(i) (a): 1^st situation can be represented as x + 7y = 650 ...(i) and
2^nd situation can be represented as x + 11y = 970 ...(ii)
(ii) (b): Subtracting equations (i) from (ii), we get 
\(4 y=320 \Rightarrow y=80\)
\(\therefore\) Proportional expense for each person is Rs 80.
(iii) (c): Puttingy = 80 in equation (i), we get
x + 7 x 80 = 650 \(\Rightarrow\) x = 650 - 560 = 90
\(\therefore\) Fixed expense for the party is Rs 90
(iv) (d): If there will be 15 guests, then amount that Mr Jindal has to pay = Rs (90 + 15 x 80) = Rs 1290
(v) (a): We have a₁ = 1, b₁ = 7, c₁ = -650 and 
\(a_{2}=1, b_{2}=11, c_{2}=-970 \)
\(\therefore \frac{a_{1}}{a_{2}}=1, \frac{b_{1}}{b_{2}}=\frac{7}{11}, \frac{c_{1}}{c_{2}}=\frac{-650}{-970}=\frac{65}{97}\)
\(\text { Here, } \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus, system of linear equations has unique solution.

(i) (a): Situation faced by Sudhir can be represented algebraically as 2x + 3y = 850
(ii) (b): Situation faced by Suman can be represented algebraically as 3x + 2y = 900
(iii) (c) : We have 2x + 3y = 850 .........(i)
and 3x + 2y = 900 .........(ii)
Multiplying (i) by 3 and (ii) by 2 and subtracting, we get
5y = 750 \(\Rightarrow\) Y = 150
Thus, price of one Physics book is Rs 150.
(iv) (d): From equation (i) we have, 2x + 3 x 150 = 850
\(\Rightarrow\) 2x = 850 - 450 = 400 \(\Rightarrow\) x = 200
Hence, cost of one Mathematics book = Rs 200
(v) (a): From above, we have
\(a_{1} =2, b_{1}=3, c_{1}=-850 \)
\(\text { and } a_{2} =3, b_{2}=2, c_{2}=-900\)
\(\therefore \quad \frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{3}{2}, \frac{c_{1}}{c_{2}}=\frac{-850}{-900}=\frac{17}{18} \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus system of linear equations has unique solution.

(i) (b): Algebraic representation of situation of day-I is 2x + y = 1600.
(ii) (a): Algebraic representation of situation of day- II is 4x + 2y = 3000 \(\Rightarrow\) 2x + y = 1500.
(iii) (c) : At x-axis, y = 0
\(\therefore\)  At y = 0, 2x + y = 1600 becomes 2x = 1600
\(\Rightarrow\) x = 800
\(\therefore\) Linear equation represented by day- I intersect the x-axis at (800, 0).
(iv) (d) : At y-axis, x = 0
\(\therefore\) 2x + Y = 1500 \(\Rightarrow\) y = 1500
\(\therefore\) Linear equation represented by day-II intersect the y-axis at (0, 1500).
(v) (b): We have, 2x + y = 1600 and 2x + y = 1500
Since \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \text { i.e., } \frac{1}{1}=\frac{1}{1} \neq \frac{16}{15}\)
\(\therefore\) System of equations have no solution.
\(\therefore\) Lines are parallel.

(i) (b):We have \(6 x^{2}+x-2=0\)
\(\Rightarrow \quad 6 x^{2}-3 x+4 x-2=0 \)
\(\Rightarrow \quad(3 x+2)(2 x-1)=0 \)
\(\Rightarrow \quad x=\frac{1}{2}, \frac{-2}{3}\)
(ii) (c): \(2 x^{2}+x-300=0\)
\(\Rightarrow \quad 2 x^{2}-24 x+25 x-300=0 \)
\(\Rightarrow \quad(x-12)(2 x+25)=0 \)
\(\Rightarrow \quad x=12, \frac{-25}{2}\)
(iii) (d): \(x^{2}-8 x+16=0\)
\(\Rightarrow(x-4)^{2}=0 \Rightarrow(x-4)(x-4)=0 \Rightarrow x=4,4\)
(iv) (d): \(6 x^{2}-13 x+5=0\)
\(\Rightarrow \quad 6 x^{2}-3 x-10 x+5=0 \)
\(\Rightarrow \quad(2 x-1)(3 x-5)=0 \)
\(\Rightarrow \quad x=\frac{1}{2}, \frac{5}{3}\)
(v) (a): \(100 x^{2}-20 x+1=0\)
\(\Rightarrow(10 x-1)^{2}=0 \Rightarrow x=\frac{1}{10}, \frac{1}{10}\)
 

