ICSE Class 10 Mathematics 2018 Solved Question Paper

Step 1 Multiply scalar 2 into the first matrix:
$$\begin{bmatrix} 2x & 14 \\ 18 & 2y - 10 \end{bmatrix} + \begin{bmatrix} 6 & -7 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}$$ [1 Mark]
Step 2 Add corresponding elements on the LHS:
$$\begin{bmatrix} 2x + 6 & 14 - 7 \\ 18 + 4 & 2y - 10 + 5 \end{bmatrix} = \begin{bmatrix} 2x + 6 & 7 \\ 22 & 2y - 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}$$
Step 3 Equate corresponding elements:
1) $2x + 6 = 10 \implies 2x = 4 \implies \mathbf{x = 2}$ [1 Mark]
2) $2y - 5 = 15 \implies 2y = 20 \implies \mathbf{y = 10}$ [1 Mark]
Answer: $x = 2$, $y = 10$.

Step 1 Given parameters:
Monthly deposit $P = ₹600$
Time $n = 2\frac{1}{2}\text{ years} = 2.5 \times 12 = 30\text{ months}$
Rate of interest $R = 10\%$ p.a.
Step 2 Calculate total simple interest earned:
$$I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{R}{100}$$ $$I = 600 \times \frac{30 \times 31}{24} \times \frac{10}{100} = 600 \times \frac{930}{24} \times \frac{1}{10} = 25 \times 93 = ₹2,325$$ [1.5 Marks]
Step 3 Total sum deposited and maturity value:
Total principal deposited $= P \times n = 600 \times 30 = ₹18,000$ [0.5 Mark]
$$\text{Maturity Value } (MV) = (P \times n) + I = 18,000 + 2,325 = \mathbf{₹20,325}$$ [1 Mark]
Answer: The maturity value of the account is ₹20,325.

Sample Space $S = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\} \implies n(S) = 10$.
(i) Prime number: The only prime in the set is $2$.
Favourable outcomes: $\{2\} \implies n(E_1) = 1$.
$$P(\text{prime}) = \frac{n(E_1)}{n(S)} = \mathbf{\frac{1}{10}}$$ [1 Mark]
(ii) Number divisible by 4:
Favourable outcomes: $\{4, 8, 12, 16, 20\} \implies n(E_2) = 5$.
$$P(\text{divisible by 4}) = \frac{5}{10} = \mathbf{\frac{1}{2}}$$ [1 Mark]
(iii) Number that is a multiple of 6:
Favourable outcomes: $\{6, 12, 18\} \implies n(E_3) = 3$.
$$P(\text{multiple of 6}) = \mathbf{\frac{3}{10}}$$ [1 Mark]
(iv) An odd number:
All numbers in the bag are even $\{2, 4, \dots, 20\}$, so there are no odd numbers.
Favourable outcomes: $\emptyset \implies n(E_4) = 0$.
$$P(\text{odd number}) = \frac{0}{10} = \mathbf{0}$$ [1 Mark]
Answers: (i) $\frac{1}{10}$, (ii) $\frac{1}{2}$, (iii) $\frac{3}{10}$, (iv) $0$.

Step 1 Base circumference $= 2\pi r = 132\text{ cm}$:
$$2 \times \frac{22}{7} \times r = 132 \implies \frac{44}{7} r = 132$$ $$r = \frac{132 \times 7}{44} = 3 \times 7 = \mathbf{21\text{ cm}}$$ [1.5 Marks]
Step 2 Volume of cylinder $= \pi r^2 h$ with $r = 21\text{ cm}, h = 25\text{ cm}$:
$$V = \frac{22}{7} \times 21 \times 21 \times 25 = 22 \times 3 \times 21 \times 25$$ $$V = 66 \times 525 = \mathbf{34,650\text{ cm}^3}$$ [1.5 Marks]
Answers: (i) Radius $= 21\text{ cm}$, (ii) Volume $= 34,650\text{ cm}^3$.

