ICSE Class 10 Mathematics 2017 Solved Question Paper

Step 1 Since $b$ is the mean proportional between $a$ and $c$:
$$b^2 = ac \implies b^4 = a^2 c^2$$ [1 Mark]
Step 2 Substitute $b^2 = ac$ and $b^4 = a^2 c^2$ into LHS:
$$\text{Numerator} = a^4 + a^2(ac) + a^2 c^2 = a^4 + a^3 c + a^2 c^2 = a^2(a^2 + ac + c^2)$$ [1 Mark]
$$\text{Denominator} = a^2 c^2 + (ac)c^2 + c^4 = a^2 c^2 + ac^3 + c^4 = c^2(a^2 + ac + c^2)$$
Step 3 Cancel common terms:
$$\text{LHS} = \frac{a^2(a^2 + ac + c^2)}{c^2(a^2 + ac + c^2)} = \mathbf{\frac{a^2}{c^2}} = \text{RHS}$$ [1 Mark]
Hence Proved.

Step 1 Compare with $ax^2 + bx + c = 0$: $a = 4, b = -5, c = -3$. [1 Mark]
Step 2 Apply quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(4)(-3)}}{2(4)}$$ $$x = \frac{5 \pm \sqrt{25 + 48}}{8} = \frac{5 \pm \sqrt{73}}{8}$$ [1.5 Marks]
Step 3 Using $\sqrt{73} \approx 8.544$:
$$x_1 = \frac{5 + 8.544}{8} = \frac{13.544}{8} = 1.693 \approx \mathbf{1.69}$$ [0.75 Mark]
$$x_2 = \frac{5 - 8.544}{8} = \frac{-3.544}{8} = -0.443 \approx \mathbf{-0.44}$$ [0.75 Mark]
Answer: $x = 1.69$ or $x = -0.44$.

Step 1 The perpendicular from the centre of a circle to a chord bisects the chord.
For chord $AB = 24\text{ cm}$: $AN = \frac{24}{2} = 12\text{ cm}$.
In right $\triangle ONA$ ($OA = 13\text{ cm}$):
$$ON = \sqrt{OA^2 - AN^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = \mathbf{5\text{ cm}}$$ [1 Mark]
Step 2 For chord $CD = 10\text{ cm}$: $CM = \frac{10}{2} = 5\text{ cm}$.
In right $\triangle OMC$ ($OC = 13\text{ cm}$):
$$OM = \sqrt{OC^2 - CM^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = \mathbf{12\text{ cm}}$$ [1 Mark]
Step 3 Since the chords lie on opposite sides of the centre $O$:
$$\text{Distance between chords } NM = ON + OM = 5 + 12 = \mathbf{17\text{ cm}}$$ [1 Mark]
Answer: Distance between the two chords $= 17\text{ cm}$.

Step 1 Complementary angles for $\sin$:
$$\sin 62^\circ = \sin(90^\circ - 28^\circ) = \cos 28^\circ$$ $$\sin^2 28^\circ + \sin^2 62^\circ = \sin^2 28^\circ + \cos^2 28^\circ = 1$$ [1 Mark]
Step 2 Complementary angles for $\cot$:
$$\cot 52^\circ = \cot(90^\circ - 38^\circ) = \tan 38^\circ$$ $$\tan^2 38^\circ - \cot^2 52^\circ = \tan^2 38^\circ - \tan^2 38^\circ = 0$$ [1 Mark]
Step 3 Standard angle $\sec 30^\circ = \frac{2}{\sqrt{3}}$:
$$\frac{1}{4} \sec^2 30^\circ = \frac{1}{4} \times \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{1}{4} \times \frac{4}{3} = \frac{1}{3}$$ $$\text{Total Value} = 1 + 0 + \frac{1}{3} = \mathbf{\frac{4}{3}} = \mathbf{1\frac{1}{3}}$$ [1 Mark]
Answer: $\frac{4}{3}$ (or $1\frac{1}{3}$).

