ICSE Class 10 Mathematics 2019 Solved Paper

Step 1 Split the inequation into two linear parts:
Part 1: $11x - 4 < 15x + 4$
$\implies -4 - 4 < 15x - 11x \implies -8 < 4x \implies x > -2$. [1 Mark]
Part 2: $15x + 4 \le 13x + 14$
$\implies 15x - 13x \le 14 - 4 \implies 2x \le 10 \implies x \le 5$. [1 Mark]
Step 2 Combine inequalities: $-2 < x \le 5$.
Since $x \in W$ (set of whole numbers $\{0, 1, 2, 3, \dots\}$):
Solution Set: $\{0, 1, 2, 3, 4, 5\}$.
Step 3 Graph the discrete elements with solid dark dots on the real number line at $0, 1, 2, 3, 4, 5$. [1 Mark]

Step 1 Find Market Value ($MV$) of each share:
Nominal Value ($NV$) = ₹ 100.
Discount = $10\%$ of ₹ 100 = ₹ 10.
$\therefore MV = 100 - 10 = \text{₹ } 90$. [1 Mark]
Step 2 (i) Number of shares purchased:
$\text{Number of shares} = \frac{\text{Total Investment}}{MV \text{ per share}} = \frac{4500}{90} = \mathbf{50\text{ shares}}$. [1 Mark]
Step 3 (ii) Annual income from shares:
$\text{Annual Income} = \text{Number of shares} \times \text{Dividend \%} \times NV$
$\text{Annual Income} = 50 \times \frac{7.5}{100} \times 100 = 50 \times 7.5 = \mathbf{\text{₹ } 375}$. [1 Mark]

Step 1 Prepare Cumulative Frequency ($cf$) Table:
Marks 1: $f=1, cf=1$
Marks 2: $f=2, cf=3$
Marks 3: $f=3, cf=6$
Marks 4: $f=3, cf=9$
Marks 5: $f=6, cf=15$
Marks 6: $f=10, cf=25$
Marks 7: $f=5, cf=30$
Marks 8: $f=4, cf=34$
Marks 9: $f=3, cf=37$
Marks 10: $f=3, cf=40 \implies N = 40$ (even). [1 Mark]
Step 2 (i) Calculate Median:
$\text{Median} = \frac{1}{2} \left[ \left(\frac{N}{2}\right)\text{th term} + \left(\frac{N}{2} + 1\right)\text{th term} \right] = \frac{20\text{th term} + 21\text{st term}}{2}$.
Looking at $cf$, both 20th and 21st terms fall in the group where $cf = 25$, corresponding to Marks = 6.
$\therefore \text{Median} = \frac{6 + 6}{2} = \mathbf{6\text{ marks}}$. [2 Marks]
Step 3 (ii) Calculate Mode:
Mode is the observation having the maximum frequency.
Here, maximum frequency is $f = 10$, which occurs for Marks = 6.
$\therefore \mathbf{\text{Mode} = 6\text{ marks}}$. [1 Mark]

Step 1 Let $f(x) = x^3 + x^2 - 4x - 4$.
To test whether $(x - 2)$ is a factor, evaluate $f(2)$:
$f(2) = (2)^3 + (2)^2 - 4(2) - 4 = 8 + 4 - 8 - 4 = 0$.
Since $f(2) = 0$, by the Factor Theorem, $(x - 2)$ is a factor of $f(x)$. [1 Mark]
Step 2 Factorise completely by grouping or dividing by $(x - 2)$:
$x^3 + x^2 - 4x - 4 = x^2(x + 1) - 4(x + 1) = (x + 1)(x^2 - 4)$. [1 Mark]
Step 3 Apply difference of squares $a^2 - b^2 = (a - b)(a + b)$:
$x^2 - 4 = (x - 2)(x + 2)$.
$\mathbf{\therefore x^3 + x^2 - 4x - 4 = (x - 2)(x + 2)(x + 1)}$. [1 Mark]

