ICSE Class 10 Mathematics 2026 Improvement Solved Paper

Solution $\text{Marked Price} = \text{₹ } 18,000$.
$\text{GST} = 18\% \times 18000 = \frac{18}{100} \times 18000 = \text{₹ } 3240$.
$\text{Price paid by customer} = 18000 + 3240 = \mathbf{\text{₹ } 21,240}$.
Correct Answer: (d) ₹ 21,240 [1 Mark]

Solution Given: $\frac{h_1}{h_2} = \frac{2}{1}$ and $\frac{r_1}{r_2} = \frac{1}{2}$.
$\text{Ratio of volumes } \frac{V_1}{V_2} = \frac{\frac{1}{3}\pi r_1^2 h_1}{\frac{1}{3}\pi r_2^2 h_2} = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right) = \left(\frac{1}{2}\right)^2 \times \left(\frac{2}{1}\right) = \frac{1}{4} \times 2 = \mathbf{\frac{1}{2}}$.
Correct Answer: (b) 1 : 2 [1 Mark]

Solution Formula for RD interest: $I = P \times \frac{n(n + 1)}{24} \times \frac{r}{100}$.
Since $I \propto r$ when $P$ and $n$ remain constant, doubling the rate from $r\%$ to $2r\%$ doubles the interest earned.
Correct Answer: (d) double [1 Mark]

Solution Dividend declared by a company is always calculated as a percentage of the nominal value (face value) of each share, regardless of market value fluctuations.
Correct Answer: (b) face value [1 Mark]

Solution $a = 2, b = 5, c = c$.
$D = b^2 - 4ac = 5^2 - 4(2)(c) = 25 - 8c$.
Given $D = 1 \implies 25 - 8c = 1 \implies 8c = 24 \implies \mathbf{c = 3}$.
Correct Answer: (d) 3 [1 Mark]

Solution For $a, b, c$ in continued proportion, $b^2 = ac$.
$(x + 6)^2 = (x + 3)(x + 10)$
$x^2 + 12x + 36 = x^2 + 13x + 30 \implies 13x - 12x = 36 - 30 \implies \mathbf{x = 6}$.
Correct Answer: (b) 6 [1 Mark]

Solution By Factor Theorem, if $(x - 1)$ is a factor, then $f(1) = 0$.
$f(1) = (1)^3 + 2(1)^2 - (1) - k = 0 \implies 1 + 2 - 1 - k = 0 \implies \mathbf{k = 2}$.
Correct Answer: (d) 2 [1 Mark]

Solution Compare corresponding matrix entries:
$m(-1) = 3 \implies \mathbf{m = -3}$, and $m(2) = -6 \implies m = -3$.
Correct Answer: (b) -3 [1 Mark]

Solution For collinear points, $\text{Slope}(AB) = \text{Slope}(BC)$:
$\frac{-1 - 0}{-1 - x} = \frac{y - (-1)}{0 - (-1)} \implies \frac{1}{x + 1} = y + 1$
$(x + 1)(y + 1) = 1 \implies xy + x + y + 1 = 1 \implies x + y = -xy$.
Divide both sides by $xy$: $\frac{x}{xy} + \frac{y}{xy} = \frac{-xy}{xy} \implies \mathbf{\frac{1}{y} + \frac{1}{x} = -1}$.
Correct Answer: (a) -1 [1 Mark]

Solution $\text{ar}(\Delta ABC) = \text{ar}(\Delta APQ) + \text{ar}(BPQC) = 2 \times \text{ar}(\Delta APQ)$.
Since $\Delta APQ \sim \Delta ABC$, $\frac{\text{ar}(\Delta APQ)}{\text{ar}(\Delta ABC)} = \left(\frac{AP}{AB}\right)^2 = \frac{1}{2} \implies \frac{AP}{AB} = \frac{1}{\sqrt{2}}$.
$AB = \sqrt{2} AP \implies PB = AB - AP = (\sqrt{2} - 1) AP$.
$\therefore \frac{AP}{PB} = \frac{1}{\sqrt{2} - 1} = \frac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \mathbf{(\sqrt{2} + 1) : 1}$.
Correct Answer: (d) (√2 + 1) : 1 [1 Mark]