Step 1: Compute A²
$$A^2 = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 1(1)+3(3) & 1(3)+3(4) \\ 3(1)+4(3) & 3(3)+4(4) \end{bmatrix} = \begin{bmatrix} 10 & 15 \\ 15 & 25 \end{bmatrix}$$ [1 Mark]
Step 2: Compute B²
$$B^2 = \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 4-3 & -2+2 \\ 6-6 & -3+4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$ [1 Mark]
Step 3: Compute A² - 5B²
$$5B^2 = 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$$ $$5C = A^2 - 5B^2 = \begin{bmatrix} 10 & 15 \\ 15 & 25 \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 5 & 15 \\ 15 & 20 \end{bmatrix}$$ [1 Mark]
Step 4: Solve for C
$$C = \frac{1}{5} \begin{bmatrix} 5 & 15 \\ 15 & 20 \end{bmatrix} = \mathbf{\begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}}$$ [1 Mark]
Answer: $C = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}$.

Year 1 Principal $P_1 = ₹50,000$, Rate $R_1 = 12\%$.
$$\text{Interest for Year 1 } (I_1) = \frac{50,000 \times 12 \times 1}{100} = ₹6,000$$ $$\text{Amount at end of Year 1} = 50,000 + 6,000 = ₹56,000$$ [1 Mark]
Repayment Repaid at end of Year 1 $= ₹33,000$.
$$\text{Principal for Year 2 } (P_2) = 56,000 - 33,000 = ₹23,000$$ [1 Mark]
Year 2 Rate $R_2 = 15\%$.
$$\text{Interest for Year 2 } (I_2) = \frac{23,000 \times 15 \times 1}{100} = ₹3,450$$ $$\text{Amount to clear debt} = 23,000 + 3,450 = \mathbf{₹26,450}$$ [1 Mark]
Answer: Jaya must pay ₹26,450 at the end of the second year.

Successive Discounts
Catalogue Price $= ₹42,000$.
First Discount $(10\%) = \frac{10}{100} \times 42,000 = ₹4,200$.
Price after 1st discount $= 42,000 - 4,200 = ₹37,800$.
Off-season Discount $(5\%) = \frac{5}{100} \times 37,800 = ₹1,890$.
Final Selling Price before tax $= 37,800 - 1,890 = ₹35,910$. [1 Mark]
(i) Sales Tax (8%)
$$\text{Sales Tax} = \frac{8}{100} \times 35,910 = \mathbf{₹2,872.80}$$ [1 Mark]
(ii) Total Amount Paid
$$\text{Total Price} = 35,910 + 2,872.80 = \mathbf{₹38,782.80}$$ [1 Mark]
Answers: (i) Sales tax $= ₹2,872.80$, (ii) Total price $= ₹38,782.80$.

Step 1 Let $A(3, -6) = (x_1, y_1)$ and $B(x, y) = (x_2, y_2)$ with ratio $m_1 : m_2 = 2 : 3$.
Step 2: Solve for x
$$x_P = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \implies 1 = \frac{2x + 3(3)}{2 + 3}$$ $$1 = \frac{2x + 9}{5} \implies 5 = 2x + 9 \implies 2x = -4 \implies \mathbf{x = -2}$$ [2 Marks]
Step 3: Solve for y
$$y_P = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \implies -2 = \frac{2y + 3(-6)}{2 + 3}$$ $$-2 = \frac{2y - 18}{5} \implies -10 = 2y - 18 \implies 2y = 8 \implies \mathbf{y = 4}$$ [2 Marks]
Answer: The coordinates of $B$ are $(-2, 4)$.

Step 1: Find x For $n = 10$ (even number of terms):
$$\text{Median} = \frac{\left(\frac{10}{2}\right)\text{th term} + \left(\frac{10}{2} + 1\right)\text{th term}}{2} = \frac{5\text{th term} + 6\text{th term}}{2}$$ $$48 = \frac{x + (x + 4)}{2} = \frac{2x + 4}{2} = x + 2$$ $$x = 48 - 2 = \mathbf{46}$$ [2 Marks]
Step 2: Find Mode Substitute $x = 46$ into the dataset:
$$13, 35, 43, \mathbf{46}, \mathbf{46}, 50, 55, 61, 71, 80$$ $46$ appears with the highest frequency (twice).
$$\mathbf{\text{Mode} = 46}$$ [1 Mark]
Answers: $x = 46$; Mode $= 46$.