Step 1 Condition for consecutive terms $a, b, c$ in A.P.:
$$b - a = c - b \iff 2b = a + c$$ [1 Mark]
Step 2 Substitute the given expressions:
$$2(2k + 1) = (k - 3) + (4k + 3)$$ $$4k + 2 = 5k$$ [1 Mark]
Step 3 Solve for $k$:
$$5k - 4k = 2 \implies \mathbf{k = 2}$$ [1 Mark]
Check: Terms are $2 - 3 = -1$, $2(2) + 1 = 5$, and $4(2) + 3 = 11$. Common difference $d = 6$.
Answer: $k = 2$.

(i) Find $\angle QRS$ Opposite angles of a cyclic quadrilateral are supplementary:
$$\angle QPS + \angle QRS = 180^\circ \implies 73^\circ + \angle QRS = 180^\circ$$ $$\angle QRS = 180^\circ - 73^\circ = \mathbf{107^\circ}$$ [1 Mark]
(ii) Find $\angle RQS$ In $\triangle PQS$:
$$\angle PSQ = 180^\circ - (\angle QPS + \angle PQS) = 180^\circ - (73^\circ + 55^\circ) = 180^\circ - 128^\circ = 52^\circ$$ Given $\angle PSR = 82^\circ$, we have:
$$\angle QSR = \angle PSR - \angle PSQ = 82^\circ - 52^\circ = 30^\circ$$ Now in $\triangle QRS$, the angle sum is $180^\circ$:
$$\angle RQS + \angle QSR + \angle QRS = 180^\circ \implies \angle RQS + 30^\circ + 107^\circ = 180^\circ$$ $$\angle RQS = 180^\circ - 137^\circ = \mathbf{43^\circ}$$ [1.5 Marks]
(iii) Find $\angle PRQ$ Angles subtended by the same chord $PQ$ in the same segment of the circle are equal:
$$\angle PRQ = \angle PSQ = \mathbf{52^\circ}$$ [1.5 Marks]
Answers: (i) $\angle QRS = 107^\circ$, (ii) $\angle RQS = 43^\circ$, (iii) $\angle PRQ = 52^\circ$.

Step 1 Let $f(x) = x^3 + ax + b$.
Since $(x + 2)$ is a factor, $f(-2) = 0$:
$$(-2)^3 + a(-2) + b = 0 \implies -8 - 2a + b = 0 \implies -2a + b = 8 \quad \text{--- (1)}$$ [1 Mark]
Step 2 Since $(x + 3)$ is a factor, $f(-3) = 0$:
$$(-3)^3 + a(-3) + b = 0 \implies -27 - 3a + b = 0 \implies -3a + b = 27 \quad \text{--- (2)}$$ [1 Mark]
Step 3 Subtract (2) from (1):
$$(-2a + b) - (-3a + b) = 8 - 27 \implies a = -19$$ Substitute $a = -19$ into (1):
$$-2(-19) + b = 8 \implies 38 + b = 8 \implies b = 8 - 38 = -30$$ [1 Mark]
Answer: $a = -19$, $b = -30$.

Method 1 Using reciprocal relations:
$$\text{LHS} = \sqrt{\sec^2 \theta + \csc^2 \theta} = \sqrt{\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}}$$ [1 Mark]
$$\text{LHS} = \sqrt{\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}} = \sqrt{\frac{1}{\sin^2 \theta \cos^2 \theta}} = \frac{1}{\sin \theta \cos \theta}$$ [1 Mark]
Express 1 in numerator as $\sin^2 \theta + \cos^2 \theta$:
$$\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{\sin^2 \theta}{\sin \theta \cos \theta} + \frac{\cos^2 \theta}{\sin \theta \cos \theta} = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \mathbf{\tan \theta + \cot \theta} = \text{RHS}$$ [1 Mark]
Hence Proved.

Step 1 Modal class identification:
The class with the highest frequency ($18$) is $4000 - 5000$. [1 Mark]
Step 2 Drawing the Histogram:
Plot class intervals on the x-axis and frequencies on the y-axis, drawing rectangles of corresponding heights. [2 Marks]
Step 3 Graphical Mode Estimation:
Inside the modal rectangle ($4000 - 5000$), connect the upper left corner to the top left of the adjacent right bar, and the upper right corner to the top right of the adjacent left bar. Drop a perpendicular from their intersection to the x-axis.
$$\text{Estimated Mode} = \mathbf{4600\text{ runs}} \quad (\text{acceptable range: } 4550 - 4650)$$ [1 Mark]
Answer: Estimated Mode $\approx 4600$ runs.