Number of pairs of shoes in 1^st, 2^nd, 3^rd row, ... are 3,5,7, ...
So, it forms an A.P. with first term a = 3, d = 5 - 3 = 2
(i) (d): Let n be the number of rows required.
\(\therefore S_{n}=120 \)
\(\Rightarrow \quad \frac{n}{2}[2(3)+(n-1) 2]=120 \)
\(\Rightarrow \quad n^{2}+2 n-120=0 \Rightarrow n^{2}+12 n-10 n-120=0\)
\(\Rightarrow \quad(n+12)(n-10)=0 \Rightarrow n=10\)
So, 10 rows required to put 120 pairs.
(ii) (b): No. of pairs in 1ih row = t₁₇ = 3 + 16(2) = 35
No. of pairs in 10th row = t₁₀ = 3 + 9(2) = 21
\(\therefore\) Required difference = 35 - 21 = 14
(iii) (c) : Here n = 15
\(\therefore\) t₁₅ = 3 + 14(2) = 3 + 28 = 31
(iv) (a): No. of pairs in 30^th row = t₃₀ = 3 +29(2) = 61
(v) (c): No. of pairs in 5^th row = t₅ = 3 + 4(2) = 11
No. of pairs in 8^th row = t₈ = 3 + 7(2) = 17
\(\therefore\) Required sum = 11 + 17 = 28

Let the numbers on the cards be a, a + d, a + Zd, ...
According to question, We have (a + 5d) + (a + 13d) = -76
\(\Rightarrow\) 2a+18d = -76\(\Rightarrow\)a + 9d= -38 ... (1)
And (a + 7d) + (a + 15d) = -96
\(\Rightarrow\) 2a + 22d = -96 \(\Rightarrow\) a + 11d = -48 ...(2)
From (1) and (2), we get
2d= -10 \(\Rightarrow\) d= -5
From (1), a + 9(-5) = -38 \(\Rightarrow\) a = 7
(i) (b): The difference between the numbers on any two consecutive cards = common difference of the A.P.=-5
(ii) (d): Number on first card = a = 7
(iii) (b): Number on 19th card = a + 18d = 7 + 18(-5) = -83
(iv) (a): Number on 23rd card = a + 22d = 7 + 22( -5) = -103
(v) (d): \(S_{15}=\frac{15}{2}[2(7)+14(-5)]=-420\)

(i) (c)
(ii) (c)
(iii) (d)
(iv) (b)
(v) (c)

Here S_n = 0.1n² + 7.9n
(i) (c): S_{n -1}= 0.1(n - 1)² + 7.9(n - 1)
= 0.1n² + 7.7n - 7.8
(ii) (b): S₁ = t₁ = a = 0.1(1)² + 7.9(1) = 8 cm = Diameter of core
So, radius of the core = 4 cm
(iii) (a): S₂ = 0.1(2)² + 7.9(2) = 16.2
(iv) (d): Required diameter = t₂ = S₂ - S₁ = 16.2 - 8 = 8.2 cm
(v) (c): As d = t₂ - t₁ = 8.2 - 8 = 0.2 cm
So, thickness of tissue = 0.2 \(\div\) 2 = 0.1 cm = 1 mm

(i) (a): Speed = 1200 km/hr
\(\text { Time }=1 \frac{1}{2} \mathrm{hr}=\frac{3}{2} \mathrm{hr}\)
\(\therefore\) Required distance = Speed x Time
\(=1200 \times \frac{3}{2}=1800 \mathrm{~km}\)
(ii) (c): Speed = 1500 km/hr
Time = \(\frac{3}{2}\) hr.
\(\therefore\) Required distance = Speed x Time
\(=1500 \times \frac{3}{2}=2250 \mathrm{~km}\)
(iii) (b): Clearly, directions are always perpendicular to each other.
\(\therefore \quad \angle P O Q=90^{\circ}\)
(iv) (a): Distance between aeroplanes after \(1\frac{1}{2}\) hour 
\(\begin{array}{l} =\sqrt{(1800)^{2}+(2250)^{2}}=\sqrt{3240000+5062500} \\ =\sqrt{8302500}=450 \sqrt{41} \mathrm{~km} \end{array}\)
(v) (d): Area of \(\Delta\)POQ= \(\frac{1}{2}\)x base x height
\(=\frac{1}{2} \times 2250 \times 1800=2250 \times 900=2025000 \mathrm{~km}^{2}\)