Step 1 Express all terms in terms of $\sin\theta$ and $\cos\theta$:
$\text{LHS} = \left(\frac{1}{\sin\theta} - \sin\theta\right) \left(\frac{1}{\cos\theta} - \cos\theta\right) \left(\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}\right)$. [1 Mark]
Step 2 Simplify each bracket:
$\text{Bracket 1} = \frac{1 - \sin^2\theta}{\sin\theta} = \frac{\cos^2\theta}{\sin\theta}$.
$\text{Bracket 2} = \frac{1 - \cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta}$.
$\text{Bracket 3} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}$. [1 Mark]
Step 3 Multiply the expressions:
$\text{LHS} = \left(\frac{\cos^2\theta}{\sin\theta}\right) \times \left(\frac{\sin^2\theta}{\cos\theta}\right) \times \left(\frac{1}{\sin\theta\cos\theta}\right) = \frac{\sin^2\theta\cos^2\theta}{\sin^2\theta\cos^2\theta} = \mathbf{1} = \text{RHS}$.
Hence Proved. [1 Mark]

Step 1 Set up linear equations in $a$ and $d$:
$T_4 = a + 3d = 8$   ...(i)
$T_6 = a + 5d = 14$   ...(ii) [1 Mark]
Step 2 Subtract (i) from (ii):
$(a + 5d) - (a + 3d) = 14 - 8 \implies 2d = 6 \implies \mathbf{d = 3}$.
$\therefore \text{Common difference } d = \mathbf{3}$. [1 Mark]
Step 3 Substitute $d = 3$ into (i):
$a + 3(3) = 8 \implies a + 9 = 8 \implies \mathbf{a = -1}$.
$\therefore \text{First term } a = \mathbf{-1}$. [1 Mark]
Step 4 Find sum of the first 20 terms ($S_{20}$):
$S_n = \frac{n}{2}[2a + (n - 1)d]$
$S_{20} = \frac{20}{2}[2(-1) + (20 - 1)(3)] = 10[-2 + 19(3)] = 10[-2 + 57] = 10(55) = \mathbf{550}$. [1 Mark]

Step 1 Multiply scalar elements into each matrix:
$\sin A \begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} = \begin{bmatrix} \sin^2 A & -\sin A\cos A \\ \sin A\cos A & \sin^2 A \end{bmatrix}$ [1 Mark]
$\cos A \begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix} = \begin{bmatrix} \cos^2 A & \sin A\cos A \\ -\sin A\cos A & \cos^2 A \end{bmatrix}$ [1 Mark]
Step 2 Add corresponding matrix entries:
$= \begin{bmatrix} \sin^2 A + \cos^2 A & -\sin A\cos A + \sin A\cos A \\ \sin A\cos A - \sin A\cos A & \sin^2 A + \cos^2 A \end{bmatrix}$
Using Pythagorean identity $\sin^2 A + \cos^2 A = 1$:
$= \mathbf{\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}}$ (or Identity matrix $I$). [1 Mark]

Step 1 Assign coordinates:
$M$ lies on X-axis $\implies M(x, 0)$.
$N$ lies on Y-axis $\implies N(0, y)$.
Ratio $m_1 : m_2 = 2 : 3$, and $P(3, 2)$ divides $MN$.
Step 2 Apply Section Formula:
$x_P = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \implies 3 = \frac{2(0) + 3(x)}{2 + 3} = \frac{3x}{5} \implies 3x = 15 \implies x = 5$.
$y_P = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \implies 2 = \frac{2(y) + 3(0)}{2 + 3} = \frac{2y}{5} \implies 2y = 10 \implies y = 5$.
$\mathbf{\therefore M(5, 0) \text{ and } N(0, 5)}$. [2 Marks]
Step 3 Calculate slope of line $MN$:
$\text{Slope } m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 0}{0 - 5} = \frac{5}{-5} = \mathbf{-1}$. [1 Mark]

Step 1 Equate volumes of metallic solids:
Radius of sphere $R = 6\text{ cm}$.
$\text{Volume of sphere} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi (216) = 288\pi\text{ cm}^3$. [1 Mark]
Step 2 (i) Determine radius of cylinder ($r$):
Height of cylinder $h = 32\text{ cm}$.
$\text{Volume of cylinder} = \pi r^2 h = \pi r^2 (32) = 32\pi r^2$.
Since the sphere is recast into the cylinder:
$32\pi r^2 = 288\pi \implies r^2 = \frac{288}{32} = 9 \implies r = \sqrt{9} = \mathbf{3\text{ cm}}$. [1 Mark]
Step 3 (ii) Calculate Curved Surface Area ($CSA$):
$CSA = 2\pi r h = 2 \times (3.1) \times 3 \times 32$
$CSA = 6 \times 3.1 \times 32 = 18.6 \times 32 = \mathbf{595.2\text{ cm}^2}$. [2 Marks]