Part 1 First inequality:
$$-2 + 10x \le 13x + 10 \implies -2 - 10 \le 13x - 10x$$ $$-12 \le 3x \implies x \ge -4$$ [1 Mark]
Part 2 Second inequality:
$$13x + 10 < 24 + 10x \implies 13x - 10x < 24 - 10$$ $$3x < 14 \implies x < \frac{14}{3} \approx 4.67$$ [1 Mark]
Solution Set Combining both: $-4 \le x < \frac{14}{3}$.
Since $x \in \mathbb{Z}$ (integers):
$$\text{Solution Set} = \mathbf{\{-4, -3, -2, -1, 0, 1, 2, 3, 4\}}$$ [0.5 Mark]
Number line Represented by solid darkened dots on integers from $-4$ to $4$ inclusive. [0.5 Mark]
Answer: Solution Set $= \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$.

Step 1 Find slope of line 1 ($3x - 5y = 7$):
$$5y = 3x - 7 \implies y = \frac{3}{5}x - \frac{7}{5} \implies m_1 = \frac{3}{5}$$ [1 Mark]
Step 2 Find slope of line 2 ($4x + ay + 9 = 0$):
$$ay = -4x - 9 \implies y = -\frac{4}{a}x - \frac{9}{a} \implies m_2 = -\frac{4}{a}$$ [1 Mark]
Step 3 Apply perpendicularity condition $m_1 \times m_2 = -1$:
$$\left(\frac{3}{5}\right) \times \left(-\frac{4}{a}\right) = -1 \implies -\frac{12}{5a} = -1$$ $$5a = 12 \implies \mathbf{a = \frac{12}{5} = 2.4}$$ [1 Mark]
Answer: $a = \frac{12}{5}$ (or $2.4$).

Step 1 Standard quadratic form: $x^2 + 7x - 7 = 0$.
Here $a = 1, b = 7, c = -7$. [1 Mark]
Step 2 Quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{7^2 - 4(1)(-7)}}{2(1)}$$ $$x = \frac{-7 \pm \sqrt{49 + 28}}{2} = \frac{-7 \pm \sqrt{77}}{2}$$ [1.5 Marks]
Step 3 Calculate decimal roots with $\sqrt{77} \approx 8.775$:
$$x_1 = \frac{-7 + 8.775}{2} = \frac{1.775}{2} = 0.8875 \approx \mathbf{0.89}$$ [0.75 Mark]
$$x_2 = \frac{-7 - 8.775}{2} = \frac{-15.775}{2} = -7.8875 \approx \mathbf{-7.89}$$ [0.75 Mark]
Answer: $x = 0.89$ or $x = -7.89$.

Step 1 General term $T_n = a r^{n-1}$:
$$T_4 = a r^3 = 16 \quad \text{--- (1)}$$ [1 Mark]
$$T_7 = a r^6 = 128 \quad \text{--- (2)}$$ [1 Mark]
Step 2 Divide (2) by (1):
$$\frac{a r^6}{a r^3} = \frac{128}{16} \implies r^3 = 8 \implies \mathbf{r = 2}$$ Step 3 Substitute $r = 2$ in (1):
$$a(2^3) = 16 \implies 8a = 16 \implies \mathbf{a = 2}$$ [1 Mark]
Answer: First term $a = 2$, Common ratio $r = 2$.

Step 1 Nominal value $NV = ₹50$.
Discount $= 10\% \text{ of } 50 = ₹5$.
Market Value per share $MV = 50 - 5 = ₹45$.
(i) Number of shares purchased:
$$n = \frac{\text{Total Investment}}{MV} = \frac{22,500}{45} = \mathbf{500\text{ shares}}$$ [1 Mark]
(ii) Annual dividend received:
$$\text{Dividend per share} = 12\% \text{ of } 50 = \frac{12}{100} \times 50 = ₹6$$ $$\text{Total Annual Dividend} = 500 \times 6 = \mathbf{₹3,000}$$ [1 Mark]
(iii) Rate of return on investment:
$$\text{Rate of return} = \frac{\text{Annual Dividend}}{\text{Total Investment}} \times 100 = \frac{3000}{22500} \times 100 = \frac{40}{3}\% = 13.33\% \approx \mathbf{13\%}$$ [1 Mark]
Answers: (i) 500 shares, (ii) ₹3,000, (iii) $13\%$.