Solution In $\Delta PAB$, $PA = PB = AB \implies \Delta PAB$ is equilateral $\implies \angle APB = 60^\circ$.
Central angle $\angle AOB = 180^\circ - \angle APB = 180^\circ - 60^\circ = 120^\circ$.
The angle subtended by arc $AB$ at the circumference is half the central angle:
$\angle ACB = \frac{1}{2}\angle AOB = \frac{1}{2}(120^\circ) = \mathbf{60^\circ}$.
Correct Answer: (c) 60° [1 Mark]

Solution When a semi-circular sheet of radius $R$ is folded into a cone, the radius of the semicircle becomes the slant height ($l$) of the cone.
$\therefore \mathbf{l = 7\text{ cm}}$.
Correct Answer: (b) 7 cm [1 Mark]

Solution $\sec\theta (1 + \sin\theta) = \sec\theta + \sec\theta\sin\theta = \sec\theta + \tan\theta$.
Then $(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = \sec^2\theta - \tan^2\theta = 1$.
Hence Assertion (A) is true, Reason (R) is true, and (R) is the exact justification for (A).
Correct Answer: (c) Both (A) and (R) are true and (R) is the correct explanation of (A). [1 Mark]

Solution The Sun rising in the west is an impossible physical event, and since $P(\phi) = 0$, (R) directly explains (A).
Correct Answer: (c) Both (A) and (R) are true and (R) is the correct explanation of (A). [1 Mark]

Solution If every observation $x_i$ is increased by a constant $k$, the arithmetic mean is also increased by $k$. Here $k = 2$, so new mean $= \mathbf{m + 2}$.
Correct Answer: (c) m + 2 [1 Mark]

Step 1 Dimensions of Cylinder:
Radius $r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$, height $h = 7\text{ cm}$.
$\text{Volume of Cylinder } V_{\text{cyl}} = \pi r^2 h = \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times 7 = \frac{22}{7} \times \frac{49}{4} \times 7 = \frac{1078}{4} = 269.5\text{ cm}^3$. [1.5 Marks]
Step 2 Dimensions of Inscribed Sphere:
Since the sphere fits completely into the container, its diameter $= h = 7\text{ cm} \implies$ radius $R = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$.
$\text{Volume of Sphere } V_{\text{sph}} = \frac{4}{3}\pi R^3 = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^3 = \frac{4}{3} \times \frac{22}{7} \times \frac{343}{8} = \frac{539}{3} \approx 179.67\text{ cm}^3$. [1.5 Marks]
Step 3 Volume of water to fill remaining space:
$V_{\text{water}} = V_{\text{cyl}} - V_{\text{sph}} = 269.5 - 179.67 = 89.83\text{ cm}^3$.
Rounding to nearest whole number: $\mathbf{V_{\text{water}} \approx 90\text{ cm}^3}$. [1 Mark]

Step 1 Let the original speed of the car be $x\text{ km/h}$.
Original time $t_1 = \frac{400}{x}\text{ hours}$.
Increased speed $= (x + 12)\text{ km/h}$.
New time $t_2 = \frac{400}{x + 12}\text{ hours}$.
Time difference $= 1\text{ hr } 40\text{ min} = 1 + \frac{40}{60} = 1 + \frac{2}{3} = \frac{5}{3}\text{ hours}$. [1 Mark]
Step 2 Formulate and solve quadratic equation:
$\frac{400}{x} - \frac{400}{x + 12} = \frac{5}{3} \implies 400 \left[\frac{x + 12 - x}{x(x + 12)}\right] = \frac{5}{3}$
$\frac{4800}{x^2 + 12x} = \frac{5}{3} \implies 5(x^2 + 12x) = 14400 \implies x^2 + 12x - 2880 = 0$. [1.5 Marks]
Step 3 Factorise quadratic equation:
$(x + 60)(x - 48) = 0 \implies x = 48 \text{ or } x = -60$.
Since speed cannot be negative, $x = 48\text{ km/h}$.
(a) $\mathbf{\text{Original speed} = 48\text{ km/h}}$. [1 Mark]
Step 4 (b) Time taken with increased speed ($48 + 12 = 60\text{ km/h}$):
$\text{Time} = \frac{400}{60} = \frac{20}{3}\text{ hours} = \mathbf{6\text{ hours } 40\text{ minutes}}$. [0.5 Mark]