Step 1 Set up proportional ratio definition:
$\frac{K + 3}{K + 2} = \frac{3K - 7}{2K - 3}$. [1 Mark]
Step 2 Cross-multiply and expand terms:
$(K + 3)(2K - 3) = (K + 2)(3K - 7)$
$2K^2 - 3K + 6K - 9 = 3K^2 - 7K + 6K - 14$
$2K^2 + 3K - 9 = 3K^2 - K - 14$. [1 Mark]
Step 3 Collect all terms into standard quadratic form:
$3K^2 - 2K^2 - K - 3K - 14 + 9 = 0$
$K^2 - 4K - 5 = 0$
$(K - 5)(K + 1) = 0 \implies \mathbf{K = 5 \text{ or } K = -1}$. [1 Mark]

Step 1 Identify coefficients and evaluate discriminant:
$a = 1, b = -4, c = -8$.
$D = b^2 - 4ac = (-4)^2 - 4(1)(-8) = 16 + 32 = 48$. [1 Mark]
Step 2 Apply quadratic formula:
$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-4) \pm \sqrt{48}}{2(1)} = \frac{4 \pm 4\sqrt{3}}{2} = 2 \pm 2\sqrt{3}$.
Using $\sqrt{3} \approx 1.73205 \implies 2\sqrt{3} \approx 3.4641$. [1 Mark]
Step 3 Calculate both roots to 3 significant figures:
$x_1 = 2 + 3.4641 = 5.4641 \approx \mathbf{5.46}$
$x_2 = 2 - 3.4641 = -1.4641 \approx \mathbf{-1.46}$
Answer: $x = 5.46$ or $-1.46$. [1 Mark]

Step 1 Draw circle of radius $r = 4\text{ cm}$ with centre $O$. [1 Mark]
Step 2 Mark point $P$ such that $OP = 7\text{ cm}$. Draw the perpendicular bisector of $OP$ to locate its midpoint $M$. [1 Mark]
Step 3 With $M$ as centre and radius $MO = MP$, draw an auxiliary circle (or semi-circles) intersecting the given circle at points $T_1$ and $T_2$. [1 Mark]
Step 4 Join $PT_1$ and $PT_2$. These are the required two tangents from $P$.
Theoretical verification:
In right $\Delta OT_1 P$: $PT_1 = \sqrt{OP^2 - OT_1^2} = \sqrt{7^2 - 4^2} = \sqrt{49 - 16} = \sqrt{33} \approx 5.74\text{ cm}$.
$\mathbf{\text{Measured length of tangent: } 5.7\text{ cm}}$ (acceptable range: $5.6\text{ cm}$ to $5.8\text{ cm}$). [1 Mark]

Step 1 Sample space $S = \{1, 2, 3, \dots, 25\} \implies n(S) = 25$.
(i) Event $A$ (odd numbers) = $\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25\} \implies n(A) = 13$.
$\mathbf{P(A) = \frac{13}{25}}$. [1 Mark]
Step 2 (ii) Event $B$ (divisible by 2 and 3 both $\implies$ divisible by $\text{LCM}(2, 3) = 6$):
$B = \{6, 12, 18, 24\} \implies n(B) = 4$.
$\mathbf{P(B) = \frac{4}{25}}$. [1 Mark]
Step 3 (iii) Event $C$ (numbers less than 16):
$C = \{1, 2, 3, \dots, 15\} \implies n(C) = 15$.
$\mathbf{P(C) = \frac{15}{25} = \frac{3}{5}}$. [1 Mark]

Step 1 State parameters and formula:
Number of months $n = 20$.
Rate of interest $r = 9\% \text{ p.a.}$, Interest $I = \text{₹ } 441$.
Formula: $I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$. [1 Mark]
Step 2 Substitute given values into equation:
$441 = P \times \frac{20(21)}{24} \times \frac{9}{100}$
$441 = P \times \frac{420}{24} \times \frac{9}{100} = P \times 17.5 \times 0.09 = P \times 1.575$. [1 Mark]
Step 3 Solve for monthly deposit $P$:
$P = \frac{441}{1.575} = \frac{441000}{1575} = \mathbf{\text{₹ } 280}$.
Answer: Rekha deposited ₹ 280 each month. [1 Mark]

