(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.

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) (d): Since 8 men and 12 women can finish the work in 10 days.
\(\therefore\left(\frac{8}{x}+\frac{12}{y}\right)=\frac{1}{10} \Rightarrow \frac{80}{x}+\frac{120}{y}=1\)
(ii) (c): Since 6 men and 8 women can finish a piece of work in 14 days
\(\therefore\left(\frac{6}{x}+\frac{8}{y}\right)=\frac{1}{14} \Rightarrow \frac{84}{x}+\frac{112}{y}=1\)
(iii) (c) : Let \(\frac{1}{x}=u, \frac{1}{y}=v\)
Thus, we have
80u + 120v = 1 and 84u + 112v = 1
Solving above two equations, we get
\(v=\frac{1}{280} \Rightarrow \frac{1}{y}=\frac{1}{280} \Rightarrow y=280\)
Thus one woman alone can finish the work in 280 days.
(iv) (a): We have \(\frac{80}{x}+\frac{120}{y}=1 \Rightarrow \frac{80}{x}+\frac{120}{280}=1\)
\(\Rightarrow \quad \frac{80}{x}=1-\frac{3}{7} \Rightarrow \frac{80}{x}=\frac{4}{7}=\Rightarrow x=140\)
Thus one man alone can finish the work in 140 days.
(v) (d): We have, x = 140 and y = 280
One day's work of 14 men and 28 women
\(=\frac{14}{140}+\frac{28}{280}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5}\)
Thus, work will be finished in 5 days.

(i) (a) : Suppose two cars meet at point Q. Then,
Distance travelled by car X = A Q,
Distance travelled by car Y = BQ.
It is given that two cars meet in 9 hours.
\(\therefore\) Distance travelled by car X in 9 hours = 9x km
\(\Rightarrow\) AQ=9x
Distance travelled by car Y in 9 hours = 9y km
\(\Rightarrow\) BQ=9y

Clearly, AQ - BQ = AB
\(\Rightarrow\) 9x - 9y = 90
\(\Rightarrow\) x-y =10
(ii) (c): Suppose two cars meet at point P. Then
Distance travelled by car X = AP and
Distance travelled by car Y = BP.
In this case, two cars meet in 917 hours.
\(\therefore\) Distance travelled by car X in 9/7 hours = \(\frac {9}{7}\) x km
\(\Rightarrow A P=\frac{9}{7} x\)
Distance travelled by car Y in 9/7 hours \(\frac {9}{7}\) y km
\(\Rightarrow B P=\frac{9}{7} y\)
Clearly, AP + BP = AB
\(\Rightarrow \quad \frac{9}{7} x+\frac{9}{7} y=90 \Rightarrow \frac{9}{7}(x+y)=90 \Rightarrow x+y=70\)
(iii) (b): We have x - y = 10
\(\Rightarrow x+y=70\)
Adding equations (i) and (ii), we get
2x = 80 \(\Rightarrow\) x = 40
Hence, speed of car X is 40 km/hr.
(iv) (c): We have x - y = 10
\(\Rightarrow\) 40 - Y = 10 \(\Rightarrow\) Y = 30
Hence, speed of car y is 30 km/hr.
(v) (d)

(i) (a): For student Anu: Fixed charge + cost of food for 25 days = Rs 4500
i.e., x + 25y = 4500 For student Bindu:
Fixed charges + cost of food for 30 days = Rs 5200
i.e., x + 30y = 5200
(ii) (b): From above, we have a₁ = 1, b₁, = 25
\(c_{1}=-4500 \text { and } a_{2}=1, b_{2}=30, c_{2}=-5200 \)
\(\therefore \frac{a_{1}}{a_{2}}=1, \frac{b_{1}}{b_{2}}=\frac{25}{30}=\frac{5}{6}, \frac{c_{1}}{c_{2}}=\frac{-4500}{-5200}=\frac{45}{52} \)
\(\Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus, system of linear equations has unique solution
(iii) (c) : We have x + 25y = 4500 .......(i)
and x + 30y = 5200                      ........(ii)
Subtracting (i) from (ii), we get
5y=700 \(\Rightarrow\) y=140
\(\therefore\) Cost of food per day is Rs 140
(iv) (c): We have, x + 25y = 4500
\(\Rightarrow\) x = 4500 - 25 x 140
\(\Rightarrow\) x = 4500 - 3500 = 1000
\(\therefore\) Fixed charges per month for the hostel is Rs 1000
(v) (d): We have, x = 1000, Y = 140 and Bindu takes food for 20 days.
\(\therefore\) Amount that Bindu has to pay = Rs (1000 + 20 x 140) = Rs 3800