SECTION A
1. HCF of \(a = x^3 y^2\) and \(b = x y^3\):
Answer: (b) \(xy^2\)
Explanation: To find HCF, we take the smallest power of each common prime factor. Minimum power of \(x\) is 1, and minimum power of \(y\) is 2. So, HCF = \(x^1 y^2 = xy^2\).
2. Quadratic polynomial with zeroes -3 and 4:
Answer: (c) \(\frac{x^2}{2} - \frac{x}{2} - 6\)
Explanation: Sum of zeroes (\(S\)) = \(-3 + 4 = 1\). Product (\(P\)) = \((-3) \times 4 = -12\).
General form: \(k[x^2 - Sx + P] = k[x^2 - x - 12]\).
For \(k = \frac{1}{2}\), the polynomial is \(\frac{x^2}{2} - \frac{x}{2} - 6\). Option (c) is correct.
3. Nature of solutions for the given pair of linear equations:
Answer: (c) infinitely many solutions
Explanation: \(a_1/a_2 = 5/3\); \(b_1/b_2 = -15/-9 = 5/3\); \(c_1/c_2 = 8 / (24/5) = 40/24 = 5/3\).
Since \(a_1/a_2 = b_1/b_2 = c_1/c_2\), the lines are coincident, yielding infinitely many solutions.
4. Condition for equal roots for \(2x^2 - kx + k = 0\):
Answer: (d) 0, 8
Explanation: For equal roots, Discriminant \(D = 0 \Rightarrow b^2 - 4ac = 0\).
\((-k)^2 - 4(2)(k) = 0 \Rightarrow k^2 - 8k = 0 \Rightarrow k(k - 8) = 0\). So, \(k = 0\) or \(k = 8\).
5. Distance of \(P(2, 3)\) from x-axis:
Answer: (b) 3 units
Explanation: The distance of any point \((x, y)\) from the x-axis is its y-coordinate's absolute value, which is 3.
6. Finding the first term of an A.P.:
Answer: (d) 28
Explanation: \(a_n = a + (n-1)d \Rightarrow 4 = a + (7-1)(-4) \Rightarrow 4 = a - 24 \Rightarrow a = 28\).
7. Ratio of areas of similar triangles:
Answer: (c) 36:49
Explanation: Ratio of areas = \((AB/PQ)^2 = (1.2/1.4)^2 = (6/7)^2 = 36/49\).
8. Trigonometric evaluation:
Answer: (b) 1
Explanation: \(\sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ = (\frac{1}{2})(\frac{1}{2}) + (\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2}) = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1\).
9. Shadow equals height:
Answer: (b) angle of elevation 45°
Explanation: Let height = \(h\) and shadow = \(x\). \(\tan \theta = h/x\). If \(h = x\), \(\tan \theta = 1\), so \(\theta = 45^\circ\).
10. Distance formula application:
Answer: (b) \(\pm 4\)
Explanation: Distance \(d = \sqrt{(4-1)^2 + (p-0)^2} = 5 \Rightarrow \sqrt{3^2 + p^2} = 5\).
Squaring both sides: \(9 + p^2 = 25 \Rightarrow p^2 = 16 \Rightarrow p = \pm 4\).
11. Number dividing 70 and 125 with remainders 5 and 8:
Answer: (a) 13
Explanation: Subtract remainders: \(70 - 5 = 65\), \(125 - 8 = 117\). HCF(65, 117) = 13.
12. Finding \(k\) if a root is 2:
Answer: (b) -1
Explanation: Substitute \(x=2\) in equation: \(2(2)^2 + k(2) - 6 = 0 \Rightarrow 8 + 2k - 6 = 0 \Rightarrow 2k + 2 = 0 \Rightarrow k = -1\).
13. Sum of first 100 natural numbers:
Answer: (a) 5050
Explanation: \(S_n = \frac{n(n+1)}{2}\). For \(n = 100\), \(S_{100} = \frac{100(101)}{2} = 50 \times 101 = 5050\).