Step 1 (i) Plot original vertices on Cartesian grid: $A(0, 5), B(3, 0), C(1, 0), D(1, -5)$. [1 Mark]
Step 2 (ii) & (iii) Reflection in the Y-axis has coordinate mapping $(x, y) \xrightarrow{M_y} (-x, y)$:
$B(3, 0) \xrightarrow{M_y} \mathbf{B'(-3, 0)}$
$C(1, 0) \xrightarrow{M_y} \mathbf{C'(-1, 0)}$
$D(1, -5) \xrightarrow{M_y} \mathbf{D'(-1, -5)}$. [2 Marks]
Step 3 (iv) Join points $A \to B \to C \to D \to D' \to C' \to B' \to A$ in order.
The resulting closed polygon has 7 sides and 7 vertices ($A, B, C, D, D', C', B'$).
Name of closed figure: Heptagon (also accepted: Arrowhead or concave heptagon). [1 Mark]

Step 1 (i) Prove $\Delta PQR \sim \Delta PST$:
In $\Delta PQR$ and $\Delta PST$:
$\angle P = \angle P$ (Common angle)
$\angle PQR = \angle PST = 90^\circ$ (Given)
$\therefore \mathbf{\Delta PQR \sim \Delta PST}$ by **AA (Angle-Angle) Similarity Criterion**. [1 Mark]
Step 2 (ii) Apply ratio of areas of similar triangles:
The ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides:
$\frac{\text{Area}(\Delta PQR)}{\text{Area}(\Delta PST)} = \left(\frac{PQ}{PS}\right)^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4}$. [1 Mark]
Step 3 Relate to quadrilateral $SRQT$:
Let $\text{Area}(\Delta PST) = 4k$, then $\text{Area}(\Delta PQR) = 25k$.
$\text{Area of quadrilateral } SRQT = \text{Area}(\Delta PQR) - \text{Area}(\Delta PST) = 25k - 4k = 21k$.
$\therefore \mathbf{\frac{\text{Area}(\Delta PQR)}{\text{Area}(SRQT)} = \frac{25k}{21k} = 25 : 21}$. [1 Mark]

Step 1 State parameters: $a = 3$, $l = T_n = 96$, $r = 2$.
Formula for $n\text{th}$ term: $T_n = a r^{n-1}$.
$96 = 3(2^{n-1}) \implies 2^{n-1} = \frac{96}{3} = 32$.
Since $32 = 2^5$: $n - 1 = 5 \implies \mathbf{n = 6}$.
$\therefore \text{Number of terms } n = \mathbf{6}$. [1.5 Marks]
Step 2 Find sum of the $n$ terms ($S_6$):
Formula: $S_n = \frac{a(r^n - 1)}{r - 1}$ for $r > 1$.
$S_6 = \frac{3(2^6 - 1)}{2 - 1} = \frac{3(64 - 1)}{1} = 3 \times 63 = \mathbf{189}$.
$\therefore \text{Sum of } 6 \text{ terms} = \mathbf{189}$. [1.5 Marks]

Step 1 Calculate individual geometric volumes in terms of $\pi$:
Volume of Cylinder: $V_{\text{cyl}} = \pi R^2 H = \pi (3)^2 (7) = 63\pi\text{ cm}^3$. [1 Mark]
Volume of Conical Hole: $V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3)^2 (3) = 9\pi\text{ cm}^3$.
Volume of Hemispherical Hole: $V_{\text{hemi}} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (3)^3 = \frac{2}{3}\pi(27) = 18\pi\text{ cm}^3$. [1 Mark]
Step 2 Subtract the scooped volumes from the cylinder:
$V_{\text{remaining}} = V_{\text{cyl}} - (V_{\text{cone}} + V_{\text{hemi}})$
$V_{\text{remaining}} = 63\pi - (9\pi + 18\pi) = 63\pi - 27\pi = 36\pi\text{ cm}^3$. [1 Mark]
Step 3 Evaluate numerically using $\pi = \frac{22}{7}$:
$V = 36 \times \frac{22}{7} = \frac{792}{7} = 113.1428\dots\text{ cm}^3$.
Rounding to nearest whole number: $\mathbf{V \approx 113\text{ cm}^3}$. [1 Mark]