Step 1 (a) Use Centroid Formula $G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$:
$x_G = \frac{a + 5 + (-5)}{3} = \frac{a}{3} = \frac{2}{3} \implies \mathbf{a = 2}$.
$y_G = \frac{b + (-3) + 3}{3} = \frac{b}{3} = \frac{1}{3} \implies \mathbf{b = 1}$.
$\therefore \mathbf{\text{Coordinates of } A = (2, 1)}$. [2 Marks]
Step 2 (b) Find midpoint $M$ of side $AB$ opposite to vertex $C$:
$A(2, 1)$ and $B(5, -3) \implies M = \left(\frac{2 + 5}{2}, \frac{1 + (-3)}{2}\right) = \left(\frac{7}{2}, -1\right)$. [1 Mark]
Step 3 Equation of median line joining $C(-5, 3)$ and $M\left(\frac{7}{2}, -1\right)$:
$\text{Slope } m = \frac{-1 - 3}{\frac{7}{2} - (-5)} = \frac{-4}{\frac{17}{2}} = -\frac{8}{17}$.
Using point-slope form with $C(-5, 3)$:
$y - 3 = -\frac{8}{17}(x + 5) \implies 17(y - 3) = -8(x + 5) \implies 17y - 51 = -8x - 40$
$\mathbf{8x + 17y - 11 = 0}$. [1 Mark]

Step 1 (a) Interest earned by Anamika:
$P = \text{₹ } 300, n = 3 \times 12 = 36\text{ months}, r = 10\%$.
$I_{\text{Anamika}} = 300 \times \frac{36 \times 37}{24} \times \frac{10}{100} = 300 \times 55.5 \times 0.1 = \mathbf{\text{₹ } 1665}$. [2 Marks]
Step 2 Interest earned by Sanjeev:
$P = \text{₹ } 500, n = 2 \times 12 = 24\text{ months}, r = 10\%$.
$I_{\text{Sanjeev}} = 500 \times \frac{24 \times 25}{24} \times \frac{10}{100} = 500 \times 25 \times 0.1 = \text{₹ } 1250$. [1 Mark]
Step 3 (b) Compare interest earned:
$I_{\text{Anamika}} - I_{\text{Sanjeev}} = 1665 - 1250 = \text{₹ } 415$.
$\mathbf{\text{Anamika got more interest by ₹ 415}}$. [1 Mark]

Step 1 (a) Simplify surd terms:
$T_1 = \sqrt{2}$, $T_2 = \sqrt{8} = 2\sqrt{2}$, $T_3 = \sqrt{18} = 3\sqrt{2}$.
$T_2 - T_1 = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$, and $T_3 - T_2 = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$.
Since the difference between consecutive terms is constant, it is an **Arithmetic Progression (A.P.)** with $a = \sqrt{2}$ and $d = \sqrt{2}$. [2 Marks]
Step 2 (b) Calculate sum of first 10 terms ($S_{10}$):
$S_{10} = \frac{10}{2}[2a + (10 - 1)d] = 5[2\sqrt{2} + 9\sqrt{2}] = 5(11\sqrt{2}) = 55\sqrt{2}$.
Given $S_{10} = p\sqrt{2} \implies \mathbf{p = 55}$. [2 Marks]