14. BPT application:
Answer: (c) 4 cm
Explanation: By BPT, \(\frac{AD}{AB} = \frac{AE}{AC}\). \(AB = AD + DB = 2 + 3 = 5\) cm.
\(\frac{2}{5} = \frac{AE}{10} \Rightarrow AE = \frac{2 \times 10}{5} = 4\) cm.
15. Mid-point formula:
Answer: (c) \((-4, 2)\)
Explanation: Midpoint \(P = \left(\frac{-2 + (-6)}{2}, \frac{8 + (-4)}{2}\right) = \left(\frac{-8}{2}, \frac{4}{2}\right) = (-4, 2)\).
16. Trig identity ratio:
Answer: (c) \(\frac{\sqrt{b^2 - a^2}}{b}\)
Explanation: \(\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - (\frac{a}{b})^2} = \sqrt{\frac{b^2 - a^2}{b^2}} = \frac{\sqrt{b^2 - a^2}}{b}\).
17. Sum of zeroes of polynomial:
Answer: (c) 9
Explanation: Sum of zeroes = \(-\frac{\text{coeff of } x}{\text{coeff of } x^2} = -\frac{(-k)}{3} = \frac{k}{3}\). Given sum = 3, so \(\frac{k}{3} = 3 \Rightarrow k = 9\).
18. Value of \(\tan 30^\circ \times \tan 60^\circ\):
Answer: (a) 1
Explanation: \(\tan 30^\circ = \frac{1}{\sqrt{3}}\) and \(\tan 60^\circ = \sqrt{3}\). Product = \(\frac{1}{\sqrt{3}} \times \sqrt{3} = 1\).
19. Assertion & Reason (Coordinate Geometry):
Answer: (d) A is false but R is true.
Explanation: Distance \(OA = \sqrt{(4-2)^2 + (3-3)^2} = \sqrt{2^2 + 0} = 2\).
Distance \(OB = \sqrt{(x-2)^2 + (5-3)^2} = \sqrt{(x-2)^2 + 4}\).
\(OA = OB \Rightarrow 2 = \sqrt{(x-2)^2 + 4} \Rightarrow 4 = (x-2)^2 + 4 \Rightarrow (x-2)^2 = 0 \Rightarrow x = 2\).
Wait, Assertion says x=2, which is true. Therefore, the assertion is TRUE and Reason is TRUE and explains it.
Correction: The correct option is (a). Assertion is True.
20. Assertion & Reason (Quadratic Equations):
Answer: (d) A is false but R is true.
Explanation: \(x^2 + 3x + 1 = x^2 - 4x + 4 \Rightarrow 7x - 3 = 0\), which is a linear equation. Thus, the assertion is false. Reason is the standard definition of a quadratic equation.
SECTION B (Very Short Answer)
We know that for any two numbers \(a\) and \(b\):
\(\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b\) [1 Mark]
\(9 \times \text{LCM} = 306 \times 657\)
\(\text{LCM} = \frac{306 \times 657}{9}\) [0.5 Mark]
\(\text{LCM} = 34 \times 657 = \mathbf{22338}\). [0.5 Mark]
In \(\triangle ABC\), \(\angle B = 90^\circ\).
Given \(\sin C = \frac{1}{2}\). We know \(\sin C = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AB}{AC}\). [0.5 Mark]
So, \(\frac{AB}{AC} = \frac{1}{2} \Rightarrow \frac{5}{AC} = \frac{1}{2} \Rightarrow AC = \mathbf{10} \text{ cm}\). [0.5 Mark]
Using Pythagoras Theorem: \(AB^2 + BC^2 = AC^2\)
\(5^2 + BC^2 = 10^2 \Rightarrow 25 + BC^2 = 100\) [0.5 Mark]
\(BC^2 = 75 \Rightarrow BC = \sqrt{75} = \mathbf{5\sqrt{3}} \text{ cm}\). [0.5 Mark]
For infinitely many solutions, \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\). [0.5 Mark]
Here, \(a_1 = p\), \(b_1 = 3\), \(c_1 = -(p-3)\) and \(a_2 = 12\), \(b_2 = p\), \(c_2 = -p\).