Step 1 Let $k$ be subtracted. The resulting polynomial is $f(x) = 16x^3 - 8x^2 + 4x + 7 - k$.
By Factor Theorem, for $(2x + 1)$ to be a factor, $f\left(-\frac{1}{2}\right) = 0$. [1 Mark]
Step 2 Substitute $x = -\frac{1}{2}$:
$$16\left(-\frac{1}{2}\right)^3 - 8\left(-\frac{1}{2}\right)^2 + 4\left(-\frac{1}{2}\right) + 7 - k = 0$$ $$16\left(-\frac{1}{8}\right) - 8\left(\frac{1}{4}\right) - 2 + 7 - k = 0$$ [1 Mark]
Step 3 Simplify:
$$-2 - 2 - 2 + 7 - k = 0 \implies 1 - k = 0 \implies \mathbf{k = 1}$$ [1 Mark]
Answer: 1 must be subtracted.

Step 1: Rectangle Dimensions Radius $r = 5\text{ cm}$.
Width $AB = 2r = 2 \times 5 = 10\text{ cm}$.
Length $BC = r + 2r + r = 4r = 4 \times 5 = 20\text{ cm}$.
$$\text{Area of Rectangle } ABCD = 20 \times 10 = 200\text{ cm}^2$$ [1.5 Marks]
Step 2: Unshaded Circular Area
One full circle + two semicircles of radius $5\text{ cm}$:
$$\text{Total circular area} = \pi r^2 + 2\left(\frac{1}{2} \pi r^2\right) = 2\pi r^2 = 2 \times 3.1416 \times 25 = 157.08\text{ cm}^2$$ [1.5 Marks]
Step 3: Shaded Region Area
$$\text{Area} = 200 - 157.08 = 42.92\text{ cm}^2$$ To three significant figures: $\mathbf{42.9\text{ cm}^2}$ [1 Mark]
Answer: Area of shaded region $= 42.9\text{ cm}^2$.

Step 1 Add $\frac{1}{2}$ throughout the inequation:
$$-8.5 + 0.5 < -4x \le 7.5 + 0.5 \implies -8 < -4x \le 8$$ [1 Mark]
Step 2 Divide throughout by $-4$ (reversing the inequality signs):
$$\frac{-8}{-4} > x \ge \frac{8}{-4} \implies 2 > x \ge -2 \iff -2 \le x < 2$$ [1 Mark]
Step 3: Solution Set Since $x \in I$ (integers):
$$\text{Solution Set} = \mathbf{\{-2, -1, 0, 1\}}$$ [0.5 Mark]
Represented by solid filled dots on $-2, -1, 0, 1$ on the number line. [0.5 Mark]
Answer: Solution Set $= \{-2, -1, 0, 1\}$.

Step 1: Compute B²
$$B^2 = \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} = \begin{bmatrix} 1+8 & 1+3 \\ 8+24 & 8+9 \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix}$$ [1 Mark]
Step 2: Compute X = B² - 4B
$$4B = \begin{bmatrix} 4 & 4 \\ 32 & 12 \end{bmatrix}$$ $$X = \begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix} - \begin{bmatrix} 4 & 4 \\ 32 & 12 \end{bmatrix} = \mathbf{\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}}$$ [1 Mark]
Step 3: Solve for a and b
$$\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 5a \\ 5b \end{bmatrix} = \begin{bmatrix} 5 \\ 50 \end{bmatrix}$$ $$5a = 5 \implies \mathbf{a = 1}$$ [1 Mark]
$$5b = 50 \implies \mathbf{b = 10}$$ [1 Mark]
Answers: $X = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$; $a = 1, b = 10$.

Step 1 $NV = ₹50$, $MV = ₹60$, Dividend Rate $= 10\%$.
Dividend per share $= \frac{10}{100} \times 50 = ₹5$.
Number of shares $= \frac{\text{Total Income}}{\text{Dividend per share}} = \frac{450}{5} = 90\text{ shares}$. [1 Mark]
Step 2: Total Investment
$$\text{Investment} = 90 \times 60 = \mathbf{₹5,400}$$ [1 Mark]
Step 3: Yield Percent
$$\text{Yield \%} = \frac{\text{Annual Income}}{\text{Investment}} \times 100 = \frac{450}{5400} \times 100 = \frac{25}{3}\% = 8.33\% \approx \mathbf{8\%}$$ [1 Mark]
Answers: Total Investment $= ₹5,400$; Yield percent $\approx 8\%$.