(i) Plotting Reflection in the y-axis maps $(x, y) \to (-x, y)$. Plot vertices $A(2,2), B(2,-2), C(0,-1), D(0,1)$ and connect to form $ABCD$. [1 Mark]
(ii) Coordinates
$$A'( -2, 2 ) \quad \text{and} \quad B'( -2, -2 )$$ [1 Mark]
(iii) Invariant Points Points that lie on the axis of reflection ($y$-axis, where $x = 0$) remain unchanged under reflection.
Hence, $C(0, -1)$ and $D(0, 1)$ are invariant points. [1 Mark]
(iv) Name of Polygon In $A'B'CD$:
$A'B'$ is vertical (from $(-2, -2)$ to $(-2, 2)$, length 4 units) and $DC$ is vertical along the y-axis (from $(0, -1)$ to $(0, 1)$, length 2 units).
Since $A'B' \parallel DC$, and non-parallel sides $A'D = B'C = \sqrt{2^2 + 1^2} = \sqrt{5}$, polygon $A'B'CD$ is an Isosceles Trapezium (or Trapezium). [1 Mark]
Answers: (ii) $A'(-2,2), B'(-2,-2)$, (iii) $C(0,-1)$ and $D(0,1)$, (iv) Isosceles Trapezium.

Step 1 Express as ratio and apply Componendo & Dividendo:
$$\frac{2x + \sqrt{4x^2 - 1}}{2x - \sqrt{4x^2 - 1}} = \frac{4}{1}$$ $$\frac{(2x + \sqrt{4x^2 - 1}) + (2x - \sqrt{4x^2 - 1})}{(2x + \sqrt{4x^2 - 1}) - (2x - \sqrt{4x^2 - 1})} = \frac{4 + 1}{4 - 1}$$ [1 Mark]
Step 2 Simplify LHS and RHS:
$$\frac{4x}{2\sqrt{4x^2 - 1}} = \frac{5}{3} \implies \frac{2x}{\sqrt{4x^2 - 1}} = \frac{5}{3}$$ [1 Mark]
Step 3 Square both sides:
$$\frac{4x^2}{4x^2 - 1} = \frac{25}{9} \implies 36x^2 = 25(4x^2 - 1) = 100x^2 - 25$$ $$100x^2 - 36x^2 = 25 \implies 64x^2 = 25 \implies x^2 = \frac{25}{64}$$ Since $x$ is positive: $x = \sqrt{\frac{25}{64}} = \mathbf{\frac{5}{8}}$ [1 Mark]
Answer: $x = \frac{5}{8}$.

Step 1: Compute AC
$$AC = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 2(1)+3(-1) & 2(0)+3(4) \\ 5(1)+7(-1) & 5(0)+7(4) \end{bmatrix} = \begin{bmatrix} -1 & 12 \\ -2 & 28 \end{bmatrix}$$ [1 Mark]
Step 2: Compute B²
$$B^2 = \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} = \begin{bmatrix} 0-4 & 0+28 \\ 0-7 & -4+49 \end{bmatrix} = \begin{bmatrix} -4 & 28 \\ -7 & 45 \end{bmatrix}$$ [1 Mark]
Step 3: Compute AC + B² - 10C
$$10C = \begin{bmatrix} 10 & 0 \\ -10 & 40 \end{bmatrix}$$ $$AC + B^2 - 10C = \begin{bmatrix} -1 - 4 - 10 & 12 + 28 - 0 \\ -2 - 7 - (-10) & 28 + 45 - 40 \end{bmatrix} = \mathbf{\begin{bmatrix} -15 & 40 \\ 1 & 33 \end{bmatrix}}$$ [1 Mark]
Answer: $\begin{bmatrix} -15 & 40 \\ 1 & 33 \end{bmatrix}$.

