Section A (1 Mark Each)

1. Which is a rational number?
   Option (c) \(\frac{3}{7}\) is rational. \(\pi\) and \(\sqrt{2}\) are irrational. Ans: (c)
2. Which is a polynomial?
   Option (b) \(3x^2 – 2x + 5\) has only whole number powers. Ans: (b)
3. If \(x+2y=10\), find \(y\) when \(x=4\).
   \(4 + 2y = 10 \Rightarrow 2y = 6 \Rightarrow y = 3\). Ans: (c) 3
4. If \(a_n = 4n-2\), find the 3rd term.
   \(a_3 = 4(3) – 2 = 12 – 2 = 10\). Ans: (a) 10
5. Ratio of opposite side to hypotenuse is:
   \(\frac{Opp}{Hyp} = \sin \theta\). Ans: (c) Sin 0
6. Midpoint of (2, 1) and (6, 3).
   \((\frac{2+6}{2}, \frac{1+3}{2}) = (4, 2)\). Ans: (c) (4, 2)
7. LCM of 12 and 30.
   \(12=2^2 \times 3\), \(30=2 \times 3 \times 5\). LCM = \(4 \times 3 \times 5 = 60\). Ans: (b) 60
8. Probability of getting a Head.
   \(P(H) = \frac{1}{2} = 0.5\). Ans: (a) 0.5
9. Volume of cube with side 7 cm.
   \(V = a^3 = 7^3 = 343 \text{ cm}^3\). Ans: (c) 343
10. Equation \(ax^2 + 2x + a = 0\) has equal roots if:
    \(D=0 \Rightarrow (2)^2 – 4(a)(a) = 0 \Rightarrow 4 = 4a^2 \Rightarrow a = \pm 1\). Ans: (b) \(\pm 1\)
11. Prime factorization of 1771 is \(7 \times 11 \times 13\).
    False. Correct is \(7 \times 11 \times 23\). Ans: False
12. Common difference of AP \(\frac{1}{2b}, \frac{1-6b}{2b} \dots\)
    \(d = \frac{1-6b-1}{2b} = \frac{-6b}{2b} = -3\). Ans: (a) -3
13. Probability of impossible event is 1.
    False. It is 0. Ans: False
14. All _______ triangles are similar.
    Ans: Equilateral
15. Angle between tangent and radius.
    Ans: \(90^\circ\)
16. First three terms of \(a_n = \frac{n(n-2)}{2}\).
    \(n=1 \to -0.5\), \(n=2 \to 0\), \(n=3 \to 1.5\). Ans: \(-1/2, 0, 3/2\)
17. \(x^2=a, y=b\) has unique solution.
    False. \(x = \pm \sqrt{a}\) gives two solutions. Ans: False
18. Cot A is product of Cot and A.
    False. It is a single ratio. Ans: False (OR: \(x+y=3\))
19. Mean of first 5 multiples of 2.
    2, 4, 6, 8, 10. Mean is 6. Ans: 6
20. Formula for Mean (Direct Method).
    Ans: \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\) (OR: Median = 19.33)

Section B (2 Marks Each)

21. Garden Dimensions: Half perimeter = 36m, \(l = w + 4\).
    \(l+w = 36 \Rightarrow (w+4)+w = 36 \Rightarrow 2w=32\).
    Width = 16m, Length = 20m
22. Roots of \(6x^2 – x – 2 = 0\):
    \(6x^2 – 4x + 3x – 2 = 0 \Rightarrow 2x(3x-2) + 1(3x-2) = 0\).
    Roots: \(x = 2/3, -1/2\)
23. Surface Area of Cuboid: Two cubes of vol \(64 cm^3\) joined.
    Side \(a=4\). New \(L=8, B=4, H=4\). TSA = \(2(LB+BH+HL)\).
    Area = \(160 \text{ cm}^2\)
24. Trigonometry: Given \(\tan A = 4/3\).
    Hypotenuse = \(\sqrt{4^2+3^2} = 5\).
    \(\sin A = 4/5, \cos A = 3/5\) etc.
25. Equidistant Point: Point on x-axis equidistant from (2, -5) and (-2, 9).
    Let \(P(x,0)\). Distance formula gives \((x-2)^2 + 25 = (x+2)^2 + 81\).
    Point: \((-7, 0)\)
26. Polynomial Zeroes: \(6x^2 – 7x – 3\).
    Factors: \(6x^2 – 9x + 2x – 3\). Zeroes are \(3/2\) and \(-1/3\).
    Sum = \(7/6\), Product = \(-1/2\) (Verified)