Sample Space $S = \{a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p\}$, $n(S) = 16$.
(i) Vowel Vowels in $S$: $\{a, e, i, o\} \implies 4$ outcomes.
$$P(\text{vowel}) = \frac{4}{16} = \mathbf{\frac{1}{4}}$$ [1 Mark]
(ii) Consonant Consonants in $S$: $16 - 4 = 12$ outcomes.
$$P(\text{consonant}) = \frac{12}{16} = \mathbf{\frac{3}{4}}$$ [1 Mark]
(iii) None of letters in 'median' Letters $\{m, e, d, i, a, n\} = 6$ cards.
Cards containing none of these letters $= 16 - 6 = 10$.
$$P(\text{none of the letters of median}) = \frac{10}{16} = \mathbf{\frac{5}{8}}$$ [1 Mark]
Answers: (i) $\frac{1}{4}$, (ii) $\frac{3}{4}$, (iii) $\frac{5}{8}$.

Step 1: Triangle ABC Draw base $AB = 7\text{ cm}$. At $A$, construct $\angle CAB = 60^\circ$. Cut arc $AC = 5\text{ cm}$ on the ray. Join $BC$. [1 Mark]
Step 2: Locus (i) Locus of points equidistant from $AB$ and $AC$ is the angle bisector of $\angle CAB$ ($30^\circ$). Construct this ray. [1 Mark]
Step 3: Locus (ii) Locus of points equidistant from $BA$ and $BC$ is the angle bisector of $\angle ABC$. Construct this ray. [1 Mark]
Step 4: Incircle Their intersection is the incentre $I$. Drop a perpendicular from $I$ to $AB$ to obtain the inradius $r$. Draw the incircle touching all three sides. [1 Mark]
Result: $\triangle ABC$ and its incircle successfully constructed.

Step 1: Volume Total volume of air $= 77 \times 16 = 1232\text{ m}^3$.
Radius $r = 7\text{ m}$.
$$V = \frac{1}{3} \pi r^2 h \implies 1232 = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times h$$ $$1232 = \frac{154}{3} h \implies h = \frac{1232 \times 3}{154} = \mathbf{24\text{ m}}$$ [1.5 Marks]
Step 2: Curved Surface Area
Slant height $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\text{ m}$.
$$\text{Curved Surface Area} = \pi r l = \frac{22}{7} \times 7 \times 25 = 22 \times 25 = \mathbf{550\text{ m}^2}$$ [1.5 Marks]
Answers: Height of tent $= 24\text{ m}$; Curved surface area $= 550\text{ m}^2$.

(i) Find m : n Apply Componendo and Dividendo:
$$\frac{(7m + 2n) + (7m - 2n)}{(7m + 2n) - (7m - 2n)} = \frac{5 + 3}{5 - 3}$$ $$\frac{14m}{4n} = \frac{8}{2} = 4 \implies \frac{7m}{2n} = 4 \implies \frac{m}{n} = \frac{8}{7}$$ $$\mathbf{m : n = 8 : 7}$$ [1.5 Marks]
(ii) Find (m² + n²) / (m² - n²) Square both sides of $\frac{m}{n} = \frac{8}{7}$:
$$\frac{m^2}{n^2} = \frac{64}{49}$$ Apply Componendo and Dividendo:
$$\frac{m^2 + n^2}{m^2 - n^2} = \frac{64 + 49}{64 - 49} = \mathbf{\frac{113}{15}}$$ [1.5 Marks]
Answers: (i) $m : n = 8 : 7$, (ii) $\frac{113}{15}$.

Qualifying Minimum Balances (10th to end of month)
• January: ₹5,600
• February: ₹4,100
• March: ₹4,100
• April: ₹2,000
• May: ₹8,500
• June: ₹10,000
$$\text{Total Qualifying Balance } (\sum P) = 5600 + 4100 + 4100 + 2000 + 8500 + 10000 = ₹34,300$$ [2.5 Marks]
(i) Calculate Interest
$$\text{Interest} = \frac{\sum P \times R \times 1}{100 \times 12} = \frac{34,300 \times 6 \times 1}{1200} = \frac{34,300}{200} = \mathbf{₹171.50}$$ [1.5 Marks]
(ii) Amount Received on Closing Final balance in June was ₹10,000.
$$\text{Total Amount} = 10,000 + 171.50 = \mathbf{₹10,171.50}$$ [1 Mark]
Answers: (i) Interest $= ₹171.50$, (ii) Amount received $= ₹10,171.50$.