Step 1 (a) Plot points: $A(2, 3)$ and $B(4, 5)$. [1 Mark]
Step 2 (b) Reflection in the x-axis has mapping $(x, y) \xrightarrow{M_x} (x, -y)$:
$\mathbf{A'(2, -3)}$ and $\mathbf{B'(4, -5)}$. [2 Marks]
Step 3 (c) Reflection of $A'(2, -3)$ in the y-axis has mapping $(x, y) \xrightarrow{M_y} (-x, y)$:
$\mathbf{A''(-2, -3)}$. [1 Mark]
Step 4 (d) Transformation mapping $A(2, 3) \to A''(-2, -3)$:
$(x, y) \to (-x, -y)$ represents a **Reflection in the Origin $(0, 0)$** (or rotation through $180^\circ$ about origin). [1 Mark]

Step 1 (a) Total discount:
Discount on Component A $= 20\% \times 5000 = \text{₹ } 1000$.
Discount on Component B $= 30\% \times 6500 = \text{₹ } 1950$.
$\mathbf{\text{Total Discount} = 1000 + 1950 = \text{₹ } 2950}$. [1.5 Marks]
Step 2 (b) Total GST amount:
Taxable amount for A $= 5000 - 1000 = \text{₹ } 4000$.
Taxable amount for B $= 6500 - 1950 = \text{₹ } 4550$.
Total taxable value $= 4000 + 4550 = \text{₹ } 8550$.
$\mathbf{\text{Total GST} = 18\% \times 8550 = 0.18 \times 8550 = \text{₹ } 1539}$. [1.5 Marks]

Step 1 (a) Evaluate $A^2$ and $5A$:
$A^2 = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 3(3) + (-5)(-4) & 3(-5) + (-5)(2) \\ -4(3) + 2(-4) & -4(-5) + 2(2) \end{bmatrix} = \mathbf{\begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix}}$. [1 Mark]
$5A = 5 \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} = \mathbf{\begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix}}$. [0.5 Mark]
Step 2 (b) Evaluate $A^2 - 5A$ and solve for $k$:
$A^2 - 5A = \begin{bmatrix} 29 - 15 & -25 - (-25) \\ -20 - (-20) & 24 - 10 \end{bmatrix} = \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix}$.
$kI = k \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} \implies \mathbf{k = 14}$. [1.5 Marks]

Step 1 Combine the fractions by taking the common denominator:
$\text{LHS} = \frac{\left(\sqrt{\sec\theta + 1}\right)^2 + \left(\sqrt{\sec\theta - 1}\right)^2}{\sqrt{(\sec\theta - 1)(\sec\theta + 1)}} = \frac{(\sec\theta + 1) + (\sec\theta - 1)}{\sqrt{\sec^2\theta - 1}}$. [1.5 Marks]
Step 2 Simplify numerator and denominator:
$\text{Numerator} = 2\sec\theta$.
Using identity $\sec^2\theta - 1 = \tan^2\theta$, $\sqrt{\sec^2\theta - 1} = \tan\theta$.
$\text{LHS} = \frac{2\sec\theta}{\tan\theta}$. [1.5 Marks]
Step 3 Express in terms of sine and cosine:
$\frac{2\sec\theta}{\tan\theta} = 2 \times \frac{1/\cos\theta}{\sin\theta/\cos\theta} = \frac{2}{\sin\theta} = \mathbf{2\text{ cosec}\,\theta} = \text{RHS}$.
Hence Proved. [1 Mark]

Step 1 (a) Number of shares purchased:
$n = \frac{\text{Investment}}{MV} = \frac{7500}{150} = \mathbf{50\text{ shares}}$. [1 Mark]
Step 2 (b) Annual income:
$\text{Income} = n \times \text{Dividend \%} \times NV = 50 \times \frac{6}{100} \times 100 = \mathbf{\text{₹ } 300}$. [1 Mark]
Step 3 (c) Gain on selling $50\%$ shares ($25$ shares) at ₹ 175:
Gain per share $= 175 - 150 = \text{₹ } 25$.
$\mathbf{\text{Total Gain} = 25 \times 25 = \text{₹ } 625}$. [1 Mark]