Step 1 Convert all terms into $\sin \theta$ and $\cos \theta$:
$$\text{LHS} = \left(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right) \left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right)$$ [1 Mark]
Step 2 Take common denominators in each bracket:
$$\text{LHS} = \left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right) \left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right)$$ [1 Mark]
Step 3 Group $(\sin \theta + \cos \theta)$ as $a$ and $1$ as $b$ using $(a - b)(a + b) = a^2 - b^2$:
$$\text{Numerator} = (\sin \theta + \cos \theta)^2 - 1^2 = \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta - 1$$ [1 Mark]
Since $\sin^2 \theta + \cos^2 \theta = 1$:
$$\text{Numerator} = 1 + 2\sin \theta \cos \theta - 1 = 2\sin \theta \cos \theta$$ $$\text{LHS} = \frac{2\sin \theta \cos \theta}{\sin \theta \cos \theta} = \mathbf{2} = \text{RHS}$$ [1 Mark]
Hence Proved.

Step 1 For equal roots, discriminant $D = b^2 - 4ac = 0$.
Here $a = 1, b = 4k, c = k^2 - k + 2$. [1 Mark]
Step 2 Form the equation in $k$:
$$(4k)^2 - 4(1)(k^2 - k + 2) = 0$$ $$16k^2 - 4k^2 + 4k - 8 = 0 \implies 12k^2 + 4k - 8 = 0$$ [1 Mark]
Step 3 Divide by 4 and factorise:
$$3k^2 + k - 2 = 0 \implies 3k^2 + 3k - 2k - 2 = 0$$ $$3k(k + 1) - 2(k + 1) = 0 \implies (3k - 2)(k + 1) = 0$$ $$\mathbf{k = \frac{2}{3}} \quad \text{or} \quad \mathbf{k = -1}$$ [1 Mark]
Answer: $k = \frac{2}{3}$ or $k = -1$.

Scale Factor $k = 50,000$. Hence $1\text{ cm on map} = 50,000\text{ cm} = 0.5\text{ km}$ in reality.
(i) Diagonal AC In right $\triangle ABC$ on the map:
$$AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\text{ cm}$$ [0.5 Mark]
$$\text{Actual length of } AC = 10\text{ cm} \times 0.5\text{ km/cm} = \mathbf{5\text{ km}}$$ [1 Mark]
(ii) Actual Area
$$\text{Area on map} = 6\text{ cm} \times 8\text{ cm} = 48\text{ cm}^2$$ [0.5 Mark]
$$\text{Actual area} = 48 \times (0.5\text{ km})^2 = 48 \times 0.25 = \mathbf{12\text{ sq km}}$$ [1 Mark]
Alternative: Actual $AB = 3\text{ km}$, Actual $BC = 4\text{ km} \implies \text{Area} = 3 \times 4 = 12\text{ km}^2$.
Answers: (i) Diagonal $AC = 5\text{ km}$, (ii) Actual area $= 12\text{ sq km}$.

Step 1 Coordinates of $M$ dividing $AB$ in ratio $1 : 2$ ($m_1=1, m_2=2$):
$$x_M = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{1(-1) + 2(2)}{1 + 2} = \frac{-1 + 4}{3} = \frac{3}{3} = 1$$ $$y_M = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{1(2) + 2(5)}{1 + 2} = \frac{2 + 10}{3} = \frac{12}{3} = 4$$ $$\mathbf{M = (1, 4)}$$ [2 Marks]
Step 2 Slope of line passing through $C(5, 8)$ and $M(1, 4)$:
$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 4}{5 - 1} = \frac{4}{4} = 1$$ [1 Mark]
Step 3 Equation of line $CM$ using point $M(1, 4)$ and $m = 1$:
$$y - 4 = 1(x - 1) \implies y - 4 = x - 1$$ $$\mathbf{x - y + 3 = 0} \quad (\text{or } y = x + 3)$$ [1 Mark]
Answers: Coordinates of $M = (1, 4)$; Equation of line $CM$: $x - y + 3 = 0$.