(i) Plotting Plot points $A(0,4), B(2,3), C(1,1), D(2,0)$ on the graph. [1.5 Marks]
(ii) Coordinates of Images Reflection across y-axis maps $(x, y) \to (-x, y)$:
$$B' = \mathbf{(-2, 3)}, \quad C' = \mathbf{(-1, 1)}, \quad D' = \mathbf{(-2, 0)}$$ [1.5 Marks]
(iii) Closed Figure & Symmetry Connecting $A - B - C - D - D' - C' - B' - A$ yields a symmetrical arrowhead figure.
The line of symmetry is the y-axis, whose equation is: $$\mathbf{x = 0}$$ [2 Marks]
Answers: (ii) $B'(-2,3), C'(-1,1), D'(-2,0)$; (iii) Equation of line of symmetry is $x = 0$.

Step 1 Let Assumed Mean $A = 35$, class width $h = 10$, $u_i = \frac{x_i - 35}{10}$.
• 0 - 10: $x_1 = 5, f_1 = 10, u_1 = -3 \implies f_1 u_1 = -30$
• 10 - 20: $x_2 = 15, f_2 = 9, u_2 = -2 \implies f_2 u_2 = -18$
• 20 - 30: $x_3 = 25, f_3 = 25, u_3 = -1 \implies f_3 u_3 = -25$
• 30 - 40: $x_4 = 35, f_4 = 30, u_4 = 0 \implies f_4 u_4 = 0$
• 40 - 50: $x_5 = 45, f_5 = 16, u_5 = 1 \implies f_5 u_5 = 16$
• 50 - 60: $x_6 = 55, f_6 = 10, u_6 = 2 \implies f_6 u_6 = 20$
$$\sum f_i = 100, \quad \sum f_i u_i = -37$$ [2 Marks]
Step 2 Apply Step-Deviation formula:
$$\bar{x} = A + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 35 + \left(\frac{-37}{100}\right) \times 10 = 35 - 3.7 = \mathbf{31.3}$$ [2 Marks]
Answer: $\text{Mean} = 31.3$.

(i) Prove BD is Diameter
Since $AB$ bisects $\angle CAQ$: $\angle CAB = \angle BAQ = 30^\circ \implies \angle CAQ = 60^\circ$.
Since $PAQ$ is a straight tangent line: $\angle PAC = 180^\circ - 60^\circ = 120^\circ$.
$AD$ bisects $\angle PAC \implies \angle DAC = \frac{120^\circ}{2} = 60^\circ$.
$$\angle DAB = \angle DAC + \angle CAB = 60^\circ + 30^\circ = 90^\circ$$ Since $\angle DAB = 90^\circ$ is an inscribed angle in the circle, chord $BD$ must subtend a right angle at the circumference.
$\therefore \mathbf{BD\text{ is a diameter of the circle}}$. [1.5 Marks]
(ii) Prove ABC is Isosceles
By Alternate Segment Theorem: Angle between tangent $AQ$ and chord $AB$ equals the angle in the alternate segment:
$$\angle ACB = \angle BAQ = 30^\circ$$ In $\triangle ABC$, $\angle CAB = 30^\circ$ and $\angle ACB = 30^\circ$.
Since two angles are equal, opposite sides $AB = BC$.
$\therefore \mathbf{\triangle ABC\text{ is an isosceles triangle}}$. [1.5 Marks]
Hence Proved.

Step 1: Wholesaler to Shopkeeper Printed Price $= ₹45,000$.
Cost price to shopkeeper $= 45,000 - 10\% \text{ of } 45,000 = ₹40,500$.
Input Tax paid by shopkeeper $= \frac{12}{100} \times 40,500 = ₹4,860$.
Step 2: Shopkeeper to Customer
Selling price to customer $= 45,000 - 5\% \text{ of } 45,000 = ₹42,750$.
Output Tax collected $= \frac{12}{100} \times 42,750 = ₹5,130$.
(i) VAT Paid by Shopkeeper
$$\text{VAT} = \text{Output Tax} - \text{Input Tax} = 5,130 - 4,860 = \mathbf{₹270}$$ [1.5 Marks]
(ii) Total Amount Paid by Customer
$$\text{Total Price} = 42,750 + 5,130 = \mathbf{₹47,880}$$ [1.5 Marks]
Answers: (i) VAT paid $= ₹270$, (ii) Total amount paid $= ₹47,880$.