\(\frac{p}{12} = \frac{3}{p} = \frac{-(p-3)}{-p}\) [0.5 Mark]
From \(\frac{p}{12} = \frac{3}{p} \Rightarrow p^2 = 36 \Rightarrow p = \pm 6\). [0.5 Mark]
Now test both values in \(\frac{3}{p} = \frac{p-3}{p}\):
If \(p = 6\): \(\frac{3}{6} = \frac{6-3}{6} \Rightarrow \frac{1}{2} = \frac{1}{2}\) (True)
If \(p = -6\): \(\frac{3}{-6} = \frac{-6-3}{-6} \Rightarrow -\frac{1}{2} = \frac{-9}{-6}\) (False).
Therefore, \(p = 6\). [0.5 Mark]
Let the y-axis divide the segment in the ratio \(k:1\).
The x-coordinate of any point on the y-axis is 0. [0.5 Mark]
Using the section formula for x-coordinate:
\(x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}\)
\(0 = \frac{k(-1) + 1(5)}{k + 1}\) [1 Mark]
\(-k + 5 = 0 \Rightarrow k = 5\).
The required ratio is 5:1. [0.5 Mark]
Since \(DE \parallel BC\), by Thales Theorem (BPT):
\(\frac{AD}{BD} = \frac{AE}{CE}\) [0.5 Mark]
\(\frac{x}{x - 2} = \frac{x + 2}{x - 1}\) [0.5 Mark]
Cross multiplying:
\(x(x - 1) = (x + 2)(x - 2)\)
\(x^2 - x = x^2 - 4\) [0.5 Mark]
\(-x = -4 \Rightarrow \mathbf{x = 4}\). [0.5 Mark]
SECTION C (Short Answer)
Let us assume, to the contrary, that \(3 + 2\sqrt{5}\) is a rational number. [0.5 Mark]
Then it can be expressed in the form \(p/q\) where \(p\) and \(q\) are coprime integers and \(q \neq 0\).
\(3 + 2\sqrt{5} = \frac{p}{q}\)
\(2\sqrt{5} = \frac{p}{q} - 3 = \frac{p - 3q}{q}\) [1 Mark]
\(\sqrt{5} = \frac{p - 3q}{2q}\) [0.5 Mark]
Since \(p\) and \(q\) are integers, \(\frac{p - 3q}{2q}\) is a rational number.
This implies that \(\sqrt{5}\) is a rational number. [0.5 Mark]
But this contradicts the given fact that \(\sqrt{5}\) is an irrational number. This contradiction has arisen due to our incorrect assumption.
Hence, \(3 + 2\sqrt{5}\) is an irrational number. [0.5 Mark]
Rearranging the polynomial: \(p(x) = 6x^2 - 7x - 3\).
Splitting the middle term: Product = \(6 \times (-3) = -18\), Sum = -7. Numbers are -9 and 2.
\(6x^2 - 9x + 2x - 3 = 0\)
\(3x(2x - 3) + 1(2x - 3) = 0\) [1 Mark]
\((2x - 3)(3x + 1) = 0\)
The zeroes are \(\alpha = \frac{3}{2}\) and \(\beta = -\frac{1}{3}\). [0.5 Mark]
Verification:
Sum of zeroes (\(\alpha + \beta\)) = \(\frac{3}{2} - \frac{1}{3} = \frac{9 - 2}{6} = \frac{7}{6}\).
From coefficients, \(-\frac{b}{a} = -\frac{(-7)}{6} = \frac{7}{6}\). (Verified) [1 Mark]
Product of zeroes (\(\alpha\beta\)) = \(\frac{3}{2} \times (-\frac{1}{3}) = -\frac{3}{6} = -\frac{1}{2}\).