Step 1 Solve linear inequation in two parts:
Part 1: $3x + \frac{14}{3} > \frac{4x}{3} - 2$
$\implies 3x - \frac{4x}{3} > -2 - \frac{14}{3} \implies \frac{5x}{3} > -\frac{20}{3} \implies 5x > -20 \implies x > -4$. [1 Mark]
Part 2: $\frac{4x}{3} - 2 \ge 3x - 4$
$\implies -2 + 4 \ge 3x - \frac{4x}{3} \implies 2 \ge \frac{5x}{3} \implies 5x \le 6 \implies x \le \frac{6}{5} = 1.2$. [1 Mark]
Step 2 Combine inequalities:
$\mathbf{\text{Solution Set: } \{x : x \in \mathbb{R}, -4 < x \le 1.2\}}$.
Graph with open circle at $-4$, filled solid dot at $1.2$, and thickened segment between them. [1 Mark]

Step 1 (a) Corresponding sides in similar triangles $\Delta ABC \sim \Delta DEC$:
$\mathbf{\frac{AB}{DE} = \frac{BC}{EC} = \frac{AC}{DC}}$. [1.5 Marks]
Step 2 (b) Angle correspondence:
Since $\Delta ABC \sim \Delta DEC$, corresponding angles are equal:
$\mathbf{\angle ABC = \angle DEC}$. [1 Mark]
Step 3 (c) Ratio of areas of similar triangles:
$\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta DEC)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{2}{3}\right)^2 = \mathbf{\frac{4}{9}}$ (or $4 : 9$). [1.5 Marks]

Step 1 (a) Find $\angle PSR$:
Since $PQRS$ is a cyclic quadrilateral, opposite angles are supplementary:
$\angle PSR + \angle PQR = 180^\circ \implies \angle PSR + 112^\circ = 180^\circ \implies \mathbf{\angle PSR = 68^\circ}$. [1 Mark]
Step 2 (b) Find $\angle BPR$:
By Alternate Segment Theorem, the angle between tangent $PB$ and chord $PR$ equals the angle subtended by chord $PR$ in the alternate segment ($\angle PSR$):
$\mathbf{\angle BPR = \angle PSR = 68^\circ}$. [1 Mark]
Step 3 (c) Find $\angle PTS$:
In cyclic quadrilateral $PRST$ or using chord $PS$ with isosceles triangle $PS = SR$:
In $\Delta PSR$, $PS = SR \implies \angle SPR = \angle SRP = \frac{180^\circ - 68^\circ}{2} = 56^\circ$.
Angles subtended by chord $PS$ in the same segment: $\angle PTS = \angle PRS = \mathbf{56^\circ}$. [1 Mark]

Step 1 Prepare step-deviation table with Assumed Mean $A = 32.5$, class width $h = 5$:
$u_i = \frac{x_i - 32.5}{5}$:
$20-25: x_1 = 22.5, u_1 = -2, f_1 = 10, f_1 u_1 = -20$
$25-30: x_2 = 27.5, u_2 = -1, f_2 = 7, f_2 u_2 = -7$
$30-35: x_3 = 32.5, u_3 = 0, f_3 = 9, f_3 u_3 = 0$
$35-40: x_4 = 37.5, u_4 = +1, f_4 = 13, f_4 u_4 = +13$
$40-45: x_5 = 42.5, u_5 = +2, f_5 = 5, f_5 u_5 = +10$
$45-50: x_6 = 47.5, u_6 = +3, f_6 = 6, f_6 u_6 = +18$
$\sum f_i = 10 + 7 + 9 + 13 + 5 + 6 = 50$.
$\sum f_i u_i = -27 + 41 = 14$. [1.5 Marks]
Step 2 Apply Step-Deviation Formula:
$\bar{x} = A + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 32.5 + \left(\frac{14}{50}\right) \times 5 = 32.5 + 1.4 = \mathbf{33.9}$.
Rounding to nearest whole number: $\mathbf{\bar{x} \approx 34\text{ marks}}$. [1.5 Marks]