(i) Find ∠BCD In cyclic quadrilateral $ABCD$, the exterior angle is equal to the interior opposite angle:
$$\angle BCD = \angle DAE = \mathbf{70^\circ}$$ [1.5 Marks]
(ii) Find ∠BOD The angle subtended by arc $BCD$ at the centre is twice the angle subtended by it at the circumference:
$$\angle BOD = 2 \times \angle BCD = 2 \times 70^\circ = \mathbf{140^\circ}$$ [1.5 Marks]
(iii) Find ∠OBD In $\triangle OBD$, $OB = OD = \text{radius } r$, so $\triangle OBD$ is isosceles:
$$\angle OBD = \angle ODB = \frac{180^\circ - \angle BOD}{2} = \frac{180^\circ - 140^\circ}{2} = \mathbf{20^\circ}$$ [1 Mark]
Answers: (i) $\angle BCD = 70^\circ$, (ii) $\angle BOD = 140^\circ$, (iii) $\angle OBD = 20^\circ$.

(i) Centroid Coordinates
$$x_G = \frac{-1 + 4 + 3}{3} = \frac{6}{3} = 2, \quad y_G = \frac{3 + 2 - 2}{3} = \frac{3}{3} = 1$$ $$\mathbf{G = (2, 1)}$$ [1.5 Marks]
(ii) Equation of Line Slope of $AC$ with $A(-1, 3)$ and $C(3, -2)$:
$$m = \frac{-2 - 3}{3 - (-1)} = -\frac{5}{4}$$ Equation through $G(2, 1)$ with slope $m = -\frac{5}{4}$:
$$y - 1 = -\frac{5}{4}(x - 2) \implies 4(y - 1) = -5x + 10$$ $$\mathbf{5x + 4y - 14 = 0}$$ [1.5 Marks]
Answers: (i) $G(2, 1)$; (ii) $5x + 4y - 14 = 0$.

Step 1 Factor out $\sin \theta$ in numerator and $\cos \theta$ in denominator:
$$\text{LHS} = \frac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta - 1)}$$ [1 Mark]
Step 2 Express $1$ as $\sin^2 \theta + \cos^2 \theta$:
$$1 - 2\sin^2 \theta = \sin^2 \theta + \cos^2 \theta - 2\sin^2 \theta = \cos^2 \theta - \sin^2 \theta$$ $$2\cos^2 \theta - 1 = 2\cos^2 \theta - (\sin^2 \theta + \cos^2 \theta) = \cos^2 \theta - \sin^2 \theta$$ [1 Mark]
Step 3 Cancel common terms:
$$\text{LHS} = \frac{\sin \theta}{\cos \theta} \times \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta - \sin^2 \theta} = \mathbf{\tan \theta} = \text{RHS}$$ [1 Mark]
Hence Proved.

Step 1 Let Vivek's age be $x$ years. Then Amit's age $= 47 - x$ years. [1 Mark]
Step 2 Formulate quadratic equation:
$$x(47 - x) = 550 \implies 47x - x^2 = 550$$ $$x^2 - 47x + 550 = 0$$ [1 Mark]
Step 3 Factorise ($25 \times 22 = 550$ and $25 + 22 = 47$):
$$(x - 25)(x - 22) = 0$$ [1 Mark]
Since Vivek is the older brother, $x = 25$, and Amit $= 47 - 25 = 22$. [1 Mark]
Answer: Vivek is 25 years old and Amit is 22 years old.