Step 1 Find $\angle B$ using properties of circle and tangent:
$AC$ is tangent at $A$ and $OA$ is radius $\implies OA \perp AC \implies \angle OAC = 90^\circ$.
In $\Delta ABD$, angle in semicircle / tangent property gives $\angle A = 90^\circ$.
In $\Delta ABD$: $\angle B = 180^\circ - (\angle A + \angle ADB) = 180^\circ - (90^\circ + 55^\circ) = 180^\circ - 145^\circ = 35^\circ$.
Since $OB = OE$ (radii of circle), in $\Delta OBE$, $\angle OEB = \angle B = 35^\circ$. [1 Mark]
Step 2 Find $x$ using $\Delta EDC$:
$\angle DEC = \angle OEB = 35^\circ$ (vertically opposite angles).
Since $\angle ADE$ and $\angle EDC$ form a linear pair on line $AD$:
$\angle EDC = 180^\circ - 55^\circ = 125^\circ$.
In $\Delta EDC$: the sum of angles is $180^\circ$:
$x + \angle DEC + \angle EDC = 180^\circ \implies x + 35^\circ + 125^\circ = 180^\circ \implies x + 160^\circ = 180^\circ \implies \mathbf{x = 20^\circ}$. [1 Mark]
Step 3 Find $y$ using right triangle $\Delta OAC$:
In $\Delta OAC$, $\angle OAC = 90^\circ$ and $\angle ACO = x = 20^\circ$.
$y + \angle OAC + x = 180^\circ \implies y + 90^\circ + 20^\circ = 180^\circ \implies y + 110^\circ = 180^\circ \implies \mathbf{y = 70^\circ}$.
Answer: $x = 20^\circ$ and $y = 70^\circ$. [1 Mark]

Step 1 Linear scale factor $k = \frac{\text{Model Length}}{\text{Actual Length}} = \frac{1}{30}$.
(i) Find actual height of building:
$\text{Actual Height} = \text{Model Height} \times 30 = 80\text{ cm} \times 30 = 2400\text{ cm}$.
Converting to meters: $\frac{2400}{100} = \mathbf{24\text{ m}}$. [1.5 Marks]
Step 2 (ii) Volume scaling factor $k^3 = \left(\frac{1}{30}\right)^3 = \frac{1}{27000}$.
$\text{Volume of model tank} = \text{Actual Volume} \times k^3 = 27\text{ m}^3 \times \frac{1}{27000} = \frac{1}{1000}\text{ m}^3 = \mathbf{0.001\text{ m}^3}$.
In $\text{cm}^3$: $0.001 \times 10^6\text{ cm}^3 = \mathbf{1000\text{ cm}^3}$ (or $1\text{ litre}$). [1.5 Marks]

Step 1 (i) Determine order of $M$:
Let $A = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$, which has order $2 \times 2$.
RHS $= 6I = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$, which has order $2 \times 2$.
For product $A_{2 \times 2} \times M_{r \times c} = (6I)_{2 \times 2}$ to be defined:
Number of rows of $M$ must equal columns of $A \implies r = 2$.
Result has order $2 \times c = 2 \times 2 \implies c = 2$.
$\mathbf{\therefore \text{Order of matrix } M = 2 \times 2}$. [1 Mark]
Step 2 (ii) Let $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$:
$\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 4a + 2c & 4b + 2d \\ -a + c & -b + d \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$. [1 Mark]
Step 3 Solve simultaneous linear systems:
From column 1:
$-a + c = 0 \implies c = a$.
$4a + 2c = 6 \implies 4a + 2a = 6 \implies 6a = 6 \implies a = 1 \implies c = 1$.
From column 2:
$4b + 2d = 0 \implies 2d = -4b \implies d = -2b$.
$-b + d = 6 \implies -b + (-2b) = 6 \implies -3b = 6 \implies b = -2 \implies d = 4$. [1 Mark]
Step 4 Form matrix $M$:
$\mathbf{M = \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}}$. [1 Mark]