Step 1 Let the original number of children be $x$.
Original share of each child $= \frac{7500}{x}$.
New share of each child with $(x - 20)$ children $= \frac{7500}{x - 20}$. [1 Mark]
Step 2 Form the quadratic equation:
$$\frac{7500}{x - 20} - \frac{7500}{x} = 100$$ Divide by 100: $\frac{75}{x - 20} - \frac{75}{x} = 1$
$$75 \left(\frac{x - (x - 20)}{x(x - 20)}\right) = 1 \implies \frac{1500}{x^2 - 20x} = 1$$ $$x^2 - 20x - 1500 = 0$$ [1 Mark]
Step 3 Factorise:
$$(x - 50)(x + 30) = 0$$ Since the number of children cannot be negative, $x = 50$. [1 Mark]
Answer: The original number of children was 50.

Step 1 Find midpoints ($x_i$) and products ($f_i x_i$):
• 0 - 10: $x_1 = 5, f_1 = 7 \implies f_1 x_1 = 35$
• 10 - 20: $x_2 = 15, f_2 = a \implies f_2 x_2 = 15a$
• 20 - 30: $x_3 = 25, f_3 = 8 \implies f_3 x_3 = 200$
• 30 - 40: $x_4 = 35, f_4 = 10 \implies f_4 x_4 = 350$
• 40 - 50: $x_5 = 45, f_5 = 5 \implies f_5 x_5 = 225$
$$\sum f_i = 30 + a, \quad \sum f_i x_i = 810 + 15a$$ [1 Mark]
Step 2 Apply Mean formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = 24$:
$$24 = \frac{810 + 15a}{30 + a} \implies 24(30 + a) = 810 + 15a$$ [1 Mark]
$$720 + 24a = 810 + 15a \implies 9a = 90 \implies \mathbf{a = 10}$$ [1 Mark]
Answer: $a = 10$.

Step 1 Construct $\triangle ABC$:
• Draw base $BC = 5\text{ cm}$.
• At $B$, construct $\angle CBX = 120^\circ$ using compass.
• Cut an arc of radius $6.5\text{ cm}$ along ray $BX$ to mark $A$. Join $AC$. [1.5 Marks]
Step 2: Circumcircle
• Draw the perpendicular bisectors of sides $AB$ and $BC$.
• Their point of intersection is the circumcentre $O$ (lying outside $\triangle ABC$ because $\angle B > 90^\circ$).
• With centre $O$ and radius $OA = OB = OC$, draw the circumcircle. [1.5 Marks]
Step 3: Point D and Quad ABCD
• The locus of points equidistant from $AB$ and $BC$ is the internal angle bisector of $\angle ABC$.
• Construct the bisector of $\angle ABC$ ($60^\circ$) and produce it to intersect the circumcircle at point $D$.
• Join $AD$ and $CD$ to complete the required cyclic quadrilateral $ABCD$. [1 Mark]
Result: $\triangle ABC$ constructed with circumcircle and cyclic quadrilateral $ABCD$.

Step 1 RD Formula for interest: $I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{R}{100}$.
Here $P = 1000, R = 10\%, I = ₹5550$. [1 Mark]
Step 2 Substitute values:
$$5550 = 1000 \times \frac{n(n + 1)}{24} \times \frac{10}{100} = \frac{100 \times n(n + 1)}{24}$$ $$5550 \times 24 = 100(n^2 + n) \implies 133200 = 100(n^2 + n)$$ $$n^2 + n - 1332 = 0$$ [1 Mark]
Step 3 Factorise the quadratic equation ($36 \times 37 = 1332$):
$$(n + 37)(n - 36) = 0$$ Since time cannot be negative, $n = 36\text{ months}$.
$$\text{Time} = \frac{36}{12} = \mathbf{3\text{ years}}$$ [1 Mark]
Answer: 36 months (or 3 years).

