(i) (d): 1^st situation can be represented algebraically as 2x + 3y = 46
(ii) (c): 2^nd situation can be represented algebraically as 3x + 5y = 74
(iii) (b): We have, 2x + 3y = 46 .........(i)
3x+5y=74........... (ii)
Multiplying (i) by 5 and (ii) by 3 and then subtracting,
we get 10x - 9x = 230 - 222 \(\Rightarrow\) x = 8
\(\therefore\) Fare from Bengaluru to Malleswaram is Rs 8.
(iv) (a): Putting the value of x in equation (i), we get
3y = 46 - 2 x 8 = 30 \(\Rightarrow\) Y = 10
\(\therefore\) Fare from Bengaluru to Yeswanthpur is Rs 10.
(v) (c): We have, a₁ = 2, b₁, = 3, c₁ = -46 and
\(a_{2}=3, b_{2}=5, c_{2}=-74 \)
\(\therefore \quad \frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{3}{5}, \frac{c_{1}}{c_{2}}=\frac{-46}{-74}=\frac{23}{37} \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus system oflinear equations has unique solution.

(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.

Speed of boat in upstream = (x - y)km/hr and speed of boat in downstream = (x + y)km/hr.
(i) (a): 1^st situation can be represented algebraically as \(\frac{24}{x-y}+\frac{36}{x+y}=6\)
(ii) (b): 2nd situation can be represented algebraically as \(\frac{36}{x-y}+\frac{24}{x+y}=\frac{13}{2}\)
(iii) (c) : Putting \(\frac{1}{x-y}=u \text { and } \frac{1}{x+y}=v\)
we get,
24u + 36v = 6 and 36u + 24v = 13/2
Solving the above equations, we get u \(=\frac{1}{8}, v=\frac{1}{12}\)
(iv) (d): \(\because u=\frac{1}{8}=\frac{1}{x-y} \Rightarrow x-y=8\) ........(i)
\(\text { and } v=\frac{1}{12}=\frac{1}{x+y} \Rightarrow x+y=12\) .........(ii)
Adding equations (i) from (ii), we get 2x = 20 \(\Rightarrow\) x = 10
\(\therefore\) Speed of boat in still water = 10 km/hr -.
(v) (c): From equation (i), 10 - y = 8 \(\Rightarrow\) y = 2
\(\therefore\) Speed of stream = 2 km/hr.

(i) (a): At x-axis, y = 0
\(\therefore\) 2x + 4y = 8 \(\Rightarrow\) x = 4
At y-axis, x = 0
\(\therefore\) 2x + 4y = 8 \(\Rightarrow\) Y = 2
\(\therefore\) Required coordinates are (4, 0), (0, 2).
(ii) (c): At x-axis, y = 0
\(\therefore\) 3x + 6y = 18 \(\Rightarrow\) 3x = 18 \(\Rightarrow\) x = 6
At y-axis, x = 0
\(\therefore\) 3x + 6y = 18 \(\Rightarrow\) 6y = 18 \(\Rightarrow\) Y = 3
\(\therefore\) Required coordinates are (6, 0), (0, 3).
(iii) (d): Since, lines are parallel. So, point of intersection of these lines does not exist.
(iv) (a)
(v) (a): Since the lines are parallel.
\(\therefore\) These equations have no solution i.e., the given system of linear equations is inconsistent.

(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.