Step 1 Let the three consecutive terms of the A.P. be $(a - d)$, $a$, and $(a + d)$.
Sum of the three terms $= (a - d) + a + (a + d) = 3a = 42 \implies \mathbf{a = 14}$. [1 Mark]
Step 2 Product of the first and third terms $= 52$:
$(a - d)(a + d) = a^2 - d^2 = 52$.
Substitute $a = 14$:
$14^2 - d^2 = 52 \implies 196 - d^2 = 52 \implies d^2 = 196 - 52 = 144 \implies \mathbf{d = \pm 12}$. [1 Mark]
Step 3 Determine the terms and solutions:
Case 1 ($d = 12$):
First term $T_1 = a - d = 14 - 12 = \mathbf{2}$.
Common difference $d = \mathbf{12}$. (Terms: $2, 14, 26$)
Case 2 ($d = -12$):
First term $T_1 = a - d = 14 - (-12) = \mathbf{26}$.
Common difference $d = \mathbf{-12}$. (Terms: $26, 14, 2$)
Both solutions are completely valid. [1 Mark]

Step 1 (i) Slope of line segment $BC$ joining $B(-1, 2)$ and $C(6, -6)$:
$m_{BC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-6 - 2}{6 - (-1)} = \frac{-8}{7} = \mathbf{-\frac{8}{7}}$. [1 Mark]
Step 2 (ii) Slope of perpendicular line ($m_\perp$):
Since $m_1 \times m_2 = -1 \implies m_\perp = -\frac{1}{m_{BC}} = -\frac{1}{-8/7} = \mathbf{\frac{7}{8}}$. [1 Mark]
Step 3 Equation of line passing through $A(3, 8)$ with slope $m = \frac{7}{8}$:
$y - y_1 = m(x - x_1) \implies y - 8 = \frac{7}{8}(x - 3)$
$8(y - 8) = 7(x - 3) \implies 8y - 64 = 7x - 21$
$\mathbf{7x - 8y + 43 = 0}$ (or $y = \frac{7}{8}x + \frac{43}{8}$). [1 Mark]

Step 1 Draw line segment $BC = 7\text{ cm}$. Bisect $BC$ to find centre $O$, then construct semicircle with radius $OB = 3.5\text{ cm}$. [1 Mark]
Step 2 Locate point $A$ equidistant from $B$ and $C$:
Locus of points equidistant from two fixed points $B$ and $C$ is the **perpendicular bisector of $BC$**.
The perpendicular bisector meets the circumference at point $A$.
(Here, $\Delta ABC$ is isosceles right-angled at $A$ with $\angle ABC = \angle ACB = 45^\circ$). [1 Mark]
Step 3 Locate point $D$ equidistant from $AB$ and $BC$:
Locus of points equidistant from two intersecting lines $AB$ and $BC$ is the **internal angle bisector of $\angle ABC$**.
Construct the bisector of $\angle ABC = 45^\circ$ to intersect the semicircle at point $D$. Join $CD$ and $AD$ to complete cyclic quadrilateral $ABCD$. [1 Mark]
Step 4 Measure $\angle ADC$:
Since $ABCD$ is a cyclic quadrilateral, opposite angles are supplementary:
$\angle ABC + \angle ADC = 180^\circ \implies 45^\circ + \angle ADC = 180^\circ \implies \mathbf{\angle ADC = 135^\circ}$. [1 Mark]

Step 1 Prepare Assumed Mean Table with $A = 45$:
[1.5 Marks]
Step 2 Apply Assumed Mean Formula:
$\text{Mean } \bar{x} = A + \frac{\sum f_i d_i}{\sum f_i} = 45 + \frac{-140}{30} = 45 - 4.6667 = \mathbf{40.33}$.
Average number of patients = 40.33. [1.5 Marks]