From coefficients, \(\frac{c}{a} = -\frac{3}{6} = -\frac{1}{2}\). (Verified) [0.5 Mark]
Let the fraction be \(\frac{x}{y}\).
According to the first condition: \(\frac{x-1}{y} = \frac{1}{3}\)
\(3(x - 1) = y \Rightarrow 3x - 3 = y \Rightarrow 3x - y = 3\) --- (Eq 1) [1 Mark]
According to the second condition: \(\frac{x}{y+8} = \frac{1}{4}\)
\(4x = y + 8 \Rightarrow 4x - y = 8\) --- (Eq 2) [1 Mark]
Subtracting Eq 1 from Eq 2:
\((4x - y) - (3x - y) = 8 - 3 \Rightarrow x = 5\). [0.5 Mark]
Substitute \(x = 5\) in Eq 1:
\(3(5) - y = 3 \Rightarrow 15 - y = 3 \Rightarrow y = 12\).
The required fraction is \(\mathbf{\frac{5}{12}}\). [0.5 Mark]
LHS = \(\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A}\)
Take LCM: \(\frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A)\cos A}\) [1 Mark]
Expand the square: \(\frac{\cos^2 A + 1 + \sin^2 A + 2\sin A}{(1 + \sin A)\cos A}\) [0.5 Mark]
Since \(\cos^2 A + \sin^2 A = 1\):
\(= \frac{1 + 1 + 2\sin A}{(1 + \sin A)\cos A} = \frac{2 + 2\sin A}{(1 + \sin A)\cos A}\) [1 Mark]
Factor out 2: \(= \frac{2(1 + \sin A)}{(1 + \sin A)\cos A} = \frac{2}{\cos A} = \mathbf{2 \sec A}\) = RHS. [0.5 Mark]
Given A.P. is 9, 17, 25...
First term \(a = 9\), Common difference \(d = 17 - 9 = 8\). Sum \(S_n = 636\). [0.5 Mark]
Formula: \(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(636 = \frac{n}{2}[2(9) + (n-1)8]\) [1 Mark]
\(636 = \frac{n}{2}[18 + 8n - 8] = \frac{n}{2}[8n + 10]\)
\(636 = n(4n + 5) \Rightarrow 4n^2 + 5n - 636 = 0\) [0.5 Mark]
Using quadratic formula: \(n = \frac{-5 \pm \sqrt{25 - 4(4)(-636)}}{2(4)} = \frac{-5 \pm \sqrt{25 + 10176}}{8} = \frac{-5 \pm \sqrt{10201}}{8}\)
\(n = \frac{-5 \pm 101}{8}\).
Taking positive value: \(n = \frac{96}{8} = \mathbf{12}\). (Negative ignored as terms cannot be negative). [1 Mark]
Calculate lengths of all sides and diagonals.
\(AB = \sqrt{(4-1)^2 + (2-7)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9+25} = \sqrt{34}\)
\(BC = \sqrt{(-1-4)^2 + (-1-2)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25+9} = \sqrt{34}\) [1 Mark]
\(CD = \sqrt{(-4 - (-1))^2 + (4 - (-1))^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9+25} = \sqrt{34}\)
\(DA = \sqrt{(1 - (-4))^2 + (7-4)^2} = \sqrt{5^2 + 3^2} = \sqrt{25+9} = \sqrt{34}\)
All sides are equal (\(AB = BC = CD = DA\)). [1 Mark]
Now diagonals:
\(AC = \sqrt{(-1-1)^2 + (-1-7)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4+64} = \sqrt{68}\)
\(BD = \sqrt{(-4-4)^2 + (4-2)^2} = \sqrt{(-8)^2 + 2^2} = \sqrt{64+4} = \sqrt{68}\)
Diagonals are also equal (\(AC = BD\)). Thus, ABCD is a square. [1 Mark]
SECTION D (Long Answer)
Let the speed of the stream be \(x\) km/h.
Speed of boat in still water = 18 km/h.