Cumulative Frequency Table
• 400-450: $cf = 2$ → $(450, 2)$
• 450-500: $cf = 8$ → $(500, 8)$
• 500-550: $cf = 20$ → $(550, 20)$
• 550-600: $cf = 38$ → $(600, 38)$
• 600-650: $cf = 62$ → $(650, 62)$
• 650-700: $cf = 75$ → $(700, 75)$
• 700-750: $cf = 80$ → $(750, 80)$
Base point: $(400, 0)$. [2 Marks]
(i) Median Wage $\frac{N}{2} = 40^{\text{th}}$ worker. From $y = 40$ on graph, drop to x-axis:
$$\text{Median} \approx \mathbf{₹605} \quad (\text{acceptable: } 600 - 610)$$ [1.5 Marks]
(ii) Lower Quartile Wage $\frac{N}{4} = 20^{\text{th}}$ worker. From $y = 20$ on graph:
$$Q_1 = \mathbf{₹550}$$ [1.5 Marks]
(iii) Workers Earning > ₹625 At $x = 625$, the cumulative frequency is $50$.
$$\text{Number of workers} = 80 - 50 = \mathbf{30\text{ workers}}$$ [1 Mark]
Answers: (i) Median $\approx ₹605$, (ii) Lower quartile $= ₹550$, (iii) 30 workers.

Step 1 Let $CD = 60\text{ m}$ be the lighthouse. $A$ and $B$ are on opposite sides of base $D$.
Angle of depression $60^\circ \implies \angle CAD = 60^\circ$.
Angle of depression $45^\circ \implies \angle CBD = 45^\circ$.
Step 2: Distance AD In right $\triangle CDA$ ($\angle D = 90^\circ$):
$$\tan 60^\circ = \frac{CD}{AD} \implies \sqrt{3} = \frac{60}{AD} \implies AD = \frac{60}{\sqrt{3}} = 20\sqrt{3}\text{ m}$$ [1.5 Marks]
Step 3: Distance BD In right $\triangle CDB$ ($\angle D = 90^\circ$):
$$\tan 45^\circ = \frac{CD}{BD} \implies 1 = \frac{60}{BD} \implies BD = 60\text{ m}$$ [1 Mark]
Step 4: Total Distance AB
$$AB = AD + BD = 20\sqrt{3} + 60 = 20(1.732) + 60 = 34.64 + 60 = 94.64\text{ m}$$ To the nearest whole number: $\mathbf{95\text{ metres}}$ [1.5 Marks]
Answer: The distance between the two ships is 95 metres.

(i) Prove Similarity In $\triangle PQR$ and $\triangle SPR$:
• $\angle QPR = \angle PSR$ (given)
• $\angle PRQ = \angle SRP$ (common angle $\angle R$)
$\therefore \mathbf{\triangle PQR \sim \triangle SPR}$ by AA similarity criterion. [1 Mark]
(ii) Find QR and PS Corresponding sides are proportional:
$$\frac{PQ}{SP} = \frac{QR}{PR} = \frac{PR}{SR} \implies \frac{8}{PS} = \frac{QR}{6} = \frac{6}{3} = 2$$ 1) $\frac{QR}{6} = 2 \implies \mathbf{QR = 12\text{ cm}}$
2) $\frac{8}{PS} = 2 \implies \mathbf{PS = 4\text{ cm}}$ [1 Mark]
(iii) Ratio of Areas
$$\frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle SPR)} = \left(\frac{PR}{SR}\right)^2 = \left(\frac{6}{3}\right)^2 = 2^2 = \mathbf{\frac{4}{1}} = \mathbf{4 : 1}$$ [1 Mark]
Answers: (i) Proved, (ii) $QR = 12\text{ cm}, PS = 4\text{ cm}$, (iii) $4 : 1$.

Step 1 Time $n = 3 \times 12 = 36\text{ months}$, Rate $R = 7.5\%$, Interest $I = ₹8,325$.
$$I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{R}{100}$$ $$8325 = P \times \frac{36 \times 37}{24} \times \frac{7.5}{100} = P \times 55.5 \times 0.075 = 4.1625 P$$ [1 Mark]
(i) Monthly Deposit P
$$P = \frac{8325}{4.1625} = \mathbf{₹2,000}$$ [1 Mark]
(ii) Maturity Value MV Total deposited $= 2000 \times 36 = ₹72,000$.
$$MV = (P \times n) + I = 72,000 + 8,325 = \mathbf{₹80,325}$$ [1 Mark]
Answers: (i) Monthly deposit $= ₹2,000$, (ii) Maturity value $= ₹80,325$.