(i) (b)
(ii) (c): Using Pythagoras theorem, we have PQ² = PR² + RQ²
\(\Rightarrow(26)^{2}=(2 x)^{2}+(2(x+7))^{2} \Rightarrow 676=4 x^{2}+4(x+7)^{2} \)
\(\Rightarrow 169=x^{2}+x^{2}+49+14 x \Rightarrow x^{2}+7 x-60=0\)
\(\Rightarrow x^{2}+12 x-5 x-60=0 \)
\(\Rightarrow x(x+12)-5(x+12)=0 \Rightarrow(x-5)(x+12)=0 \)
\(\Rightarrow x=5, x=-12\)
\(\therefore \quad x=5\) [Since length can't be negative]
(iii) (a) : PR = 2x = 2 x 5 = 10 km
(iv) (b): RQ= 2(x + 7) = 2(5 + 7) = 24 km
(v) (d): Since, PR + RQ = 10 + 24 = 34 km Saved distance = 34 - 26 = 8 km

(i) (b): If \(\Delta\)AED and \(\Delta\)BEC, are similar by SAS similarity rule, then their corresponding proportional sides are \(\frac{B E}{A E}=\frac{C E}{D E}\)
(ii) (c): By Pythagoras theorem, we have
\(\begin{array}{l} B C=\sqrt{C E^{2}+E B^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{16+9} \\ =\sqrt{25}=5 \mathrm{~cm} \end{array}\)
(iii) (a): Since \(\Delta\)ADE and \(\Delta\)BCE are similar.
\(\therefore \quad \frac{\text { Perimeter of } \triangle A D E}{\text { Perimeter of } \Delta B C E}=\frac{A D}{B C} \)
\(\Rightarrow \frac{2}{3}=\frac{A D}{5} \Rightarrow A D=\frac{5 \times 2}{3}=\frac{10}{3} \mathrm{~cm}\)
(iv) (b):\(\frac{\text { Perimeter of } \triangle A D E}{\text { Perimeter of } \Delta B C E}=\frac{E D}{C E} \)
\(\Rightarrow \frac{2}{3}=\frac{E D}{4} \Rightarrow E D=\frac{4 \times 2}{3}=\frac{8}{3} \mathrm{~cm}\)
(v) (d) : \(\frac{\text { Perimeter of } \Delta A D E}{\text { Perimeter of } \Delta B C E}=\frac{A E}{B E} \Rightarrow \frac{2}{3} B E=A E\)
\(\Rightarrow A E=\frac{2}{3} \sqrt{B C^{2}-C E^{2}} \)
\(\text { Also, in } \triangle A E D, A E=\sqrt{A D^{2}-D E^{2}}\)

(i) (a): Here, CD = \(\sqrt{(7-3)^{2}+(7-4)^{2}}\)
\(=\sqrt{4^{2}+3^{2}}\) = 5 units
Also, it is given that CE = 10 units
Thus, DE = CE - CD = 10 - 5 = 5 units (\(\because\) A, B, C, E are a line)
(ii) (b): Since, CD = DE = 5 units
\(\therefore\) Dis the midpoint of CE.
\(\therefore \quad \frac{x+3}{2}=7 \text { and } \frac{y+4}{2}=7 \)
\(\Rightarrow \quad x=11 \text { and } y=10 \Rightarrow x+y=21\)
(iii) (b)
(iv) (d): Let B divides AC in the ratio k:1, then

\(\frac{7}{4}=\frac{4 k+0}{k+1} \)
\(\Rightarrow 7 k+7=16 k \)
\(\Rightarrow 7=9 k \)
\(\Rightarrow k=\frac{7}{9}\)
Thus, the required ratio is 7 : 9
(v) (c): It can be easily verify that all the given points lie on the line represented by 3x - 4y + 7 = 0.

