x - 3y + 2 = 0$): $m_1 = \frac{5}{3}$.
Slope of Line 2 ($6x - py + 7 = 0$): $m_2 = \frac{6}{p}$.
Since lines are perpendicular, $m_1 \times m_2 = -1$:
$$\frac{5}{3} \times \frac{6}{p} = -1 \implies \frac{10}{p} = -1 \implies \mathbf{p = -10}$$. [1.5 Marks]
Step 2 Slope of Line 2 with $p = -10$ is $m_2 = \frac{6}{-10} = -\frac{3}{5}$.
Any parallel line has the same slope $m = -\frac{3}{5}$. Passing through $(-2, -1)$:
$$y - y_1 = m(x - x_1) \implies y - (-1) = -\frac{3}{5}(x - (-2))$$
$$5(y + 1) = -3(x + 2) \implies 5y + 5 = -3x - 6 \implies \mathbf{3x + 5y + 11 = 0}$$. [1.5 Marks]

Method Factoring numerators and denominators:
$$\frac{x(x + 2)}{2(x + 2)} = \frac{y(y + 3)}{3(y + 3)}$$ [1.5 Marks]
Cancelling common non-zero terms $(x + 2)$ and $(y + 3)$:
$$\frac{x}{2} = \frac{y}{3} \implies \frac{x}{y} = \frac{2}{3} \implies \mathbf{x : y = 2 : 3}$$ [1.5 Marks]
(Alternative by Componendo & Dividendo also accepted).

Tangent radius theorem: $OA \perp TP \implies \angle OAP = 90^\circ$ and $OC \perp TQ \implies \angle OCQ = 90^\circ$.
(i) In $\triangle OAB$: $OA = OB$ (radii).
$\angle OAB = 90^\circ - \angle BAP = 90^\circ - 60^\circ = 30^\circ \implies \mathbf{\angle OBA = 30^\circ}$.
In $\triangle OBC$: $OC = OB$ (radii).
$\angle OCB = 90^\circ - \angle BCQ = 90^\circ - 55^\circ = 35^\circ \implies \mathbf{\angle OBC = 35^\circ}$. [1.5 Marks]
(ii) Total angle $\angle ABC = \angle OBA + \angle OBC = 30^\circ + 35^\circ = 65^\circ$.
Angle at centre is double the angle at the circumference:
$\mathbf{\angle AOC = 2 \times \angle ABC = 2 \times 65^\circ = 130^\circ}$. [1.5 Marks]
(iii) In tangent quadrilateral $OATC$: $\angle OAT = \angle OCT = 90^\circ$.
$\mathbf{\angle ATC = 180^\circ - \angle AOC = 180^\circ - 130^\circ = 50^\circ}$. [1 Mark]

Step 1 Let $k$ be added to the polynomial. Then $P(x) = 2x^3 - 3x^2 - 8x + k$.
By Remainder Theorem, when $P(x)$ is divided by $2x + 1$, remainder is $P\left(-\frac{1}{2}\right) = 10$. [1 Mark]
Step 2 Substitute $x = -\frac{1}{2}$ into $P(x)$:
$$P\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{8}\right) - 3\left(\frac{1}{4}\right) - 8\left(-\frac{1}{2}\right) + k = -\frac{1}{4} - \frac{3}{4} + 4 + k = -1 + 4 + k = 3 + k$$ [1 Mark]
Step 3 Equating remainder to 10:
$$3 + k = 10 \implies \mathbf{k = 7}$$
Answer: 7 must be added. [1 Mark]

Monthly deposit $P = ₹750$, Time $n = 2 \text{ years} = 24 \text{ months}$.
Step 1 Total deposit $= P \times n = 750 \times 24 = ₹18,000$.
Maturity Value $= ₹19,125 \implies$ Total Interest $I = 19,125 - 18,000 = \mathbf{₹1,125}$. [1 Mark]
Step 2 Apply RD interest formula:
$$I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{R}{100} \implies 1125 = 750 \times \frac{24 \times 25}{24} \times \frac{R}{100}$$ [1 Mark]
$$1125 = 750 \times 25 \times \frac{R}{100} = 187.5 R \implies R = \frac{1125}{187.5} = \mathbf{6\% \text{ p.a.}}$$
Answer: Rate of interest = 6% per annum. [1 Mark]