Step 1 Write RHS as fraction $\frac{4}{1}$ and apply Componendo and Dividendo:
$\frac{(\sqrt{5x} + \sqrt{2x - 6}) + (\sqrt{5x} - \sqrt{2x - 6})}{(\sqrt{5x} + \sqrt{2x - 6}) - (\sqrt{5x} - \sqrt{2x - 6})} = \frac{4 + 1}{4 - 1}$. [1 Mark]
Step 2 Simplify both sides:
$\frac{2\sqrt{5x}}{2\sqrt{2x - 6}} = \frac{5}{3} \implies \frac{\sqrt{5x}}{\sqrt{2x - 6}} = \frac{5}{3}$. [1 Mark]
Step 3 Square both sides and solve for $x$:
$\frac{5x}{2x - 6} = \frac{25}{9} \implies \frac{x}{2x - 6} = \frac{5}{9}$
$9x = 5(2x - 6) \implies 9x = 10x - 30 \implies 10x - 9x = 30 \implies \mathbf{x = 30}$.
Answer: $x = 30$. [1 Mark]

Step 1 Initial Investment Analysis:
Investment = ₹ 8500, $NV_1 = \text{₹ } 100, MV_1 = \text{₹ } 170, D_1 = 10\%$.
$\text{Number of shares } n_1 = \frac{8500}{170} = 50\text{ shares}$.
$\text{Initial Annual Income } I_1 = 50 \times 10\% \times 100 = \text{₹ } 500$. [1 Mark]
Step 2 (i) Determine sale proceeds:
Price rises by ₹ 30 $\implies \text{New Selling Price} = 170 + 30 = \text{₹ } 200$.
$\mathbf{\text{Sale Proceeds} = 50 \times 200 = \text{₹ } 10,000}$. [1 Mark]
Step 3 (ii) Second investment in ₹ 125 shares:
Proceeds available = ₹ 10,000, $NV_2 = \text{₹ } 100, MV_2 = \text{₹ } 125, D_2 = 12\%$.
$\mathbf{\text{Number of new shares bought } n_2 = \frac{10000}{125} = 80\text{ shares}}$. [1 Mark]
Step 4 (iii) Change in annual income:
$\text{New Annual Income } I_2 = 80 \times 12\% \times 100 = \text{₹ } 960$.
$\text{Change in Annual Income} = I_2 - I_1 = 960 - 500 = \mathbf{\text{₹ } 460\text{ increase}}$. [1 Mark]

Step 1 Cumulative Frequency Table ($N = 120$):
Upper Limits & Cumulative Frequencies:
$(10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117), (100, 120)$.
Plot points on graph paper and join with a smooth, continuous freehand S-curve. [2 Marks]
Step 2 (i) Determine Median Marks:
$\text{Median position} = \frac{N}{2} = \frac{120}{2} = 60\text{th term}$.
Draw a horizontal line from $y = 60$ on the vertical axis to meet the curve, then drop a vertical perpendicular to the x-axis.
$\mathbf{\text{Median Marks} \approx 43\text{ marks}}$ (accepted range: $42 - 44$). [1.5 Marks]
Step 3 (ii) Number of students who did not pass ($< 50\text{ marks}$):
Pass mark $= 50$. At $x = 50$ on the horizontal axis, draw a vertical line up to meet the ogive.
From that intersection point, project horizontally to the y-axis, which corresponds to $y = 78$.
$\mathbf{\text{Number of students who failed} = 78\text{ students}}$. [1.5 Marks]
Step 4 (iii) Determine Upper Quartile ($Q_3$):
$Q_3\text{ position} = \frac{3N}{4} = \frac{3 \times 120}{4} = 90\text{th term}$.
From $y = 90$ on the y-axis, draw a horizontal line to meet the curve, then drop a vertical line to the x-axis.
$\mathbf{Q_3 \approx 57\text{ marks}}$ (accepted range: $56 - 58$). [1 Mark]

Step 1 Let $AB = h$ be the height of the vertical tower and $B$ be its base.
Let observer start at $C$ ($\angle ACB = 45^\circ$) and walk $20\text{ m}$ to $D$ ($\angle ADB = 60^\circ$).
Let $DB = x$, then $CB = x + 20$. [1 Mark]
Step 2 In right triangle $\Delta ABC$:
$\tan 45^\circ = \frac{AB}{CB} \implies 1 = \frac{h}{x + 20} \implies x + 20 = h \implies x = h - 20$. [1 Mark]
Step 3 In right triangle $\Delta ABD$:
$\tan 60^\circ = \frac{AB}{DB} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3}$.
Substitute $x = h - 20$:
$h = (h - 20)\sqrt{3} \implies h = h\sqrt{3} - 20\sqrt{3} \implies h(\sqrt{3} - 1) = 20\sqrt{3}$.
$h = \frac{20\sqrt{3}}{\sqrt{3} - 1} = \frac{20\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = 10\sqrt{3}(\sqrt{3} + 1) = 10(3 + \sqrt{3})$. [1 Mark]
Step 4 Evaluate numerically with $\sqrt{3} \approx 1.732$:
$h = 10(3 + 1.732) = 10(4.732) = 47.32\text{ m}$.
Rounding to 2 significant figures: $\mathbf{h \approx 47\text{ m}}$.
Height of tower = 47 m. [1 Mark]