Step 1 (a) Apply Remainder Theorem for both polynomials:
Let $f(x) = mx^2 - 3x + 6$. Dividing by $(x - 2)$ gives remainder $R_1 = f(2)$:
$R_1 = m(2)^2 - 3(2) + 6 = 4m - 6 + 6 = 4m$. [1 Mark]
Let $g(x) = 3x^2 + 5x - m$. Dividing by $(x + 3)$ gives remainder $R_2 = g(-3)$:
$R_2 = 3(-3)^2 + 5(-3) - m = 27 - 15 - m = 12 - m$. [1 Mark]
Step 2 Given condition $R_1 = 2 R_2$:
$4m = 2(12 - m) \implies 4m = 24 - 2m \implies 6m = 24 \implies \mathbf{m = 4}$. [1 Mark]
Step 3 (b) Calculate remainder $R_1$ using $m = 4$:
$\mathbf{\text{Remainder } R_1 = 4m = 4(4) = 16}$. [1 Mark]

Step 1 Sample Space:
$S = \{\text{'B'}, \text{'I'}, \text{'N'}, \text{'O'}, \text{'C'}, \text{'U'}, \text{'L'}, \text{'A'}, \text{'R'}, \text{'S'}\} \implies n(S) = 10$. [1 Mark]
Step 2 (a) Event $E_1$ (vowels in the word):
$E_1 = \{\text{'I'}, \text{'O'}, \text{'U'}, \text{'A'}\} \implies n(E_1) = 4$.
$\mathbf{P(E_1) = \frac{4}{10} = \frac{2}{5}}$. [1 Mark]
Step 3 (b) Event $E_2$ (first 8 letters of English alphabet $\{A, B, C, D, E, F, G, H\}$ present in word):
$E_2 = \{\text{'B'}, \text{'C'}, \text{'A'}\} \implies n(E_2) = 3$.
$\mathbf{P(E_2) = \frac{3}{10}}$. [1 Mark]

Step 1 Write RHS as $\frac{3}{1}$ and apply Componendo and Dividendo:
$\frac{(\sqrt{2x + 3} + \sqrt{2x - 3}) + (\sqrt{2x + 3} - \sqrt{2x - 3})}{(\sqrt{2x + 3} + \sqrt{2x - 3}) - (\sqrt{2x + 3} - \sqrt{2x - 3})} = \frac{3 + 1}{3 - 1}$. [1 Mark]
Step 2 Simplify both sides:
$\frac{2\sqrt{2x + 3}}{2\sqrt{2x - 3}} = \frac{4}{2} \implies \frac{\sqrt{2x + 3}}{\sqrt{2x - 3}} = 2$. [1 Mark]
Step 3 Square both sides and solve for $x$:
$\frac{2x + 3}{2x - 3} = 4 \implies 2x + 3 = 4(2x - 3) = 8x - 12$
$8x - 2x = 3 + 12 \implies 6x = 15 \implies x = \frac{15}{6} = \mathbf{2.5}$ (or $\frac{5}{2}$). [1 Mark]

Step 1 (a) Prove $\Delta PTR \sim \Delta QTP$:
In $\Delta PTR$ and $\Delta QTP$:
$\angle T = \angle T$ (Common angle)
$\angle TPR = \angle TQP$ (Angles in alternate segment by Alternate Segment Theorem)
$\therefore \mathbf{\Delta PTR \sim \Delta QTP}$ by **AA Similarity Criterion**. [2 Marks]
Step 2 (b) Determine length of $PT$ using ratio of corresponding sides:
$\frac{PT}{QT} = \frac{RT}{PT} \implies PT^2 = QT \times RT$.
Given $QR = 9\text{ cm}$ and $RT = 16\text{ cm} \implies QT = QR + RT = 9 + 16 = 25\text{ cm}$.
$PT^2 = 25 \times 16 = 400 \implies PT = \sqrt{400} = \mathbf{20\text{ cm}}$. [2 Marks]