(i) Points $A(-4, 0), B(-3, 2), C(0, 4), D(4, 1), E(7, 3)$ plotted accurately on graph sheet. [1 Mark]
(ii) Coordinates of reflected points across the x-axis ($M_x(x, y) = (x, -y)$):
$B'(-3, -2), \quad C'(0, -4), \quad D'(4, -1), \quad E'(7, -3)$. [1 Mark]
(iii) Join in order: $A \to B \to C \to D \to E \to E' \to D' \to C' \to B' \to A$. [1 Mark]
(iv) The figure has 9 sides and 9 vertices.
Name of closed figure: Nonagon (or Arrowhead Nonagon / 9-sided polygon). [1 Mark]

Cumulative Table (Upper Limit, cf): $(12, 0), (13, 3), (14, 12), (15, 24), (16, 33), (17, 37), (18, 39), (19, 40)$.
Smooth ogive plotted on graph paper with given scale. [3 Marks]
(i) Median $= \frac{N}{2} = \frac{40}{2} = 20\text{th term}$. Horizontal line from $y = 20$ meets curve at $x \approx \mathbf{14.7\text{ m}}$ (acceptable: $14.6 - 14.8$). [1 Mark]
(ii) Upper Quartile $Q_3 = \frac{3N}{4} = \frac{3 \times 40}{4} = 30\text{th term}$. Horizontal line from $y = 30$ meets curve at $x \approx \mathbf{15.7\text{ m}}$ (acceptable: $15.6 - 15.8$). [1 Mark]
(iii) Above 16.5 m Vertical line from $x = 16.5$ meets curve at $cf \approx 35$.
Number of students covering $> 16.5\text{ m} = 40 - 35 = \mathbf{5\text{ students}}$ (acceptable: $4 - 6$). [1 Mark]

Step 1 Express $x$ as $\frac{x}{1}$ and apply Componendo and Dividendo:
$$\frac{x + 1}{x - 1} = \frac{(\sqrt{2a + 1} + \sqrt{2a - 1}) + (\sqrt{2a + 1} - \sqrt{2a - 1})}{(\sqrt{2a + 1} + \sqrt{2a - 1}) - (\sqrt{2a + 1} - \sqrt{2a - 1})} = \frac{2\sqrt{2a + 1}}{2\sqrt{2a - 1}} = \frac{\sqrt{2a + 1}}{\sqrt{2a - 1}}$$ [1.5 Marks]
Step 2 Squaring both sides:
$$\frac{(x + 1)^2}{(x - 1)^2} = \frac{2a + 1}{2a - 1} \implies \frac{x^2 + 2x + 1}{x^2 - 2x + 1} = \frac{2a + 1}{2a - 1}$$ [1 Mark]
Step 3 Applying Componendo and Dividendo again:
$$\frac{(x^2 + 2x + 1) + (x^2 - 2x + 1)}{(x^2 + 2x + 1) - (x^2 - 2x + 1)} = \frac{(2a + 1) + (2a - 1)}{(2a + 1) - (2a - 1)}$$
$$\frac{2(x^2 + 1)}{4x} = \frac{4a}{2} \implies \frac{x^2 + 1}{2x} = 2a \implies x^2 + 1 = 4ax \implies \mathbf{x^2 - 4ax + 1 = 0}$$
Hence Proved. [1.5 Marks]

Step 1 Let the first term be $a$ and common difference be $d$.
$T_6 = a + 5d$. Given $T_6 = 4a$:
$$a + 5d = 4a \implies 5d = 3a \implies a = \frac{5}{3}d$$ [1 Mark]
Step 2 Given $S_6 = 75$:
$$S_6 = \frac{6}{2}[2a + (6 - 1)d] = 3(2a + 5d) = 75 \implies 2a + 5d = 25$$ [1 Mark]
Step 3 Substitute $5d = 3a$:
$$2a + 3a = 25 \implies 5a = 25 \implies \mathbf{a = 5}$$
Then $5d = 3(5) = 15 \implies \mathbf{d = 3}$.
Answer: First term $a = 5$, common difference $d = 3$. [1 Mark]