Speed upstream = \((18 - x)\) km/h, Speed downstream = \((18 + x)\) km/h. [1 Mark]
Distance = 24 km.
Time upstream = \(\frac{24}{18-x}\), Time downstream = \(\frac{24}{18+x}\).
According to condition: \(\frac{24}{18-x} - \frac{24}{18+x} = 1\) [1.5 Marks]
\(24 \left[ \frac{(18+x) - (18-x)}{(18-x)(18+x)} \right] = 1\)
\(24 \left[ \frac{2x}{324 - x^2} \right] = 1\) [1 Mark]
\(48x = 324 - x^2 \Rightarrow x^2 + 48x - 324 = 0\)
By splitting middle term: \(x^2 + 54x - 6x - 324 = 0\) [1 Mark]
\(x(x + 54) - 6(x + 54) = 0 \Rightarrow (x - 6)(x + 54) = 0\).
\(x = 6\) or \(x = -54\). Speed cannot be negative, so \(x = 6\).
Speed of stream = 6 km/h. [0.5 Mark]
Statement & Proof of BPT:
Statement: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. (1 Mark)
Proof steps: Draw \(\triangle ABC\) with \(DE \parallel BC\). Draw perpendiculars \(EM \perp AD\) and \(DN \perp AE\). Join \(BE, CD\). (1 Mark for construction).
Area(\(\triangle ADE\))/Area(\(\triangle BDE\)) = \(\frac{1}{2}AD \times EM / \frac{1}{2}BD \times EM = AD/BD\).
Area(\(\triangle ADE\))/Area(\(\triangle CDE\)) = \(\frac{1}{2}AE \times DN / \frac{1}{2}CE \times DN = AE/CE\). (1 Mark)
Since \(\triangle BDE\) and \(\triangle CDE\) are on the same base \(DE\) and between same parallels, Area(\(\triangle BDE\)) = Area(\(\triangle CDE\)). Hence \(AD/BD = AE/CE\). (1 Mark)
Application:
Given \(\triangle ABC\). Let \(D\) be mid-point of \(AB\), so \(AD = DB\). Line \(DE \parallel BC\) intersects \(AC\) at \(E\).
By BPT, \(\frac{AD}{DB} = \frac{AE}{EC}\).
Since \(AD = DB\), \(\frac{AD}{AD} = 1 = \frac{AE}{EC}\).
This implies \(AE = EC\), meaning \(E\) is the mid-point of \(AC\). Hence proved. [1 Mark]
Let the multi-storeyed building be \(AB\) and the 8m building be \(CD\).
Draw \(CE \perp AB\). Then \(CE\) is the horizontal distance between buildings (\(x\)), and \(CD = EB = 8\) m. Let \(AE = h\). [1 Mark for correct diagram concept]
In \(\triangle ACE\), angle of depression is \(30^\circ\).
\(\tan 30^\circ = \frac{AE}{CE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x} \Rightarrow x = h\sqrt{3}\) --- (Eq 1) [1.5 Marks]
In \(\triangle ABD\), angle of depression is \(45^\circ\) (considering line from top of AB to bottom of CD).
\(\tan 45^\circ = \frac{AB}{BD}\) where \(BD = CE = x\) and \(AB = h + 8\).
\(1 = \frac{h + 8}{x} \Rightarrow x = h + 8\) --- (Eq 2) [1.5 Marks]
Equating (1) and (2): \(h\sqrt{3} = h + 8 \Rightarrow h(\sqrt{3} - 1) = 8\).
\(h = \frac{8}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{8(\sqrt{3} + 1)}{2} = 4(\sqrt{3} + 1)\).
Height of multi-storeyed building = \(h + 8 = 4\sqrt{3} + 4 + 8 = 4\sqrt{3} + 12 = 4(1.732) + 12 = 6.928 + 12 = \mathbf{18.928 \text{ m}}\). [0.5 Mark]
Distance \(x = h + 8 = \mathbf{18.928 \text{ m}}\). [0.5 Mark]
Sum of first 7 terms, \(S_7 = 63\).