(i) (a): We have, OA = 2\(\sqrt{2}\) km
\(\Rightarrow \sqrt{2^{2}+y^{2}}=2 \sqrt{2} \)
\(\Rightarrow 4+y^{2}=8 \Rightarrow y^{2}=4 \)
\(\Rightarrow y=2 \quad(\because y=-2 \text { is not possible })\)
(ii) (c): We have OB = 8\(\sqrt{2}\)
\(\Rightarrow \sqrt{x^{2}+8^{2}}=8 \sqrt{2} \)
\(\Rightarrow x^{2}+64=128 \Rightarrow x^{2}=64 \)
\(\Rightarrow x=8 \quad(\because x=-8 \text { is not possible })\)
(iii) (c) : Coordinates of A and Bare (2, 2) and (8, 8) respectively, therefore coordinates of point M are
\(\left(\frac{2+8}{2}, \frac{2+8}{2}\right)\)i.e .,(5.5)
(iv) (d): Let A divides OM in the ratio k: 1.Then
\(2=\frac{5 k+0}{k+1} \Rightarrow 2 \mathrm{k}+2=5 k \Rightarrow 3 k=2 \Rightarrow k=\frac{2}{3}\)
\(\therefore\) Required ratio = 2 : 3
(v) (b): Since M is the mid-point of A and B therefore AM = MB. Hence, he should try his luck moving towards B.

(i) (c): Required coordinates are \(\left(0, \frac{4}{3}\right)\)
(ii) (c)
(iii) (a): Radius = Distance between (0,0) and \(\left(\frac{4}{3}, 0\right)\)
\(=\sqrt{\left(\frac{4}{3}\right)^{2}+0^{2}}=\frac{4}{3} \text { units }\)
(iv) (b): Area of circle = \(\pi\)(radius)²
\(=\pi\left(\frac{4}{3}\right)^{2}=\frac{16}{9} \pi \text { sq. units }\)
(v) (d): Let the coordinates of the other end be (x,y).
Then (0,0) will bethe mid-point of \(\left(1, \frac{\sqrt{7}}{3}\right)\) and (x, y).
\(\therefore\left(\frac{1+x}{2}, \frac{\frac{\sqrt{7}}{3}+y}{2}\right)=(0,0) \)
\(\Rightarrow \frac{1+x}{2}=0 \text { and } \frac{\frac{\sqrt{7}}{3}+y}{2}=0 \)
\(\Rightarrow x=-1 \text { and } y=-\frac{\sqrt{7}}{3}\)
Thus, the coordinates of other end be \(\left(-1, \frac{-\sqrt{7}}{3}\right)\)

Coordinates of A, Band Care (-2, -3), (2, 3) and (3,2).
(i) (d): Required distance \(=\sqrt{(2+2)^{2}+(3+3)^{2}}\)
\(=\sqrt{4^{2}+6^{2}}=\sqrt{16+36}=2 \sqrt{13} \mathrm{~km} \approx 7.2 \mathrm{~km}\)
(ii) (d): Required distance \(=\sqrt{(3+2)^{2}+(2+3)^{2}}\)
\(=\sqrt{5^{2}+5^{2}}=5 \sqrt{2} \mathrm{~km}\)
(iii) (b): Distance between Band C
\(=\sqrt{(3-2)^{2}+(2-3)^{2}}=\sqrt{1+1}=\sqrt{2} \mathrm{~km}\)
Thus, distance travelled by first bus to reach to B
\(=A C+C B=5 \sqrt{2}+\sqrt{2}=6 \sqrt{2} \mathrm{~km} \approx 8.48 \mathrm{~km}\)
and distance travelled by second bus to reach to B
\(=A B=2 \sqrt{13} \mathrm{~km} \approx 7.2 \mathrm{~km}\)
\(\therefore\) Distance of first bus is greater than distance of the
second bus, therefore second bus should be chosen.
(iv) (d): Distance travelled by first bus = 8.48 km
\(\therefore\) Total fare = 8.48 x 10 = Rs 84.80
(v) (b): Distance travelled by second bus = 7. 2 km
\(\therefore\) Total fare = 7.2 x 15 = Rs 108