Step 1 Let the smaller natural number be $x$. Then the larger number is $x + 7$.
Product equation: $x(x + 7) = 450 \implies x^2 + 7x - 450 = 0$. [1 Mark]
Step 2 Factorising $x^2 + 7x - 450 = 0$:
Product $= -450$, Sum $= 7 \implies +25$ and $-18$.
$x^2 + 25x - 18x - 450 = 0 \implies x(x + 25) - 18(x + 25) = 0 \implies (x + 25)(x - 18) = 0$. [1 Mark]
Step 3 Since $x$ is a natural number ($x \in \mathbb{N}$), $x = 18$ ($x = -25$ rejected).
Larger number $= 18 + 7 = 25$.
Answer: The numbers are 18 and 25. [1 Mark]

(i) Construct circle of radius $4.5\text{ cm}$. Draw chord $AB = 6\text{ cm}$.
The locus of points equidistant from two points $A$ and $B$ is the perpendicular bisector of $AB$.
Construct the perpendicular bisector of $AB$; it passes through the centre $O$ and intersects the circle at $D$. [1.5 Marks]
(ii) Join $AD$. The locus of points equidistant from two intersecting lines $AD$ and $AB$ is the internal angle bisector of $\angle DAB$.
Bisect $\angle DAB$; let the bisector intersect the circle at $C$. [1.5 Marks]
(iii) Join $BC$ and $CD$ to complete cyclic quadrilateral $ABCD$.
Measure the side $CD$:
Length of side $CD \approx 5.2\text{ cm}$ (acceptable tolerance: $5.0\text{ cm}$ to $5.4\text{ cm}$). [1 Mark]

Scale factor $k = \frac{\text{Model dimension}}{\text{Actual dimension}} = \frac{1}{50}$.
(i) Actual height $= \text{Model height} \times 50 = 0.8\text{ m} \times 50 = \mathbf{40\text{ m}}$. [1.5 Marks]
(ii) Area scale factor $= k^2 = \left(\frac{1}{50}\right)^2 = \frac{1}{2500}$.
$$\text{Floor area of model} = \frac{\text{Actual Area}}{2500} = \frac{20\text{ m}^2}{2500} = \frac{1}{125}\text{ m}^2 = \mathbf{0.008\text{ m}^2}$$
In $\text{cm}^2$: $0.008 \times 10,000 = \mathbf{80\text{ cm}^2}$. [1.5 Marks]

Radius of cylinder and cones $r = \frac{6}{2} = 3\text{ cm}$.
Height of cylinder $H = 28\text{ cm}$. Height of each cone $h = 10.5\text{ cm} = \frac{21}{2}\text{ cm}$.
Step 1 Volume of cylinder:
$$V_{\text{cylinder}} = \pi r^2 H = \frac{22}{7} \times 3^2 \times 28 = \frac{22}{7} \times 9 \times 28 = 22 \times 9 \times 4 = \mathbf{792\text{ cm}^3}$$ [1 Mark]
Step 2 Volume of two identical conical cavities:
$$2 \times V_{\text{cone}} = 2 \times \left(\frac{1}{3} \pi r^2 h\right) = \frac{2}{3} \times \frac{22}{7} \times 3^2 \times \frac{21}{2} = \frac{2}{3} \times \frac{22}{7} \times 9 \times \frac{21}{2} = 22 \times 9 = \mathbf{198\text{ cm}^3}$$ [1 Mark]
Step 3 Volume of remaining solid:
$$V_{\text{remaining}} = V_{\text{cylinder}} - 2 \times V_{\text{cone}} = 792 - 198 = \mathbf{594\text{ cm}^3}$$
Answer: Volume of remaining solid = $594\text{ cm}^3$. [1 Mark]

Step 1 Express $\cot \theta$ as $\frac{1}{\tan \theta}$:
$$\text{LHS} = \left(\frac{1 - \tan \theta}{1 - \frac{1}{\tan \theta}}\right)^2$$ [1.5 Marks]
Step 2 Simplify the denominator fraction:
$$1 - \frac{1}{\tan \theta} = \frac{\tan \theta - 1}{\tan \theta} = -\frac{1 - \tan \theta}{\tan \theta}$$ [1.5 Marks]
Step 3 Substitute back into LHS:
$$\text{LHS} = \left(\frac{1 - \tan \theta}{-\frac{1 - \tan \theta}{\tan \theta}}\right)^2 = \left(-\tan \theta\right)^2 = \mathbf{\tan^2 \theta} = \text{RHS}$$
Hence Proved. [1 Mark]