(i) Find MN / QR Given $\frac{PM}{MQ} = \frac{2}{3} \implies \frac{PM}{PQ} = \frac{2}{2 + 3} = \frac{2}{5}$.
In $\triangle PMN$ and $\triangle PQR$, $MN \parallel QR \implies \triangle PMN \sim \triangle PQR$ by AA similarity.
$$\frac{MN}{QR} = \frac{PM}{PQ} = \mathbf{\frac{2}{5}}$$ [1 Mark]
(ii) Prove Similarity In $\triangle OMN$ and $\triangle ORQ$:
• $\angle OMN = \angle ORQ$ (alternate interior angles, since $MN \parallel QR$)
• $\angle ONM = \angle OQR$ (alternate interior angles)
• $\angle MON = \angle ROQ$ (vertically opposite angles)
$\therefore \mathbf{\triangle OMN \sim \triangle ORQ}$ by AA similarity. [1 Mark]
(iii) Ratio of Areas Ratio of areas of similar triangles is equal to the square of their corresponding side ratio:
$$\frac{\text{Area}(\triangle OMN)}{\text{Area}(\triangle ORQ)} = \left(\frac{MN}{QR}\right)^2 = \left(\frac{2}{5}\right)^2 = \mathbf{\frac{4}{25}}$$ [1 Mark]
Answers: (i) $\frac{MN}{QR} = \frac{2}{5}$, (ii) Proved, (iii) $4 : 25$.

Given Common radius $r = 7\text{ cm}$, Cylinder height $h_{\text{cyl}} = 4\text{ cm}$, Cone height $h_{\text{cone}} = 4\text{ cm}$.
1. Volume of Cone
$$V_1 = \frac{1}{3} \pi r^2 h_{\text{cone}} = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 4 = \frac{616}{3}\text{ cm}^3$$ [1 Mark]
2. Volume of Cylinder
$$V_2 = \pi r^2 h_{\text{cyl}} = \frac{22}{7} \times 7^2 \times 4 = 616\text{ cm}^3$$ [1 Mark]
3. Volume of Hemisphere
$$V_3 = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7^3 = \frac{2156}{3}\text{ cm}^3$$ [1 Mark]
Total Volume
$$V = V_1 + V_2 + V_3 = \frac{616}{3} + 616 + \frac{2156}{3} = \frac{2772}{3} + 616 = 924 + 616 = \mathbf{1540\text{ cm}^3}$$ [1 Mark]
Answer: Volume of the solid is $1540\text{ cm}^3$.

Step 1 Let $f(x) = 2x^3 + 3x^2 - 9x - 10$.
Testing $x = -1$:
$$f(-1) = 2(-1)^3 + 3(-1)^2 - 9(-1) - 10 = -2 + 3 + 9 - 10 = 0$$ $\therefore (x + 1)$ is a factor of $f(x)$ by Factor Theorem. [1 Mark]
Step 2 Divide $f(x)$ by $(x + 1)$:
$$2x^3 + 3x^2 - 9x - 10 = (x + 1)(2x^2 + x - 10)$$ [1 Mark]
Step 3 Factorise the quadratic expression $2x^2 + x - 10$ ($2 \times -10 = -20$, factors $5, -4$):
$$2x^2 + 5x - 4x - 10 = x(2x + 5) - 2(2x + 5) = (2x + 5)(x - 2)$$ $$\mathbf{(x + 1)(x - 2)(2x + 5)}$$ [1 Mark]
Answer: $(x + 1)(x - 2)(2x + 5)$.

Step 1 Since $O$ is the centre, $OP = OQ = \text{radius } r$.
Given $QR = OP$, we have $QR = OQ = r$.
In $\triangle OQR$, $OQ = QR \implies \triangle OQR$ is isosceles.
$$\angle QOR = \angle QRO = 20^\circ$$ [1 Mark]
Step 2 In $\triangle OQR$, exterior angle at $Q$ is $\angle OQP$:
$$\angle OQP = \angle QOR + \angle QRO = 20^\circ + 20^\circ = 40^\circ$$ In $\triangle OPQ$, $OP = OQ = r \implies \triangle OPQ$ is also isosceles.
$$\angle OPQ = \angle OQP = 40^\circ$$ [1 Mark]
Step 3 For $\triangle OPR$, the exterior angle at vertex $O$ on straight line $TOR$ is $\angle POT = x$:
$$x = \angle OPR + \angle ORP = 40^\circ + 20^\circ = \mathbf{60^\circ}$$ [1 Mark]
Alternative: In $\triangle OPQ$, $\angle POQ = 180^\circ - (40^\circ + 40^\circ) = 100^\circ$.
On straight line $TOR$: $x + \angle POQ + \angle QOR = 180^\circ \implies x + 100^\circ + 20^\circ = 180^\circ \implies x = 60^\circ$.
Answer: $x = 60^\circ$.