Section C (3 Marks Each)

27. Section Formula: \(AP = \frac{3}{7}AB \Rightarrow AP:PB = 3:4\).
    Using section formula \(m=3, n=4\) on \(A(-2,-2), B(2,-4)\).
    P coordinates: \((-2/7, -20/7)\)
28. Area of Segment: \(r=12\), \(\theta=120^\circ\).
    Area = Sector – Triangle. Sector = \(150.72\). Triangle = \(36\sqrt{3} \approx 62.28\).
    Segment Area \(\approx 88.44 \text{ cm}^2\)
29. Chord Length: Concentric circles \(r=5, r=3\).
    Half chord = \(\sqrt{5^2 – 3^2} = 4\). Total Length = \(2 \times 4\).
    Length = 8 cm
30. Similar Triangles: Prove \(CA^2 = CB \cdot CD\).
    Compare \(\Delta ABC\) and \(\Delta DAC\). By AA Similarity (\(\angle C\) common, \(\angle D = \angle A\)).
    Ratio \(\frac{CA}{CD} = \frac{CB}{CA}\) proves the result.
31. Theorem: Prove Basic Proportionality Theorem (BPT).
    Standard proof using area of triangles. See Textbook Theorem 6.1
32. Irrationality: Prove \(3 + 2\sqrt{5}\) is irrational.
    Assume rational \(a/b\). Isolate \(\sqrt{5} = \frac{1}{2}(\frac{a}{b} – 3)\). RHS is rational, LHS irrational. Contradiction.
33. AP Sum: How many terms of 9, 17, 25… sum to 636?
    \(S_n = \frac{n}{2}[18 + (n-1)8] = 636\). Forms \(4n^2 + 5n – 636 = 0\).
    \(n = 12\)
34. Probability: Die thrown once.
    (i) Prime (2,3,5) = 3/6 = 0.5
    (ii) Between 2 & 6 (3,4,5) = 3/6 = 0.5

Section D (4 Marks Each)

35. Age Problem: Mother is 26 older. Product 3 years later is 360.
    \((x+3)(x+29) = 360 \Rightarrow x^2 + 32x – 273 = 0\).
    Rohan = 7 years, Mother = 33 years
36. Solid Surface Area: Cube side 7cm surmounted by Hemisphere.
    Max Diameter = 7cm. Area = \(6a^2 – \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2\).
    TSA = \(332.5 \text{ cm}^2\)
37. Height & Distance: Tower on 20m building. Angles 45° & 60°.
    Base = 20m (from \(\tan 45\)). Total Height \(H = 20\sqrt{3}\). Tower = \(H – 20\).
    Tower Height = \(20(\sqrt{3}-1) \approx 14.64\) m
38. Trigonometry Proof: Prove \(\frac{\sin\theta – \cos\theta + 1}{\sin\theta + \cos\theta – 1} = \frac{1}{\sec\theta – \tan\theta}\).
    Divide by \(\cos\theta\). Use identity \(\sec^2\theta – \tan^2\theta = 1\) in numerator. Factorize and cancel.
39. Shadow Problem: Girl (90cm) walking from Lamp (3.6m) at 1.2m/s.
    Distance = \(4.8\)m. Similar Triangles: \(\frac{3.6}{0.9} = \frac{4.8+x}{x}\).
    Shadow Length = 1.6 m
40. Statistics: Find Median Weight.
    \(N=30\). Median Class 55-60. \(l=55, f=6, cf=13, h=5\).
    Formula: \(l + \frac{N/2 – cf}{f} \times h\).
    Median = 56.67 kg

Asterisk Classes: Attempt Strategy

1. Attempt Count: The paper note says “Attempt any 68 Marks”. Ensure you prioritize questions you are 100% sure about to maximize this limit.

2. Diagrams: For the Surface Area (Q36) and Height/Distance (Q37) questions, a diagram carries 0.5-1 mark. Do not skip it.

3. Calculations: In Statistics (Q40), check your Cumulative Frequency (cf) addition twice. A single error there ruins the 4-mark answer.