Step 1 Identify Modal Class:
Highest frequency is $18$, corresponding to modal class $50 - 60$. [1 Mark]
Step 2 Draw histogram and criss-cross lines:
Join top corners of modal bar to opposite top corners of adjacent bars to locate intersection point. [1 Mark]
Step 3 Drop perpendicular from intersection to x-axis:
$\mathbf{\text{Estimated Mode} = 54}$ (acceptable range: $53 - 55$). [1 Mark]

Step 1 Identify coefficients and evaluate discriminant:
$a = 1, b = -6, c = -8$.
$D = b^2 - 4ac = (-6)^2 - 4(1)(-8) = 36 + 32 = 68$. [1 Mark]
Step 2 Apply quadratic formula:
$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-6) \pm \sqrt{68}}{2(1)} = \frac{6 \pm 8.2462}{2} = 3 \pm 4.1231$. [1 Mark]
Step 3 Calculate roots to 3 significant figures:
$x_1 = 3 + 4.1231 = 7.1231 \approx \mathbf{7.12}$
$x_2 = 3 - 4.1231 = -1.1231 \approx \mathbf{-1.12}$
Answer: $x = 7.12$ or $-1.12$. [1 Mark]

Step 1 (a) Find radius of sphere ($R$):
$\text{Surface Area} = 4\pi R^2 = 616 \implies 4 \times \frac{22}{7} \times R^2 = 616$
$R^2 = \frac{616 \times 7}{88} = 7 \times 7 = 49 \implies \mathbf{R = 7\text{ cm}}$. [1.5 Marks]
Step 2 Calculate Volume of Sphere:
$V_{\text{sph}} = \frac{4}{3}\pi R^3 = \frac{4}{3} \times \frac{22}{7} \times (7)^3 = \frac{4312}{3}\text{ cm}^3$. [1 Mark]
Step 3 Calculate Volume of One Cone ($r = \frac{3.5}{2} = \frac{7}{4}\text{ cm}, h = \frac{7}{2}\text{ cm}$):
$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times \left(\frac{7}{4}\right)^2 \times \frac{7}{2} = \frac{539}{48}\text{ cm}^3$. [1 Mark]
Step 4 (b) Calculate number of cones formed:
$\text{Number of cones} = \frac{V_{\text{sph}}}{V_{\text{cone}}} = \frac{4312/3}{539/48} = \frac{4312}{3} \times \frac{48}{539} = 8 \times 16 = \mathbf{128\text{ cones}}$. [0.5 Mark]

Step 1 (a) Freehand sketch with angles of depression $35^\circ$ and $55^\circ$ converting to alternate angles of elevation. [1.5 Marks]
Step 2 (b) In right triangle $\Delta ABD$ ($\angle ADB = 55^\circ$):
$\tan 55^\circ = \frac{AB}{BD} \implies BD = \frac{80}{\tan 55^\circ}$.
Using mathematical tables: $\tan 55^\circ \approx 1.4281 \implies BD = \frac{80}{1.4281} \approx 56.02\text{ m}$. [1.5 Marks]
Step 3 In right triangle $\Delta ABC$ ($\angle ACB = 35^\circ$):
$\tan 35^\circ = \frac{AB}{BC} \implies BC = \frac{80}{\tan 35^\circ}$.
Using mathematical tables: $\tan 35^\circ \approx 0.7002 \implies BC = \frac{80}{0.7002} \approx 114.25\text{ m}$. [1 Mark]
Step 4 Distance travelled by ship ($CD$):
$CD = BC - BD = 114.25 - 56.02 = 58.23\text{ m}$.
Rounding to nearest meter: $\mathbf{CD \approx 58\text{ m}}$. [1 Mark]