Step 1 Let distance between the two towers $PQ = d$, and height of tower $PT = h$.
In right $\triangle PQR$ ($\angle PQR = 90^\circ$):
$$\tan 60^\circ = \frac{QR}{PQ} = \frac{50}{d}$$ $$\sqrt{3} = \frac{50}{d} \implies d = \frac{50}{\sqrt{3}}\text{ m}$$ [1.5 Marks]
Step 2 In right $\triangle QPT$ ($\angle QPT = 90^\circ$):
$$\tan 30^\circ = \frac{PT}{PQ} = \frac{h}{d}$$ $$\frac{1}{\sqrt{3}} = \frac{h}{d} \implies h = \frac{d}{\sqrt{3}}$$ [1.5 Marks]
Step 3 Substitute $d = \frac{50}{\sqrt{3}}$:
$$h = \frac{\frac{50}{\sqrt{3}}}{\sqrt{3}} = \frac{50}{3} = 16.67\text{ m}$$ [0.5 Mark]
Rounding to the nearest metre:
$$h \approx \mathbf{17\text{ metres}}$$ [0.5 Mark]
Answer: The height of the tower $PT$ is 17 metres.

Step 1 Let first term be $a$ and common difference be $d$.
$$T_4 = a + 3d = 22 \quad \text{--- (1)}$$ [1 Mark]
$$T_{15} = a + 14d = 66 \quad \text{--- (2)}$$ [1 Mark]
Step 2 Subtract (1) from (2):
$$(a + 14d) - (a + 3d) = 66 - 22 \implies 11d = 44 \implies \mathbf{d = 4}$$ [0.5 Mark]
Substitute $d = 4$ into (1):
$$a + 3(4) = 22 \implies a + 12 = 22 \implies \mathbf{a = 10}$$ [0.5 Mark]
Step 3 Sum of first 8 terms ($S_8$):
$$S_n = \frac{n}{2}[2a + (n - 1)d]$$ $$S_8 = \frac{8}{2}[2(10) + (8 - 1)4] = 4[20 + 28] = 4 \times 48 = \mathbf{192}$$ [1 Mark]
Answers: First term $a = 10$, Common difference $d = 4$, Sum of 8 terms $S_8 = 192$.

Cumulative Frequency Table
• 135 – 140: $f = 4, cf = 4$ → Plot $(140, 4)$
• 140 – 145: $f = 8, cf = 12$ → Plot $(145, 12)$
• 145 – 150: $f = 20, cf = 32$ → Plot $(150, 32)$
• 150 – 155: $f = 14, cf = 46$ → Plot $(155, 46)$
• 155 – 160: $f = 7, cf = 53$ → Plot $(160, 53)$
• 160 – 165: $f = 6, cf = 59$ → Plot $(165, 59)$
• 165 – 170: $f = 1, cf = 60$ → Plot $(170, 60)$
Base point: $(135, 0)$. [2 Marks for table & ogive]
(i) Median Total $N = 60 \implies \frac{N}{2} = 30^{\text{th}}$ term.
From $y = 30$ on y-axis, draw horizontal line to meet curve, then drop vertically to x-axis:
$$\text{Median} \approx \mathbf{149.5\text{ cm}} \quad (\text{accept } 149 - 150\text{ cm})$$ [1.5 Marks]
(ii) Lower Quartile Q₁ $Q_1 = \frac{N}{4} = \frac{60}{4} = 15^{\text{th}}$ term.
From $y = 15$ on y-axis, drop vertically to x-axis:
$$Q_1 \approx \mathbf{146\text{ cm}} \quad (\text{accept } 145.5 - 146.5\text{ cm})$$ [1.5 Marks]
(iii) Tall Boys (> 158 cm) At height $x = 158\text{ cm}$, the cumulative frequency from the graph is $50$.
Number of boys with height $> 158\text{ cm} = 60 - 50 = \mathbf{10\text{ boys}}$. [1 Mark]
Answers: (i) Median $\approx 149.5\text{ cm}$, (ii) Lower Quartile $Q_1 \approx 146\text{ cm}$, (iii) Number of tall boys $= 10$.