Step 1 Let $f(x) = x^3 + (kx + 8)x + k = x^3 + kx^2 + 8x + k$.
By the Remainder Theorem:
When divided by $(x + 1)$, remainder $R_1 = f(-1)$:
$R_1 = (-1)^3 + k(-1)^2 + 8(-1) + k = -1 + k - 8 + k = \mathbf{2k - 9}$. [1 Mark]
Step 2 When divided by $(x - 2)$, remainder $R_2 = f(2)$:
$R_2 = (2)^3 + k(2)^2 + 8(2) + k = 8 + 4k + 16 + k = \mathbf{5k + 24}$. [1 Mark]
Step 3 Use condition $R_1 + R_2 = 1$:
$(2k - 9) + (5k + 24) = 1$
$7k + 15 = 1 \implies 7k = 1 - 15 = -14 \implies \mathbf{k = -2}$.
Answer: $k = -2$. [1 Mark]

Step 1 Let the two consecutive natural numbers which are multiples of 3 be $3x$ and $3x + 3 = 3(x + 1)$, where $x \in \mathbb{N}$. [1 Mark]
Step 2 Set up quadratic equation from their product:
$(3x)(3x + 3) = 810$
$9x(x + 1) = 810 \implies x(x + 1) = \frac{810}{9} = 90$
$x^2 + x - 90 = 0$. [1 Mark]
Step 3 Factorise the quadratic equation:
$x^2 + 10x - 9x - 90 = 0 \implies (x + 10)(x - 9) = 0$
$\implies x = 9$ or $x = -10$.
Since $x$ must be a natural number ($x \in \mathbb{N}$), reject $x = -10$.
$\therefore x = 9$.
First number $= 3(9) = \mathbf{27}$.
Second number $= 27 + 3 = \mathbf{30}$.
Answer: The two numbers are 27 and 30. [1 Mark]

Step 1 (i) Find $\angle ACB$:
Since $AC$ is a diameter, the angle subtended by it in the semicircle is a right angle:
$\angle ABC = 90^\circ$ (Angle in a semi-circle).
In $\Delta ABC$: $\angle BAC + \angle ABC + \angle ACB = 180^\circ$
$50^\circ + 90^\circ + \angle ACB = 180^\circ \implies \mathbf{\angle ACB = 40^\circ}$. [1 Mark]
Step 2 (ii) Find $\angle EDC$:
Given $BC \parallel AE$ with transversal $AC$:
$\angle EAC = \angle ACB = 40^\circ$ (Alternate interior angles).
Since vertices $A, C, D, E$ lie on the circle, $ACDE$ is a cyclic quadrilateral.
Opposite angles of a cyclic quadrilateral are supplementary:
$\angle EAC + \angle EDC = 180^\circ \implies 40^\circ + \angle EDC = 180^\circ \implies \mathbf{\angle EDC = 140^\circ}$. [1 Mark]
Step 3 (iii) Find $\angle BEC$:
Angles subtended by the same chord $BC$ at the circumference are equal:
$\mathbf{\angle BEC = \angle BAC = 50^\circ}$ (Angles in the same segment of a circle). [1 Mark]
Step 4 Prove that $BE$ is also a diameter:
Consider $\angle BAE = \angle BAC + \angle EAC = 50^\circ + 40^\circ = 90^\circ$.
Since chord $BE$ subtends a right angle ($90^\circ$) on the circumference of the circle, by the converse of the semicircle angle theorem, **$BE$ must be a diameter of the circle**.
(Alternatively, in $\Delta BCE$, $\angle BCE = \angle BCA + \angle ACE = 40^\circ + 50^\circ = 90^\circ$, proving $BE$ is a diameter).
Hence Proved. [1 Mark]