Step 1 Prepare Cumulative Frequency Table ($N = 80$):
$(155, 8), (160, 23), (165, 47), (170, 64), (175, 75), (180, 80)$.
Plot cumulative frequencies against upper limits on graph paper and join with smooth S-curve. [2 Marks]
Step 2 (a) Median height ($N/2 = 40\text{th student}$):
Horizontal line from $y = 40$ to ogive, then drop perpendicular to x-axis $\implies \mathbf{\text{Median} \approx 163.5\text{ cm}}$ (acceptable range: $163 - 164\text{ cm}$). [1 Mark]
Step 3 (b) Lower-quartile height $Q_1$ ($N/4 = 20\text{th student}$):
From $y = 20$ on y-axis to curve, drop perpendicular $\implies \mathbf{Q_1 \approx 159\text{ cm}}$ (acceptable range: $158.5 - 159.5\text{ cm}$). [1 Mark]
Step 4 (c) Number of students taller than $173\text{ cm}$:
From $x = 173$ on x-axis, vertical line to curve meets at $y \approx 71$.
Number of students with height $> 173\text{ cm} = 80 - 71 = \mathbf{9\text{ students}}$. [1 Mark]

Step 1 Let the three terms of G.P. be $\frac{a}{r}, a, ar$.
Product $= \left(\frac{a}{r}\right)(a)(ar) = a^3 = 27 \implies \mathbf{a = 3}$. [1 Mark]
Step 2 Sum of three terms $= \frac{21}{2}$:
$\frac{3}{r} + 3 + 3r = \frac{21}{2} \implies 3\left(\frac{1}{r} + 1 + r\right) = \frac{21}{2} \implies \frac{r^2 + r + 1}{r} = \frac{7}{2}$
$2r^2 + 2r + 2 = 7r \implies 2r^2 - 5r + 2 = 0 \implies (2r - 1)(r - 2) = 0$.
$\implies r = \frac{1}{2}$ or $r = 2$.
Since given $r < 1$, choose $\mathbf{r = \frac{1}{2}}$. [1 Mark]
Step 3 Determine the three terms:
$T_1 = \frac{a}{r} = \frac{3}{1/2} = \mathbf{6}$
$T_2 = a = \mathbf{3}$
$T_3 = ar = 3 \times \frac{1}{2} = \mathbf{\frac{3}{2}}$ (or $1.5$).
Answer: Common ratio $r = \frac{1}{2}$, terms are $6, 3, \frac{3}{2}$. [1 Mark]

Step 1 (a) Convert line to slope-intercept form $y = mx + c$:
$3x - 4y + 7 = 0 \implies 4y = 3x + 7 \implies y = \frac{3}{4}x + \frac{7}{4}$.
$\mathbf{\text{Slope } m_1 = \frac{3}{4}}$. [1 Mark]
Step 2 (b) Find perpendicular slope $m_2$:
$m_1 \times m_2 = -1 \implies m_2 = -\frac{1}{m_1} = -\frac{1}{3/4} = -\frac{4}{3}$. [1 Mark]
Step 3 Form perpendicular line equation with y-intercept $c = 2$:
$y = m_2 x + c \implies y = -\frac{4}{3}x + 2$
$3y = -4x + 6 \implies \mathbf{4x + 3y - 6 = 0}$. [1 Mark]

Step 1 (a) Construct $\Delta ABC$:
Draw base $AB = 5\text{ cm}$. At $B$, construct $\angle ABX = 60^\circ$ using compass arcs.
With $A$ as centre and radius $6.3\text{ cm}$, draw an arc cutting ray $BX$ at point $C$. Join $AC$ to complete $\Delta ABC$. [1.5 Marks]
Step 2 (b) Construct Circumcircle of $\Delta ABC$:
Draw the perpendicular bisectors of sides $AB$ and $AC$ (or $BC$).
Their point of intersection is the circumcentre $O$.
With $O$ as centre and radius $OA = OB = OC$, draw the circumcircle passing through all three vertices. [1.5 Marks]
Step 3 (c) Construct Locus of points equidistant from $AB$ and $BC$:
The locus of points equidistant from two intersecting lines $AB$ and $BC$ is the **internal angle bisector of $\angle ABC$**.
Construct the bisector of $\angle ABC = 60^\circ$. [1 Mark]