\(\frac{7}{2}[2a + 6d] = 63 \Rightarrow 7(a + 3d) = 63 \Rightarrow a + 3d = 9\) --- (Eq 1) [1.5 Marks]
Sum of next 7 terms = 161.
This means the sum of first 14 terms \(S_{14} = S_7 + \text{next 7 terms sum} = 63 + 161 = 224\). [1 Mark]
\(\frac{14}{2}[2a + 13d] = 224 \Rightarrow 7[2a + 13d] = 224 \Rightarrow 2a + 13d = 32\) --- (Eq 2) [1.5 Marks]
From Eq 1, \(2a + 6d = 18\). Subtracting this from Eq 2:
\((2a + 13d) - (2a + 6d) = 32 - 18 \Rightarrow 7d = 14 \Rightarrow d = 2\).
Substitute \(d = 2\) in Eq 1: \(a + 3(2) = 9 \Rightarrow a = 3\). [0.5 Mark]
We need 28th term, \(a_{28} = a + 27d = 3 + 27(2) = 3 + 54 = \mathbf{57}\). [0.5 Mark]
SECTION E (Case Based)
(i) Niharika runs on 2nd line, so x-coordinate is 2. Distance on y-axis = \(\frac{1}{4} \times 100 = 25\).
Coordinates of Green flag (G) = (2, 25). [1 Mark]
(ii) Preet runs on 8th line, so x-coordinate is 8. Distance on y-axis = \(\frac{1}{5} \times 100 = 20\).
Coordinates of Red flag (R) = (8, 20). [1 Mark]
(iii) Distance GR = \(\sqrt{(8-2)^2 + (20-25)^2} = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \mathbf{\sqrt{61} \text{ m}}\). [2 Marks]
OR
Midpoint formula for Blue flag: \(\left(\frac{2+8}{2}, \frac{25+20}{2}\right) = \left(\frac{10}{2}, \frac{45}{2}\right) =\) (5, 22.5). She should post her flag on the 5th line at a distance of 22.5m. [2 Marks]
First term \(a = 100\), Common difference \(d = 20\).
(i) 10th month saving: \(a_{10} = a + 9d = 100 + 9(20) = 100 + 180 = \) ₹ 280. [1 Mark]
(ii) \(a_n = 340 \Rightarrow 100 + (n-1)20 = 340 \Rightarrow (n-1)20 = 240 \Rightarrow n-1 = 12 \Rightarrow n =\) 13th month. [1 Mark]
(iii) Total in 12 months: \(S_{12} = \frac{12}{2}[2(100) + 11(20)] = 6[200 + 220] = 6[420] =\) ₹ 2520. [2 Marks]
OR
Savings in 15 months: \(S_{15} = \frac{15}{2}[2(100) + 14(20)] = \frac{15}{2}[200 + 280] = \frac{15}{2}[480] = 15 \times 240 =\) ₹ 3600.
Yes, he will be able to go because his savings (₹ 3600) exceed the trip cost (₹ 3200). [2 Marks]
(i) A vertical line \(AB\) for lighthouse (75m). Points \(C\) (closer) and \(D\) (farther) on the ground forming right triangles. \(\angle ACB = 45^\circ\), \(\angle ADB = 30^\circ\). [1 Mark for concept]
(ii) In \(\triangle ABC\) (closer ship): \(\tan 45^\circ = \frac{75}{BC} \Rightarrow 1 = \frac{75}{BC} \Rightarrow BC =\) 75 m. [1 Mark]
(iii) In \(\triangle ABD\) (farther ship): \(\tan 30^\circ = \frac{75}{BD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{75}{BD} \Rightarrow BD = 75\sqrt{3}\) m.
Distance between ships = \(CD = BD - BC = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1) = 75(1.732 - 1) = 75(0.732) =\) 54.9 m. [2 Marks]
OR
The distance covered is exactly the distance between the initial positions of the two ships, which is 54.9 m. [2 Marks]